velocity profiles in laminar boundary layers often are approximated...

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Worked Out Examples(External Flows)

Example 1 (Velocity Profiles):

Velocity profiles in laminar boundary layers often are approximated by the equations

Linear:

Sinusoidal:

Parabolic:

Compare the shapes of these velocity profiles by plotting y/ (on the ordinate) versus u/U (on the abscissa).

1. Statement of the Problema) Given

Three approximated velocity profiles in laminar boundary layers, linear, sinusoidal, and parabolic.

b) Find Compare these three approximated velocity profiles by plotting.

2. System DiagramIt is not necessary for this particular problem.

3. Assumptions Laminar boundary layer

4. Governing EquationsNone. Just plot them.

5. Detailed SolutionThere is no detailed discussion for this problem. Just plot and compare them.Using MatLab the plots look like:

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0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

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1Approximated velocity profiles in laminar boundary layer

u/U

y/

LinearSinusoidalParabolic

By common sense, it can be said that the linear profile is not very close approximation to the actual shape of boundary layer velocity profile. The plot for the parabolic profile is the closest approximation to the Blasius solution for velocity profile.

6. Critical AssessmentNote that the velocity profiles are only useful for 0<y<. Although the given equations hold outside this range, the curves have no meaning for y> in boundary layer theory.

Examp le 2. (Boundary Layer Thicknesses):

Velocity profiles in laminar boundary layers often are approximated by the equations

Linear:

Sinusoidal:

Parabolic:

Calculate * (displacement thickness) and (momentum thickness) for these velocity profiles and compare the result for each case.

7. Statement of the Problemc) Given

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Three approximated velocity profiles, linear, sinusoidal, and parabolic, in laminar boundary layers.

d) Find * (displacement thickness) for three approximated velocity profiles (momentum thickness) for three approximated velocity profiles Compare the result for each case

8. System Diagram

9. Assumptions Steady state condition Laminar boundary layer

10. Governing Equations Displacement thickness definition

since u U at y = , the integrand is essentially zero for y .

Momentum thickness definition

Again, the integrand is essentially zero for y .

11. Detailed Solution

Let , then dy = d because = (x).

0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 10

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1Approximated velocity profiles in laminar boundary layer

u/U

y/

LinearSinusoidalParabolic

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* (Displacement thickness)

Linear velocity profile

Sinusoidal velocity profile

Parabolic velocity profile

(Momentum thickness)

Linear velocity profile

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Sinusoidal velocity profile

Parabolic velocity profile

Comparison*/ /

Linear 1/2 = 50% of B.L. 1/6 = 16.7% of B.L.Sinusoidal 1-2/ = 36.3% of B.L. -1/2+2/ = 13.7% of B.L.Parabolic 1/3 = 33.3% of B.L. 2/15 = 13.3% of B.L.

*** Note: B.L. means "Boundary Layer."

12. Critical AssessmentUnderstand the concepts of displacement thickness (*) and momentum thickness (). This problem illustrates how to calculate them from velocity profiles. The above table shows that <*< for most types of velocity profiles.

Examp le 3. (Use of the displacement thickness):

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Air flows in the entrance region of a square duct, as shown. The velocity is uniform, V1 = 30 m/s, and the duct is 80 mm square. At a section 0.3 m downstream from the entrance, the displacement thickness, *, on each wall measures 1.0 mm. Determine pressure change between sections and .


Working fluid is air which has air = 1.23 kg/m3 at T = 15C. Uniform flow at the entrance, V1 = 30 m/s. Duct is H = 80 mm square. Displacement thickness, *

2 = 1.0 mm, on each wall at a section L = 0.3 mdownstream from the entrance.

b) Find Pressure change between sections and .

2. System Diagram

3.

Assumptions Steady state

condition Incompressible fluid flow No frictional effects in freestream Flow uniform at each section outside *

2

Flow along a streamline between sections and Negligible elevation changes

4. Governing Equations … Integral version of mass conservation

Incompressible fluid flow problem, the equation above

*

2 = 1.0 mm

V1

80 mm

80 mm

*

2 = 1.0 mm

V1

H = 80 mm

H = 80 mm

L = 0.3 m

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1 inlet () and 1 outlet ()

Bernoulli's Equation:

Restrictions:(1) Steady flow(2) Incompressible flow(3) Frictionless flow(4) Flow along a streamline


Use the displacement-thickness concept to find the effective flow area for the freestream flow outside the thin wall boundary layers. Replace the actual boundary-layer velocity profiles with uniform velocity profiles as sketched in the following figures.

Apply the continuity and Bernoulli equations to freestream flow outside the boundary-layer displacement thickness, where viscous effects are negligible.

