chapter 6 laminar external flow - mathworks · dr. m. azzazy proprietary notes egme 526 fall 2017...

Dr. M. Azzazy Proprietary Notes EGME 526 Fall 2017

Chapter 6 Laminar External Flow

Contents 1 Thermal Boundary Layer 1

2 Scale analysis 2

2.1 Case 1: 𝛿𝑡 > 𝛿 (Thermal B.L. is larger than the velocity B.L.) 3

2.2 Case 2: 𝛿𝑡 < 𝛿 (Thermal B.L. is smaller than the velocity B.L.) 4

3 Summary for B.L. equations 5

4 Semi-infinite Plate: constant free stream velocity, constant surface temperature 6

5 Pohlhausen solution 7

6 Examples 12

6.1 Example 1. 12

6.2 Example 2 12

6.3 Example 3 13

7 Variable Surface Temperature 13

8 Faulkner-Skan flows 16

9 Problems 19

1 Thermal Boundary Layer This chapter deals with heat transfer between a body and a fluid flowing with steady laminar motion over that body. We assume that: (1) all body forces are negligible (flow is forced over the body by some external means that are not related to the temperature field in the fluid), (2) idealized constant fluid properties, (3) no mass diffusion (concentration gradients are negligible), and (4) free stream velocity is sufficiently low that the viscous dissipation term in the energy equation could be neglected (flows with high values of dissipation terms will be treated later). Consider the flow over the semi-infinite heated surface shown in Figure 1. As discussed in Chapter 4 the foundation of the boundary layer concept is that the effect of viscosity is confined to a thin region near the surface (boundary layer); 𝛿. Similarly, the effect of thermal interaction between the surface and the moving fluid is confined to a thin region near the surface defined as 𝛿𝑡 (thermal boundary layer). As we discussed in Chapter 4, the condition for the velocity boundary layer model is that the Reynolds number is very high (remember that we neglected axial momentum diffusion relative to normal momentum

diffusion (𝜕2𝑢

𝜕𝑥2 ≪𝜕2𝑢

𝜕𝑦2). The condition for thermal boundary layer is that the product of Reynolds number

(𝑅𝑒) and Prandtl number (𝑃𝑟) (called Peclet number – 𝑃𝑒) must be very high (𝑃𝑒 > 100).


𝑃𝑒 = 𝑅𝑒. 𝑃𝑟 =𝑈𝐿

𝛼 (1)

Where 𝑈 is the free stream velocity, 𝐿 is a characteristic length and 𝛼 is the thermal diffusivity.

d dt

x

y

u

U

Ts

Tf

Figure 1. Velocity and Thermal Boundary Layer

The boundary conditions are:

1) No slip velocity at the surface (𝑢 = 0 𝑎𝑡 𝑦 = 0), 2) At the edge of the boundary layer, the velocity is equal to the free stream velocity

(𝑢 = 𝑈 𝑎𝑡 𝑦 = 𝛿), 3) The fluid temperature at the surface is equal to the surface temperature, 𝑇𝑓 = 𝑇𝑠 𝑎𝑡 𝑦 = 0, 4) The fluid temperature at the edge of the thermal boundary layer is equal to the free stream

temperature.

2 Scale analysis The simplified energy equation using steady incompressible flow without heat generation, for a two dimensional Cartesian flow is

𝑢𝜕𝑇

𝜕𝑥+ 𝑣

𝜕𝑇

𝜕𝑦= 𝛼 (

𝜕2𝑇

𝜕𝑥2+

𝜕2𝑇

𝜕𝑦2) (2)

Since in the velocity boundary layer we neglected 𝜕2𝑢

𝜕𝑥2 relative to 𝜕2𝑢

𝜕𝑦2, we wish to perform an analysis that

would allow us to neglect axial conduction 𝜕2𝑇

𝜕𝑥2 relative to conduction in the normal direction. If we

examine Figure 1, we find that the value of the thermal boundary layer thickness is much smaller than the axial length of the plate. Therefore thermal gradients along the normal direction will be much larger than the axial direction, and therefore axial conduction could be neglected, i.e.

𝜕2𝑇

𝜕𝑥2 ≪𝜕2𝑇

𝜕𝑦2 (3)

And the energy equation (equation 2) becomes

𝑢𝜕𝑇

𝜕𝑥+ 𝑣

𝜕𝑇

𝜕𝑦= 𝛼

𝜕2𝑇

𝜕𝑦2 (4)


2.1 Case 1: 𝜹𝒕 > 𝜹 (Thermal B.L. is larger than the velocity B.L.) Figure 2 is a schematic illustration of this case. In this case, the velocity component u inside the thermal B.L. could be approximated as

𝑢~𝑈 (5)

d

dt

x

y

u

Tf

U

Ts

Figure 2. Case 1, Thermal B.L. larger than velocity B.L.

And from the continuity equation, the velocity component 𝑣 could be approximated as given in equation

6. Since, 𝜕𝑣

𝜕𝑦= −

𝜕𝑢

𝜕𝑥 then 𝑣 = − ∫

𝜕𝑢

𝜕𝑥𝑑𝑦 but since from equation 5 𝑢 is constant, then

𝑣~𝑈𝛿𝑡

𝐿 (6)

Therefore the convective terms in the energy equation (4) become

𝑢𝜕𝑇

𝜕𝑥~𝑈

∆𝑇

𝐿 (7)

And

𝑣𝜕𝑇

𝜕𝑦~𝑈

∆𝑇

𝐿 (8)

In equation 8, notice that 𝑦 is of the order of 𝛿𝑡 and relationship 6 was used to derive relationship 8. Since 𝛿𝑡 is much less than 𝐿, then 𝑣 could be neglected relative to 𝑢, and the energy equation 4 could be simplified to

𝑢𝜕𝑇

𝜕𝑥~𝛼

𝜕2𝑇

𝜕𝑦2

Therefore

𝑈∆𝑇

𝐿~𝛼

∆𝑇

𝛿𝑡2

And

𝛿𝑡

𝐿= √

𝛼

𝑈𝐿= √

𝜌𝛼

𝜇√

𝜇

𝜌𝑈𝐿=

1

√𝑃𝑟

1

√𝑅𝑒=

1

√𝑃𝑒 (9)

Therefore, for large 𝑃𝑒 number, the thermal boundary layer thickness is small.


