§6.2 Laminar incompressible boundary layer on a flat surface Figure 6.11 Comparison of the third-degree polynomial fit with the exact b.l velocity profile (Notice that the approximate result has been forced to u/u∞ = instead of 0.99 at y = δ.) We integrate this using the b.c δ2 = at x = 0: δ2 = 280 νx 13 u∞ or δ 4.64 = x Rex (6.31) This b.l thickness is of the correct functional form, and the constant is low by only 5.6% The skin friction coefficient The fact that the function f (η) gives all information about flow in the b.l must be stressed For example, the shear stress can be obtained from it 289 Laminar and turbulent boundary layers 290 §6.2 by using Newton’s law of viscous shear: τw =µ ∂u ∂y =µ y=0 ∂ u∞ f ∂y √ u ∞ d2 f =µu∞ √ νx dη2 = µu∞ y=0 df ∂η dη ∂y y=0 η=0 But from Fig 6.10 and Table 6.1, we see that (d2 f /dη2 )η=0 = 0.33206, so µu∞ Rex (6.32) τw = 0.332 x The integral method that we just outlined would have given 0.323 for the constant in eqn (6.32) instead of 0.332 (Problem 6.6) The local skin friction coefficient, or local skin drag coefficient, is defined as Cf ≡ 0.664 τw = Rex ρu∞ /2 (6.33) The overall skin friction coefficient, C f , is based on the average of the shear stress, τw , over the length, L, of the plate ⌠L ⌠L ρu2 ρu2  1 0.664 ν ∞ ∞ ⌡ dx = 1.328 τ w = ⌡ τw dx = L 2L u∞ x/ν u∞ L so Cf = 1.328 ReL (6.34) As a matter of interest, we note that Cf (x) approaches infinity at the leading edge of the flat surface This means that to stop the fluid that first touches the front of the plate—dead in its tracks—would require infinite shear stress right at that point Nature, of course, will not allow such a thing to happen; and it turns out that the boundary layer analysis is not really valid right at the leading edge In fact, the range x 5δ is too close to the edge to use this analysis with accuracy because the b.l is relatively thick and v is no longer u With eqn (6.2), this converts to x > 600 ν/u∞ for a boundary layer to exist Laminar incompressible boundary layer on a flat surface §6.2 or simply Rex 600 In Example 6.2, this condition is satisfied for all x’s greater than about mm This region is usually very small Example 6.3 Calculate the average shear stress and the overall friction coefficient for the surface in Example 6.2 if its total length is L = 0.5 m Compare τ w with τw at the trailing edge At what point on the surface does τw = τ w ? Finally, estimate what fraction of the surface can legitimately be analyzed using boundary layer theory Solution Cf = 1.328 = Re0.5 1.328 = 0.00607 47, 893 and τw = ρu2 1.183(1.5)2 ∞ 0.00607 = 0.00808 kg/m·s2 Cf = 2 N/m2 (This is very little drag It amounts only to about 1/50 ounce/m2 ) At x = L, τw (x) τw x=L = ρu2 /2 ∞ ρu2 /2 ∞ 0.664 1.328 ReL ReL = and τw (x) = τ w where 1.328 0.664 √ = √ x 0.5 so the local shear stress equals the average value, where x= m or x = L Thus, the shear stress, which is initially infinite, plummets to τ w onefourth of the way from the leading edge and drops only to one-half of τ w in the remaining 75% of the plate The boundary layer assumptions fail when x < 600 1.566 × 10−5 ν = 0.0063 m = 600 u∞ 1.5 Thus, the preceding analysis should be good over almost 99% of the 0.5 m length of the surface 291 292 Laminar and turbulent boundary layers 6.3 §6.3 The energy equation Derivation We now know how fluid moves in the b.l Next, we must extend the heat conduction equation to allow for the motion of the fluid This equation can be solved for the temperature field in the b.l., and its solution can be used to calculate h, using Fourier’s law: h= Tw q k ∂T =− − T∞ Tw − T∞ ∂y (6.35) y=0 To predict T , we extend the analysis done in Section 2.1 Figure 2.4 shows an element of a solid body subjected to a temperature field We allow this volume to contain fluid with a velocity field u(x, y, z) in it, as shown in Fig 6.12 We make the following restrictive approximations: • The fluid is incompressible This means that ρ is constant for each tiny parcel of fluid; we shall make the stronger approximation that ρ is constant for all parcels of fluid This approximation