From Bernoulli's equation, we obtain

From the continuity equation, we have

Substituting this expression into Bernoulli's equation,

*

V V

H - 2*

H - 2*

(a) Actual velocity profile (b) Hypothetical velocity profile (c) Cross section of duct

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Areas, A1 and A2, are

… (only effective flow area)

Thus,

After plugging in values,

6. Critical AssessmentTo solve this problem, it is critical to understand the meaning and physical interpretation of displacement thickness concept.

Examp le 4. (Use of the Momentum Integral Method):

The velocity profile in a laminar boundary-layer flow at zero pressure gradient is

approximated by the linear expression, . Use the momentum integral equation with

this profile to obtain expressions for /x and Cf.


Laminar boundary-layer flow Zero pressure gradient

Velocity profile is approximated by the linear expression,

b) FindUsing the momentum integral equation, obtain expression for /x Cf

2. System Diagram

y

x, u

U

y

Uu

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3. Assumptions Steady state condition Incompressible fluid flow

4. Governing Equations Momentum integral equation

where

… displacement thickness

… momentum thickness


For the special case of flow over a flat plate, U = constant. From Bernoulli's equation, we see that for this case, p = constant, and thus dp/dx = 0.

The momentum integral equation then reduces to

Define

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On the other hand, the shear stress can be calculated by .

And , thus

Comparing (equating) this shear stress equation with the previous shear stress expression,

When x = 0, = 0 const = 0.

Skin friction coefficient is defined as .

Using the result obtained above,

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6. Critical AssessmentThis problem dealt with linear velocity profile as an approximate solution. The results obtained are rough. However the exercise illustrates the use of the momentum integral method. Practice this method with other types of approximated velocity profile, such as parabolic, sinusoidal, … etc.

Examp le 5. (Friction Drag Calculation):

Water at 15 C flows over a flat plate at a speed of 1 m/s. The plate is 0.4 m long and 1 m wide. The boundary layer on each surface of the plate is laminar. Assume that the velocity profile may be approximated as linear. Determine the drag force on the plate.


Working fluid is water at T = 15 C = 999 kg/m3 & = 1.14 10-3 Ns/m2

U = 1 m/s L = 0.4 m W = 1 m The boundary layer on each surface of the plate is laminar Velocity profile is linear (assuming approximately)

b) Find

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Drag force on the plate

2. System Diagram

3. Assumptions Steady state conditionIncompressible fluid flow Laminar boundary layer

4. Governing Equations

Skin friction coefficient definition:

Reynolds number definition for a flat plate:


We know that for a linear velocity profile ,

Equating this result and the definition of skin friction coefficient,

Drag force on one side of the plate is given by .

Since dA = w dx and 0 x L, .

U = 1 m/s

L = 0.4 m

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Plug in values into this expression obtained above for FD, N.For both sides of the plate N.

6. Critical AssessmentProblem says, "the boundary layer on each surface of the plate is laminar." Let us double check that this is true.

<< 500,000 = Recr Obviously it is a laminar flow.

(Note: This problem could be solved by first obtaining the Overall Skin Friction Coefficient, . In that case, the calculation will proceed by obtaining

, where the integration limits will be set at x = 0 and x = L.

Then , where, Aw (=W.L) indicates the wet area

on each face of the plate.)

Examp le 6. (Power Calculation using Friction Drag):

A flat-bottomed barge, 25 m long and 10 m wide, submerged to a depth of 1.5 m, is to be pushed up a river at the rate of 8 km/hr. Estimate the power required to overcome skin friction if the water temperature is 15 C.


L = 25 m

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W = 10 m D = 1.5 m V = 8 km/hr = 2.222 m/s Working fluid is water at T = 15 C = 999 kg/m3 & = 1.14 10-3 Ns/m2

b) Find Power required to overcome skin friction

2. System Diagram

3. Assumptions Model a flat-bottomed barge as a flat plate Steady state condition Incompressible fluid flow Neglect separation


Drag Coefficient Definition:


First of all, calculate Reynolds number:

Transition of laminar to turbulent flow occurs at

Water line

V

W

L

D

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<< 25 m

This xcr shows that the effect of laminar flow is negligible. It can be said that the flow is turbulent from the leading edge.

For ReL < 109, the empirical equation given by Schlichting

fits experimental data very well.

Friction force is (from the definition of drag coefficient)

, where A is the wetted area, A = LW+2(LD).

Finally,

Plug in values into this expression = 4200.8 W = 4.20 kW

6. Critical AssessmentDrag coefficient must be chosen depending upon the value of Reynolds number for a particular flow condition. Some of CD expressions are derived by analytical calculation, and others are empirical formulas.

