This is question is related to the Momentum energy. The course of this question

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This is question is related to the Momentum energy. The course of this question is Advanced Transport Processes / Transport Phenomena. Please, help me to solve this question. Thank you.

2. A steel pipe (K 45.0 W/m.K) g a 0.05m O.D is covered with a 0.042 m thick layer of magnesia (K-0.07W/m.K) which in turn covered with a 0.024 m layer of fiberglass insulation (K-0.048 W/m.K)·The pipe wall outside temperature is 370 K and the outer surface temperature of the fiberglass is 305K. b) Calculate the steady state heat transfer.

Explanation / Answer

2. given steelmpipe, d = 0.05 m

Ks = 45 W/mK

magnesia, t = 0.042 m

Km = 0.07 W/mK

fiberglass, tf = 0.024 m

Kf = 0.048 W/mK

To = 305 K

Ti = 370 K

rate of heat flow, q' = k2*pi*r*l*dT/dr

integrating

q' = k*2*pi*l*(T' - T)/ln(r'/r)

hence

let temperature bw fiberglass and magnesia be T1

then

q' = 2*pi*l*Km(Ti - T1)/ln((d + 2t)/d) = 2*pi*l*Kf(T1 - To)/ln((d + 2t + 2t" )/(d + 2t))

0.07(370 - T1)/ln((0.05 + 2*0.042)/0.05) = 0.048(T1 - 305)/ln((0.05 + 2*0.042 + 2*0.024 )/(0.05 + 2*0.042))

0.4529171946951(370 - T1) = T1 - 305

167.579362037 - 1.4529171946951T1 = - 305

T1 = 323.9063482090 C

b. stady state heat transfer per unit length

q' = 2*pi*km(Ti - T1)/ln((d + 2t)/d) = 20.564875 W/m

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