Chapter 4.5, Problem 139RP

A piston–cylinder device initially contains 0.35 kg of steam at 3.5 MPa, superheated by 7.4°C. Now the steam loses heat to the surroundings and the piston moves down, hitting a set of stops at which point the cylinder contains saturated liquid water. The cooling continues until the cylinder contains water at 200°C. Determine (a) the final pressure and the quality (if mixture), (b) the boundary work, (c) the amount of heat transfer when the piston first hits the stops, and (d) the total heat transfer.

FIGURE P4–141

To determine

The final pressure of the piston-cylinder device.

The quality at the final state of the piston-cylinder device.

Answer to Problem 139RP

The final pressure of the piston-cylinder device is $\underset{\_}{1555\text{kPa}}$.

The quality at the final state of the piston-cylinder device is $\underset{\_}{0.00062}$.

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 $\Delta E_{\text{system}}$.

Simplify Equation (I) and write energy balance relation of piston-cylinder device.

$W_{\text{b,in}}-Q_{\text{out}}=\Delta U$ (III)

Here, the work to be done into the system is $W_{\text{b,in}}$, the heat to be transfer by the system is $Q_{\text{out}}$, and the change in the internal energy is $\Delta U$.

Simplify the Equation (III), write energy balance relation when the piston first hits the stops state(1-2).

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

Here, the mass of the piston-cylinder device is $m$, the specific internal energy at state 2 for steam is $u_2$, and the specific internal energy at state 1 for steam is $u_1$.

Similarly the Equation (IV), when the piston first hits and the final state(1-3).

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

Here, the mass of the piston-cylinder device is $m$, the specific internal energy at state 3 for steam is $u_3$, and the specific internal energy at state 1 for steam is $u_1$.

Determine the state 1 temperature of the piston-cylinder device.

$T_1=T_{\text{sat@3500}\text{kPa}}+\Delta T_{\text{superheated}}$ (VI)

Here, the saturated temperature at 3500 kPa is $T_{\text{sat@3500}\text{kPa}}$ and the superheated of the piston-cylinder device is $\Delta T_{\text{superheated}}$.

Conclusion:

Write the unit conversion pressure from MPa to kPa for piston-cylinder device.

$\begin{array}{c}{P}_1=3.5\text{MPa}\times \left(\frac{1000\text{kPa}}{1\text{MPa}}\right)\\ =3500\text{kPa}\end{array}$

From the Table A-5 “Saturated water-Pressure table”, obtain the value of saturated temperature at 3500 kPa pressure as $242.56°\text{C}$.

Substitute $242.56°\text{C}$ for $T_{\text{sat@3500}\text{kPa}}$ and $7.4°\text{C}$ for $\Delta T_{\text{superheated}}$ in Equation (VI).

$\begin{array}{c}{T}_1=242.56°\text{C}+7.4°\text{C}\\ =250\text{°C}\end{array}$

From the Table A-4 through A-6 “Saturated water”, obtain the value of steam at various states for piston-cylinder device.

At state 1 pressure and temperature of steam as $3.5\text{MPa}$ and $250°\text{C}$.

$v_1=0.05875{\text{m}}^{\text{3}}/\text{kg}$.

$u_1=2624.0\text{kJ}/\text{kg}$.

At state 1-2 pressure and quality of state of steam as $3.5\text{MPa}$ and 0.

$v_2=0.001235{\text{m}}^{\text{3}}/\text{kg}$.

$u_2=1045.4\text{kJ}/\text{kg}$.

At state 2-3 specific volume and temperature of steam as $0.001235{\text{m}}^{\text{3}}/\text{kg}$ and $200°\text{C}$.

$x_3=0.00062$.

$P_3=1555\text{kPa}$.

$u_3=851.55\text{kJ}/\text{kg}$.

Thus, the final pressure of the piston-cylinder device is $\underset{\_}{1555\text{kPa}}$ and quality at the final state of the piston-cylinder device is $\underset{\_}{0.00062}$.

