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.

Answer: 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

Total heat loss = 102.8 W/m

Explanation:

We are given that;

he = 16.5 W/m^ (2)

d = 5cm = 0.05m

surface emissivity (ε) = 0.8

T1 = 50°C

T2 = 20°C

Now, the total heat loss is the sum of convection and radiation.

Also, surface area is πdL, and thus

the convection loss per unit length is; (q/L) conv = h (πd) [ (Tw - T∞) ]

= 16.5 (π x 0.05) [ (50 - 20) ] =

2.592 x 30 = 77.76W/m

From the question, Since the pipe is a body surrounded by a large enclosure, the radiation heat transfer can be calculated from;

(q/L) rad = (ε1) (πd1) (σ) [ (T1) ⁴ - (T2) ⁴]

σ is Stefan Boltzmann constant and its = 5.67 x 10^ (-8) W m^ (-2) K^ (-4)

We have to convert temperature now to kelvins.

Thus, T1 = 50°C and when converted to kelvins = 273 + 50 = 323°K and also in the same way;

T2 = 20°C = 293°K,

So; (q/L) rad = (0.8) (π x 0.05) (5.67 x 10^ (-8)) [ (323) ⁴ - (293) ⁴] = 7.13 x 10^ (-6) ((1.088x 10^ (10)) - (0.737 x 10^ (10)) = 25.04W/m

Therefore, total heat loss = (q/L) conv + (q/L) rad = 77.76 + 25.04 = 102.8W/m