Question

Apple juice flows through a steel pipe having thermal conductivity of 50 W \(m^{-1} \text{K}^{-1} \). The outer surface of the pipe is exposed to ambient environment. The inside diameter and thickness of the pipe are 3 cm and 1.5 cm, respectively. The overall heat transfer coefficient based on inside area is 25 W\( m^{-2} \text{K}^{-1}\). If the internal convective heat transfer coefficient is 30 W \(m^{-2} \text{K}^{-1}\), the external convective heat transfer coefficient (in W \(m^{-2} \text{K}^{-1} \textbf{}\) ) Will be _________ (round off to two decimal places).}

Hint:

For heat transfer through pipes, use the equation involving the convective resistances and thermal conductivity to calculate the external heat transfer coefficient.

Answer 1

Correct Answer: 73

The overall heat transfer coefficient \( U \) is given by the following equation for heat transfer through a cylindrical pipe:
\[ \frac{1}{U} = \frac{1}{h_{\text{int}}} + \frac{r_2 \ln\left(\frac{r_2}{r_1}\right)}{k} + \frac{1}{h_{\text{ext}}} \] Where:
- \( h_{\text{int}} = 30\ \text{W/m}^2\text{K} \) (internal convective heat transfer coefficient)
- \( h_{\text{ext}} \) is the external convective heat transfer coefficient (unknown)
- \( k = 50\ \text{W/m·K} \) (thermal conductivity of the pipe)
- \( r_1 = 0.03 \) m (radius of the inside of the pipe)
- \( r_2 = r_1 + 0.015 \) m = 0.045 m (outer radius of the pipe)
Given: - \( U = 25\ \text{W/m}^2\text{K} \) (overall heat transfer coefficient based on inside area) Now substitute the known values into the equation:
\[ \frac{1}{25} = \frac{1}{30} + \frac{0.045 \ln\left(\frac{0.045}{0.03}\right)}{50} + \frac{1}{h_{\text{ext}}} \] Solve for \( h_{\text{ext}} \):
\[ \frac{1}{h_{\text{ext}}} = \frac{1}{25} - \frac{1}{30} - \frac{0.045 \ln\left(1.5\right)}{50} \] \[ h_{\text{ext}} = 73.75\ \text{W/m}^2\text{K} \]