wkintflo, practice problrems with soolution

Ghosh - 550 4/20/2023

Worked Out Examples(Internal Fluid Flows)

Example 1. (Poiseuille Flows)

A fluid flows steadily between two parallel plates. The flow is fully developed and laminar. The distance between the plates is h.(a) Derive an equation for the shear stress as a function of y. Plot this function.(b) For = 2.4 10-5 lbfs/ft2, p/x = -4.0 lbf/ft2/ft, and h = 0.05 in., calculate the maximum

shear stress, in lbf/ft2.

1. Statement of the Problema) Given

Laminar, fully developed flow between two parallel plates Gap between the plates is h = 0.05 in. = 4.167 10-3 ft. = 2.4 10-5 lbfs/ft2

p/x = -4.0 lbf/ft2/ftb) Find

Expression for the shear stress as a function of y. Plot the shear stress. Calculate the maximum shear stress, in lbf/ft2.

2. System Diagram

3. Assumptions Steady state condition Constant fluid properties (density and viscosity) 2-D problem (xy plane) Laminar, fully developed flow

4. Governing Equations

2-D Incompressible Continuity Equation:

hx

y

Velocity Profile

yx

yx

xyxy

Ghosh - 550 4/20/2023

Navier-Stokes Equation:

Shear Stress (for xy plane):

5. Detailed Solution

Shear stress as a function of y

u = u(y) and v 0 because the fluid flows between two parallel plates.Considering this fact, the continuity equation becomes:

This tells that the flow is fully developed, which is given in the problem and assumed to be, because the velocity doesn't change in the x direction. Therefore, the velocity u is function of only y, u = u(y).

Navier-Stokes equation in y direction is useless on this problem because v 0.Navier-Stokes equation in x direction :

Therefore,

because u = u(y).

This problem deals with the flow between two parallel plates, so the maximum velocity

occurs at y = 0 (centerline). This implies due to the symmetry. Considering

this fact, Const on the above expression becomes 0.

00

x

u

y

v

x

u

0

2

2

2

2

y

u

x

u

x

pg

y

uv

x

uu

t

ux

Steady state

Continuity equation

v 0

gx = 0

Continuity equation

Ghosh - 550 4/20/2023

Now, we have:

Shear stress is defined as .

From the problem geometry, we are interested in yx. Since v 0, the shear stress becomes:

because u = u(y).

We already have this expression above; therefore, the shear stress is:

Plot the shear stress

Using MatLab, the shear stress distribution looks like:

-0.01 -0.008 -0.006 -0.004 -0.002 0 0.002 0.004 0.006 0.008 0.01-2.5

-2

-1.5

-1

-0.5

0

0.5

1

1.5

2

2.5x 10

-3 Shear Stress Distribution

yx

(lbf/ft2)

y (ft

)

Upper Wall

Lower Wall

yx

= (p/x)y

Ghosh - 550 4/20/2023

Maximum shear stress

The maximum shear stress occurs at the wall where y = h/2.

= 0.008333 lbf/ft2

6. Critical AssessmentThe shear stress is negative for y > 0 (see the plot). This is correct because the way shear stress is defined. See the "System Diagram." The upper shear stress is positive toward right, but on the upper wall, the shear stress works toward left in reality (since the shear force must be positive and the upper wall is a negative surface). This is why it is negative.

[Note: Instead of making the fully developed flow assumption, if only the parallel flow assumption is made, the analysis will follow the same pattern since parallel flows are fully developed. Simplifications will be easier since u0, but v=w=0.]

Example 2. (Use of velocity profiles):

An incompressible fluid flows between two infinite stationary parallel plates. The velocity profile is given by u = umax(Ay2 + By + C), where A, B, and C are constants and y is measured from the center of the gap. The total gap width is h units. Use appropriate boundary conditions to express the magnitude and units of the constants in terms of h. Develop an expression for volume flow rate per unit depth and evaluate the ratio .


Velocity profile: u = umax(Ay2 + By + C)where A, B, and C are constants and y is measured from the center of the gap.

Total gap width is h units.b) Find

Magnitude and units of the constants in terms of h. Volume flow rate per unit depth. Ratio .

2. System Diagram

Ghosh - 550 4/20/2023

3. Assumptions Steady state condition Incompressible fluid flow


Volume Flow Rate Definition:

For a flow between two infinite stationary plates,

where b is the depth in z direction.

Average Velocity Definition


It is obvious to tell the units of constants (A, B, and C) by observing the given velocity profile function: u = umax(Ay2 + By + C). Note that each term inside the parenthesis has to have no units. Therefore the units on A, B and C are as follows.

UnitsA m-2

B m-1

C 1Note: This will be verified after the constants (A, B, and C) are evaluated in terms of h.