is reasonable for most liquid flows and for gas flows moving at speeds less than about 1/3 the speed of sound We have seen in Sect 6.2 that ∇· u = for incompressible flow • Pressure variations in the flow are not large enough to affect thermodynamic properties From thermodynamics, we know that the ˆ ˆ ˆ specific internal energy, u, satisfies du = cv dT + (∂ u/∂p)T dp ˆ ˆ ˆ and that the specific enthalpy, h = u + p/ρ, satisfies dh = cp dT + ˆ (∂ h/∂p)T dp We shall neglect the dp contributions to both energies We have already neglected the effect of p on ρ • Temperature variations in the flow are not large enough to change k significantly; we have already neglected temperature effects on ρ • Potential and kinetic energy changes are negligible in comparison to thermal energy changes Since the kinetic energy of a fluid can change owing to pressure gradients, this again means that pressure variations may not be too large • The viscous stresses not dissipate enough energy to warm the fluid significantly The energy equation §6.3 293 Figure 6.12 Control volume in a heat-flow and fluid-flow field Just as we wrote eqn (2.7) in Section 2.1, we now write conservation of energy in the form d dt R ˆ ρ u dR = − rate of internal energy increase in R S ˆ (ρ h) u · n dS rate of internal energy and flow work out of R − S (−k∇T ) · n dS + net heat conduction rate out of R ˙ q dR (6.36) R rate of heat generation in R In the third integral, u · n dS represents the volume flow rate through an element dS of the control surface The position of R is not changing in time, so we can bring the time derivative inside the first integral If we then we call in Gauss’s theorem [eqn (2.8)] to make volume integrals of the surface integrals, eqn (6.36) becomes ρ R ˆ ∂u ˆ ˙ + ρ∇ · (u h) − ∇ · k∇T − q dR = ∂t Because the integrand must vanish identically (recall the footnote on pg 55 in Chap 2) and because k depends weakly on T , ρ ˆ ∂u ˆ + ∇ · uh ∂t ˙ − k∇2 T − q = ˆ ˆ = u · ∇h + h ∇ · u = 0, by continuity 294 Laminar and turbulent boundary layers §6.3 Since we are neglecting pressure effects and density changes, we can approximate changes in the internal energy by changes in the enthalpy: p ˆ ˆ du = dh − d ρ ˆ ≈ dh ˆ Upon substituting dh ≈ cp dT , it follows that ρcp ∂T ∂t + u · ∇T energy storage = k∇2 T + heat conduction enthalpy convection ˙ q (6.37) heat generation This is the energy equation for an incompressible flow field It is the same as the corresponding equation (2.11) for a solid body, except for the enthalpy transport, or convection, term, ρcp u · ∇T Consider the term in parentheses in eqn (6.37): ∂T ∂T ∂T ∂T DT ∂T + u · ∇T = +u +v +w ≡ ∂t ∂t ∂x ∂y ∂z Dt (6.38) DT /Dt is exactly the so-called material derivative, which is treated in some detail in every fluid mechanics course DT /Dt is the rate of change of the temperature of a fluid particle as it moves in a flow field In a steady two-dimensional flow field without heat sources, eqn (6.37) takes the form u ∂T ∂T +v =α ∂x ∂y Furthermore, in a b.l., ∂ T /∂x is u ∂2T ∂2T + ∂x ∂y (6.39) ∂ T /∂y , so the b.l energy equation ∂2T ∂T ∂T =α +v ∂y ∂y ∂x (6.40) Heat and momentum transfer analogy Consider a b.l in a fluid of bulk temperature T∞ , flowing over a flat surface at temperature Tw The momentum equation and its b.c.’s can be The energy equation §6.3 295 written as u u ∂ ∂x u∞ +v u ∂ ∂y u∞ =ν ∂2 u u ∂y ∞  u   =0    u∞ y=0     u =1   u∞ y=∞    ∂  u   =0  ∂y u∞ y=∞ (6.41) And the energy equation (6.40) can be written in terms of a dimensionless temperature, Θ = (T − Tw )/(T∞ − Tw ), as u ∂Θ ∂Θ ∂2Θ +v =α ∂x ∂y ∂y   Θ(y = 0) =      Θ(y = ∞) =  ∂Θ     =0  ∂y (6.42) y=∞ Notice that the problems of predicting u/u∞ and Θ are identical, with one exception: eqn (6.41) has ν in it whereas eqn (6.42) has α If ν and α should happen to be equal, the temperature distribution in the b.l is for ν = α : T − Tw = f (η) T∞ − T w derivative of the Blasius function since the two problems must have the same solution In this case, we can immediately calculate the heat transfer coefficient using eqn (6.5): h= ∂(T − Tw ) k T∞ − T