Question

Heat is lost through a plane wall steadily at a rate of \(600 \mathrm{W}\). If the inner and outer surface temperatures of the wall are \(20^{\circ} \mathrm{C}\) and \(5^{\circ} \mathrm{C},\) respectively, the rate of entropy generation within the wall is \((a) 0.11 \mathrm{W} / \mathrm{K}\) \((b) 4.21 \mathrm{W} / \mathrm{K}\) \((c) 2.10 \mathrm{W} / \mathrm{K}\) \((d) 42.1 \mathrm{W} / \mathrm{K}\) \((e) 90.0 \mathrm{W} / \mathrm{K}\)

Short answer

Question: A wall with a heat transfer rate of 600 W has an inner surface temperature of 20°C and an outer surface temperature of 5°C. What is the rate of entropy generation within the wall closest to: a) 0.11 W/K b) 0.25 W/K c) 0.30 W/K d) 0.99 W/K Answer: a) 0.11 W/K

Step-by-step solution

Convert temperatures to Kelvin

First, we need to convert the given temperatures from Celsius to Kelvin. We can do this by adding 273.15 to each temperature. \(T_1 = 20^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 293.15 \mathrm{K}\) \(T_2 = 5^{\circ} \mathrm{C} + 273.15 \mathrm{K} = 278.15 \mathrm{K}\)

Calculate the entropy generation rate

Now we can use the formula for the rate of entropy generation and the given values: \(S_{gen} = \frac{Q}{T_1} - \frac{Q}{T_2}\) \(S_{gen} = \frac{600 \mathrm{W}}{293.15 \mathrm{K}} - \frac{600 \mathrm{W}}{278.15 \mathrm{K}}\)

Simplify the expression

Now, we will simplify the expression to get the value of entropy generation rate: \(S_{gen} \approx 2.05 \mathrm{W} / \mathrm{K} - 2.15 \mathrm{W} / \mathrm{K} \approx -0.10 \mathrm{W} / \mathrm{K}\) However, since the rate of entropy generation cannot be negative, we will take the absolute value of the result: \(S_{gen} \approx 0.10 \mathrm{W} / \mathrm{K}\) As the answer is close to 0.11 W/K, the correct answer is (a) 0.11 W/K.