Chapter 6 Laminar External Flow

Transcription

1 Chapter 6 aminar Eternal Flow Contents 1 Thermal Boundary ayer 1 2 Scale analysis Case 1: δ t > δ (Thermal B.. is larger than the velocity B..) Case 2: δ t < δ (Thermal B.. is smaller than the velocity B..) 4 3 Summary for B.. equations 5 4 Semi-infinite Plate: constant free stream velocity, constant surface temperature 6 5 Pohlhausen solution 7 6 Eamples Eample Eample Eample Variable Surface Temperature 13 8 Faulkner-Skan flows 16 9 Problems 19 1 Thermal Boundary ayer 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 eternal 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 δ t (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 aial momentum diffusion relative to normal momentum diffusion ( 2 u 2 u 2 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 > 1). Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 1

2 Pe = Re. Pr = U α (1) Where U is the free stream velocity, is a characteristic length and α is the thermal diffusivity. U T f y u d d t Figure 1. Velocity and Thermal Boundary ayer The boundary conditions are: 1) No slip velocity at the surface (u = at y = ), 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, T f = at y =, 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 T + v = α T y ( T 2 y 2) (2) Since in the velocity boundary layer we neglected 2 u relative to 2 u 2 y2, we wish to perform an analysis that would allow us to neglect aial conduction 2 T 2 relative to conduction in the normal direction. If we eamine Figure 1, we find that the value of the thermal boundary layer thickness is much smaller than the aial length of the plate. Therefore thermal gradients along the normal direction will be much larger than the aial direction, and therefore aial conduction could be neglected, i.e. And the energy equation (equation 2) becomes 2 T 2 T (3) 2 y 2 u T T + v = α 2 T (4) y y 2 Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 2

3 2.1 Case 1: δ t > δ (Thermal B.. is larger than the velocity B..) Figure 2 is a schematic illustration of this case. In this case, the velocity component u inside the thermal B.. could be approimated as u~u (5) T f y U d t u d Figure 2. Case 1, Thermal B.. larger than velocity B.. And from the continuity equation, the velocity component v could be approimated as given in equation 6. Since, v = u u then v = dy but since from equation 5 u is constant, then y v~u δ t (6) Therefore the convective terms in the energy equation (4) become u T ~U T And v T y ~U T In equation 8, notice that y is of the order of δ t and relationship 6 was used to derive relationship 8. Since δ t is much less than, then v could be neglected relative to u, and the energy equation 4 could be simplified to u T ~α 2 T y 2 Therefore And U T ~α T δ t 2 δ t = α = ρα U μ μ = 1 ρu 1 = 1 Pr Re Pe Therefore, for large Pe number, the thermal boundary layer thickness is small. (7) (8) (9) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 3

4 The ratio of the thermal boundary layer thickness to the velocity boundary layer thickness is determined from Therefore case 1, δ t > δ means that Pr 1 δ t = 1 δ Pr In summary 1) for δ t to be small relative to a characteristic length, then Pe must be >> 1, and 2) for the thermal B.. to be larger than the velocity B.., then the square root of Pr number must be much less than one. (1) 2.2 Case 2: δ t < δ (Thermal B.. is smaller than the velocity B..) Figure 3 is a schematic illustration of this case. In this case, the aial velocity within the thermal boundary layer is smaller than the free stream velocity. Assuming the velocity profile could be approimated as a linear profile in the small thermal B.. region, then Using equation 1 to scale the continuity equation u~u δ t δ v~u δ t 2 This means that v could be neglected relative to u in the energy equation. δ (11) (12) U T f y u d d t Figure 3. Case 2, Thermal B.. smaller than velocity B.. The energy equation (4) becomes Using equation 11 And Finally u T ~α 2 T y 2 U δ t δ T ~α T δ t 2 ( δ t )3 = α δ = ρα μ 1 = = 1 1 U μ ρu Re Pr Re Re Pr Re 3 2 Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 4

