Question

A piston-cylinder device with a set of stops initially contains \(0.6 \mathrm{kg}\) of steam at \(1.0 \mathrm{MPa}\) and \(400^{\circ} \mathrm{C}\). The location of the stops corresponds to 40 percent of the initial volume. Now the steam is cooled. Determine the compression work if the final state is \((a) 1.0 \mathrm{MPa}\) and \(250^{\circ} \mathrm{C}\) and (b) 500 kPa. (c) Also determine the temperature at the final state in part \((b)\).

Short answer

Question: Determine the compression work done in a piston-cylinder device containing steam when it is cooled to different final states, and determine the final temperature in the second case. Answer: The compression work done in the first case (a) is -41.6 kJ, and in the second case (b) is -111.54 kJ. The final temperature in the second case (b) is 80.79°C.

Step-by-step solution

Calculate the initial volume

Steam behaves like an ideal gas, so we can use the ideal gas law to find the initial volume, \(V_1\). The specific gas constant for steam is \(R = 0.4615\, \mathrm{kJ/(kg\cdot K)}\). $$V_1 = \frac{mRT_1}{P_1} = \frac{0.6\, \mathrm{kg} \cdot 0.4615\, \mathrm{kJ/(kg\cdot K)} \cdot 673\, \mathrm{K}}{1000\, \mathrm{kPa}} = 0.1859\, \mathrm{m^3}$$ Now we'll determine the final state properties and then the compression work for each case.

Case (a): \(P_{2a} = 1.0 \, \mathrm{MPa}\) and \(T_{2a} = 250^{\circ} \mathrm{C}\)

Final pressure, \(P_{2a} = 1.0 \, \mathrm{MPa} = 1000 \, \mathrm{kPa}\) Final temperature, \(T_{2a} = 250^{\circ} \mathrm{C} = 523 \, \mathrm{K}\) First, we'll find the final volume \(V_{2a}\) using the ideal gas law and then calculate the compression work, \(W_a\).

Calculate the final volume for case (a)

Using the ideal gas law, we can find the final volume, \(V_{2a}\): $$V_{2a} = \frac{mRT_{2a}}{P_{2a}} = \frac{0.6 \, \mathrm{kg} \cdot 0.4615\, \mathrm{kJ/(kg\cdot K)} \cdot 523\, \mathrm{K}}{1000\, \mathrm{kPa}} = 0.1443\, \mathrm{m^3}$$ Since \(V_{2a} < V_\mathrm{stop}\), the stops didn't need to be engaged.

Calculate the compression work for case (a)

Now, we can calculate the compression work, \(W_a\), using the formula: $$W_{a} = -P_1(V_1 - V_{2a}) = -1000 \, \mathrm{kPa}(0.1859\, \mathrm{m^3} - 0.1443\, \mathrm{m^3}) = -41.6 \, \mathrm{kJ}$$ So, the compression work in case (a) is \(W_a = -41.6 \, \mathrm{kJ}\).

Case (b): \(P_{2b} = 500 \, \mathrm{kPa}\)

Final pressure, \(P_{2b} = 500 \, \mathrm{kPa}\) Now, we'll find the final volume \(V_{2b}\) and then the compression work \(W_b\).

Calculate the final volume for case (b)

The stops will be engaged in this case, so the final volume, \(V_{2b} = V_\mathrm{stop}\): $$V_\mathrm{stop} = 0.4 V_1 = 0.4 \cdot 0.1859\, \mathrm{m^3} = 0.07436\, \mathrm{m^3}$$

Calculate the compression work for case (b)

Now, we can calculate the compression work, \(W_b\), using the formula: $$W_{b} = -P_1(V_1 - V_\mathrm{stop}) = -1000 \, \mathrm{kPa}(0.1859\, \mathrm{m^3} - 0.07436\, \mathrm{m^3}) = -111.54 \, \mathrm{kJ}$$ So, the compression work in case (b) is \(W_b = -111.54 \, \mathrm{kJ}\).

Case (c): Determine the final temperature in case (b)

Now, we'll use the ideal gas law to find the final temperature for case (b), \(T_{2b}\): $$T_{2b} = \frac{P_{2b}V_{2b}}{mR} = \frac{500\, \mathrm{kPa} \cdot 0.07436\, \mathrm{m^3}}{0.6\, \mathrm{kg} \cdot 0.4615\, \mathrm{kJ/(kg\cdot K)}} = 353.79\, \mathrm{K}$$ We need to convert the temperature in Kelvin to Celsius: $$T_{2b} = 353.79\, \mathrm{K} - 273 = 80.79^{\circ} \mathrm{C}$$ So, the final temperature in case (b) is \(T_{2b} = 80.79^{\circ} \mathrm{C}\). Summary: - Compression work in case (a): \(W_a = -41.6 \, \mathrm{kJ}\) - Compression work in case (b): \(W_b = -111.54 \, \mathrm{kJ}\) - Final temperature in case (b): \(T_{2b} = 80.79^{\circ} \mathrm{C}\)