# A piston?cylinder device contains steam that undergoes a reversible thermodynamic cycle....

## Question:

A piston?cylinder device contains steam that undergoes a reversible thermodynamic cycle. Initially the steam is at 400 kPa and 350{eq}^{\circ} {/eq}C with a volume of 0.5 {eq}m^3 {/eq} . The steam is first expanded isothermally to 150 kPa, then compressed adiabatically to the initial pressure, and finally compressed at the constant pressure to the initial state. Determine the net work and heat transfer for the cycle after you calculate the work and heat interaction for each process.

## Net heat transfer:

Net heat transfer is the change of thermal energy between the system and its surroundings. Heat transfer is done through three modes. With the help of thermodynamic's first law, we can calculate heat transfer.

## Answer and Explanation:

**Given data**

- Initial pressure {eq}{P_1} = 400\;{\rm{kPa}}. {/eq}

- Initial temperature {eq}{T_1} = 350\;{\rm{^\circ C}}. {/eq}

- Initial volume {eq}{V_1} = 0.3\;{{\rm{m}}^{\rm{3}}}. {/eq}

- Pressure after isothermal expansion {eq}{P_2} = 150\;{\rm{kPa}}. {/eq}

- Temperature after isothermal expansion {eq}{T_2} = 350\;{\rm{^\circ C}}. {/eq}

- Pressure after adiabatic compression {eq}{P_3} = 400\;{\rm{kPa}}. {/eq}

- At constant pressure compression pressure {eq}{P_4} = 400\;{\rm{kPa}}. {/eq}

- At constant pressure compression temperature {eq}{T_4} = 350\;{\rm{^\circ C}}. {/eq}

- At constant pressure compression volume {eq}{V_4} = 0.3\;{{\rm{m}}^{\rm{3}}}. {/eq}

The expression for the mass of steam is,

{eq}m = \dfrac{{{V_1}}}{{{v_1}}} {/eq}

From steam table,

At {eq}{T_1} = 350\;{\rm{^\circ C}} {/eq} and {eq}{P_1} = 400\;{\rm{kPa}}, {/eq}

{eq}\begin{align*} {u_1} &= 2884.5\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {{\rm{kg}}}}} \right. } {{\rm{kg}}}}\\ {s_1} &= 7.7399\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {{\rm{kg}} \cdot {\rm{K}}}}} \right. } {{\rm{kg}} \cdot {\rm{K}}}}\\ {v_1} &= 0.7139\;{{{{\rm{m}}^{\rm{3}}}} {\left/ {\vphantom {{{{\rm{m}}^{\rm{3}}}} {{\rm{kg}}}}} \right. } {{\rm{kg}}}} \end{align*} {/eq}

At {eq}{T_2} = 350\;{\rm{^\circ C}} {/eq} and {eq}{P_2} = 150\;{\rm{kPa}}, {/eq}

{eq}\begin{align*} {u_2} &= 2888\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {{\rm{kg}}}}} \right. } {{\rm{kg}}}}\\ {s_2} &= 8.198\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {{\rm{kg}} \cdot {\rm{K}}}}} \right. } {{\rm{kg}} \cdot {\rm{K}}}} \end{align*} {/eq}

At {eq}{P_3} = 400\;{\rm{kPa}} {/eq}

{eq}\begin{align*} {u_3} &= 3132.9\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {{\rm{kg}}}}} \right. } {{\rm{kg}}}}\\ {s_3} &= 8.1983\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {{\rm{kg}} \cdot {\rm{K}}}}} \right. } {{\rm{kg}} \cdot {\rm{K}}}}\\ {v_3} &= 0.89148\;{{{{\rm{m}}^{\rm{3}}}} {\left/ {\vphantom {{{{\rm{m}}^{\rm{3}}}} {{\rm{kg}}}}} \right. } {{\rm{kg}}}} \end{align*} {/eq}