w ∂y =k y=0 ∂f ∂η ∂η ∂y η=0 but (∂ f /∂η2 )η=0 = 0.33206 (see Fig 6.10) and ∂η/∂y = u∞ /νx, so hx = Nux = 0.33206 Rex k for ν = α (6.43) Normally, in using eqn (6.43) or any other forced convection equation, properties should be evaluated at the film temperature, Tf = (Tw +T∞ )/2 296 Laminar and turbulent boundary layers §6.4 Example 6.4 Water flows over a flat heater, 0.06 m in length, under high pressure at 300◦ C The free stream velocity is m/s and the heater is held at 315◦ C What is the average heat flux? Solution At Tf = (315 + 300)/2 = 307◦ C: ν = 0.124 × 10−6 m2 /s α = 0.124 × 10−6 m2 /s Therefore, ν = α and we can use eqn (6.43) First we must calculate the average heat flux, q To this, we call Tw − T∞ ≡ ∆T and write q= L L h∆T dx = k∆T L L k∆T Nux dx = 0.332 L x so q = 2∆T 0.332 L u∞ dx νx k ReL = 2qx=L L Thus, h = 2hx=L = 0.664 0.520 0.06 2(0.06) = 5661 W/m2 K 0.124 × 10−6 and q = h∆T = 5661(315 − 300) = 84, 915 W/m2 = 84.9 kW/m2 Equation (6.43) is clearly a very restrictive heat transfer solution We now want to find how to evaluate q when ν does not equal α 6.4 The Prandtl number and the boundary layer thicknesses Dimensional analysis We must now look more closely at the implications of the similarity between the velocity and thermal boundary layers We first ask what dimensional analysis reveals about heat transfer in the laminar b.l We know by now that the dimensional functional equation for the heat transfer coefficient, h, should be h = fn(k, x, ρ, cp , µ, u∞ ) The Prandtl number and the boundary layer thicknesses §6.4 We have excluded Tw − T∞ on the basis of Newton’s original hypothesis, borne out in eqn (6.43), that h ≠ fn(∆T ) during forced convection This gives seven variables in J/K, m, kg, and s, or − = pi-groups Note that, as we indicated at the end of Section 4.3, there is no conversion between heat and work so it we should not regard J as N·m, but rather as a separate unit The dimensionless groups are then: Π1 = hx ≡ Nux k Π2 = ρu∞ x ≡ Rex µ and a new group: Π3 = µcp ν ≡ ≡ Pr, Prandtl number k α Thus, Nux = fn(Rex , Pr) (6.44) in forced convection flow situations Equation (6.43) was developed for the case in which ν = α or Pr = 1; therefore, it is of the same form as eqn (6.44), although it does not display the Pr dependence of Nux To better understand the physical meaning of the Prandtl number, let us briefly consider how to predict its value in a gas Kinetic theory of µ and k Figure 6.13 shows a small neighborhood of a point of interest in a gas in which there exists a velocity or temperature gradient We identify the mean free path of molecules between collisions as and indicate planes at y ± /2 which bracket the average travel of those molecules found at plane y (Actually, these planes should be located closer to y ± for a variety of subtle reasons This and other fine points of these arguments are explained in detail in [6.4].) The shear stress, τyx , can be expressed as the change of momentum of all molecules that pass through the y-plane of interest, per unit area: τyx = mass flux of molecules change in fluid · from y − /2 to y + /2 velocity The mass flux from top to bottom is proportional to ρC, where C, the mean molecular speed of the stationary fluid, is u or v in incompressible flow Thus, τyx = C1 ρC du dy N du and this also equals µ m dy (6.45) 297 Laminar and turbulent boundary layers 298 §6.4 Figure 6.13 Momentum and energy transfer in a gas with a velocity or temperature gradient By the same token, qy = C2 ρcv C dT dy and this also equals − k dT dy where cv is the specific heat at constant volume The constants, C1 and C2 , are on the order of unity It follows immediately that µ = C1 ρC so ν = C1 C k = C2 ρcv C so α = C2 and C γ where γ ≡ cp /cv is approximately a constant on the order of unity for a given gas Thus, for a gas, Pr ≡ ν = a constant on the order of unity α More detailed use of the kinetic theory of gases reveals more specific information as to the value of the Prandtl number, and these points are borne out reasonably well experimentally, as you can determine from Appendix A: • For