Question 9 of 9: Isentropic Efficiency of an Air Turbine

Question 9: Isentropic Efficiency of an Air Turbine (15 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. Air turbine, steady state, adiabatic, negligible $\Delta$KE/$\Delta$PE. Inlet $P_1=3$ MPa, $T_1=390$ K; exit $P_2=1$ MPa. Actual work developed $w_{actual}=74$ kJ/kg. Ideal-gas air model with temperature-dependent specific heats.

Find. Isentropic efficiency $\eta_t$ of the turbine.

Approach

Find the isentropic (ideal) exit temperature $T_{2s}$ from the ideal-gas isentropic relation $s^\circ(T_{2s})-s^\circ(T_1)=R\ln(P_2/P_1)$, giving the ideal specific work $w_s=h_1-h_{2s}$; the isentropic efficiency is then the ratio of actual to ideal work.

1. Inlet enthalpy and entropy function. At $T_1=390$ K: $h_1=517.12$ kJ/kg, $s^\circ_1=4.1514$ kJ/kg·K.
2. Isentropic exit state. Solving $s^\circ(T_{2s})-s^\circ_1=R\ln(P_2/P_1)=0.287\ln(1/3)=-0.3153\ \text{kJ/kg}\cdot\text{K}$ for $T_{2s}$ (using the actual variable-$c_p$ air property function, not a constant-$k$ shortcut): $$T_{2s}=285.29\ \text{K}\quad\Rightarrow\quad h_{2s}=411.50\ \text{kJ/kg}.$$
3. Ideal (isentropic) work. $$w_s=h_1-h_{2s}=517.12-411.50=\boxed{105.620\ \text{kJ/kg}}.$$
4. Isentropic efficiency. $$\eta_t=\frac{w_{actual}}{w_s}=\frac{74}{105.620}=\boxed{70.1\%}.$$

Where the real exit state sits. The actual exit enthalpy is $h_2=h_1-w_{actual}=517.12-74=443.12$ kJ/kg, i.e. $T_2=316.7$ K against the isentropic $T_{2s}=285.3$ K — the real exhaust is about 31 K hotter than the reversible one at the same 1 MPa back-pressure. That temperature rise is exactly the energy the irreversibilities kept in the stream instead of delivering to the shaft, and it is why the comparison in $\eta_t$ must be made at equal exit pressure rather than at equal exit state. The exit entropy rise is $s_2-s_1=[s^\circ(316.7)-s^\circ(390)]-R\ln(1/3)=+0.1052$ kJ/kg·K, positive as the second law requires for an adiabatic device.

Sanity check against the cold-air-standard shortcut. With constant $k=1.4$ the isentropic exit would be $T_{2s}=T_1(P_2/P_1)^{(k-1)/k}=390(1/3)^{0.2857}=284.9$ K and $w_s=c_p(T_1-T_{2s})=1.005(390-284.9)=105.59$ kJ/kg, giving $\eta_t=70.1\%$ — within 0.03% of the variable-property answer. At this modest temperature span ($\approx$285–390 K) the constant-$c_p$ approximation is essentially exact for air, so quoting the shortcut alongside the table method is a cheap way to confirm no property-lookup or root-solving error has crept in; the two only diverge once the expansion spans several hundred kelvin.