The ratio of the thermal boundary layer thickness to the velocity boundary layer thickness is determined from

𝛿𝑡

𝛿=

1

√𝑃𝑟 (10)

Therefore case 1, 𝛿𝑡 > 𝛿 means that √𝑃𝑟 ≪ 1 In summary

1) for 𝛿𝑡 to be small relative to a characteristic length 𝐿, then 𝑃𝑒 must be >> 1, and 2) for the thermal B.L. to be larger than the velocity B.L., then the square root of 𝑃𝑟 number must

be much less than one.

2.2 Case 2: 𝜹𝒕 < 𝜹 (Thermal B.L. is smaller than the velocity B.L.) Figure 3 is a schematic illustration of this case. In this case, the axial velocity within the thermal boundary layer is smaller than the free stream velocity. Assuming the velocity profile could be approximated as a linear profile in the small thermal B.L. region, then

𝑢~𝑈𝛿𝑡

𝛿 (11)

Using equation 10 to scale the continuity equation

𝑣~𝑈𝛿𝑡

2

𝐿𝛿 (12)

This means that 𝑣 could be neglected relative to 𝑢 in the energy equation.

d

dt

x

yu

Tf

U

Ts

Figure 3. Case 2, Thermal B.L. smaller than velocity B.L.

The energy equation (4) becomes

𝑢𝜕𝑇

𝜕𝑥~𝛼

𝜕2𝑇

𝜕𝑦2

Using equation 11

𝑈𝛿𝑡

𝛿

∆𝑇

𝐿~𝛼

∆𝑇

𝛿𝑡2

And

(𝛿𝑡

𝐿)

3=

𝛼

𝑈𝐿

𝛿

𝐿=

𝜌𝛼

𝜇

𝜇

𝜌𝑈𝐿

1

√𝑅𝑒=

1

𝑃𝑟

1

𝑅𝑒

1

√𝑅𝑒=

1

𝑃𝑟

1

𝑅𝑒3 2⁄

Finally


𝛿𝑡

𝐿=

1

𝑃𝑟1 3⁄

1

√𝑅𝑒 (13)

The ratio of the thermal B.L. to the velocity B.L. in this case is

𝛿𝑡

𝛿=

1

𝑃𝑟1 3⁄ (14)

In summary,

1) For the thermal B.L. to be small compared to a characteristic length 𝐿, the product 𝑃𝑟1 3⁄ √𝑅𝑒 must be >> 1.

2) The second condition, for the thermal B.L. to be smaller than the velocity B.L. then 𝑃𝑟1 3⁄ ≫ 1.

3 Summary for B.L. equations Assumptions

1. Continuum 2. Newtonian fluid 3. 2-D case 4. Constant properties 5. No body forces 6. No potential flow singularities (slender body) 7. High Reynolds number (𝑅𝑒 > 100) 8. High Peclet number (𝑃𝑒 > 100) 9. Steady state 10. Laminar flow 11. No dissipation

Continuity equation

𝜕𝑢

𝜕𝑥+

𝜕𝑣

𝜕𝑦= 0 (15)

X-momentum

𝑢𝜕𝑢

𝜕𝑥+ 𝑣

𝜕𝑢

𝜕𝑦= −

1

𝜌

𝑑𝑝

𝑑𝑥+

𝜇

𝜌

𝜕2𝑢

𝜕𝑦2 (16)

Energy

𝑢𝜕𝑇

𝜕𝑥+ 𝑣

𝜕𝑇

𝜕𝑦= 𝛼

𝜕2𝑇

𝜕𝑦2 (17)


4 Semi-infinite Plate: constant free stream velocity, constant surface temperature

Assume that the plate is maintained at a constant temperature 𝑇𝑠, and the fluid temperature in the free stream is 𝑇𝑓 as schematically shown in Figure 4.

d dt

x

y

u

U

Ts

Tf

Figure 4. Momentum and Thermal Boundary Layers

The continuity and momentum equations (equations 15 and 16) were solved in Chapter 4 using the exact Blasius solution such that: Similarity parameter

𝜂 = 𝑦√𝑈∞

𝜈𝑥 (18)

Velocity 𝑢 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

𝑢

𝑈∞= 𝑓′(𝜂) (19)

Velocity 𝑣 − 𝑐𝑜𝑚𝑝𝑜𝑛𝑒𝑛𝑡

𝑣

𝑈∞=

1

2√

𝜇

𝜌𝑈∞𝑥(𝜂𝑓′ − 𝑓) (20)

And the governing equation is

𝑓𝑓′′ + 2𝑓′′′ = 0 (21)

Equation 21 was solved by Blasius and the results are tabulated (Table 1)


Table 1. Blasius Solution

Blasius Solution

Eta f f' f''

