Chapter 12.6, Problem 95RP

(a)

To determine

The power output of the turbine.

Answer to Problem 95RP

The power output of the turbine is $922\text{hp}$.

Explanation of Solution

Write the energy rate balance equation for one inlet and one outlet system.

$\left[{\dot{Q}}_{\text{1}}+{\dot{W}}_{\text{1}}+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+g{z}_1\right)\right]-\left[{\dot{Q}}_{\text{2}}+{\dot{W}}_{\text{2}}+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+g{z}_2\right)\right]=\Delta {\dot{E}}_{\text{system}}$ (I)

Here, the rate of heat transfer is $\dot{Q}$, the rate of work transfer is $\dot{W}$, the enthalpy is $h$ and the velocity is $V$, the gravitational acceleration is $g$, the elevation from the datum is $z$ and the rate of change in net energy of the system is $\Delta {\dot{E}}_{\text{system}}$; the suffixes 1 and 2 indicates the inlet and outlet of the system.

The argon flows at steady state through the turbine. Hence, the rate of change in net energy of the system becomes zero.

$\Delta {\dot{E}}_{\text{system}}=0$

Heat loss occurs to the surrounding at the exit. Neglect the potential energy changes. The work done is by the system (turbine) and the work done on the system is zero i.e. ${\dot{W}}_1=0$.

The Equations (II) reduced as follows to obtain the work input.

$\begin{array}{l}\left[0+0+\dot{m}\left(h_1+\frac{V_1{}^2}{2}+0\right)\right]-\left[{\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}+0\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-\left[{\dot{Q}}_2+{\dot{W}}_2+\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)\right]=0\\ \dot{m}\left(h_1+\frac{V_1{}^2}{2}\right)-{\dot{Q}}_2-{\dot{W}}_2-\dot{m}\left(h_2+\frac{V_2{}^2}{2}\right)=0\\ {\dot{W}}_2=-\dot{m}\left[\left(h_2-h_1\right)+\frac{V_2{}^2-V_1{}^2}{2}\right]-{\dot{Q}}_2\end{array}$ (II)

Here, the ${\dot{W}}_1$ is the power input of the turbine.

Write formula for enthalpy departure factor $\left(Z_h\right)$.

$Z_h=\frac{{\left(h_{\text{ideal}}-h\right)}_{T,P}}{R{T}_{\text{cr}}}$ (III)

Here, the enthalpy at ideal gas state is $h_{\text{ideal}}$, the enthalpy and normal state is $h$, the gas constant of argon is $R$, and the critical temperature is $T_{\text{cr}}$; The subscripts $T,P$ indicates the correspondence of given temperature and pressure.

Rearrange the Equation (III) to obtain $h$.

$h=h_{\text{ideal}}-Z_{h}R{T}_{\text{cr}}$ (IV)

Refer Equation (IV) express as two states of enthalpy difference (initial and final).

$h_2-h_1={\left(h_2-h_1\right)}_{\text{ideal}}-\left(Z_{h2}-Z_h{}_1\right)R{T}_{\text{cr}}$ (V)

The change in enthalpy at ideal state is expressed as follow.

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

Here, the specific heat is $c_p$, the exit temperature is $T_2$ and the inlet temperature is $T_1$.

Substitute $c_p\left(T_2-T_1\right)$ for $h_2-h_1$ in Equation (V).

$h_2-h_1=c_p\left(T_2-T_1\right)-\left(Z_{h2}-Z_h{}_1\right)R{T}_{\text{cr}}$ (VI)

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

The critical temperature and pressure of argon gas is as follows.

$\begin{array}{l}{T}_{\text{cr}}=272\text{R}\\ P_{\text{cr}}=705\text{psia}\end{array}$

The reduced pressure $\left(P_{R1}\right)$ and temperature $\left(T_{R1}\right)$ at initial state is expressed as follows.

