Chapter 8, Problem 164RQ

To determine

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

Explanation of Solution

Given:

The initial pressure of steam $\left(P_1=P_3\right)$ is 400 kPa.

The initial temperature of steam $\left(T_1\right)$ is $350°\text{C}$.

The initial volume of steam $\left(v_1\right)$ is $0.3{\text{ m}}^3$.

The pressure at expansion $\left(P_2\right)$ is 150 kPa.

Calculation:

Refer Table A-6, “Superheated water” to obtain the value of internal energy state 1 $\left(u_1\right)$ at the initial pressure $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature $\left(T_1\right)$ of $350°\text{C}$ by using interpolation method.

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

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

Write the formula of interpolation method of two variables.

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

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

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

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

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 Table A-6, “Superheated water”.to obtain the value of initial specific volume $\left(v_1\right)$ at the initial pressure $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature $\left(T_1\right)$ of $350°\text{C}$ by using interpolation method.

Show temperature and specific volume values from the Table A-6 as below.

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

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

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

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

Calculate the mass of the steam in the cylinder.

  $\begin{array}{c}m=\frac{{\nu }_1}{v_1}\\ m=\frac{0.3{\text{m}}^3}{0.713770{\text{m}}^3/\text{kg}}\\ =0.4203\text{kg}\end{array}$

Refer Table A-6, “Superheated water”, to obtain the value of entropy at state1 $\left(s_1\right)$ at the initial pressure $\left(P_1\right)$ 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 as below.

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 $x_1=300$, $x_2=350$, $x_3=400$, $y_1=7.5677$, and $y_3=7.9003$ in Equation (I).

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

The value of entropy at state 1 $\left(s_1\right)$ at the initial pressure $\left(P_1\right)$ of $400\text{kPa}$ and at the initial temperature $\left(T_1\right)$ 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 $\left(P_2\right)$ of $150\text{kPa}$ and at the initial temperature $\left(T_1\right)$ 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.

Calculate the heat transfer in for the isothermal expansion process 1-2.

  $\begin{array}{c}{Q}_{in,1-2}=m{T}_1\left(s_2-s_1\right)\\ Q_{in,1-2}=\left(0.4203\text{kg}\right)\left(350° \text{C}\right)\left(\begin{array}{l}8.1983\text{kJ}/\text{kg}\cdot \text{K}\\ -7.734\text{kJ}/\text{kg}\cdot \text{K}\end{array}\right)\\ =\left(0.4203\text{kg}\right)\left(350+273\right)\text{K}\left(\begin{array}{l}8.1983\text{kJ}/\text{kg}\cdot \text{K}\\ -7.734\text{kJ}/\text{kg}\cdot \text{K}\end{array}\right)\\ =120\text{kJ}\end{array}$

Obtain the values of internal energy at state 3 $\left(u_3\right)$, specific volume at state 3 $\left(v_3\right)$ at the pressure $\left(P_3\right)$ 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 $3132.9\text{kJ}/\text{kg}$ and $0.89148{\text{m}}^3/\text{kg}$ respectively by interpolation method.

Calculate the volume at state 3.

  $\begin{array}{c}{\nu }_3=m{v}_3\\ {\nu }_3=\left(0.4203\text{kg}\right)\left(0.89148{\text{m}}^3/\text{kg}\right)\\ =0.375{\text{m}}^3\end{array}$

Calculate the work done out for the isothermal expansion process 1-2.

  $\begin{array}{c}{W}_{out,1-2}=Q_{in,1-2}-m\left(u_2-u_1\right)\\ W_{out,1-2}=120\text{kJ}-\left(0.4203\text{kg}\right)\left(2888\text{kJ}/\text{kg}-2885\text{kJ}/\text{kg}\right)\\ =118.74\text{kJ}\end{array}$

Calculate the work done in for the isentropic compression process 2-3.

  $\begin{array}{c}{W}_{in,2-3}=m\left(u_3-u_2\right)\\ W_{in,2-3}=\left(0.4203\text{kg}\right)\left(3132.9\text{kJ}/\text{kg}-2888\text{kJ}/\text{kg}\right)\\ =102.93\text{kJ}\end{array}$

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

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

  $\begin{array}{c}{W}_{in,3-1}=P_3\left({\nu }_3-{\nu }_1\right)\\ W_{in,3-1}=\left(400\text{kPa}\right)\left(0.375{\text{m}}^{\text{3}}-0.3{\text{m}}^{\text{3}}\right)\\ =30\text{kJ}\end{array}$

Calculate the heat transfer out for the constant pressure compression process 3-1.

  $\begin{array}{c}{Q}_{out,3-1}=W_{in,3-1}-m\left(u_1-u_3\right)\\ Q_{out,3-1}=30\text{kJ}-\left(0.4203\text{kg}\right)\left(2885\text{kJ}/\text{kg}-3132.9\text{kJ}/\text{kg}\right)\\ =134.2\text{kJ}\end{array}$

Calculate the net work done by piston cylinder device.

  $\begin{array}{c}{W}_{net,in}=W_{in,3-1}+W_{in,2-3}-W_{out,1-2}\\ W_{net,in}=30\text{kJ}+102.93\text{kJ}-118.74\text{kJ}\\ =14.2\text{kJ}\\ \text{=14}\text{.2}\text{kJ}\end{array}$

Thus, the net work done by piston cylinder device is $14.2\text{kJ}$.

Calculate the net heat transfer by piston cylinder device.

  $\begin{array}{c}{Q}_{net,in}=Q_{in,1-2}-Q_{out,3-1}\\ Q_{net,in}=\text{120}\text{kJ}-134.2\text{kJ}\\ =-14.2\text{kJ}\end{array}$

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

Thus, the net heat transfer by piston cylinder device is $14.2\text{kJ}$.