Question

The inner and outer surfaces of a \(5-m \times 6-m\) brick wall of thickness \(30 \mathrm{cm}\) are maintained at temperatures of \(20^{\circ} \mathrm{C}\) and \(5^{\circ} \mathrm{C},\) respectively, and the rate of heat transfer through the wall is 900 W. Determine the rate of exergy destruction associated with this process. Take \(T_{0}=0^{\circ} \mathrm{C}\).

Short answer

Answer: The rate of exergy destruction associated with the heat transfer process in the given brick wall is 900 W.

Step-by-step solution

Finding the temperature difference

We are given the temperatures at the inner and outer surfaces of the wall. First, find the temperature difference between the two surfaces ΔT = T_inner - T_outer = 20°C - 5°C = 15°C.

Finding the heat transfer rate

The rate of heat transfer through the wall is given as 900 W in the problem, so we can use this information directly.

Finding the average temperature

Next, we need to find the average temperature between the inner and outer surfaces. The heat transfer rate depends on the average wall temperature: T_avg = (T_inner + T_outer)/2 = (20°C + 5°C)/2 = 12.5°C.

Calculating the exergy destruction

Since T₀ = 0°C, we can calculate the exergy destruction using the following formula: Exergy destruction = Q * (1 - (T₀ / T_avg)) First, we need to convert T_avg from °C to Kelvin. T_avg = 12.5°C + 273.15 = 285.65 K. And now, we can calculate the exergy destruction: Exergy destruction = 900 W * (1 - (0°C / 285.65 K)) = 900 W * (1 - 0) = 900 W So, the rate of exergy destruction associated with this process is 900 W.