an overview of undergrad fluids · this material was inspired by prof. marcus hultmark’s class...

$: An Overview of Undergrad Fluids · This material was inspired by Prof. Marcus Hultmark’s class lectures, Prof. Alexander Smits’ book \A Physical Introduction to Fluid Mechanics,"$
An Overview of Undergrad Fluids

Notes developed for MAE 222

Written and compiled by:

Clayton P. Byers

Princeton University, 2016

Foreword

These notes are a compilation and edit of my precept notes I developed for MAE 222 inSpring 2016 at Princeton University. The goal here was to combine them into a sort ofstudy guide that can be used by future students. They assume the reader is either famil-iar with, or concurrently studying, basic fluids mechanics. These notes will not standalone as a way to learn fluids from scratch! Some sections go into fine detail aboutderivations, which aren’t necessary for the basic undergrad education. Other sections as-sume you have some base level of knowledge, and may be complicated if you haven’t seenthe material before. They will serve well for an undergrad reviewing the material for yourfinal, or even for a graduate student in the early preparation for your general examination.

The flow and order of the material follows the general outline in which it was presentedin the course. In general, each section should have a number of fully worked examplesto demonstrate the core concepts. I go through each step in the math and logic to helpdevelop your understanding of the application of the material.

I have tried to edit the content so that it is consistent in the internal equation and figurereferences, but there might be an equation or two that accidentally got cross-referenced.If you see any mistakes in notation (or more egregious errors in the content!) please letme know: claytonb (at) princeton (dot) edu.

This material was inspired by Prof. Marcus Hultmark’s class lectures, Prof. AlexanderSmits’ book “A Physical Introduction to Fluid Mechanics,” and the book “EngineeringFluid Mechanics” by Crowe, Elger, Roberson, and Williams. You’ll find similarities tothose texts or classes for good reason, but my goal was to present them in a new light orwith a different approach. I hope you find this useful and enlightening!

Intro to Fluids Notes 1 C. P. Byers 2016

Contents

1 Introductory Concepts 51.1 Dimensional homogeneity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 51.2 The unique characteristics of pressure in fluids . . . . . . . . . . . . . . . . . . . 51.3 Example: breaking a barrel . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 71.4 Example: an airplane door . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81.5 Statics and moments . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10

1.5.1 Example: a dam wall with one fluid . . . . . . . . . . . . . . . . . . . . . 101.5.2 Example: a dam wall with two fluids . . . . . . . . . . . . . . . . . . . . . 13

1.6 Rigid body motion summary . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 151.7 Specific weight and specific gravity . . . . . . . . . . . . . . . . . . . . . . . . . . 181.8 Buoyancy . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18

1.8.1 Example: floating some aluminum . . . . . . . . . . . . . . . . . . . . . . 191.8.2 Example: floating in a different fluid . . . . . . . . . . . . . . . . . . . . . 21

1.9 Control Volumes . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 221.9.1 Example: accumulating mass . . . . . . . . . . . . . . . . . . . . . . . . . 221.9.2 Example: non-uniform flow . . . . . . . . . . . . . . . . . . . . . . . . . . 23

1.10 Material Derivative . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25

2 Continuity and Momentum 272.1 Continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27

2.1.1 Differential form of the continuity equation (bonus material) . . . . . . . 292.1.2 Example: continuity on a nozzle . . . . . . . . . . . . . . . . . . . . . . . 30

2.2 Momentum Balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.1 Example: force on a nozzle . . . . . . . . . . . . . . . . . . . . . . . . . . 322.2.2 Example: Balance with angled velocities . . . . . . . . . . . . . . . . . . . 34

3 The Momentum Equation 383.1 Integral & Differential momentum equation . . . . . . . . . . . . . . . . . . . . . 38

3.1.1 Example: momentum practice . . . . . . . . . . . . . . . . . . . . . . . . 413.1.2 Example: differential momentum practice . . . . . . . . . . . . . . . . . . 45

3.2 Bernoulli’s equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 463.2.1 Rigorous derivation of Bernoulli’s equation (bonus material) . . . . . . . . 473.2.2 Pressure and streamlines . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49

3.3 Example: flow over a bump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 49


CONTENTS

4 Introducing Potential Flow 534.1 Some flow definitions and characteristics . . . . . . . . . . . . . . . . . . . . . . . 53

4.1.1 Vorticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 534.1.2 Velocity Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 544.1.3 Stream functions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55

4.2 Example: a potential vortex . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 564.3 The power of the velocity potential . . . . . . . . . . . . . . . . . . . . . . . . . . 594.4 Example: superposition . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60

5 Non-Dimensionalization 635.1 Non-dimensionalization . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63

5.1.1 Similarity and non-dimensional parameters . . . . . . . . . . . . . . . . . 635.1.2 Non-dimensional equations . . . . . . . . . . . . . . . . . . . . . . . . . . 665.1.3 Example: non-dimensionalizing the momentum equation . . . . . . . . . . 665.1.4 Example: flow in a channel . . . . . . . . . . . . . . . . . . . . . . . . . . 68

5.2 Prandtl’s Boundary Layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695.2.1 x-momentum equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 695.2.2 y-momentum equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 725.2.3 Combining the x- and y-momentum equations . . . . . . . . . . . . . . . 735.2.4 Bonus: similarity solution . . . . . . . . . . . . . . . . . . . . . . . . . . . 74

5.3 Bonus: Energy Equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75

6 Analysis of the Boundary Layer 776.1 Summary of the boundary layer . . . . . . . . . . . . . . . . . . . . . . . . . . . . 776.2 Zero pressure gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 786.3 Displacement thickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78

6.3.1 Displacement thickness and the zero pressure gradient assumption . . . . 806.4 Momentum thickness . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81

6.4.1 Momentum thickness and drag . . . . . . . . . . . . . . . . . . . . . . . . 846.5 Example: approximate boundary layer . . . . . . . . . . . . . . . . . . . . . . . . 846.6 Example: finding the skin friction . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

7 Drag, Separation, and Shedding 897.1 Drag force from friction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 89

7.1.1 Example: drag on a plate . . . . . . . . . . . . . . . . . . . . . . . . . . . 907.2 Separation of the boundary layer . . . . . . . . . . . . . . . . . . . . . . . . . . . 91

7.2.1 Favorable pressure gradient . . . . . . . . . . . . . . . . . . . . . . . . . . 937.2.2 Zero pressure gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 937.2.3 Adverse pressure gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . 94

7.3 Resistance to separation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 947.4 Case: a curved surface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 947.5 Strouhal number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 95

7.5.1 Example: Tacoma Narrows Bridge . . . . . . . . . . . . . . . . . . . . . . 967.5.2 Example: car antenna . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 98


CONTENTS

8 Internal Flows and Losses 998.1 Internal flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 99

8.1.1 Continuity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1008.1.2 Inertial terms . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1018.1.3 The x-momentum equation for a channel (and pipe) . . . . . . . . . . . . 1018.1.4 The pressure gradient . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 102

8.2 The velocity profile in a pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1038.3 The friction factor in a laminar pipe . . . . . . . . . . . . . . . . . . . . . . . . . 1058.4 Example: a vertical pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1068.5 Example: force to hold a pipe . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1098.6 Losses in pipe flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1108.7 Example: flow from a reservoir . . . . . . . . . . . . . . . . . . . . . . . . . . . . 113

9 Supercritical and Supersonic Flows 1159.1 Comparison of Mach number and Froude number . . . . . . . . . . . . . . . . . . 1159.2 Hydraulic jumps . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 116

9.2.1 Example: hydraulic jumps on a dam . . . . . . . . . . . . . . . . . . . . . 1189.2.2 Example: jump in a channel . . . . . . . . . . . . . . . . . . . . . . . . . . 118

9.3 Compressible flow relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1209.3.1 Example: space shuttle re-entry . . . . . . . . . . . . . . . . . . . . . . . . 121

9.4 Normal shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1229.4.1 Example: revisiting the shuttle . . . . . . . . . . . . . . . . . . . . . . . . 123


Chapter 1

Introductory Concepts

1.1 Dimensional homogeneity

This concept is one of the most important concepts in science and engineering, yet is oftenglazed over and not actively discussed as early as (I think) it should be. Whenever we writean equation, we aren’t just looking to plug in number, but are also dealing with dimensionsor quantities of some sort. Something as simple as Newton’s Second Law,

∑ ~F = m~a has nomeaning or usefulness unless we use the proper quantities and dimensions.

m→[mass] a fundamental dimension

a→ [acceleration]length

time2

F → [force]mass * length

time2

From this information, we can see that both sides of the equation balance in terms of fundamentaldimensions - we have [mass ∗ length]/time2 on both sides. However, we must also make surewe use consistent units. It doesn’t make sense to have meters on one side and feet on the other.All equations we work with (which are based on some physical interaction in the world aroundus) will be dimensionally homogeneous.

This seems simplistic, but keep it in mind and always take a look at it to make sure you’reapproaching your equation correctly. In fluids, we can sometimes get information in units likepsi but need Pa. The easiest way to make sure you’re doing things right is to write down theunits when doing math. Not only will it help you keep track of what you’re doing, but it mayclarify where a mistake could be.

1.2 The unique characteristics of pressure in fluids

We like to make analogies of pressure with weight balancing on an area. However, this can bemisleading, and requires a little more thought to grasp how pressure really works.


CHAPTER 1. INTRODUCTORY CONCEPTS

Remember that we classify pressure as a “scalar,” which means it has a value, but nodirection. This is analogous to temperature - does it have a direction? Heat transfer does, buttemperature itself does not. It simply has a value at a location. Pressure is the same, while theforce we feel (or calculate) due to pressure isn’t prescribed by the pressure field, but instead bythe physical boundaries in which pressure acts. The force due to pressure, which is a vector,gets the magnitude from the pressure and area it acts on, but the direction comes from theprescribed boundaries on which the pressure is acting.

One way to think about the force of pressure is to think of a person sitting on the bottomof the ocean. Assuming the depth this person is at stays constant, then we can say the pressureacting on him is constant. Recall the equation for hydrostatic balance:

dP

dz= −ρg (1.1)

where the z-direction is defined as positive going up. Integration of the equation then will giveus:

P − P0 = −ρg(z − z0)

Let’s think for a moment - what are the initial conditions, and what are the variables here?Since we’re dealing with the ocean, ρ will be the density of sea water, which we can wave ourhands and say is 1000 kg/m3. Notice the units we’re using? Keep it consistent throughout!Let’s also assume that gravity isn’t going to change on us too much, which is usually a safe bet,and say g ∼ 10m/s2. Again, note the unit’s we’re using! Now think on the initial conditions.

We can specify the pressure at the surface of the water as atmospheric pressure, Patm. Inreal life, this can change a bit day to day, but we usually are interested in departures fromthis pressure, not the actual value of it. So sometimes we’ll just leave it as a variable, Patm,while other times we will just set it to zero and note that we are looking for the departure fromatmospheric pressure. As for the height, let’s set the surface of the ocean as zero. We could alsoset it as any other value, but why make it hard on ourselves?

OK, here’s where we stand now with the equation:

P − Patm = −1000kg

m3∗ 10

m

s2∗ (z − 0)

or

∆P = −10000N

m3∗ (z)

All we’ve done is simply call the pressure difference ∆P since we’re talking about a departurefrom atmospheric. Now, let’s say our person is 33 feet below the surface. What is the pressuredifference from atmospheric pressure?

∆P = −10000N

m3∗ (−33ft)

Note that we’ve set the depth as a negative value, as we defined the positive z-direction as up.But look at the units! We shouldn’t be using feet! Recall dimensional homogeneity! We needto convert this to meters, which is about 3.3 feet to a meter. Therefore, we would find:

∆P = −10000N

m3∗ (−33ft ∗ 1m

3.3ft) = 100000

N

m2



The pressure at 33 ft below the surface is approximately 100 kPa higher than the atmosphericpressure! So even though our equation had the minus sign, we found the pressure increased withdepth (which our intuition should agree with).

Now why did I say that pressure can be tricky or unique? Let’s think about the water abovethe person. We specified them to be at the bottom of the ocean, 33 ft down (ok I’m tired offeet, lets use meters like real scientists). What if instead it was the bottom of a 10 m pool? Ora 10 m dunk tank? The person would still experience an additional 100 kPa of pressure. Youcould even shrink this down to a tube that encapsulates the person, but with 10 m of waterabove, and it would still be the same pressure. Even more crazy, you could use a 10 m longstraw, place it on top of their head, and fill it with water, and the pressure at the top of theirhead would be 100 kPa higher than atmospheric pressure! So pressure isn’t so much about theweight of fluid above the point, because the straw full of water and the ocean exert the samepressure at a depth of 10 meters, even though their volumes are vastly different. Remember thatthe pressure is a scalar value that exerts itself on the surface, and the surface is what determinesthe direction of the resultant force.

Overall, just be careful when doing the integration, and define your zero point and positivedirection clearly.

1.3 Example: breaking a barrel

This example highlights the “strangeness” of pressure while showing how powerful it can be.Recall how we just said a straw 10 meters tall and full of water exerts the same pressure as theocean at 10 meters below the surface. This concept can be used to exert large forces throughsimple setups. Imagine a sealed barrel, expertly drawn in figure 1.1, with a pipe connected tothe top. The pressure in the barrel, assuming it is completely sealed, is then dependent on theheight of the fluid in the tube. Think: if the tube were disconnected, the barrel empty, andthere was just the hole at the top, then that would be exposed to the atmosphere. Assuming thebarrel isn’t absurdly large, the pressure within would be the same as atmospheric pressure. Ifit were filled with water to the hole, it would have increasing pressure towards the bottom, butagain assuming it isn’t too big it wouldn’t be much different from Patm. Now, we put the tubeon, but don’t fill it. What’s the pressure? Have we really changed anything? No, we haven’t!Sure, there is a column of air in the tube, but the pressure at the opening of the barrel will stillbe Patm, as that is the pressure of the air at the bottom of the tube. This air pressure theninterfaces with the water in the barrel, so the pressure exerted is no different than if we didn’thave the tube.

Now, let’s start filling the tube up with water. Again, the barrel is perfectly sealed. Whathappens as the water level in the tube increases? You can integrate equation 1.1 and set it allup, but your intuition should tell you that pressure will rise. What happens when the waterlevel in the tube is 10m? (hint, we just talked about a straw that had 10m of water in it!) Sincethat pressure from the water column exerts itself on the water in the barrel, we would find thatthe pressure inside is suddenly 100 kPa higher than when there was no water column. Increasethe water height to 20 meters and suddenly the water pressure in the barrel is 200 kPa aboveatmosphere! If you then integrated this pressure over the area inside the barrel, you would find



Figure 1.1: a totally legit drawing of a barrel with some pipe contraption coming out of the top.

a tremendous amount of force on the walls, which at some point would lead to it bursting. Ofcourse that depends on the material properties and exact geometries.

The point from this thought exercise is to help you grasp how powerful but simple pressurecan be. A very tall but thin column of water can end up exerting very large amounts of pressure,even though the actual weight of the water itself is small.

1.4 Example: an airplane door

Airplane doors open inward before swinging out. The inside cabin air is pressurized relative tothe cruising altitude pressure, but still lower than atmospheric. This example will help illustratewhy.

Let’s say you’re cruising at 10,000m in a commercial airliner with your friend. You’re flying



international, so they’ve fed you dinner. You decided to have the beef, while your friend hadthe fish. Now, 1 hour later, your friend is entering a hallucinogenic state brought about by thefish. He decides that the airplane is too confining, so he goes up to the cabin door to open itand leave. Should you try to stop him?

Let’s see what pressure says you should do! We know there is a pressure difference betweenthe outside and inside of the plane, so we should calculate those values to determine the directionand magnitude of the resultant force. Lets start by assuming that the pressure inside the cabinis half that of regular atmospheric pressure, or 50kPa. Let’s also assume the cabin door isaround 1 meter wide by 2 meters tall. Now what’s the outside pressure?

You could look it up in a table, or online, or any number of sources, but let’s make it a littlemore work. The function describing the density in the troposphere (the first 10,000m or so ofthe atmosphere) is:

ρ =P

286.9(288.1− 0.0065 ∗ z)kg

m3(1.2)

This means we have density as a function of height in the atmosphere. This can then be pluggedinto our hydrostatic relation, equation 1.1, to get:

dP

dz=

−P ∗ g286.9(288.1− 0.0065 ∗ z)

or, by rearranging and plugging in 9.81m/s2 for g, we have:

dP

P=

−0.0342

(288.1− 0.0065 ∗ z)dz

I hope that you’ve done some basic differential equations, or at least separation of variables. Ifnot, then at this point just remember that you want all of one variable on one side, and theother variable on the other side. Here, you can see that all P variables are on the LHS, whileall z variables are on the RHS. Then you can integrate both sides with respect to their ownvariables. ∫ P (z=10,000)

P (z=0)

1

PdP =

∫ z=10,000

z=0

−0.0342

(288.1− 0.0065 ∗ z)dz

ln (P )

∣∣∣∣P?Patm

= − 0.0342 ln (288.1− 0.0065 ∗ z) ∗ 1

−0.0065

∣∣∣∣10,000

0

ln (P?)− ln (100kPa) = 5.26

[ln (223.1)− ln (288.1)

]ln (

P?100kPa

) = − 1.345

P? = 26, 000Pa

So the math was a little fun, but otherwise it was a straightforward plug and chug to get thepressure at 10,000m elevation. If you look up the value in a table, you’d see that we’re withina few percent, so all is well!



Now, we have the outside pressure, and we know the inside pressure, so the force exerted onthe cabin door will be:∑

Forces = Finside + Foutside = (−Pinside ni +−Poutside no) ∗Adoor

Remember how the definition of the force due to pressure has the minus sign in it. This meansyou also need to pay attention to the wall normal vectors. It’s pretty straightforward in thisexample, but for cases where the walls or surfaces are sloped, it becomes very important to payattention to n. Assuming ni (inside wall normal direction) is directed in the positive direction,and no (the outside wall normal direction) is in the negative direction, then the above equationcan be written to find:

(−50kPa−−26kPa) ∗ 2m2 = −48kN

This means the pressure difference across the two sides of the door leads to a net force of 48kNfrom the inside (pointing in the negative direction). In other words, your crazy friend wouldhave to pull inwards with 48kN of force to overcome the pressure difference on the door. Thisis equivalent to lifting ∼5000kg off the ground. Don’t think it’s going to happen.

1.5 Statics and moments

This utilizes the concepts of hydrodynamics and a force/moment balance. These problems endup being a good exercise in your calculus skills.

When a surface is horizontal (or equivalently at uniform depth) in a quiescent fluid, we canshow that the pressure acting on it is uniform. Likewise, when the fluid is a gas you can usuallyassume the pressure acting on a non-horizontal wall as uniform. However, when we depart fromthis, like an angled wall in water, we find that the pressure acting on the surface is not uniform.We must then perform a more general type of analysis.

1.5.1 Example: a dam wall with one fluid

Lets say we have a dam that is 100m high, 200m wide, holding up 100m of water, as seen infigure 1.2. What’s the total force exerted on the wall? Where would the equivalent point forceoccur to produce the same moment on the wall? This is a simple application of forces andmoments, so let’s start by writing an expression for force. We know that the pressure varieswith depth, since that’s exactly what the hydrostatic equation tells us! Water has a density of1000kg/m3, so we start by writing:

dP

dz= −ρg = −1000

kg

m3∗ 9.8

m

s2= −9800

N

m3(1.3)

This is how fast the pressure changes with height. Note we have assumed a positive z-directionas upward, so the negative value would mean increasing pressure with decreasing height. Wealso define zero elevation at the surface, so going into the water yields negative height.

By taking the reference pressure at the water surface as zero (in other words, looking at thedeparture from atmospheric pressure, or other-other words: gage pressure) we can calculate theforce on the wall by decomposing into infinitesimally small portions:

dF = PdA



Figure 1.2: schematic of a dam supporting a single fluid.

where dA is simply a small area, or the width (which is a constant 200m) times dz. Why dz?well, think on what is the variable here. Does the pressure vary with any other direction? No!Again, your intuition helps here!

Now, P varies in z based on equation 1.3. We can integrate to find the exact relationship:∫ top

bottomdP =

∫ top

bottom−9800 dz →

∫ P (top)

P (z)dP ′ =

∫ 0

z−9800 dz′

P (top)− P (z) = −9800(0− z) → P (z) = −9800zN

m2

So this tells us that the value of the pressure at a given elevation below the surface of the water.Note that you can see at 10m below the surface, the pressure is about 100kPa, one about 1atmosphere! So we’re doing OK thus far.

Now, the fractional area dA is simply 200 ∗ dz, so we can now write:

F =

∫ top

bottomdF

=

∫ top

bottomP dA

=

∫ top

bottom−9800z ∗ 200 dz

=

∫ 0

−100−1960000z dz

=− 980000z2

∣∣∣∣0−100

=0−−980000(−100)2

F =9.8GN

That’s a lot of force. Notice that we got a positive value, which is good news considering wedidn’t prescribe a direction the force is applied in! Even though it can be kind of annoying to



carry through all these minus signs, it will still get you to the correct spot if you are diligent inyour definitions.

How about the moment about the base of the wall? Instead of an infinitesimal force, let’scalculate the infinitesimal moment:

d ~M = ~r × d~F → dM = −(z − (−100)

)dF → dM = −

(z + 100

)dF

I’m using the right-handed coordinate system, so the forces on the dam will produce a positivemoment about the base of the dam. Also note that there is a z + 100 term, not just z. Thatsdue to the moment being about the base, so the greater the depth, the shorter the moment-arm.Therefore, we need to be better about our normal and define the force on the dam as acting inthe negative direction.

dF = −PdA

This means the force will be acting towards the left on the page. Then, working with our crossproduct, we get positive moments about the base. Plugging this in gives:

M =

∫ top

bottomdM

M =

∫ 0

−100−(z + 100) dF

=

∫ 0

−100(z + 100)P dA

=

∫ 0

−100(z + 100) ∗ (−9800 ∗ z) ∗ 200 dz

=

∫ 0

−100−1960000(z2 + 100z) dz

=− 1960000(z3

3+ 50z2)

∣∣∣∣0−100

=0−−1960000

(−1003

3+ 50(−100)2

)M =326GN ·m

Note that I was better with the directions of the forces and moments. The force from the fluidon the dam is applied in the negative direction, while the moment about the base is positive.The calculus worked how we had hoped it to!

Next, we can find where the equivalent point force would be applied. By taking the moment,and dividing by the total force, we find:

zcop =M

F

=326GN ·m

9.8GNzcop =33.3m



From this, we got a distance of 33.3m, which means the location we could apply the point forcewould be 1/3rd of the way from the moment axis, or 2/3rds of the distance below the surface.This is the location of our center of pressure, and since the pressure distribution is triangular,this makes sense! This simple setup is giving a result that matches what we should have seenin our mechanics classes.

1.5.2 Example: a dam wall with two fluids

Now we will change the setup ever so slightly to see what happens. Imagine we have two layersof fluid, where fresh water is floating on top of salt water, seen in figure 1.3. Let’s calculate outthe forces, moments, and center of pressure again.

Figure 1.3: schematic of a dam supporting two different fluids.

The pressure distribution will be slightly different. Let’s first look at the top 50 meters:∫ top

zdP =

∫ top

z−ρg dz

P (0)− P (z) =

∫ 0

z−9800 dz

P (z) =− 9800zN

m2if 0 ≥ z ≥ −50 (1.4)

So for the first 50 meters, we wouldn’t see anything different than the previous example. How-ever, once we cross to the next fluid, we need to account for the different density as well as thepressure at the fluid interface.



∫ P (−50)

P (z)dP ′ =

∫ −50

z−ρg dz′

P (−50)− P (z) =

∫ −50

z−10094 dz′

490000− P (z) =− 10094 ∗ z∣∣∣∣−50

z

P (z) =− 10094z − 14700N

m2if −50 > z ≥ −100 (1.5)

Notice that the second layer of a fluid has lead to a more complicated expression for our pressure.We have equation 1.4 for the first 50m in the fluid, then equation 1.5 for the second 50m in thefluid. That is how we found the value for P (−50).

To calculate the total force, we perform the integration as was done before. We will notethat the force is in the negative direction (putting the minus sign on it) and keep our expressionsin mind for the different z-locations:

F =

∫ 0

−100−PdA

=

∫ 0

−50−(−9800 ∗ z) ∗ 200 dz +

∫ −50

−100−(−10094z − 14700) ∗ 200 dz

=980000z2

∣∣∣∣0−50

+ (1009400z2 + 2940000z)

∣∣∣∣−50

−100

F =− 9.87GN

Notice that this force is just a little bit higher than the previous example? That’s due to thedenser fluid on the bottom increasing the overall pressure on the bottom half! Also notice howthe sign is negative? That’s due to us correctly accounting for the fact that the pressure actson the wall to move it in the negative x-direction.

The moment will be done in a similar manner, where the axis which we calculate the momentabout is located on the bottom of the wall, and making sure to account for the change in theexpression for pressure at the z = −50m point.



M =

∫ top

bottomdM

=

∫ 0

−100−(z + 100) dF

=

∫ 0

−100(z + 100)P dA

=

∫ 0

−50(z + 100) ∗ (−9800 ∗ z) ∗ 200 dz +

∫ −50

−100(z + 100) ∗ (−10094z − 14700) ∗ 200 dz

=

∫ 0

−50−1960000(z2 + 100z) dz +

∫ −50

−100−2018800z2 − 204820000z − 294000000 dz

=− 1960000(z3

3+ 50z2)

∣∣∣∣0−50

− (672933z3 + 102410000z2 + 294000000z)

∣∣∣∣−50

−100

=0−−1960000

(−503

3+ 50(−50)2

)+−

(672933(−50)3 + 102410000(−50)2 + 294000000(−50)

)−−

(672933(−100)3 + 102410000(−100)2 + 294000000(−100)

)M =328GN ·m

Just like the force, the moment is ever so slightly higher than the previous example was.Again this intuitively make sense, as we have a larger force acting on the wall. If we thencalculate out the center of pressure, we find:

zcop =M

F

=328GN ·m

9.87GNzcop =33.2m

This is only 0.1m lower than the previous center of pressure. Not very significant, but thenagain the density difference isn’t very significant either. However, this isn’t the centroid of atriangle now! This goes to show that you need to take the time to calculate out the pres-sures/forces/moments in order to accurately find where the center of pressure is.

Overall, this is to show you how being diligent with your signs and writing out your integralsis an important thing when calculating the center of pressure.

1.6 Rigid body motion summary

The previous examples illustrated the power of pressure, but we weren’t considering fluid flow.That is why we used the “Hydrostatic Equation” - the fluids were static! (Yes, the plane moves,but for the analysis we assume the outside pressure is constant for a given altitude). We can alsouse a similar analysis on fluid that is in motion, but only under a simplified case. Rigid body



motion is when the fluid particles do not move relative to one another. When this occurs, theforces on a fluid particle are only the normal and body forces, not shear stresses. So performingthe same type of analysis in both the x- and y-directions as was done in the z-direction (fordeveloping the hydrostatic equation) will give you expressions for pressure gradients in all threecoordinate directions:

∂P

∂x= −ρax (1.6)

∂P

∂y= −ρay (1.7)

∂P

∂z= −ρ(g + az) (1.8)

These equations of course depend on the way you define your coordinate system, but here weuse the standard positive z-direction being up. Also note that the derivatives are now partials,because pressure is a function of multiple variables.

