Chapter 12.6, Problem 97RP

a)

To determine

The entropy changes and the work input of the compressor per unit mass by treating the propane as an ideal gas variable specific heats.

Answer to Problem 97RP

The work input of the compressor per unit mass by treating the propane as an ideal gas is $1121\text{kJ/kg}$.

The change in entropy per unit mass by treating the propane as an ideal gas is $1.5873\text{kJ/kg}\cdot \text{K}$.

Explanation of Solution

Refer the table A-2 (c), “Ideal gas specific heats of various common gases”.

The general empirical correlation is ${\overline{c}}_P=a+bT+c{T}^2+d{T}^3$.

Write the formula for enthalpy change in molar basis at ideal gas state ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}$.

$\begin{array}{c}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}{c}_{P}dT}\\ ={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(a+bT+c{T}^2+d{T}^3\right)dT}\\ ={\left[aT+\frac{b}{2}{T}^2+\frac{c}{3}{T}^3+\frac{d}{4}{T}^4\right]}_{T_1}^{T_2}\\ =a\left(T_2-T_1\right)+\frac{b}{2}\left(T_2^2-T_1^2\right)+\frac{c}{3}\left(T_2^3-T_1^3\right)+\frac{d}{4}\left(T_2^4-T_1^4\right)\end{array}$ (I)

Here, specific heat capacity at constant pressure is $c_P$, initial temperature is $T_1$, final temperature is $T_2$, the empirical constants are $a$, $b$, $c$, and $d$.

Write the formula for work input to the compressor $\left(w_{in}\right)$.

$w_{in}=\frac{{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}}{M}$ (II)

Here, molar mass of propane is $M$.

Write the formula for entropy change in molar basis at ideal gas state ${\left({\overline{s}}_2-{\overline{s}}_1\right)}_{\text{ideal}}$.

$\begin{array}{c}{\left({\overline{s}}_2-{\overline{s}}_1\right)}_{\text{ideal}}={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\frac{c_P}{T}dT}-R_u\ln \left(\frac{P_2}{P_1}\right)\\ =={\displaystyle \underset{T_1}{\overset{T_2}{\int }}\left(\frac{a}{T}+b+cT+d{T}^2\right)dT}-R_u\ln \left(\frac{P_2}{P_1}\right)\\ ={\left[a\ln \left(T\right)+bT+\frac{c}{2}{T}^2+\frac{d}{3}{T}^3\right]}_{T_1}^{T_2}-R_u\ln \left(\frac{P_2}{P_1}\right)\\ =a\ln \left(\frac{T_2}{T_1}\right)+b\left(T_2^{}-T_1^{}\right)+\frac{c}{2}\left(T_2^2-T_1^2\right)+\frac{d}{3}\left(T_2^3-T_1^3\right)-R_u\ln \left(\frac{P_2}{P_1}\right)\end{array}$ (III)

Here, universal gas constant is $R_u$, initial pressure is $P_2$, and final pressure is $P_2$.

Write the formula for entropy change in mass basis ${\left(s_2-s_1\right)}_{\text{ideal}}$.

${\left(s_2-s_1\right)}_{\text{ideal}}=\frac{{\left({\overline{s}}_2-{\overline{s}}_1\right)}_{\text{ideal}}}{M}$ . (IV)

Refer table A-1, “Molar mass, gas constant and critical properties table”.

The molar mass $\left(M\right)$ propane is $44.097\text{kg/kmol}$.

Refer the table A-2 (c), “Ideal gas specific heats of various common gases”.

Obtain the empirical constants as follows.

$\begin{array}{l}a=-4.04\\ b=30.48\times {10}^{-2}\\ c=-15.72\times {10}^{-5}\\ d=31.74\times {10}^{-9}\end{array}$

The universal gas constant $\left(R_u\right)$ is $8.314\text{kJ/kmol}\cdot \text{K}$.

Conclusion:

Convert the temperature $T_1,T_2$ from degree Celsius to Kelvin.

$\begin{array}{c}{T}_1=100°\text{C}+273\\ =373\text{K}\end{array}$

$\begin{array}{c}{T}_2=500°\text{C}+273\\ =773\text{K}\end{array}$

Substitute $-4.04$ for $a$, $773\text{K}$ for $T_2$, $373\text{K}$ for $T_1$, $30.48\times {10}^{-2}$ for $b$, $-15.72\times {10}^{-5}$ for $c$, and $31.74\times {10}^{-9}$ for $d$ in Equation (I).

