Chapter 5, Problem 38P

To determine

The heat transferred and the work produced during the process.

Explanation of Solution

Given:

Initial pressure of steam is $75\text{ kPa}$.

Initial quality is $8\%$.

Initial volume is $2{\text{ m}}^3$.

Final volume is $5{\text{ m}}^3$.

Final pressure of the steam is $225\text{ kPa}$.

Calculation:

From the Table A-5, to obtain the value of the specific volume of saturated liquid is $v_f$, the specific volume of saturated vapour is $v_g$, the specific internal energy of saturated liquid is $u_f$, the specific internal energy change upon vaporization is $v_{fg}$ at initial pressure of $250\text{kPa}$.

  $\begin{array}{l}{v}_{f1}=0.001037{\text{m}}^{\text{3}}/\text{kg}\\ v_{g1}=2.2172{\text{m}}^{\text{3}}/\text{kg}\\ u_{f1}=384.36\text{kJ}/\text{kg}\\ u_{fg1}=2111.8\text{kJ}/\text{kg}\end{array}$

The specific volume of the spring-loaded piston cylinder device is,

  $v_1=v_{f1}+x{v}_{fg1}$

  $\begin{array}{c}{v}_1=\left(0.001037{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.08\right)\times \left(2.2172{\text{m}}^{\text{3}}/\text{kg}-0.001037{\text{m}}^{\text{3}}/\text{kg}\right)\\ =\left(0.001037{\text{m}}^{\text{3}}/\text{kg}\right)+\left(0.08\right)\times \left(2.216163{\text{m}}^{\text{3}}/\text{kg}\right)\\ =0.1783{\text{m}}^{\text{3}}/\text{kg}\end{array}$

The initial specific internal energy of the spring-loaded piston cylinder device is,

  $u_1=u_{f1}+x{u}_{fg1}$

  $\begin{array}{c}{u}_1=\left(384.36\text{kJ}/\text{kg}\right)+\left(0.08\right)\times \left(2111.8\text{kJ}/\text{kg}\right)\\ =\left(384.36\text{kJ}/\text{kg}\right)+\left(168.94\text{kJ}/\text{kg}\right)\\ =553.30\text{kJ}/\text{kg}\end{array}$

The mass of the system is,

  $m=\frac{{\nu }_1}{v_1}$

  $\begin{array}{c}m=\frac{2{\text{m}}^{\text{3}}}{0.1783{\text{m}}^{\text{3}}/\text{kg}}\\ =11.217\text{kg}\end{array}$

Determine the final specific volume of the piston cylinder device.

  $v_2=\frac{V_2}{m}$

  $\begin{array}{c}{v}_2=\frac{\left(5{\text{m}}^{\text{3}}\right)}{\left(11.217\text{kg}\right)}\\ =0.44575{\text{m}}^{\text{3}}/\text{kg}\\ \cong 0.4458{\text{m}}^{\text{3}}/\text{kg}\end{array}$

From the Table A-5, to obtain the value of the specific volume of saturated liquid is $v_f$, the specific volume of saturated vapour is $v_g$, the specific internal energy of saturated liquid is $u_f$, the specific internal energy change upon vaporization is $v_{fg}$ at final pressure of $225\text{kPa}$.

  $\begin{array}{l}{v}_{f2}=0.001064{\text{m}}^{\text{3}}/\text{kg}\\ v_{g2}=0.79329{\text{m}}^{\text{3}}/\text{kg}\\ u_{f2}=520.47\text{kJ}/\text{kg}\\ u_{fg2}=2012.7\text{kJ}/\text{kg}\end{array}$

Determine the quality of final state for the spring-loaded piston-cylinder device.

  $x_2=\frac{v_2-v_{f2}}{\left(v_{g2}-v_{f2}\right)}$

  $\begin{array}{c}{x}_2=\frac{\left(0.4458{\text{m}}^{\text{3}}/\text{kg}\right)-\left(0.001064{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.79329{\text{m}}^{\text{3}}/\text{kg}-0.001064{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =\frac{\left(0.444736{\text{m}}^{\text{3}}/\text{kg}\right)}{\left(0.792226{\text{m}}^{\text{3}}/\text{kg}\right)}\\ =0.561375\end{array}$

The final specific internal energy of the spring-loaded piston cylinder device is,

  $u_2=u_{f2}+x_2{u}_{fg2}$

  $\begin{array}{c}{u}_2=\left(520.47\text{kJ}/\text{kg}\right)+\left(0.561375\right)\times \left(2012.7\text{kJ}/\text{kg}\right)\\ =\left(520.47\text{kJ}/\text{kg}\right)+\left(1129.75\text{kJ}/\text{kg}\right)\\ =1650.35\text{kJ}/\text{kg}\\ \cong 1650.4\text{kJ}/\text{kg}\end{array}$

The work done during the constant pressure process is,

  $\begin{array}{c}{W}_{b,out}={\displaystyle {\int }_1^{2}Pd\nu }\\ =\frac{P_1+P_2}{2}\left({\nu }_2-{\nu }_1\right)\end{array}$

  $\begin{array}{c}{W}_{\text{b,out}}=\frac{\left(75+225\right)\text{kPa}}{2}\left(5-2\right){\text{m}}^{\text{3}}\\ =\frac{300\text{kPa}}{2}\times 3{\text{m}}^{\text{3}}\\ =150\text{kPa}\times 3{\text{m}}^{\text{3}}\\ =450\text{kJ}\end{array}$

Thus, the work produced during the process is $450\text{kJ}$.

The energy balance equation is,

  $\begin{array}{l}{E}_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}\\ Q_{in}-W_{b,out}=\Delta U\\ Q_{in}=m\left(u_2-u_1\right)+W_{b,out}\end{array}$

  $\begin{array}{c}{Q}_{\text{in}}=\left(11.217\text{kg}\right)\left(1650.4\text{kJ}/\text{kg}-553.30\text{kJ}/\text{kg}\right)+450\text{kJ}\\ =\left(11.217\text{kg}\right)\left(1097.1\text{kJ}/\text{kg}\right)+450\text{kJ}\\ =12756\text{kJ}\end{array}$

Thus, the heat transferred during the process is $12,756\text{kJ}$.