0 0 0 0.33206

0.4 0.02656 0.13277 0.33147

0.8 0.10611 0.26471 0.32739

1.2 0.23795 0.39378 0.31659

1.6 0.42032 0.51676 0.29667

2 0.65003 0.62977 0.26675

2.4 0.9223 0.72899 0.22809

2.8 1.23099 0.81152 0.18401

3.2 1.56911 0.87609 0.13913

3.6 1.92954 0.92333 0.09809

4 2.30576 0.95552 0.06424

4.4 2.69238 0.97587 0.03897

4.8 3.08534 0.98779 0.02187

5 3.28329 0.99155 0.01591

5.2 3.48189 0.99425 0.01134

5.4 3.68094 0.99616 0.00793

5.6 3.88031 0.99748 0.00543

6 4.27964 0.99898 0.0024

7 5.27926 0.99992 0.00022

8 6.27923 1 0.00001

5 Pohlhausen solution Let 𝜃 be defined as

𝜃 =𝑇𝑠−𝑇

𝑇𝑠−𝑇𝑓 (22)

Where 𝑇𝑠 is the surface temperature and 𝑇𝑓 is the free stream fluid temperature as shown in Figure 4.

Substituting equation 22 into the energy equation (17)

𝑢𝜕𝜃

𝜕𝑥+ 𝑣

𝜕𝜃

𝜕𝑦= 𝛼

𝜕2𝜃

𝜕𝑦2 (23)

The boundary conditions are:

𝜃(𝑥, 0) = 0 (24)

𝜃(𝑥, ∞) = 1 (25)

𝜃(0, 𝑦) = 1 (26)


In order to solve equation 23 using the similarity method, the two independent variables 𝑥 and 𝑦 are combined to form the similarity variable 𝜂(𝑥, 𝑦) as defined by equation 18. The solution of equation 23 then becomes 𝜃(𝑥, 𝑦) = 𝜃(𝜂) From calculus

𝜕𝜃

𝜕𝑥= −

𝜂

2𝑥

𝑑𝜃

𝑑𝜂 (27)

𝜕𝜃

𝜕𝑦= √

𝜌𝑈

𝜇𝑥

𝑑𝜃

𝑑𝜂 (28)

And

𝜕2𝜃

𝜕𝑦2 =𝜌𝑈

𝜇𝑥

𝑑2𝜃

𝑑𝜂2 (29)

Substituting equations 27-29 into equation 23 then

𝜃′′ +𝑃𝑟

2𝑓(𝜂)

𝑑𝜃

𝑑𝜂= 0 (30)

Subject to the boundary conditions

𝜃(0) = 0 (31)

𝜃(∞) = 1 (32)

Notice that the three boundary conditions are now two since we used the similarity variable to combine the 𝑥 and 𝑦 independent variables. Equation 30 could be written as

𝜃′′

𝜃′ = −𝑃𝑟

2𝑓(𝜂)𝑑𝜂 (33)

Which upon integration becomes

𝜃′ = 𝐶1𝐸𝑥𝑝 (−𝑃𝑟

2∫ 𝑓(𝜂)𝑑𝜂

𝜂

0) (34)

Integrating equation 34 one more time, the temperature distribution becomes

𝜃 = {𝐶1 ∫ [𝐸𝑥𝑝 (−𝑃𝑟


𝜂

0)] 𝑑𝜂

𝜂

0} + 𝐶2 (35)

Using the boundary condition, equation 31, then 𝐶2 = 0. Using the boundary condition, equation 32, then

𝐶1 =1

∫ [𝐸𝑥𝑝(−𝑃𝑟


𝜂

0)]𝑑𝜂

∞

0

(36)

Therefore the temperature distribution becomes

𝜃 =∫ [𝐸𝑥𝑝(−

𝑃𝑟


𝜂

0)]𝑑𝜂

𝜂

0

∫ [𝐸𝑥𝑝(−𝑃𝑟


𝜂

0)]𝑑𝜂

∞

0

(37)

Equation 37 has the integral ∫ 𝑓(𝜂)𝑑𝜂𝜂

0 in it. Using the momentum equation (equation 21), then

𝑓 = −2𝑓′′′

𝑓′′

Which upon integration becomes


∫ 𝑓(𝜂)𝑑𝜂𝜂

0= −2𝑙𝑛 [

𝑓′′(𝜂)

𝑓′′(0)] (38)

Substituting equation 38 into equation 37

𝜃(𝜂) =∫ [𝐸𝑥𝑝{(𝑃𝑟)𝑙𝑛[

𝑓′′(𝜂)

𝑓′′(0)]}]𝑑𝜂

𝜂

0

∫ [𝐸𝑥𝑝{(𝑃𝑟)𝑙𝑛[𝑓′′(𝜂)

𝑓′′(0)]}]𝑑𝜂

∞

0

(39)

Which upon further manipulation becomes

𝜃(𝜂) =∫ [𝑓′′(𝜂)]

𝑃𝑟𝑑𝜂

𝜂

0

∫ [𝑓′′(𝜂)]𝑃𝑟𝑑𝜂∞

0

(40)

Numerical calculations of the temperature profile using equation 40 often suffers inaccuracies due to round-off errors. In order to calculate the temperature profile accurately the following equation is often used

𝜃(𝜂) = 1 −∫ [𝑓′′(𝜂)]

𝑃𝑟𝑑𝜂

∞

𝜂


0

(40,a)

Equation (40,a) is integrated using the trapezoidal rule. Results for different 𝑃𝑟 numbers are shown in Figure 5, and tabulated in an excel file accompanying this Chapter “Pohlhausen_Table.xlsx”.

