A pipe of length L=25m and diameter d=10cm, carrying steam, loses heat to the am
ID: 2961456 • Letter: A
Question
A pipe of length L=25m and diameter d=10cm, carrying steam, loses heat to the ambient air and surrounding surfaces by convection and radiation. If the total flow of heat per unit time Q, emanating from the surface of the pipe is measured, then the surface temperature of the pipe Ts is determined by the following equation:
Q = pidL[h(Ts-Tair)+e*sigmaSB(Ts^4-Tsurr^4)]
e = 0.8 - radiative emissivity
sigmaSB = 5.67x10^-8 W/m2/K4
Q = 18405 W
h = 10W/m^2/K
T air=Tsurr = 298K
Find the surface temperature (Ts) using the bisection method.
Explanation / Answer
18405 = pi*0.1*25*[ 10*(T-298) + 0.8*5.67*10^-8*(T^4-298^4)]
18405=7.85398(4538*10^-8(T^4-7886150418)+10*(T-10)
18405=3.56*10^-4*(T-298)(T+698.1)*(T^2-391.2T+358.3841)
-3.56*10^-7T^4-78.53T+44619.4=0
u get answers
T=422.95 K
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