$\begin{array}{c}{\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}=\left\{\begin{array}{l}-4.04\left(773\text{K}-373\text{K}\right)+\frac{30.48\times {10}^{-2}}{2}\left[{\left(773\text{K}\right)}^2-{\left(373\text{K}\right)}^2\right]+\\ \frac{-15.72\times {10}^{-5}}{3}\left[{\left(773\text{K}\right)}^3-{\left(373\text{K}\right)}^3\right]+\frac{31.74\times {10}^{-9}}{4}\left[{\left(773\text{K}\right)}^4-{\left(373\text{K}\right)}^4\right]\end{array}\right\}\\ =-1616+69860.16-21524.727+2679.5227\\ =49398.9557\text{kJ/kmol}\\ \simeq 49440\text{kJ/kmol}\end{array}$

Substitute $49440\text{kJ/kmol}$ for ${\left({\overline{h}}_2-{\overline{h}}_1\right)}_{\text{ideal}}$ and $44.097\text{kg/kmol}$ for $M$ in Equation (II)

$\begin{array}{c}{w}_{in}=\frac{49440\text{kJ/kmol}}{44.097\text{kg/kmol}}\\ =1121.1647\text{kJ/kg}\\ \simeq 1121\text{kJ/kg}\end{array}$

Thus, the work input of the compressor per unit mass by treating the propane as an ideal gas variable specific heats is $1121\text{kJ/kg}$.

Substitute $-4.04$ for $a$, $773\text{K}$ for $T_2$, $373\text{K}$ for $T_1$, $30.48\times {10}^{-2}$ for $b$, $-15.72\times {10}^{-5}$ for $c$, $31.74\times {10}^{-9}$ for $d$, $4000\text{kPa}$ for $P_2$, $500\text{kPa}$ for $P_1$, and $8.314\text{kJ/kmol}\cdot \text{K}$ for $R_u$ in

Equation (III).

$\begin{array}{c}{\left({\overline{s}}_2-{\overline{s}}_1\right)}_{\text{ideal}}=\left\{\begin{array}{l}\left(-4.04\right)\ln \left(\frac{773\text{K}}{373\text{K}}\right)+30.48\times {10}^{-2}\left[\left(773\text{K}\right)-\left(373\text{K}\right)\right]\\ +\frac{-15.72\times {10}^{-5}}{2}\left[{\left(773\text{K}\right)}^2-{\left(373\text{K}\right)}^2\right]\\ +\frac{31.74\times {10}^{-9}}{3}\left[{\left(773\text{K}\right)}^3-{\left(373\text{K}\right)}^3\right]\\ -\left(8.314\text{kJ/kmol}\cdot \text{K}\right)\ln \left(\frac{4000\text{kPa}}{500\text{kPa}}\right)\end{array}\right\}\\ =-2.9439+121.92-36.0302+4.3377-17.2885\\ =69.9951\text{kJ/kmol}\cdot \text{K}\end{array}$

Substitute $69.9951\text{kJ/kmol}\cdot \text{K}$ for ${\left({\overline{s}}_2-{\overline{s}}_1\right)}_{\text{ideal}}$ and $44.097\text{kg/kmol}$ for $M$ in Equation (IV).

$\begin{array}{c}{\left(s_2-s_1\right)}_{\text{ideal}}=\frac{69.9951\text{kJ/kmol}\cdot \text{K}}{44.097\text{kg/kmol}}\\ =1.5873\text{kJ/kg}\cdot \text{K}\end{array}$

Thu, the change in entropy per unit mass by treating the propane as an ideal gas variable specific heats is $1.5873\text{kJ/kg}\cdot \text{K}$.

b)

To determine

The entropy changes and the work input of the compressor per unit mass by using departure charts.

Answer to Problem 97RP

The work input of the compressor per unit mass by using enthalpy departure chart is $664.2\text{kJ/kg}$.

The change in entropy per unit mass by using entropy departure chart is $0.8278\text{kJ/kg}\cdot \text{K}$.

Explanation of Solution

Calculate the reduced temperature $\left(T_{R1}\right)$ at initial state.