simple monatomic gases, Pr = §6.4 The Prandtl number and the boundary layer thicknesses • For diatomic gases in which vibration is unexcited (such as N2 and O2 at room temperature), Pr = • As the complexity of gas molecules increases, Pr approaches an upper value of unity • Pr is most insensitive to temperature in gases made up of the simplest molecules because their structure is least responsive to temperature changes In a liquid, the physical mechanisms of molecular momentum and energy transport are much more complicated and Pr can be far from unity For example (cf Table A.3): • For liquids composed of fairly simple molecules, excluding metals, Pr is of the order of magnitude of to 10 • For liquid metals, Pr is of the order of magnitude of 10−2 or less • If the molecular structure of a liquid is very complex, Pr might reach values on the order of 105 This is true of oils made of long-chain hydrocarbons, for example Thus, while Pr can vary over almost eight orders of magnitude in common fluids, it is still the result of analogous mechanisms of heat and momentum transfer The numerical values of Pr, as well as the analogy itself, have their origins in the same basic process of molecular transport Boundary layer thicknesses, δ and δt , and the Prandtl number We have seen that the exact solution of the b.l equations gives δ = δt for Pr = 1, and it gives dimensionless velocity and temperature profiles that are identical on a flat surface Two other things should be easy to see: • When Pr > 1, δ > δt , and when Pr < 1, δ < δt This is true because high viscosity leads to a thick velocity b.l., and a high thermal diffusivity should give a thick thermal b.l • Since the exact governing equations (6.41) and (6.42) are identical for either b.l., except for the appearance of α in one and ν in the other, we expect that ν δt = fn only δ α 299 Laminar and turbulent boundary layers 300 §6.5 Therefore, we can combine these two observations, defining δt /δ ≡ φ, and get φ = monotonically decreasing function of Pr only (6.46) The exact solution of the thermal b.l equations proves this to be precisely true The fact that φ is independent of x will greatly simplify the use of the integral method We shall establish the correct form of eqn (6.46) in the following section 6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface The integral method for solving the energy equation Integrating the b.l energy equation in the same way as the momentum equation gives δt u ∂T dy + ∂x δt v ∂T dy = α ∂y δt ∂2T dy ∂y And the chain rule of differentiation in the form xdy ≡ dxy − ydx, reduces this to δt ∂uT dy − ∂x δt δt ∂u dy + T ∂x ∂vT dy − ∂y δt δt ∂T ∂v dy = α T ∂y ∂y or δt ∂uT dy + ∂x δt vT − δt ∂v ∂u + ∂x ∂y T =T∞ v|y=δt −0 = 0, eqn (6.11) dy  = α  ∂T ∂y − δt ∂T ∂y  =0 We evaluate v at y = δt , using the continuity equation in the form of eqn (6.23), in the preceeding expression: δt ∂ u(T − T∞ ) dy = ∂x ρcp −k ∂T ∂y = fn(x only) §6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface or d dx δt u(T − T∞ ) dy = qw ρcp (6.47) Equation (6.47) expresses the conservation of thermal energy in integrated form It shows that the rate thermal energy is carried away by the b.l flow is matched by the rate heat is transferred in at the wall Predicting the temperature distribution in the laminar thermal boundary layer We can continue to paraphrase the development of the velocity profile in the laminar b.l., from the preceding section We previously guessed the velocity profile in such a way as to make it match what we know to be true We also know certain things to be true of the temperature profile The temperatures at the wall and at the outer edge of the b.l are known Furthermore, the temperature distribution should be smooth as it blends into T∞ for y > δt This condition is imposed by setting dT /dy equal to zero at y = δt A fourth condition is obtained by writing eqn (6.40) at the wall, where u = v = This gives (∂ T /∂y )y=0 = These four conditions take the following dimensionless form:  T − T∞   = at y/δt = 0   Tw − T ∞        T − T∞  = at y/δt = 1    Tw − T ∞ (6.48)  