5.5k views
5 votes
A long, uninsulated steam line with a diameter of 89 mm and a surface emissivity of 0.8 transports steam at 200C and is exposed to atmospheric air and large surroundings at an equivalent temperature of 20C. (a) Calculate the heat loss per unit length for a calm day. (b) Calculate the heat loss on a breezy day when the wind speed is 8 m/s. (c) For the conditions of part (a), calculate the heat loss with a 20-mm-thick layer of insulation (k 0.08 W/m K). Would the heat loss change significantly with an appreciable wind speed

1 Answer

5 votes

Answer:


Q_net=534.67(w)/(m)

Step-by-step explanation:

From the question we are told that:

Steam line diameter
D=89mm \approx 0.089m

Surface emissivity
\mu=0.8

Steam temp
T_s=200\textdegree C \approx 200+273=473k

Surrounding temp
T_a=20 \textdegree C \aprrox 20+273= 293k

Generally the equation for heat loss per unit length due to radiation
Q_(net) is mathematically given by


Q_net=\sigma*\mu>(\pi *d)*(T_s^4-T_a^4)


Q_net=5.6*10^8*0.8*(\pi *0.089)*(473^4-293^4)


Q_net=534.67(w)/(m)

User Sudeep Cv
by
4.9k points