$T_{R1}=\frac{T_1}{T_{\text{cr}}}$ (V)

Here, critical temperature is $T_{\text{cr}}$ and initial temperature is $T_1$.

Calculate the reduced pressure $\left(P_{R1}\right)$ at initial state.

$P_{R1}=\frac{P_1}{P_{\text{cr}}}$ (VI)

Here, critical pressure is $P_{\text{cr}}$ and initial pressure is $P_1$.

Calculate the reduced temperature $\left(T_{R2}\right)$ at final state.

$T_{R2}=\frac{T_2}{T_{\text{cr}}}$ (VII)

Here, critical temperature is $T_{\text{cr}}$ and final temperature is $T_2$.

Calculate the reduced pressure $\left(P_{R2}\right)$ at final state.

$P_{R2}=\frac{P_2}{P_{\text{cr}}}$ (VIII)

Here, critical pressure is $P_{\text{cr}}$ and final pressure is $P_2$.

Write the formula for change in enthalpy and change in entropy at ideal state.

${\left(h_2-h_1\right)}_{\text{ideal}}=c_p\left(T_2-T_1\right)$ (IX)

${\left(s_2-s_1\right)}_{\text{ideal}}=c_p\ln \frac{T_2}{T_1}-R\ln \frac{P_2}{P_1}$ (X)

Write the formula for change in enthalpy $\left(h_2-h_1\right)$ using generalized enthalpy departure chart relation.

$h_2-h_1={\left(h_2-h_1\right)}_{\text{ideal}}-R{T}_{\text{cr}}\left(Z_{h2}-Z_{h1}\right)$ (XI)

Here, change in enthalpy of ideal gas is ${\left(h_2-h_1\right)}_{\text{ideal}}$, gas constant of propane is $R$, the enthalpy departure factor is $Z_h$, and the subscripts 1 and 2 indicates initial and final states.

Write the formula for change in entropy $\left(s_2-s_1\right)$ using generalized entropy departure chart relation.

$\left(s_2-s_1\right)={\left(s_2-s_1\right)}_{\text{ideal}}-R{T}_{\text{cr}}\left(Z_{s2}-Z_{s1}\right)$ (XII)

Here, change in entropy of ideal gas is ${\left(s_2-s_1\right)}_{\text{ideal}}$, and the entropy departure factor is $Z_s$.

Refer Table A-1, “Molar mass, gas constant, and critical-point properties”.

The critical temperature and critical pressure of the propane is as follows.

$\begin{array}{l}{T}_{\text{cr}}=370\text{K}\\ P_{\text{cr}}=4.26\text{MPa}\end{array}$

Refer Table A-2 (a), “Ideal-gas specific heats of various common gases”.

The gas constant $\left(R\right)$ of propane is $0.1885\text{kJ/kg}\cdot \text{K}$.

The specific heat at constant pressure $\left(c_p\right)$ of propane is $1.6794\text{kJ/kg}\cdot \text{K}$.

Conclusion:

Substitute $373\text{K}$ for $T_1$ and $370\text{K}$ for $T_{\text{cr}}$ in Equation (V).

$\begin{array}{c}{T}_{R1}=\frac{373\text{K}}{370\text{K}}\\ =1.008\end{array}$

Substitute $500\text{kPa}$ for $P_1$ and $4.26\text{MPa}$ for $P_{\text{cr}}$ in Equation (VI).

$\begin{array}{c}{P}_{R1}=\frac{500\text{kPa}}{4.26\text{MPa}}\\ =\frac{0.5\text{MPa}}{4.26\text{MPa}}\\ =0.117\end{array}$

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h1}\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.124$.

Refer Figure A-30, “Generalized entropy departure chart”.

The entropy departure factor $\left(Z_{s1}\right)$ corresponding the reduced pressure $\left(P_{R1}\right)$ and reduced temperature $\left(T_{R1}\right)$ is $0.0837$.

Substitute $773\text{K}$ for $T_2$ and $370\text{K}$ for $T_{cr}$ in Equation (VII).

$\begin{array}{c}{T}_{R2}=\frac{773\text{K}}{370\text{K}}\\ =2.089\end{array}$

Substitute $4000\text{kPa}$ for $P_2$ and $4.26\text{MPa}$ for $P_{\text{cr}}$ in Equation (VIII).