Examp le 7. (Flow Separation Characteristics):

Two hypothetical boundary-layer velocity profiles are shown. Obtain an expression for the momentum flux of each profile. If the two profiles were subjected to the same pressure gradient conditions, which would be more likely to separate first? Why?

U

(a)

U

u

(b)

2

2

yy

Uu

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Two hypothetical boundary-layer velocity profilesb) Find

Expression for the momentum flux of each profile Which would be more likely to separate first if the two profiles were subjected to the

same pressure gradient conditions? And why?

2. System Diagram


4. Governing Equations Definition of Momentum Flux (mf)

U

(a)

U

u

(b)

2

2

yy

Uu

y

Uu

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Since the flow is 1 - D (positive x direction) and dA = w dy, the momentum equation can be written as

The integrand is essentially zero for y .

Linear Velocity Profile

Finally, .

Parabolic Velocity Profile

Let . Then and .

Now,

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Finally, .

Which separates first?Separation occurs when the momentum of fluid layers near the surface is reduced to zero by the combined action of pressure and viscous forces.

As shown in this figure below, the momentum of the fluid near the surface is greater for the parabolic velocity profile.

Our previous calculation also shows < .

Consequently, the parabolic velocity profile is better able to resist separation in the same pressure gradient condition.

Linear velocity profile would separate first.

6. Critical AssessmentReview and understand how the flow separation occurs. Flow separation occurs only when there exists an adverse pressure gradient.

Examp le 8. (Terminal Velocity Calculation):

A small sphere (D = 6 mm) is observed to fall through caster oil at a terminal speed of 60 mm/s. The temperature is 20 C. Compute the drag coefficient for the sphere. Determine the density of the sphere. If dropped in water, would the sphere fall slower or faster? Why?

0 0.5 10

0.2

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1Velocity Profiles

Linear

Parabolic

u/U

y/

0 0.5 10

0.2

0.4

0.6

0.8

1Momentun-Flux Profiles

Linear

Parabolic

(u/U)2

y/

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D = 6 mm = 0.006 m Working fluid is caster oil at T = 20 C S.G.oil = 0.969 & oil = 0.9 Ns/m2

Vt = 60 mm/s = 0.06 m/sb) Find

Drag coefficient for the sphere Density of the sphere If dropped in water, would the sphere fall slower or faster? Why?

2. System Diagram



Drag Coefficient Definition:

Newton's Second Law: , where is momentum.

When the mass is constant, 1 - D in y direction

Reynolds Number for Sphere:

D

Vt

Caster oil at T = 20C

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Drag CoefficientFirst of all, calculate Reynolds number:

ReD = 0.3876 < 1 There is no flow separation from a sphere. The wake is laminar and the drag is predominantly friction drag.

Stokes has shown analytically, for very low Reynolds number flows where inertia forces may be neglected, that drag force on a sphere of diameter D, moving speed V, through a fluid of viscosity , is given by

The drag coefficient, CD, is then

(Note: For sphere, the area, A, is just a cross-sectional area, which is .)

Thus,

Density of the Sphere

Free Body Diagram

The sphere reached the terminal speed ay = 0

W

FD FB

y

gVmgW ss

DVF toilD 3

gVWF soildisplaced

oilB

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After plug in values into this expression, s = 3721 kg/m3.

If dropped in water …

If the working fluid is water at T = 20 C w = 998 kg/m3 & w = 1 10-3 Ns/m2

Because w = 1 10-3 Ns/m2 << oil = 0.9 Ns/m2, the author guesses the sphere drops faster in water than in caster oil.

If it's faster and w << oil , .

FD = 3VD cannot be used because the equation works only for very low Reynolds number which we don't know whether this is appropriate or not any more for this case.

Now, guess a value of CD from Figure 9.11 (Drag coefficient of a smooth sphere as a function of Reynolds number) and calculate VtW. Then calculate ReD and verify the chosen CD was appropriate or not.

Guess CD = 0.4

Free Body Diagram (again)

The sphere reaches a new terminal speed, VtW ay = 0

W

FD FB

y

gVmgW ss

DtWwD CAVF 2

21

gVWF swdisplaced

waterB

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After plugging values into this expression, VtW = 0.732 m/s.

With this new terminal speed, Reynolds number is .

Figure 9.11 says when CD = 0.4, ReD 4 103, which is about right for this case. This shows the new terminal speed is a valid number.

VtW = 0.732 m/s > Vt = 0.06 m/s The sphere drops faster in water than in caster oil.

6. Critical AssessmentDrag coefficient depends upon the value of Reynolds number. Be careful with choosing a right CD depending on a particular flow condition.

velocity profiles in laminar boundary layers often are approximated...

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