Chapter 7.13, Problem 175RP

A piston–cylinder device contains steam that undergoes a reversible thermodynamic cycle. Initially the steam is at 400 kPa and 350°C with a volume of 0.5 m³. 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.

To determine

The net work done and net heat transfer by piston cylinder device.

Answer to Problem 175RP

The net work done by piston cylinder device is $20.8\text{kJ}$.

The net heat transfer by piston cylinder device is $20.8\text{kJ}$.

Explanation of Solution

Write the expression to calculate the mass of the steam in the cylinder.

$m=\frac{{\nu }_1}{v_1}$ (I)

Here, mass of the steam is m, initial volume is ${\nu }_1$, and initial molar volume is $v_1$.

Write the expression for the volume at state 3.

${\nu }_3=m{v}_3$ (II)

Here, volume at state 3 is ${\nu }_3$, mass of the steam is m, and molar volume at state 3 is $v_3$.

Write the expression to calculate the heat transfer in for the isothermal expansion process 1-2.

$Q_{in,1-2}=m{T}_1\left(s_2-s_1\right)$ (III)

Here, heat transfer in for process 1-2 is $Q_{in,1-2}$, entropy at state1 is $s_1$ and entropy at state 2 is $s_2$.

Write the expression to calculate the work done out for the isothermal expansion process 1-2.

$W_{out,1-2}=Q_{in,1-2}-m\left(u_2-u_1\right)$ (IV)

Here, work done out for process 1-2 is $W_{out,1-2}$, internal energy state 1 is $u_1$ and internal energy state 2 is $u_3$.

Write the expression to calculate the work done in for the isentropic compression process 2-3.

$W_{in,2-3}=m\left(u_3-u_2\right)$ (V)

Here, work done in for process 2-3 is $W_{in,2-3}$ and internal energy state 3 is $u_3$.

Write the expression to calculate the work done in for the constant pressure compression process 3-1.

$W_{in,3-1}=P_3\left({\nu }_3-{\nu }_1\right)$ (VI)

Here, work done in for process 3-1 is $W_{in,3-1}$, volume at state 1 is ${\nu }_1$ and volume at state 3 is ${\nu }_3$.

Write the expression to calculate the heat transfer out for the constant pressure compression process 3-1.

$Q_{out,3-1}=W_{in,3-1}-m\left(u_1-u_3\right)$ (VII)

Here, heat transfer out for the process 3-1.

Write the expression to calculate the net work done by piston cylinder device.

$W_{net,in}=W_{in,3-1}+W_{in,2-3}-W_{out,1-2}$ (VIII)

Here, the net work done is $W_{net,in}$

Write the expression to calculate the net heat transfer by piston cylinder device.

$Q_{net,in}=Q_{in,1-2}-Q_{out,3-1}$ (IX)

Here, the net heat transfer is $Q_{net,in}$

Conclusion:

Refer to Table A-6, “Superheated water”.

Obtain the value of internal energy state 1 $\left(u_1\right)$ at the initial pressure of $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature of $\left(T_1\right)$ of $350°\text{C}$ by using interpolation method.

Write the formula of interpolation method of two variables.

$y_2=\frac{\left(x_2-x_1\right)\left(y_3-y_1\right)}{\left(x_3-x_1\right)}+y_1$ (X)

Here, the variables denoted by x and y are temperature and internal energy.

Show temperature and initial internal energy values from the Table A-6.

Temperature $\left(T_1\right)$,in $°\text{C}$	Internal energy $\left(u_1\right)$, in $\text{kJ}/\text{kg}$
300	2805.1
350	?
400	2964.9

Substitute $300$ for $x_1$, $350$ for $x_2$, $400$ for $x_3$, $2805.1$ for $y_1$, and $2964.9$ for $y_3$ in Equation (X).

$\begin{array}{c}{y}_2=\frac{\left(350-300\right)\left(2964.9-2805.1\right)}{\left(400-300\right)}+2805.1\\ =2885\end{array}$

The value of internal energy state 1 $\left(u_1\right)$ at the initial pressure of $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature of $\left(T_1\right)$ of $350°\text{C}$ is $2885\text{kJ}/\text{kg}$.

