Chapter 4.5, Problem 114RP

(a)

To determine

The initial temperature of the piston cylinder device.

The final temperature of the piston cylinder device.

Answer to Problem 114RP

The initial temperature of the piston cylinder device is $\underset{\_}{113.2°\text{C}}$.

The final temperature of the piston cylinder device is $\underset{\_}{679.6°\text{C}}$.

Explanation of Solution

Determine the total initial volume of piston cylinder device.

${\nu }_1=m_f{v}_f+m_g{v}_g$ (I)

Here, the mass of the liquid phase is $m_f$, the specific volume of saturated liquid is $v_f$, the mass of the vapour phase is $m_g$, and the specific volume of saturated vapour is $v_g$.

Determine the total volume of the piston cylinder device at final state.

${\nu }_3=1.2{\nu }_1$ (II)

Determine the specific volume of the piston cylinder device at final state.

$v_3=\frac{{\nu }_3}{m}$ (III)

Here, the mass of the saturated liquid vapour mixture of water is contained in a piston cylinder device is $m$.

Conclusion:

Write the formula of interpolation method of two variables.

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

Here, the variables denote by x and y is saturated pressure and saturated temperature.

For initial temperature of the piston cylinder device.

Show the temperature at pressure of 150 kPa, 160 kPa, and 175 kPa as in Table (1).

Pressure, kPa $\left(x\right)$	Temperature,  C $\left(y\right)$
150 kPa	111.35
160 kPa	$y_2=?$
175 kPa	116.04

Substitute the value of x and y from Table (1) in Equation (IV) to calculate the value of initial temperature $\left(y_2\right)$ at pressure 160 kPa as $113.2°\text{C}$.

Thus, the initial temperature of the piston cylinder device is $\underset{\_}{113.2°\text{C}}$.

For specific volume of saturated liquid of the piston cylinder device.

Show the specific volume of saturated liquid at pressure of 150 kPa, 160 kPa, and 175 kPa as in Table (2).

Pressure, kPa $\left(x\right)$	Specific volume of saturated liquid, ${\text{m}}^{\text{3}}/\text{kg}$ $\left(y\right)$
150 kPa	0.001053
160 kPa	$y_2=?$
175 kPa	0.001057

Substitute the value of x and y from Table (2) in Equation (IV) to calculate the value of specific volume of saturated liquid $\left(y_2\right)$ at pressure 160 kPa as $0.001055{\text{m}}^{\text{3}}/\text{kg}$.

For specific volume of saturated vapour of the piston cylinder device.

Show the specific volume of saturated vapour at pressure of 150 kPa, 160 kPa, and 175 kPa as in Table (3).

Pressure, kPa $\left(x\right)$	Specific volume of saturated vapour, ${\text{m}}^{\text{3}}/\text{kg}$ $\left(y\right)$
150 kPa	1.1594
160 kPa	$y_2=?$
175 kPa	1.0037

Substitute the value of x and y from Table (3) in Equation (IV) to calculate the value of specific volume of saturated vapour $\left(y_2\right)$ at pressure 160 kPa as $1.097{\text{m}}^{\text{3}}/\text{kg}$.

Substitute $1\text{kg}$ for $m_f$, $0.001055{\text{m}}^{\text{3}}/\text{kg}$ for $v_f$, $2\text{kg}$ for $m_g$, and $1.097{\text{m}}^{\text{3}}/\text{kg}$ for $v_g$ in Equation (I).

$\begin{array}{c}{\nu }_1=\left(1\text{kg}\right)\left(0.001055{\text{m}}^{\text{3}}/\text{kg}\right)+\left(2\text{kg}\right)\left(1.097{\text{m}}^{\text{3}}/\text{kg}\right)\\ =\left(0.001055{\text{m}}^{\text{3}}\right)+\left(2.194{\text{m}}^{\text{3}}\right)\\ =2.195{\text{m}}^{\text{3}}\end{array}$

Substitute $2.195{\text{m}}^{\text{3}}$ for ${\nu }_1$ in Equation (II).

