Fundamentals of heat and mass transfer frank p incropera david p dewitt solution manual ch09

PROBLEM 9.1 KNOWN: Tabulated values of density for water and definition of the volumetric thermal expansion coefficient, β FIND: Value of the volumetric expansion coefficient at 300K; compare with tabulated values -3 3 PROPERTIES: Table A-6, Water (300K): ρ = 1/vf = 1/1.003 × 10 m /kg = 997.0 kg/m , β -6 -1 -3 3 = 276.1 × 10 K ; (295K): ρ = 1/vf = 1/1.002 × 10 m /kg = 998.0 kg/m ; (305K): ρ = 1/vf -3 3 = 1/1.005 × 10 m /kg = 995.0 kg/m ANALYSIS: The volumetric expansion coefficient is defined by Eq 9.4 as β =−  ∂ρ  ρ  ∂ T  p The density change with temperature at constant pressure can be estimated as  ρ1 − ρ2   ∂ρ   ∂ T  ≈  T −T   p  p where the subscripts (1,2) denote the property values just above and below, respectively, the condition for T = 300K denoted by the subscript (o) That is, βo ≈ −  ρ1 − ρ    ρo  T1 − T2 p Substituting numerical values, find βo ≈ −1 997.0 kg/m ( 995.0 − 998.0) kg/m3 ( 305 − 295 ) K = 300.9 × 10−6 K −1 < -6 -1 K , to find our estimate is 8.7% high Compare this value with the tabulation, β = 276.1 × 10 COMMENTS: (1) The poor agreement between our estimate and the tabulated value is due to the poor precision with which the density change with temperature is estimated The tabulated values of β were determined from accurate equation of state data (2) Note that β is negative for T < 275K Why? What is the implication for free convection? PROBLEM 9.2 KNOWN: Object with specified characteristic length and temperature difference above ambient fluid FIND: Grashof number for air, hydrogen, water, ethylene glycol for a pressure of atm SCHEMATIC: ASSUMPTIONS: (1) Thermophysical properties evaluated at Tf = 350K, (2) Perfect gas behavior, (β = 1/Tf) PROPERTIES: Evaluate at atm, Tf = 350K: -6 -6 Table A-4, Air: ν = 20.92 × 10 m /s; Hydrogen: ν = 143 × 10 m /s -6 -3 -1 Table A-6, Water (Sat liquid): = àf vf = 37.5 ì 10 m /s, β f = 0.624 × 10 K -6 -3 -1 Table A-5, Ethylene glycol: ν = 3.17 × 10 m /s, β = 0.65 × 10 K ANALYSIS: The Grashof number is given by Eq 9.12, GrL = gβ ( Ts − T∞ ) L3 ν2 Substituting numerical values for air with β = 1/Tf, find GrL,air = 9.8m/s × (1/350K ) ( 25K ) ( 0.25m )3 20.92 × 10−6 m / s ( ) GrL,air = 2.50 ×107 < Performing similar calculations for the other fluids, find GrL,hyd = 5.35 ×105 < GrL,water = 1.70 × 106 < GrL,eth = 2.48 ×108 < COMMENTS: Higher values of GrL imply increased free convection flows However, other properties affect the value of the heat transfer coefficient PROBLEM 9.3 KNOWN: Relation for the Rayleigh number FIND: Rayleigh number for four fluids for prescribed conditions SCHEMATIC: ASSUMPTIONS: (1) Perfect gas behavior for specified gases -6 PROPERTIES: Table A-4, Air (400K, atm): ν = 26.41 × 10 -3 1/T = 1/400K = 2.50 × 10 -6 10 -6 10 -1 -6 -3 -7 2 -1 -3 -3 m /s, β = m /s, α = 295 × -6 K ; Table A-5, Glycerin (12°C = 285K): ν = 2830 × 10 m /s, β = 0.475 × 10 N⋅s/m × 1.007 × 10 -6 K ; Table A-4, Helium (400K, atm): ν = 199 × 10 m /s, β = 1/T = 2.50 × 10 0.964 × 10 m /s, α = 38.3 × 10 -1 K ; Table A-6, Water (37°C = 310K, sat