5 moles per second of n-butane enter a turbine at P=15 bar and T=500 K and leave the turbine at P=1 bar. The turbine has an efficiency of 80%. Find the rate at which work is done using the following methods:
A) The Lee-Kesler approach.
B) The Peng-Robinson equation of state.
A) Throughout the solution we will use “2” to indicate the state of the exiting fluid and “1” to indicate the state of the entering fluid. The entropy balance for a reversible turbine is:
\underline{S}_2-\underline{S}_1=0
Using residuals:
0=\underline{S}_2-\underline{S}_1=\left(\underline{S}_2-\underline{S}_2^{i g}\right)+\left(\underline{S}_2^{i g}-\underline{S}_1^{i g}\right)-\left(\underline{S}_1-\underline{S}_1^{i g}\right)
Assume the exiting fluid ( P=1 bar) is an ideal gas:
0=\underline{S}_2-\underline{S}_1=\left(\underline{S}_2^{i g}-\underline{S}_1^{i g}\right)-\left(\underline{S}_1-\underline{S}_1^{i g}\right)
According to Appendix C, for butane, \mathrm{T}_{\mathrm{c}}=425.12 \mathrm{~K}, \mathrm{P}_{\mathrm{c}}=37.96 bar and \omega=0.2. Thus:
\begin{aligned}& T_{1, r}=\frac{500 \mathrm{~K}}{425.12 \mathrm{~K}}=1.18 \\& P_{1, r}=\frac{15 \,\mathrm{bar}}{37.96 \,\mathrm{bar}}=0.40\end{aligned}
From Figures 7-18 and 7-19:
\begin{gathered}\left(\underline{S}_1-\underline{S}_1^{i g}\right)=\left(\underline{S}_1-\underline{S}_1^{i g}\right)^0+\omega\left(\underline{S}_1-\underline{S}_1^{i g}\right)^1 \\\frac{\left(\underline{S}_1-\underline{S}_1^{i g}\right)}{R}=-0.2+(0.2)(-0.1)\\\left(\underline{S}_1-\underline{S}_1^{i g}\right)=-1.8 \frac{{J}}{{mol} \,K}\end{gathered}
The ideal gas component is determined starting with equation 4.54.
\begin{aligned}d \underline{S} & =\frac{C_V^*}{T} d T+\frac{R}{\underline{V}} d \underline{V} \\\left(\underline{S}_2^{i g}-\underline{S}_1^{i g}\right) & =\int_{T_1=500 K}^{T_2} \frac{C_V^*}{T} d T+R \ln \frac{\underline{V}_2}{\underline{V}_1}\end{aligned}
Noting \mathrm{CV}^*=\left(\mathrm{CP}^*-\mathrm{R}\right) and expressing \mathrm{CP}^* in the form of Appendix D :
\begin{aligned}\left(\underline{S}_2^{i g}-\underline{S}_1^{i g}\right)= & R \int_{T_1=500 K}^{T_2} \frac{(A-1)+B T+C T^2+D T^3+E T^4}{T} d T+R \ln \frac{\frac{R T_2}{P_2}}{\frac{R T_1}{P_1}} \\\left(\underline{S}_2^{i g}-\underline{S}_1^{i g}\right)= & R\left[(A-1) \ln \frac{T_2}{T_1}+B\left(T_2-T_1\right)+\frac{C\left(T_2^2-T_1^2\right)}{2}+\frac{D\left(T_2^3-T_1^3\right)}{3}\right. \\& \left.+\frac{E\left(T_2^4-T_1^4\right)}{4}\right]+R \ln \frac{P_1 T_2}{P_2 T_1}\end{aligned}
Return to the entropy balance:
0=\underline{S}_2-\underline{S}_1=\left(\underline{S}_2^{i g}-\underline{S}_1^{i g}\right)-\left(\underline{S}_1-\underline{S}_1^{i g}\right)
We now have a numerical value for \left(\underline{S}_1-\underline{S}_1^{i g}\right) and an expression for \left(\underline{S}_2^{i g}-\underline{S}_1^{i g}\right) in terms of temperatures and pressures. While it is a long expression, everything is known \left(\mathrm{P}_1=15\right. bar, \mathrm{T}_1=500 \mathrm{~K}, \mathrm{P}_2=1 bar) or available in Appendix D (A, B, C, D and E for butane). Thus, the equation can be solved for T_2, and the result is T_2=419.5 \mathrm{~K}. Two points:
• This is actually the temperature of fluid leaving a reversible turbine- we need to calculate the reversible work and apply the efficiency of 80%.
• Butane is indeed a vapor at 419.5 K and P=1 bar. If we had obtained a temperature that was below the boiling point of butane at P=1 bar, that would tell us we needed to re-do the problem modeling the material exiting the reversible turbine as a liquid-vapor mixture, rather than an ideal gas. Chapter 8 presents a detailed look at vapor pressure.
