hol29362_Ch07 11/3/2008 15:0 336 7-4 Free Convection from Vertical Planes and Cylinders Figure 7-6 Free-convection heat transfer from horizontal isothermal cylinders. 10000 1000 100 d Nu 10 1 0.1 10 −10 10 −7 10 −4 1 103 106 109 1012 GrdPr Constant-Heat-Flux Surfaces Extensive experiments have been reported in References 25, 26, and 39 for free convection from vertical and inclined surfaces to water under constant-heat-flux conditions. In such experiments, the results are presented in terms of a modified Grashof number, Gr∗: 4 ∗ gβq x Gr = Gr Nu = w [7-30] x x x kν2 where qw = q/A is the heat flux per unit area and is assumed constant over the entire plate surface area. The local heat-transfer coefficients were correlated by the following relation for the laminar range: = hx = ∗ 1/5 5 ∗ 11; = Nuxf 0.60(Grx Prf ) 10 < Grx Pr < 10 qw const [7-31] kf ∗ It is to be noted that the criterion for laminar flow expressed in terms of Grx is not the same as that expressed in terms of Grx. Boundary-layer transition was observed to begin between ∗ = × 12 × 13 × 13 14 Grx Pr 3 10 and 4 10 and to end between 2 10 and 10 . Fully developed ∗ = 14 ∗ turbulent flow was present by Grx Pr 10 , and the experiments were extended up to Grx Pr = 1016. For the turbulent region, the local heat-transfer coefficients are correlated with = ∗ 1/4 × 13 ∗ 16; = Nux 0.17(Grx Pr) 2 10 < Grx Pr < 10 qw const [7-32] # 101675 Cust: McGraw-Hill Au: Holman Pg. No.336 K/PMS 293 DESIGN SERVICES OF S4CARLISLE Title: Heat Transfer 10/e Server: Short / Normal / Long Publishing Services hol29362_Ch07 11/3/2008 15:0 CHAPTER7 Natural Convection Systems 337 All properties in Equations (7-31) and (7-32) are evaluated at the local film temperature. Although these experiments were conducted for water, the resulting correlations are shown to work for air as well. The average heat-transfer coefficient for the constant-heat-flux case may not be evaluated from Equation (7-24) but must be obtained through a separate applica- tion of Equation (7-23). Thus, for the laminar region, using Equation (7-31) to evaluate hx, 1 L h = hxdx L 0 = 5 = h 4 hx=L qw const At this point we may note the relationship between the correlations in the form of ∗ = Equation (7-25) and those just presented in terms of Grx Grx Nux. Writing Equation (7-25) as a local heat-transfer form gives m Nux = C(Grx Pr) [7-33] = ∗ Inserting Grx Grx/Nux gives 1+m = ∗ m Nux C(Grx Pr) or = 1/(1+m) ∗ m/(1+m) Nux C (Grx Pr) [7-34] Thus, when the “characteristic” values of m for laminar and turbulent flow are compared ∗ to the exponents on Grx, we obtain 1 m 1 Laminar, m = : = 4 1 + m 5 1 m 1 Turbulent, m = : = 3 1 + m 4 While the Gr∗ formulation is easier to employ for the constant-heat-flux case, we see that the characteristic exponents fit nicely into the scheme that is presented for the isothermal surface correlations. It is also interesting to note the variation of hx with x in the two characteristic regimes. = 1 For the laminar range m 4 , and from Equation (7-25) 1 h ∼ (x3)1/4 = x−1/4 x x = 1 In the turbulent regime m 3 , and we obtain 1 h ∼ (x3)1/3 = const with x x x So when turbulent free convection is encountered, the local heat-transfer coefficient is essentially constant with x. Churchill and Chu [71] show that Equation (7-28) may be modified to apply to the constant-heat-flux case if the average Nusselt number is based on the wall heat flux and the temperature difference at the center of the plate (x = L/2). The result is ∗ 1/4 1/4 0.67(GrL Pr) Nu (NuL − 0.68) = [7-35] L [1 + (0.492/Pr)9/16]4/9 where NuL = qwL/(kT ) and T = Tw − T∞ at L/2 − T∞. # 101675 Cust: McGraw-Hill Au: Holman Pg. No.337 K/PMS 293 DESIGN SERVICES OF S4CARLISLE Title: Heat Transfer 10/e Server: Short / Normal / Long Publishing Services hol29362_Ch07 11/3/2008 15:0 338 7-4 Free Convection from Vertical Planes and Cylinders EXAMPLE 7-1 Constant Heat Flux from Vertical Plate In a plant location near a furnace, a net radiant energy flux of 800 W/m2 is incident on a vertical metal surface 3.5 m high and 2 m wide. The metal is