To determine

The boundary work done of the piston-cylinder device.

Answer to Problem 139RP

The boundary work done of the piston-cylinder device is $\underset{\_}{70.46\text{kJ}}$.

Explanation of Solution

Substitute 0 for $Q_{\text{out}}$ in Equation (III) and write energy balance relation of boundary work for piston-cylinder device.

$\begin{array}{c}{W}_{\text{b,in}}=\Delta U\\ =m{P}_1\left(v_1-v_2\right)\end{array}$ (VII)

Here, the mass of piston-cylinder device is $m$, the pressure at state 1 is $P_1$, the specific volume of the state 1 is $v_1$, and the specific volume of the state 2 is $v_2$.

Conclusion:

Substitute $0.35\text{kg}$ for $m$, $3500\text{kPa}$ for $P_1$, $0.05875{\text{m}}^{\text{3}}/\text{kg}$ for $v_1$, and $0.001235{\text{m}}^{\text{3}}/\text{kg}$ in Equation (VII).

$\begin{array}{c}{W}_{\text{b,in}}=\left(0.35\text{kg}\right)\left(3500\text{kPa}\right)\left(0.05875-0.001235\right){\text{m}}^{\text{3}}/\text{kg}\\ =\left(0.35\text{kg}\right)\left(3500\text{kPa}\right)\left(0.057515\right){\text{m}}^{\text{3}}/\text{kg}\\ =70.45588\text{kPa}\cdot {\text{m}}^{\text{3}}\times \left(\frac{1\text{kJ}}{1\text{kPa}\cdot {\text{m}}^{\text{3}}}\right)\\ =70.46\text{kJ}\end{array}$

Thus, the boundary work done of the piston-cylinder device is $\underset{\_}{70.46\text{kJ}}$.

To determine

The amount of heat transfer when the piston first hits the stops.

Answer to Problem 139RP

The amount of heat transfer when the piston first hits the stops is $\underset{\_}{620\text{kJ}}$.

Explanation of Solution

Conclusion:

Substitute $70.45\text{kJ}$ for $W_{\text{b,in}}$, $0.35\text{kg}$ for $m$, $1045.4\text{kJ}/\text{kg}$ for $u_2$, and $2624.0\text{kJ}/\text{kg}$ in Equation (IV).

$\begin{array}{c}{Q}_{\text{out,1-2}}=\left(70.45\text{kJ}\right)-\left(0.35\text{kg}\right)\left(1045.4-2624.0\right)\text{kJ}/\text{kg}\\ =\left(70.45\text{kJ}\right)-\left(0.35\text{kg}\right)\left(-1569.6\right)\text{kJ}/\text{kg}\\ =619.81\text{kJ}\\ \cong \text{620}\text{kJ}\end{array}$

Thus, the amount of heat transfer when the piston first hits the stops is $\underset{\_}{620\text{kJ}}$.

To determine

The total amount heat transfer in the piston-cylinder device.

Answer to Problem 139RP

The total amount heat transfer in the piston-cylinder device is $\underset{\_}{690.8\text{kJ}}$.

Explanation of Solution

Conclusion:

Substitute $70.45\text{kJ}$ for $W_{\text{b,in}}$, $0.35\text{kg}$ for $m$, $851.55\text{kJ}/\text{kg}$ for $u_3$, and $2624.0\text{kJ}/\text{kg}$ in Equation (IV).

$\begin{array}{c}{Q}_{\text{out,1-2}}=\left(70.45\text{kJ}\right)-\left(0.35\text{kg}\right)\left(851.55-2624.0\right)\text{kJ}/\text{kg}\\ =\left(70.45\text{kJ}\right)-\left(0.35\text{kg}\right)\left(-1772.45\right)\text{kJ}/\text{kg}\\ =690.8\text{kJ}\end{array}$

Thus, the total amount heat transfer in the piston-cylinder device is $\underset{\_}{690.8\text{kJ}}$.