Chapter 5, Problem 120RQ

(a)

To determine

The final pressure and quality of the piston-cylinder device.

Explanation of Solution

Given:

The mass of the steam $\left(m\right)$ is $0.35\text{kg}$.

The pressure of the steam $\left(P\right)$ is 3.5 MPa.

The change in temperature $\left(\Delta T_{\text{superheated}}\right)$ is $7.4°\text{C}$.

The final temperature of the steam $\left(T_2\right)$ is .

Calculation:

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 $242.56°\text{C}$.

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

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

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

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 $\underset{\_}{0.00062}$.

(b)

To determine

The boundary work done of the piston-cylinder device.

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 $u_1$.

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

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

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 $v_2$.

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 (VI).

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

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

(c)

To determine

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

Explanation of Solution

Substitute $W_{\text{b,in}}=70.45\text{kJ}$, $m=0.35\text{kg}$, $u_2=1045.4\text{kJ}/\text{kg}$, and $u_1=2624.0\text{kJ}/\text{kg}$ 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 $\underset{\_}{620\text{kJ}}$.

(d)

To determine

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

Explanation of Solution

Substitute $W_{\text{b,in}}=70.45\text{kJ}$, $m=0.35\text{kg}$, $u_3=851.55\text{kJ}/\text{kg}$, and $u_2=2624.0\text{kJ}/\text{kg}$ 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 $\underset{\_}{690.8\text{kJ}}$.