Figure 5. Pohlhausen temperature profile as a function of Pr number

The value of 𝑑𝜃(0)

𝑑𝜂 is important in the calculation the Nusselt Modulus. Differentiating equation 40 one

obtains

𝑑𝜃(0)

𝑑𝜂=

[𝑓′′(0)]𝑃𝑟


0

=(0.332)𝑃𝑟


0

(41)


The integrals in equation 41 are evaluated numerically and shown in Table 2.

Table 2

𝑃𝑟 𝑑𝜃(0)

𝑑𝜂

0.005 0.03766

0.01 0.0516

0.1 0.14

0.5 0.259

0.7 0.292

1.0 0.332

7.0 0.645

10.0 0.73

15.0 0.835

50.0 1.247

100 1.572

From the table, the following equations give a good approximation of 𝑑𝜃(0)

𝑑𝜂

𝑑𝜃(0)

𝑑𝜂= 0.5𝑃𝑟1 2⁄ 0.005 < 𝑃𝑟 < 0.05 (42)

𝑑𝜃(0)

𝑑𝜂= 0.332𝑃𝑟1 3⁄ 0.6 < 𝑃𝑟 < 10 (43)

𝑑𝜃(0)

𝑑𝜂= 0.339𝑃𝑟1 3⁄ 𝑃𝑟 > 10 (44)

The heat transfer coefficient is determined from

ℎ(𝑇𝑠 − 𝑇∞) = −𝑘𝜕𝑇(𝑥,0)

𝜕𝑦 (45)

Since

𝜕𝑇(𝑥,0)

𝜕𝑦=

𝑑𝑇

𝑑𝜃

𝑑𝜃(0)

𝑑𝜂

𝜕𝜂

𝜕𝑦 (46)

Then

𝜕𝑇(𝑥,0)

𝜕𝑦= (𝑇∞ − 𝑇𝑠)√

𝜌𝑈∞

𝜇𝑥

𝑑𝜃(0)

𝑑𝜂 (47)


ℎ(𝑥) = 𝑘√𝜌𝑈∞

𝜇𝑥

𝑑𝜃(0)

𝑑𝜂 (48)

The average heat transfer coefficient for a plate of length 𝐿 is calculated from

ℎ =1

𝐿∫ ℎ(𝑥)𝑑𝑥

𝐿

0 (49)

Substituting equation 48 into equation 49 and integrating


ℎ = 2𝑘

𝐿√𝑅𝑒𝐿

𝑑𝜃(0)

𝑑𝜂 (50)

The local Nusselt number is obtained from

𝑁𝑢(𝑥) =ℎ𝑥

𝑘= √𝑅𝑒𝑥

𝑑𝜃(0)

𝑑𝜂 (51)

And the average Nusselt number is

𝑁𝑢𝐿 = 2√𝑅𝑒𝐿

𝑑𝜃(0)

𝑑𝜂 (52)


6 Examples

6.1 Example 1. Water at 25 °𝐶 flows over a flat plate with uniform velocity of 2 𝑚 𝑠𝑒𝑐⁄ . The plate is maintained at 85 °𝐶. Determine the following:

a) Heat flux at 8 𝑐𝑚 from leading edge b) Total heat transfer from the first 8 𝑐𝑚 of the plate c) Can Pohlhausen solution be used to find the heat flux at 80 𝑐𝑚 from leading edge

The average temperature is 𝑇𝑎𝑣𝑒 =85+25

2= 55 °C

Water properties at the average temperature are: 𝑘 = 0.6507 𝑊 𝑚. °𝐶⁄

𝑃𝑟 = 3 𝜈 = 0.4748𝑥10−6 𝑚2 𝑠𝑒𝑐⁄

At 8 𝑐𝑚, the Reynolds number is calculated such that 𝑅𝑒𝐿 =𝑈𝐿

𝜈=

2(0.08)

0.4748x10−6 = 336,984

For 𝑃𝑟 = 3, the value of 𝑑𝜃(0)

𝑑𝜂 is calculated from:

𝑑𝜃(0)

𝑑𝜂= 0.332𝑃𝑟1 3⁄ = 0.4788

The heat transfer coefficient is calculated from

ℎ = 𝑘√𝑈∞𝑥

𝜈

𝑑𝜃(0)

𝑑𝜂= 0.6507√

2(0.08)

0.4748x10 − 60.4788 = 2,260.86

The heat flux at 8 𝑐𝑚 is calculated from: 𝑞 = ℎ(𝑇𝑠 − 𝑇𝑓) = 2,260.8(85 − 25) = 135,652 𝑊

𝑚2

The total heat transfer coefficient for the first 8 𝑐𝑚 is ℎ = 2𝑘

𝐿√𝑅𝑒𝐿

𝑑𝜃(0)

𝑑𝜂= 4,521.72

𝑊

𝑚2℃

Therefore, total heat transfer for the first 8 𝑐𝑚 is calculated from �� = ℎ(𝑇𝑠 − 𝑇𝑓) = 271,303.38 𝑊

𝑚2

At 80 𝑐𝑚 from the leading edge, the Reynolds number becomes 3,369,840 which means that the flow is turbulent (Reynolds number for transition between laminar and turbulent flow is 500,000). Pohlhausen solution is only valid for laminar flow.

