Answer to Question #232136 in Molecular Physics | Thermodynamics for Unknown346307

Question

A 150 mm steam pipe has inside dimater of 120 mm and outside diam￾eter of 160 mm. It is insulated at the outside with asbestos. The steam temperature is 150°C and

the air temperature is 20°C. h (steam side) = 100 W/m2°C, h (air side) = 30 W/m2°C, k (asbestos)

= 0.8 W/m°C and k (steel) = 42 W/m°C. How thick should the asbestos be provided in order to

limit the heat losses to 2.1 kW/m2 ?

Answer 1

$r_1=\frac{120}{2} = 60 \;mm = 0.06 \;m \\ r_2 = \frac{160}{2}=80 \;mm = 0.08 \;m \\ k_A= 42 \;W/m°C \\ k_B= 0.8 \;W/m°C \\ t_{hf}=150 \;°C \\ t_{cf}= 20 \; \\ h_{hf}= 100 \; W/m^2°C \\ h_{cf}= 30 \;W/m^C \\ Heat \;loss= 2.1 \;kW/m^2$

Thickness of insulation (asbestos), ($r_3 – r_2$) :

Area for heat transfer $= 2 \pi r L$

(where L = length of the pipe)

$Heat \; loss = 2.1 \times 2 \pi r L \;kW \\ = 2.1 \times 2 \pi \times 0.075 \times L = 0.989 \;L \;kW \\ = 0.989 \;L \times 10^3 \;watts$

Where

$r= \frac{150}{2}=75 \;mm = 0.075 \;m$

Heat transfer rate in such a case is given by

$Q= \frac{2 \pi L (t_{hf} -t_{cf}}{[\frac{1}{h_{hf} \times r_1} + \frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} + \frac{1}{h_{cf} \times r_3} ]} \\ 0.989 \;L \times 10^3 = \frac{2 \pi L (150 -20)}{[\frac{1}{100 \times 0.06} + \frac{ln(0.08/0.06)}{42} + \frac{ln(r_3/0.08)}{0.8} + \frac{1}{30 \times r_3} ]} \\ 0.989 \times 10^3 = \frac{816.81}{[0.16666 + 0.00685 + \frac{ln(r_3/0.08)}{0.8} + \frac{1}{30 \times r_3} ]} \\ \frac{ln(r_3/0.08)}{0.8} + \frac{1}{30 \times r_3} = \frac{816.81}{0.989 \times 10^3} -(0.16666+0.00685)=0.6524 \\ 1.25 ln(r_3/0.08) + \frac{1}{30r_3}- 0.6524 =0 \\ r_3= 0.105 \;m = 105 \;mm \\ Thickness \; of\; insulation = r_3-r_2 = 105-80=25 \;mm$