liq.): = àf vf = 695ì -6 m /kg = 0.700 × 10 -6 m /kg/4178 J/kg⋅K = 0.151 × 10 m /s, α = 2 -3 m /s, α = kf vf/cp,f = 0.628 W/m⋅K × 1.007 × 10 -6 m /s, βf = 361.9 × 10 -1 K ANALYSIS: The Rayleigh number, a dimensionless parameter used in free convection analysis, is defined as the product of the Grashof and Prandtl numbers Ra L ≡ Gr ⋅ Pr = gβ∆TL3 µ c p gβ∆TL3 (νρ ) c p gβ∆TL3 = ⋅ = k k να ν2 ν2 where α = k/ρcp and ν = µ/ρ The numerical values for the four fluids follow: Air (400K, atm) Ra L = 9.8m/s (1/400K ) 30K ( 0.01m )3 /26.41×10 −6 m / s × 38.3 × 10−6 m / s = 727 < Helium (400K, atm) Ra L = 9.8m/s (1/400K ) 30K ( 0.01m )3 /199 × 10−6 m / s × 295 × 10−6m / s = 12.5 Glycerin (285K) Ra L = 9.8m/s Water (310K) Ra L = 9.8m/s < ) < ) < ( 0.475 × 10 −3 −1 −6 −7 K 30K ( 0.01m )3 /2830 × 10 m / s × 0.964 × 10 m / s = 512 ( 0.362 × 10 −3 −1 −6 −6 K 30K ( 0.01m )3 /0.700 ×10 m / s × 0.151 × 10 m / s = 9.35 × 10 COMMENTS: (1) Note the wide variation in values of Ra for the four fluids A large value of Ra implies enhanced free convection, however, other properties affect the value of the heat transfer coefficient PROBLEM 9.4 KNOWN: Form of the Nusselt number correlation for natural convection and fluid properties FIND: Expression for figure of merit FN and values for air, water and a dielectric liquid -1 -5 PROPERTIES: Prescribed Air: k = 0.026 W/m⋅K, β = 0.0035 K , ν = 1.5 × 10 m /s, Pr = 0.70 -4 -1 -6 Water: k = 0.600 W/m⋅K, β = 2.7 × 10 K , ν = 10 m /s, Pr = 5.0 Dielectric liquid: k = 0.064 -1 -6 W/m⋅K, β = 0.0014 K , ν = 10 m /s, Pr = 25 ANALYSIS: With NuL ~ R a n , the convection coefficient may be expressed as h~ n k  gβ∆TL   ~ L  αν  ( g∆TL3 L ) n  k βn   α nν n      The figure of merit is therefore FN = kβn < α nν n and for the three fluids, with n = 0.33 and α = ν /Pr , ( FN W ⋅ s / / m / ⋅ K / ) Air 5.8 Water Dielectric 663 209 < Water is clearly the superior heat transfer fluid, while air is the least effective COMMENTS: The figure of merit indicates that heat transfer is enhanced by fluids of large k, large β and small values of α and ν PROBLEM 9.5 KNOWN: Heat transfer rate by convection from a vertical surface, 1m high by 0.6m wide, to quiescent air that is 20K cooler FIND: Ratio of the heat transfer rate for the above case to that for a vertical surface that is 0.6m high by 1m wide with quiescent air that is 20K warmer SCHEMATIC: ASSUMPTIONS: (1) Thermophysical properties independent of temperature; evaluate at 300K; (2) Negligible radiation exchange with surroundings, (3) Quiescent ambient air -6 PROPERTIES: Table A-4, Air (300K, atm): ν = 15.89 × 10 -6 m /s, α = 22.5 × 10 m /s ANALYSIS: The rate equation for convection between the plates and quiescent air is q = h L A s∆T (1) where ∆T is either (Ts - T∞) or (T∞ - Ts ); for both cases, As = Lw The desired heat transfer ratio is then q1 h L1 = q2 hL2 (2) To determine the dependence of hL on geometry, first calculate the Rayleigh number, Ra L = g β ∆ TL3 / να (3) and substituting property values at 300K, find, -6 Case 1: RaL1 = 9.8 m/s (1/300K) 20K (1m) /15.89 × 10 -6 m /s × 22.5 × 10 m /s = 1.82 × 10 Case 2: RaL2 = RaL1 (L2/L1) = 1.82 ×10 (0.6m/1.0m) = 3.94 × 10 Hence, Case is turbulent and Case is laminar Using the correlation of Eq 9.24, h L Nu L = L = C ( Ra L )n k hL = k C Ra nL L (4) where for Case 1: C1 = 0.10, n1 = 1/3 and for Case 2: C2 = 0.59, n2 = 1/4 Substituting Eq (4) into the ratio of Eq (2) with numerical values, find 1.82 × 109 ) ( q1 = = = 0.881 n2 1/4 q2 ( C2 / L2 ) Ra L2 ( 0.59/0.6m ) ( 3.94 ×10 ) n1 ( C1 / L1 ) Ra L1 ( 0.10/1m) 1/3 < COMMENTS: Is this result to be expected? How you explain this effect of plate orientation on the heat rates? PROBLEM 9.6 KNOWN: Large vertical plate with uniform surface temperature of 130°C suspended in quiescent air at 25°C and atmospheric pressure FIND: (a) Boundary layer thickness at 0.25 m from lower edge, (b) Maximum velocity in boundary layer at this location and position of maximum, (c) Heat transfer coefficient at this location, (d) Location where boundary layer becomes turbulent SCHEMATIC: ASSUMPTIONS: (1) Isothermal, vertical surface in an extensive, quiescent medium, (2) Boundary layer assumptions valid ( ) -6 PROPERTIES: Table A-4, Air Tf = ( Ts + T∞ ) / = 350K, atm : ν = 20.92 × 10 m /s, k = 0.030 W/m⋅K, Pr = 0.700 ANALYSIS: (a) From the similarity solution results, Fig 9.4 (see above right), the boundary layer thickness corresponds to a value of η ≈ From Eqs 9.13 and 9.12, y = η x ( Grx / 4) −1 / (1) ( ) m Grx = g β ( Ts − T∞ ) x / ν = 9.8 × (130 − 25 ) K x3 / 20.92 × 10−6m / s = 6.718 × 109 x 350K s −1 / y ≈ ( 0.25m )  6.718 ×109 ( 0.25 )3 /  = 1.746 ×10−2m = 17.5 mm   (2) (3) < (b) From the similarity solution shown above, the maximum velocity occurs at η ≈ with f ′ (η ) = 0.275 From Eq.9.15, find u= 2ν 1/2 × 20.92 ×10 −6 m / s  1 / Grx f ′ (η ) = × 0.275 = 0.47 m/s  6.718 ×10 ( 0.25 )  x 0.25m   < The maximum velocity occurs at a value of η = 1; using Eq (3), it follows that this corresponds to a position in the boundary layer given as ymax = 1/5 (17.5 mm ) = 3.5 mm < (c) From Eq 9.19, the local heat transfer coefficient at x = 0.25 m is 1/4 Nu x = h x x/k = ( Grx / )1 / g ( Pr ) =  6.718 × 109 ( 0.25)3 /  0.586 = 41.9   h x = Nu xk/x = 41.9 × 0.030 W/m ⋅ K/0.25 m = 5.0 W/m ⋅ K < The value for g(Pr) is determined from Eq 9.20 with Pr = 0.700 (d) According to Eq 9.23, the boundary layer becomes turbulent at xc given as Ra x,c = Gr x,c Pr ≈ 109 1/3 x c ≈ 10 /6.718 ×10 (0.700 )  = 0.60 m   COMMENTS: Note that β = 1/Tf is a suitable approximation for air < PROBLEM 9.7 KNOWN: Thin, vertical plates of length 0.15m at 54°C being cooled in a water bath at 20°C FIND: Minimum spacing between plates such that no interference will occur between freeconvection boundary layers SCHEMATIC: ASSUMPTIONS: (a) Water in bath is quiescent, (b) Plates are at uniform temperature PROPERTIES: Table A-6, Water (Tf = (Ts + T∞)/2 = (54 + 20)°C/2 = 310K): ρ = 1/vf = 993.05 -6 kg/m , µ = 695 ì10 -7 Ns/m , = à/ = 6.998 × 10 -6 m /s, Pr = 4.62, β = 361.9 × 10 -1 K ANALYSIS: The minimum separation distance will be twice the thickness of the boundary layer at the trailing edge where x = 0.15m Assuming laminar, free convection boundary layer conditions, the