Answer:
The answers to the question are
(i) The rate of heat loss per-unit-length (W/m) from the pipe is 131.62 W
(ii) The temperature of the outer surface of the insulation is 49.89 °C
Explanation:
To solve the question, we note that the heat transferred is given by
![Q = \frac{2\pi L(t_{hf} - t_{cf}) }{\frac{1}{h_{hf}r_1}+\frac{ln(r_2/r_1)}{k_A} + \frac{ln(r_3/r_2)}{k_B} +\frac{1}{h_{cf}r_3}}](https://tex.z-dn.net/?f=Q%20%3D%20%5Cfrac%7B2%5Cpi%20L%28t_%7Bhf%7D%20-%20t_%7Bcf%7D%29%20%7D%7B%5Cfrac%7B1%7D%7Bh_%7Bhf%7Dr_1%7D%2B%5Cfrac%7Bln%28r_2%2Fr_1%29%7D%7Bk_A%7D%20%2B%20%5Cfrac%7Bln%28r_3%2Fr_2%29%7D%7Bk_B%7D%20%2B%5Cfrac%7B1%7D%7Bh_%7Bcf%7Dr_3%7D%7D)
Where
= Temperature at the inside of the pipe = 300 °C
= Temperature at the outside of the pipe = 20 °C
r₁ =internal radius of pipe = 4.0 cm
r₂ = Outer radius of pipe = 4.5 cm
r₃ = Outer radius of the insulation = r₂ + 2.5 = 7.0 cm
= 15 W/m·K
= 0.038 W/m·K
= 75 W/m²·K
= 10 W/m²·K
Plugging in the values in the above equation where for a unit length L = 1 m, we have
Q = 131.32 W
From which we have, for the film of air at the pipe outer boundary layer
Where
for the air film on the pipe outer surface is given by
where A =area of the outside of the pipe
=
= 0.227 K/W
Therefore
131.32 W =
which gives
= 49.89 °C
Heat transferred by radiation = q' = ε×σ×(T₁⁴ - T₂⁴)
Where ε = 0.9, σ, = 5.67×10⁻⁸W/m²·(K⁴)
T₁ = Surface temperature of the pipe = 49.89 °C and
T₂ = Temperature of the surrounding = 20.00 °C
Plugging in the values gives, q' = 0.307 W per m²
Total heat lost per unit length = 131.32 + 0.307 =131.62 W