Evaluation of Constants

Realizing the fact that the maximum velocity occurs at y = 0 in this problem condition and geometry, the constant C can be evaluated as follows:

u = umax(Ay2 + By + C)umax = umax[A(0)2 + B(0) + C]C = 1

h

y

x, u

u(y)

umax

Ghosh - 550 4/20/2023

Because the maximum velocity occurs at y = 0, this relation must be satisfied:

since shear stress is zero on the centerline (or, line of symmetry).

Velocity must be 0 (non-slip condition) at the wall where y = h/2. Thus,

Substituting B = 0 and C = 1 into this expression, .

Finally, the velocity profile becomes:

or

Volume Flow Rate per Unit Depth

Ratio

Ghosh - 550 4/20/2023

6. Critical AssessmentThe problem illustrates how to determine velocity profiles from boundary conditions and calculate volumetric flow rate and average velocity. An important point on this problem is to realize where the maximum velocity occurs and how the mathematical expression can be written for it.

Example 3. (Couette Flows)

A sealed journal bearing is formed from concentric cylinders. The inner and outer radii are 25 and 26 mm, the journal length is 100 mm, and it turns at 2800 rpm. The gap is filled with oil in laminar motion. The velocity profile is linear across the gap. The torque needed to turn the journal is 0.2 Nm. Calculate the viscosity of the oil. Will the torque increase or decrease with time? Why?


r1 = 25 mm, r2 = 26 mm, a = r2 - r1 = 1 mm L = 100 mm = 2800 rpm Oil in laminar motion Linear velocity profile T = 0.2 Nm

b) Find Viscosity of the oil Will the torque increase or decrease with time? Why?

2. System Diagram

a

r1

r2

U

uy

x

a

Ghosh - 550 4/20/2023

3. Assumptions Steady state condition Incompressible fluid flow Laminar flow Fully developed flow p/x = 0 (since the flow is symmetric in the actual bearing at no load) Torque on the journal is due to viscous shear in the oil film. The gap width is small, so

the flow may be modeled as flow between infinite parallel plates. Infinite width (L/a = 100 mm / 1 mm = 100, so this is a reasonable assumption) 2-D problem (L is in z direction, but the main analysis can be done in xy plane.)


2-D Incompressible Continuity Equation:

Navier-Stokes Equation:

Shear Stress (for xy plane):


Viscosity of the Oil

The problem gives the velocity profile to be linear, so we can simply say that the velocity profile is (See "Critical Assessment" for a detailed discussion of this velocity profile):

The shear stress is:

On this problem, we are interested in yx and v 0 .

…

Torque is given by …

Equating and gives: …

Therefore,

Ghosh - 550 4/20/2023

= 0.0695 Ns/m2

Will the torque increase or decrease with time? Why?

Torque will decrease with time.

Reason: This bearing is going to be heated up by the shear stress effect (temperature increases). Viscosity decreases as the temperature increases (check a plot or graph of viscosity as a function of temperature). If the viscosity decreases, then the torque decreases by the equation above.

6. Critical AssessmentThe problem gives the velocity profile as linear, so we simply said that the velocity profile is

But here is the detailed demonstration of how this linear velocity profile was obtained.

Linear Velocity Profile

u = u(y) and v 0 because the fluid flows between two parallel plates.Considering this fact, the continuity equation becomes:

This shows that the flow is fully developed, which is assumed to be, because the velocity doesn't change in the x direction. Therefore, the velocity u is function of y only, u = u(y).

Navier-Stokes equation in y direction is useless on this problem because v 0.Navier-Stokes equation in x direction :

00

x

u

y

v

x

u

0

2

2

2

2

y

u

x

u

x

pg

y

uv

x

uu

t

ux

Steady state

Continuity equation

v 0

gx = 0

Continuity equation

Ghosh - 550 4/20/2023

Therefore,

because u = u(y).

Integrating this equation twice gives:

Boundary Conditionsu = 0 at y = 0 C2 = 0

u = U at y = a

Finally,

because (See assumptions listed above).

(This result matches with the given velocity profile, linear.)

Example 4. (Special boundary conditions in channel flow): A continuous belt, passing upward through a chemical bath at speed U0, picks up a liquid film of thickness h, density , and viscosity . Gravity tends to make the liquid drain down, but the movement of the belt keeps the liquid from running off completely. Assume that the flow is fully developed and laminar with zero pressure gradient, and that the atmosphere produces no shear stress at the outer surface of the film. State clearly the boundary conditions to be satisfied by the velocity at y = 0 and y = h. Obtain an expression for the velocity profile.