$\begin{array}{c}{\nu }_3=1.2\times \left(2.195{\text{m}}^{\text{3}}\right)\\ =2.634{\text{m}}^{\text{3}}\end{array}$

Substitute $2.634{\text{m}}^{\text{3}}$ for ${\nu }_3$ and $3\text{kg}$ for $m$ in Equation (III).

$\begin{array}{c}{\nu }_3=\frac{2.634{\text{m}}^{\text{3}}}{3\text{kg}}\\ =0.87802{\text{m}}^{\text{3}}/\text{kg}\end{array}$

The unit conversion of pressure from kPa to MPa.

$\begin{array}{c}{P}_3=500\text{kPa}\times \left(\frac{{10}^{-3}\text{MPa}}{1\text{kPa}}\right)\\ =0.5\text{MPa}\end{array}$

For temperature of the piston cylinder device at final state.

Show the temperature at specific volume of the piston cylinder device at final state at $0.80409{\text{m}}^{\text{3}}/\text{kg}$, $0.87802{\text{m}}^{\text{3}}/\text{kg}$, and $0.89696{\text{m}}^{\text{3}}/\text{kg}$ as in Table (4).

specific volume of the piston cylinder device at final state, ${\text{m}}^{\text{3}}/\text{kg}$ $\left(x\right)$	Temperature, $°\text{C}$ $\left(y\right)$
$0.80409$	600
$0.87802$	$y_2=?$
$0.89696$	700

Substitute the value of x and y from Table (4) in Equation (IV) to calculate the value of temperature of the piston cylinder device at final state $\left(y_2\right)$ at specific volume of saturated vapour $0.87802{\text{m}}^{\text{3}}/\text{kg}$ as $679.6°\text{C}$.

Thus, the final temperature of the piston cylinder device is $\underset{\_}{679.6°\text{C}}$.

(b)

To determine

The mass of liquid water when the piston first starts moving.

Answer to Problem 114RP

The mass of liquid water when the piston first starts moving is $\underset{\_}{0\text{kg}}$.

Explanation of Solution

Determine the specific volume of the piston cylinder device at this state.

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

Here, the mass of the saturated liquid vapour mixture of water is contained in a piston cylinder device is $m$.

Conclusion:

Since, ${\nu }_1={\nu }_2$ the volume of piston cylinder device at state 1 is equal to volume of piston cylinder device at state 2.

Substitute $2.195{\text{m}}^{\text{3}}$ for ${\nu }_2$ and $3\text{kg}$ for $m$ in Equation (V).

$\begin{array}{c}{\nu }_3=\frac{2.195{\text{m}}^{\text{3}}}{3\text{kg}}\\ =0.731667{\text{m}}^{\text{3}}/\text{kg}\end{array}$

Therefore, the value of specific volume of the piston cylinder device at this state is greater than $v_g=0.3748{\text{m}}^{\text{3}}/\text{kg}$ at 500 kPa. However, the mass of the liquid will be converted into gaseous phase.

Thus, the mass of liquid water when the piston first starts moving is $\underset{\_}{0\text{kg}}$.

(c)

To determine

The work done during the process state 2 and 3.

Answer to Problem 114RP

The work done during the process state 2 and 3 is $\underset{\_}{220\text{kJ}}$.

Explanation of Solution

Determine the work done in constant pressure process.

$\begin{array}{c}{W}_b={\displaystyle {\int }_2^{3}Pd\nu }\\ =P_2\left({\nu }_3-{\nu }_2\right)\end{array}$ (VI)

Conclusion:

Substitute $500\text{kPa}$ for $P_2$, $2.634{\text{m}}^{\text{3}}$ for ${\nu }_3$, and $2.195{\text{m}}^{\text{3}}$ for ${\nu }_2$ in Equation (VI).

$\begin{array}{c}{W}_b=\left(500\text{kPa}\right)\left(\left(2.634{\text{m}}^{\text{3}}\right)-\left(2.195{\text{m}}^{\text{3}}\right)\right)\\ =\left(500\text{kPa}\right)\left(0.439{\text{m}}^{\text{3}}\right)\\ =219.5\text{kJ}\\ \cong \text{220}\text{kJ}\end{array}$

Thus, the work done during the process state 2 and 3 is $\underset{\_}{220\text{kJ}}$.

Show the P-v diagram of this process.