Answer to Question #232138 in Molecular Physics | Thermodynamics for Unknown346307

Question

Hot air at a temperature of 65°C is flowing through a steel pipe of 120 mm

diameter. The pipe is covered with two layers of different insulating materials of thickness 60 mm

and 40 mm, and their corresponding thermal conductivities are 0.24 and 0.4 W/m°C. The inside

and outside heat transfer coefficients are 60 and 12 W/m°C. The atmosphere is at 20°C. Find the

rate of heat loss from 60 m length of pipe.

Answer 1

$r_1= \frac{120}{2}= 60 \;mm = 0.06 \;m \\ r_2= 60 +60 = 120 \;mm = 0.12 \;mm \\ r_3= 60 +60 + 40 = 160 \;mm = 0.16 \;mm \\ k_A=0.24 \;W/m°C \\ k_B= 0.4 \;W/m°C \\ h_{hf}=60 \;W/m^2°C \\ h_{cf}= 12 \;W/m^2°C \\ t_{hf}= 60\;°C \\ t_{cf}=20 \;°C Length \; of \; pipe, L= 60 \;m$

Rate of heat loss, Q :

Rate of heat loss is given by

$Q= \frac{2 \pi L (t_{hf} -t_{cf}}{[\frac{1}{h_{hf} \times r_1} + \frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} + \frac{1}{h_{cf} \times r_3} ]} \\ = \frac{2 \pi \times 60 (65 -20}{[\frac{1}{60 \times 0.06} + \frac{ln(0.12/0.06)}{0.24} + \frac{ln(0.16/0.12)}{0.4} + \frac{1}{12 \times 0.16} ]} \\ = \frac{16964.6}{0.2777 + 2.8881 + 0.7192 + 0.5208} = 3850.5 \;W$

Answer: 3850.5 W