6.2 Example 2 An isosceles triangle is drawn on a semi-infinite flat plate at a uniform surface temperature 𝑇𝑠. Consider laminar uniform flow of constant properties fluid over the plate. Determine the rate of heat transfer between the triangle area and the fluid. The local heat transfer coefficient is calculated from

ℎ(𝑥) = 𝑘√𝜌𝑈∞

𝜇𝑥

𝑑𝜃(0)

𝑑𝜂

The elemental area is 𝑏(𝑥)𝑑𝑥 = 𝑥𝐻

𝐿

The local rate of heat transfer is then given by 𝑞(𝑥) = ℎ(𝑥)(𝑇𝑠 − 𝑇∞)𝑏(𝑥)𝑑𝑥

= (𝑇𝑠 − 𝑇∞)𝑘√𝜌𝑈∞

𝜇𝑥

𝑑𝜃(0)

𝑑𝜂𝑥

𝐻

𝐿𝑑𝑥

U∞

T∞

L

H

dx

x

b(x)


Which upon rearranging becomes 𝑞(𝑥) = (𝑇𝑠 − 𝑇∞)𝑘√𝜌𝑈∞

𝜇

𝑑𝜃(0)

𝑑𝜂

𝐻

𝐿𝑥1 2⁄ 𝑑𝑥 which could be integrated to

determine the rate of heat transfer between the triangle area and the fluid such that

�� = (𝑇𝑠 − 𝑇∞)𝑘√𝜌𝑈∞

𝜇

𝑑𝜃(0)

𝑑𝜂

𝐻

𝐿∫ 𝑥1 2⁄ 𝑑𝑥

𝐿

0

=2

3(𝑇𝑠 − 𝑇∞)𝑘𝐻

𝑑𝜃(0)

𝑑𝜂√𝑅𝑒𝐿

6.3 Example 3 The cap of an electronic package is cooled by forced convection. The free stream temperature is 25 °𝐶. The Reynolds number at the downstream end of the cap is 𝑅𝑒 = 110,000. Surface temperature was found to be 145 °𝐶. However, reliability requires that surface temperature does not exceed 83 °𝐶. One possible solution is to increase the free stream velocity by a factor of 3. You are asked to determine if the surface temperature under this plan will meet design specifications. The heat transfer coefficient is calculated from

ℎ = 2𝑘

𝐿√𝑅𝑒𝐿

𝑑𝜃(0)

𝑑𝜂

The amount of heat removed by the original design is

𝑄 = ℎ1 𝐴(145 − 25) = 120ℎ1

𝐴 When the velocity is multiplied by a factor of 3, the Reynolds number becomes 𝑅𝑒𝐿2 = 3𝑅𝑒𝐿1 The heat transfer coefficient becomes

ℎ2 = 2

𝑘

𝐿√𝑅𝑒𝐿2

𝑑𝜃(0)

𝑑𝜂= √3 ℎ1

The amount of heat to be removed from the package is the same. Therefore

𝑄 = ℎ2 𝐴(𝑇𝑠2 − 25) = 120ℎ1

𝐴 Therefore, the surface temperature under the new design is

𝑇𝑠2 − 25 =120

√3= 69.28

The surface temperature under the new design is 𝑇𝑠2 = 94.28 °𝐶 which exceeds that of the reliability requirement.

7 Variable Surface Temperature A self similar class of solutions exists for problems where the surface temperature varies as

𝑇𝑠(𝑥) = 𝑇∞ + 𝐶𝑥𝑛 (53)

Where 𝐶 and the exponent 𝑛 are constants. The self-similar boundary layer solution is summarized as

𝑢

𝑈∞= 𝑓′(𝜂); 𝜂 = 𝑦√

𝑈∞

𝜈𝑥;

v

𝑈∞=

1

2√

𝜈

𝑈∞𝑥(𝜂𝑓′ − 𝑓) (54)

The energy equation is given


𝑢𝜕𝑇

𝜕𝑥+ v

𝜕𝑇

𝜕𝑦= 𝛼

𝜕2𝑇

𝜕𝑦2 (55)


𝑇(𝑥, 0) = 𝑇𝑠 (56)

𝑇(𝑥, ∞) = 𝑇∞ (57)

𝑇(0, 𝑦) = 𝑇𝑠 (58)

Introducing the non-dimensional temperature variable


𝑇𝑠−𝑇∞ (59)


𝑇 = 𝑇∞ + 𝐶𝑥𝑛 − 𝐶𝑥𝑛𝜃 (60)

Consider the variation of the temperature along the x-direction; 𝜕𝑇

𝜕𝑥

𝜕𝑇

𝜕𝑥= 𝐶𝑛𝑥𝑛−1 − 𝐶𝑛𝑥𝑛−1𝜃 − 𝐶𝑥𝑛 𝑑𝜃

𝑑𝜂

𝑑𝜂

𝑑𝑥 (61)

From equation 54 𝑑𝜂

𝑑𝑥=

−𝜂

2𝑥. Therefore equation 61 becomes

𝜕𝑇

𝜕𝑥= 𝐶𝑛𝑥𝑛−1 − 𝐶𝑛𝑥𝑛−1𝜃 + 𝐶𝑥𝑛−1 𝑑𝜃

𝑑𝜂

𝜂

2= 𝐶𝑛𝑥𝑛−1 (1 − 𝜃 +

𝜂

2𝑛𝜃′) (62)

The y-derivative of the temperature is

𝜕𝑇

𝜕𝑦= −𝐶𝑥𝑛 𝑑𝜃

𝑑𝜂

𝑑𝜂

𝑑𝑦= −𝐶𝑥𝑛𝜃′√

𝑈∞

𝜈𝑥 (63)

The second derivative of the temperature relative to y is

𝜕2𝑇

𝜕𝑦2 =𝜕

𝜕𝑦(−𝐶𝑥𝑛𝜃′√

𝑈∞

𝜈𝑥) = −𝐶𝑥𝑛𝜃′′ 𝑈∞

𝜈𝑥= −𝐶𝑥𝑛−1𝜃′′

𝑈∞

𝜈 (64)