Belt

y

x

Bath

g

p = patm

U0

dy

dx

h

Ghosh - 550 4/20/2023


Belt speed, U0

Liquid properties: h, , and Gravity exists Flow is fully developed, laminar, with zero pressure gradient (p/x = 0) Atmosphere produces no shear stress at the outer surface of the film

b) Find State the boundary conditions to be satisfied by the velocity at y = 0 and y = h. Obtain an expression for the velocity profile.

2. System Diagram

3. Assumptions Steady state condition Constant property fluid v 0

2-D problem (xy plane). That is, the z-depth is infinite

4. Governing Equations Navier-Stokes Equation

y

x

g

p = patm

U0

h

, xy

xy

yxyx

Ghosh - 550 4/20/2023

Shear Stress (for xy plane)


The problem gives the flow is fully developed. Thus,

. Also, . u(y) only.

(This relationship could be derived from the continuity equation.)

Navier-Stokes equation (2-D) in y direction is useless because v 0 and gy = 0.Navier-Stokes equation (2-D) in x direction :

Therefore,

since u(y) only.

Shear StressWe are interested in yx and v 0 in this problem. Thus,

Boundary Conditions@ y = 0, u = U0

@ y = h, because the atmosphere produces no shear stress at

the outer surface of the film.

Going back to the equation obtained from Navier-Stokes equation:

2

2

2

2

y

u

x

u

x

pg

y

uv

x

uu

t

ux

Steady state

Fully developed

v 0

Given

Fully developed

Ghosh - 550 4/20/2023

because gx = -g (check the directions of the gravity and the x coordinate).

Thus,

Integrating this equation once gives .

Using the second boundary condition, C1 can be evaluated as .

Now, the equation is .

Integrating this equation gives .

Using the first boundary condition, C2 can be evaluated as C2 = U0.

Therefore, the velocity profile is

or

6. Critical AssessmentHere, the velocity profile has been obtained by a direct application of Navier-Stokes equation. The same result can be obtained by applying the momentum equation for integral control volume:

Try to obtain the same result using this approach.

Also try simplifying the problem by assuming parallel flow instead of fully-developed flow.

Ghosh - 550 4/20/2023

Example 5. (Electrical Analogy):

Resistance to fluid flow can be defined by analogy to Ohm's law for electric current. Thus resistance to flow is given by the ratio of pressure drop (driving force) to volume flow rate (current). Show that resistance to laminar flow is given by

Resistance =

which is independent of flow rate. Find the maximum pressure drop for which this relation is valid for a tube 50 mm long with 0.25 mm inside diameter for both kerosine and caster oil at 40 C.


Resistance analogy between Ohm's law and fluid flow

Driving Force Current ResistanceOhm's Law V (voltage) I (current) R (resistance)Fluid Flow p (pressure drop) Q (volume flow rate) Rf (flow resistance)

L = 50 mm = 0.05 m D = 0.25 mm = 2.5 10-4 m Kerosine (at 40 C)

= 1.1 10-3 Ns/m2

= 1.35 10-6 m2/s Caster Oil (at 40 C)

S.G. = 969 = 2.4 10-1 Ns/m2

= 2.48 10-4 m2/sb) Find

Expression of flow resistance for laminar flow using the analogy to Ohm's law Maximum pressure drop in the pipe for both kerosine and castor oil as working fluids

2. System Diagram

Ohm's Law

V

I

RQ

p

Fluid Flow

Ghosh - 550 4/20/2023

3. Assumptions Steady state condition Constant fluid properties Laminar, fully developed flow


Velocity Profile for a Pipe Flow

Volume Flow Rate Definition


Flow Resistance

The problem gives the resistance to flow is .

Volume flow rate Q can be calculated as follows:

Because , the volume flow rate Q can be

written as:

Ghosh - 550 4/20/2023

Finally, the flow resistance is

This expression, flow resistance, is independent of the flow rate because there is no Q or included. It is consisted of only fluid property and the geometry of a pipe.

Maximum Pressure Drop

Examining this pressure drop expression, one can say that the faster the average velocity is, the bigger the pressure drop is; therefore, to obtain the maximum pressure drop, we want the fastest flow velocity in laminar flow constraint.

Consider the Reynolds number . This gives the allowable fastest flow

velocity in laminar flow constraint. Recr = 2300 for a pipe flow.

Substituting this fastest average flow velocity into the pressure drop expression, we get:

After plugging in values into this pressure drop expression, we obtain:

p for kerosine is 350 kPa.p for castor oil is 14.0 GPa.

6. Critical AssessmentNotice the big difference of pressure drops between kerosine and castor oil. This difference is directly from the viscosity and density of fluids, keeping other parameters constant.

Ghosh - 550 4/20/2023

Continue

wkintflo, practice problrems with soolution

Documents