The minus signs again can be confusing, but a little intuition helps us. If we were acceleratingin the positive x-direction, then we would see a negative pressure gradient with respect to x.This signifies that the pressure decreases as you move in the positive x-direction. Think of avery long car, completely sealed. If it accelerated at 1g, then the fluid within would have thelargest pressure at the back, and the smallest pressure up front. This is a negative pressuregradient, because the value drops as you move in the positive direction!

Likewise with the z−direction, we can think about what the equation is telling us. If wewere to be in an enclosed box with a fluid, then we accelerated in the negative z-direction at1g, we would find there is no pressure gradient. That is because accelerating downward at 1gis simply freefall! If we didn’t accelerate in the z-direction, then we’re back to hydrostatics. Ifinstead we accelerate in the positive z-direction, we would increase the pressure gradient, andthe pressure at the bottom of the box would increase, proportional to the amount we accelerate.

If you need any clarification on the derivation of these equations, draw a fluid particle andperform a force balance. Don’t ask me, I’m just a set of notes.

Additionally, under rigid body motion, the pressure from a sloped or curved surface can befound. Alternatively, the acceleration needed to produce certain surfaces can be found as well.For instance, an example taken from Prof, Smits’ book asks us to find the acceleration requiredto make water spill out of a cart, as seen in figure 1.4. However, getting to the equation thatdetermines this isn’t always the most straightforward.

We start by writing the total derivative of pressure:

dP (x, y, z) =∂P

∂xdx+

∂P

∂ydy +

∂P

∂zdz

This is a general expression for the total derivative of any function that depends on threevariables. Now this example doesn’t have any y-dependence, so we can instead write:

dP (x, z) =∂P

∂xdx+

∂P

∂zdz



Figure 1.4: schematic of an accelerated cart, taken from Prof. Smits’ book.

Next, we can plug in our expressions for the pressure gradients:

dP (x, z) = −ρax dx− ρ(g + az) dz

Finally, we want to find the slope of a surface (or specify the accelerations to achieve a particularslope). If we are on the surface of water in the atmosphere, we can say the pressure is constant.Think about a flat body of water - the pressure all along the surface is Patm. If the body of wateris then accelerated, it will slope somehow. However, the surface is still exposed to atmosphericpressure, so all along that surface the pressure is still constant. Therefore, for us, we can saysurface of the water will have a constant pressure, even when under acceleration! Therefore wecan write the following for the surface of water:

0 = −ρax dx− ρ(g + az) dz

or

ρax dx = −ρ(g + az) dz

or

− axg + az

=dz

dx

In this example, there is no vertical acceleration besides gravity. Therefore, the slope issimply −ax/g. Based on the dimensions given, the water level at the back of the cart needs toincrease in height by h/3, and the location in which the surface will “pivot” is at the halfwaypoint, or an x-distance of 3h/2. This leads to a negative slope, so we can then write:

−axg

=dz

dx

−axg

=−h

33h2

ax =2

9g

Note that the height needed to increase and the pivot point of the free surface were straight-forward to find here since the container is nearly full. If there were only a thin layer of water ofthe bottom, you’d have to get creative with defining the shape of the fluid in the container.



1.7 Specific weight and specific gravity

Specific weight, γ, is simply the gravitational force of a fluid per unit volume. In other words,it is the weight of the fluid divided by the volume. It can be expressed simply as:

γ =weight

volume=ρV g

V= ρg (1.9)

This assumed a constant density throughout the fluid volume, which we can assume true for afluid particle. We can then use the specific weight to re-write our hydrostatic relation:

dP

dz= −γ

Specific gravity is simply a ratio of the specific weight of a fluid to that of water. Thisrequires some reference state for water, but you can choose that.

SG =γfluidγwater

=ρfluid g

ρwater g=ρfluidρwater

(1.10)

So a specific gravity greater than 1 means the fluid (or object) has a density greater than water,and less than 1 means the density is less than water.

1.8 Buoyancy

Buoyancy is intimately related to the hydrostatic relation. It also factors in specific gravity,since the ratio of densities will determine just how buoyant force the object can exert. Astraightforward example is to look at a block of aluminum and a block of styrofoam.

SGAl = 2.7 and SGsty = 0.04

We can then show how each of these would behave in water. Your intuition will tell you that thealuminum should sink, and the styrofoam float. We can mathematically show that! UtilizingArchimedes principle, we can solve for the amount of water that each would displace when atequilibrium. Starting with styrofoam: ∑

Forces = 0 in equilibrium

Fb − Fweight = 0

ρwater ∗ g ∗ Vdisplaced − ρsty ∗ g ∗ Vblock = 0

ρwater ∗ g ∗ Vdisplaced =SGsty ∗ ρwater ∗ g ∗ VblockVdisplaced =SGsty ∗ Vblock

= 0.04 ∗ Vblock

This tells us that the displaced volume of water will be less than the volume of the styrofoamblock. In layman’s terms, that means the block will float! In general, the expression Vdisp =SG ∗ Vobject will tell you how much water the material will want to displace. So long as thespecific gravity (ratio of densities) of the material is less than 1, then the object will want to



displace less volume than it has. This means some portion of it will remain above the surfaceof the water. If the SG is greater than 1, then it will want to displace more water than it can.How does something displace more volume than it has itself? It can’t, which means it sinks!

Going through the math for the amount of volume the aluminum block would want todisplace, we would find Vdisp = 2.7 ∗Vblock. Now, just like we said, it can’t displace more volumethan it has, so the block would sink.

This methodology is how you can determine the size of a ship needed to move cargo. If youwant to move 1000 metric tons of cargo across the ocean, you need to build a ship that candisplace 1000 metric tons of water and not sink!

1.8.1 Example: floating some aluminum

Lets imagine a container that is 2x2 meters square, and tall enough to contain the followingproblem (poorly drawn in figure 1.5). We want to float a 1x1x1 meter cube of aluminum in thiscontainer. What is the minimum amount of styrofoam that would be required to keep the blockfrom sinking, and how much does the fluid depth increase in the container?

Figure 1.5: the basic schematic for floating an aluminum block in water.

In order to float the block, we need to displace an amount of water equal to its weight.

FAl =ρAl ∗ g ∗ VAl

=2.7 ∗ 1000kg

m3∗ 9.8

m

s2∗ 1m3

=26460N



We thus need to displace an equivalent amount of water, or:

Fwater =26460

ρwater ∗ g ∗ Vwater =26460N

Vwater =26460N

9.8ms2∗ 1000 kg

m3

=2.7m3

This seems familiar right? That’s because we’re putting the material in water, which is whata specific gravity is defined with! For every unit volume of aluminum, it requires 2.7 units ofwater to balance the force. That’s exactly what specific gravity is!

Now we need to figure out how much additional buoyancy force we require to float. Think:the aluminum block will sink, but since it still displaces some water, it will have an upwardbuoyancy force acting on it. Ever moved rocks in a lake, or an anchor perhaps? It’s easierwhen it’s in the water than when it’s in air. That’s buoyancy working for you! So lets see whatremaining force we need to overcome:

Fremaining =Fbuoyancy − Fweight=ρwater ∗ g ∗ Vdisplaced − 26460N

=9800N − 26460N

=− 16660N

This means there is still a remaining 16660N of force that needs to be overcome for the aluminumblock to be neutrally buoyant. Now we look into the amount of styrofoam needed to overcomethis force. We know it floats, as was seen previously. We can then see how much force it willexert. From our previous calculations, we were able to show that the volume of water displacedby styrofoam will be 0.04 times the volume of the styrofoam itself. That was because there issome amount of weight to the object, which means there must be some displacement to resultin a buoyancy force. Therefore, for every 1.00m3 of styrofoam, we get an effective 0.96m3 ofdisplacement.

To obtain the 16660N of buoyancy force, we thus need:

16660N =ρwater ∗ g ∗ Vdisplaced16660N =1000 ∗ 9.8 ∗ Vdisplaced1.70m3 =Vdisplaced

This amount of displaced water would yield the appropriate buoyancy force. However, we knowthat there is only 96% of the volume of styrofoam that contributes to additional buoyancybeyond what is required for the weight of the material. Therefore, we need:

0.96 ∗ Vstyrofoam =1.70m3

Vstyrofoam =1.77m3

So to float the 1m3 block of aluminum, we would need 1.77m3 of styrofoam to attach to it.



Let’s do this in a different manner. I had gone through and methodically laid out each andevery step to see what volume of styrofoam was needed. Instead, we can start with a basic forcebalance! This is a hydrostatics problem, so all the forces should balance. Therefore:

∑~F =0

Fbuoyancy − Fweight =0

Fb,Al + Fb,st − Fw,Al − Fw,st =0

ρwater ∗ g ∗ VAl + ρwater ∗ g ∗ Vst − ρAl ∗ g ∗ VAl − ρst ∗ g ∗ Vst =0

ρwater ∗ g ∗ (VAl + Vst)− SGAl ∗ ρwater ∗ g ∗ VAl − SGst ∗ ρwater ∗ g ∗ Vst =0

ρwater ∗ g(VAl + Vst − SGAl ∗ VAl − SGst ∗ Vst) =0

(1− SGAl) ∗ VAl + (1− SGst) ∗ Vst =0

−1.7 ∗ 1m3 + 0.96 ∗ Vst =0

Vst =1.77m3

We were able to get the exact same value using a different method. Intuition should tell you thateach block will have a force due to it’s weight, and a buoyancy force due to the displaced water.Summing the contributions of these forces will balance out to zero, since it’s hydrostatics. Theonly unknown is the size of the styrofoam block, so no additional equations are needed.

Now, we know the volume of styrofoam needed to float the aluminum, and we know thevolume of aluminum, so let’s calculate how much the water level has increased in the tank. Weknow the tank has a 2m x 2m footprint, so the calculation is pretty straightforward:

h =V

A

=1m3 + 1.77m3

4m2

=0.69m

The height of the water then increases 0.69m due to the additional material in the water.

1.8.2 Example: floating in a different fluid

Let’s do the exact same setup, but now the fluid is honey! Assuming honey has SG = 1.42,what is the new volume of styrofoam needed?

Doing the force balance again, we find:

Fb,Al + Fb,st − Fw,Al − Fw,st =0

ρhoney ∗ g ∗ VAl + ρhoney ∗ g ∗ Vst − ρAl ∗ g ∗ VAl − ρst ∗ g ∗ Vst =0

SGhoney ∗ ρwater ∗ VAl + SGhoney ∗ ρwater ∗ Vst − SGAl ∗ ρwater ∗ VAl − SGst ∗ ρwater ∗ Vst =0

(SGhoney − SGAl) ∗ VAl + (SGhoney − SGst) ∗ Vst =0

(1.42− 2.7) ∗ 1m3 + (1.42− 0.04) ∗ Vst =0

Vst =1.08m3



The volume of styrofoam needed to float the aluminum in honey has decreased compared tothat needed in water. That’s due to the honey being a more dense fluid! Every unit volume ofhoney displaced has a higher buoyancy force it exerts compared to water.

1.9 Control Volumes

Now we’ll look into some basic applications of mass flux and control volumes. The basic conceptwe will be sticking to is the conservation of mass: all mass must be accounted for in one way oranother. This can be expressed as:

min −mout = ∆msys (1.11)

In other words, the difference between the mass entering the system and the mass leaving thesystem is the amount of mass accumulated in the system.

1.9.1 Example: accumulating mass

A quick demonstration of equation 1.11 can be shown through the following setup. A tank ofwater with cross-sectional area of 10m2 has an inflow of 10kg/s and outflow of 5kg/s, sketchedin figure 1.6. Find the rate in which the water level in the tank changes.

Figure 1.6: a tank with differing flow rates on the inlet and outlet.

We start by looking to equation 1.11 and modifying it for a flow rate rather than a fixedmass. in other words, we differentiate the equation with respect to time. Plugging in values:

min − mout =msys

10kg

s− 5

kg

s=msys

5kg

s=msys



So the system accumulates 5kg/s of water. We can then look to how this affects the height.The mass accumulation is simply a rate of change of the total mass of the system:

msys =d

dtmsys =

d

dtρV =

d

dtρAh = ρA

dh

dt

Since the density isn’t changing, and the cross sectional area is a constant, the height is the onlyparameter that changes with time. Therefore, we can find the change in height with time:

5kg

s=1000

kg

m3∗ 10m2 ∗ dh

dt

0.0005m

s=dh

dt

The water level in the tank will increase by 0.0005m/s, or 1.8m/hr.

1.9.2 Example: non-uniform flow

The previous example only gave the total mass flux into and out of the tank. In general, manyof these types of problems assume a uniform inflow and outflow condition. Let’s now look at acase of non-uniform flow in the inlet and outlet conditions.

Assume a parabolic inlet flow with max velocity of 5m/s, and triangular outflow as shownin figure 1.7. If the inlet has an area of 1m x 1m, and the outlet has an area of 0.75m x 1m,what is the max outlet velocity, assuming an incompressible fluid?

Figure 1.7: nozzle with incompressible flow (no mass accumulation).

Once again, this is an application of equation 1.11. We want the mass flux version, sodifferentiating gives:

min − mout = msys

But in this case, we are assuming there is no mass accumulating in the system (because it’s asteady flow problem with an incompressible fluid). All mass influx must be balanced by mass



outflux.

min − mout = 0

min = mout

To get the max velocity at the outlet, we need to find the total mass flux into the system. Weare given a parabolic velocity profile on the inlet, so we can integrate to find min.

min =

∫∫surface

ρUdA

First, let’s determine what the expression for U is. It is parabolic, reaching a maximum at thecenterline, and being zero at the edges. We can therefore write it as:

Uin = Umax ∗[1−

( y

0.5 ∗ h)2]

In this, y is the vertical coordinate. We define the centerline as 0, so the top and bottom of theinlet are at 0.5h and −0.5h, respectively. Now, plugging in our known values, the velocity canbe expressed as:

Uin = 5m

s∗[1−

( y0.5

)2]We next write dA based off it having a unit width (into the page). Since the only variation is inthe y-direction, this gives dA = 1 ∗ dy. Then, integrating from −0.5m to 0.5m, since the heightis 1 meter, we get:

min =

∫∫surface

ρUdA

=ρkg

m3∗ 1m ∗

∫ 0.5

−0.55m

s∗[1−

( y0.5

)2]dy

=5ρ

[y − 4

3

(y)3]∣∣∣∣0.5

−0.5

kg

m · s

=10

3ρkg

s

Now that we have the mass flow in, we know it has to be the mass flow out. Also, we knowthe area out, and the velocity profile shape. Therefore, we can put these together to find themaximum velocity at the outlet. Since it’s triangular in shape, we can write the profile as:

Uout =Umax(1− y

0.5 ∗ h)

for positive y

=Umax(1 +

y

0.5 ∗ h)

for negative y



Here, the zero point is on the centerline again, and the total height is 0.75m, so we can writethis as:

Uout =Umax(1− 8y

3

)for positive y

=Umax(1 +

8y

3

)for negative y

Now, since the outflow is symmetric, we could always just integrate one half and double (whichcould have also been done on the inlet, but here since it’s discontinuous it makes more sense).Then, writing out the total mass outflux, which must be equal to its influx, we get:

min = mout =

∫∫surface

ρUdA

10

3ρkg

s=ρ

kg

m3∗ 1m ∗ 2 ∗

∫ 0.375

0Umax ∗

(1− 8y

3

)dy

5

3

m2

s=Umax

(y − 4

3(y)2

)∣∣∣∣0.375

0

5

3

m2

s=Umax ∗

3

16m

80

9

m

s=Umax

So we find that the max velocity at the outlet is 8.89m/s.

Again, as with the buoyancy example, this was stepped through methodically. If instead yousaw that the mass fluxes must balance, and therefore equated the integrals, you would get thesame answer in fewer steps:

min = mout

ρ ∗ 1 ∗ 2 ∗∫ 0.5

05 ∗[1−

( y0.5

)2]dy =ρ ∗ 1 ∗ 2 ∗

∫ 0.375

0Umax ∗

(1− 8y

3

)dy

5 ∗[y − 4

3

(y)3]∣∣∣∣0.5

0

=Umax ∗(y − 4

3(y)2

)∣∣∣∣0.375

0

5

3=Umax ∗

3

1680

9=Umax

1.10 Material Derivative

One final introductory topic that needs to be addressed is the material derivative. Hopefullyyou’ve already seen it, but the basic idea is that the material derivative is a way to relatechanges in the Lagrangian reference frame to changes in the Eulerian referenceframe. To refresh your memory, the Lagrangian reference frame is a way to look at a fluid



particle where you follow along with it. Think about being a small fluid particle in the air thattravels around, and everything you experience is because of the surroundings acting on you asyou move through them. The Eulerian reference frame is a way to look at a single fixed pointin the flow. Now you are an observer looking at one location, seeing many fluid particles passthrough that point.

The world of classical mechanics operates within the Lagrangian frame, but with fluid me-chanics, we like to think in terms of the Eulerian frame. This stems from the fact that it becomesextremely difficult to track each and every fluid particle in a system - we’re talking orders ofmagnitude around 1020 particles just in a moderately sized room! Our current computational ca-pabilities cannot handle this, let alone people working it out by hand back in the early foundingdays of fluid mechanics. So instead, we desire to be able to look at a single point and evaluatewhat happens at that location.

A simple example would be looking at a rocket nozzle. Do you care about the dynamics ofeach infinitesimally small fluid particle exiting the nozzle, or is it more productive to have anunderstanding of the flow at the exit of the nozzle? The Eulerian reference frame allows us tolook at the nozzle exit and understand the flowfield without tracking each particle.

Our problem is that mechanics works with Lagrangian concepts. We desire to take theEulerian reference frame and apply the Lagrangian concepts we are more familiar with. Ingeneral, any property of a fluid particle can be described as a function of time and space whenwe utilize the Eulerian reference frame. We observe a location (x, y, z) and watch it changein time t. For example, the temperature, density, or any property Φ at that position can bedescribed as:

Φ = Φ(t, x, y, z)

Since the Lagrangian frame follows a particle that can move, its position will change in time.Therefore:

x = x(t), y = y(t), z = z(t)

thus

Φparticle = Φ(t, x(t), y(t), z(t)

)Looking at the time derivative of the particle (and making sure we use the chain rule), we thensee:

d

dtΦ =

∂Φ

∂t+∂Φ

∂x

∂x

∂t+∂Φ

∂y

∂y

∂t+∂Φ

∂z

∂z

∂t

=∂Φ

∂t+∂Φ

∂xU +

∂Φ

∂yV +

∂Φ

∂zW

This can be compactly expressed as:

d

dtΦ︸︷︷︸

Lagrangian

=DΦ

Dt=∂Φ

∂t+(~U · ∇

)Φ (1.12)

This is our material derivative of the quantity Φ, which will be seen throughout the course.Note that we will reference it by utilizing D/Dt, since that is a more distinct nomenclaturecompared to the Lagrangian d/dt.


Chapter 2

Continuity and Momentum

2.1 Continuity equation

After utilizing the relatively simplistic equation for mass balance (equation 1.11), we want todevelop a“fluid dynamics” version of it, which we call the continuity equation. Recall thisapplies in the Eulerian frame, where we fix a control volume and look at the flow through it.The simple way to explain the continuity equation is taking the concept of conservation of massand re-wording it:

mass in−mass out = change in mass

−or−rate of change of mass in C.V + total mass flux out of C.V = 0

Mathematically, we expressed this as:

d

dt

∫∫∫volume

ρ dV +

∫∫surface

ρ~U · d~S = 0

or, using the unit normal, we can equivalently state:

d

dt

∫∫∫volume

ρ dV +

∫∫surface

n · ρ~U dS = 0 (2.1)

It is important to realize that both density and velocity can be a function of position and time,or ρ = ρ(x, y, z, t) and ~U = ~U(x, y, z, t). This makes things considerably more complicated inthe most general sense of things, but lucky for us we usually get problems where one or bothare constants. Just realize there is no restriction on their temporal or spacial dependence.

Also note that the integral over the surface represents a net outflux of mass. This is becausethe dot-product of n with ~U will be positive if they are in the same direction, and n is positiveoutwards. Therefore, a positive surface integral will have to be balanced with a negative volumeintegral, implying a net outflux of mass and decrease in system volume.

Now, there is another way to obtain the continuity equation. We can utilize a wonderfulmathematical tool called Reynolds Transport Theorem. If we have an extensive property


CHAPTER 2. CONTINUITY AND MOMENTUM

B, and intensive property b, the RTT allows us to relate the change of the extensive propertyin a Lagrangian view to the Euerian view through:

d

dt

∫∫∫V (t)

b dV =d

dtBsys

︸︷︷︸Lagrangian

=d

dt

∫∫∫Vs

b dV +

∫∫Ss

n · b~U dS

︸︷︷︸Eulerian

(2.2)

You may have seen this equation before. It is sometimes re-cast in different forms, where insteadof b, there is b ∗ ρ. I am defining b here as an intensive property (per unit volume) where thelater case has b defined per unit mass. If this gets confusing, don’t fret. We won’t be focusingon RTT so much, mainly on the results that it can provide us.

For our convenience, let’s review the Lagrangian and Eulerian viewpoints:

1. Lagrangian: Following the fluid particle/system. Boundaries can deform so there is noflux through them.

2. Eulerian: Fixed Control Volume that allows a flux through boundaries. You look at thewhole flow field rather than individual fluid particles.

Therefore, looking to the LHS of equation 2.2, we can see that this is the classical mechanicsviewpoint, where as the RHS is the continuum mechanics approach. The RTT is the bridgebetween fluids and classical mechanics!

Now let’s look at an intensive property, such as density. if we set b = ρ, we plug it into theLHS of equation 2.2 and get:

d

dt

∫∫∫V (t)

b dV =d

dtBsys

d

dt

∫∫∫V (t)

ρ dV =d

dtBsys

d

dtm =

d

dtBsys

So, by defining our intensive property b as density, we found the extensive property Bsys to bemass. Is mass an extensive property? Yes! Think about what happens if you cut a volume inhalf: the mass changes. Think about the intensive property density: it isn’t dependent on thevolume of material (in the strictest of senses, of course internal density gradients change things,but let’s not get too complicated here).

In the classical mechanics point of view (Lagrangian), what happens to the mass of a closedsystem? It is conserved! Therefore:

d

dtm = 0



Now let’s look to the RHS of equation 2.2 and plug in our intensive property:

d

dt

∫∫∫Vs

b dV +

∫∫Ss

n · b~U dS

d

dt

∫∫∫Vs

ρ dV +

∫∫Ss

n · ρ~U dS

and equating the RHS to the LHS, we get:

d

dt

∫∫∫Vs

ρ dV +

∫∫Ss

n · ρ~U dS = 0

This is simply equation 2.1! There are many ways to derive the continuity equation, but ReynoldsTransport Theorem is just one way to quickly obtain it. There are long-winded proofs on howto obtain RTT, but let’s not worry about that.

2.1.1 Differential form of the continuity equation (bonus material)

We can re-cast the continuity equation in differential form. The integral form allows us to lookat a control volume and understand the whole system while “glossing over” the details withinit. The differential form is a powerful tool for understanding the flow field itself.

An important point to remember is that the volume/surface in which the integrals occurover in the Eulerian frame are fixed in space. They don’t move or deform (we can sometimeshave a moving frame if needed, but lets not worry about that for now). Because they are fixed,the time derivative can be moved inside the limits of integration. That’s because the boundariesaren’t dependent on time! So the continuity equation can be re-written as:∫∫∫

Vs

∂

∂tρ dV +

∫∫Ss

n · ρ~U dS = 0 (2.3)

Note that when moving the time derivative within the integral, we change it to a partial deriva-tive with respect to time. That’s because the integral “kills” the spacial dependence densitymay have when you integrate in space, so there is only a (possible) time dependency remaining.By moving into the integral, there still may be a spatial dependency on the density, so we mustbe clear in that it’s a partial derivative.

Another tool we can use is the divergence theorem, also known as Gauss’s theorem. Itrelates the flux of a vector field through a surface to the change of the vector field within thevolume bound by that surface. In mathematical terms, we can express this as:∫∫∫

V

(∇ · ~F

)dV =

∫∫S

(n · ~F

)dS (2.4)

By applying equation 2.4 to the equation 2.3, we can now write it as:∫∫∫Vs

d

dtρ dV +

∫∫∫Vs

∇ ·(ρ~U)dV = 0



Since both integrals apply over the same volume, they can be combined:∫∫∫Vs

[∂ρ

∂t+∇ ·

(ρ~U)]dV = 0

We see that this integral will be zero. Since the volume in which we integrate over is arbitrary(think: you can define a control volume however you want) then it will be zero no matter whatboundary of integration is chosen. This in turn means the integrand itself must be zero, givingus:

∂ρ

∂t+∇ ·

(ρ~U)

= 0

Expanding out the dot product (a good exercise to practice your vector calculus skills) gives us:

∂ρ

∂t+(~U · ∇

)ρ+ ρ

(∇ · ~U

)= 0

We can make one more simplification by utilizing the material derivative (which we introducedin equation 1.12) to get the final differential form of the continuity equation:

Dρ

Dt+ ρ(∇ · ~U

)= 0 (2.5)

−with−D

Dt=

∂

∂t+ ~U · ∇

2.1.2 Example: continuity on a nozzle

We know how to do a simple mass balance from the previous chapter. Let’s slightly modifya previous example (shown in figure 2.1) and apply the continuity equation to investigate thebehavior of the system.

We previously assumed a steady, constant density flow to solve for the max velocity at theoutlet. Instead, we now have values prescribed at the inlet and outlet, then apply equation 2.1to gain some insight into this system. First, let’s assume we know the velocity profiles at theinlet (parabolic) and exit (triangular), the depth is 1m into the page, and we’re given the densityat the inlet and exit. What is happening to the mass of the system?

d

dt

∫∫∫volume

ρ dV +

∫∫surface

n · ρ~U dS = 0

The first term is unknown, while the second term is made of variables we have. The sum of thetwo should be zero, so it seems we’re good to go (one equation, one unknown). Plugging in thevelocity profiles and densities:

d

dt

∫∫∫volume

ρ dV =−∫∫

surface

n · ρ~U dS

=−∫∫inlet

n · ρ~U dS −∫∫outlet

n · ρ~U dS

=−∫ 1

02 ∗∫ 0.5

0(−i) · 5ρin

[1−

(4y2)]i dy dz −

∫ 1

02 ∗∫ 0.375

0(i) · 10ρout

[1− 8

3y

]i dy dz



Figure 2.1: modified example for applying continuity equation

Let’s note a few things at this point. First, I’ve performed the integration in y over half thearea because it is a symmetric profile, thus there is an additional 2 in each integral. Second,I’ve put in the unit normals, −i in the inlet, and i in the outlet. Third, I’ve specified the unitvector direction of the velocity, which is i in both the inlet and outlet. Continuing the math:

d

dt

∫∫∫volume

ρ dV =1 ∗ 2 ∗ 5 ∗ 1.125 ∗∫ 0.5

0

[1−

(4y2)]dy − 1 ∗ 2 ∗ 10 ∗ 1 ∗

∫ 0.375

0

[1− 8

3y

]dy

=11.25

[y − 4

3y3

]∣∣∣∣0.50

− 20

[y − 4

3y2

]∣∣∣∣0.375

0

=3.75− 3.75

=0

So we find that the time rate of change in the overall mass in the system is zero. The system isnot accumulating or losing mass! We refer to this type of situation as a steady flow, where thesystem isn’t changing in time. In this instance, we had a density that was higher at the inletthan the outlet, so there is obviously some sort of density gradient within the system. However,because we’re doing a control volume analysis, we don’t care about the exact details of thesystem within the boundaries. In fact we can’t even solve for the details within the boundaries -we only know the overall change within the system (in this example, 0). Using a control volume,the only detail we have (or could potentially solve for) is the fluxes at the boundaries!



