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. Determine the initial specific volume $v_1$

Using the initial conditions $P_1 = 5 \text{ bar} = 0.5 \text{ MPa}$ and $T_1 = 240^\circ\text{C}$, we consult saturated steam tables or superheated steam tables to find the specific volume $v_1$. Since $240^\circ\text{C}$ is higher than the saturation temperature at $5 \text{ bar}$ ($151.83^\circ\text{C}$), we know the water is in the superheated state. From the superheated steam tables, we find $v_1 = 0.46462 , \text{m}^3/\text{kg}$.

2. Apply the First Law of Thermodynamics

Since kinetic and potential energy changes are negligible, the first law of thermodynamics simplifies to: $Q - W = \Delta U = m(u_2 - u_1)$, where $Q$ is heat transfer, $W$ is work done, $m$ is mass, and $u$ is specific internal energy.

3. Calculate the work done $W$

For a constant pressure process, the work done is $W = P(V_2 - V_1) = mP(v_2 - v_1)$.

4. Apply the First Law and Work equation

Substituting the work equation into the first law gives $Q = m(u_2 - u_1) + mP(v_2 - v_1) = m[(u_2 + Pv_2) - (u_1 + Pv_1)] = m(h_2 - h_1)$, where $h$ is the specific enthalpy.

5. Find the initial specific enthalpy $h_1$

From the superheated steam tables at $P_1 = 5 \text{ bar}$ and $T_1 = 240^\circ\text{C}$, we find $h_1 = 2939.9 , \text{kJ/kg}$.

6. Calculate the final specific enthalpy $h_2$

Using the equation $Q = m(h_2 - h_1)$, we can solve for $h_2$: $h_2 = h_1 + \frac{Q}{m} = 2939.9 , \text{kJ/kg} + \frac{2960 , \text{kJ}}{5 , \text{kg}} = 2939.9 , \text{kJ/kg} + 592 , \text{kJ/kg} = 3531.9 , \text{kJ/kg}$.

7. Determine the final specific volume $v_2$

We know $P_2 = P_1 = 5 \text{ bar} = 0.5 \text{ MPa}$ and $h_2 = 3531.9 , \text{kJ/kg}$. Using the superheated steam tables at $P_2 = 0.5 \text{ MPa}$, we look for the enthalpy value closest to $3531.9 , \text{kJ/kg}$. We find that this enthalpy corresponds to a temperature of $600^\circ\text{C}$. At this state, the specific volume is $v_2 = 0.71873 , \text{m}^3/\text{kg}$.

8. Calculate the final volume $V_2$

The final volume is $V_2 = m \cdot v_2 = 5 , \text{kg} \cdot 0.71873 , \text{m}^3/\text{kg} = 3.59365 , \text{m}^3$.

9. Calculate the work done $W$

The work done is $W = mP(v_2 - v_1) = 5 , \text{kg} \cdot 500 , \text{kPa} \cdot (0.71873 , \text{m}^3/\text{kg} - 0.46462 , \text{m}^3/\text{kg}) = 5 , \text{kg} \cdot 500 , \text{kPa} \cdot 0.25411 , \text{m}^3/\text{kg} = 635.275 , \text{kJ}$.