Question

Consider a one-dimensional steady heat conduction process through a solid slab of thickness 0.1 m. The higher temperature side A has a surface temperature of 80°C, and the heat transfer rate per unit area to low temperature side B is 4.5 kW/m². The thermal conductivity of the slab is 15 W/m·K. The rate of entropy generation per unit area during the heat transfer process is ________________ W/m²·K (round off to 2 decimal places).

Hint:

Entropy generation is related to the temperature difference and the heat transfer rate. Use the temperatures in Kelvin for accurate calculations.

Answer 1

Correct Answer: 1.12

Given Data

• Slab thickness, \( L = 0.1 \, \text{m} \)
• Hot side temperature, \( T_A = 80^\circ\text{C} = 353 \, \text{K} \)
• Heat transfer rate per unit area, \( q = 4.5 \, \text{kW/m}^2 = 4500 \, \text{W/m}^2 \)
• Thermal conductivity, \( k = 15 \, \text{W/m·K} \)

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Step 1: Find the temperature at the cold side (B)

Using Fourier’s law of heat conduction:

\[ q = \frac{k (T_A - T_B)}{L} \]

Rearranging,

\[ T_A - T_B = \frac{qL}{k} \]

\[ T_A - T_B = \frac{4500 \times 0.1}{15} = 30 \, \text{K} \]

\[ T_B = 353 - 30 = 323 \, \text{K} \]

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Step 2: Entropy generation per unit area

For steady one-dimensional heat conduction, entropy generation per unit area is:

\[ \dot{S}_{gen} = q \left( \frac{1}{T_B} - \frac{1}{T_A} \right) \]

\[ \dot{S}_{gen} = 4500 \left( \frac{1}{323} - \frac{1}{353} \right) \]

\[ \dot{S}_{gen} = 4500 \times 0.000263 \approx 1.18 \, \text{W/m}^2\cdot\text{K} \]

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Final Answer

Rate of entropy generation per unit area = 

\[\boxed{1.18  \text{W/m}^.K}\]