Substituting equations 54, 60, 62, 63, and 64 into the energy equation 55

𝑈∞𝑓′𝐶𝑛𝑥𝑛−1 (1 − 𝜃 +𝜂

2𝑛𝜃′) −

𝑈∞

2√

𝜈

𝑈∞𝑥(𝜂𝑓′ − 𝑓)𝐶𝑥𝑛𝜃′√

𝑈∞

𝜈𝑥= −

𝛼

𝜈𝑈∞𝐶𝑥𝑛−1𝜃′′

Upon rearranging

𝑓′𝑛 (1 − 𝜃 +𝜂

2𝑛𝜃′) −

1

2(𝜂𝑓′ − 𝑓)𝜃′ = −

1

𝑃𝑟𝜃′′

Which could be further reduced to

𝑓′𝑛(1 − 𝜃) +1

2𝑓′𝜂𝜃′ −

1

2𝜂𝑓′𝜃′ +

1

2𝑓𝜃′ = −

1

𝑃𝑟𝜃′′

Which finally becomes

𝜃′′ + 𝑛𝑃𝑟(1 − 𝜃)𝑓′ +𝑃𝑟

2𝑓𝜃′ = 0 (65)


𝜃(0) = 0 (66)

𝜃(∞) = 1 (67)

The value of the heat transfer coefficient is of special interest. Again applying the Nusselt Modulus one obtains


ℎ(𝑇𝑠 − 𝑇∞) = −𝑘𝜕𝑇(𝑥,0)

𝜕𝑦 (68)

Substituting equations 53, and 63 into equation 68

ℎ𝐶𝑥𝑛 = 𝑘𝐶𝑥𝑛√𝑈∞

𝜈𝑥

𝑑𝜃(0)

𝑑𝜂

Therefore

ℎ(𝑥) =𝑘

𝑥√𝑅𝑒𝑥

𝑑𝜃(0)

𝑑𝜂 (69)

and the local Nusselt number is

𝑁𝑢(𝑥) = √𝑅𝑒𝑥𝑑𝜃(0)

𝑑𝜂 (70)

The average heat transfer coefficient over a distance L is given by

ℎ =2𝑘

𝐿√𝑅𝑒𝐿

𝑑𝜃(0)

𝑑𝜂 (71)

And the average Nusselt number is given by

𝑁𝑢 = 2√𝑅𝑒𝐿𝑑𝜃(0)

𝑑𝜂 (72)

The values of 𝑑𝜃(0)

𝑑𝜂 for different 𝑛 and 𝑃𝑟 are given in Figure 6.

Figure 6. Surface temperature gradient for plate with varying surface Temperature, 𝑇𝑠(𝑥) − 𝑇∞ =

𝐶𝑥𝑛

0.0

1.0

2.0

0 0.5 1 1.5

d_

Thet

a(0

)/d

_Et

a

n


8 Faulkner-Skan flows This class of fluid flows has a self similar solution for the velocity profile. The free stream velocity in this class of flows is given by:

𝑈∞ = 𝑐𝑥𝑚 (73)

Where m is related to the cone angle b through the relationship

𝑚 =𝛽

2−𝛽 (74)

Special cases 1. 𝛽 = 0 → 𝑚 = 0 → flow over a flat plate (75)

x=0

2. 𝛽 = 1 → 𝑚 = 1 → stagnation flow

x

=1

3. 𝛽 < 1 → flow over a wedge (external flow)

p

x

4. 𝛽 > 1 → flow over a corner (internal flow)

pU∞=Cxm

T∞

xy

Ts


x

>1

The continuity equation may be written as

𝜕𝑢

𝜕𝑥+

𝜕v

𝜕𝑦= 0 (76)

And the momentum equation is

𝑢𝜕𝑢

𝜕𝑥+ v

𝜕𝑢

𝜕𝑦= 𝑈∞

𝑑𝑈∞

𝑑𝑥+ 𝜈

𝜕2𝑢

𝜕𝑦2 (77)

Where

𝑈∞ = 𝑐𝑥𝑚 (78)

The boundary conditions are

𝑢(𝑥, 0) = 0 (79)

v(𝑥, 0) = 0 (80)

𝑢(𝑥, ∞) = 𝑐𝑥𝑚 (81)

The solution of equations 77-81 is obtained by the method of similarity. The similarity variable is defined as

𝜂 = 𝑦√𝑈∞

𝜈𝑥= 𝑦√

𝑐

𝜈𝑥

𝑚−1

2 (82)

The velocity distribution is given by

𝑢

𝑈∞= 𝐹′(𝜂) (83)

From the continuity, the velocity component v is given by

v

𝑈∞= −√

𝜈

𝑈∞𝑥(

𝑚+1

2) (𝐹 −

1−𝑚

1+𝑚𝜂𝐹′) (84)

Substituting equations 78-84 into equation 77, one obtains

𝐹′′′ +𝑚+1

2𝐹𝐹′′ − 𝑚𝐹′2 + 𝑚 = 0 (85)


𝐹′(0) = 0 (86)

𝐹(0) = 0 (87)

𝐹′(∞) = 1 (88)


To determine the temperature distribution, consider the energy equation

𝑢𝜕𝜃

𝜕𝑥+ v

𝜕𝜃

𝜕𝑦= 𝛼

𝜕2𝜃

𝜕𝑦2 (89)


𝜃(𝑥, 0) = 0 (90)

𝜃(𝑥, ∞) = 1 (91)

𝜃(0, 𝑦) = 1 (92)

Where the non-dimensional temperature variable


𝑇𝑠−𝑇∞ (93)

Using the similarity variable, the energy equation becomes

𝑑2𝜃

𝑑𝜂2 +𝑃𝑟

2(𝑚 + 1)𝐹(𝜂)

𝑑𝜃

𝑑𝜂= 0 (94)

As usual, we are interested in the value of 𝑑𝜃(0)

𝑑𝜂 to calculate the heat transfer coefficient and the Nusselt

number. These values are given in Table 3.