d[(T − T∞ )/(Tw − T∞ )]   = at y/δt = 1   d(y/δt )       [(T − T )/(T − T )]   ∂ ∞ w ∞  = at y/δt = 0  ∂(y/δt ) Equations (6.48) provide enough information to approximate the temperature profile with a cubic function y y T − T∞ =a+b +c Tw − T ∞ δt δt +d y δt (6.49) Substituting eqn (6.49) into eqns (6.48), we get a=1 −1=b+c+d = b + 2c + 3d = 2c 301 302 Laminar and turbulent boundary layers §6.5 which gives b = −3 a=1 c=0 d= so the temperature profile is 3y T − T∞ + =1− Tw − T ∞ δt y δt (6.50) Predicting the heat flux in the laminar boundary layer Equation (6.47) contains an as-yet-unknown quantity—the thermal b.l thickness, δt To calculate δt , we substitute the temperature profile, eqn (6.50), and the velocity profile, eqn (6.29), in the integral form of the energy equation, (6.47), which we first express as u∞ (Tw − T∞ ) d dx δt u u∞ T − T∞ y d Tw − T ∞ δt α(Tw − T∞ ) =− δt d T − T∞ Tw − T ∞ d(y/δt ) (6.51) y/δt =0 There is no problem in completing this integration if δt < δ However, if δt > δ, there will be a problem because the equation u/u∞ = 1, instead of eqn (6.29), defines the velocity beyond y = δ Let us proceed for the δ will be satisfied moment in the hope that the requirement that δt Introducing φ ≡ δt /δ in eqn (6.51) and calling y/δt ≡ η, we get  d  δt δt dx  1 ηφ − η3 φ3 2 3α  − η + η3 dη  = 2 2u∞ (6.52) 3 = 20 φ− 280 φ3 Since φ is a constant for any Pr [recall eqn (6.46)], we separate variables: 2δt dδ2 dδt t = = dx dx 3α/u∞ 3 φ− φ3 20 280 §6.5 Heat transfer coefficient for laminar, incompressible flow over a flat surface Figure 6.14 The exact and approximate Prandtl number influence on the ratio of b.l thicknesses Integrating this result with respect to x and taking δt = at x = 0, we get δt = 3αx u∞ 3 φ− φ3 20 280 (6.53) But δ = 4.64x/ Rex in the integral formulation [eqn (6.31)] We divide by this value of δ to be consistent and obtain δt ≡ φ = 0.9638 δ Pr φ − φ2 /14 Rearranging this gives δt = δ 1.025 Pr1/3 − (δ2 /14δ2 ) t 1/3 1.025 Pr1/3 (6.54) The unapproximated result above is shown in Fig 6.14, along with the results of Pohlhausen’s precise calculation (see Schlichting [6.3, Chap 14]) It turns out that the exact ratio, δ/δt , is represented with great accuracy 303 Laminar and turbulent boundary layers 304 §6.5 by δt = Pr−1/3 δ 0.6 Pr 50 (6.55) So the integral method is accurate within 2.5% in the Prandtl number range indicated Notice that Fig 6.14 is terminated for Pr less than 0.6 The reason for doing this is that the lowest Pr for pure gases is 0.67, and the next lower values of Pr are on the order of 10−2 for liquid metals For Pr = 0.67, δ, but only by a δt /δ = 1.143, which violates the assumption that δt small margin For, say, mercury at 100◦ C, Pr = 0.0162 and δt /δ = 3.952, which violates the condition by an intolerable margin We therefore have a theory that is acceptable for gases and all liquids except the metallic ones The final step in predicting the heat flux is to write Fourier’s law: ∂T q = −k ∂y y=0 Tw − T∞ = −k δt ∂ T − T∞ Tw − T ∞ ∂(y/δt ) (6.56) y/δt =0 Using the dimensionless temperature distribution given by eqn (6.50), we get q = +k Tw − T∞ δt or h≡ q 3k 3k δ = = ∆T 2δt δ δt (6.57) and substituting eqns (6.54) and (6.31) for δ/δt and δ, we obtain Nux ≡ Rex hx 1/2 = 1.025 Pr1/3 = 0.3314 Rex Pr1/3 k 4.64 Considering the various approximations, this is very close to the result of the exact calculation, which turns out to be 1/2 Nux = 0.332 Rex Pr1/3 0.6 Pr 50 (6.58) This expression gives very accurate results under the assumptions on which it is based: a laminar two-dimensional b.l on a flat surface, with Tw = constant and 0.6 Pr 50 Heat transfer coefficient for laminar, incompressible flow over a flat surface §6.5 Figure 6.15 A laminar b.l in a low-Pr liquid The velocity b.l is so thin that u u∞ in the thermal b.l Some other laminar boundary layer heat transfer equations High Pr is At high Pr, eqn (6.58) is still close to correct The exact solution 1/2 Nux → 0.339 Rex Pr1/3 , Pr → ∞ (6.59) Low Pr Figure 6.15 shows a low-Pr liquid flowing over a flat plate In δ, and for all practical