Chapter 7.13, Problem 173RP

To determine

The work and heat transfer for each process.

Answer to Problem 173RP

The heat transfer for the isothermal process 1–2 is $117.7\text{kJ}$.

The work done during the process 1-2 is $117.7\text{kJ}$.

The work done during the isentropic compression process 2-3 is $97.1\text{kJ}$.

The heat transfer for the isentropic process 2–3 is $0\text{kJ}$.

The work done during constant pressure compression process 3-1 is $37\text{kJ}$.

The heat transfer during constant pressure compression process 3-1 is $135.8\text{kJ}$.

Explanation of Solution

Write the expression to calculate the enthalpy change in process 1-2.

$\Delta S_{1-2}=-mR\ln \frac{P_2}{P_1}$        (I)

Here, pressure at process 1 is $P_1$ , mass of the air is m, gas constant of air is R and pressure at process 2 is $P_2$.

Write the expression to calculate the ideal gas equation, to find mass of the air.

$m=\frac{P_1{\nu }_1}{R{T}_1}$        (II)

Here, mass of the air is m , volume at process 1 is ${\nu }_1$ and temperature at process1 is $T_1$

Write the expression to calculate the heat transfer for the isothermal process 1–2.

$Q_{in,1-2}=T_1\Delta S_{1-2}$             (III)

Here, enthalpy change in process 1-2 is $\Delta S_{1-2}$, heat transfer for the isothermal process 1–2 is $Q_{in,1-2}$ and temperature at 1 stage is $T_1$,

Write the expression to calculate the work done during the process 1-2 $\left(W_{out,1-2}\right)$.

$W_{out,1-2}=Q_{in,1-2}$      (IV)

Write the expression to calculate the work done during the isentropic compression process 2-3 $\left(W_{in,2-3}\right)$.

$W_{in,2-3}=m\left(u_3-u_2\right)$        (V)

Here, mass of the air is m, internal energy at process 3 is $u_3$ and internal energy at process 2 is $u_2$.

Write the expression to calculate the relative pressure at process 3 $\left(P_{r3}\right)$.

$P_{r3}=\frac{P_3}{P_2}{P}_{r2}$        (VI)

Here, relative pressure at process 2 is $P_{r2}$.

Write the expression to calculate the volume at process 3$\left({\nu }_3\right)$

${\nu }_3=\frac{mR{T}_3}{P_3}$      (VII)

Write the expression to calculate the work done during constant pressure compression process 3-1 $\left(W_{in,3-1}\right)$.

$W_{in,3-1}=P_3\left({\nu }_3-{\nu }_1\right)$      (VIII)

Here, volume at process 3 is ${\nu }_3$ and volume at process 1 is ${\nu }_1$.

Write the expression to calculate the heat transfer during constant pressure compression process 3-1$\left(Q_{out,3-1}\right)$.

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

Here, heat transfer during constant pressure compression process 3-1 is $Q_{out,3-1}$ and internal energy at process 1 is $u_1$

Conclusion:

From Table A-1 “the molar mass, gas constant and critical point properties table”, obtain the gas constant $\left(R\right)$ of air as $0.287\text{kJ}/\text{kg}\cdot \text{K}$.

Substitute $\text{400}\text{kPa}$ for $P_1$, $0.3{\text{m}}^3$ for ${\nu }_1$, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for $R$ and 300 K for $T_1$ in Equation (II).

$\begin{array}{c}m=\frac{\left(400\text{ kPa}\right)\left(0.3{\text{m}}^3\right)}{\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(300\text{K}\right)}\\ =1.394\text{kg}\end{array}$

Substitute 1.394 kg for m, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, $\text{400}\text{kPa}$ for $P_1$ and $\text{150}\text{kPa}$ for $P_2$ in Equation (I).

$\begin{array}{c}\Delta S_{1-2}=-\left(1.394\text{kg}\right)\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\ln \frac{150\text{kPa}}{400\text{kPa}}\\ =0.3924\text{kJ}/\text{K}\end{array}$

Substitute $27°\text{C}$ for $T_1$ and $0.3924\text{kJ}/\text{K}$ for $\Delta S_{1-2}$ in Equation (III).

$\begin{array}{c}{Q}_{in,1-2}=27°\text{C}\left(0.3924\text{kJ}/\text{K}\right)\\ =\left(27+273\right)\text{K}\left(0.3924\text{kJ}/\text{K}\right)\\ =117.7\text{kJ}\end{array}$

Thus, the heat transfer for the isothermal process 1–2 is $117.7\text{kJ}$.

Substitute $117.7\text{kJ}$ for $Q_{in,1-2}$ in Equation (IV).

$W_{out,1-2}=117.7\text{kJ}$

Thus, the work done during the process 1-2 is $117.7\text{kJ}$.

