Answered

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

Answer :

okpalawalter8

Answer:

[tex]Q_net=534.67\frac{w}{m}[/tex]

Explanation:

From the question we are told that:

Steam line diameter [tex]D=89mm \approx 0.089m[/tex]

Surface emissivity [tex]\mu=0.8[/tex]

Steam temp [tex]T_s=200\textdegree C \approx 200+273=473k[/tex]

Surrounding temp [tex]T_a=20 \textdegree C \aprrox 20+273= 293k[/tex]

 

Generally the equation for heat loss per unit length due to radiation [tex]Q_{net}[/tex] is mathematically given by

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

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

[tex]Q_net=534.67\frac{w}{m}[/tex]

Other Questions