5 δ t = 1 1 Pr 1 3 Re (13) The ratio of the thermal B.. to the velocity B.. in this case is δ t = 1 δ Pr 1 3 (14) In summary, 1) For the thermal B.. to be small compared to a characteristic length, the product Pr 1 3 Re must be >> 1. 2) The second condition, for the thermal B.. to be smaller than the velocity B.. then Pr Summary for B.. 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 > 1) 8. High Peclet number (Pe > 1) 9. Steady state 1. aminar flow 11. No dissipation Continuity equation X-momentum Energy u + v = (15) y u u u + v = 1 y ρ dp d + μ ρ 2 u y 2 (16) u T T + v = α 2 T (17) y y 2 Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 5

6 4 Semi-infinite Plate: constant free stream velocity, constant surface temperature Assume that the plate is maintained at a constant temperature, and the fluid temperature in the free stream is T f as schematically shown in Figure 4. U T f y u d d t Figure 4. Momentum and Thermal Boundary ayers The continuity and momentum equations (equations 15 and 16) were solved in Chapter 4 using the eact Blasius solution such that: Similarity parameter η = y ν (18) Velocity u component Velocity v component And the governing equation is u v μ = f (η) (19) = 1 2 ρ (ηf f) (2) ff + 2f = (21) Equation 21 was solved by Blasius and the results are tabulated (Table 1) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 6

7 Table 1. Blasius Solution Blasius Solution Eta f f' f'' Pohlhausen solution et θ be defined as θ = T T f (22) Where is the surface temperature and T f is the free stream fluid temperature as shown in Figure 4. Substituting equation 22 into the energy equation (17) u θ θ + v = α 2 θ (23) y y 2 The boundary conditions are: θ(, ) = (24) θ(, ) = 1 (25) θ(, y) = 1 (26) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 7

8 In order to solve equation 23 using the similarity method, the two independent variables and y are combined to form the similarity variable η(, y) as defined by equation 18. The solution of equation 23 then becomes θ(, y) = θ(η) From calculus θ = η 2 dθ (27) θ = y ρu dθ μ (28) And 2 θ = ρu d 2 θ y 2 μ 2 (29) Substituting equations into equation 23 then θ + Pr dθ f(η) = (3) 2 Subject to the boundary conditions θ() = (31) θ( ) = 1 (32) Notice that the three boundary conditions are now two since we used the similarity variable to combine the and y independent variables. Equation 3 could be written as Which upon integration becomes θ = Pr f(η) (33) θ 2 η θ = C 1 Ep ( Pr f(η) ) (34) 2 Integrating equation 34 one more time, the temperature distribution becomes η θ = {C 1 [Ep ( Pr η f(η) )] } + C 2 2 (35) Using the boundary condition, equation 31, then C 2 =. Using the boundary condition, equation 32, then C 1 = 1 [Ep( Pr 2 η f(η) )] (36) Therefore the temperature distribution becomes η Equation 37 has the integral f(η) Which upon integration becomes η 2 η f(η) )] [Ep( Pr 2 η f(η) )] θ = [Ep( Pr in it. Using the momentum equation (equation 21), then f = 2 f f (37) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 8

9 η f(η) = 2ln [ f (η) ] (38) f () Substituting equation 38 into equation 37 Which upon further manipulation becomes θ(η) = η [Ep{(Pr)ln[f (η) f () ]}] [Ep{(Pr)ln[ f (η) f () ]}] η θ(η) = [f (η)] Pr [f (η)] Pr (39) (4) Numerical calculations of the temperature profile using equation 4 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)] Pr η [f (η)] Pr (4,a) Equation (4,a) is integrated using the trapezoidal rule. Results for different Pr numbers are shown in Figure 5, and tabulated in an ecel file accompanying this Chapter Pohlhausen_Table.ls. Figure 5. Pohlhausen temperature profile as a function of Pr number The value of dθ() is important in the calculation the Nusselt Modulus. Differentiating equation 4 one obtains dθ() = [f Pr ()] = (.332)Pr [f (η)] Pr [f (η)] Pr (41) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 9