$\begin{array}{c}{T}_{R1}=\frac{T_1}{T_{\text{cr}}}\\ =\frac{1000\text{}\text{R}}{272\text{R}}\\ =3.68\end{array}$

$\begin{array}{c}{P}_{R1}=\frac{P_1}{P_{\text{cr}}}\\ =\frac{1000\text{psia}}{705\text{psia}}\\ =1.418\end{array}$

The reduced pressure $\left(P_{R2}\right)$ and temperature $\left(T_{R2}\right)$ at final state is expressed as follows.

$\begin{array}{c}{T}_{R2}=\frac{T_2}{T_{\text{cr}}}\\ =\frac{500\text{R}}{272\text{R}}\\ =1.838\end{array}$

$\begin{array}{c}{P}_{R2}=\frac{P_2}{P_{\text{cr}}}\\ =\frac{150\text{psia}}{705\text{psia}}\\ =0.213\end{array}$

At initial:

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 $0$.

$Z_{h1}\cong 0$

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 $0$.

$Z_{s1}\cong 0$

At final:

Refer Figure A-29E, “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 $0.04$.

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 $0.02$.

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

The specific heat at constant pressure of argon is $0.1253\text{Btu/lbm}\cdot \text{R}$.

The gas constant of argon is $0.04971\text{Btu/lbm}\cdot \text{R}$.

Conclusion:

Substitute 0.04 for $Z_{h2}$, 0 for $Z_{h1}$, $500\text{R}$ for $T_2$, $1000\text{R}$ for $T_1$, $0.04971\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$, $0.1253\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$ and 272 R for $T_{\text{cr}}$ in Equation (VI).

$\begin{array}{c}{h}_2-h_1=\left\{\begin{array}{l}\left[0.1253\text{Btu}/\text{lbm}\cdot \text{R}\left(500\text{R}-1000\text{R}\right)\right]\\ -\left[\left(0.04-0\right)\left(0.04971\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(272\text{R}\right)\right]\end{array}\right\}\\ =-62.65\text{Btu}/\text{lbm}-0.5408\text{Btu}/\text{lbm}\\ =-63.1608\text{Btu}/\text{lbm}\\ \simeq -63.2\text{Btu}/\text{lbm}\end{array}$

Substitute $12\text{lbm}/\text{s}$ for $\dot{m}$, $-63.2\text{Btu}/\text{lbm}$ for $h_2-h_1$, $80\text{Btu}/\text{s}$ for ${\dot{Q}}_{out}$, $450\text{ft}/\text{s}$ for $V_2^{}$ and $300\text{ft}/\text{s}$ for $V_1^{}$ in Equation (II).

$\begin{array}{c}{\dot{W}}_2=-\left(12\text{lbm}/\text{s}\right)\left(\begin{array}{l}-63.2\text{Btu}/\text{lbm}\\ +\left(\frac{{\left(450\text{ft}/\text{s}\right)}^2-{\left(300\text{ft}/\text{s}\right)}^2}{2}\right)\left(\frac{1\text{Btu}/\text{lbm}}{25,037{\text{ft}}^{\text{2}}/{\text{s}}^{\text{2}}}\right)\end{array}\right)-80\text{Btu}/\text{s}\\ =731.4399\text{Btu}/\text{s}-80\text{Btu}/\text{s}\\ =651.4\text{Btu}/\text{s}\left(\frac{1\text{hp}}{0.7065\text{Btu}/\text{s}}\right)\\ =922\text{hp}\end{array}$

Thus, the power output of the turbine is $922\text{hp}$.

(b)

To determine

The exergy destruction associated with the process.

Answer to Problem 95RP

The exergy destruction associate with process is $124.5\text{Btu}/\text{s}$.

Explanation of Solution

Write the entropy balance equation for closed system.

$S_{in}+S_{gen}-S_{out}=\Delta S_{system}$ (VII)

Here, the entropy input is $S_{in}$, the entropy output is $S_{out}$, the entropy generation in the system is $S_{gen}$ and the change in entropy of system is $\Delta S_{system}$.

Rewrite the Equation (VII) as follows by substituting 0 for $S_{in}$, $\frac{Q_{out}}{T_{b,out}}$ for $S_{out}$ and $\dot{m}\left(s_2-s_1\right)$ for $\Delta S_{system}$, for this case.

