Answer:
(a) Rate of heat transfer assuming there is no wind is 152.70J/s
(b) Rate of heat transfer assuming the stack is exposed to 20km/h winds is 151.49J/s
Step-by-step explanation:
Q/t = KA(T2 - T1)/d
Range of heat transfer coefficient (K) for flue gases is 60-180W/mk
Assuming K=180W/mk, diameter (d) =0.6m, A=3.142×0.6^2/4 = 028278m^2, distance (d) = 10m, T2=40°C, T1=10°C
(a) Q/t = 180×0.28278×(40-10)/10 = 180×0.28278×30/10 = 152.70J/s
(b) V= 20km/h=20×1000/3600m/s= 5.6m/s, t=d/v=10/5.6=1.8s
Q = KA(T2 - T1)/V = 180×0.28278×(40 - 10)/5.6 = 180×0.28278×30/5.6 = 272.68J
Q/t = 272.68J/1.8s = 151.48J/s