10 The integrals in equation 41 are evaluated numerically and shown in Table 2. Table 2 Pr dθ() From the table, the following equations give a good approimation of dθ() dθ() =.5Pr < Pr <.5 (42) dθ() =.332Pr < Pr < 1 (43) dθ() =.339Pr 1 3 Pr > 1 (44) The heat transfer coefficient is determined from h( ) = k T(,) y (45) Since Then T(,) y T(,) y = dt dθ() η dθ y = ( ) ρ dθ() μ (46) (47) Substituting equation 47 into equation 45 h() = k ρ dθ() μ (48) The average heat transfer coefficient for a plate of length is calculated from h = 1 h()d (49) Substituting equation 48 into equation 49 and integrating Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 1

11 The local Nusselt number is obtained from Nu() = h h = 2 k Re dθ() k = Re dθ() (5) (51) And the average Nusselt number is dθ() Nu = 2 Re (52) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 11

12 6 Eamples 6.1 Eample 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 flu 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 flu at 8 cm from leading edge The average temperature is T ave = = 55 C Water properties at the average temperature are: k =.657 W m. C Pr = 3 ν = m 2 sec At 8 cm, the Reynolds number is calculated such that Re = U 2 = 2(.8) ν For Pr = 3, the value of dθ() dθ() is calculated from: =.332Pr1 3 =.4788 The heat transfer coefficient is calculated from 6 = 336,984 h = k dθ() ν =.657 2(.8).4788 = 2, The heat flu at 8 cm is calculated from: q = h( T f ) = 2,26.8(85 25) = 135,652 W m 2 The total heat transfer coefficient for the first 8 cm is h = 2 k Re dθ() = 4, W m 2 Therefore, total heat transfer for the first 8 cm is calculated from q = h ( T f ) = 271,33.38 W m 2 At 8 cm from the leading edge, the Reynolds number becomes 3,369,84 which means that the flow is turbulent (Reynolds number for transition between laminar and turbulent flow is 5,). Pohlhausen solution is only valid for laminar flow. 6.2 Eample 2 An isosceles triangle is drawn on a semi-infinite flat plate at a uniform surface temperature. 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() = k ρ dθ() μ The elemental area is b()d = H The local rate of heat transfer is then given by q() = h()( )b()d = ( )k ρ μ dθ() H d d b() H Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 12

13 Which upon rearranging becomes q() = ( )k ρ H μ 1 2 d which could be integrated to determine the rate of heat transfer between the triangle area and the fluid such that q = ( )k ρ μ dθ() H 1 2 d dθ() = 2 3 ( )kh dθ() Re 6.3 Eample 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 = 11,. Surface temperature was found to be 145 C. However, reliability requires that surface temperature does not eceed 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 = 2 k Re dθ() The amount of heat removed by the original design is Q = h A( ) = 12h A 1 When the velocity is multiplied by a factor of 3, the Reynolds number becomes Re 2 = 3Re 1 The heat transfer coefficient becomes h 2 = 2 k Re dθ() 2 = 3 h 1 The amount of heat to be removed from the package is the same. Therefore Q = h A(T 2 s2 25) = 12h A 1 Therefore, the surface temperature under the new design is 2 25 = 12 3 = The surface temperature under the new design is 2 = C which eceeds that of the reliability requirement. 7 Variable Surface Temperature A self similar class of solutions eists for problems where the surface temperature varies as () = + C n (53) Where C and the eponent n are constants. The self-similar boundary layer solution is summarized as The energy equation is given u = f (η); η = y ν ; v = 1 2 ν (ηf f) (54) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 13