similarity parameter, η, given by Eq 9.13, is η= y ( Grx / 4)1 / x where y is measured normal to the plate (see Fig 9.3) According to Fig 9.4, the boundary layer thickness occurs at a value η ≈ It follows then that, ybl = η x ( Grx / )−1 / where Grx = g β ( Ts − T∞ ) x3 ν2 ( ) Grx = 9.8 m/s2 × 361.9 × 10 −6 K −1 ( 54 − 20 ) K × ( 0.15m )3 / 6.998 ×10−7 m2 / s = 8.310 ×108 ( < ) −1 / Hence, ybl = × 0.15m 8.310 × 108 / = 6.247 × 10−3 m = 6.3 mm and the minimum separation is d = y bl = × 6.3 mm = 12.6 mm < COMMENTS: According to Eq 9.23, the critical Grashof number for the onset of turbulent conditions in the boundary layer is Grx,c Pr ≈ 10 For the conditions above, Grx Pr = 8.31 × 10 × 4.62 = 3.8 × 10 We conclude that the boundary layer is indeed turbulent at x = 0.15m and our calculation is only an estimate which is likely to be low Therefore, the plate separation should be greater than 12.6 mm PROBLEM 9.8 KNOWN: Square aluminum plate at 15°C suspended in quiescent air at 40°C FIND: Average heat transfer coefficient by two methods – using results of boundary layer similarity and results from an empirical correlation SCHEMATIC: ASSUMPTIONS: (1) Uniform plate surface temperature, (2) Quiescent room air, (3) Surface radiation exchange with surroundings negligible, (4) Perfect gas behavior for air, β = 1/Tf -6 PROPERTIES: Table A-4, Air (Tf = (Ts + T∞)/2 = (40 +15)°C/2 = 300K, atm): ν = 15.89 × 10 -6 m /s, k = 0.0263 W/m⋅K, α = 22.5 × 10 m /s, Pr = 0.707 ANALYSIS: Calculate the Rayleigh number to determine the boundary layer flow conditions, Ra L = g β ∆T L3 /ν α Ra L = 9.8 m/s (1/300K ) ( 40 − 15 ) °C ( 0.2m )3 / ( 15.89 × 10 −6 m /s )( 22.5 × 10 ) −6 m / s = 1.827 × 10 where β = 1/Tf and ∆T = T∞ - Ts Since RaL < 10 , the flow is laminar and the similarity solution of Section 9.4 is applicable From Eqs 9.21 and 9.20, h L Nu L = L = ( GrL / )1 / g ( Pr ) k 0.75 Pr1/2 g ( Pr ) = 1/4  0.609 + 1.221 Pr1/2 + 1.238 Pr   and substituting numerical values with GrL = RaL/Pr, find 1/4 1/2  1/2 g ( Pr ) = 0.75 ( 0.707 ) / 0.609 + 1.22 ( 0.707 ) + 1.238 × 0.707  = 0.501   1/4 /0.707   0.0263 W/m ⋅ K 1.827 × 10  ×   hL =  × 0.501 = 4.42 W/m2 ⋅ K    0.20m     < The appropriate empirical correlation for estimating hL is given by Eq 9.27, 0.670 Ra1/4 hL L L Nu L = = 0.68 + 4/9 k 1 + ( 0.492/Pr )9/16    1/4 4/9  h L = ( 0.0263 W/m ⋅ K/0.20m)  0.68 + 0.670 1.827 × 10 / 1 + ( 0.492/0.707 )9/16       ( h L = 4.42 W/m2 gK ) < COMMENTS: The agreement of hL calculated by these two methods is excellent Using the Churchill-Chu correlation, Eq 9.26, find hL = 4.87 W/m ⋅ K This relation is not the most accurate for the laminar regime, but is suitable for both laminar and turbulent regions PROBLEM 9.9 KNOWN: Dimensions of vertical rectangular fins Temperature of fins and quiescent air FIND: (a) Optimum fin spacing, (b) Rate of heat transfer from an array of fins at the optimal spacing SCHEMATIC: ASSUMPTIONS: (1) Fins are isothermal, (2) Radiation effects are negligible, (3) Air is quiescent -6 PROPERTIES: Table A-4, Air (Tf = 325K, atm): ν = 18.41 × 10 0.703 m /s, k = 0.0282 W/m⋅K, Pr = ANALYSIS: (a) If fins