2.2 Momentum Balance

Just as we did with density, lets utilize another quantity, specific momentum b = ρ~U . Substi-tuting this for our intensive property in the RTT then gives the following:

d

dt

∫∫∫V (t)

b dV =d

dt

∫∫∫Vs

b dV +

∫∫Ss

n · b~U dS

d

dt

∫∫∫V (t)

ρ~U dV

︸︷︷︸Lagrangian

=d

dt

∫∫∫Vs

(ρ~U)dV +

∫∫Ss

n ·(ρ~U)~U dS

d

dt

(m~U

)︸︷︷︸

Newton’s 2nd

=d

dt

∫∫∫Vs

(ρ~U)dV +

∫∫Ss

(n · ρ~U

)~U dS

∑~F =

d

dt

∫∫∫Vs

(ρ~U)dV +

∫∫Ss

(n · ρ~U

)~U dS (2.6)

This is a simplified form of the momentum equation. Some notes:

1. The LHS of equation 2.6 was obtained by noting that the integral of ρ~U over the volumein the Lagrangian view is the linear momentum of the entire system. That’s once againbecause no mass enters or exits the Lagrangian control volume. The rate of change oflinear momentum is the true form of Newton’s Second Law, so we can re-write the LHSas the sum of the forces applied on the system.

2. The RHS of equation 2.6 has two parts: the first being the time rate of change of mo-mentum within the volume, the second being the net flux of momentum out of the controlvolume.

We will expand on equation 2.6 in later chapters, and express the LHS in terms of pressure,surface, and body forces. Either way, we can see that RTT is a powerful tool. Without goinginto the details of deriving the forces on the system, we’ll use an example to show how theseforces arise.

2.2.1 Example: force on a nozzle

Let’s explore equation 2.6 by applying it to the previous problem! We know the inlet and outletconditions, and we also know that the system is steady in time. This allows us to simplifyequation 2.6 even further:

∑~F =

��

��

��*0

d

dt

∫∫∫Vs

(ρ~U)dV +

∫∫Ss

(n · ρ~U

)~U dS

∑~F =

∫∫Ain

(n · ρ~U

)~U dS +

∫∫Aout

(n · ρ~U

)~U dS



Figure 2.2: it’s the previous problem again!

We split the surface integral to show there is a momentum flux in and momentum flux out,and they must balance with possible forces on the system. To determine what forces we mighthave, let’s draw the control volume and list out what could be acting on it. I’ll do this C.V. forconvenience:

By defining the C.V. in this fashion, it allows us to not worry about any forces that couldbe acting on the surface of the top and bottom of the C.V. If it coincided with the inner wallsof the nozzle, there could be frictional forces from the fluid! However, by “cutting through” thewalls, we now have some sort of external force on our system. Think: if there was a net forcefrom the flow, the walls would want to move. By having the C.V go through the wall, we mustaccount for that force acting on the wall to hold it in place. Also, there could be pressure forcesat the inlet and outlet, but let’s just assume there isn’t for now. Finally, we will assume gravityis down, so no gravitational forces will be affecting the flow.



With all this info, we can finally write our momentum balance as:

~Fext =

∫∫Ain

(n · ρ~U

)~U dA+

∫∫Aout

(n · ρ~U

)~U dA

Now we need to plug in what we know and solve:

~Fext =

∫∫Ain

(− i · 1.125 ∗ 5

[1−

(4y2)]i

)︸︷︷︸

n·ρ~Ui

[1−

(4y2)]i︸︷︷︸

~Ui

dA+

∫∫Aout

(i · 1.0 ∗ 10

[1− 8

3y

]i

)︸︷︷︸

n·ρ~Uo

[1− 8

3y

]i︸︷︷︸

~Uo

dA

=−∫ 1

02

∫ 0.5

01.125 ∗ 5

[16y4 − 8y2 + 1

]i dy dz +

∫ 1

02

∫ 0.375

01 ∗ 10

[64

9y2 − 16

3y + 1

]i dy dz

=− 11.25

∫ 0.5

0

[16y4 − 8y2 + 1

]i dy + 20

∫ 0.375

0

[64

9y2 − 16

3y + 1

]i dy

=− 11.25

[16

5y5 − 8

3y3 + y

]∣∣∣∣0.50

i+ 20

[64

27y3 − 8

3y2 + y

]∣∣∣∣0.375

0

i

=− 3i+ 2.5i

~Fext =(− 0.5 i, 0 j, 0 k

)N

It looks like the overall force balance was only in the x-direction, and has a negative value.Also note the units; they’re a force as expected. We didn’t carry it through each step, but youshould see that they would work out from (kg/m3)∗ (m/s)∗ (m/s)∗ (m2) to give you kg ∗m/s2,or Newtons.

We should expect the force to only be in the x-direction, as that’s the only direction flowis in. What about the negative value? That signifies the external force on the control volumemust be applied in the negative x-direction, or towards the left side of the page. Does that makesense? Let’s think about a nozzle on the end of a hose. The flow is slower in the body of thehose, then reaches the entrance of the nozzle, accelerates, and exits out the end of the nozzle.If there weren’t threading holding the nozzle into place, it would want to pop off and fly awayfrom the hose. It therefore requires an external force in the opposite direction of the flow tohold it on. This is exactly what we found in this example!

The most complicated part (for me) when it comes to a momentum balance is rememberingthe dot products. This was a quasi-one dimensional problem (no balance needed for the y- orz-components), so it was pretty straightforward. Introducing additional directions can makethings a little complicated.

2.2.2 Example: Balance with angled velocities

Let’s do one last example with multiple directions of velocity. The previous example was astraightforward application of the momentum equations because there was no ambiguity overthe direction of flux or the component of momentum. This example will show how you need tothink about what components of the vector apply to what terms in the equation.



Suppose we have a simple square duct as shown in figure 2.3. We desire to determine the

Figure 2.3: flow where the velocity has two components.

outflow velocity and balance of forces on the system. We will assume that the flow is steady,has uniform inflow and outflow conditions, and has no contributions from pressure.

Start by drawing a control volume:

Once again, I’ve drawn the C.V. in a way that allows us to not worry about possible surfaceforces. If the C.V. had coincided with the inner walls, we would need to think about what ishappening on the surface of the C.V. where the fluid is interacting with it.

Starting with the continuity equation, we should apply the integral form, since this is a C.V.application. Recall, the integral approach allows us to look at inflow and outflow conditionswhile “glossing over” the details of the flow. The differential form would require us to know theflow at all locations in the problem.



Since we are steady, we can simplify equation 2.1 to get:

��

��*0

d

dt

∫∫∫volume

ρ dV +

∫∫surface

n · ρ~U dS =0

∫∫Ain

n · ρ~U dA+

∫∫Aout

n · ρ~U dA =0

Fortunately, we have constant areas, constant density, and velocities that are constant. Thisallows considerable simplification. However, we have to be careful about the unit normal for thesurface and the direction of flow. For the outlet:

~Uo =(Uo i, 0 j, 0 k

)ms

and n = i

Therefore, at the outlet, n · ~Uo = Uo. The inlet is a little more complicated. However, we justneed to break the flow out into its components:

~Ui =(|Ui| cos 60 i, |Ui| sin 60 j, 0 k

)ms

and n = −i

So at the inlet, n · ~Ui = −Ui cos 60. We then can solve:

−ρ ∗ 10 ∗ cos 60 ∗ 1 + ρ ∗ Uo ∗ 1 =0

Uo =5m

s

Continuity tells us that the mass flow in must balance the mass flow out (because it is steady),and only the x-component of velocity contributed to this mass flux. That’s because there isn’ta surface in which the y-component of velocity could enter the system.

Now to momentum, we simply apply the same treatment to the steady momentum equation.We know the inflow and outflow velocities, so let’s plug them in:

∑~F =

��

��

��*0

d

dt

∫∫∫Vs

(ρ~U)dV +

∫∫Ss

(n · ρ~U

)~U dS

~Fext =

∫∫Ai

(n · ρ~Ui

)~Ui dA+

∫∫Ao

(n · ρ~Uo

)~Uo dA

~Fext =ρ ∗[− i ·

(5 i, 5√

3 j, 0 k)]

︸︷︷︸n· ~Ui

(5 i, 5√

3 j, 0 k)︸︷︷︸

~Ui

∗1 + ρ ∗[i ·(5 i, 0 j, 0 k

)]︸︷︷︸

n· ~Uo

(5 i, 0 j, 0 k

)︸︷︷︸~Uo

∗1

~Fext =ρ ∗ −5 ∗(5 i, 5√

3 j, 0 k)

+ ρ ∗ 5 ∗(5 i, 0 j, 0 k

)~Fext =

(0 i,−25

√3ρ j, 0 k

)All we have is a force in the y-direction! The momentum flux in the x-direction balances out sothere is no resultant force (think - same area, density, and velocity at the inlet and outlet in the



x-direction). However, the y-direction has a force holding the duct in place. That is becausethere is a net flux into the C.V. that carries y-momentum. Being diligent with your dot productis very important when dealing with momentum fluxes.

Here’s another way to think about momentum vs. mass flux.

• MASS: The velocity dotted with the surface normal provides the flux itself. It carrieswith it certain properties. In the case of continuity, we’re looking at a mass per unitvolume, which is a scalar. That means it has a value at a point. Therefore, for continuity,it’s like we’re carrying an amount of mass through the surface, and it’s dependent on thevelocity going into the surface.

• MOMENTUM: We use the same velocity dotted with the surface normal to providethe flux. No matter what it is we are “fluxing,” it is given a magnitude of n · ~U . Formomentum, we’re fluxing some momentum per unit volume (ρ~U now), which is a vectorquantity. If there was no velocity normal to the surface which we pass through, then therewould be no flux, thus no momentum being carried into the C.V. However, with a flux intothe surface, it carries the momentum of the fluid into it. Think: a small particle just onthe outside of the C.V can have x-, y-, and z-momentum. An infinitesimally small amountof time later, it will have crossed the control surface (thanks to the flux into it), but stillcarry its momentum.

Big wall of text, I know, but I hope there is a bit of insight in it for you. The point I’m trying toconvey is that for our example, only the x-component of velocity carried mass into the system.But that same component is responsible for the flux of any parameter we want to look at. Thismeans that, for momentum, a y-component of velocity will contribute to the force balance eventhough it doesn’t contribute to the flux.


Chapter 3

The Momentum Equation

3.1 Integral & Differential momentum equation

Our work with the momentum equation thus far has been limited to the integral form, and asimplified one at that. There are a number of forces that can be applied to a fluid particle, solets break down the few that are common in fluids.

We know the pressure acting on a face is the magnitude of the pressure integrated over thearea, and acts in the opposite direction of the surface normal. We represent this as:

~Fpressure = −∫∫S

nP dS (3.1)

Note that the pressure is a scalar, but when acting on the normal n, we get a vector. Additionally,the minus sign indicates that a pressure is pushing on a surface, since it’s direction is oppositethe surface normal. Another force to consider is the gravitational force, which is just a bodyforce. This means it acts throughout the body (not just on the surface like pressure did). Weknow it is proportional to the weight, so we take an infinitesimal volume, calculate the weight,and sum them up to get the total gravitational force:

~Fgravity =

∫∫∫V

ρ~g dV (3.2)

Note that the direction of the force is set by the gravitational vector. This is why we ignoregravitational effects when considering flow in the x-direction: gravity often applies in the y- orz-direction (under traditional coordinate systems).

The last force we should consider (for now) is the viscous force. We’ve been neglecting itthus far, and for good reason. It likes to make things a bit more complex. However, it is a veryimportant component in most fluid flow applications. For now, we can incorporate equations3.1 and 3.2 into the simplified momentum equation (equation 2.6) to get:

d

dt

∫∫∫V

ρ~U dV +

∫∫S

(n · ρ~U)~U dS = −∫∫S

nP dS +

∫∫∫V

ρ~g dV + ~Fviscous (3.3)

I’ve neglected the external force that acts through control volumes. That’s because this analysisis only going to pertain to small fluid particles, and not through solid objects in the system.


CHAPTER 3. THE MOMENTUM EQUATION

The viscous component is one that we could spend weeks discussing at a graduate level,going into the form of the viscous stress tensor, whether the fluid is isotropic or not, andconsidering compressibility issues. That’s not productive at the introductory level, and onlyserves to confuse. For the most part, we will allow for a simple treatment of viscous effects. Ifwe consider that viscous effects are shears and stresses on a surface, we can conceptually thinkof the viscous component of force as being the sum of these forces over the surface of the controlvolume. In other words, we would have:

~Fviscous =

∫∫S

n · T v dS (3.4)

I’ve not said anything about the form of T v, but lets just call it some viscous force tensor thatacts on the surface.

Now we substitute equation 3.4 into 3.3 to get:

d

dt

∫∫∫V

ρ~U dV +

∫∫S

(n · ρ~U)~U dS = −∫∫S

nP dS +

∫∫∫V

ρ~g dV +

∫∫S

(n · T v) dS (3.5)

Equation 3.5 is the integral form of the momentum equation in a fluid flow! In theprevious chapter there was some bonus material that took the integral form of the continuityequation to the differential form, and we will do the same here. We apply a lovely math trickcalled the divergence theorem, which relates surface fluxes of a vector to a change in thevolume of that vector.

We start by re-writing the pressure term in the form of a dot product:

−∫∫S

nP dS = −∫∫S

(n · PI) dS

where I is the identity matrix. I’ve written it in this fashion to help show that it too is a tensor,but a simple one at that. Think about what happens if you take the unit normal n and dot itwith the identity matrix: you get a vector of each component of the normal, or simply just getn back! Do note that the magnitude of n below is not unity, but the idea is all the same.

n · PI =

(i, j, k

)·

P 0 00 P 00 0 P

=

(P i, P j, P k

)= nP

By recasting the pressure term in this form, we can now apply the divergence theorem to thesecond, third, and fifth terms in our momentum equation. The divergence theorem for a unitnormal n and vector ~F can be expressed as:∫∫

S

(n · ~F

)dS =

∫∫∫V

(∇ · ~F

)dV

so equation 3.5 can be written:∫∫∫V

∂

∂tρ~U dV +

∫∫∫V

∇ ·((ρ~U)~U

)dV = −

∫∫∫V

∇ · PI dV +

∫∫∫V

ρ~g dV +

∫∫∫V

∇ · T v dV



Because we’re using a fixed control volume, the derivative on the first term was able to bebrought inside (the limits of integration don’t depend on time, so there are no additional termsintroduced by moving it in). Now, since these integrals are all evaluated over the same controlvolume, we can group them together:∫∫∫

V

[∂

∂tρ~U +∇ ·

((ρ~U)~U

)+∇ · PI − ρ~g −∇ · T v

]dV = 0

Just like we saw with the continuity equation, we see that the integral is zero. Since the volumeof integration is arbitrary, the integrand must always be zero no matter what CV is chosen.This means:

∂

∂tρ~U +∇ ·

((ρ~U)~U

)+∇ · PI − ρ~g −∇ · T v = 0

We can simplify the pressure term much in the same way we had made it a tensor:

∇ · PI =

(∂

∂x,∂

∂y,∂

∂z

)·

P 0 00 P 00 0 P

=

(∂P

∂x,∂P

∂y,∂P

∂z

)= ∇P

So rearranging the terms, we get:

∂

∂tρ~U +∇ ·

((ρ~U)~U

)= −∇P + ρ~g +∇ · T v (3.6)

Equation 3.6 is the differential form of the momentum equation in a fluid flow! This isalmost the same as the Navier-Stokes equations, but we need to do a little bit of simplifyingfirst. Starting by expanding out the LHS derivatives, we get:

~U∂

∂tρ+ ρ

∂

∂t~U + (ρ~U · ∇)~U + (∇ · ρ~U)~U = −∇P + ρ~g +∇ · T v

This expansion may seem confusing, but if you write it all out in terms of each derivative andeach component of velocity, you’ll find this is indeed the correct form. Do it yourself to practiceyour vector calculus skills! Then, grouping together some terms:

~U

[∂

∂tρ+ (∇ · ρ~U)

]+ ρ

∂

∂t~U + (ρ~U · ∇)~U = −∇P + ρ~g +∇ · T v

Do you see something unique about the term in brackets? This is just the differential form ofthe continuity equation! Since the continuity equation tells us the bracketed term is always zero,we can drop it to achieve

ρ∂~U

∂t+ ρ(~U · ∇)~U = −∇P + ρ~g +∇ · T v (3.7)

We can further simplify this by utilizing the material derivative, DDt = ∂

∂t+~U ·∇, and by assuming

a form for the viscous stresses. By assuming the fluid is incompressible, we can represent the



viscous forces as T v = µ(∇~U + (∇~U)T

), and by further assuming viscosity µ is constant, we get

the viscous term as:

∇ · T v = ∇ · µ(∇~U + (∇~U)T

)= µ∇2~U

with

∇2~U =

(∂2U

∂x2+∂2U

∂y2+∂2U

∂z2

)i+

(∂2V

∂x2+∂2V

∂y2+∂2V

∂z2

)j +

(∂2W

∂x2+∂2W

∂y2+∂2W

∂z2

)k

This is another great exercise in your vector calculus skills. First, note that ∇~U is a vectoroperation on each component of another vector, which results in a tensor. Then we add thetranspose of that tensor to itself. When we take the divergence of this tensor, we are applyinga dot product across each row, resulting in a vector, as shown above. Finally, recall that weassumed the fluid was incompressible (∇ · ~U = 0), which means that there will be a number ofterms that cancel out, leaving us with the final form of µ∇2~U .

While there have been a number of simplifying assumptions to get to this point, we don’tneed to worry about it at this level. The point is, we have a nice concise expression now forthe viscous term. Putting this and the material derivative into equation 3.7 then gives the theincompressible Navier-Stokes Equations:

ρD~U

Dt= −∇P + ρ~g + µ∇2~U (3.8)

3.1.1 Example: momentum practice

Before we get into what the Navier-Stokes Equations can tell us about a flow when we knowthe velocity profile, lets do a classic CV analysis. Assume for a rectangular channel that thevelocity profile can be described by the equation:

U(y) = 5(1− y2

) ms

where the geometry is defined in figure 3.1. Note that we will assume it is infinitely long in the

Figure 3.1: rectangular channel with a parabolic velocity profile

z-direction, so we can neglect any velocities or gradients in z. This is equivalent to calling it a



2D flow. Let’s also assume gravity is acting downwards in the y-direction. What would happenif you did a CV approach to this problem? Let’s start with analyzing CV1 in figure 3.2. Thisproblem is steady, and there are no fluxes in the y-direction. We can then simplify equation 3.5to get our integral form:

0 +

∫∫Ain

(−ρUx)~U dA+

∫∫Aout

(ρUx)~U dA = −∫∫S

nP dS +

∫∫∫V

ρ~g dV +

∫∫S

(n · T v) dS + ~Fext

Figure 3.2: two different control volumes applied to the problem

At this point, we can simplify the LHS by plugging in our velocity profile. If we assume aunit depth into the page for simplicity, we get:

−ρ∫ 1

−1U2 dy + ρ

∫ 1

−1U2 dy = RHS

You can see that the LHS is now zero! This is because the velocity profile doesn’t change fromthe inlet to the outlet - the flux in matches the flux out, so the sum of the two is zero. Then,we’re left with the RHS terms:

x-direction: 0 =

∫ 1

−1Pin dy −

∫ 1

−1Pout dy +

∫∫A

(n · T v) dA

︸︷︷︸just x-component

+Fext,x

y-direction: 0 = −Wfluid +

∫∫A

(n · T v) dA

︸︷︷︸just y-component

+Fext,y



These simplifications take place because gravity doesn’t act in the x-direction, and the grav-itational force integrated with density over the volume gives the weight of the fluid in they-direction. The pressures in the y-direction balance over the control surface, so they can beneglected. We then need to know what the viscous forces are doing. By writing out the 2Dviscous force tensor, we can find:

T v = µ ∗

(∂U∂x + ∂U

∂x∂U∂y + ∂V

∂x∂V∂x + ∂U

∂y∂V∂y + ∂V

∂y

)=

(0 −10µy

−10µy 0

)Now keep in mind that this tensor is only valid in the flow. Therefore, we can immediately seethat the top and bottom surfaces of the CV will not have any viscous contributions, as thosefaces lie outside the flow. The inlet and outlet surfaces have the unit normal as ±(1i, 0j), sowhen we dot this with the viscous tensor we get:

n · T v = (±1 0) ·(

0 −10µy−10µy 0

)=

(0

±10µy

)Looking in the x-direction, we end up getting zero! In the y-direction, we have an antisymmetricfunction (positive on the inlet, negative on the outlet) that is integrated from −1 to 1 at boththe inlet and outlet, giving us zero contribution from the viscous forces! Therefore, for thiscontrol volume, there are no viscous forces!

Overall, we find for CV1:

x-direction:

∫ 1

−1∆Px dy = Fext,x

y-direction: Wfluid = Fext,y

Now let’s look at CV2. It should be apparent that the influx and outflux of momentumdoesn’t change, so the LHS of the equation is still zero. However, there is no external forcenow because we’re not cutting through a solid object! Then, our expressions in each directionbecomes:

x-direction: 0 =

∫ 1

−1Pin dy −

∫ 1

−1Pout dy +

∫∫A

(n · T v) dA

︸︷︷︸just x-component

y-direction: 0 = −Wfluid −∫ L

0Pbottom dy +

∫ L

0Ptop dy +

∫∫A

(n · T v) dA

︸︷︷︸just y-component

Notice how we need to account for the pressure along the top and bottom of the CV, becausethe surface is in the flow and may not have the same pressure on them anymore! Additionally,because the new top and bottom surfaces are in the flow, we need to include these normals inour viscous integrals. Lets look:∫∫

A

(n · T v) dA =

∫∫inlet

(n · T v) dA+

∫∫outlet

(n · T v) dA+

∫∫top

(n · T v) dA+

∫∫bottom

(n · T v) dA



With our viscous stress tensor being the same as it was before, we can now apply the followingnormals:

ninlet = (−1, 0) noutlet = (1, 0) ntop = (0,−1) nbottom = (0, 1)

and then the viscous integral becomes:∫∫A

(n · T v) dA =

∫∫inlet

[0

10µy

]dA+

∫∫outlet

[0

−10µy

]dA+

∫∫top

[10µy

0

]dA+

∫∫bottom

[−10µy

0

]dA

Once again, the integration over the inlet and outlet goes to zero, since it is an antisymmetricfunction that is integrated from −1 to 1. You should hopefully then see that there is onlyan x-contribution to the force balance from viscous forces! This should have been your initialthought as well, since we know viscous stresses are caused by shearing forces on a surface. Sincethere are only going to be shears on y-normal surfaces, and they are acting in the x-direction,we would expect there to only be a viscous force in the x-direction.

We can then write our integrals out along the top and bottom, and simplify our forcebalances:

x-direction:

∫ 1

−1∆Px dy =

∫ L

010µ(1) dx+

∫ L

0−10µ(−1) dx

y-direction: 0 = −Wfluid −∫ L

0Pbottom dy +

∫ L

0Ptop dy

Note that we’ve plugged in the y-coordinate in which the force takes place for each of theintegrals, and then we integrate along the length of the wall in which the CV applies. Simplifyingthis gives us:

x-direction:

∫ 1

−1∆Px dy = 20µL

y-direction: Wfluid =

∫ L

0∆Py dy

So in the x-direction, we now see that the pressure at the inlet will be higher than the pressureat the outlet, proportional to the length of the control volume.

Using a different control volume lead us to getting a different force balance. If we lookto the two different results, we find that the viscous force is that external force we initiallyfound in the x-direction, and the pressure difference is the external force we found in the y-direction! However, we still don’t know the fine details, just that some sort of pressure drop inthe x-direction will have to balance the viscous forces on the wall, and that some pressure dropchange in the y-direction will have to balance the weight of the fluid.



3.1.2 Example: differential momentum practice

If we were to plug the velocity profile into equation 3.8 rather than doing the integral approach,we’d find the following:

ρ∂~U

∂t+ ρ(~U · ∇

)~U = −∇P + ρ~g + µ∇2~U

x-direction: ρ∂U

∂t+ ρ

(U∂U

∂x+ V

∂U

∂y

)= −∂P

∂x+ µ

(∂2U

∂x2+∂2U

∂y2

)y-direction: ρ

∂V

∂t+ ρ

(U∂V

∂x+ V

∂V

∂y

)= −∂P

∂y− ρg + µ

(∂2V

∂x2+∂2V

∂y2

)Take a moment to look at how we broke the equation into the two vector components. Themost difficult part is the “advective” term, or the non-linear term. The best way to think ofhow this works is to treat the term

(~U · ∇

)as an operator. This operator acts the same on all

three components of velocity. It’s much the same as the ∇2 operator, in that the form looks thesame for all three components of velocity:(

~U · ∇)

= U∂

∂x+ V

∂

∂y+W

∂

∂z

and

∇2 =∂2

∂x2+

∂2

∂y2+

∂2

∂z2

So hopefully you can see that the 2D form of these operators have been applied in both the x-and y-directions, and that they are identical in their form, regardless of which component it actson!

Since U = U(y), and V = 0, these equations reduce to:

x-direction: 0 + 0 + 0 = −∂P∂x

+ µ

(0− 10

)y-direction: 0 + 0 + 0 = −∂P

∂y− ρg + µ

(0 + 0

)which we then can rearrange to show:

∂P

∂x= −10µ

Pa

m∂P

∂y= −ρg Pa

m

The first equation is telling us the magnitude of the pressure change in the x-direction. We seethat it is negative in value, implying that pressure is higher at the inlet and lower at the outlet,exactly as we found in the CV analysis! There’s actually a lot of physical meaning in this:

1. The Navier-Stokes Equation has a viscous component, which we found was non-zero inthe x-direction.



2. The LHS was zero, implying there was no net change of momentum due to the flow velocityitself. Recall this side of the equation was from the fluxes in the integral equation. It’salso equivalent to the d

dt(m~v) term of Newton’s 2nd law.

3. Since there wasn’t a net force from a momentum change in the flow, the viscous forcesmust be balanced by another force, which we find is the pressure forces!