The energy balance for a reversible turbine is:
\frac{\dot{W}_{s, \text { rev }}}{\dot{n}}=\underline{H_2}-\underline{H_1}
Applying residuals:
\underline{H}_2-\underline{H}_1=\left(\underline{H}_2-\underline{H}_2^{i g}\right)+\left(\underline{H}_2^{i g}-\underline{H}_1^{i g}\right)-\left(\underline{H}_1-\underline{H}_1^{i g}\right)
Again assume the butane leaving at 1 bar is an ideal gas:
\begin{gathered}\underline{H_2}-\underline{H_1}=\left(\underline{H}_2^{i g}-\underline{H}_1^{i g}\right)-\left(\underline{H}_1-\underline{H}_1^{i g}\right) \\\underline{H_2}-\underline{H_1}=\int_{T_1=500 \mathrm{~K}}^{T_2=419.5 \mathrm{~K}} C_P^* d T-\left(\underline{H}_1-\underline{H}_1^{i g}\right)\end{gathered}
The reduced properties for state 1 were determined previously. Using Figures 6-16 and 6-17:
\begin{gathered}\frac{\left(\underline{H}_1-H_1^{i g}\right)}{R T_c}=\frac{\left(\underline{H}_1-\underline{H}_1^{i g}\right)^0}{R T_c}+\frac{\omega\left(\underline{H}_1-\underline{H}_1^{i g}\right)^1}{R T_c} \\\frac{\left(\underline{H}_1-H_1^{i g}\right)}{R T_c}=-0.3+(0.2)(-0.1)=-1131 \frac{\mathrm{J}}{\mathrm{mol}}\end{gathered}
For the ideal gas portion:
\begin{gathered}\left(\underline{H}_2^{i g}-\underline{H}_1^{i g}\right)=\int_{T_1=500 K}^{T_2=419.5 K} C_P^* d T \\\left(\underline{H}_2^{i g}-\underline{H}_1^{i g}\right)=R \int_{T_1=500 K}^{T_2=419.5 \mathrm{~K}} A+B T+C T^2+D T^3+E T^4 d T=-11,170 \frac{\mathrm{J}}{\mathrm{mol}}\end{gathered}
The residual is ~10% the magnitude of the ideal gas component; it is not negligible and it would not have been valid to assume ideal gas behavior at T=500 K and P=15 bar.
\underline{H}_2-\underline{H}_1=\left(\underline{H}_2^{i g}-\underline{H}_1^{i g}\right)-\left(\underline{H}_1-\underline{H}_1^{i g}\right)=-10,039 \frac{\mathrm{J}}{\mathrm{mol}}
To the level of accuracy of the figures and data we used, this is essentially -10 kJ/mol produced by the reversible turbine. Applying the efficiency of 80% gives an actual work of -8 kJ/mol.
B) The fundamental model equations for a turbine- entropy balance, energy balance- do not depend upon the model used for the fluid. Therefore the approach to part B is substantially the same as the approach to part A. We used Lee-Kesler only to calculate numerical values for the residual enthalpy and residual entropy; those are the steps for which we use Peng-Robinson.
As in part A, we assume the residual is 0 for the outlet stream (ideal gas) and calculate the residuals for the inlet stream (T=500 K, P=15 bar) for butane using the expressions in Section 7.2.8. The results are as follows:
\begin{aligned}& \mathrm{a}=1.34 \times 10^7 \mathrm{bar~cm}^6 / \mathrm{mol}^2 \\& \mathrm{~b}=72.4 \mathrm{~cm}^3 / \mathrm{mol} \\& \mathrm{Z}=0.901 \\& \mathrm{~A}=0.116 \\& \mathrm{~B}=0.026 \\\\& \frac{\left(\underline{s}_1-\underline{s}_1^{i g}\right)}{R}=-0.221\end{aligned}
This is essentially identical to the residual entropy in part A, so our outlet temperature from the reversible turbine, as in part A, is T=419.5 K
\begin{aligned}& \frac{\left(\underline{H}_1-\underline{H}_1^{i g}\right)}{R T}=-0.312 \\& \left(\underline{H}_1-\underline{H}_1^{i g}\right)=-1297 \frac{J}{mol}\end{aligned}
The energy balance and ideal gas calculations are the same as in part A, since these do not depend upon the equation of state model used. Introducing the Peng-Robinson value for the residual enthalpy into these calculations:
\begin{gathered}\underline{H}_2-\underline{H}_1=\left(\underline{H}_2^{i g}-\underline{H}_1^{i g}\right)-\left(\underline{H}_1-\underline{H}_1^{i g}\right) \\\underline{H_2}-\underline{H}_1=-11,170 \frac{\mathrm{J}}{\mathrm{mol}}-\left(-1297 \frac{\mathrm{J}}{\mathrm{mol}}\right) \\\underline{H}_2-\underline{H}_1=-9873 \frac{\mathrm{J}}{\mathrm{mol}}\end{gathered}
This is for the reversible turbine. Applying the 80 % efficiency:
\frac{\dot{W}_{s, a c t}}{\dot{n}}=0.8\left(-9873 \frac{\mathrm{J}}{\mathrm{mol}}\right)\left(\frac{1 \mathrm{~kJ}}{1000 \mathrm{~J}}\right)=\bf -7.9 \frac{kJ}{mol}