Chapter 4.5, Problem 40P

Steam at 75 kPa and 8 percent quality is contained in a spring-loaded piston–cylinder device, as shown in Fig. P4–40, with an initial volume of 2 m³. Steam is now heated until its volume is 5 m³ and its pressure is 225 kPa. Determine the heat transferred to and the work produced by the steam during this process.

FIGURE P4–40

To determine

The heat transfer of the spring-loaded piston cylinder device.

The work done of the spring-loaded piston cylinder device.

Answer to Problem 40P

The heat transfer of the spring-loaded piston cylinder device is $\underset{\_}{450\text{kJ}}$.

The work done of the spring-loaded piston cylinder device is $\underset{\_}{12,756\text{kJ}}$.

Explanation of Solution

Write the expression for the energy balance equation.

$E_{\text{in}}-E_{\text{out}}=\Delta E_{\text{system}}$ (I)

Here, the total energy entering the system is $E_{\text{in}}$, the total energy leaving the system is $E_{\text{out}}$, and the change in the total energy of the system is $\Delta E_{\text{system}}$.

Substitute $Q_{\text{in}}$ for $E_{\text{in}}$, $W_{b,out}$ for $E_{\text{out}}$, and $\Delta U$ for $\Delta E_{\text{system}}$ in Equation (I)

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

Here, the mass of the piston cylinder device is $m$, the final specific internal energy is $u_2$, the initial specific internal energy is $u_1$, and the work done during the process is $W_{b,out}$.

Calculate the specific volume of the spring-loaded piston cylinder device.

$v=v_f+x{v}_{fg}$ (III)

Here, the specific volume of saturated liquid is $v_f$, the specific volume of saturated vapour is $v_g$.

Calculate the specific internal energy of the spring-loaded piston cylinder device.

$u=u_f+x{u}_{fg}$ (IV)

Here, the specific internal energy of saturated liquid is $u_f$, the specific internal energy change upon vaporization is $u_{fg}$.

Write the expression for the mass of the system.

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

Here, the initial volume of the system is ${\nu }_1$ and the specific volume of the system is $v_1$.

Determine the final specific volume of the piston cylinder device.

$v_2=\frac{{\nu }_2}{m}$ (VI)

The final volume of the piston cylinder device is ${\nu }_2$.

Determine the work done during the constant pressure process.

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

Here, the initial pressure is $P_1$, the final pressure is $P_2$, the initial volume is ${\nu }_1$, and the final volume is ${\nu }_2$.

Conclusion:

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 $250\text{kPa}$.

$\begin{array}{l}{v}_f=0.001037{\text{m}}^{\text{3}}/\text{kg}\\ v_g=2.2172{\text{m}}^{\text{3}}/\text{kg}\\ u_f=384.36\text{kJ}/\text{kg}\\ u_{fg}=2111.8\text{kJ}/\text{kg}\end{array}$

Substitute 0.08 for $x$, $0.001037{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, and $2.2172{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$ in Equation (III).

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

Substitute 0.08 for $x$, $384.36\text{kJ}/\text{kg}$ for $u_f$, and $2111.8\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (IV).

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

Substitute $2{\text{m}}^{\text{3}}$ for ${\nu }_1$ and $0.1783{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$ in Equation (V)

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

Substitute $5{\text{m}}^{\text{3}}$ for ${\nu }_2$ and $11.217\text{kg}$ for $m$ in the Equation (VI).

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

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 $225\text{kPa}$.

$\begin{array}{l}{v}_f=0.001064{\text{m}}^{\text{3}}/\text{kg}\\ v_g=0.79329{\text{m}}^{\text{3}}/\text{kg}\\ u_f=520.47\text{kJ}/\text{kg}\\ u_{fg}=2012.7\text{kJ}/\text{kg}\end{array}$

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

$x_2=\frac{v_2-v_f}{\left(v_g-v_f\right)}$ (VIII)

Here, the specific volume of saturated liquid is $v_f$ and the specific volume of saturated vapour is $v_g$.

Substitute $0.001064{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $0.79329{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$, and $0.4458{\text{m}}^{\text{3}}/\text{kg}$ for $v_2$ in Equation (III).

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

Substitute 0.561375 for $x$, $520\text{kJ}/\text{kg}$ for $u_f$, and $2012.7\text{kJ}/\text{kg}$ for $u_{fg}$ in Equation (IV).

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

Substitute $75\text{kPa}$ for $P_1$, $225\text{kPa}$ for $P_2$, $2{\text{m}}^{\text{3}}$ for ${\nu }_1$, and $5{\text{m}}^{\text{3}}$ for ${\nu }_2$ in Equation (VII)

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

Thus, the heat transfer of the spring-loaded piston cylinder device is $\underset{\_}{450\text{kJ}}$.

Substitute $11.217\text{kg}$ for $m$, $450\text{kJ}$ for $W_{b,out}$, $1650.4\text{kJ}/\text{kg}$ for $u_2$, and $553.30\text{kJ}/\text{kg}$ for $u_1$ in Equation (II).

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

Thus, the work done of the spring-loaded piston cylinder device is $\underset{\_}{12,756\text{kJ}}$.