Therefore, we can see that this expression in the x-direction means that the pressure within thechannel must drive the flow to oppose the viscous stresses, and this results in a pressure dropas you move towards the right. This is quite literally what a pump is doing - increasing thepressure of the system to drive the flow and overcome viscous forces.

The second equation is simply the hydrostatic equation but in terms of y, not z. But how isthis possible? The fluid isn’t static, right? That’s indeed right, the fluid is in motion! However,this is something important to notice and realize:

The hydrostatic relation is always affect-ing the fluid, even when it is in motion!

This means that even with the flow, the pressure will increase from the top of the channel tothe bottom of the channel. Now, in a fluid like air, and with relatively small distances, thispressure increase can be extremely small. Even in water, with small distances we can neglectthis. However, over the atmosphere, or into a lake or ocean, the pressure change with elevation isstill governed by the hydrostatic equation. However, any additional flow in the vertical directionmay change this pressure gradient.

Overall, we see that the Navier-Stokes Equations can give us a lot of information about aflow field. In our case, since the parabolic velocity profile is assumed to be all throughout thisflow, we can say that the pressure gradient relations we calculated apply everywhere in the flow.That’s the power of the differential fluid equations - they apply to the flow field itself. A controlvolume analysis would lose all these fine details.

3.2 Bernoulli’s equation

1

2ρU2 + P + ρgz = Const. (3.9)

This is a favorite equation for most undergrad fluids courses. It’s quite powerful in it’s use,but only for a certain class of flows. It can only be applied if one of the the following set ofconditions are true:

1. steady, inviscid, constant density, irrotational flow

2. steady, inviscid, constant density flow along a streamline

If either of these sets of rules isn’t true, then equation 3.9 cannot be used. However, we willusually apply some assumptions or simplifications to many of our fluids problems to allow us theuse of Bernoulli’s equation. You’ll even find out that there are situations that we use it where



the assumptions aren’t strictly true, but give us a close enough approximation. That’s the heartof engineering for you.

Since the LHS of equation 3.9 is equal to a constant, then it should hold for anywhere in aflow that Bernoulli’s equation applies. That means, for any two points in a flow:

1

2ρU2

1 + P1 + ρgz1 =1

2ρU2

2 + P2 + ρgz2

This becomes useful when you can pick points in a flow where you know some reference values.For example, a good point to pick would be at the surface of a fluid, where P = Patm, orperhaps at the top of a large body of water where V = 0. Also, we can see that rearranging thegravitational terms gives us a difference in height. Therefore, it is advantageous to define z = 0at one of the locations, since we’re only concerned about the relative distance between the two.

3.2.1 Rigorous derivation of Bernoulli’s equation (bonus material)

This is a nice look into how you can derive equation 3.9 from the momentum equation.

First, we start with the Navier-Stokes equation, which we were able to get earlier:

∂~U

∂t+(~U · ∇

)~U = −1

ρ∇P + ~g +

1

ρ∇ · T v

This is the most general form, where we haven’t assumed anything about the variables involvedin the equation. The continuity equation has been applied, allowing us to write the LHS as itis. Also, we have divided through by ρ, but we haven’t assumed anything about ρ. Let’s startby re-writing some terms by using some vector identities:(

~U · ∇)~U = ∇

(1

2~U2)− ~U ×

(∇× ~U

)Plugging this in and rearranging gives:

∂~U

∂t+∇

(1

2~U2)

+1

ρ∇P − ~g = ~U ×

(∇× ~U

)+

1

ρ∇ · T v

Next we look to the gravity vector. It is a conservative body force because the net work doneby it is independent of the path taken. From this property, we can write any conserved force asa gradient of a potential:

~g = −g∇zHere, g is a constant, and z is the elevation. Additionally, we can re-write ~U2 as U2, becausethe square of a vector is simply the dot product of that vector with itself. Therefore, we canwrite it as the scalar U2.

Now, plugging in these expressions, we get:

∂~U

∂t+∇

(1

2U2)

+1

ρ∇P + g∇z = ~U ×

(∇× ~U

)+

1

ρ∇ · T v (3.10)

This equation is simply the momentum equation with a few re-arrangements. Again, this is allrigorous without any assumptions to this point. However, now we can simplify equation 3.10 ifwe assume:



1. Inviscid: ∇ · T v = 0

2. Steady flow: ∂∂t → 0

3. Constant density: 1ρ∇P = ∇P

ρ

The first assumption allows us to neglect the stress tensor. This term is where viscosity lies, soif we assume an inviscid solution, then we don’t have it! The second assumption allows us toget rid of the first term in equation 3.10, and the third assumption allows us to move densityinside the gradient operator on the pressure. We then have:

∇(1

2U2)

+∇(Pρ

)+ g∇z =~U ×

(∇× ~U

)∇[

1

2U2 +

P

ρ+ gz

]=~U ×

(∇× ~U

)At this point, you should notice that the three assumptions we’ve made all appear in the twosets of assumptions we made at the beginning of section 3.2. However, there is still two moresimplifications we can make: either irrotational flow, or flow along a streamline.

1. Irrotational Flow:In irrotational flow, we can represent the velocity as a gradient of a potential: ~U = ∇Φ. Ifthis is then plugged into the RHS, we get

(∇×∇Φ

), which is the curl of a gradient. This

is always zero! The expression is now:

∇[

1

2U2 +

P

ρ+ gz

]= 0

Since the terms in the brackets are scalar values, and the gradient of their sum is zero,that implies that everything in the brackets is constant. That returns us to equation 3.9!

2. Flow Along a Streamline:When moving along a streamline, we follow the trajectory of the velocity at a point. Inother words, an infinitesimal distance along a streamline d~ is parallel with the velocityfield. To represent movement along the streamline, you take the dot product of our equa-tion with d~. However, the expression ~U ×

(∇ × ~U

)gives a vector perpendicular to the

velocity (think about what a cross product does). Therefore, ~U ×(∇ × ~U

)· d~ = 0. So

now we are at:

∇[

1

2U2 +

P

ρ+ gz

]· d~= 0

The gradient of any scalar, let’s say B, dotted with a streamline can be represented as:

∇B · d~=

(∂B

∂x,∂B

∂y,∂B

∂z

)· d~

=

(∂B

∂x,∂B

∂y,∂B

∂z

)·(dx, dy, dz

)=∂B

∂xdx+

∂B

∂ydy +

∂B

∂zdz

=dB



This tells us that the LHS dotted with a streamline gives us the perfect differential of thequantity in the brackets:

d

[1

2U2 +

P

ρ+ gz

]= 0

Since that equals zero, integration once will give a constant, once again bringing us backto equation 3.9.

So we can see that there are two different sets of assumptions that both give us Bernoulli’sequation. However, there are still more ways to go about deriving equation 3.9 that utilizedifferent approaches! We don’t need to worry about that here, and in all reality this derivationis probably more rigorous than necessary for an undergrad course. However, it’s always a goodthing to see how and why we use the equations we use. It also helps reinforce the conditions inwhich we can apply Bernoulli’s equation and the limitations involved.

3.2.2 Pressure and streamlines

I just want to give a quick summary of an important concept in fluids: pressure and streamlines.From the rigorous derivation of Bernoulli’s equation along a streamline (case 2 above), we knowthat the force balance along a streamline is simply Bernoulli’s equation! That’s because wetook the momentum equation (a continuum representation of force balance), dotted it with aninfinitesimal streamline d~, and found equation 3.9.

A similar process can be done to look at what happens across streamlines, which is outlinedin a number of undergrad books. I won’t step through the derivation here, but I do wish toreiterate the summary we get from that math:

The pressure across straight streamlines isa constant if you can neglect gravitationalforces!

3.3 Example: flow over a bump

Determine the magnitude and direction of the horizontal component of force exerted by thefluid on the bump (figure 3.3) in terms of ρ, V1, H1, and width w. Assume uniform inflow andoutflow over the areas, and that H1/H2 = 2.

As with most fluids problems, the first step we can take is to utilize the continuity equation.Assuming the flow is steady, let’s determine the value of V2 in terms of our inflow parameters:

��

��*0

∂

∂t

∫∫∫Vs

ρ dV +

∫∫Ss

n · ρ~U dS =0

∫∫Ai

n · ρ~U dA+

∫∫Ao

n · ρ~U dS =0

−ρV1wH1 + ρV2wH2 =0

ρV2wH1

2=ρV1wH1

V2 =2V1



Figure 3.3: Image credit thanks to A. Smits, “A Physical Introduction to Fluid Mechanics.”John Wiley & Sons Inc. 2000

Therefore, the velocity out is twice the velocity in. This could be intuited since all fluid propertiesare constant, and the area out is half the area in, but it’s nice to see the math prove it.

Next, we need to look into what the pressures in the system are. There are a few toolswe can utilize to figure this out. First, note at the exit how the streamlines look. The flow isuniform out, which implies straight streamlines. That means there are no pressure gradientsacross the streamlines (assuming we neglect gravity). Therefore, the pressure across the outletis the same as the pressure below the outlet. We also see the same case on the inlet, so thepressure is uniform at that location as well.

Let’s investigate what the system would look like with no flow:

We can see that without any flow, we just have a constant atmospheric pressure everywhere.Then, when we move back to the case of flow, since the outlet streamlines imply the pres-sure doesn’t change across the outlet, we can assume the pressure in that region is just Patm.Additionally, the pressure at region 1 is just Patm plus whatever additional flow pressure mayexist.

Since we already know the flow is steady and has constant density, let’s further assume it’s



inviscid (a safe bet in this case) and irrotational. That allows us to apply Bernoulli’s equation!Utilizing the locations below, we know the velocities, and now the pressures. Additionally, we’ve

neglected gravitational effects in our assumption that straight streamlines mean no pressuregradient, so we can neglect the gravitational contribution in the Bernoulli equation. Therefore,we have:

1

2ρU2 + P +��

�*0ρgz = const.

1

2ρV 2

1 + P1 + Patm =1

2ρV 2

2 + Patm

1

2ρV 2

1 + P1 =1

2ρ(2V1)2

P1 =3

2ρV 2

1

So now we have the pressure in terms of the velocity and density that we’re given. The final taskis to find the force on the step, so we can perform a CV analysis using the momentum equation.There are a number of ways to draw the CV, but I’ll just choose one that cuts through thebump:

Looking to what is acting on the CV, we have flow into it on face 1, flow out of it on face2, pressure acting on both the left and right faces (and that’s all of the right face, not just the



portion where the flow exits), and a cut through a solid body on the bottom. The savvy studentwould ask: “what about the fluid interacting with the walls on the top and bottom?” The evenmore savvy student will say: “we are treating it as inviscid, so there aren’t any frictional forces!”Hooray for being savvy!

Then, applying the momentum equation, we only have an external force and pressure forces.This gives us:

∑~F =

��

��

��*

0d

dt

∫∫∫Vs

(ρ~U)dV +

∫∫Ss

(n · ρ~U

)~U dS

~Fext +

∫∫Ss

(− nP

)dS =

∫∫Ai

(n · ρ ~V1

)~V1 dA+

∫∫Ao

(n · ρ ~V2

)~V2 dA

~Fext +

∫∫Aleft

(− n(P1 + Patm)

)dA+

∫∫Aright

(− nPatm

)dA =

∫∫Ai

(n · ρ ~V1

)~V1 dA+

∫∫Ao

(n · ρ ~V2

)~V2 dA

Fext,x + P1wH1 + PatmwH1︸︷︷︸left side

−PatmwH1︸︷︷︸right side

=− ρV 21 wH1 + ρV 2

2 wH2 (in x-direction)

Fext,x + P1wH1 =− ρV 21 wH1 + ρ(2V1)2w

H1

2

Fext,x +3

2ρV 2

1 wH1 =ρV 21 wH1

Fext,x =− 1

2ρV 2

1 wH1

So we find that the external force on the control volume is in the negative x-direction. Since thisforce required to hold the CV in place is negative, the force exerted by the flow is thus in thepositive x-direction. Therefore, the overall force on the step (which is what the external force

must be acting through) is1

2ρV 2

1 wH1 .


Chapter 4

Introducing Potential Flow

4.1 Some flow definitions and characteristics

This is just a review of a few key definitions for fluid flows. We want to somehow combine theseto solve for different types of flow fields under different assumptions. The idea is to see whenwe can combine these, and how they work together.

4.1.1 Vorticity

Vorticity, ~ω, is an important concept in fluid mechanics, and we use it to measure rotation. Itarises in many flows, and can make a headache of some problems. In fact, vorticity is one ofthe defining characteristics of turbulent flows, and we can build transport equations for vortic-ity, much in the same way we build transport equations for mass (continuity) and momentum(Navier-Stokes). We’re not so interested in those details at this level, so for now we simply desireto define ~ω as the curl of the velocity field:

∇× ~U = ~ω (4.1)

We can evaluate the curl by taking the determinant of the following matrix:

∇× ~U =

∣∣∣∣∣∣i j k∂∂x

∂∂y

∂∂z

u v w

∣∣∣∣∣∣ = i

(∂w

∂y− ∂v

∂z

)− j(∂w

∂x− ∂u

∂z

)+ k

(∂v

∂x− ∂u

∂y

)So then we have vorticity respresented as:

~ω =

(∂w

∂y− ∂v

∂z,∂u

∂z− ∂w

∂x,∂v

∂x− ∂u

∂y

)(4.2)

Notice that each component of vorticity contains the other components of velocities and gradi-ents. For example, the x-component of vorticity has the y- and z-components of velocities, andy-and z-derivatives. The same holds for the other components of vorticity. This means that a2D flow that has vorticity will have it in the neglected dimension. For example, consider theparabolic velocity profile:

U(y) = Ucl

(1−

( yD

)2)Intro to Fluids Notes 53 C. P. Byers 2016

CHAPTER 4. INTRODUCING POTENTIAL FLOW

We we say this velocity profile represents a 2D flow. That’s because we have a velocity in thex-direction, and it varies in y. There is no velocity or variation in z, thus only 2 dimensions arerelevant! Now, when we compute the vorticity we get:

~ω =

(0, 0,

2Ucly

D

)So even though the flow is 2D and in the x-y plane, the vorticity is in the z-direction. This isakin to a moment in statics. That’s just a little bonus information to keep in mind.

Why do we care about vorticity? When we can neglect it, we are dealing with considerablysimplified flows! In the parabolic velocity profile above, we found there was a component ofvorticity due to the velocity gradient. This gradient will also give rise to viscous forces in theflow. In general, a viscous shearing stress requires a gradient in the flow. Another property ofa viscous fluid flow is that any contact with a surface results in rotational effects. Since thereare rotational effects due to viscosity, which in turn means there is vorticity, we can say that anirrotational flow can be treated as inviscid.

Let’s see how this actually works. We look to the vector curl identity:

∇× (∇× ~U) = ∇(∇ · ~U)−∇2~U

This is a vector identity, and is always true for any vector ~U , not just our velocity field. Now, ifwe have an incompressible flow (or like we said for Bernoulli, ρ is constant), then the differentialform of the continuity equation takes the form ∇ · ~U = 0. Then, if we say that the flow isirrotational, we have ~ω = ∇× ~U = 0. These two properties combined gives us:

∇× (0) = ∇(0)−∇2~U → ∇2~U = 0

The viscous term in the Navier-Stokes Equations is of the form µ∇2~U , so we can see thatincompressible, irrotational flow will imply an inviscid flow as well!

4.1.2 Velocity Potential

If we’ve assumed a flow is irrotational, we’ve decided to neglect viscous effects. Mathematically,we can represent that as:

∇× ~U = 0 (4.3)

Since this is a vector, each component of vorticity must be zero:

∂w

∂y− ∂v

∂z= 0 → ∂w

∂y=∂v

∂z

∂u

∂z− ∂w

∂x= 0 → ∂u

∂z=∂w

∂x∂v

∂x− ∂u

∂y= 0 → ∂v

∂x=∂u

∂y

one possible solution to these three equations would be a function φ such that:

u =∂φ

∂x, v =

∂φ

∂y, w =

∂φ

∂z(4.4)



We can then represent this as:

~U = ∇φ (4.5)

Plugging equation 4.5 into 4.3 then gives us:

∇×∇φ = 0

Which is always true, since this is a vector identity. This confirms that equation 4.5 is indeeda valid solution to the irrotational flow field. We call this a potential flow, since the velocityis represented as the gradient of a potential. Keep in mind this is only valid if the flow field isassumed irrotational. We also can express this in cylindrical coordinates, which is given in anumber of books.

The way I like to think about velocity potentials is to compare it to other “potential-type”quantities. One that was mentioned in a previous chapter was gravity. It’s a conservativeforce, which means we can express it as a gradient of some potential function. This lead to theexpression:

~g = ∇(−gz) = −g∇z = (0, 0,−g)

This means that the gravity vector is dependent on some gravitational constant g and the changein height z, with the negative sign meaning it acts in the minus z direction. We think of thegravitational potential energy as the height in which a weight is elevated, or equivalently, thechange in potential energy as the change in height:

∆PE = mg∆z

Therefore, with gravity, we think of equal heights as having equal potential. Equivalently, wecan think of gravity as having the same action at all elevations, as ∇z = (0, 0, 1), which meansgravity is a constant −g and acts only in the z-direction. This same reasoning can be appliedto a velocity potential. If we’re given a velocity potential of the form φ = Ax, then we can findthe velocity field based on this potential:

~U = ∇φ → (u, v, w) = ∇(Ax) = (A, 0, 0)

Thus, for the given potential, we found that the velocity field has a constant action in the x-direction only. This is quite analogous to the gravity potential!

Potential functions become really useful when constructing different types of flow fields, andwe will see that in an example later.

4.1.3 Stream functions

Just as we defined a function that satisfies the requirement of irrotational flow, we can definea function to satisfy the continuity equation. For a 2D, incompressible flow, our continuityequation can be expressed as:

∂u

∂x+∂v

∂y= 0



Just as we previously introduced a function to satisfy the irrotational requirement for potentialflow, we can introduce a function to satisfy the incompressible requirement here. We define afunction ψ such that:

u =∂ψ

∂yand v = −∂ψ

∂x(4.6)

By substituting this into the 2D incompressible continuity equation, we find:

∂u

∂x+∂v

∂y=∂

∂x

(∂ψ

∂y

)+

∂

∂y

(− ∂ψ

∂x

)=∂

∂x

(∂ψ

∂y

)− ∂

∂x

(∂ψ

∂y

)=0

Therefore, our stream function ψ satisfies the continuity equation. Again, there is also acylindrical coordinate representation for the stream function.

One thing to keep an eye out for is the way we defined the stream function. For this course,and hopefully for the rest of your undergrad work, we put the minus sign on the v differentialexpression. Some authors and courses like to put the minus sign on the u differential. So longas only one of the derivatives has a negative sign, then the stream function will satisfy the 2Dcontinuity equation. We will be consistent with our definition of equation 4.6.

An important property of stream functions is that lines of constant ψ are streamlines (it’salmost as if it were named this way for a reason). This can be seen by writing the total differentialof a streamline:

dψ =∂ψ

∂xdx+

∂ψ

∂ydy = −v dx+ u dy

For a constant ψ, dψ = 0, so we can rearrange the equation to get:

v dx = u dy → dx

dy=u

v

This expression is saying that a differential change in position is in the direction of the velocity.In other words, lines of constant ψ have a direction that is tangential to the velocity field.Therefore we can say streamlines are represented by a stream function of constant value.

4.2 Example: a potential vortex

One type of flow we can analyze with stream functions and velocity potentials is called thepotential vortex. A name like “vortex” seems to imply some sort of vorticity, doesn’t it?Wouldn’t that violate the irrotational flow requirement for potential flow? We’ll find out if thatis indeed true in this example. Let’s look at the stream function we can define for a potentialvortex:

ψ = − Γ

2πln(r)

Note that we’ve use the coordinate r, which implies this is in cylindrical coordinates. Usingthis stream function, let’s draw the streamlines and define a corresponding potential function,



if one exists. Then, let’s explain the relationship between the streamlines and lines of constantpotential.

We can start by drawing out a few lines of constant ψ: The exact value of the stream function

Figure 4.1: coordinate system and three lines of constant ψ

itself isn’t so important, but what we see is that for a constant r, the stream function has aconstant value, meaning we get streamlines that are circles. The velocity is tangential to theselines, so we see that the flow is in a circular direction. We can then look to find the exactmagnitude of the velocity components:

ur =1

r

∂ψ

∂θ=

1

r

∂

∂θ

(− Γ

2πln(r)

)= 0

uθ =− ∂ψ

∂r= − ∂

∂r

(− Γ

2πln(r)

)=

Γ

2πr

This says that there is no radial component of velocity, only an angular component, which decaysin magnitude with increasing radius. This matches the streamlines we had drawn!

Since this was a stream function, we know that it is a 2D incompressible flow. Incompress-ibility can be confirmed however:

∇ · ~U =1

r

∂(rur)

∂r+

1

r

∂uθ∂θ

= 0 +1

r

∂

∂θ

Γ

2πr= 0

So indeed the incompressible continuity equation (in cylindrical coordinates) has been satisfied.Can we write a velocity potential for this flow? That requires the flow to be irrotational. Wecan check by taking the curl of the velocity field. In cylindrical coordinates, this can be writtenas:

∇× ~U =

(1

r

∂uz∂θ− ∂uθ

∂z

)er +

(∂ur∂z− ∂uz

∂r

)eθ +

(1

r

∂(ruθ)

∂r− 1

r

∂ur∂θ

)ez

We don’t have any uz component, and no gradients in z, therefore the first two vorticity com-ponents will be zero. This should make sense, as we are only a 2D problem, and the curl givesyou a vector perpendicular to the two components in the operation. Since we have r- and θ-components, the curl should give a z-component of vorticity, if any. Therefore, plugging in ourexpressions for ur and uθ gives us:

∇× ~U =

(1

r

∂

∂r

(r

Γ

2πr

)− 1

r

∂

∂θ(0)

)ez =

(1

r

∂

∂r

( Γ

2π

)− 0

)ez = 0 ez



The curl of the velocity field of a potential vortex is zero, meaning we can define a velocitypotential for the flow field! Perhaps that’s why it’s called a potential vortex! We then use thecylindrical coordinate version of a potential flow to find the function:

ur =∂φ

∂r= 0 → φ = const. + f(θ)

uθ =1

r

∂φ

∂θ=

Γ

2πr→ φ =

Γ

2πθ + const. + f(r)

By inspection, we can see the potential function has no r-dependence. Also, since we define thepotential function so that we can extract the velocity through the gradient of it, it stands toreason that we don’t care about the possible additive constant. Therefore, we can set it to zeroto get:

φ =Γ

2πθ

We can then plot lines of constant potential on our graph with the streamlines, as shown infigure 4.2. Once again, the actual values of the lines of constant potential aren’t so important

Figure 4.2: Potential vortex with lines of constant ψ and φ

here. What we see is that φ is constant for a constant angle, giving these radial lines. Also ofinterest is the fact that the lines of constant velocity potential are perpendicular to the lines ofconstant stream function.



In general, if we can define both a stream function and velocity potential, we find that linesof constant φ and lines of constant ψ are orthogonal. This can be shown by taking thetotal differential of a velocity potential (now back in Cartesian coordinates):

dφ =∂φ

∂xdx+

∂φ

∂ydy = u dx+ v dy

For a constant φ, dφ = 0, so rearranging gives:

−v dx = u dy → dx

dy= −v

u

This most certainly isn’t the same case as the streamlines, where the instantaneous slope coin-cided with the velocity at that point. We can see the two different slopes are orthogonal becausetheir product, u/v×−v/u, is −1 indicating perpendicular lines. This is a general result for anyfield with both a potential function and stream function!

4.3 The power of the velocity potential

We’ve seen that we can define stream functions if the flow is 2D and incompressible, and velocitypotentials if the flow is irrotational, but what benefit does this provide us? Let’s focus on thevelocity potential for now to see how we can use it to our advantage.

If, in addition to irrotational flow, we say a flow is incompressible, we can then apply thesimplified continuity equation to our velocity potential:

∇ · ~U = 0 → ∇ · ∇φ = 0

However, just like we did with the viscous term in the Navier-Stokes equations, we can writethis using the Laplacian operator, ∇2:

∇2φ = 0 (4.7)

Likewise, since we have an incompressible, irrotational flow, if we say it is 2D, we can do thesame for the stream function. With the velocity components from a stream function defined as:

u =∂ψ

∂yand v = −∂ψ

∂x

we then plug this into our irrotational definition:

∇× ~U =

(∂v

∂x− ∂u

∂y

)k =0(

∂

∂x

(− ∂ψ

∂x

)− ∂

∂y

(∂ψ∂y

))k =0

−(∂2ψ

∂x2+∂2ψ

∂y2

)k =0

This too is simply represented as:

∇2ψ = 0 (4.8)



What equations 4.7 and 4.8 represent is something we call Laplace’s Equation. This is a verycompact and powerful equation, and is a nice simplification for our fluids equations. Before, wehad to solve the Navier-Stokes equations to know the flow field. This was a tough thing to do,as it’s a non-linear partial differential equation. Even if we say the flow field is inviscid, theequation is still difficult. We called this form Euler’s Equation:

ρ∂~U

∂t+ ρ~U · ∇~U = −∇P + ρ~g

Note how it is still a non-linear partial differential equation with 4 dimensions (x, y, z, t), possiblynon-constant coefficient ρ, and 4 variables (u, v, w, P ). Solving an equation like this is no easytask, and in fact we only have a small number of exact solutions. However, Laplace’s Equationis a different story. It is much more simple to solve, and more importantly, it is linear! Thatmeans that for any two solutions to the equation, their sum (or difference) is also a solution.

∇2φ1 = 0 and ∇2φ2 = 0 → ∇2(φ1 + φ2) = 0

This means we can build flow fields using multiple velocity potential solutions! This methodologyis called superposition.

4.4 Example: superposition

Let’s look at a flow field where we have a source in a uniform flow. Where is the stagnationpoint? What is the pressure at the stagnation point?