Table 3.Values for 𝑑𝜃(0)

𝑑𝜂 for various 𝑃𝑟 numbers

Surface temperature gradient 𝒅𝜽(𝟎)

𝒅𝜼 and surface velocity gradient 𝑭′′(𝟎) for flow over an

isothermal wedge

m Wedge

angle 𝜋𝛽 𝐹′′(0)

𝑑𝜃(0)

𝑑𝜂 at five values of 𝑃𝑟

0.7 0.8 1.0 5.0 10.0

0 0 0.326 0.292 0.307 0.332 0.585 0.73

0.111 𝜋 5⁄ (36°) 0.5120 0.331 0.348 0.378 0.669 0.851

0.333 𝜋 2⁄ (96°) 0.7575 0.384 0.403 0.440 0.792 1.013

1.0 𝜋 (180°) 1.2326 0.496 0.523 0.570 1.043 1.344


9 Problems 1. Fluid flows between two parallel plates. It enters with uniform velocity 𝑈∞ and temperature 𝑇∞.

The plates are maintained at uniform surface temperature 𝑇𝑠. Assume laminar boundary layer flow at the entrance. Can Pohlhausen solution be applied to determine the heat transfer coefficient? Explain

y

x

Ts

T∞

U∞

2. Two identical rectangles, 𝐴 and 𝐵, of dimensions 𝐿1𝑥𝐿2 are drawn on the surface of a semi-infinite flat plate as shown. Rectangle 𝐴 is oriented with side 𝐿1 along the leading edge while rectangle 𝐵 is oriented with side 𝐿2 along the edge. The plate is maintained at uniform surface temperature.

a. If the flow over rectangle 𝐴 is laminar, what is it for 𝐵? b. If the heat transfer rate from plate 𝐴 is 435 W, what is the rate from plate 𝐵?

A

B

L1

L1

L2

L2

Top View

TsT∞

U∞

3. A semi-infinite plate is divided into four equal sections of one centimeter long each. Free stream temperature and velocity are uniform and the flow is laminar. The surface is maintained at uniform temperature. Determine the ratio of the heat transfer rate from the third section to that from the second section.

T∞

U∞

1 2 3 4

x

Ts


4. A fluid at a uniform velocity and temperature flows over a semi-infinite flat plate. The surface

temperature is uniform. Assume laminar boundary layer flow. a. What will be the percent change in the local heat transfer coefficient if the free stream

velocity is reduced by a factor of two? b. What will be the percent change in the local heat transfer coefficient if the distance from

the leading edge is reduced by a factor of two?

5. Use Pohlhausen's solution to derive an expression for the ratio of the thermal boundary layer thickness for two fluids. The Prandtl number of one fluid is 1.0 and its kinematic viscosity is 0.12𝑥10−6 𝑚2 𝑠𝑒𝑐⁄ . The Prandtl number of the second fluid is 100 and its kinematic viscosity is 6.8𝑥10−6 𝑚2 𝑠𝑒𝑐⁄ .

6. Water at 25°𝐶 flows over a flat plate with a uniform velocity of 2 𝑚 𝑠𝑒𝑐⁄ . The plate is maintained

at 85°𝐶. Determine the following: a. The thermal boundary layer thickness at a distance of 8 𝑐𝑚 from the leading edge. b. The heat flux at this location. c. The total heat transfer from the first 8 𝑐𝑚 of the plate. d. Whether Pohlhausen's solution can be used to find the heat flux at a distance of 80 𝑐𝑚

from the leading edge.

7. The cap of an electronic package is cooled by forced convection. The free stream temperature is 25°𝐶. The Reynolds number at the downstream end of the cap is 110,000. Surface temperature was found to be 145°𝐶. However, reliability requires that surface temperature does not exceed 83°𝐶. One possible solution to this design problem is to increase the free stream velocity by a factor of 3. You are asked to determine if surface the temperature under this plan will meet design specification.

T∞

U∞ Ts

cap

8. A fluid with Prandtl number 0.098 flows over a semi-infinite flat plate. The free stream temperature is 𝑇∞ and the free stream velocity is 𝑈∞. The surface of the plate is maintained at uniform temperature 𝑇𝑠. Assume laminar flow.

a. Derive an equation for the local Nusselt number. b. Determine the heat transfer rate from a section of the plate between 𝑥1 and 𝑥2. The width

of the plate is 𝑊. c. Derive an equation for the thermal boundary layer thickness 𝛿𝑡(𝑥).


x1

U∞

T∞ yx

x2

W

9. Two identical triangles are drawn on the surface of a flat plate as shown. The plate, which is maintained at uniform surface temperature, is cooled by laminar forced convection. Determine the ratio of the heat transfer rate from the two triangles, 𝑞1 𝑞2⁄ .

L

Top View

Ts

T∞

U∞

1

2H

10. An isosceles triangle is drawn on a semi-infinite flat plate at a uniform surface temperature 𝑇𝑠. Consider laminar uniform flow of constant properties fluid over the plate. Determine the rate of heat transfer between the triangular area and the fluid.

U∞

T∞ Ts

L

H

11. Determine the total heat transfer rate from a half circle drawn on a semi-infinite plate as shown. Assume laminar two-dimensional boundary layer flow over the plate.