$\begin{array}{l}{S}_{gen}-\frac{Q_{out}}{T_{b,out}}=m\left(s_2-s_1\right)\\ S_{gen}=\dot{m}\left(s_2-s_1\right)+\frac{Q_{out}}{T_{b,out}}\end{array}$ (VIII)

Here, mass flow rate is $\dot{m}$, amount of heat transfer is $q_{out}$ and surrounding temperature is $T_{b,out}$.

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

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

Here, the gas constant is R, the specific heat at constant pressure is $c_p$, the initial pressure is $P_1$, the final pressure is $P_2$, the final temperature is $T_2$ and the initial temperature is $T_1$.

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

$s_2-s_1=R\left(Z_{s1}-Z_{s2}\right)+{\left(s_2-s_1\right)}_{ideal}$ (X)

Here, the entropy departure factor is $Z_s$.

Write the formula for exergy destruction associate with process.

${\dot{X}}_{\text{destruction}}=T_{b,out}{S}_{gen}$

Substitute $\dot{m}\left(s_2-s_1\right)+\frac{Q_{out}}{T_{b,out}}$ for $S_{gen}$.

${\dot{X}}_{\text{destruction}}=T_{b,out}\left(\dot{m}\left(s_2-s_1\right)+\frac{Q_{out}}{T_{b,out}}\right)$ (XI)

Conclusion:

Substitute $500\text{R}$ for $T_2$, $1000\text{R}$ for $T_1$, $0.04971\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$, $0.1253\text{Btu}/\text{lbm}\cdot \text{R}$ for $c_p$, $150\text{psia}$ for $P_2$ and $1000\text{psia}$ for $P_1$ in Equation (IX).

$\begin{array}{c}{\left(s_2-s_1\right)}_{ideal}=\left(\begin{array}{l}\left(0.1253\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \left(\frac{500\text{R}}{1000\text{R}}\right)\\ -\left(0.04971\text{Btu}/\text{lbm}\cdot \text{R}\right)\ln \frac{150\text{psia}}{1000\text{psia}}\end{array}\right)\\ =-0.08685\text{Btu}/\text{lbm}\cdot \text{R}+0.094306\\ =0.007455\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $0.007455\text{Btu}/\text{lbm}\cdot \text{R}$ for ${\left(s_2-s_1\right)}_{ideal}$, 0.02 for $Z_{s2}$, 0 for $Z_{s1}$ and $0.04971\text{Btu}/\text{lbm}\cdot \text{R}$ for $R$.

$\begin{array}{c}{s}_2-s_1=\left(0.04971\text{Btu}/\text{lbm}\cdot \text{R}\right)\left(0-0.02\right)+0.007455\text{Btu}/\text{lbm}\cdot \text{R}\\ =-0.0009942\text{Btu}/\text{lbm}\cdot \text{R}+0.007455\text{Btu}/\text{lbm}\cdot \text{R}\\ =0.00646\text{Btu}/\text{lbm}\cdot \text{R}\end{array}$

Substitute $12\text{lbm}/\text{s}$ for $\dot{m}$, $75\text{°F}$ for $T_{b,out}$, $0.00646\text{Btu}/\text{lbm}\cdot \text{R}$ for $s_2-s_1$, $80\text{Btu}/\text{s}$ for $Q_{out}$ and 535 R for $T_{surr}$.

$\begin{array}{c}{\dot{X}}_{\text{destruction}}=\left(75\text{°F}\right)\left(\left(12\text{lbm}/\text{s}\right)\left(0.00646\text{Btu}/\text{lbm}\cdot \text{R}\right)+\frac{80\text{Btu}/\text{s}}{535\text{R}}\right)\\ =\left(75+460\text{R}\right)\left(0.07752\text{Btu}/\text{s}\cdot \text{R}+0.1495\text{Btu}/\text{s}\cdot \text{R}\right)\\ =535\text{R}\left(0.22705\text{Btu}/\text{s}\cdot \text{R}\right)\\ =121.47\text{Btu}/\text{s}\end{array}$

$\simeq 124.5\text{Btu}/\text{s}$

Thus, the exergy destruction associate with process is $124.5\text{Btu}/\text{s}$.