From Table A-17, “Ideal-gas properties of air”, obtain the internal energy $\left(u_1\right)$ or $\left(u_2\right)$, entropy $\left(s_1\right)$ or $\left(s_2\right)$, and relative pressure $\left(P_{r1}\right)$ or $\left(P_{r2}\right)$, at temperature of $\text{300}\text{K}$ as

$214.07\text{kJ}/\text{kg}$, $1.70203\text{kJ}/\text{kg}\cdot \text{K}$ and 1.3860 respectively.

Substitute $\text{400}\text{}\text{kPa}$ for $P_3$, $\text{150}\text{kPa}$ for $P_2$ and 1.3860 for $P_{r2}$ in Equation (VI).

$\begin{array}{c}{P}_{r3}=\frac{400\text{kPa}}{150\text{kPa}}\left(1.3860\right)\\ =3.696\end{array}$

Refer to Table A-17, “Ideal-gas properties of air”.

Obtain the select the internal energy $\left(u_3\right)$ and temperature $\left(T_3\right)$ at the relative pressure of 3.696 by using interpolation method.

Write the formula of interpolation method of two variables.

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

Here, the variables denoted by x and y are relative pressure and internal energy.

Show relative pressure and internal energy values from the Table A-17.

Relative pressure $\left(P_{r3}\right)$	Internal energy $\left(u_3\right)$, in $\text{kJ}/\text{kg}$
3.481	278.93
3.696	?
3.806	286.16

Substitute $3.481$ for $x_1$, $3.696$ for $x_2$, $3.806$ for $x_3$, $278.93$ for $y_1$, and $286.16$ for $y_3$ in Equation (X).

$\begin{array}{c}{y}_2=\frac{\left(3.696-3.481\right)\left(286.16-278.931\right)}{\left(3.806-3.481\right)}+278.93\\ =283.71\end{array}$

The value of internal energy process 1 $\left(u_3\right)$ at the relative pressure of 3.696 is $283.71\text{kJ}/\text{kg}$.

Show temperature and initial internal energy values from the Table A-17.

Temperature $\left(T_3\right)$	Internal energy $\left(u_3\right)$, in $\text{kJ}/\text{kg}$
3.481	390
3.696	?
3.806	400

Substitute $3.481$ for $x_1$, $3.696$ for $x_2$, $3.806$ for $x_3$, $390$ for $y_1$, and $400$ for $y_3$ in Equation (X).

$\begin{array}{c}{y}_2=\frac{\left(3.696-3.481\right)\left(400-390\right)}{\left(3.806-3.481\right)}+390\\ =396.615\end{array}$

The value of Temperature $\left(T_3\right)$ at the relative pressure of 3.696 is $396.615\text{K}$.

Substitute 1.394 kg for m, $283.71\text{kJ}/\text{kg}$ for $u_3$ and $214.07\text{kJ}/\text{kg}$ for $u_2$ in Equation (V).

$\begin{array}{c}{W}_{in,2-3}=\left(1.394kg\right)\left(283.71\text{kJ}/\text{kg}-214.07\text{kJ}/\text{kg}\right)\\ =97.1\text{kJ}\end{array}$

Thus, the work done during the isentropic compression process 2-3 is $97.1\text{kJ}$.

The heat transfer for the isentropic process 2–3 is zero when entropy change remains unchanged for the isentropic compression process.

Thus, the heat transfer for the isentropic process 2–3 is $0\text{kJ}$.

Substitute 1.394 kg for m, $0.287\text{kJ}/\text{kg}\cdot \text{K}$ for R, 396.6 K for $T_3$ and 400 kPa for $P_3$ in Equation (VII).

$\begin{array}{c}{\nu }_3=\frac{\left(1.394\text{ kg}\right)\left(0.287\text{kJ}/\text{kg}\cdot \text{K}\right)\left(396.6\text{K}\right)}{\left(400\text{kPa}\right)}\\ =0.3967{\text{m}}^{\text{3}}\end{array}$

Substitute $\text{400}\text{kPa}$ for $P_3$, $0.3967{\text{m}}^{\text{3}}$ for $v_3$ and $0.3{\text{m}}^3$ for $v_1$ in Equation (VIII).

$\begin{array}{c}{W}_{in,3-1}=\left(400\text{ kPa}\right)\left(0.3967{\text{m}}^{\text{3}}-0.3{\text{m}}^3\right)\\ =37\text{kJ}\end{array}$

Thus, the work done during constant pressure compression process 3-1 is $37\text{kJ}$.

Substitute 37 kJ for $W_{in,3-1}$, 1.394 kg for m, $214.07\text{kJ}/\text{kg}$ for $u_1$ and $283.71\text{kJ}/\text{kg}$ for $u_3$ in Equation (IX).

$\begin{array}{c}{Q}_{out,3-1}=37\text{kJ}-\left(1.394\text{ kg}\right)\left(214.07\text{kJ}/\text{kg}-283.71\text{kJ}/\text{kg}\right)\\ =135.8\text{kJ}\end{array}$

Thus, the heat transfer during constant pressure compression process 3-1 is $135.8\text{kJ}$.