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Question 12.9: Forced Airflow on Roof of Building† Consider a 10 m × 6 m at......

Forced Airflow on Roof of Building^†

Consider a 10 m × 6 m at roof of a building at sea level over which wind blows at 10 m/s. The air temperature is 20°C while the temperature of the roof is 67°C. Calculate the convective heat transfer coefficient when the wind blows along the 10 m long side.

Given: v = 10 m/s, characteristic length D = 10 m
Find: h_{con} using Equation 12.22 for forced external flow over plates
Lookup values: From property tables of air (Table A1) at a mean temperature of (20°C + 74°C)/2 = 47°C: \rho = 1.110  kg/m^{3}, \mu = 1.94 × 10^{−5}  Pa ⋅ s, c_{p} = 1007  J/(kg ⋅ K), conductivity k = 0.0275 W/(m·K)

^†Same inputs as Example 2.10 solved using a dimensional correlation.

TABLE A.1
(IP Units) Transport Properties of Standard, Dry Air (14.7 psia)
°R °F ρ (lbm/ft³) cp (Btu/lb · °R) k (c_p/c_v) μ x 104  (lbm/ft · s)
180 -280 0.2247 0.2456 0.0466
198 -262 0.2033 0.2440 1.4202 0.0513
216 -244 0.1858 0.2430 1.4166 0.0559
234 -226 0.1711 0.2423 1.4139 0.0604
252 -208 0.1586 0.2418 1.4119 0.0648
270 -190 0.1478 0.2414 1.4102 0.0691
288 -172 0.1384 0.2411 1.4089 0.0733
306 -154 0.1301 0.2408 1.4079 0.0774
324 -136 0.1228 0.2406 1.4071 0.0815
342 -118 0.1163 0.2405 1.4064 0.0854
360 -100 0.1104 0.2404 1.4057 0.0892
369 -91 0.1078 0.2403 1.4055 0.0911
378 -82 0.1051 0.2403 1.4053 0.0930
387 -73 0.1027 0.2403 1.4050 0.0949
396 -64 0.1003 0.2402 1.4048 0.0967
405 -55 0.0981 0.2402 1.4046 0.0986
414 -46 0.0959 0.2402 1.4044 0.1004
423 -37 0.0939 0.2402 1.4042 0.1022
432 -28 0.0919 0.2401 1.4040 0.1039
441 -19 0.0901 0.2401 1.4038 0.1057
450 -10 0.0882 0.2401 1.4036 0.1074
459.7 0 0.0865 0.2401 1.4034 0.1092
468 8 0.0848 0.2401 1.4032 0.1109
486 26 0.0817 0.2492 1.4029 0.1143
504 44 0.0787 0.2402 1.4024 0.1176
522 62 0.0760 0.2403 1.4020 0.1208
540 80 0.0735 0.2404 1.4017 0.1241
558 98 0.0711 0.2405 1.4013 0.1272
576 116 0.0689 0.2406 1.4008 0.1303
594 134 0.0668 0.2407 1.4004 0.1334
612 152 0.0648 0.2409 1.3999 0.1364
630 170 0.0630 0.2411 1.3993 0.1394
648 188 0.0612 0.2412 1.3987 0.1423
666 206 0.0595 0.2415 1.3981 0.1452
684 224 0.0580 0.2417 1.3975 0.1479
702 242 0.0565 0.2420 1.3968 0.1508
720 260 0.0551 0.2422 1.3961 0.1536
738 278 0.0537 0.2425 1.3953 0.1563
756 296 0.0525 0.2428 1.3946 0.1590
774 314 0.0512 0.2432 1.3938 0.1617
792 332 0.0501 0.2435 1.3929 0.1643
810 350 0.0490 0.2439 1.3920 0.1670
100 -173.15 3.598 1.028 6.929
110 -163.15 3.256 1.022 1.420 7.633
120 -153.15 2.975 1.017 1.416 8.319
130 -143.15 2.740 1.014 1.413 8.990
140 -133.15 2.540 1.012 1.411 9.646
150 -123.15 2.367 1.010 1.410 10.28
160 -113.15 2.217 1.009 1.408 10.91
170 -103.15 2.085 1.008 1.407 11.52
180 -93.15 1.968 1.007 1.407 12.12
190 -83.15 1.863 1.007 1.406 12.71
200 -73.15 1.769 1.006 1.405 13.28
210 -63.15 1.684 1.006 1.405 13.85
220 -53.15 1.607 1.006 1.404 14.40
230 -43.15 1.537 1.006 1.404 14.94
240 -33.15 1.473 1.005 1.404 15.47
250 -23.15 1.413 1.005 1.403 15.99
260 -13.15 1.359 1.005 1.403 16.50
270 -3.15 1.308 1.006 1.402 17.00
280 6.85 1.261 1.006 1.402 17.50
290 16.85 1.218 1.006 1.402 17.98
300 26.85 1.177 1.006 1.401 18.46
310 36.85 1.139 1.007 1.401 18.93
320 46.85 1.103 1.007 1.400 19.39
330 56.85 1.070 1.008 1.400 19.85
340 66.85 1.038 1.008 1.399 20.30
350 76.85 1.008 1.009 1.399 20.75
360 86.85 0.980 1.010 1.398 21.18
370 96.85 0.953 1.011 1.398 21.60
380 106.85 0.928 1.012 1.397 22.02
390 116.85 0.905 1.013 1.396 22.44
400 126.85 0.882 1.014 1.396 22.86
410 136.85 0.860 1.015 1.395 23.27
420 146.85 0.840 1.017 1.394 23.66
430 156.85 0.820 1.018 1.393 24.06
440 166.85 0.802 1.020 1.392 24.85
450 176.85 0.784 1.021 1.392 24.85
Source:    CRC Press, Handbook of Tables for Applied Engineering Science, CRC Press, Inc., Boca Raton, FL, 1973.
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First, the Reynolds number is determined from Equation 12.14 as

Re =\frac{\rho \text{v} D}{\mu} = \frac{1.11  kg/m^{3} \times 10  m/s \times 10  m}{1.94 \times 10^{-5}  Pa \cdot s} = 5.72 \times 106{6}

The flow is clearly turbulent.
Next, the Prandtl number is computed from Equation 12.15 as

Pr = \frac{\mu c_{p}}{k}            (12.15)

Pr = \frac{1007  J/kg \times 1.94 \times 10^{-5}  Pa \cdot s}{0.0275  W/(m \cdot K)} = 0.71

We then use Equation 12.22 and find

Nu_{L} = 0.037 {Re_{L}}^{0.8} Pr^{1/3}            (12.22)

Nu_{L} = 0.037 \times (5.72 \times 10^{6})^{0.8} \times 0.71^{1/3} = 8405

from which the convective heat transfer coefficient is deduced from Equation 12.16 as

Nu = \frac{h_{con} D}{k}                (12.16)

h_{con} = \frac{0.0275  W/(m \cdot K) \times 8405}{10  m} = 23.11  W/(m^{2} \cdot K)

Comments
This value is very close (to within 6%) of that found in Example 2.10 using the simplified dimensional equation.

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