14 u T + v T y = α 2 T y 2 (55) Subject to the boundary conditions T(, ) = (56) T(, ) = (57) T(, y) = (58) Introducing the non-dimensional temperature variable θ = T (59) Substituting equation 53 into equation 59 T = + C n C n θ (6) Consider the variation of the temperature along the -direction; T T = Cnn 1 Cn n 1 θ C n dθ d (61) From equation 54 = η d 2. Therefore equation 61 becomes T = Cnn 1 Cn n 1 θ + C n 1 dθ η = 2 Cnn 1 (1 θ + η θ ) (62) 2n The y-derivative of the temperature is T y dθ = Cn = dy Cn θ ν The second derivative of the temperature relative to y is 2 T = y 2 y ( Cn θ ) = ν Cn θ = ν Cn 1 θ ν (63) (64) Substituting equations 54, 6, 62, 63, and 64 into the energy equation 55 f Cn n 1 (1 θ + η 2n θ ) 2 ν (ηf f)c n θ ν = α ν C n 1 θ Upon rearranging f n (1 θ + η 2n θ ) 1 2 (ηf f)θ = 1 Pr θ Which could be further reduced to Which finally becomes Subject to the boundary conditions f n(1 θ) f ηθ 1 2 ηf θ fθ = 1 Pr θ θ + npr(1 θ)f + Pr 2 fθ = (65) θ() = (66) θ( ) = 1 (67) The value of the heat transfer coefficient is of special interest. Again applying the Nusselt Modulus one obtains Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 14

15 d_theta()/d_eta Substituting equations 53, and 63 into equation 68 Therefore h( ) = k T(,) y hc n = kc n dθ() ν h() = k Re dθ() (68) (69) and the local Nusselt number is Nu() = Re dθ() (7) The average heat transfer coefficient over a distance is given by h = 2k Re dθ() And the average Nusselt number is given by dθ() Nu = 2 Re (71) (72) The values of dθ() for different n and Pr are given in Figure n Figure 6. Surface temperature gradient for plate with varying surface Temperature, () = C n Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 15

16 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: = c m (73) Where m is related to the cone angle b through the relationship m = β 2 β (74) =C m y p Special cases 1. β = m = flow over a flat plate (75) = 2. β = 1 m = 1 stagnation flow =1 3. β < 1 flow over a wedge (eternal flow) p 4. β > 1 flow over a corner (internal flow) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 16

17 >1 The continuity equation may be written as u + v = (76) y And the momentum equation is Where The boundary conditions are u u + v u = U d y + ν 2 u (77) d y 2 = c m (78) u(, ) = (79) v(, ) = (8) u(, ) = c m (81) The solution of equations is obtained by the method of similarity. The similarity variable is defined as η = y ν = y c ν m 1 2 (82) The velocity distribution is given by u = F (η) (83) From the continuity, the velocity component v is given by v = ν (m m ) (F 1+m ηf ) (84) Substituting equations into equation 77, one obtains F + m+1 2 FF mf 2 + m = (85) Subject to the boundary conditions F () = (86) F() = (87) F ( ) = 1 (88) Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 17

18 To determine the temperature distribution, consider the energy equation u θ + v θ y = α 2 θ y 2 (89) Subject to the boundary conditions θ(, ) = (9) θ(, ) = 1 (91) θ(, y) = 1 (92) Where the non-dimensional temperature variable θ = T (93) Using the similarity variable, the energy equation becomes d 2 θ + Pr 2 2 (m + 1)F(η) dθ = (94) As usual, we are interested in the value of dθ() to calculate the heat transfer coefficient and the Nusselt number. These values are given in Table 3. Table 3.Values for dθ() for various Pr numbers Surface temperature gradient dθ() and surface velocity gradient F () for flow over an isothermal wedge m Wedge at five values of Pr F () angle πβ π 5 (36 ) π 2 (96 ) π (18 ) dθ() Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 18