purposes u = u∞ everywhere within this case δt the thermal b.l It is as though the no-slip condition [u(y = 0) = 0] and the influence of viscosity were removed from the problem Thus, the dimensional functional equation for h becomes h = fn x, k, ρcp , u∞ (6.60) There are five variables in J/K, m, and s, so there are only two pi-groups They are Nux = hx k and Π2 ≡ Rex Pr = u∞ x α The new group, Π2 , is called a Péclét number, Pex , where the subscript identifies the length upon which it is based It can be interpreted as follows: Pex ≡ ρcp u∞ ∆T heat capacity rate of fluid in the b.l u∞ x = = (6.61) α k∆T axial heat conductance of the b.l 305 306 Laminar and turbulent boundary layers §6.5 So long as Pex is large, the b.l assumption that ∂ T /∂x ∂ T /∂y will be valid; but for small Pex (i.e., Pex 100), it will be violated and a boundary layer solution cannot be used The exact solution of the b.l equations gives, in this case:   Pex ≥ 100 and    1/2 or Pr 100 (6.62) Nux = 0.565 Pex     Re ≥ 104 x General relationship Churchill and Ozoe [6.5] recommend the following empirical correlation for laminar flow on a constant-temperature flat surface for the entire range of Pr: 1/2 Nux = 0.3387 Rex Pr1/3 + (0.0468/Pr)2/3 1/4 Pex > 100 (6.63) This relationship proves to be quite accurate, and it approximates eqns (6.59) and (6.62), respectively, in the high- and low-Pr limits The calculations of an average Nusselt number for the general case is left as an exercise (Problem 6.10) Boundary layer with an unheated starting length Figure 6.16 shows a b.l with a heated region that starts at a distance x0 from the leading edge The heat transfer in this instance is easily obtained using integral methods (see Prob 6.41) 1/2 Nux = 0.332 Rex Pr1/3 − (x0 /x)3/4 1/3 , x > x0 (6.64) Average heat transfer coefficient, h The heat transfer coefficient h, is the ratio of two quantities, q and ∆T , either of which might vary with x So far, we have only dealt with the uniform wall temperature problem Equations (6.58), (6.59), (6.62), and (6.63), for example, can all be used to calculate q(x) when (Tw − T∞ ) ≡ ∆T is a specified constant In the next subsection, we discuss the problem of predicting [T (x) − T∞ ] when q is a specified constant This is called the uniform wall heat flux problem Heat transfer coefficient for laminar, incompressible flow over a flat surface §6.5 Figure 6.16 A b.l with an unheated region at the leading edge The term h is used to designate either q/∆T in the uniform wall temperature problem or q/∆T in the uniform wall heat flux problem Thus, uniform wall temp.: h≡ q = ∆T ∆T L L = q dx L L h(x) dx (6.65) uniform heat flux: h≡ q = ∆T L q (6.66) L ∆T (x) dx The Nusselt number based on h and a characteristic length, L, is designated NuL This is not to be construed as an average of Nux , which would be meaningless in either of these cases Thus, for a flat surface (with x0 = 0), we use eqn (6.58) in eqn (6.65) to get h= L L 0.332 k Pr1/3 h(x) dx = L k x u∞ ν L √ x dx x Nux 1/2 = 0.664 ReL Pr1/3 k L (6.67) Thus, h = 2h(x = L) in a laminar flow, and NuL = hL 1/2 = 0.664 ReL Pr1/3 k (6.68) Likewise for liquid metal flows: 1/2 NuL = 1.13 PeL (6.69) 307 308 Laminar and turbulent boundary layers §6.5 Some final observations The preceding results are restricted to the two-dimensional, incompressible, laminar b.l on a flat isothermal wall at velocities that are not too high These conditions are usually met if: • Rex or ReL is not above the turbulent transition value, which is typically a few hundred thousand • The Mach number of the flow, Ma ≡ u∞ /(sound speed), is less than about 0.3 (Even gaseous flows behave incompressibly at velocities well below sonic.) A related condition is: • The Eckert number, Ec ≡ u2 /cp (Tw − T∞ ), is substantially less than ∞ unity (This means that heating by viscous dissipation—which we have neglected—does not play any role in the problem This assumption was included implicitly when we treated J as an independent unit in the dimensional analysis of this problem.) It is worthwhile to notice how h and Nu depend on their independent