Question

A two-phase, liquid-vapor mixture of H2O, initially at x = 30% and a pressure of 100
kPa, is contained in a piston-cylinder assembly, as shown in Fig. The mass of the piston
is 10kg, and its diameter is 15 cm. The pressure of the surroundings is 100kPa. As the
water is heated, the pressure inside the cylinder remains constant until the piston hits the stops. Heat transfer to the water continues at constant volume untilthe pressure is 150 kPa. Friction between the piston and the cylinder wall and the kinetic and potential
energy effects are negligible. For the overall process of the water, determine the work and heat transfer, each in kJ

Answer 1

Given

Quality \(x=0.3\)

Mass of piston, \(m_{\text {piston }}=10 \mathrm{~kg}\)

Diameter \(D=0.15 \mathrm{~m}\)

Area of piston, \(A_{\text {piston }}=0.018 \mathrm{~m}^{2}\)

Pressure, \(p_{\text {atm }}=100 \mathrm{kPa}\)

\(g=9.81 \mathrm{~m} / \mathrm{s}^{2}\)

We know that

\(p A_{\mathrm{piston}}=\left[p_{\text {atn }} A_{\text {piston }}+m_{\mathrm{piston}} g\right]\)

\(P_{1}=p_{c t m}+\frac{m_{\text {piston }} g}{A_{\text {pitton }}}\)

\(P_{1}=100+\frac{10 \times 9.81}{0.018 \times 10^{3}}\)

\(P_{1}=105.45 \mathrm{kPa}\)

From saturated water table

At \(P_{1}=105.45 \mathrm{kPa}\)

\(u_{f}=417.33 \mathrm{~kJ} / \mathrm{kg}\)

\(u_{f g}=2088.72 \mathrm{~kJ} / \mathrm{kg}\)

\(u_{1}=u_{f}+x u_{f g}\)

\(u_{1}=417.33+0.3(2088.72)\)

\(u_{1}=1043.946 \mathrm{~kJ} / \mathrm{kg}\)

Now at \(P_{2}=130 \mathrm{kPa}\)

\(u_{2}=2440 \mathrm{~kJ} / \mathrm{kg}\)

From energy balance equation \(\Delta Q=\Delta U-\Delta W\)

At constant volume process

\(\Delta W=0\)

Then

\(\Delta Q=\Delta U\)

\(\Delta Q=u_{2}-u_{1}\)

\(\Delta Q=2440-1043.946\)

\(\Delta Q=1396.054 \mathrm{~kJ} / \mathrm{kg}\)