To solve this problem, we need to determine the overall heat-transfer coefficient ($U$) and the heat loss per unit length ($Q$) to the surrounding air. Here’s a step-by-step solution.

1. Given Information

• Inside heat transfer coefficient, $h_i = 3500 \, \text{W/m}^2 \cdot ^\circ \text{C}$
• Inside diameter of the tube, $D_i = 2.5 \, \text{cm} = 0.025 \, \text{m}$
• Tube wall thickness, $t = 0.8 \, \text{mm} = 0.0008 \, \text{m}$
• Thermal conductivity of the tube material, $k = 16 \, \text{W/m} \cdot ^\circ \text{C}$
• Outside heat transfer coefficient, $h_o = 7.6 \, \text{W/m}^2 \cdot ^\circ \text{C}$
• Temperature of water inside the tube, $T_i = 50^\circ \text{C}$
• Surrounding air temperature, $T_o = 20^\circ \text{C}$

2. Determine Outside Diameter of the Tube

The outside diameter $D_o$ of the tube is calculated as: $D_o = D_i + 2 \times t = 0.025 + 2 \times 0.0008 = 0.0266 \, \text{m}$

3. Overall Heat-Transfer Coefficient (U)

The overall heat-transfer coefficient $U$ is given by the expression for heat transfer through a cylindrical wall: $\frac{1}{U} = \frac{1}{h_i} + \frac{D_i \ln \left( \frac{D_o}{D_i} \right)}{2k} + \frac{D_i}{D_o h_o}$

Step-by-Step Calculation

1. Inside Convective Resistance: $\frac{1}{h_i} = \frac{1}{3500} = 2.857 \times 10^{-4} \, \text{m}^2 \cdot ^\circ \text{C/W}$

2. Conduction Resistance through the Tube Wall: $\frac{D_i \ln \left( \frac{D_o}{D_i} \right)}{2k} = \frac{0.025 \times \ln \left( \frac{0.0266}{0.025} \right)}{2 \times 16}$

   Calculating the logarithmic term: $\ln \left( \frac{0.0266}{0.025} \right) = \ln (1.064) \approx 0.0621$ Thus, $\frac{D_i \ln \left( \frac{D_o}{D_i} \right)}{2k} = \frac{0.025 \times 0.0621}{2 \times 16} = 4.83 \times 10^{-5} \, \text{m}^2 \cdot ^\circ \text{C/W}$

3. Outside Convective Resistance: $\frac{D_i}{D_o h_o} = \frac{0.025}{0.0266 \times 7.6} = 1.233 \times 10^{-3} \, \text{m}^2 \cdot ^\circ \text{C/W}$

4. Calculate $U$: $\frac{1}{U} = 2.857 \times 10^{-4} + 4.83 \times 10^{-5} + 1.233 \times 10^{-3} = 1.566 \times 10^{-3} \, \text{m}^2 \cdot ^\circ \text{C/W}$ $U = \frac{1}{1.566 \times 10^{-3}} = 638.7 \, \text{W/m}^2 \cdot ^\circ \text{C}$

4. Heat Loss per Unit Length (Q)

The heat loss per unit length $Q$ can be calculated using: $Q = U \times \pi \times D_o \times (T_i - T_o)$ Substitute the values: $Q = 638.7 \times \pi \times 0.0266 \times (50 - 20)$ $Q = 638.7 \times \pi \times 0.0266 \times 30$ $Q \approx 1602.3 \, \text{W/m}$

Summary of Results

1. Overall Heat-Transfer Coefficient: $U \approx 638.7 \, \text{W/m}^2 \cdot ^\circ \text{C}$
2. Heat Loss per Unit Length: $Q \approx 1602.3 \, \text{W/m}$

Would you like further explanation on any step, or do you have additional questions?

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Here are some related questions that expand on this topic:

1. How would changes in tube material affect the overall heat-transfer coefficient?
2. How does the wall thickness impact heat loss in this type of setup?
3. What factors contribute to the difference between inside and outside heat-transfer coefficients?
4. How can insulation improve heat retention in tubes with high-temperature fluid flow?
5. What methods exist to enhance convective heat transfer on the inside or outside of tubes?

Tip: Always consider the thermal conductivity of materials and the thickness of walls when evaluating heat transfer across barriers, as these factors significantly influence the overall heat transfer efficiency.