Question

Superheated water vapor at 180 psia and \(500^{\circ} \mathrm{F}\) is allowed to cool at constant volume until the temperature drops to \(250^{\circ} \mathrm{F}\). At the final state, determine \((a)\) the pressure, (b) the quality, and ( \(c\) ) the enthalpy. Also, show the process on a \(T\) -v diagram with respect to saturation lines.

Short answer

Answer: After cooling at constant volume, the final pressure is approximately 28 psia, the quality is approximately 17.6%, and the enthalpy is approximately 351.8 Btu/lb.

Step-by-step solution

Identify the initial and final states

We are given the initial state as superheated water vapor with a pressure of 180 psia and a temperature of \(500^{\circ} \mathrm{F}\). The final state is reached when the water vapor cools at constant volume until the temperature drops to \(250^{\circ} \mathrm{F}\).

Determine the initial specific volume

We can use a steam table or Mollier diagram to find the specific volume at the given initial state. At 180 psia and \(500^{\circ} \mathrm{F}\), the specific volume (\(v_1\)) is approximately 2.413 ft³/lb. Since the volume is constant throughout the process, \(v_2\) (the specific volume at the final state) will also be 2.413 ft³/lb.

Determine the final pressure

To find the final pressure of the water vapor at \(250^{\circ} \mathrm{F}\), we can refer to the steam tables or a Mollier diagram. Looking up the values for steam at \(250^{\circ} \mathrm{F}\) and a specific volume of 2.413 ft³/lb, we find that the final pressure (\(P_2\)) is approximately 28 psia.

Determine the final quality

Since we have the specific volume at both states, we can determine the quality (\(x\)) of the final state. At \(250^{\circ} \mathrm{F}\), the specific volume of saturated liquid \(v_f\) is approximately 0.01677 ft³/lb and the specific volume of saturated vapor \(v_g\) is approximately 13.821 ft³/lb. The quality can be determined by the equation: \(x = \frac{v_2 - v_f}{v_g - v_f}\) Substituting the values, we get: \(x = \frac{2.413 - 0.01677}{13.821 - 0.01677} \approx 0.176\) So, the quality at the final state is approximately 0.176 or 17.6%.

Determine the final enthalpy

Now that we know the final state's pressure and quality, we can determine the enthalpy (\(h_2\)) using the steam tables. At a temperature of \(250^{\circ} \mathrm{F}\) and a quality of 0.176, the enthalpy can be calculated as: \(h_2 = h_f + x \cdot h_{fg}\) Where \(h_f\) is the enthalpy of saturated liquid, and \(h_{fg}\) is the enthalpy of vaporization. Using steam tables, we get the values of \(h_f \approx 180.20 \,\mathrm{Btu/lb}\) and \(h_{fg} \approx 970.8 \,\mathrm{Btu/lb}\). Calculating \(h_2\), we get: \(h_2 \approx 180.20 + 0.176 \cdot 970.8 \approx 351.8 \,\mathrm{Btu/lb}\)

Plot the process on a T-v diagram

To plot the process on a T-v diagram, first, draw the saturation lines. Then, plot the initial state as a point above the saturation line (superheated region) at the given initial temperature and specific volume. Plot the final state point inside the saturation dome (two-phase region) at the final temperature and same specific volume as the initial state. Draw an arrow linking the two states representing the cooling process at constant volume.