Refer to Table A-6, “Superheated water”.

Obtain the value of initial molar volume $\left(v_1\right)$ at the initial pressure of $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature of $\left(T_1\right)$ of $350°\text{C}$ by using interpolation method.

Show temperature and molar volume values from the Table A-6.

Temperature $\left(T_1\right)$,in $°\text{C}$	Molar volume $\left(v_1\right)$ , in ${\text{m}}^3/\text{kg}$
300	0.65489
350	?
400	0.77265

Substitute $300$ for $x_1$, $350$ for $x_2$, $400$ for $x_3$, $0.65489$ for $y_1$, and $0.77265$ for $y_3$ in Equation (X).

$\begin{array}{c}{y}_2=\frac{\left(350-300\right)\left(0.77265-0.65489\right)}{\left(400-300\right)}+0.65489\\ =0.713770\end{array}$

The value of initial molar volume $\left(v_1\right)$ at the initial pressure of $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature of $\left(T_1\right)$ of $350°\text{C}$ is $0.713770{\text{m}}^3/\text{kg}$.

Refer to Table A-6, “Superheated water”.

Obtain the value of entropy at state1 $\left(s_1\right)$ at the initial pressure of $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature of $\left(T_1\right)$ of $350°\text{C}$ by using interpolation method.

Show temperature and entropy values from the Table A-6.

Temperature $\left(T_1\right)$,in $°\text{C}$	Entropy $\left(s_1\right)$ , in $\text{kJ}/\text{kg}\cdot \text{K}$
300	7.5677
350	?
400	7.9003

Substitute $300$ for $x_1$, $350$ for $x_2$, $400$ for $x_3$, $7.5677$ for $y_1$, and $7.9003$ for $y_3$ in Equation (X).

$\begin{array}{c}{y}_2=\frac{\left(350-300\right)\left(7.9003-7.5677\right)}{\left(400-300\right)}+7.5677\\ =7.734\end{array}$

The value of entropy at state1 $\left(s_1\right)$ at the initial pressure of $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature of $\left(T_1\right)$ of $350°\text{C}$ is $7.734\text{kJ}/\text{kg}\cdot \text{K}$.

Similarly,

obtain the values of internal energy at state 2 $\left(u_2\right)$ , entropy at state 2 $\left(s_2\right)$ at the pressure of $\left(P_2\right)$ of $150\text{kPa}$ and at the initial temperature of $\left(T_1\right)$ of $350°\text{C}$ as $2888.0\text{kJ}/\text{kg}$ and $8.19683\text{kJ}/\text{kg}\cdot \text{K}$ respectively by interpolation method.

Obtain the values of internal energy at state 3 $\left(u_3\right)$ , molar volume at state 3 $\left(v_3\right)$ at the pressure of $\left(P_3\right)$ of $400\text{kPa}$ and at the entropy at state 2 $\left(s_2\right)$ of $8.19683\text{kJ}/\text{kg}\cdot \text{K}$ as $0.89148{\text{m}}^3/\text{kg}$ respectively by interpolation method.

Substitute $0.5{\text{m}}^3$ for ${\nu }_1$ and $0.713770{\text{m}}^3/\text{kg}$ for $v_1$ in Equation (I).

$\begin{array}{c}m=\frac{0.5{\text{m}}^3}{0.713770{\text{m}}^3/\text{kg}}\\ =0.700506\text{kg}\end{array}$

Substitute $0.700506\text{kg}$ for m and $0.89148{\text{m}}^3/\text{kg}$ for $v_3$ in Equation (II).

$\begin{array}{c}{\nu }_3=\left(0.700506\text{kg}\right)\left(0.89148{\text{m}}^3/\text{kg}\right)\\ =0.624487{\text{m}}^3\end{array}$

Substitute $0.700506\text{kg}$ for m, $350° \text{C}$ for $T_1$, $8.1983\text{kJ}/\text{kg}\cdot \text{K}$ for $s_2$ and $7.734\text{kJ}/\text{kg}\cdot \text{K}$ for $s_1$ in Equation (III).

