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).

The correct answer is 1.184.

Based on the text and diagram in the provided image, we have the following parameters for the one-dimensional steady heat conduction process:
- Thickness of the slab (L) = 0.1 m
- Surface temperature of side A (higher temperature, T_A) = 80 °C = 80 + 273 = 353 K
- Heat transfer rate per unit area (q) = 4.5 kW/m² = 4500 W/m²
- Thermal conductivity of the slab (k) = 15 W/m·K

First, we use Fourier's law of heat conduction to find the temperature of the lower temperature side B (T_B):

$$q=\frac{k(T_A-T_B)}{L}$$
Substituting the given values:

$$4500=\frac{15\times (80-T_B)}{0.1}$$
Rearranging to solve for the temperature difference:

$$80-T_B=\frac{4500\times 0.1}{15}$$
$$80-T_B=30$$
$$T_B=50°\mathrm{C}$$
Converting the temperature of side B to Kelvin:

$$T_B=50+273=323K$$

During steady heat transfer, the rate of entropy generation per unit area (ṡ_gen) is calculated by performing an entropy balance across the control volume of the slab:

$${\dot{s}}_{\mathrm{gen}}=\frac{q}{T_B}-\frac{q}{T_A}$$
Substituting the values of heat flux and absolute temperatures:

$${\dot{s}}_{\mathrm{gen}}=\frac{4500}{323}-\frac{4500}{353}$$
Calculating each term:

$$\frac{4500}{323}\approx 13.93189{\text{W/m}}^2\text{·K}$$
$$\frac{4500}{353}\approx 12.74788{\text{W/m}}^2\text{·K}$$
Subtracting the entropy flow out from the entropy flow in:

$${\dot{s}}_{\mathrm{gen}}=13.93189-12.74788=1.184{\text{W/m}}^2\text{·K}$$
Thus, the rate of entropy generation per unit area is 1.184 W/m²·K.