Water flows at 50°C inside a 2.5 cm inside diameter tube such that hi = 3500
W/m² °C. The tube has a wall thickness of 0.8 mm with a thermal conductivity
of 16 W/m² °C. The outside of the tube loses heat by free convection with ho=
7.6 W/m2 °C. Calculate the overall heat-transfer coefficient and heat loss per
unit length to surrounding air at 20°C.

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.

Water flows at 50°C inside a 2.5 cm inside diameter tube such that hi = 3500 W/m² °C. The tube has a wall thickness of 0.8 mm with a thermal conductivity of 16 W/m² °C. The outside of the tube loses heat by free convection with ho = 7.6 W/m² °C. Calculate the overall heat-transfer coefficient and heat loss per unit length to surrounding air at 20°C.

Explore the calculation of the overall heat-transfer coefficient and heat loss per unit length for water flowing at 50°C inside a tube with an inside diameter of 2.5 cm. The solution involves thermal conductivity, convection, and detailed logarithmic calculations, making it suitable for university-level thermodynamics.

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