If we have a uniform flow described by the potential φ = Ur cos(θ), and the source describedby φ = q

2π ln(r), with U = 2m/s and q = 4m2/s, we can assemble the total potential functionof the flow:

φ = φ1 + φ2 = 2r cos(θ) +4

2πln(r)

We could express this in Cartesian coordinates, but that would just give complicated expressionswith ln

(√x2 + y2

)that we don’t want to deal with. If we plot this out (preferably with a

computer) we would find the find the lines of constant velocity potential. However, velocityacts across the gradient of the potentials, which means the plot wouldn’t look so meaningful!Instead, let’s look to the stream function, since the velocity is tangential to streamlines. First,let’s start by decomposing the potential into the radial and angular velocity components:

ur =∂φ

∂r= 2 cos θ +

2

πr

uθ =1

r

∂φ

∂θ= −2 sin θ

So we have the velocity field, and we know the flow is 2D and irrotational. To use a stream



function, we need to know if the flow is incompressible. Apply continuity and check to see:

∇ · ~U =0

1

r

∂

∂r(rur) +

1

r

∂

∂θ(uθ) =0

1

r

∂

∂r

(r(2 cos θ +

2

πr))

+1

r

∂

∂θ(−2 sin θ) =0

1

r

(2 cos θ + 0)− 1

r(2 cos θ) =0

2 cos θ − 2 cos θ

r=0

The incompressible continuity equation is indeed satisfied, so we can define a stream function!

ur =1

r

∂ψ

∂θ= 2 cos θ +

2

πr

uθ =− ∂ψ

∂r= −2 sin θ

Integration of each equation then gives us:

∂ψ

∂θ= 2r cos θ +

2

π→ ψ = 2r sin θ +

2θ

π+ f(r) + C1

∂ψ

∂r= 2 sin θ → ψ = 2r sin θ + f(θ) + C2

Again, the additive constant isn’t important in this situation. We can see from the two equationsthat our stream function ψ is defined as:

ψ = 2r sin θ +2θ

π(4.9)

Plotting out equation 4.9 through the aid of a computer gives us the following plot:

Figure 4.3: streamlines for a point source in a uniform flow

To get the stagnation point, we look for when the velocity has stagnated, meaning we have0 velocity in both the radial and angular directions. Therefore, setting ur = 0 and uθ = 0 gives:

0 = 2 cos θ +2

πr0 = −2 sin θ



The second expression constrains our θ value to 0 or π. If we chose θ = 0, then we would haveto have a negative r value, which doesn’t make sense. Therefore, we find stagnation occurs atthe point:

(r, θ) = (1

π, π)

To find the pressure at the stagnation point, we need some sort of reference pressure tostart with. Let’s assume the pressure far upstream is just atmospheric pressure, Patm. Theassumptions we’ve already used for this problem (incompressible, irrotational, inviscid, steady)allows us to use Bernoulli’s equation. Therefore, picking one point far upstream (r = ∞ andθ = π), and the second point at our stagnation point, we find:

1

2ρU2

1 + P1 +��:0ρgz1 =

1

2ρU2

2 + P2 +��:0ρgz2

1

2ρ

((2 cos(π) +

2

π ∗∞)er + (−2 sinπ)eθ

)2

+ Patm =1

2ρ

((2 cos(π) +

2

π ∗ 1π

)er + (−2 sinπ)eθ

)2

+ P2

1

2ρ

(2er + 0eθ

)2

+ Patm =1

2ρ

(0er + 0eθ

)2

+ P2

2ρ+ Patm = P2

This simply says the pressure at the stagnation point is 2ρ Pascals higher than the free streampressure. This matches up with the expression we have from a pitot tube:

U =

√2∆P

ρ

If you plug in the freestream velocity of 2m/s, you find that ∆P is 2ρ!


Chapter 5

Non-Dimensionalization

5.1 Non-dimensionalization

In fluids, we love to use non-dimensional numbers. One primary reason for this is the simplifi-cation of data analysis that is possible due to non-dimensionalization. Another reason is to findthe leading order terms in the equations of motion. I’ll go through an example of both of theseto show just how they can be used and their implications.

5.1.1 Similarity and non-dimensional parameters

Data analysis can be simplified through the use of non-dimensional parameters by a conceptwe call similarity. By matching all relevant non-dimensional parameters, we achieve flowsimilarity. This is especially relevant when we want to do scale modeling, which is the engineersplayground.

We can construct non-dimensional parameters by investigating what sort of terms show upin the problem itself. Let’s start by looking at a simple cylinder in a flow (figure 5.1) anddetermine what terms are important. The flow itself has a velocity, U∞, which we would expectto influence the drag. The density ρ and viscosity µ will also probably have an effect on thedrag. We could also surmise that the size of the object, or diameter D, will influence the amountof drag. Since we want to know the drag FD, it’s also an important parameter. Now, we couldalso include other things that may influence the flow, such as surface roughness ks, or maybefreestream temperature T∞, or even perhaps what phase the moon is in, φmoon. However, wehave to decide we have sufficient parameters at some point. In our case, we’ll just limit ourselvesto U∞, ρ, µ, D, and FD.

With this, we can then say that the drag is a function of these parameters:

FD = f(U∞, ρ, µ,D) (5.1)

The next step is to see what fundamental units are involved in this expression. These aredimensions such as length, time, mass, temperature, etc. So for us, we have:[

FD]

=ML

T 2

[U∞]

=L

T

[ρ]

=M

L3

[µ]

=M

LT

[D]

= L


CHAPTER 5. NON-DIMENSIONALIZATION

Figure 5.1: Cylender in a flow with relevant parameters

So in our example, we have units of mass, length, and time, or three fundamental units.We decided that there were five relevant parameters. Therefore, we will have 5 − 3 = 2 non-dimensional groupings. The power of non-dimensional numbers is that we can define the problemwith these 2 non-dimensional parameters! This means we’ll have one non-dimensional parameter,Π1 that is a function of the second, Π2, giving us a pretty simple relationship:

Π1 = f(Π2) (5.2)

This is the basis of Buckingham-Pi theory, where we build up a number of non-dimensional“Pi groups”. Notice how equation 5.2 is much more simplistic than equation 5.1. I’d ratherdeal with a function of one variable than a function of four! I encourage you to read up onBuckingham-Pi a bit, but hopefully the basics here get you on your way.

There are a few ways to build our non-dimensional parameters. One way is to multiplyterms together and give them unknown powers, then solving the powers so that the groupingis dimensionless. This is a classic way to find our “Pi groups,” but I like to do it in a differentfashion. Let’s instead pick a parameter and work on getting rid of its dimensions. We knowthe drag is what we’re solving for, so it should make up one of the dimensionless parameters.Starting with it, we look at what it’s fundamental dimensions are:[

FD]

=ML

T 2

So we need to somehow eliminate the M , L, and two T ’s. Let’s start by dividing by ρ, whichwill kill off the M dimension: [

FDρ

]=L4

T 2

So mass is no longer in this grouping. Let’s next eliminate T by dividing by U2∞:[

FDρU2∞

]= L2



Finally, we can divide by D2 to get rid of the length dependence:[FD

ρU2∞D

2

]= dimensionless (5.3)

So we’ve found one non-dimensional parameter, CD = FDρU2D2 , which is similar to the classic

definition of the drag coefficient!However, what would have happened if we instead used µ to get rid of the mass term rather

than ρ? Let’s calculate it through: [FDµ

]=L2

T

We can then eliminate time by dividing by velocity:[FDµU∞

]= L

And finally, to get rid of length we divide by D:[FD

µU∞D


So there is a completely different non-dimensional force grouping we could have come up with.Either grouping from equations 5.3 or 5.4 are valid, it’s just our experience and actual datacollection that tells us which is a better one to use.

So let’s select equation 5.3 so we have one non-dimensional parameter. We still need to buildour second parameter. Let’s start with density and get rid of the mass component by dividingwith viscosity: [

ρ

µ

]=

T

L2

Now, we eliminate time by multiplying with velocity:[ρU∞µ

]=

1

L

Finally we can get rid of the length by multiplying with D:[ρU∞D

µ


Equation 5.5 is once again the Reynold’s number, which we’ve seen before. Then, using our twonon-dimensional parameters, we can simplify equation 5.1 to:

FDρU2∞D

2= f

(ρU∞D

µ

)With non-dimensionalization, you’ll often find it’s your intuition and experience that drives theselection of parameters. It’s definitely more of an art, but yields some powerful science!



5.1.2 Non-dimensional equations

Rather than list out terms like we did in the last section, we can instead look to the governingequations to tell us what the flow is doing. We select each variable and assign a scalingparameter and scaling function. This sounds complicated, but in practice it’s quite simple.For example, let’s say that the ~U velocity is in our equation, then we can re-define it as:

~U = Uscale~U∗

The idea here is we pick an appropriate value for Uscale based on our flow, and ~U∗ is the scalingfunction. Another way to think of this is to take an appropriate scale, like Uscale, and divideour relevant variable by it. This will then give us a non-dimensional variable. So we see:

~U

Uscale= ~U∗

This is just a re-arrangement of our previous relation. Likewise, we can re-define derivatives ina similar fashion:

d

dx=

d

d(Xscalex∗)

Where our x-coordinate is a non-dimensional x∗ scaling function times the scaling parameterXscale. Again, we must pick Xscale to be appropriate to our flow. So, let’s look to an exampleto see what these terms mean.

5.1.3 Example: non-dimensionalizing the momentum equation

If we look to the steady, incompressible momentum equation neglecting gravity, we can non-dimensionalize it using the scaling parameters and functions. This means our momentum equa-tion can be written as:

ρ(~U · ∇)~U = ∇P + µ∇2~U

Now, if we say we have some velocity scale U , a length scale L, and pressure scale Ps, we candefine:

~U = U ~U∗ and ∇ =1

L∇∗ and P = PsP

∗

Note that derivatives deal with changes over a direction, so that’s why the L is on the bottomof the fraction. We don’t know what the pressure scale Ps is, but we can figure that out later.Plugging in our parameters then yields:

ρ(U ~U∗ · 1

L∇∗)U ~U∗ =

1

L∇∗(PsP ∗) + µ(

1

L∇∗)2U ~U∗

ρU2

L(~U∗ · ∇∗)~U∗ =

PsL∇∗P ∗ +

µU

L2∇∗2~U∗

The second line has all the scaling parameters (which are just constants!) pulled out front. Wethen have to choose what parameter we feel is the most important. For instance, in many flows



we know the acceleration term (or advective term) is important in our flow. Therefore, we canthen divide the whole equation by the constants on the first term (ρU2/L) to find:

(~U∗ · ∇∗)~U∗ =PsρU2∇∗P ∗ +

µ

ρUL∇∗2~U∗ (5.6)

Look at what we found - two non-dimensional numbers we’ve dealt with before! The firstone looks very familiar to our coefficient of pressure, and the second one is the reciprocal of theReynolds number. Recall:

Cp =P − P0

12ρU

2and ReL =

ρUL

µ

The difference with Cp is a multiplicative constant of 1/2, and since Ps is just some scale pressure,setting it to P − P0 is perfectly valid. As for the Reynolds number, a body of 1m length in airat 1m/s will have a Reynolds number of:

ReL =ρUL

µ

=1.2 kg

m3 ∗ 1ms ∗ 1m

1.846 ∗ 10−5 kgms

∼66000

So if we look to equation 5.6, we see that the last term has a factor of 1/65000 in front of it.That means that term is negligible, leaving us with:

(~U∗ · ∇∗)~U∗ =PsρU2∇∗P ∗ (5.7)

This equation is simply a non-dimensional form of Euler’s Equation without the gravity term!Since the LHS is all non-dimensional scaling functions, then it should have an order of magnitudearound 1. This means the non-dimensional parameter on the RHS, Cp, should have an orderof magnitude around 1 as well. If you look at a plot of Cp over a cylinder, you’ll find that itvaries from -3 to 1, which is what we expected! If it were to vary up to 100, then we’d havedone something wrong.

Therefore, from equation 5.7 we can see that the appropriate scaling for pressure isPs = ρU2

s . However, this is under the assumption that the advective terms were the importantparameters, (or equivalently stated, the inertial effects are large), and viscous effects small.

This process is exactly how we had issues with viscosity and the equations of motions upto the early 1900s. Since the Reynolds number was so high, the viscous term was negligible,so people would argue that Euler’s equation is the way to go. However, we know viscosity isimportant, so we can’t just neglect it! That second derivative allows us to utilize a secondboundary condition in our PDE (because it was a second order equation), which is the no-slipcondition. It took a trick with how we think about scaling to finally solve our problem. We willinvestigate that in a later example.



5.1.4 Example: flow in a channel

Imagine a 2-D rectangular channel, length 100m and height of 1m, with a centerline velocity ofU = 5ms . Assuming the flow is incompressible, let’s non-dimensionalize the continuity equationand see what it tells us:

Continuity:∂U

∂x+∂V

∂y= 0

If we non-dimensionalize the x- and y- directions with the lengths we’re give, we can write:

∂

∂x=

∂

∂(100x∗)=

1

100

∂

∂x∗and

∂

∂y=

∂

∂(1y∗)=

∂

∂y∗

Likewise, we can non-dimensionalize the velocities with what we have:

U = 5U∗ and V = VsV∗

Since we don’t have a V velocity scale, we’ll just use some constant Vs as a placeholder. Pluggingeverything in, we then find our continuity equation:

5

100

∂U∗

∂x∗+ Vs

∂V ∗

∂y∗= 0

Here is where non-dimensionalization become powerful and important. We’ve chosen scalingvalues for each variable, and we determined those to be important based on the flow. Thenon-dimensional functions are now the same order of magnitude, that being order 1. Thinkabout it this way, we took changes in x and divided by 100. Since our geometry in this exampleis 100 m long in the x-direction, then the change in x∗ would be at most 1. This is how wenon-dimensionalized all the variables!

Since the continuity equation is only made of 2 terms, we must always have a balance betweenthose two terms. Therefore, with ∂U∗

∂x∗ and ∂V ∗

∂y∗ both being order 1, their coefficients must alsobalance. The coefficient of the first term is 1/20, so that tells us that:

Vs =1

20

m

s

This means that in our channel, the V -velocity will be small compared to the U -velocity. Thisdoesn’t tell us the actual value of V -velocity, just its relative magnitude. Therefore, in this pipe,we see that velocities in the y-direction are very small compared to velocities in the x-direction.

Imagine next the balance of momentum for our equation. For a 2-D, steady, incompressibleflow, neglecting gravity, the x-momentum equation can be written as:

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P

∂x+ µ

∂2U

∂x2+ µ

∂2U

∂y2

We can non-dimensionalize this with the scaling parameters we previously used for both velocityand length. We don’t have a parameter for pressure, so let’s use Ps again. Plugging in the valueswill yield:

(5U∗)∂(5U∗)

∂(100x∗)+

(1

20V ∗)∂(5U∗)

∂(1y∗)= −1

ρ

∂(PsP∗)

∂(100x∗)+µ

ρ

∂2(5U∗)

∂(100x∗)2+µ

ρ

∂2(5U∗)

∂(1y∗)2



Pulling out all constants from their derivatives and re-arranging yields:(5 ∗ 5

100

)U∗

∂U∗

∂x∗+

(5

20V ∗)∂U∗

∂y∗=−

(Ps

100ρ

)∂P ∗

∂x∗+

(5µ

1002ρ

)∂2U∗

∂x∗2+

(5µ

ρ

)∂2U∗

∂y∗2

U∗∂U∗

∂x∗+ V ∗

∂U∗

∂y∗=−

(Ps25ρ

)∂P ∗

∂x∗+

(µ

500ρ

)∂2U∗

∂x∗2+

(20µ

ρ

)∂2U∗

∂y∗2(5.8)

Now we can see a little more about what terms are important in our x-momentum equation.Recall how the non-dimensional terms have an order of magnitude of 1. This means the LHSof this equation is order 1. Then, looking to the RHS, we can conclude that the second term(first viscous term) is smaller than the third term. This conclusion is similar to what we foundwith equation 5.6, where the whole viscous term was deemed negligible. However, now we seethat the second viscous term in equation 5.8 is more important than the first, so it isn’t strictlycorrect to neglect the whole µ∇2~U term when non-dimensionalizing.

In a future chapter, we will investigate internal flow (pipe and channel) equations in furtherdetail. For now, just focus on how applying physical dimensions from an example resulted ina change in the balance in our non-dimensional relation. We see that different components ofviscosity are more important than others. (In this case by four orders of magnitude!)

5.2 Prandtl’s Boundary Layer

Hopefully the last few sections were able to show you how non-dimensionalization can be per-formed, and how it can tell you what terms in an equation are important. However, as wasshown with equation 5.6, a traditional scaling analysis led us to believe that the viscous termwas negligible, and thus Euler’s equation is the valid representation of a flow, even around orthrough a body with solid walls. Granted, we can get decent approximations in some cases, butwe’ve lost the physics of what’s really going on - we violate the no-slip condition.

We then saw in equation 5.8 that one of the two viscous terms was smaller than the other(by a factor of 104 in that example). This should lead you to believe that we need to think

about how we non-dimensionalize the equations. We saw the µ∂2U∂y2

term ended up being more

significant than the µ∂2U∂x2

term. This can be explained by thinking about the distance over whichthese gradients were occurring. The x-direction had a characteristic length scale of 100m, wherethe y-direction had a characteristic length scale of 1m. So gradients would be much strongerin the y-direction compared to the x-direction. Prandtl had made this same conclusion whenthinking about viscous effects on solid boundaries.

5.2.1 x-momentum equation

To re-create Prandtl’s analysis, let’s look at a flat plate, 2-D, incompressible, steady flow withno gravitational effects. The physical setup can be seen in figure 5.2.

This is actually a popular setup to have in a lab, so these simplifications aren’t that far-fetched. The x-momentum equation will then look like:

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P

∂x+µ

ρ

∂2U

∂x2+µ

ρ

∂2U

∂y2



Figure 5.2: Flow over a flat plate, vertical and horizontal distances are not to scale.

We then say that the U velocity has a characteristic velocity of U∞, which is the free-streamvelocity. We’ll set the pressure scale as Ps, then refine it further when have non-dimensionalized.Based off what we’ve seen before, it’s likely that it will be ρU2

∞. The changes in the x-directionwill occur over some length, let’s say L. The changes in the y-direction are then argued tooccur over a very small distance, δ, where δ < L. We constrain the vertical length scale in thismanner to make the viscous term important, as we know that there must be viscous effects inthe near-wall region.

We don’t worry about scaling ρ or µ because they are constant in this problem. If they werevarying throughout the flow, then yes, we would need to use a scaling parameter for them.

Next, to find the appropriate Vs scale, we turn to the continuity equation:

∂U

∂x+∂V

∂y=0

∂(U∞U∗)

∂(Lx∗)+∂(VsV

∗)

∂(δy∗)=0[

U∞L

]∂U∗

∂x∗+

[Vsδ

]∂V ∗

∂y∗=0

We then argue that the terms in brackets must be the same order of magnitude, since theyare the only two terms in this equation. The non-dimensional terms again have an order ofmagnitude of 1, so the order of magnitude of the bracketed terms have to match. Therefore, ourappropriate velocity scale in y, or Vs, is:

Vs =U∞δ

L

Look at what this is saying - the ratio of δ/L is small, so our characteristic velocity in they-direction is small compared to the free-stream velocity. Now, using this with our other scaling



parameters, we turn to the momentum equation:

U∂U

∂x+ V

∂U

∂y=− 1

ρ

∂P

∂x+µ

ρ

∂2U

∂x2+µ

ρ

∂2U

∂y2

(U∞U∗)∂(U∞U

∗)

∂(Lx∗)+

(U∞δ

L

)V ∗

∂(U∞U∗)

∂(δy∗)=− 1

ρ

∂(PsP∗)

∂(Lx∗)+µ

ρ

∂2(U∞U∗)

∂(Lx∗)2+µ

ρ

∂2(U∞U∗)

∂(δy∗)2[U2∞L

]U∗

∂U∗

∂x∗+

[U2∞L

]V ∗

∂U∗

∂y∗=−

[PsρL

]∂P ∗

∂x∗+

[µU∞ρL2

]∂2U∗

∂x∗2+

[µU∞ρδ2

]∂2U∗

∂y∗2

U∗∂U∗

∂x∗+ V ∗

∂U∗

∂y∗=−

[PsρU2∞

]∂P ∗

∂x∗+

[µ

ρU∞L

]∂2U∗

∂x∗2+

[µL

ρU∞δ2

]∂2U∗

∂y∗2

Once again, we’re able to look at the bracketed terms to determine what it important in thisequation. The LHS should all have an order of magnitude around 1. The first bracketed termon the RHS is like the Cp term we’ve seen before. We can then scale pressure as Ps = ρU2

∞, orthe dynamic pressure. The factor of 1

2 that classically comes from Bernoulli’s equation doesn’treally affect the order of magnitude, so we don’t worry about including it.

The second bracket on the RHS is 1/Re, and we know Reynolds number tends to be largein a boundary layer (1m/s velocity with a 1m long plate in air gives us Re ∼ 66000). So wecan conclude this first viscous term is negligible. What about the second viscous term on theLHS - is it negligible? If it is, then we’ve once again lost viscosity! Prandtl argued that cannotbe true, especially since we’re saying viscous effects need to be important in order to satisfy theno-slip condition. Therefore, we need the bracketed parameter on our second viscous term toalso have an order of magnitude of 1:

µL

ρU∞δ2∼1

δ2 ∼ νLU∞

δ ∼√νL

U∞

Note that we’ve used ν = µ/ρ. We can divide both sides by the L scale to re-cast in a differentlight:

δ

L∼√

ν

U∞L=

1√ReL

(5.9)

This tells us that the length scale in the y-direction, which is our boundary layer thickness,is very small compared to the x-direction length scale. We assumed this to be the case by sayingδ < L, and of course we constrained it by solving for δ assuming it is small enough to make thelast viscous term important. Nonetheless, by using some dimensional analysis, we’ve been ableto understand how the boundary layer thickness is related to flow parameters!

Lastly, we can then go back to our non-dimensional x-momentum equation to find the final



form. Plugging in our expression for the boundary layer thickness, we get:

U∗∂U∗

∂x∗+ V ∗

∂U∗

∂y∗=−

[PsρU2∞

]∂P ∗

∂x∗+��>

0[1

ReL

]∂2U∗

∂x∗2+

[µL

ρU∞δ2

]∂2U∗

∂y∗2

U∗∂U∗

∂x∗+ V ∗

∂U∗

∂y∗=−

[ρU2∞

ρU2∞

]∂P ∗

∂x∗+

[µL

ρU∞

(√νLU∞

)2

]∂2U∗

∂y∗2

U∗∂U∗

∂x∗+ V ∗

∂U∗

∂y∗=−

[1

]∂P ∗

∂x∗+

[µ

ρν

]∂2U∗

∂y∗2

U∗∂U∗

∂x∗+ V ∗

∂U∗

∂y∗= −∂P

∗

∂x∗+∂2U∗

∂y∗2

We see that the non-dimensional form of the x-momentum equation has four terms that are allof order 1. This means the four original terms (not non-dimensional form) can be retained inour equation, while we then neglect all other terms. Therefore, the x-momentum equation forour 2D, steady, incompressible, negligible gravity boundary layer is:

U∂U

∂x+ V

∂U

∂y= −1

ρ

∂P

∂x+ ν

∂2U

∂y2(5.10)

Even though viscosity is small, the last term remains because the gradients in the verticaldirection are extremely strong in the near wall region.

5.2.2 y-momentum equation

The next step in this analysis is to scale the y-momentum equation. There is one importantfactor when doing the scaling analysis on this equation - it is just one of the components of avector equation! This means that it is directly tied to the other components of the momentumequation. Think on this: if you scale one component of a vector in a certain way, you have toscale the other components of that vector in the same manner in order to not “skew” the vector.The magnitude may have changed, but the direction should be the same.

Since we have to scale in the same fashion, we re-use all the previous scaling parametersfrom the x-momentum equation on the y-momentum equation, which is written as:

U∂V

∂x+ V

∂V

∂y= −1

ρ

∂P

∂y+µ

ρ

∂2V

∂x2+µ

ρ

∂2V

∂y2

Notice that this looks extremely similar to the x-momentum equation, but instead is acting inthe y-direction. The same assumptions apply as before, but now we’re concerned about thevertical components.

Now, applying the same scalings as before, we find:

(U∞U∗)∂(U∞δL V ∗

)∂(Lx∗)

+

(U∞δ

LV ∗)∂(U∞δL V ∗

)∂(δy∗)

=− 1

ρ

∂(ρU2∞P

∗)

∂(δy∗)+µ

ρ

∂2(U∞δL V ∗

)∂(Lx∗)2

+µ

ρ

∂2(U∞δL V ∗

)∂(δy∗)2[

U2∞δ

L2

]U∗

∂V ∗

∂x∗+

[U2∞δ

L2

]V ∗

∂V ∗

∂y∗=−

[U2∞δ

]∂P ∗

∂y∗+

[µU∞δ

ρL3

]∂2V ∗

∂x∗2+

[µU∞ρδL

]∂2V ∗

∂y∗2



At this point, it would be tempting to divide by the bracketed term from the LHS of the equation.However, we had argued that we must scale both equations in the same manner! Since the x-

momentum equation was re-scaled by dividing it by U2∞L , we must do the same here so our vector

equation isn’t “skewed.” This will result in:[δ

L

]U∗

∂V ∗

∂x∗+

[δ

L

]V ∗

∂V ∗

∂y∗= −

[L

δ

]∂P ∗

∂y∗+

[µ

ρU∞L

δ

L

]∂2V ∗

∂x∗2+

[µ

ρU∞L

L

δ

]∂2V ∗

∂y∗2

The grouping here is for us to see how each of the bracketed terms evolve. Recall that ReL =ρU∞Lµ , and from equation 5.9 we know that δ

L ∼ 1/√ReL. Therefore, we can plug these into our

above relation to get:[1√ReL

]U∗

∂V ∗

∂x∗+

[1√ReL

]V ∗

∂V ∗

∂y∗= −

[√ReL

]∂P ∗

∂y∗+

[1

(ReL)3/2

]∂2V ∗

∂x∗2+

[1√ReL

]∂2V ∗

∂y∗2

As was argued with the x-momentum equation, certain terms become less and less importantas the Reynolds number grows. In this case, we see only one term remains important withincreasing ReL! Therefore, all other terms are negligible and we can write our y-momentumequation as:

0 =∂P

∂y(5.11)

This is a very important result for us that study fluid mechanics. It says that through aboundary layer, there is no gradient in pressure in the y-direction. That means that whateverthe pressure is outside the boundary layer, it will be the same all the way through to the wall.We can show this through integrating this equation:

freestream∫wall

0 dy =

freestream∫wall

∂P

∂ydy

0 =P∞ − Pwall

In other words, there is no difference between the freestream pressure and the wall pressure.Therefore, for a given x-location, we can treat the static pressure as a constant!

5.2.3 Combining the x- and y-momentum equations

Since we’ve seen that the pressure isn’t changing in the wall-normal direction, then P inside theboundary layer is the same as P∞, which is defined as the “inviscid” region that lies outsidethe boundary layer. Since this outer flow is inviscid, incompressible, steady, and irrotational,Bernoulli’s equation applies! Therefore, we can say for the outer flow:

1

2ρU2∞ + P∞ = const.



Differentiating this with respect to the x-direction gives us:

d

dx

(1

2ρU2∞ + P∞ =const.

)ρU∞

dU∞dx

+dP∞dx

=0

U∞dU∞dx

=− 1

ρ

dP∞dx

From the y-momentum equation, we know P isn’t a function of y, so the x-momentum equation’spressure gradient can be a total derivative with respect to x rather than a partial derivative.This allows us to substitute the above expression into equation 5.10 to get:

U∂U

∂x+ V

∂U

∂y= U∞

dU∞dx

+ ν∂2U

∂y2(5.12)

This is our final expression for the x-momentum equation in a 2-D, incompressible boundarylayer. If the flow outside the boundary layer weren’t changing, then U∞ is a constant (this isanother way of saying there is no imposed pressure gradient on the flow!), and we get:

U∂U

∂x+ V

∂U

∂y= ν

∂2U

∂y2(5.13)

5.2.4 Bonus: similarity solution

One of Prandtl’s students, Blasius, was able to perform a coordinate transformation on thezero pressure gradient equation. By utilizing a streamfunction to simplify the non-linear partialdifferential equation above into a non-linear ordinary differential equation, he found:

f ′′′ + f ′′f = 0

The primes indicate differentiation with respect to the new coordinate. The math is very in-volved, but it nonetheless reduced the equation to a form that can be solved numerically. Ofcourse in 1908 we didn’t have computers, so the numerical solution was done by hand. Com-pletely doable, but time intensive.

