Answer:
a) The temperature on the outer surface of the pipe is approximately 179.97 °C
b) The thickness of the insulation is approximately 0.857 m
Step-by-step explanation:
We have;
![(1)/(U) = (1)/(\alpha _A)](https://img.qammunity.org/2022/formulas/engineering/college/grqt9fz1rb7w952rnofib34sp3165mza8c.png)
αA = 200 W/(m²·K)
= (T₂ - T₁) × U
= (200 - 180) × 200 = 4,000
For the pipe, we have;
![(1)/(U) =(x)/(kc )](https://img.qammunity.org/2022/formulas/engineering/college/xsbrk298r1nf5lfc5eyrdyyk6k6cr02ghs.png)
/U= (T₂ - T₁)
∴ 4000×
= (180 - T₂)
T₂ ≈ 179.97 °C
The temperature on the outer surface of the pipe, T₂ ≈ 179.97 °C
b) For the insulation, we have;
![(1)/(U) = (x)/(ki ) = (x)/(0.03)](https://img.qammunity.org/2022/formulas/engineering/college/3w9hzb0oeuhyxc6r48sym6n4a4wdv8hp30.png)
T₂ - T₃ = 179.97 °C - 40°C ≈ 139.97°C
/U= (T₂ - T₃)
![x = (\dot q \cdot kc)/(T_2 - T_3) = (4,000 * 0.03)/(139.97) \approx 0.857](https://img.qammunity.org/2022/formulas/engineering/college/8v8e2fijbrl5r4u28ef1b7rsq4v01p0lzc.png)
The thickness of the insulation, x ≈ 0.857 m