variables: 1 h or h ∝ √ or √ , x L Nux or NuL ∝ x or L, √ u∞ , ν −1/6 , (ρcp )1/3 , k2/3 √ u∞ , ν −1/6 , (ρcp )1/3 , k−1/3 (6.70) Thus, h → ∞ and Nux vanishes at the leading edge, x = Of course, an infinite value of h, like infinite shear stress, will not really occur at the leading edge because the b.l description will actually break down in a small neighborhood of x = In all of the preceding considerations, the fluid properties have been assumed constant Actually, k, ρcp , and especially µ might all vary noticeably with T within the b.l It turns out that if properties are all evaluated at the average temperature of the b.l or film temperature Tf = (Tw + T∞ )/2, the results will normally be quite accurate It is also worth noting that, although properties are given only at one pressure in Appendix A; µ, k, and cp change very little with pressure, especially in liquids Example 6.5 Air at 20◦ C and moving at 15 m/s is warmed by an isothermal steamheated plate at 110◦ C, ½ m in length and ½ m in width Find the average heat transfer coefficient and the total heat transferred What are h, δt , and δ at the trailing edge? Heat transfer coefficient for laminar, incompressible flow over a flat surface §6.5 Solution We evaluate properties at Tf = (110+20)/2 = 65◦ C Then Pr = 0.707 and ReL = 15(0.5) u∞ L = = 386, 600 ν 0.0000194 so the flow ought to be laminar up to the trailing edge The Nusselt number is then 1/2 NuL = 0.664 ReL Pr1/3 = 367.8 and h = 367.8 367.8(0.02885) k = = 21.2 W/m2 K L 0.5 The value is quite low because of the low conductivity of air The total heat flux is then Q = hA ∆T = 21.2(0.5)2 (110 − 20) = 477 W By comparing eqns (6.58) and (6.68), we see that h(x = L) = ½ h, so h(trailing edge) = (21.2) = 10.6 W/m2 K And finally, δ(x = L) = 4.92L ReL = 4.92(0.5) = 0.00396 m 386, 600 = 3.96 mm and 3.96 δ = 4.44 mm δt = √ = √ 3 Pr 0.707 The problem of uniform wall heat flux When the heat flux at the heater wall, qw , is specified instead of the temperature, it is Tw that we need to know We leave the problem of finding Nux for qw = constant as an exercise (Problem 6.11) The exact result is 1/2 Nux = 0.453 Rex Pr1/3 for Pr 0.6 (6.71) 309 Laminar and turbulent boundary layers 310 §6.5 where Nux = hx/k = qw x/k(Tw − T∞ ) The integral method gives the same result with a slightly lower constant (0.417) We must be very careful in discussing average results in the constant heat flux case The problem now might be that of finding an average temperature difference (cf (6.66)): Tw − T ∞ = L L (Tw − T∞ ) dx = L L dx qw x √ 1/3 k(0.453 u∞ /ν Pr ) x or Tw − T ∞ = qw L/k 1/2 0.6795 ReL (6.72) 1/3 Pr 1/2 1/3 (although the which can be put into the form NuL = 0.6795 ReL Pr Nusselt number yields an awkward nondimensionalization for Tw − T∞ ) Churchill and Ozoe [6.5] have pointed out that their eqn (6.63) will describe (Tw − T∞ ) with high accuracy over the full range of Pr if the constants are changed as follows: 1/2 Nux = 0.4637 Rex Pr1/3 + (0.02052/Pr)2/3 1/4 Pex > 100 (6.73) Example 6.6 Air at 15◦ C flows at 1.8 m/s over a 0.6 m-long heating panel The panel is intended to supply 420 W/m2 to the air, but the surface can sustain only about 105◦ C without being damaged Is it safe? What is the average temperature of the plate? Solution In accordance with eqn (6.71), ∆Tmax = ∆Tx=L = qL qL/k = 1/2 k Nux=L 0.453 Rex Pr1/3 or if we evaluate properties at (85 + 15)/2 = 50◦ C, for the moment, ∆Tmax = 420(0.6)/0.0278 0.453 1/2 (0.709)1/3 0.6(1.8)/1.794 × 10−5 = 91.5◦ C This will give Twmax = 15 + 91.5 = 106.5◦ C This is very close to 105◦ C If 105◦ C is at all conservative, q = 420 W/m2 should be safe— particularly since it only occurs over a very small distance at the end of the plate The Reynolds analogy §6.6 311 From eqn (6.72) we find that ∆T = 0.453 ∆Tmax = 61.0◦ C 0.6795 so Tw = 15 + 61.0 = 76.0◦ C 6.6 The Reynolds analogy The analogy between heat and momentum transfer can now be generalized to provide a very useful result We begin by recalling eqn (6.25), which is restricted to a flat surface with no pressure gradient: d dx δ u u∞ y u −1 d u∞ δ =− Cf (6.25) and by rewriting eqns (6.47) and (6.51), we obtain for the constant wall temperature case: d dx φδ u u∞ T − T∞ y d Tw − T ∞ δt = qw ρcp u∞ (Tw − T∞ ) (6.74) But the similarity of temperature and flow boundary layers to one another [see, e.g., eqns (6.29) and (6.50)], suggests the following approximation, which becomes exact only when Pr = 1: u T − T∞ δ= 1− δt Tw − T ∞ u∞ Substituting this result in eqn (6.74) and comparing it to eqn (6.25), we get − d dx δ u u∞ y u −1 d u∞ δ =− Cf =− qw ρcp u∞ (Tw − T∞ )φ2 (6.75) Finally, we substitute eqn (6.55) to eliminate φ from eqn (6.75) The result is one instance of the Reynolds-Colburn analogy:8 Cf h Pr2/3 = ρcp u∞ (6.76) Reynolds [6.6] developed the analogy in 1874 Colburn made important use of it in this century The form given is for flat plates with 0.6 ≤ Pr ≤ 50 The Prandtl number factor is usually a little different for other flows or other ranges of Pr 312 Laminar and turbulent boundary layers §6.6 For use in Reynolds’ analogy, Cf must be a pure skin friction coefficient The profile drag that results from the variation of pressure around the body is unrelated to heat transfer The analogy does not apply when profile drag is included in Cf The dimensionless group h/ρcp u∞ is called the Stanton number It is defined as follows: St, Stanton number ≡ Nux h = ρcp u∞ Rex Pr The physical significance of the Stanton number is St = actual heat flux to the fluid h∆T = ρcp u∞ ∆T heat flux capacity of the fluid flow (6.77) The group St Pr2/3 was dealt with by the chemical engineer Colburn, who gave it a special symbol: j ≡ Colburn j-factor = St Pr2/3 = Nux Rex Pr1/3 (6.78) Example 6.7 Does the equation for the Nusselt number on an isothermal flat surface in laminar flow satisfy the Reynolds analogy? Solution If we rewrite eqn (6.58), we obtain 0.332 Nux 2/3 = 1/3 = St Pr Rex Rex Pr (6.79) But comparison with eqn (6.33) reveals that the left-hand side of eqn (6.79) is precisely Cf /2, so the analogy is satisfied perfectly Likewise, from eqns (6.68) and (6.34), we get Cf 0.664 NuL 2/3 = = 1/3 ≡ St Pr ReL ReL Pr (6.80) The Reynolds-Colburn analogy can be used directly to infer heat transfer data from measurements of the shear stress, or vice versa It can also be extended to turbulent flow, which is much harder to predict analytically We shall undertake that problem in Sect 6.8 Turbulent boundary layers §6.7 Example 6.8 How much drag force does the air flow in Example 6.5 exert on the heat transfer surface? Solution From eqn (6.80) in Example 6.7, we obtain Cf = NuL ReL Pr1/3 From Example 6.5 we obtain NuL , ReL , and Pr1/3 : Cf = 2(367.8) = 0.002135 (386, 600)(0.707)1/3 so τyx = (0.002135) (0.002135)(1.05)(15)2 ρu2 = ∞ 2 = 0.2522 kg/m·s2 and the force is τyx A = 0.2522(0.5)2 = 0.06305 kg·m/s2 = 0.06305 N = 0.23 oz 6.7 Turbulent boundary layers Turbulence Big whirls have little whirls, That feed on their velocity Little whirls have littler whirls, And so on, to viscosity This bit of doggerel by the English fluid mechanic, L F Richardson, tells us a great deal about the nature of turbulence Turbulence in a fluid can be viewed as a spectrum of coexisting vortices in which kinetic energy from the larger ones is dissipated to successively smaller ones until the very smallest of these vortices (or “whirls”) are damped out by viscous shear stresses The next time the weatherman shows a satellite photograph of North America on the 10:00 p.m news, notice the cloud patterns There will be 313 ... 6.5 Air at 20 ◦ C and moving at 15 m/s is warmed by an isothermal steamheated plate at 110◦ C, ½ m in length and ½ m in width Find the average heat transfer coefficient and the total heat transferred... Table 6.1, we see that (d2 f /d? ?2 )η=0 = 0 .3 320 6, so µu∞ Rex (6. 32 ) τw = 0 .3 32 x The integral method that we just outlined would have given 0. 32 3 for the constant in eqn (6. 32 ) instead of 0 .3 32. .. Boundary layer with an unheated starting length Figure 6.16 shows a b.l with a heated region that starts at a distance x0 from the leading edge The heat transfer in this instance is easily obtained