insulated on the back side and painted black ◦ so that all the incoming radiation is lost by free convection to the surrounding air at 30 C. What average temperature will be attained by the plate? Solution We treat this problem as one with constant heat flux on the surface. Since we do not know the surface temperature, we must make an estimate for determining Tf and the air properties. An ◦ approximate value of h for free-convection problems is 10 W/m2 · C, and so, approximately, qw 800 ◦ T = ≈ = 80 C h 10 Then 80 ◦ T ≈ + 30 = 70 C = 343 K f 2 ◦ At 70 C the properties of air are − 1 − − ν = 2.043 × 10 5 m2/s β = = 2.92 × 10 3 K 1 Tf ◦ k = 0.0295 W/m · CPr= 0.7 From Equation (7-30), with x = 3.5m, 4 −3 4 ∗ gβqwx (9.8)(2.92 × 10 )(800)(3.5) Gr = = = 2.79 × 1014 x kν2 (0.0295)(2.043 × 10−5)2 We may therefore use Equation (7-32) to evaluate hx: k ∗ 1/4 hx = (0.17)(Gr Pr) x x 0.0295 = (0.17)(2.79 × 1014 × 0.7)1/4 3.5 ◦ ◦ = 5.36 W/m2 · C [0.944 Btu/h · ft2 · F] In the turbulent heat transfer governed by Equation (7-32), we note that hx ∗ 1/4 4 1/4 Nux = ∼ (Gr ) ∼ (x ) k x or hx does not vary with x, and we may take this as the average value. The value of 2 ◦ h = 5.41 W/m · C is less than the approximate value we used to estimate Tf . Recalculating T , we obtain qw 800 ◦ T = = = 149 C h 5.36 Our new film temperature would be 149 ◦ T = 30 + = 104.5 C f 2 ◦ At 104.5 C the properties of air are − 1 − ν = 2.354 × 10 5 m2/s β = = 2.65 × 10 3/K Tf ◦ k = 0.0320 W/m · CPr= 0.695 # 101675 Cust: McGraw-Hill Au: Holman Pg. No.338 K/PMS 293 DESIGN SERVICES OF S4CARLISLE Title: Heat Transfer 10/e Server: Short / Normal / Long Publishing Services hol29362_Ch07 11/3/2008 15:0 CHAPTER7 Natural Convection Systems 339 Then −3 4 ∗ (9.8)(2.65 × 10 )(800)(3.5) Gr = = 1.75 × 1014 x (0.0320)(2.354 × 10−5)2 and hx is calculated from k ∗ 1/4 hx = (0.17)(Gr Pr) x x (0.0320)(0.17) = [(1.758 × 1014)(0.695)]1/4 3.5 ◦ ◦ = 5.17 W/m2 · C [−0.91 Btu/h · ft2 · F] Our new temperature difference is calculated as qw 800 ◦ T = (Tw − T∞)av = = = 155 C h 5.17 The average wall temperature is therefore ◦ Tw,av = 155 + 30 = 185 C Another iteration on the value of Tf is not warranted by the improved accuracy that would result. Heat Transfer from Isothermal Vertical Plate EXAMPLE 7-2 ◦ ◦ A large vertical plate 4.0 m high is maintained at 60 C and exposed to atmospheric air at 10 C. Calculate the heat transfer if the plate is 10 m wide. Solution We first determine the film temperature as 60 + 10 ◦ T = = 35 C = 308 K f 2 The properties of interest are thus 1 − β = = 3.25 × 10 3 k = 0.02685 308 − ν = 16.5 × 10 6 Pr = 0.7 and − (9.8)(3.25 × 10 3)(60 − 10)(4)3 Gr Pr = 0.7 (16.5 × 10−6)2 = 2.62 × 1011 We then may use Equation (7-29) to obtain 11 1/6 1/2 (0.387)(2.62 × 10 ) Nu = 0.825 + [1 + (0.492/0.7)9/16]8/27 = 26.75 Nu = 716 The heat-transfer coefficient is then (716)(0.02685) ◦ h = = 4.80 W/m2 · C 4.0 # 101675 Cust: McGraw-Hill Au: Holman Pg. No.339 K/PMS 293 DESIGN SERVICES OF S4CARLISLE Title: Heat Transfer 10/e Server: Short / Normal / Long Publishing Services hol29362_Ch07 11/3/2008 15:0 340 7-5 Free Convection from Horizontal Cylinders The heat transfer is q = hA(Tw − T∞) = (4.80)(4)(10)(60 − 10) = 9606 W As an alternative, we could employ the simpler relation Nu = 0.10(Gr Pr)1/3 = (0.10)(2.62 × 1011)1/3 = 639.9 which gives a value about 10 percent lower than Equation (7-29). 7-5 FREE CONVECTION FROM HORIZONTAL CYLINDERS The values of the constants C and m are given in Table 7-1 according to References 4 and 76. The predictions of Morgan (Reference 76 in Table 7-1) are the most reliable for Gr Pr of approximately 10−5. A more complicated expression for use over a wider range of Gr Pr is given by Churchill and Chu [70]: 1/6 1/2 Gr Pr − Nu = 0.60 + 0.387 for 10 5 < Gr Pr [1 + (0.559/Pr)9/16]16/9 < 1012 [7-36] A simpler equation is available from Reference 70 but is restricted to the laminar range of 10−6 < Gr Pr < 109: 1/4 0.518(Grd Pr) Nud = 0.36 + [7-37] [1 + (0.559/Pr)9/16]4/9 Properties in Equations (7-36) and (7-37) are evaluated at the film temperature.