Substituting the values,

{eq}\begin{align*} m &= \dfrac{{0.3}}{{0.7139}}\\ m &= 0.4202\;{\rm{kg}} \end{align*} {/eq}

Volume of steam at stage 3 is,

{eq}{V_3} = m \times {v_3} {/eq}

Substituting the values,

{eq}\begin{align*} {V_3} &= 0.4202 \times 0.89048\\ {V_3} &= 0.3746\;{{\rm{m}}^{\rm{3}}} \end{align*} {/eq}

a) For isothermal expansion,

The expression for entropy change is,

{eq}\Delta s = m \times \left( {{s_2} - {s_1}} \right) {/eq}

Substituting the values,

{eq}\begin{align*} \Delta s &= 0.4202 \times \left( {8.1983 - 7.7399} \right)\\ \Delta s &= 0.1926\;{{{\rm{kJ}}} {\left/ {\vphantom {{{\rm{kJ}}} {\rm{K}}}} \right. } {\rm{K}}} \end{align*} {/eq}

The expression for heat input,

{eq}{Q_1} = {T_1} \times \Delta s {/eq}

Substituting the values,

{eq}\begin{align*} {Q_1} &= \left( {350 + 273} \right) \times 0.1926\\ {Q_1} &= 120\;{\rm{kJ}} \end{align*} {/eq}

The work output is,

{eq}{W_1} = {Q_1} - m \times \left( {{u_2} - {u_1}} \right) {/eq}

Substituting the values,

{eq}\begin{align*} {W_1} &= 120 - 0.4202 \times \left( {2888 - 2884.5} \right)\\ {W_1} &= 118.5\;{\rm{kJ}} \end{align*} {/eq}

Thus, the work input is {eq}118.5\;{\rm{kJ}} {/eq} and heat transfer is {eq}120\;{\rm{kJ}}. {/eq}

b) For isentropic process,

Heat transfer is,

{eq}{Q_2} = 0 {/eq}

Work input is,

{eq}{W_2} = m \times \left( {{u_3} - {u_2}} \right) {/eq}

Substituting the values,

{eq}\begin{align*} {W_2} &= 0.4202 \times \left( {3132.9 - 2888} \right)\\ {W_2} &= 102.9\;{\rm{kJ}} \end{align*} {/eq}

Thus the heat transfer is {eq}0 {/eq} and the work input is {eq}102.9\;{\rm{kJ}}. {/eq}

c) For constant pressure compression,

Work input is,

{eq}{W_3} = {P_3} \times \left( {{V_3} - {V_1}} \right). {/eq}

Substituting the values,

{eq}\begin{align*} {W_3} &= 400 \times \left( {0.3746 - 0.3} \right)\\ {W_3} &= 29.8\;{\rm{kJ}} \end{align*} {/eq}

The expression for heat transfer,

{eq}{Q_3} = {W_3} - m \times \left( {{u_1} - {u_3}} \right) {/eq}

Substituting the values,

{eq}\begin{align*} {Q_3} &= 29.8 - 0.4202 \times \left( {2884.5 - 3132.9} \right)\\ {Q_3} &= 134.2\;{\rm{kJ}} \end{align*} {/eq}

Thus, the work input is {eq}29.8\;{\rm{kJ}} {/eq} and heat transfer is {eq}134.2\;{\rm{kJ}}. {/eq}

The net work input is,

{eq}W = {W_1} + {W_2} + {W_3} {/eq}

Substituting the values,

{eq}\begin{align*} W &= 29.8 + 102.9 - 118.5\\ W &= 14.2\;{\rm{kJ}} \end{align*} {/eq}

The net heat transfer is,

{eq}Q = {Q_3} - {Q_1} {/eq}

Substituting the values,

{eq}\begin{align*} Q &= 134.2 - 120\\ Q &= 14.2\;{\rm{kJ}} \end{align*} {/eq}

Thus the net work and the net heat transfer is {eq}14.2\;{\rm{kJ}}. {/eq}