From his analysis, Blasius was able to show:

δ

x=

5√Rex

Here, x is the downstream position. This is just like we found in equation 5.9! We said the ratioof δ/L was proportional to 1/

√ReL, and the “exact” solution of Blasius found the constant of

proportionality is 5! This goes to show that a scaling analysis and non-dimensionalization areextremely powerful and useful for us.



5.3 Bonus: Energy Equation

This section is a nice exercise of non-dimensionalization. Additionally, it shows a few morenon-dimensional numbers that come up in fluids. Let’s take the differential form of the energyequation:

ρcpDT

Dt= k∇2T +

DP

Dt+(τv · ∇

)~U (5.14)

In this equation, k is the thermal conductivity of the fluid, cp is the specific heat of the fluid,DDt is the material derivative, and τv is our viscous stress tensor, defined as:

τv = µ∇ · ~U + µ(∇ · ~U)T (5.15)

We’ve assumed an incompressible flow and used continuity and a number of other relationsto simplify this expression. It can be found in a similar manner to how you found the Navier-Stokes equation, but the math isn’t important here. Let’s just look to non-dimensionalize thisequation and see what pops out! First, let’s assume a steady flow, so all ∂

∂t terms go to zero.Then we write our equation as:

ρcp(~U · ∇)T = k∇2T + (~U · ∇)P +(τv · ∇

)~U

Next we introduce appropriate scaling parameters. We start by using T = TsT∗ for the temper-

ature scaling. Next, for τv, we look to equation 5.15. Note that it’s a gradient of velocity timesµ, so we can scale it with our velocity and length scales. Therefore, we’ll write τv = µUs

L τ∗v .Plugging these into our energy equation gives:

ρcp(Us~U∗ · 1

L∇∗)(TsT ∗) = k(

1

L∇∗)2(TsT

∗) + (Us~U∗ · 1

L∇∗)(PsP ∗) +

(µUsLτ∗v ·

1

L∇∗)(Us~U

∗)

Moving all constants out of the derivatives gives:[ρcpUsTs

L

](~U∗ · ∇∗)T ∗ =

[kTsL2

]∇∗2T ∗ +

[UsPsL

](~U∗ · ∇∗)P ∗ +

[µU2

s

L2

](τ∗v · ∇∗

)~U∗

Note here that the bracketed terms are our scaling parameters and physical constants. Theseare all constants and set by the flow we’re looking at. If we determine that the inertial termsdominate (which is the LHS, and is usually the case), then we can divide by that parametergrouping. This gives us:

(~U∗ · ∇∗)T ∗ =

[k

ρcpUsL

]∇∗2T ∗ +

[Ps

ρcpTs

](~U∗ · ∇∗)P ∗ +

[µUs

ρcpTsL

](τ∗v · ∇∗

)~U∗

Now we can start defining some new non-dimensional parameters. Starting with the first termon the RHS, we have:

k

ρcpUsL=

µ

ρUsL

k

cpµ=

1

ReL

1

Pr

So we’ve once again found the Reynolds number, but we’ve also discovered the Prandtl number,defined as Pr =

cpµk , which is a measure of viscous diffusion to thermal diffusion. Next, the

second term on the RHS has Ps. From our previous scaling of the momentum equation, we



determined an appropriate scale for Ps is ρU2s . This comes from the fact that pressure and

inertial effects tend to balance. If we had a different scenario, we would have to scale differently.However, setting Ps = ρU2

s , we get:

PsρcpTs

=ρU2

s

ρcpTs=

U2s

cpTs= Ec

Here, we’ve defined a new non-dimensional number called the Eckert number. This is can beshown to be proportional to the Mach number squared in a perfect gas. The last term is then:

µUsρcpTsL

=µ

ρUsL

U2s

cpTs=

Ec

ReL

So by just re-arranging the terms, we came up with two non-dimensional numbers we’ve seenbefore. All together now, we finally get:

(~U∗ · ∇∗)T ∗ =

[1

ReL · Pr

]∇∗2T ∗ +

[Ec

](~U∗ · ∇∗)P ∗ +

[Ec

ReL

](τ∗v · ∇∗

)~U∗

While you likely won’t be using the full energy equation in an undergrad fluids class (atleast not in this form), we can see that we are able to get a lot of information through non-dimensionalizing. We find non-dimensional parameters that are relevant to the flow withouthaving to make a list. The equations hold a lot more information than they first seem to!

Also recall that we can analyze and compare flows that are similar, so long as we matchnon-dimensional parameters. The energy equation (eq 5.14) is telling us how the temperaturefield in a flow changes due to heat, pressure, and viscous energy contributions. So if we want toanalyze a flow where we have changes in temperature, we should be looking to match not justthe Reynolds number, but also the Prandtl and Eckart numbers too.


Chapter 6

Analysis of the Boundary Layer

6.1 Summary of the boundary layer

After discovering the concept of the boundary layer, it is important to try and understand someparameters that we use to describe and characterize these types of flow. We previously learnedthat the boundary layer itself is a thin region near the wall where viscosity plays an importantrole in the flow. Outside the boundary layer, the fluid can be described as inviscid, and thusthe flow is governed by Euler’s Equation.

Within the boundary layer, we cannot neglect viscosity (which is the whole reason we havedefined a boundary layer). We then utilized an order of magnitude analysis to reduce thex-momentum equation to:

U∂U

∂x+ V

∂U

∂y= −1

ρ

dP∞dx

+ ν∂2U

∂y2(6.1)

Note that the pressure gradient is not a partial derivative. This was found from the y-momentumequation, where the gradient in the y-direction is zero, thus pressure only varies in x and is set bythe outer flow. Note that this is the static pressure we speak of, and the velocity still contributesa dynamic pressure to the system.

In order to find this equation, we had to set the y-direction scale as δ, which was assumedto be much smaller than the x-direction scale, L. From this assumption, we found:

δ

L∼ 1√

ReL(6.2)

The “exact” Blasius solution (which assumed the pressure gradient was zero) found the unknownpre-factor, giving the relation as:

δ

x=

5√Rex

(6.3)

One quick note: we have switched L and x between the two expressions above. This isn’t anissue, since L represents some downstream length scale, which is occurring in the x-direction.Therefore, we can just use x as a variable instead.


CHAPTER 6. ANALYSIS OF THE BOUNDARY LAYER

6.2 Zero pressure gradient

The laminar boundary layer analysis is performed under the assumption that the pressure gra-dient, dP/dx, is zero. This allows a considerable simplification to the equations of motion, andwas what allowed Blasius to find a similarity solution. How is it that a flow can exist without apressure gradient? It does seem counter intuitive, doesn’t it?

We can consider it from a few perspectives. One way is to imagine a flat plate being pushedor moved through still air. There is no external pressure gradient on the flow, since it is all stilland quiescent except for the immediate region around the plate. Another way to imagine thisis to appeal to the fact that there is inertia to the flow, and the inertia carries this flow. Sure,some pressure gradient likely had to get the flow moving, but then we argue that the flow is atits set speed, and we no longer are forcing it, but letting it continue. Since the outer portion ofthe flow (outside the boundary layer) is inviscid, there is no resistance to its inertia, so it willcontinue to move at the free stream velocity.

Here’s a mathematical way to think of it. Outside the boundary layer, the viscous effectsare negligible. This allows us to use Euler’s equation:

ρD~U

Dt= −∇P − ρg∇h

Now let’s think about what assumptions go into analyzing the flow in and around the bound-ary layer. We neglect gravity, say it’s steady, 2-D, and constant density. This means ourx-momentum equation for the outer flow simplifies down to:

ρU∞∂U∞∂x

+ ρV∞∂U∞∂y

= −dP∞dx

Recall how in the last chapter we found the pressure to not change in the y-direction. Thisallows us to write the pressure derivative as a total derivative, not a partial. Now let’s thinkon what’s happening in this outer flow - the velocity is constant! Of course, there is some smallV∞ that changes with downstream position (think about continuity for that, or look ahead toour integral analysis in the next section). However, there isn’t a change in our external U∞, thex-component of the free stream velocity. This means:

ρU∞��

0∂U∞∂x

+ ρV∞��

0∂U∞∂y

=− dP∞dx

0 =dP∞dx

The pressure gradient in the external flow is zero! We’ve shown that a constant external flowvelocity means the problem can be considered a ZPG boundary layer. Let’s discuss this a littlemore in the next section.

6.3 Displacement thickness

One parameter we use to describe a boundary layer is the displacement thickness. This canbe thought of as a measurement of the displacement of an ideal (inviscid) flow due to the decrease



in velocity in the boundary layer. If we account for the “lost mass” due to the no-slip conditionslowing the flow down, we can then equate it to an idealized flow that is displaced from the wall.

Figure 6.1: concept of displacement thickness

Figure 6.1 shows the conceptual way to think about this. We can see that there is a velocitydeficit from the free-stream, which represents a departure from the idealized case of perfect slipat the wall. Assuming a width w into the page, the left side of figure 6.1 shows this lost massflow, and can be expressed as:

lost mass flow = total mass flow with slip − total mass flow without slip

lost mass flow = w

∞∫0

ρU∞ dy − w∞∫

0

ρu(y) dy

lost mass flow = ρw

∞∫0

U∞ − u(y) dy

The next step is to define this lost mass flow as a “chunk” removed from the ideal slip case,which we define on the right side of figure 6.1. This lost mass flow is simply ρU∞wδ

∗. In otherwords, the ideal no-slip velocity is displaced a distance δ∗ from the wall. If we equate these two“lost mass flows” we then get:

ρU∞wδ∗ =ρw

∞∫0

U∞ − u(y) dy

δ∗ =1

U∞

∞∫0

U∞ − u(y) dy

Since the free stream velocity U∞ is a constant, we can move this inside the integral to get:

δ∗ =

∞∫0

(1− u(y)

U∞

)dy (6.4)



The expression in equation 6.4 gives the displacement from the wall that an ideal flow wouldbe to give an equivalent loss of mass flow that the viscous case has. This is another parameterto characterize a boundary layer, but is distinctly different from the boundary layer thickness.Will δ∗ be larger or smaller than δ? Think about figure 6.1: we will always find that δ∗ < δ.

6.3.1 Displacement thickness and the zero pressure gradient assumption

Remember the whole ZPG boundary layer assumption? What happens when we enclose ourflow in a wind tunnel? The mass flow into the entrance of the tunnel must equal the mass flowat the exit of a tunnel. However, we just came up with this concept of a displacement thicknessto account for the “lost mass flow” in the near wall region. At the exit of the tunnel, we canapproximate the flow as an inviscid external flow displaced from the wall by a distance δ∗. Thismeans we’ve “modified” the exit area, as there won’t be any flow in the displaced area. Sincemass must be conserved, this then means we’ve modified the free-stream velocity! Let’s take aquick look at why this is happening.

Figure 6.2: Wind tunnel - viscous effects at wall give a displacement thickness at the exit

Figure 6.2 shows the schematic setup of a wind tunnel. We have an inviscid flow into thetunnel, then the boundary layer at the top and bottom walls results in an equivalent displacementthickness at the exit. With boundary layers, we assume a 2-D, steady, incompressible flow, whichmeans we can write our continuity equation as:

∫∫A

n · ρ~U dA = 0

−ρU∞,inwh+ ρU∞,outw(h− 2δ∗) = 0

U∞,out = U∞,in

(h

h− 2δ∗

)This is saying the free stream velocity at the exit is larger than the free stream velocity at theinlet. In other words, we have some dU∞/dx. Based off our analysis for the inviscid outer flow,this means we don’t really have zero pressure gradient, as we end up with:

ρU∞dU∞dx

= −dP∞dx



This is non-zero in the wind tunnel! That means there’s some sort of pressure gradient due tothe change in the free stream velocity.

In general, boundary layers are small, and thus δ∗ is very small. So long as the heighth >> δ∗, this should have little effect on the free stream, and the assumption of ZPG will hold.

6.4 Momentum thickness

Utilizing the same methodology we used to calculate the displacement thickness, we can findanother length-based parameter associated with the boundary layer. We will look to find thelost momentum flux from the flow, just as we found the lost mass flow. This lost momentumwill be related to a momentum thickness that the inviscid flow is moved from the wall. Onceagain, we look to see how much momentum is lost from the ideal no-slip case. Why do we careabout lost momentum from the flow? Think about Newton’s second law - rate of change ofmomentum is equal to the total force on the system. Therefore, a loss of momentum meanssome force was imparted on the flow (we will see this is the viscous drag)!

The math involved is a little more complicated compared to the displacement thickness. Let’slook to a control volume analysis of the boundary layer. We want to find the lost momentum

Figure 6.3: control volume used to find lost momentum from ideal case

from the ideal no-slip case, so we can just do a CV analysis on the red box in figure 6.3. Wewill compare the fluxes of momentum in the x-direction, and find their overall balance. Thisvalue will be the difference in momentum from the ideal no-slip case. Since this is a steady, 2-D,constant density problem, we will find:

Net x-momentum change = sum of x-momentum into and out of the control volume

Net x-momentum change =

∫∫A

(n · ρ~U)~U dA

We need to determine what is being carried in and out through the surfaces. The inflow ofmomentum is simply the left surface, which has a velocity equal to the free stream U∞, and inthis 2-D problem occurs over a height of H and some width w. This corresponds to the idealcase, where if there were slip at the wall we would have this momentum across the entire inlet.

The outflow of momentum occurs through the top and right surface. For the top, we have ahorizontal freestream velocity U∞ and a vertical component v over a length of L and width w.



However, for the x-momentum balance, we see that the n · ~U component is occurring through asurface where the normal is vertical. Therefore, n · ~U = j · ~U = v, the vertical velocity out.

The right outflow is like the left inflow, except the velocity is u(y), not the free stream. Now,the velocity profile u(y) will reach the free stream value before the end of the control volume,but that’s just accounted for in the function (which we don’t need to know the exact value atthis time).

Therefore, using these values, we can write our momentum loss as:

Net x-momentum change = ρ

∫∫Aleft

(−i · ~U)Ux dA+ ρ

∫∫Atop

(j · ~U)Ux dA+ ρ

∫∫Aright

(i · ~U)Ux dA

Net x-momentum change = ρw

H∫0

−U2∞ dy + ρw

L∫0

vU∞ dx+ ρw

H∫0

u2 dy

Net x-momentum change = −ρwHU2∞ + ρwU∞

[ L∫0

v dx

]+ ρw

H∫0

u2 dy

The next step is to replace the currently unknown v expression (which is bracketed for ourconvenience) with known parameters. Utilizing continuity, we simply balance the mass flow inwith the mass flow out to find v as:

∫∫Aleft

n · ρ~U dA+

∫∫Atop

n · ρ~U dA+

∫∫Aright

n · ρ~U dA = 0

−ρwHU∞ + ρw

L∫0

v dx+ ρw

H∫0

u dy = 0

L∫0

v dx = HU∞ −H∫

0

u dy

This expression for the integral of v is exactly the bracketed parameter in the momentumexpression above! We then substitute it back into the momentum equation:

Net x-momentum change = −ρwHU2∞ + ρwU∞

[HU∞ −

H∫0

u dy

]+ ρw

H∫0

u2 dy

Net x-momentum change = −ρwU∞

H∫0

u dy + ρw

H∫0

u2 dy

Net x-momentum change = −ρwH∫

0

uU∞ − u2 dy



A little re-arranging was done for the last line. Now, we know that U∞ ≥ u(y), so the integralwill always be positive. This means our net x-momentum change is negative, implying a lossof momentum! We expected that, since we can think of this value as the momentum lost fromthe ideal wall-slip case. Now, we will define a new thickness θ that represents the displaced

Figure 6.4: concept of the momentum thickness, similar to the displacement thickness

momentum. This can be visualized in figure 6.4. The missing momentum from the ideal con-dition would simply be ρU2

∞wθ. Equating this “idealized lost momentum” with the actual lostmomentum from the flow, we find:

ρU2∞wθ =ρw

H∫0

uU∞ − u2 dy

θ =1

U2∞

H∫0

uU∞ − u2 dy

Once again, U∞ is a constant, and can be brought into the integral. Additionally, the limit ofintegration can be set to ∞ rather than some arbitrary H. Think about why that works: u/U∞goes to 1 as y → δ, so the integrand goes to zero as the limit goes to δ. This means there will beno additional contribution to our equation for an increase in the upper integrating limit beyondy = δ. Some rearranging gives us the following result:

θ =

∞∫0

u

U∞

(1− u

U∞

)dy (6.5)

The expression in equation 6.5 gives us the equivalent distance from the wall that an idealinviscid flow would be moved to account for the lost momentum due to viscous effects. Like δ∗,this is a new parameter to characterize a boundary layer. Also, we can see that the momentumthickness will be smaller than the displacement thickness (compare the integrands of the twodefinitions to see).



6.4.1 Momentum thickness and drag

We calculated the momentum thickness by doing a control volume analysis and finding thechange in x-momentum. We saw this was a net loss, and equated it to the “loss” in x-momentumfrom displacing an inviscid flow from the wall. However, you can also see that this net loss ofx-momentum will just be

∑Fx from the Navier-Stokes Equations:

d

dt

∫∫∫V

ρ~U dV +

∫∫A

(n · ρ~U)~U dA =∑

~F

When we look in the x-direction for our flow, like in figure 6.4, this simplifies down to:

��

��

��*0d

dt

∫∫∫V

ρUx dV +

∫∫A

(n · ρ~U)~U dA

︸︷︷︸our momentum deficit

=∑

Fx

This simplifies to ρU2∞wθ = Fvisc. The forces reduced to only viscous forces because we like to

deal with a zero pressure gradient flow, which means forces from pressure are 0. Additionally,there are no x-direction gravity forces, and the CV doesn’t cut through a surface. Therefore,we can re-express our momentum thickness (for some given width w into the page) in terms ofsurface drag:

θ =FviscρU2∞w

=

∞∫0

u

U∞

(1− u

U∞

)dy (6.6)

6.5 Example: approximate boundary layer

We know from class that the Blasius solution isn’t an analytical solution, so we don’t have adirect functional relationship between velocity and position. However, we can approximate thevelocity profile in a boundary layer with different functional forms. One way to do it is to specifythe boundary conditions and pick a function that satisfies them (and at least looks like a velocityprofile). We will assume a set x-position, so there isn’t any functional dependence on x for thiscase. We have the following boundary conditions:

u(0) = 0 and u(δ) = U∞ and∂u

∂y

∣∣∣∣y=δ

= 0

These three BCs are the no-slip condition, the definition of the boundary layer height, and no-shear at the outer boundary layer edge. You can think of any number of functions that satisfiesthese conditions, but let’s just use a simple parabolic relation:

u = U∞

(2y

δ− y2

δ2

)(6.7)

You can see that this function satisfies all three of our necessary boundary layer BCs. Giventhis function, let’s calculate out the displacement thickness and the momentum thickness.



δ∗ =

∞∫0

(1− u(y)

U∞

)dy

=

δ∫0

(1−

(2y

δ− y2

δ2

))dy

=y − y2

δ+

y3

3δ2

∣∣∣∣δ0

Therefore, we see our displacement thickness from this velocity profile is:

δ∗ =δ

3

This is saying the displacement thickness is a third the height of the boundary layer thickness,which lines up with our assumption that δ∗ < δ. Next, the momentum thickness can be found:

θ =

∞∫0

u

U∞

(1− u

U∞

)dy

=

δ∫0

(2y

δ− y2

δ2

)(1−

(2y

δ− y2

δ2

))dy

=− y5

5δ4+y4

δ3− 5y3

3δ2+y2

δ

∣∣∣∣δ0

And our momentum thickness comes out to be:

θ =2δ

15

Again, this matches our intuition that θ < δ∗ < δ. These approximations actually turn out tonot be too terrible compared to the Blasius solution, which gives δ∗B = 0.34δ and θB = 0.132δ.

6.6 Example: finding the skin friction

A laminar boundary layer velocity profile may be described approximately as u = U∞ sin(πy/2δ).Find the shear stress at the wall, τw, and express both the total and local skin friction coeffi-cients, CF and Cf in terms of the Reynolds number. (Problem adapted from Lex’s book.)

Let’s start by listing our definitions of the total and local skin friction coefficients:

CF =Fvisc

12ρU

2∞A

and Cf =τw

12ρU

2∞



Note that CF is based on the total drag force Fvisc that occurs over an area A, and Cf is basedthe wall shear τw at a single location. In order to calculate these, we turn to the velocity profile:

u = U∞ sin

(πy

2δ

)We can start by finding τw through the velocity gradient:

τw =µ∂u

∂y

∣∣∣∣y=0

=µπU∞

2δcos

(π ∗ 0

2δ

)=µπU∞

2δ

So we have the local shear stress as a function of delta. Next, we need to find the total dragforce, Fvisc, which we know is related to the momentum thickness through equation 6.6:

FviscρU2∞w

=

∞∫0

u

U∞

(1− u

U∞

)dy

We then plug in our velocity profile to get the total viscous drag:

Fvisc =ρU2∞w

δ∫0

sin

(πy

2δ

)(1− sin

(πy

2δ

))dy

=ρU2∞w

[δ

2πsin

(πy

δ

)− 2δ

πcos

(πy

2δ

)− y

2

]δ0

=ρU2∞wδ

(2

π− 1

2

)Now we want the non-dimensional parameters, CF and Cf , as a function of the Reynolds number,ρU∞x/µ. This means we need to find some sort of x-dependence that is in this system. Oneparameter here that would have an x-dependence is the boundary layer thickness, δ. Therefore,we should hope to find a way to get that functional dependence!

Lucky for us, there are other terms that depend on x, namely the wall shear τw and thetotal drag Fdrag. Think about it this way - since the boundary layer thickness grows, the veloc-ity profile will change, thus the shear will change downstream. Additionally, the total drag isdependent on how far down you integrate over the surface!

So let’s look at the total drag and relate it to the local friction. We can get the total drag(due to friction) by integrating the local friction over the area of the surface exposed to flow:

Fdrag(x) = w

∫ x

0τw dx



Here we assume the shear isn’t changing over the width w. Then, we can differentiate thisexpression to get:

1

w

d

dxFdrag = τw

We have expressions for both Fdrag and τw that have the boundary layer thickness δ in them!We can then plug those into the above relation and solve:

1

w

d

dx

[ρU2∞wδ

(2

π− 1

2

)]=µπU∞

2δ

ρU2∞

(4− π

2π

)dδ

dx=µπU∞

2δ

δ dδ =µπ2

ρU∞(4− π)dx

1

2δ2 =

µπ2x

ρU∞(4− π)

δ =

√2π2

(4− π)·√

µx

ρU∞

δ =4.8

√µx

ρU∞

This relation for δ should look pretty familiar. If we were to divide by x, we would get:

δ

x=

4.8√Rex

This is nearly the same as the result from the Blasius solution, which instead had a constantof 5! That’s pretty remarkable that we can just use an approximate shape and get nearly thesame result as the “exact” solution. So, using our expression for δ, we can now find expressionsfor our CF and Cf :

Cf =τw

12ρU

2∞

=1

12ρU

2∞∗ µπU∞

2δ

=µπ

ρU∞∗ 1

4.8√

µxρU∞

=0.655

√µ

√ρU∞x

Cf =0.655√Rex

Our skin friction coefficient, Cf , has a pre-factor of 0.655 from this velocity profile estimate.The Blasius skin friction coefficient has a prefactor of 0.664 for comparison.



Performing the same math with the total drag coefficient, we find:

CF =Fvisc

12ρU

2∞A

=1

12ρU

2∞wL

∗ ρU2∞wδ

(2

π− 1

2

)=δ

L∗ 2

(2

π− 1

2

)

=4.8√

µLρU∞

L∗ 0.273

CF =1.311√ReL

Note that when plugging in the expression for δ, we assumed the x-location was the length L inwhich the area was defined. Our drag coefficient CF has a pre-factor of 1.311, which comparedto the Blasius prefactor of 1.328 is again a good estimate.

So while there was a bit of math, we were able to find skin friction and drag coefficients for agiven boundary layer velocity profile! We had utilized the concept of the momentum thickness toprovide a relationship between drag force, boundary layer thickness, and downstream location.We see that this seemingly strange concept ended up providing a powerful tool for our analysis.


Chapter 7

Drag, Separation, and Shedding

7.1 Drag force from friction

In the previous chapter, we were able to relate the boundary layer thickness, drag force, andskin friction through the momentum thickness. We were able to take different approximationsto the velocity profile in a boundary layer (such as a quadratic function, or a sine function)and build these relations. In general, for a flat plate, zero pressure gradient boundary layer, theBlasius solution results in the following:

δ

x=

5√Rex

Cf =τw

12ρU

2∞

=0.664√Rex

CF =Ffriction12ρU

2∞A

=1.328√ReL

And with respect to the flat plate boundary layer, Ffriction was simply the total frictional forcedue to wall shear stress, which for some downstream length L and uniform width w means:

Ffriction = w

∫ L

0τw dx

Note that CF is not the same as CD, which is the drag coefficient. CD comprises of both thefrictional and pressure forces that make up the total drag force:

Fdrag = Ffriction + Fpressure

So make sure you pay attention to what it is you’re calculating!

Another important point to note is the A in CF (as well as CD). This is the characteristicarea of the object, which can be different depending on what you’re looking at. In general, it’sthe area that is responsible for the main contribution of the force. It’s likely to be the frontalarea, or projected area to the flow. However, depending on the setup, it may be defined dif-ferently, such as using the surface area. Just be mindful of how it’s being use in your specific


CHAPTER 7. DRAG, SEPARATION, AND SHEDDING

application.

These differences in non-dimensional force coefficients help us to understand why we call abody either “streamlined” or “bluff.” If the primary contribution to the drag is due to pressureforces, then we call the object a bluff body. If the primary contribution is instead from frictionalforces, we call it a streamlined body. Think of an airfoil: this is the classic streamlined body.We try not to have it at too high of an angle of attack, so it is primarily subjected to frictionalforces. Conversely, a ball has a majority of its drag come from the pressure differences betweenthe front and back, so it would be a bluff body.

7.1.1 Example: drag on a plate

Find the drag force on a thin, 1m wide by 5m long plate in a flow of air at 1 m/s.

First, checking the Reynolds number, we find that for this plate, ReL = 1∗50.000015 ≈ 300, 000,

which is less than the critical reynolds number. Therefore, we can assume this flow remainslaminar, and thus can use the Blasius relations.

For the total drag force, we can assume it’s only due to frictional forces, as this body isassumed thin. Therefore, there are little to no pressure forces acting on the plate.

