Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constant pressure to a final state. The heat transfer for t... Five kg of water is contained in a piston-cylinder assembly, initially at 5 bar and 240°C. The water is slowly heated at constant pressure to a final state. The heat transfer for the process is 2960 kJ and kinetic and potential energy effects are negligible. Determine the final volume, in m³, and the work for the process, in kJ. Read more

Steps to Solve

1. Find the specific enthalpy and specific volume at the initial state

Given $P_1 = 5 \text{ bar} = 0.5 \text{ MPa}$ and $T_1 = 240^\circ\text{C}$, we can find the specific enthalpy $h_1$ and specific volume $v_1$ from the steam tables. At this state, water is superheated vapor so we look at the superheated steam tables.

$h_1 = 2939.9 \text{ kJ/kg}$ $v_1 = 0.46462 \text{ m}^3\text{/kg}$

2. Apply the first law of thermodynamics to find the specific enthalpy at the final state

Since kinetic and potential energy effects are negligible, the first law of thermodynamics for a constant pressure process simplifies to:

$Q = m(h_2 - h_1)$

Where: $Q$ is the heat transfer = 2960 kJ $m$ is the mass = 5 kg $h_1$ is the initial specific enthaply $h_2$ is the final specific enthaply

Solving for $h_2$: $h_2 = \frac{Q}{m} + h_1$ $h_2 = \frac{2960 \text{ kJ}}{5 \text{ kg}} + 2939.9 \text{ kJ/kg} = 592 \text{ kJ/kg} + 2939.9 \text{ kJ/kg} = 3531.9 \text{ kJ/kg}$

3. Find the specific volume at the final state

Since the pressure is constant, $P_2 = P_1 = 5 \text{ bar} = 0.5 \text{ MPa}$. We have $P_2 = 0.5 \text{ MPa}$ and $h_2 = 3531.9 \text{ kJ/kg}$. Using the superheated steam tables again, we find the specific volume $v_2$ at this state. By interpolation:

At 0.5 MPa: h = 3484.5 kJ/kg, v = 0.6821 m^3/kg h = 3635.6 kJ/kg, v = 0.7415 m^3/kg

$v_2 = 0.6821 + \frac{3531.9-3484.5}{3635.6-3484.5}(0.7415-0.6821) = 0.6821 + \frac{47.4}{151.1}(0.0594) = 0.6821+0.0186 = 0.7007 \text{ m}^3\text{/kg}$

4. Calculate the final volume

The final volume $V_2$ is given by: $V_2 = m \cdot v_2 = 5 \text{ kg} \cdot 0.7007 \text{ m}^3\text{/kg} = 3.5035 \text{ m}^3$