are too close, boundary layers on adjoining surfaces will coalesce and heat transfer will decrease If fins are too far apart, the surface area becomes too small and heat transfer decreases Sop ≈ δ x=H From Fig 9.4, the edge of boundary layer corresponds to η = ( δ / H ) ( GrH / )1 / ≈ gβ ( Ts − T∞ ) H3 9.8 m/s ( 1/325K) 50K ( 0.15m )3 Hence, GrH = = = 1.5 ×10 ν H2 18.41× 10−6 m / s ( δ ( H ) = ( 0.15m ) / 1.5 ×10 / ) ( 1/4 ) = 0.017m = 17mm Sop ≈ 34mm < (b) The number of fins N can be found as ( ) N = W/ Sop + t = 355/35.5 = 10 and the rate is q = N h ( H ⋅ L ) ( Ts − T∞ ) For laminar flow conditions 4/9 1 + ( 0.492/Pr )9/16  Nu H = 0.68 + 0.67 Ra1/4 / L   ( Nu H = 0.68 + 0.67 1.5 ×107 × 0.703 ) 4/9 1/4  / 1 + ( 0.492/0.703)9/16  = 30   h = k Nu H / H = 0.0282 W/m ⋅ K ( 30 ) /0.15 m = 5.6 W/m ⋅ K q = (10 ) 5.6 W/m ⋅ K ( 0.15m × 0.02m ) ( 350 − 300 ) K = 16.8 W < COMMENTS: Part (a) result is a conservative estimate of the optimum spacing The increase in area resulting from a further reduction in S would more than compensate for the effect of fluid entrapment due to boundary layer merger From a more rigorous treatment (see Section 9.7.1), Sop ≈ 10 mm is obtained for the prescribed conditions PROBLEM 9.10 KNOWN: Interior air and wall temperatures; wall height FIND: (a) Average heat transfer coefficient when T∞ = 20°C and Ts = 10°C, (b) Average heat transfer coefficient when T∞ = 27°C and Ts = 37°C SCHEMATIC: ASSUMPTIONS: (a) Wall is at a uniform temperature, (b) Room air is quiescent -3 PROPERTIES: Table A-4, Air (Tf = 298K, atm): β = 1/Tf = 3.472 × 10 -6 m /s, k = 0.0253 W/m⋅K, α = 20.9 × 10 -3 10 -1 -6 K , ν = 16.39 × 10 -1 -6 K , ν = 14.82 × 10 m /s, Pr = 0.710; (Tf = 305K, atm): β = 1/Tf = 3.279 × -6 m /s, k = 0.0267 W/m⋅K, α = 23.2 × 10 m /s, Pr = 0.706 ANALYSIS: The appropriate correlation for the average heat transfer coefficient for free convection on a vertical wall is Eq 9.26  0.1667 hL  0.387 Ra L Nu L = = 0.825 + 0.296 k  1 + ( 0.492/Pr )0.563          where RaL = g β ∆T L /να, Eq 9.25, with ∆T = Ts - T∞ or T∞ - Ts (a) Substituting numerical values typical of winter conditions gives Ra L = −3 −1 9.8 m/s × 3.472 × 10 K ( 20 − 10 ) K ( 2.5m ) 14.82 × 10 −6 -6 m / s × 20.96 × 10 m / s ( ) 0.1667  10 0.387 1.711 × 10  Nu = 0.825 + L 0.296  1 + ( 0.492/0.710 )0.563     10 = 1.711 × 10    = 299.6   Hence, h = Nu L k/L = 299.6 ( 0.0253 W/m ⋅ K) /2.5m = 3.03 W/m2 ⋅ K < (b) Substituting numerical values typical of summer conditions gives Ra L = −3 −1 9.8 m/s × 3.279 × 10 K ( 37 − 27 ) K (2.5 m ) 23.2 × 10 −6 −6 m / s × 16.39 × 10 m / s ( ) 10 = 1.320 × 10 0.1667   0.387 1.320 × 1010   Nu L = 0.825 + = 275.8 0.296    1 + ( 0.492/0.706 )0.563      Hence, h = Nu L k/L = 275.8 × 0.0267 W/m ⋅ K/2.5m = 2.94 W/m ⋅ K < COMMENTS: There is a small influence due to Tf on h for these conditions We should expect radiation effects to be important with such low values of h PROBLEM 9.106 KNOWN: Concentric spheres with prescribed surface temperatures FIND: Convection heat transfer rate SCHEMATIC: ASSUMPTIONS: (1) Quiescent air in void space, (2) Density ~ pressure; other properties