Discussion
A rigid closed tank having a volume of 2 m3 contains 0.1 m3 of saturated liquid water and 1.9 m3 of saturated water vapor at 100 kPa. Heat is transferred to the tank until the final tank pressure reaches 2 MPa.
Following data for water is given:
At 100 kPa: \(v_{f}\) = 0.001043 m3/kg, \(v_{g}\) = 1.694 m3/kg, \(u_{f}\) = 417.33 kJ/kg, \(u_{g}\) = 2506.06 kJ/kg
At 2 MPa: \(v_{f}\) = 0.001177 m3/kg, \(v_{g}\) = 0.09963 m3/kg, \(u_{f}\) = 906.42 kJ/kg, \(u_{g}\) = 2600.26 kJ/kg
The magnitude of heat transfer in this process is
Answer: D
The heat transfer is about 77470 kJ.
The tank is rigid, so no work is done and \(Q = U_{2} - U_{1}\). Find the masses first:
\[ \begin{aligned} m_{f} &= \frac{0.1}{0.001043} = 95.88\ \text{kg} \\ m_{g} &= \frac{1.9}{1.694} = 1.12\ \text{kg} \\ m &= 97.00\ \text{kg} \end{aligned} \]
\[ \begin{aligned} U_{1} &= 95.88(417.33) + 1.12(2506.06) \\ &\approx 42823\ \text{kJ} \end{aligned} \]
The specific volume stays \(v = 2/97.00 = 0.02062\) m3/kg. At 2 MPa:
\[ \begin{aligned} x_{2} &= \frac{0.02062 - 0.001177}{0.09963 - 0.001177} = 0.1975 \\ u_{2} &= 906.42 + 0.1975(1693.84) = 1240.9 \\ U_{2} &= 97.00 \times 1240.9 \approx 120367\ \text{kJ} \end{aligned} \]
So \(Q \approx 120367 - 42823 \approx 77500\) kJ, which matches the keyed 77470 kJ within the rounding of the table values. The other options are well short of the energy needed to take nearly 97 kg of mostly liquid water from 100 kPa to a wet mixture at 2 MPa, as the working shows.
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