$\begin{array}{c}{P}_{R2}=\frac{4000\text{kPa}}{4.26\text{MPa}}\\ =\frac{4\text{MPa}}{4.26\text{MPa}}\\ =0.939\end{array}$

Refer Figure A-29, “Generalized enthalpy departure chart”.

The enthalpy departure factor $\left(Z_{h2}\right)$ corresponding the reduced pressure $\left(P_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $0.233$.

Refer Figure A-30, “Generalized entropy departure chart”.

The entropy departure factor $\left(Z_{s2}\right)$ corresponding the reduced pressure $\left(P_{R2}\right)$ and reduced temperature $\left(T_{R2}\right)$ is $0.105$.

Substitute $1.6794\text{kJ/kg}\cdot \text{K}$ for $c_p$, $773\text{K}$ for $T_2$, and $373\text{K}$ for $T_1$ in Equation (IX).

$\begin{array}{c}{\left(h_2-h_1\right)}_{\text{ideal}}=1.6794\text{kJ/kg}\cdot \text{K}\left(773\text{K}-373\text{K}\right)\\ =671.76\text{kJ/kg}\\ \simeq 671.8\text{kJ/kg}\end{array}$

Substitute $671.8\text{kJ/kg}$ for ${\left(h_2-h_1\right)}_{\text{ideal}}$, $0.233$ for $Z_{h2}$, $0.124$ for $Z_{h1}$, $0.1885\text{kJ/kg}\cdot \text{K}$ for $R$, and $370\text{K}$ for $T_{\text{cr}}$ in Equation (XI).

$\begin{array}{c}{h}_2-h_1=671.8\text{kJ/kg}-\left(0.1885\text{kJ/kg}\cdot \text{K}\right)\left(370\text{K}\right)\left(0.233-0.124\right)\\ =671.8\text{kJ/kg}-7.6022\text{kJ/kg}\\ =664.19779\\ \simeq 664.2\text{kJ/kg}\end{array}$

Here, the work input of the compressor is equal to the enthalpy difference.

$\begin{array}{c}{w}_{\text{in}}=h_2-h_1\\ =664.2\text{kJ/kg}\end{array}$

Thus, the work input of the compressor per unit mass by using enthalpy departure chart is $664.2\text{kJ/kg}$.

Substitute $1.6794\text{kJ/kg}\cdot \text{K}$ for $c_p$, $773\text{K}$ for $T_2$, $373\text{K}$ for $T_1$, $0.1885\text{kJ/kg}\cdot \text{K}$ for $R$, $4000\text{kPa}$ for $P_2$, and $500\text{kPa}$ for $P_1$ in Equation (X).

$\begin{array}{c}{\left(s_2-s_1\right)}_{\text{ideal}}=\left(\begin{array}{l}\left(1.6794\text{kJ/kg}\cdot \text{K}\right)\ln \left(\frac{773\text{K}}{373\text{K}}\right)\\ -\left(0.1885\text{kJ/kg}\cdot \text{K}\right)\ln \left(\frac{4000\text{kPa}}{500\text{kPa}}\right)\end{array}\right)\\ =1.2238\text{kJ/kg}\cdot \text{K}-0.3919\text{kJ/kg}\cdot \text{K}\\ =0.8318\text{kJ/kg}\cdot \text{K}\end{array}$

Substitute $0.8318\text{kJ/kg}\cdot \text{K}$ for ${\left(s_2-s_1\right)}_{\text{ideal}}$, $0.105$ for $Z_{s2}$, $0.0837$ for $Z_{s1}$ and $0.1885\text{kJ/kg}\cdot \text{K}$ for $R$ in Equation (XII).

$\begin{array}{c}{s}_2-s_1=0.8318\text{kJ/kg}\cdot \text{K}-\left(0.1885\text{kJ/kg}\cdot \text{K}\right)\left(0.105-0.0837\right)\\ =0.8318\text{kJ/kg}\cdot \text{K}-0.004015\text{kJ/kg}\cdot \text{K}\\ =0.82778\text{kJ/kg}\cdot \text{K}\\ \simeq 0.8278\text{kJ/kg}\cdot \text{K}\end{array}$

Thus, the change in entropy per unit mass by using entropy departure chart is $0.8278\text{kJ/kg}\cdot \text{K}$.