independent, (3) Perfect gas behavior -3 -1 -6 PROPERTIES: Table A-4, Air (Tf = 300K, atm): β = 3.33 × 10 K , ν = 1/3 × 15.89 × 10 -6 m /s, k = 0.263 W/m⋅K, α = 1/3 × 22.5 × 10 m /s, Pr = 0.707 ANALYSIS: The heat transfer rate due to free convection is q = k eff π ( D iDo / L )( Ti − To ) where (9.61) ( ) 1/4 k eff Pr   * 1/4 = 0.74  Ra s  k  0.861 + Pr      L Ra L Ra *s =    ( Do D i ) D−7 / + D−7 / 5  o i   ( (9.63) ) g β ( Ti − To ) L3 Ra L = (9.62) να (9.25) Substituting numerical values in the above expressions, find that Ra L = ( 9.8m/s2 × 3.333 × 10 −3 K −1 ( 325 − 375 ) K 12.5 ×10 −3 × 10− m2 / s (1 / ) 22.5 × 10− m2 / s (1/315.89 ) )m 3 = 80,928     −3 12.5 × 10 m 80,928  *  Ra s =  ⋅ = 330.1 4 5 − − − / − /    100 ×10 × 75 ×10 m  75 × 10− m + 100 × 10− m         ( k eff ( ) ) ( 1/4 0.707     0.861 + 0.707  = 0.74  k Hence, the heat rate becomes ) ( ) ( 330.1)1 / = 2.58 W   75 × 10−3 m × 100 × 10−3 m    ( 325 − 275) K = 64.0W q =  2.58 × 0.263 π  m ⋅ K    12.5 ×10 −3m  < COMMENTS: Note the manner in which the thermophysical properties vary with pressure Assuming perfect gas behavior, ρ ~ p Also, k, µ and cp are independent of pressure Hence, Pr is -1 -1 independent of pressure, but ν = µ/ρ ~ p and α = k/ρc ~ p PROBLEM 9.107 KNOWN: Cross flow over a cylinder with prescribed surface temperature and free stream conditions FIND: Whether free convection will be significant if the fluid is water or air SCHEMATIC: ASSUMPTIONS: (1) Constant properties, (2) Combined free and forced heat transfer -6 PROPERTIES: Table A-6, Water (Tf = (T∞ + Ts)/2 = 300K): ν = µ vf = 855 × 10 N⋅s/m × 1.003 -3 -7 -6 -1 × 10 m /kg = 8.576 × 10 m /s, β = 276.1 × 10 K ; Table A-4, Air (300K, atm): ν = 15.89 × 10 -3 -1 m /s, β = 1/Tf = 3.333 × 10 K ANALYSIS: Following the discussion of Section 9.9, the general criterion for delineating the relative significance of free and forced convection depends upon the value of Gr/Re If free convection is significant GrD /Re 2D ≥ where GrD = g β ( T∞ − Ts )D / ν and (1) Re D = VD/ ν (2,3) (a) When the surrounding fluid is water, find ( GrD = 9.8m/s × 276.1 ×10−6 K −1 ( 35 − 20 ) K ( 0.05m )3 / 8.576 ×10−7 m /s Re D = 0.05m/s × 0.05m/8.576 ×10−7 m / s = 2915 ) = 68, 980 GrD / Re 2D = 68,980/29152 = 0.00812 < We conclude that since GrD /Re2D > 1, free convection dominates the heat transfer process COMMENTS: Note also that for the air flow situation, surface radiation exchange is likely to be significant < PROBLEM 9.108 KNOWN: Parallel air flow over a uniform temperature, heated vertical plate; the effect of free convection on the heat transfer coefficient will be 5% when GrL /Re2L = 0.08 FIND: Minimum vertical velocity required of air flow such that free convection effects will be less than 5% of the heat rate SCHEMATIC: ASSUMPTIONS: (1) Steady-state conditions, (2) Criterion for combined free-forced convection determined from experimental results -6 PROPERTIES: Table A-4, Air (Tf = (Ts + T∞)/2 = 315K, atm): ν = 17.40 × 10 m /s, β = 1/Tf ANALYSIS: To delineate flow regimes, according to Section 9.9, the general criterion for predominately forced convection is that GrL /Re2L