Fdrag = Ffriction +��:0

Fpressure

Since we can assume that the total drag force is from the viscous friction, we say CD ≈ CF .Then, we can use our flat plate relations to calculate the drag force:

CD ≈ CF =Ffriction12ρU

2∞A

=1.328√ReL

where the last equality came from the Blasius flat plate drag relation. So we plug in our Reynoldsnumber to the relation:

CF =1.328√300000

= 0.00242

Then our total drag force can be found:

Fdrag ≈ Ffriction =CF

(1

2ρU2∞A

)=0.00242 ∗

(1

2∗ 1.2

kg

m3∗(

1m

s

)2

∗ (2 ∗ 1m ∗ 5m)

)Fdrag = 0.015N



Note that we used the total surface area of the plate, which is the two sides. Overall, the dragfrom the flow is very small!

What would happen if this was oriented vertically so the flow was perpendicular against it?Then we’d have a bluff body, and pressure forces would dominate. To get the total drag here,we turn to tables or charts that list out the drag coefficient as a function of Reynolds number.In this instance, a flat plate perpendicular to flow has a CD = 2.0. Then, we find:

Fdrag =CD

(1

2ρU2∞A

)=2 ∗

(1

2∗ 1.2

kg

m3∗(

1m

s

)2

∗ (1m ∗ 5m)

)Fdrag = 6N

Note here that the area was now the frontal area, which is just 1m by 5m rather than doublethat for the streamlined body scenario. Overall, we see that bluff bodies tend to have more dragforce!

7.2 Separation of the boundary layer

We performed a variety of analyses on the boundary layer under the assumption of zero pressuregradient. However, we were able to see that the concept of the displacement thickness lead toa change in the free stream velocity in a wind tunnel. That was due to the streamlines beingdisplaced, and continuity leading to an increase in free stream velocity at the exit. Looking tothe momentum equation in the free stream, we saw it was Euler’s Equation that governed itsince there weren’t viscous effects.

U∞∂U∞∂x

+ V∞∂U∞∂y

= −1

ρ

dP∞dx

(7.1)

We then argued that U∞ doesn’t change with y, but continuity showed that U∞ was larger atthe exit than the entrance. This means we had:

U∞∂U∞∂x

= −1

ρ

dP∞dx

(7.2)

Since the LHS is a positive value, it means the RHS is also positive, indicating the pressuregradient must be negative:

dP∞dx

< 0

This negative pressure gradient was associated with an acceleration of the flow. We call thisa favorable pressure gradient. If the velocity had slowed down, we would have the LHSbe negative, thus our pressure gradient would be positive, indicating an adverse pressuregradient.

The naming convention comes from the effect this pressure gradient has on the flow field.Recall that we found ∂P/∂y = 0 in the boundary layer, so the static pressure within the bound-ary layer is set by the external pressure. If there is a favorable pressure gradient in the external



Figure 7.1: external pressure gradient acting within the boundary layer

flow, then this same pressure gradient must apply through the boundary layer as well. Sincethis static pressure in the outer flow is the same at the wall, the negative pressure gradient isfelt as the wall as well, where we find Pw,1 > Pw,2. Note in figure 7.1 that it’s the displacementthickness δ∗ that’s shown, since it was what led us to conclude there could be an increase in freestream velocity.

Let’s then look at the momentum equation at the wall to try and understand what this“imposed pressure gradient” might be doing to our flow. Including the pressure gradient, ourx-momentum equation for the boundary layer is:

U∂U

∂x+ V

∂U

∂y= −1

ρ

dP∞dx

+ ν∂2U

∂y2(7.3)

Note that we can still use P∞ since it is imposed throughout the boundary layer. Now, at thewall both the U and V velocities are zero, so the momentum equation reduces down to:

0 = −1

ρ

dP∞dx

+ ν∂2U

∂y2

∣∣∣∣y=0

1

µ

dP∞dx

=∂2U

∂y2

∣∣∣∣y=0

Recall from calculus what a second derivative is telling you. It’s an indication of curvature. So,with µ always positive, the curvature of the velocity profile at the wall is determinedby the value of the pressure gradient. This is illustrated in figure 7.2, where three valuesof a pressure gradient are shown. Note that this is the inflection at the wall, while near theboundary layer edge the inflection must be negative.



Figure 7.2: Inflection (at the wall) of the velocity profiles due to the pressure gradient

7.2.1 Favorable pressure gradient

Imagine we are back in the wind tunnel, and the displacement of the streamlines has lead to anincrease of freestream velocity. This in turn causes the free stream pressure to drop. Since theexternal pressure is imposed on the wall, the pressure gradient along the wall is negative. Thiscorresponds to case (3) in figure 7.2, in which we can find:

1

µ

dP∞dx

< 0 → ∂2U

∂y2

∣∣∣∣y=0

< 0

A negative value for curvature at the wall then indicates that the curvature is negative through-out the boundary layer. This results in no flow separation and all the world is happy.

7.2.2 Zero pressure gradient

This is the classic laminar boundary layer case, which is shown in (2) in figure 7.2. The curvatureof the velocity profile at the wall is zero, then it goes negative as it moves towards the free stream.Take a look at the actual Blasius boundary layer velocity profile, and you’ll see that the shapedoes indeed have zero curvature at the wall! Like the favorable pressure gradient, this flow willnot separate from the wall.



7.2.3 Adverse pressure gradient

If we have a situation where the velocity in the free stream is decreasing, then the pressuregradient must be positive. This is what is shown in case (1) in figure 7.2. Think about whenthis situation might happen - it would have to be the opposite of the favorable case. We sawthat a favorable pressure gradient resulted from the flow “necking down”, thus increasing freestream velocity. Therefore, if we have the flow “opening up,” like in a diffuser, then we couldexpect some free stream deceleration.

With a positive pressure gradient, we find the curvature of the velocity profile at the wallis positive. Since the velocity profile has negative curvature towards the boundary layer edge,we must have an inflection point somewhere in the flow field. This inflection point is howwe characterize if flow separation will occur. If the adverse pressure gradient is sufficientenough to drive the wall shear to zero (∂U/∂y = 0), then we would have separation. This ischaracterized by the flow becoming detached from the surface and mixing, increasing systempressure, and being more susceptible to turning turbulent. This region of flow is far more difficultto characterize and measure.

In addition to this increased mixing, we usually see a significant increase in the pressurefield, which can lead to an increase in drag. This is problematic for the engineer who is tryingto overcome drag forces, and thus is why we care about pressure gradients in a flow.

Note that we can only separate if there is an inflection point in the velocity profile, and evenfurther, we must have zero shear at the wall. That is why the favorable pressure gradient andzero pressure gradient boundary layers do not undergo separation.

7.3 Resistance to separation

From the three different types of flows (favorable, zero, and adverse pressure gradient), we sawthat it’s only under certain circumstances in the adverse pressure gradient that we see separation.Once exposed to this adverse PG, the flow near the wall will be affected by the pressure actingagainst the motion. We can visualize this effect on two different flows:

A turbulent boundary layer has more momentum in the near wall region compared to thelaminar boundary layer. This results in the pressure having to apply more force to overcomethis momentum. Additionally, turbulence is characterized by its ability to mix the flow, bringinglow momentum fluid from the wall to the outer region, and bringing high momentum fluid fromthe outer region towards the wall. This means a turbulent profile has a way of “healing” theadverse pressure gradients effect, and thus delaying when separation may occur. This conceptis illustrated in figures 7.3 and 7.4

7.4 Case: a curved surface

To demonstrate the importance of separation, let’s look to an airfoil. Imagine a nice, slenderairfoil in flow at a low angle of attack, as pictured in figure 7.5. As the streamlines diverge, theflow rate between them must remain constant, so their velocity decreases. Bernoulli’s equationtells us this will lead to an increase in pressure, thus we have a positive pressure gradient inthe flow. This was defined as an adverse pressure gradient, meaning it can induce separation!



Figure 7.3: adverse PG acting against laminar and turbulent velocity profiles

Figure 7.4: turbulence moving momentum towards the surface

However, just because we have an adverse pressure gradient, it doesn’t mean we have to separate.Now, let’s look at a higher angle of attack:

Look how the velocity profile in figure 7.6 not only has a positive inflection at the wall, butit also has backwards flow (relative to the incoming flow). This is exactly what flow separa-tion is! The pressure gradient is very positive, and this high pressure ends up “pushing” theboundary layer away from the wall, causing separation. This results in decreased lift, and manycomplications for anyone trying to fly.

7.5 Strouhal number

Once separation occurs, a region of recirculation (which may or may not be turbulent) buildsup. If the recirculation is behind a bluff body, then these recirculation zones can shed in



Figure 7.5: airfoil at slight angle of attack - note increase in pressure in direction of flow fromstreamline divergence, giving an adverse pressure gradient (yes I drew this in paint, don’t askhow long it took)

Figure 7.6: airfoil at large angle of attack - note that there is separation now

alternating patterns. There will be a frequency f associated with this shedding. If you performa dimensional analysis, you can find a non-dimensional number based around this frequency,which we call the Strouhal number:

St =fD

U∞(7.4)

In figure 7.7, you see nice groupings of vorticity. However, even if the wake were to be turbulent,a fundamental frequency would still exist for the body. There are charts of Strouhal numbervs. Reynolds number for a cylinder in cross flow. For a wide range of Reynolds number, theStrouhal number remains constant at around 0.21!

7.5.1 Example: Tacoma Narrows Bridge

It’s likely that you’ve seen a video of the Tacoma Narrows bridge oscillating in the wind (if youhaven’t take a moment to look it up on YouTube). On the day it collapsed, the winds were upto 20m/s, and the height of the roadway is approximately 3m. Was vortex shedding the cause



Figure 7.7: von Karman vortex street behind a cylinder at ReD = 100 (top) and 300 (bottom)

of the collapse?

We start by calculating the Reynolds number of the flow. We have the velocity and “diam-eter,” and because it’s in air, we know the kinematic viscosity is 0.000015m2/s.

Re =20 ∗ 3

0.000015= 4× 106

For the most part, the Strouhal number can be considered to be a near constant St = 0.21 onceReynolds number is greater than 100. Of course it is more complicated, especially with differinggeometries, but let’s stick to this for now. Using this estimation, we then calculate the frequencyof shedding as:

St = 0.21 =f ∗ 3

20→ f = 1Hz

If you watch the footage, the bridge is oscillating at around 0.25Hz, which isn’t the frequencyfrom the shedding. This means a more complex interaction occurred to cause the collapse, butI’d bet the wind didn’t help.



7.5.2 Example: car antenna

You’re gifted an older car with an antenna on it. When traveling at highway speeds, you hearan low, throbbing noise. Should you attribute the noise to the antenna, or get your car checkedup?

Let’s do a quick check on the Reynolds number, and thus Strouhal number, for the antenna.It’s likely around 1/8in in diameter, and you follow the speed limit in NJ (right?) and do 65mphon the highway. We then calculate the Reynolds number:

Red =UD

ν=

65mihr ∗ 1600 mmi ∗

13600

hrsec ∗

18 in ∗ 0.0254min

0.000015m2

s

≈ 6100

From a chart relating St to ReD for a cylinder, we find St = 0.21, which matches what we’vepreviously said! Then, we can calculate the frequency associated with this antenna:

St = 0.21 =f ∗ 0.0032m

29m/s→ f ≈ 1900Hz

A frequency of 1900Hz is not a low throbbing noise, but more of a high pitched whine. Infact, if you’ve ever been to a very loud concert, you’ve probably left with your ears ringing.That ringing sound is somewhat similar in pitch to this frequency! So a low throbbing noise inyour car isn’t from the antenna, and you should probably get it looked at.


Chapter 8

Internal Flows and Losses

8.1 Internal flows

The study of boundary layers is classified under external flows, where the flow goes over theoutside of a body. Other examples are flow over a cylinder or an airfoil, or the flow in theatmosphere. We are going to now look at internal flows, where the flow is surrounded by thebody. In other words, we’ll be looking to channel and pipe flows.

We like to simplify the analysis by considering flows that are fully developed. This meansthat the velocity profile within the pipe has developed to it’s fullest extent, and will not changeas it moves downstream. But what do we mean by “developed?” If we look to the entrance offluid in a 2-D channel, we can see:

Figure 8.1: development of the velocity profile in a channel entrance

As the fluid flows in the channel, a boundary layer starts to develop and grow. Note thatthere will be a boundary layer on the top of the channel as well as the bottom. These will growtowards each other, until at some downstream location they merge. Since the velocity profilechanges with x in a boundary layer, we expect the velocity profiles to change up to this merging.At this point, the flow has developed, and there is no more growth of these boundary layers orchange to the velocity profile with further movement in x. We now call the flow fully developed.


CHAPTER 8. INTERNAL FLOWS AND LOSSES

Imagine that you were to treat the two walls independent from one another. You could thensolve for the distance into the channel that the boundary layer develops. In other words, howfar downstream must you go until the boundary layer thickness is half the height of the channel?If the flow is laminar, we can just use the Blasius relation:

δ

x=

5√Rex

If the height of the channel is H, then we want to see when the boundary layers develop tohalf that height (they both grow from opposite sides, so they only need to meet in the middle).Therefore, we can estimate the “development length” from plugging this into the Blasius relation:

δ =H

2→

H2

x=

5√Rex

H

10=

x√ρU∞xµ

x =H2U∞100ν

x

H= 0.01 ∗ReH (8.1)

Equation 8.1 is an approximate expression for the entrance length for a 2-D channel, assumingthe flow is laminar. This same expression can be found for a pipe as well. The exact solutionsfor the two cases wouldn’t be this expression, but they are very similar.

Why do we say approximate solution? Think on the laminar flow equations - we assumedzero pressure gradient to get the Blasius solution. We know there is some sort of acceleration tothe outer flow here, as we discussed with respect to displacement thickness. This accelerationmeans there is a pressure gradient, thus our Blasius solution is no longer valid. However, thisat least gives us a decent approximation to understand how long it takes to turn fully developed.

Stating that a flow is fully developed has two primary implications.

• Viscosity is important everywhere. This is because the two boundary layers havemerged, and the boundary layer is the region in a flow where viscous effects are important.Therefore, viscosity will play a roll all throughout the flow in a channel (or pipe)

• There are no velocity gradients in the x-direction. Since it is fully developed, thereis no more change in the shape of the velocity profile that needs to occur. This means Ubecomes a function of y only, not x.

Let’s investigate the result of these effects.

8.1.1 Continuity

By turning to the incompressible, 2-D continuity equation, we can immediately find a fascinatingresult for fully developed flows:

∂U

∂x+∂V

∂y= 0



We just said that a fully developed flow has no changes to U in the x-direction, which meansthe first term is zero! Therefore, we can show:

��7

0∂U

∂x+∂V

∂y= 0∫

∂V

∂ydy = 0

V (y) + const = 0

This says that our vertical velocity V (y) is actually just a constant. However, applying thekinematic boundary condition (no flow through the wall) means V = 0 at the wall. Since V isjust a constant, then:

V = 0 everywhere in fully developed pipe/channel flow

8.1.2 Inertial terms

Using this result, let’s look at the inertial terms in the x-momentum equation:

ρU∂U

∂x+ ρV

∂U

∂y= · · ·

We just saw that V is zero, so the second term is gone. However, we’ve also said that the Uvelocity doesn’t change in the x-direction, so the first term is also gone! This means that theinertial terms do not have a contribution to the x-momentum equation! If you were to look atthe y-momentum, you could probably guess that the inertial terms there are also zero!

8.1.3 The x-momentum equation for a channel (and pipe)

Due to there not being any inertial effects, we can write the x-momentum equation as:

0 = −dPdx

+ µd2U

dy2

Or equivalently, we can write:

dP

dx= µ

d2U

dy2(8.2)

Equation 8.2 is saying that in a channel, the pressure gradient is balanced by viscous effects.In other words, the viscous force balances the pressure force. The inertial terms are negligiblecompared to these two forces! If we were to do the same analysis in cylindrical coordinates, wewould find the x-momentum equation in a pipe is:

dP

dx=µ

r

d

dr

(rdU

dr

)(8.3)



Overall, the flow is viscously dominated throughout, and inertia doesn’t play an effect tothe dynamics of the fluid. So a pressure gradient is the only thing left to overcome the effect ofviscosity slowing the fluid down. This is what equations 8.2 and 8.3 are telling us, since it’s thepressure term balancing the viscous term.

8.1.4 The pressure gradient

One aspect of pipe and channel flow that can be confusing stems back to the “fully developed”description. We said that the boundary layers from the entrance of the pipe converge, thus theflow has developed to its final profile. After that point, the velocity profile will not change withincreasing downstream distance. Put another way, there are no changes in the x-direction. How-ever, we see there is a pressure gradient, where we have an x-derivative. How do we explain that?

Let’s step away from the equations and try to use intuition for a moment. If you wantto push fluid through a pipe, we do that using a pump. This is just a way to increase thepressure at one end of the pipe. The increased pressure at one end drives the fluid out theother. However, viscous effects will work to resist the flow down and induce “energy losses” inthe system. Therefore, this pressure must be decreasing throughout the pipe. That’s exactlywhat a pressure gradient is - a change in pressure over a distance.

Now, this explains why dP/dx can exist, but we said there aren’t any changes to the flowvelocity in the x-direction. This means that the gradient can’t be changing in the x-directioneither. So while the pressure changes in x, the gradient itself is constant!

Let’s do a little math to show this has to be true. Start by taking a derivative of ourmomentum equation to show the result:

d

dx

[dP

dx= µ

d2U

dy2

]We can interchange x- and y-derivatives (this is a calculus thing, hopefully you are convincedthis is true), so then we can write:

d

dx

(dP

dx

)= µ

d2

dy2

(dU

dx

)Since U is not changing in the x-direction, the RHS of this equation is zero.

d

dx

(dP

dx

)= 0

dP

dx= const.

So in addition to our “gut check” for the pressure gradient, the math also showed thatthe pressure gradient must be constant. The value of the constant will set the magnitude ofthe velocity, with increasingly large magnitudes corresponding to increasing velocity. Whatabout the sign of the constant - is the pressure gradient positive or negative? Think about thediscussion we had with separation. The second derivative of velocity indicates the curvature.



We should have negative curvature throughout the velocity profile for an internal flow, so wewould expect the value of the pressure gradient to be negative. This means the pressure is fallingwith increasing distance, which again falls in line with the gut check!

8.2 The velocity profile in a pipe

This problem is worked out in any undergrad textbook, but perhaps this will be another wayto think about it. We want to find the laminar velocity profile in a pipe.

We start by looking to the momentum equation for a pipe and applying some boundaryconditions:

dP

dx=µ

r

d

dr

(rdU

dr

)We know there is no slip at the wall, so the velocity at the wall is U = 0 at radial positionr = R. Additionally, the velocity profile will be symmetric (or equivalently, there is no shear onthe centerline), so the gradient should be zero at r = 0, or dU/dy|r=0 = 0.

Using the equation and boundary conditions, we can start rearranging:

µ

r

d

dr

(rdU

dr

)=dP

dx

d

dr

(rdU

dr

)=r

µ

dP

dx

Now, we had just done some math and shown that the pressure gradient itself is a constant, sowe can integrate both sides with respect to r:∫

d

dr

(rdU

dr

)dr =

∫r

µ

dP

dxdr

rdU

dr=

1

µ

dP

dx

∫r dr

rdU

dr=

1

µ

dP

dx

r2

2+ C1

dU

dr=

1

µ

dP

dx

r

2+C1

r

Applying the no-shear boundary condition, we find:

0 = 0 +C1

0

We can’t have a singularity in the flow! We know that the flow exists, so this means we have to



set C1 = 0 to keep the flow physical. Then, we can integrate the expression again:

dU

dr=

1

µ

dP

dx

r

2+��0

C1

r∫dU

drdr =

1

µ

dP

dx

∫r

2dr

U =1

µ

dP

dx

r2

4+ C2

Now, setting U = 0 at r = R as our second boundary condition, we can find C2:

0 =1

µ

dP

dx

R2

4+ C2

− 1

µ

dP

dx

R2

4=C2

With a little rearranging, we now have an expression for our velocity profile in laminar pipeflow:

U(r) =1

4µ

dP

dx

(r2 −R2

)(8.4)

If this expression doesn’t quite look like what you have in your book, that’s OK: it’s justbased off how we defined our boundary conditions. Also, it looks like the velocity will benegative, since we have r2 − R2 for all r ≤ R. But remember what we said about the sign ofthe pressure gradient. The pressure falls as we move down the pipe, as it’s the pressure forceovercoming the viscous force. Therefore, in addition to knowing that the pressure gradient is aconstant, we can write:

dP

dx= −K

We do this simply because there will be a given pressure gradient. We know it’s negative, andwe know it’s constant, so defining it as −K just makes the following expression easy to look at.Then we write the velocity profile as:

U(r) =K

4µ

(R2 − r2

)Notice how we can see this is a parabola that has a maximum at r = 0 and is zero at r = R.We can re-cast this in terms of the pipe diameter by substituting R = D/2 to get:

U(r) =K

4µ

(R2 − r2

)U(r) =

K

4µ

((D2

)2 − r2

)

U(r) =KD2

16µ

(1−

(2r

D

)2)Overall, we see that the velocity profile in a laminar pipe flow is parabolic!



8.3 The friction factor in a laminar pipe

The skin friction is a non-dimensional parameter that we had defined in the boundary layer:

cf =τw

12ρU

2s

Where τw is the local wall shear stress, and Us is a scaling velocity. In the boundary layer, weuse U∞, but in a pipe we will have to think about what the appropriate velocity scale is. Now,this isn’t exactly like the friction factor f , which is defined as:

f =∆PL D

12ρU

2s

Let’s turn to the laminar pipe flow equation and see why f and cf are a bit different:

dP

dx=µ

r

d

dr

(rdU

dr

)integrating, rearranging, evaluating at wall

R

2

dP

dx= µ

dU

dr

∣∣∣∣r=R

= τw

We see that the wall shear stress is balanced by the pressure gradient times a length scale in theradial direction. If we plug this relation for τw into our expression for cf , we find:

cf =τw

12ρU

2s

=R2dPdx

12ρU

2s

Now, we know that the pressure gradient is a constant, and is negative in magnitude. Therefore,we can define this pressure gradient as some pressure drop over a length:

dP

dx=

∆P

L

Plugging in, we find:

cf =R2

∆PL

12ρU

2s

=1

4

∆PL D

12ρU

2s

=f

4

So the friction factor f is just four times the skin friction coefficient cf . It doesn’t matter thatwe have this constant 4 involved, it’s just scaling the results up by a factor of 4. If we plottedcf and f on the same chart, we’d find they are the exact same shape. For pipes, we just use ffor historical purposes, so we’ll stick to that

Now, let’s see if we can find an actual relationship for the friciton factor f based off of flowparameters. We need to chose an appropriate velocity scale Us. Let’s pick an easy velocity to



measure, which would be the bulk velocity U , or area averaged velocity:

U =1

A

∫U(r) dA

=4

πD2

∫ D/2

0

KD2

16µ

(1−

(2r

D

)2)2πr dr

=K

2µ

∫ D/2

0

(1−

(2r

D

)2)r dr

=K

2µ

[r2

2−(r4

D2

)]D/20

=KD2

32µ

We can then plug this expression for our bulk velocity in to the friction factor expression:

f =∆PL D

12ρU

2

We know that ∆P/L is just the pressure gradient, and since the pressure drop is negative butwe want a positive value, we plug in K for it. Also, rather than plug in U2, let’s plug in one Uand leave the second as a variable. We’ll see why:

f =KD

12ρU

KD2

32µ

=1

ρUD64µ

You can see that leaving one of the U has allowed us to get a familiar non-dimensional number!So for laminar pipe flow, we find the friction factor f is:

f =64

ReD(8.5)

The friction factor is inversely related to the Reynolds number. That means for increasing ReD,the friction factor decreases! Does that make sense? Let’s think - friction factor falls as f ∼ 1/U ,but it is related to drag by FD ∼ f ∗ U2. Therefore, even though the friction factor is falling,the total drag is increasing. That makes more sense, so that’s good to see! Remember thatthis relationship is only good for laminar pipe flow, so ReD < 2300 or so. If you are in theturbulent regime, you’ll have to use a table, chart, or empirical formula instead.

8.4 Example: a vertical pipe

Water of density ρ and viscosity µ flows steadily down a vertical circular pipe of diameter D.The flow is fully developed and has a parabolic velocity profile described by:

U

Ucl= 1−

(2r

D

)2



where Ucl is the centerline velocity. Find the wall shear stress and express the kinematic viscos-ity ν in terms of flow parameters.

Figure 8.2: fully developed gravity-driven flow

The flow here is a fully developed pipe flow, but we’re driven by gravity. Rather than have apressure gradient in the z, we will have the gravitational force. Let’s think on why - if the flowwere stationary, we’d have a pressure gradient balanced by gravitational forces. This is exactlythe hydrostatic equation!

no flow:dP

dz= ρg

Notice that there isn’t a minus sign. Look at figure 8.2 and think about what direction the z isdefined as positive, and what direction g is pointing. Therefore, with the pressure increasing inthe direction of gravity, the pressure gradient is positive. Now, by allowing flow to occur, thefrictional forces will oppose the motion. However, we see that the pressure gradient is just setby the gravitational forces, so we can instead say our z-momentum equation is:

0 =µ

r

d

dr

(rdU

dr

)+ ρg

So the gravitational term has replaced our pressure gradient as the “flow driver.” Also, we haveU in the z-direction. Yeah, that’s a pretty sloppy move on our part since it should really be W



in the z-direction, but let’s just go with it for now. Now, integrate and solve for the shear stress:

−ρgr = µd

dr

(rdU

dr

)−ρg r

2

2

∣∣∣∣r=D/2

= µrdU

dr

∣∣∣∣r=D/2

−ρgD2

8= µ

D

2

dU

dr

∣∣∣∣w

−ρgD4

= µdU

dr

∣∣∣∣w

We then find that the wall shear stress τw is:

τw = −ρgD4

Note that the wall shear is a negative value. That’s because it is acting in the negative z-direction to oppose the flow in the positive z-direction.

Now we want to solve for the kinematic viscosity, ν = µ/ρ. We’re given the shape of thevelocity profile, so we can plug that in for our expression of the wall shear stress:

τw = −ρgD4

µdU

dr

∣∣∣∣w

= −ρgD4

µd

dr

(Ucl ∗

[1−

(2r

D

)2])∣∣∣∣w

= −ρgD4

µUcl

(− 8

D2r

)∣∣∣∣r=D/2

= −ρgD4

µUcl

(− 4

D

)= −ρgD

4

Now, we can simply re-arrange to get the kinematic velocity:

ν =gD2

16Ucl

Therefore, if we have a vertical pipe that we know is in steady laminar flow, but we don’t knowthe fluid viscosity, we can extract its value by measuring the bulk velocity! This is just one wayto measure the viscosity of a fluid.



8.5 Example: force to hold a pipe

On a very long pipeline, large sections are held in place by a series of flanges. If we assumethese flanges act independently from one another, what would be the force a flange exerts on asection of pipe that is 0.5m in diameter, 100m long, carrying crude oil at a velocity of 1m/s?