$\begin{array}{c}{Q}_{in,1-2}=\left(0.700506\text{kg}\right)\left(350° \text{C}\right)\left(\begin{array}{l}8.1983\text{kJ}/\text{kg}\cdot \text{K}\\ -7.734\text{kJ}/\text{kg}\cdot \text{K}\end{array}\right)\\ =\left(0.700506\text{kg}\right)\left(350+273\right)\text{K}\left(\begin{array}{l}8.1983\text{kJ}/\text{kg}\cdot \text{K}\\ -7.734\text{kJ}/\text{kg}\cdot \text{K}\end{array}\right)\\ =202.6\text{kJ}\end{array}$

Substitute $202.6\text{kJ}$ for $Q_{in,1-2}$, $0.700506\text{kg}$ for $m$ , $2888\text{kJ}/\text{kg}$ for $u_2$ and $2885\text{kJ}/\text{kg}$ for $u_1$ in Equation (IV).

$\begin{array}{c}{W}_{out,1-2}=202.6\text{kJ}-\left(0.700506\text{kg}\right)\left(2888\text{kJ}/\text{kg}-2885\text{kJ}/\text{kg}\right)\\ =200.49\text{kJ}\end{array}$

Substitute $0.700506\text{kg}$ for $m$, $3132.9\text{kJ}/\text{kg}$ for $u_3$ and $2888\text{kJ}/\text{kg}$ for $u_2$ in Equation (V).

$\begin{array}{c}{W}_{in,2-3}=\left(0.700506\text{kg}\right)\left(3132.9\text{kJ}/\text{kg}-2888\text{kJ}/\text{kg}\right)\\ =171.55\text{kJ}\end{array}$

The heat transfer during the process is zero, since isentropic compression process, entropy remains constant.

Substitute $\text{400}\text{kPa}$ for $P_3$, $0.6243{\text{m}}^{\text{3}}$ for ${\nu }_3$ and $0.5{\text{m}}^{\text{3}}$ for ${\nu }_1$ in Equation (VI).

$\begin{array}{c}{W}_{in,3-1}=\left(400\text{kPa}\right)\left(0.6243{\text{m}}^{\text{3}}-0.5{\text{m}}^{\text{3}}\right)\\ =49.72\text{kJ}\end{array}$

Substitute $49.72\text{kJ}$ for $W_{in,3-1}$, $\text{0}\text{.700506}\text{kg}$ for m, $2885\text{kJ}/\text{kg}$ for $u_1$ and $3132.9\text{kJ}/\text{kg}$ for $u_3$ in Equation (VII).

$\begin{array}{c}{Q}_{out,3-1}=49.72\text{kJ}-\left(\text{0}\text{.700506}\text{kg}\right)\left(2885\text{kJ}/\text{kg}-3132.9\text{kJ}/\text{kg}\right)\\ =223.38\text{kJ}\end{array}$

Substitute $49.72\text{kJ}$ for $W_{in,3-1}$, $171.55\text{kJ}$ for $W_{in,2-3}$ and $200.49\text{kJ}$ for $W_{out,1-2}$ in Equation (VIII).

$\begin{array}{c}{W}_{net,in}=49.72\text{kJ}+171.55\text{kJ}-200.49\text{kJ}\\ =20.78\text{kJ}\\ \text{=20}\text{.8}\text{kJ}\end{array}$

Thus, the net work done by piston cylinder device is $20.8\text{kJ}$.

Substitute $\text{202}\text{.6}\text{kJ}$ for $Q_{in,1-2}$, and $223.38\text{kJ}$ for $Q_{out,3-1}$ in Equation (IX).

$\begin{array}{c}{Q}_{net,in}=\text{202}\text{.6}\text{kJ}-223.38\text{kJ}\\ =-20.78\text{kJ}\\ \text{=}-20.8\text{kJ}\end{array}$

The negative sign indicates that the heat transfer occurs from system to surroundings.

Thus, the net heat transfer by piston cylinder device is $20.8\text{kJ}$.