A horizontal steel pipe having a diameter of 5 cm is maintained at a temperature of 50◦C in a large room where the air and wall temperature are at 20◦C. The surface emissivity of the steel may be taken as 0.8 and the heat-transfer coefficient for free convection with this geometry and air is he=16.5 W/m2 C. Calculate the total heat lost by the pipe per unit length.

1,580.75 w/m

Explanation:

The total heat loss will be the sum of the heat lost by radiation and convention.

Now lets find the heat lost by convention using the formula

q/L]conv =h(πd )( Tw - T∞ )= (16.5)( π)(50-20) = 1,555.71 w/m

But this pipe is a body surrounded by a large enclosure so the radiation heat

transfer can be calculated from this equation

q/L]rad =ε1(πd1)σ (T14– T24)

Here both T1 and T2 are raised to the power of 4, therefore

= (0.8 )( π )( 0.05) ( 5.669× 10raised to power of -8 ) (3234 – 2934) = 25.04 w/m

the total heat loss is therefore

q/L]tot = q/L]conv + q/L]rad =1,555.71 + 25.04 =1,580.75 w/m

answer:

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