Crude oil has a wide range of viscosity and density values, so let’s just use an average valueof ν = 5× 10−6m2

s and ρ = 900 kgm3 . We then calculate the Reynolds number for this flow:

ReD =UD

ν=

1 ∗ 0.5

0.000005= 105

So we are most definitely in the turbulent regime. To figure out the force exerted, we turn tothe friction factor, f , which we can look up in a chart. Assuming the pipe is smooth, we findfor ReD = 105 that f = 0.018. We then use the definition of the friction factor to extract thepressure drop:

f =∆PL D

12ρU

2s

=dPdxD

12ρU

2s

We also know from our previous math with the momentum equation that the pressure drop isrelated to the shear stress by:

τw =D

4

dP

dxTherefore, we have:

f =4τw

12ρU

2s

0.018 =8τw

900 ∗ 12

τw = 2.025Pa

To find the total force, we need to integrate over the area of the pipe this acts on. That’s theinner cylindrical surface area A = πDL. Since the flow is fully developed, the velocity profileisn’t changing, and thus the shear force is a constant! So we simply multiply the shear over thearea to find:

FD =

∫∫τw dA

FD =τw ∗AFD =2.025Pa ∗ π ∗ 0.5m ∗ 100m

FD = 318N

This doesn’t seem to be too large of a value for the large amount of oil the pipe is carrying.However, let’s think on what would happen if we tried to have the same volume flow rate througha smaller pipe. Let’s start by calculating the volume flux (since density is constant, we don’tworry about it):

V = UA = 1m

s∗ π0.52m2

4= 0.196

m3

s



If you were to try and pass this much fluid through a pipe with half the diameter, we wouldhave a higher average velocity:

V = 0.196 = U2 ∗A2 = U2 ∗ π0.252m2

4→ U2 = 4

m

s

With four times the velocity but half the diameter, the Reynolds number will be twice as high.The friction factor for ReD = 2× 105 is f = 0.016. Then, we can find the force needed to exerton 100m of pipe at this velocity:

f =4τw

12ρU

2s

0.016 =8τw

900 ∗ 42

τw = 28.8Pa

And once again, multiplying by the internal area over which this shear occurs, we find:

FD =28.8Pa ∗ π ∗ 0.25m ∗ 100m

FD = 2262N

This is why pipelines use very large pipe diameters - it considerably reduces the total force theymust overcome for a given flow rate for a nominal increase in construction costs!

8.6 Losses in pipe flow

We’ve used Bernoulli’s equation to solve many flow problems before. It’s a powerful tool, solong as you can satisfy the assumptions necessary for it to be valid. However, we can still usea type of Bernoulli’s equation that is modified to account for losses in the flow that occur dueto a departure from our initial assumptions. Let’s start by looking at Bernoulli’s equation in ahorizontal pipe:

P

ρ+ gz +

1

2V 2 = H

Note that we’ve divided by ρ to simply re-cast the equation in a slightly different form, and His our constant. If we were to apply this at two points in a horizontal pipe flow that is fullydeveloped, we’d find:

P1

ρ+��gz1 +

1

2V 2

1 =P2

ρ+��gz2 +

1

2V 2

2

P1 − P2

ρ=

1

2

(V 2

2 − V 21

)In fully developed pipe flow, the velocity isn’t changing downstream, so we would get:

P1 − P2

ρ= 0



This indicates that the pressure in the pipe is constant, but we know better! This is an exampleof why Bernoulli is restrictive in its use - it cannot account for the viscous effects.

We can instead modify Bernoulli’s equation to account for these losses. If you recall fromChapter 3, we did a rigorous derivation for Bernoulli’s equation starting from the Navier-Stokesequation. Through some vector identities and assuming a steady flow and constant density, wegot:

∇[

1

2U2 +

P

ρ+ gz

]= ~U ×

(∇× ~U

)+

1

ρ∇ · T v

The next step for Bernoulli’s equation is to assume irrotational and inviscid flow, then the RHSis zero, implying the bracketed term is a constant. Instead, we know there are viscous effectsand geometric losses (like valves, orifices, elbows, etc.) that can affect the flow. We want to tryand account for those losses.

Now, we usually utilize this as a quasi-1-D type equation. In a pipe, the velocity is mostcertainly not constant. However, we usually use the area averaged velocity, such as V . Whenusing this averaged value, we utilize what is called a kinetic energy coefficient, α, which isdefined as:

α =1

AV 3

∫V 3 dA

This comes from the integrated flux of kinetic energy. Overall, this coefficient α is a way toaccount for the different type of flow. In laminar flow, α = 2, while in turbulent pipe flow wecan approximate α ≈ 1. Now, using this coefficient and the bulk velocity (or area averagedvelocity), we can write our modified Bernoulli equation as:

∇[

1

2αV 2 +

P

ρ+ gz

]= ~U ×

(∇× ~U

)+

1

ρ∇ · T v

Looking at the first term on the RHS, that has a rotational component. If we were to look atthis function along a streamline, then we would see the first term on the RHS is zero (because~U ×

(∇× ~U

)is perpendicular to the flow).

∇[

1

2αV 2 +

P

ρ+ gz

]· dl =

��

��:0

~U ×(∇× ~U

)· dl +

1

ρ

(∇ · T v

)· dl

The remaining term on the RHS is the viscous contribution. We will say this can be representedby some sort of parameter to characterize the losses due to viscous effects. For now, let’s not tryto simplify it. Instead, we integrate both sides. Recall from chapter 3 that the LHS turns intoa perfect differential when you take the dot product along a streamline. Therefore, integrationbetween two points along a streamline will give us:

∫ 2

1d

[1

2αV 2 +

P

ρ+ gz

]=

∫ 2

1

1

ρ

(∇ · T v

)· dl[

1

2αV 2 +

P

ρ+ gz

]2

1

=

∫ 2

1

1

ρ

(∇ · T v

)· dl



Now our LHS is back to the original Bernoulli expression: the difference in the Bernoulli constantbetween two points! However, we have a RHS that is not zero, so long as there are viscouscontributions. We can expect that this value is negative, as it is a viscous force that is workingto reduce the momentum in the flow. Let’s simplify the expression by using the following:∫ 2

1

1

ρ

(∇ · T v

)· dl = −g ∗ hl

Why do we chose to represent this loss as −g∗hl? Look at the units on the LHS of our Bernoulliexpression - they are all units of m2/s2. We could divide this equation through by g to get allterms in the dimension of some length scale. This is often referred to some sort of head, suchas pressure head, or velocity head. Think about a pipe flow with liquid manometers on top toshow the pressure as a column of water - that’s the “head” at that point. Now (taking noticeof the minus sign) we have the following expression:(

1

2αV 2

1 +P1

ρ+ gz1

)−(

1

2αV 2

2 +P2

ρ+ gz2

)= g ∗ hl (8.6)

The two bracketed terms on the LHS are just like the original Bernoulli equation, and theright hand side is our losses in the flow. Now let’s think about what happens in a horizontalpipe flow. Before, the classic Bernoulli equation found that the pressure would remain the same,but here we see something different. The elevation is the same, so gz cancels in both brackets.The flow is fully developed, so the velocity terms cancel. You are then left with:

P1 − P2 = ρ ∗ g ∗ hl

This is saying that there is a pressure difference due to losses in the system! That’s what we knowhappens! The actual values of the head loss coefficients can be found from tables. There aredifferent expressions for different types of losses. Major losses are characterized by the viscousdrag. Minor losses are characterized by fittings, valves, changes in diameter, etc. The equationswe use are defined as:

major losses: hl = fL

D

V 2

2g

minor losses (a): hl = KV 2

2g

minor losses (b): hl = fLeD

V 2

2g

The values of K and Le can be looked up in the a table for specific flow situations, and f is ourfriction factor. We can find that in a table, or in laminar flow we can calculate it. We’ll usethese in an example to get an idea of what is going on.

This equation still holds for turbulent flow! That’s because we derived this from the Navier-Stokes equations, which also hold for turbulent flows. The factors that change will be α, whichyou set as 1 in turbulent flows, and f , which you can look up for different ReD.



8.7 Example: flow from a reservoir

Lets imagine we have a large reservoir of fluid that empties through a pipe, as pictured in thefigure below. The entrance can be approximated as a square-edged entrance. The pipe material

is galvanized iron, and the fluid is water. If the flow rate is 0.05m3

s , what is the height of thereservoir?

We start by first recognizing the types of losses that can occur. First, the entrance to thepipe has an abrupt square edge, so we have a minor loss there. Then, we have frictional lossesin the flow, which is a major loss. We need the friction factor, and have to account for thesurface roughness. The exit is an abrupt opening, so that also has a minor loss associated withit.

Turning to Bernoulli, let’s start with the balance from the top of the reservoir to the entranceof the pipe:

Patmρ

+ gH =Penterρ

+1

2αV 2

Here we’ve assumed the pipe entrance is the zero height reference point. We don’t have anylosses up to this point, so we’re set for this expression. Next, let’s balance from the pipe entranceto the pipe exit:

Penterρ

+1

2αV 2 =

Patmρ

+1

2αV 2 − g(200) + g ∗ hl

We have the losses we need to account for, so we include the g ∗ hl term. Note that the velocityis assumed the same at the entrance and exit, since it’s some sort of average velocity. Thatmeans the balance on this second equation can cancel out velocities, but we won’t do that yet.Instead, let’s substitute this into our first expression to get:

Patmρ

+ gH =Patmρ

+1

2αV 2 − g(200) + g ∗ hl

1

2αV 2 = g(200 +H − hl) (8.8)



If we would have done the balance between the surface of the reservoir and the exit of the pipe,we would have obtained this expression from the start! Bernoulli is still working for us, so longas we account for the losses.

Before we even do any math, we can see something intriguing by this expression. Just solvingfor velocity (and ignoring α for now) gives the expression:

V =√

2g(H + 200− hl)

When ignoring the head loss term, this is exactly the same expression Bernoulli’s equation givesfor the velocity at the exit of a tank of height H + 200. The losses in the system, which areaccounted for in the hl term, will work to reduce this exit velocity. That should hopefully makesense, so we are still on the right track!

Now, let’s figure out the velocity. We have a flow rate and pipe diameter, so we can get V :

V =Q

A=

0.05m3

s

π 0.12

4 m2= 6.4

m

s

Then with this, we find the Reynolds number in the pipe is:

ReD =6.4ms ∗ 0.1m

10−6m2

s

= 640000

This is most certainly turbulent! First, this means the kinetic energy coefficient α is just one.Then, using this Reynolds number, we can find the friction factor for the pipe. Recall that it’sgalvanized iron, which means it has a surface roughness of k = 0.15mm, meaning k/D = 0.00075.Then, looking up on a Moody plot for this roughness and Reynolds number, we can find thefriction factor f ≈ 0.019.

We next look to the entrance condition, which we said was square edged. This gives aminor loss coefficient of Kentry = 0.5. Finally, there is an abrupt exit, which gives a minor losscoefficient of Kexit = 1. We can now put together our relationship for equation 8.8:

1

2αV 2 = g(200 +H − hl)

1

2αV 2 = g(200 +H − hmajor − hentry − hexit)

1

2(α)V 2 = g

(200 +H − f L

D

V 2

2g−Kentry

V 2

2g−Kexit

V 2

2g

)1

2(1)(6.4)2 = (9.8)

(200 +H − 0.019

2000

0.1

(6.4)2

2 ∗ 9.8− 0.5

(6.4)2

2 ∗ 9.8− 1

(6.4)2

2 ∗ 9.8

)

Solving for H, we then find:H = 599m

This is telling us that the losses in the system require the tank height to be 599m to get thegiven flow rate through the pipe. If it were ideal, this height would result in a bulk pipe velocityof 125m/s, which most certainly higher than the 6.4m/s we have here.


Chapter 9

Supercritical and Supersonic Flows

9.1 Comparison of Mach number and Froude number

We group the discussion of water waves and compressible flow together due to their similarities.The striking analogy that can be made deals with the two non-dimensional numbers we formfor the respective flows. These two numbers are the Mach number M and Froude numberFr:

M =U

aand Fr =

U√gy

These two numbers are a ratio of the flow velocity U and a characteristic velocity. In compressibleflow, we use the speed of sound a, and in shallow water waves we use the wave speed

√gy, where

g is gravity, and y is the water depth. Another way to think about the Froude number is tosquare it and re-arrange:

Fr2 =U2

gy= 2 ·

12ρU

2

ρgy

This shows that the Froude number is a ratio of kinetic energy to potential energy.The similarity between the two non-dimensional numbers can be shown in a simple schematic.

In figure 9.1, we illustrate the propagation of a disturbance in a flow field. In this case, thereis some disturbance (the black dot) which is subjected to a flow of speed U , resulting in therespective Froude number. From the point of the disturbance, waves will propagate outwardat the wave speed,

√gy. However, they are also convected along at the flow speed. When

there is no flow we have Fr = 0, and waves propagate outward equally. When 0 < Fr < 1,then the waves are convected downstream at a velocity which is slower than their wave speed.The wavefront then moves more quickly downstream, but more slowly upstream. Think onthe doppler effect for this one. When Fr > 1, then the waves cannot move upstream, sincethey are convected away at a speed greater than their traveling speed, so a “cone” is formedthat “knows” about the disturbance. Upstream of the cone, no information is known about thedisturbance. Downstream, the wake is the only region that has experienced or been influencedby the disturbance.

The angle of this cone can be calculated by relating the distance traveled by the flow vs. thedistance the disturbance travels. This simplifies down to:

sinα =1

Fr


CHAPTER 9. SUPERCRITICAL AND SUPERSONIC FLOWS

Figure 9.1: shallow water waves from a disturbance in different flow speeds

Now, if we were to do the same analysis with sound waves, we would have the schematic as seenin figure 9.2. Notice that this is exactly the same setup as was with the shallow water waves!

Figure 9.2: sound waves from a disturbance in different flow speeds

The primary difference here is the wavefront buildup in the sound wave case leads to shocks,where in the shallow water waves we get a breaking wave.

9.2 Hydraulic jumps

A hydraulic jump occurs when a high velocity fluid undergoes a rapid deceleration and suddenrise in depth. This often occurs due to a downstream condition that requires the flow to slowdown. One way you can see this in every day life is in a sink. When you turn the faucet on, thewater comes out the spout and hits the bottom of the sink. From there, it rapidly spreads in a



radial fashion, but at some distance it will undergo a jump. This is exactly a hydraulic jump!The downstream condition in this case is likely an interaction of unstable flow with surfaceroughness.

Utilizing the continuity and momentum equations, you can extract an expression for a hy-draulic jump. Figure 9.3 illustrates the simplified flow scenario. The math to find the ratio ofheights isn’t too terribly complicated, but the derivation is a little tedious so we’ll just utilizethe result:

h1

h0=

1

2

[√1 + 8Fr2 − 1

]Note here that Fr is the Froude number of the incoming flow. We have three options for Fr:

1. Fr = 1 → h1 = h0, no change in height

2. Fr > 1 → h1 > h0, a jump in height

3. Fr < 1 → h1 < h0, a drop in height

Figure 9.3: schematic for a hydraulic jump (thanks to wikipedia for the image)

The third option of a hydraulic drop turns out to be unphysical. To understand why, comparethe Bernoulli constants for the two flows. The jump is a turbulent mess of flow, so we wouldexpect losses to occur across it. Along the surface, we would have the same pressure upstreamand downstream, so it’s just a comparison of 1

2ρU2 and ρgh on each side of the jump. Once

again, an expression can be built, and it is once again a tedious process. The math is done inLex Smits’ book, but the result shows that if a hydraulic drop were to occur, there would bemore energy after the drop than before. Since losses occur through the drop, this turns out tobe non-physical. Therefore, we can only have hydraulic jumps, which occur for Fr > 1.

This once again turns out to be analogous to compressible flow, where a hydraulic jump islike a shock. Shocks are regions of intense change in fluid properties that occur when M > 1.Across a shock, the flow will go from supersonic to subsonic, and there are losses that can beaccounted for by an increase in entropy. Do you think there could be shocks where a flow goesfrom subsonic to supersonic? No! that would be an incredible phenomenon to see in the firstplace, but more importantly that process would result in a net decrease in entropy. Since thiscannot happen, then subsonic shocks do not occur. This is quite similar to sub-critical flow(Fr < 1) being unable to experience a hydraulic drop.



9.2.1 Example: hydraulic jumps on a dam

One example of the importance of hydraulic jumps lies in the creation of dams. Many damshave some sort of spillway for when the water level is too high to maintain. An example can beseen in figure 9.4. As the water flows down the spillway, it’s velocity increases, and the heightof the water level narrows due to continuity. This means the Froude number is increasing fromzero at the top, until it goes supercritical (Fr > 1). At some point, a hydraulic jump will occur.

The issue with jumps is the intense mixing that occur in the jump itself. This region is highlyturbulent, promoting mixing and large gradients in the velocity field. Imagine if this mixingwere to occur over the riverbed - it would work to quickly significantly erode the material it isover. This would end up compromising the structural stability of the dam/riverbed system.

Figure 9.4: picture of a dam spillway with a hydraulic jump (thanks wikipedia)

Due to these concerns, engineers go out of their way to assure the hydraulic jump occurson the spillway itself. Since it is made of concrete, it will be more structurally sound than theearthen floor. There are a few ways to assure this jump occurs early enough - either make thespillway long enough, or putting flow disturbance devices (like baffles, sills, blocks, etc.) towardsthe end of the spillway to assure a jump occurs.

9.2.2 Example: jump in a channel

Water is flowing in a rectangular channel at a depth of 30cm with a velocity of 15m/s. Ifdownstream conditions force a hydraulic jump to occur, what will be the depth and velocity



downstream? What is the head loss across the jump?

We start by using the relation for a hydraulic jump. For this, we need the Froude number:

Fr =U√gy

=15m/s√

0.3m ∗ 9.8m/s2= 8.75

This flow is most certainly supercritical. Therefore, we can use the Froude number and theincoming height to calculate the height of the flow after the hydraulic jump:

h1

h0=

1

2

[√1 + 8Fr2 − 1

]→ h1 =

0.3m

2

[√1 + 8 ∗ 8.752 − 1

]h1 = 3.56m

The height of the water increased by (around) tenfold! We then can calculate the downstreamvelocity by utilizing continuity. We know water is pretty much incompressible, so we have:

mout =min

ρUoutAout =ρUinAin

Uout ∗ 3.56m ∗ w =15m/s ∗ 0.3m ∗ wUout =1.26m/s

The velocity dropped by (around) tenfold. That should have been expected from the result withthe height increase. Checking the Froude number, we see that Fr2 = 1.26/

√9.81 ∗ 3.56 = 0.21,

or subcritical. Now, to calculate the head loss, we can use the modified form of Bernoulli’sequation from the previous chapter:(

1

2αV 2

1 +P1

ρ+ gz1

)−(

1

2αV 2

2 +P2

ρ+ gz2

)= g ∗ hl

If we look along the surface of the water, we can neglect the pressure contribution since it’s justatmospheric pressure at both locations. Then, assuming the coefficient α = 1, we can plug inthe velocity and heights to get:

hl =

(1

2gαV 2

1 + z1

)−(

1

2gαV 2

2 + z2

)hl =

152

2 ∗ 9.8+ 0.3− 1.262

2 ∗ 9.8− 3.56

hl = 8.13m

This is stating that there was an “energy loss” of 8.13m of the working fluid, which in thiscase is water. Think about the pressure exerted by water - we know that 10m is approximatelyone atmosphere of pressure. That’s a lot of lost potential working energy in the jump!



9.3 Compressible flow relations

The behavior of compressible flow is unique and quite unlike what we’ve seen in undergradfluids. We’ve for the most part treated any flow as incompressible, and even further we’ve saidthat density, ρ, is constant. We’re now entering a regime where density is most certainly notconstant. We’ll find a number of relations for the flow, but first let’s tie this back to the shallowwater waves.

Recall figure 9.2, where we saw that flows with M > 1 result in a shockwave. We can finda relationship for the angle α of the wave by relating the distance traveled for the differentregimes:

sinα =distance traveled by wave

distance traveled by flow=a ·∆tU ·∆t

=a

U→ 1

M= sinα

Note that this is exactly analogous to the angle made with the shallow water waves. Also notewhat happens with M < 1: you cannot get a solution. This means that the soundwaves do notbuild up to a shock unless M > 1, as you probably would have expected.

There are many relations for compressible flow, and it involves a lot of thermo with fluids.We will once again skip over a ton of these derivations and instead just start with the resultsthemselves. If you wish, you can find the actual math in any compressible fluids book, includingSmits’ book.

We’ll start with a few basic relations. For a perfect gas (which is an ideal gas with constantspecific heats cp and cv) with a temperature T , density ρ, pressure P , and heat capacity ratioγ = cp/cv, we have:

P = c · ργ andρ0

ρ=

(T0

T

) 1γ−1

andP0

P=

(T0

T

) γγ−1

In these expressions, c is a constant, and the subscript 0 refers to a reference condition, whichwe will choose as the stagnation quantities. In other words these expressions show how thedensity or pressure in a compressible flow relate to the stagnation quantities of that flow. Forexample, if you were to smoothly decelerate a flow to rest (and thus stagnate it), then you wouldrecover ρ0, P0, and T0.

Additionally, the speed of sound is defined as the square root of the change in pressure withrespect to density at constant entropy:

a2 =∂P

∂ρ

∣∣∣∣s=const

We can then use our above relationship for the perfect gas (and ideal gas) to get an expression



for the speed of sound.

a2 =∂P

∂ρ

∣∣∣∣s=const

a2 =∂

∂ρ

(c · ργ

)a2 =cγργ−1 = cγ

ργ

ρ

a2 =cγP/c

ρ= γ

P

ρ

a2 =γRT

a =√γRT

We can then plug in some values to see what the speed of sound is! At sea level, T = 293K,γair = 1.4, and Rair = 287 J

kgK , giving us a = 343m/s. If you’re in the upper atmosphere, thingscan be quite cold, so T = 100K then gives a = 200m/s. Does this make sense? As you getcloser to space, the temperature drops, but we also have less and less atmosphere. Sound issimply pressure waves, which is a molecular transmission of kinetic energy. If the temperature isvery low, then there is less kinetic energy in the molecules, and thus pressure waves move moreslowly!

Another important relationship that can be found for compressible flow relates the temper-ature to the Mach number:

T0

T= 1 +

γ − 1

2M2

9.3.1 Example: space shuttle re-entry

This is a simplistic way to see why the space shuttle (and other re-entry vehicles) have heatshielding on them. If the shuttle enters the atmosphere at M = 15, where the temperature isT = 100K, approximately how hot does the shuttle get?

If we assume γ = 1.4 for the upper most reaches of the atmosphere (which it probably doesn’tbut let’s not worry about it), then we can calculate:

T0

T= 1 +

γ − 1

2M2

We pick the temperature of the flow to be T = 100K, and we solve for the stagnation temper-ature. Why isn’t the stagnation temperature 100K if the atmosphere is at rest? We must bemindful of the reference frame! The shuttle sees this fluid coming towards it at M = 15, whichthen will stagnate on the surface of the shuttle. Of course there is a shock, but let’s forget aboutthat at this time. We then find:

T0

100K= 1 +

1.4− 1

2152 → T ≈ 4500K

That’s pretty hot. It’s an over-approximation since the sparse gas turns into plasma, but it helpsexplain why there are special ceramic tiles on the shuttle. This also illustrates the explanation



of a common misconception about the shuttle - it’s the compressibility of the air that causes theheating, not friction!

9.4 Normal shocks

Much like hydraulic jumps, shocks can occur when a supersonic flow encounters a downstreamcondition or obstruction that requires it to decelerate. There are different types of shocks, butwe will concern ourselves with normal shocks for now. The name implies the shock will beoriented normal to the flow direction. This means shockwaves from a plane moving at M > 1,which are usually angled, are not normal shocks. Once again, there are many different relations

Figure 9.5: bow shock off a blunt object. The area near the centerline can be thought of as anormal shock, while the outer edges are oblique shocks (thanks NASA)

and derivations for normal shocks. It involves more thermo, usually using the energy equationand the second law. We don’t want to concern ourselves with the derivations, so instead we willlook into the results!

One important relation is the temperature across the shock:

T2

T1=

1 + γ−12 M2

1

1 + γ−12 M2

2

Notice that this is just the ratio of the previous temperature relation we had used? That’s



because the stagnation temperature across a shock is constant. This again is a result of somefancy work with equations.

Another important relation would be the relation of the Mach numbers across a shock. Forsome upstream Mach number M1, the downstream Mach number M2 is found by:

M22 =

M21 + 2

γ−12γγ−1M

21 − 1

This relationship shows that for any given M1 > 1, the downstream M2 is going to be less than1. If M1 < 1, then we would get M2 > 1. Does that make sense? No! This would violate thesecond law of thermodynamics, and can be shown mathematically. Another relationship we’lluse is the strength of the shock, which is a ratio of pressures across the shock:

P2

P1= 1 +

2γ

γ + 1

(M2

1 − 1)

We can see that the higher the Mach number of the incoming flow, the stronger the shock. AtM1 = 1, the pressure ratio is zero, indicating no shock occurs.

We can utilize these different relations to calculate other properties, such as density ratios,by using the ideal gas law.

9.4.1 Example: revisiting the shuttle

Because of the high Mach number of the shuttle at re-entry, we should expect a shock to form.We see that a blunt object, such as shown in figure 9.5, will create a normal shock in front ofit. Let’s calculate the change in air properties across this shock.

From the previous problem, we assumed T1 = 100K, M1 = 15. Let’s also assume the pres-sure in this upper atmosphere is 100Pa, and the adiabatic gas constant is γ = 1.4. What’s themach number, temperature, and pressure after the shock (but before it stagnates on the surface)?

We can start by calculating the Mach number after the normal shock:

M22 =

M21 + 2

γ−12γγ−1M

21 − 1

=152 + 2

1.4−12∗1.41.4−1 ∗ 152 − 1

M2 = 0.38

The Mach number after the shock is less than 1. This is much like the hydraulic jump, wherethe Froude number after a jump is less than 1. Then, using this Mach number, we can calculatethe temperature of the flow after the shock:

T2

T1=

1 + γ−12 M2

1

1 + γ−12 M2

2

→ T2 = 100K ∗1 + 1.4−1

2 ∗ 152

1 + 1.4−12 ∗ 0.382

T2 ≈ 4500K



This is the same result we had found in the stagnation case! Think about why - the Mach numberof the flow after the shock is 0.38, which isn’t all too fast. The temperature ratio compared tostagnation temperature due to that mach number is T0/T = 1.02, indicating that it will onlyheat up by 2% from post-shock to stagnation.

Finally, we can calculate the pressure after the shock:

P2

P1= 1 +

2γ

γ + 1

(M2

1 − 1)→ P2 = 100Pa ∗

(1 +

2 ∗ 1.4

1.4 + 1

(152 − 1

))

P2 = 26kPa

The pressure rises an absurd amount! This is well over a 100 fold increase. Imagine then if youwere to try and have a hypersonic vehicle near sea level, where Patm = 100kPa. Even movingat M = 5 gives a pressure after a normal shock of P ≈ 2.9MPa, which would result in a crazyamount of drag. This is why supersonic vehicles are slender and fly in the upper atmosphere.

Hopefully this gives you an idea of some of the things that need to be taken into considerationwhen dealing with compressible flow!


an overview of undergrad fluids · this material was inspired by prof. marcus hultmark’s class...

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