Adiabatic Nozzle with Isentropic Efficiency: 04-BS-10, December 2015, Q7

Question 7: Adiabatic Nozzle with Isentropic Efficiency (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. $P_1=180$ kPa, $T_1=707\ ^\circ$C (980.15 K), $V_1\approx0$. $P_2=70$ kPa. $\eta_N=0.90$. Variable specific heats.

Find. (a) $V_2$ [m/s]; (b) $T_2$ [$^\circ$C].

Fig. Q7 — T–s state points for the adiabatic nozzle: 1→2s isentropic reference expansion to 70 kPa, 1→2 the actual (higher-entropy) expansion to the same exit pressure, reaching a higher exit temperature than the isentropic case. No saturation dome for an ideal-gas working fluid.

Approach

Find the isentropic reference exit state at the same exit pressure using the variable-cp root-finding method, giving the maximum possible kinetic-energy gain. The nozzle's isentropic efficiency scales that ideal kinetic-energy gain down to the actual value; the actual exit enthalpy (and hence temperature) follows from the energy balance, and the actual exit velocity follows directly from the actual kinetic-energy gain (inlet velocity negligible, no work, adiabatic).

Isentropic reference exit state. At $s^\circ(T_{2s})=s^\circ(T_1)+R\ln(P_2/P_1)$ (180→70 kPa): $T_{2s}=768.47$ K, giving an isentropic enthalpy drop $$\Delta h_s=h(T_1)-h(T_{2s})=236.03\ \text{kJ/kg}.$$
Actual enthalpy drop and exit temperature (part b). $$\Delta h_a=\eta_N\,\Delta h_s=0.90\times236.03=212.42\ \text{kJ/kg}.$$ $$h_2=h(T_1)-\Delta h_a\ \Rightarrow\ T_2=790.05\ \text{K}=\boxed{516.90\ ^\circ\text{C}}.$$
Exit velocity (part a). Energy balance on the nozzle (adiabatic, no work, negligible inlet KE): $V_1^2/2+h_1=V_2^2/2+h_2\ \Rightarrow\ V_2^2/2=\Delta h_a$: $$V_2=\sqrt{2\,\Delta h_a}=\sqrt{2\times212.42\times1000}=\boxed{651.8\ \text{m/s}}.$$

Quantity	Result
(a) $V_2$	651.8 m/s
(b) $T_2$	516.90°C (790.05 K)

Concept & Theory

Topic. Adiabatic nozzle with a stated isentropic efficiency, variable specific heats, converting enthalpy drop into kinetic energy.

Key relations. Nozzle isentropic efficiency $\eta_N=(h_1-h_2)/(h_1-h_{2s})$ (ratio of actual to isentropic enthalpy/KE gain at the same exit pressure); steady-flow energy balance with negligible inlet velocity, $V_2=\sqrt{2(h_1-h_2)}$.

Why this works. A real (irreversible) nozzle still conserves energy exactly (adiabatic, no shaft work) — the efficiency only affects how much of the *ideal* enthalpy drop converts to kinetic energy, not the energy balance itself. The actual exit state necessarily has higher entropy and a higher temperature (at the same exit pressure) than the isentropic reference state, which is why $T_2>T_{2s}$.

Common pitfall. Computing the exit velocity from the isentropic enthalpy drop $\Delta h_s$ instead of the efficiency-scaled actual value $\Delta h_a$ — that would overstate $V_2$ by more than 5% here.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 5 (Mass and Energy Analysis) & Ch. 7 (Entropy, Isentropic Efficiencies).

Question 7 of 9: Adiabatic Nozzle with Isentropic Efficiency

Question 7: Adiabatic Nozzle with Isentropic Efficiency (15 marks)

Approach