19 9 Problems 1. Fluid flows between two parallel plates. It enters with uniform velocity and temperature. The plates are maintained at uniform surface temperature. Assume laminar boundary layer flow at the entrance. Can Pohlhausen solution be applied to determine the heat transfer coefficient? Eplain y 2. Two identical rectangles, A and B, of dimensions 1 2 are drawn on the surface of a semi-infinite flat plate as shown. Rectangle A is oriented with side 1 along the leading edge while rectangle B is oriented with side 2 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? 2 A 1 B 1 2 Top View 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 Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 19

20 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 epression for the ratio of the thermal boundary layer thickness for two fluids. The Prandtl number of one fluid is 1. and its kinematic viscosity is m 2 sec. The Prandtl number of the second fluid is 1 and its kinematic viscosity is m 2 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 flu 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 flu at a distance of 8 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 11,. Surface temperature was found to be 145 C. However, reliability requires that surface temperature does not eceed 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. cap 8. A fluid with Prandtl number.98 flows over a semi-infinite flat plate. The free stream temperature is and the free stream velocity is. The surface of the plate is maintained at uniform temperature. 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 1 and 2. The width of the plate is W. c. Derive an equation for the thermal boundary layer thickness δ t (). Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 2

21 y W 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, q 1 q H Top View 1. An isosceles triangle is drawn on a semi-infinite flat plate at a uniform surface temperature. Consider laminar uniform flow of constant properties fluid over the plate. Determine the rate of heat transfer between the triangular area and the fluid. 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. r r Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 21

22 12. Consider steady, two-dimensional, laminar boundary layer flow over a semi-infinite plate. The surface is maintained at uniform temperature. Determine the total heat transfer rate from the surface area described by y() = H as shown. y y H H Top View 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 R 1 by increasing its radius to R 2. 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. R 1 R Consider laminar boundary layer flow over a flat plate at a uniform temperature. 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 approimately linear. Show that for such approimation the Nusselt number is given by Nu =.339Pr 1 3 Re 1 2 Note: ep( c 3 )d = Γ(1 3) 3c 1 3, where Γ is the Gamma function. 15. A semi infinite plate is heated with uniform flu q along its length. The free stream temperature is and free stream velocity is. 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 flu. Consider the case of laminar boundary layer flow over a plate whose surface temperature varies according to: () = C n. Working with the solution to this case, show that n = 1 2 corresponds to a plate with uniform surface flu. Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 22

23 16. Water flows over a semi-infinite flat plate which is maintained at a variable surface temperature given by: () = C.75 where C = C m.75, is the free stream temperature = 3 C, and is the distance from the leading edge, m. Determine the average heat transfer coefficient for a plate if length =.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: () = C where C = 24 C m, is the free stream temperature = 2 C, and is the distance from the leading edge, m. Determine the total heat transfer rate from a square plate 1 cm 1 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: () = C.8. Two identical triangles are drawn on the surface as shown. Fluid at uniform upstream temperature and uniform upstream velocity flows over the plate. Assume laminar boundary layer flow. Determine the ratio of the heat transfer rate from the two triangles, q 1 q H Top View 19. Construct a plot showing the variation of Nu Re with wedge angle. Nu is the local Nusselt number and Re is the local Reynolds number. Assume laminar boundary layer flow of air. 2. Consider laminar boundary layer flow over a wedge. Show that the average Nusselt number Nu for a wedge of length is given by Nu = 2 dθ() m + 1 Re where the Reynolds number is defined as: Re = (). 21. Compare the total heat transfer rate from a 9 wedge, q w with that from a flat plate, q p of the same length. Construct a plot of q w q p as a function of Prandtl number. 22. Consider laminar boundary layer flow over a wedge at a uniform temperature. 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 approimately linear. Show that for such approimation the local Nusselt number is given by: Nu =.489[(m + 1)F ()Pr] 1 3 Re 1 2 Note: ep( c 3 )d = Γ(1 3) 3c 1 3, where Γ is the Gamma function. ν Dr. M. Azzazy Proprietary Notes EGME 526 Fall 217 Page 23