Ts

T∞

U∞

r0

r0


12. Consider steady, two-dimensional, laminar boundary layer flow over a semi-infinite plate. The surface is maintained at uniform temperature 𝑇𝑠. Determine the total heat transfer rate from the

surface area described by 𝑦(𝑥) = 𝐻√𝑥 𝐿⁄ as shown.

Ts

T∞

U∞

Top View

H

L

y

x

L

xHy

13. Fluid flows over a semi-infinite flat plate which is maintained at uniform surface temperature. It is desired to double the rate of heat transfer from a circular area of radius 𝑅1 by increasing its radius to 𝑅2. Determine the percent increase in radius needed to accomplish this change. In both cases the circle is tangent to the leading edge. Assume laminar boundary layer flow with constant properties.

U∞

T∞

TsR1

R2

14. Consider laminar boundary layer flow over a flat plate at a uniform temperature 𝑇𝑠. When the Prandtl number is very high the viscous boundary layer is much thicker than the thermal boundary layer. Assume that the thermal boundary layer is entirely within the part of the velocity boundary layer in which the velocity profile is approximately linear. Show that for such approximation the Nusselt number is given by

𝑁𝑢 = 0.339𝑃𝑟1 3⁄ 𝑅𝑒1 2⁄

Note: ∫ 𝑒𝑥𝑝(−𝑐𝑥3)𝑑𝑥∞

0=

𝛤(1 3⁄ )

3𝑐1 3⁄ , where 𝛤 is the Gamma function.

15. A semi infinite plate is heated with uniform flux 𝑞′′ along its length. The free stream temperature

is 𝑇∞ and free stream velocity is 𝑈∞. Since the heat transfer coefficient varies with distance along the plate, Newton’s law of cooling requires that surface temperature must also vary to maintain uniform heat flux. Consider the case of laminar boundary layer flow over a plate whose surface temperature varies according to: 𝑇𝑠(𝑥) − 𝑇∞ = 𝐶𝑥𝑛. Working with the solution to this case, show that 𝑛 = 1 2⁄ corresponds to a plate with uniform surface flux.


16. Water flows over a semi-infinite flat plate which is maintained at a variable surface temperature𝑇𝑠

given by: 𝑇𝑠(𝑥) − 𝑇∞ = 𝐶𝑥0.75 where 𝐶 = 54.27 °𝐶 𝑚0.75⁄ , 𝑇∞ is the free stream temperature =3°𝐶, and 𝑥 is the distance from the leading edge, 𝑚. Determine the average heat transfer coefficient for a plate if length 𝐿 = 0.3 𝑚. The free stream velocity is 1.2 𝑚 𝑠𝑒𝑐⁄ .

17. Air flows over a plate which is heated non-uniformly such that its surface temperature increases

linearly as the distance from the leading edge is increased according to: 𝑇𝑠(𝑥) − 𝑇∞ = 𝐶𝑥 where 𝐶 = 24 °𝐶 𝑚⁄ , 𝑇∞ is the free stream temperature = 20°𝐶, and 𝑥 is the distance from the leading edge, 𝑚. Determine the total heat transfer rate from a square plate 10 𝑐𝑚 𝑥 10 𝑐𝑚. The free stream velocity is 3.2 𝑚 𝑠𝑒𝑐⁄ .

18. The surface temperature of a plate varies with distance from the leading edge according to the

relationship: 𝑇𝑠(𝑥) − 𝑇∞ = 𝐶𝑥0.8. Two identical triangles are drawn on the surface as shown. Fluid at uniform upstream temperature 𝑇∞ and uniform upstream velocity 𝑈∞ flows over the plate. Assume laminar boundary layer flow. Determine the ratio of the heat transfer rate from the two triangles, 𝑞1 𝑞2⁄ .

L

Top View

Ts

T∞

U∞

1

2H

19. Construct a plot showing the variation of 𝑁𝑢𝑥 √𝑅𝑒𝑥⁄ with wedge angle. 𝑁𝑢𝑥 is the local Nusselt

number and 𝑅𝑒𝑥 is the local Reynolds number. Assume laminar boundary layer flow of air. 20. Consider laminar boundary layer flow over a wedge. Show that the average Nusselt number 𝑁𝑢

for a wedge of length 𝐿 is given by

𝑁𝑢 =2

𝑚 + 1

𝑑𝜃(0)

𝑑𝜂𝑅𝑒𝐿

where the Reynolds number is defined as: 𝑅𝑒𝐿 =𝐿𝑈∞(𝐿)

𝜈.

21. Compare the total heat transfer rate from a 90° wedge, 𝑞𝑤 with that from a flat plate, 𝑞𝑝 of the

same length. Construct a plot of 𝑞𝑤 𝑞𝑝⁄ as a function of Prandtl number.

22. Consider laminar boundary layer flow over a wedge at a uniform temperature 𝑇𝑠. When the

Prandtl number is very high the viscous boundary layer is much thicker than the thermal boundary layer. Assume that the velocity profile within the thermal boundary layer is approximately linear. Show that for such approximation the local Nusselt number is given by:

𝑁𝑢𝑥 = 0.489[(𝑚 + 1)𝐹′′(0)𝑃𝑟]1 3⁄ 𝑅𝑒1 2⁄

Note: ∫ 𝑒𝑥𝑝(−𝑐𝑥3)𝑑𝑥∞

0=

𝛤(1 3⁄ )

3𝑐1 3⁄ , where 𝛤 is the Gamma function.

chapter 6 laminar external flow - mathworks · dr. m. azzazy proprietary notes egme 526 fall 2017...

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