Adiabatic Nozzle, Variable Specific Heats: 04-BS-10, December 2017, Q5

Question 5: Adiabatic Nozzle, Variable Specific Heats (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, $P_1=180$ kPa, $T_1=707\ ^\circ$C $=980.15$ K, negligible inlet velocity ($V_1\approx0$). Adiabatic expansion to $P_2=70$ kPa. Nozzle isentropic efficiency $\eta_n=0.90$.

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

Fig. Q5 — T–s sketch of the nozzle expansion. The dashed line 1→2s is the (unattainable) isentropic path; the solid line 1→2 is the actual 90%-efficient path, landing at higher entropy and higher enthalpy (hence higher exit temperature) than the isentropic case.

Approach

Solve the isentropic exit temperature $T_{2s}$ by root-finding the variable-cp isentropic condition at the given pressure ratio, take 90% of the resulting isentropic enthalpy drop as the actual kinetic-energy gain (nozzle isentropic efficiency is defined on enthalpy drop, i.e. kinetic energy, not temperature), then back out the actual exit temperature from the actual exit enthalpy.

Isentropic exit state. $h_1=h(980.15\,\text{K})=1149.91$ kJ/kg. Root-solving $s^\circ(T_{2s})-s^\circ(T_1)=R\ln(P_2/P_1)$ for $P_2/P_1=70/180$: $$T_{2s}=768.47\ \text{K}\ (495.32\ ^\circ\text{C}),\qquad h(T_{2s})=913.88\ \text{kJ/kg}.$$ Isentropic enthalpy drop: $\Delta h_s=h_1-h(T_{2s})=1149.91-913.88=236.03$ kJ/kg.
Actual kinetic energy gain and exit velocity (part a). Nozzle isentropic efficiency is defined as the ratio of actual to isentropic kinetic-energy gain (enthalpy drop), with negligible inlet velocity: $$\Delta h_a=\eta_n\,\Delta h_s=0.90\times236.03=\boxed{212.42\ \text{kJ/kg}}.$$ $$V_2=\sqrt{2\,\Delta h_a}=\sqrt{2\times212{,}423\ \text{J/kg}}=\boxed{651.80\ \text{m/s}}.$$
Actual exit enthalpy and temperature (part b). $$h_2=h_1-\Delta h_a=1149.91-212.42=937.48\ \text{kJ/kg}.$$ Root-solving $h(T_2)=937.48$ kJ/kg gives $$T_2=790.05\ \text{K}=\boxed{516.90\ ^\circ\text{C}}.$$ As expected, the actual exit temperature is higher than the isentropic value (516.90°C vs 495.32°C): the irreversibility inside the nozzle converts some of the kinetic-energy gain back into internal energy (friction/turbulence), so less enthalpy converts to velocity and more remains as sensible heat.

Quantity	Result
(a) $V_2$	651.80 m/s
(b) $T_2$	516.90 °C

Concept & Theory

Topic. Adiabatic, non-isentropic nozzle flow with variable specific heats.

Key relations. Nozzle isentropic efficiency $\eta_n=\Delta h_a/\Delta h_s$ (actual over isentropic enthalpy drop). Steady-flow energy balance with negligible inlet velocity: $V_2=\sqrt{2(h_1-h_2)}$.

Why this works. A nozzle converts enthalpy (thermal energy) into kinetic energy; friction and flow separation inside a real nozzle destroy some of the available enthalpy drop, so the actual exit velocity is always below the isentropic (ideal) value for the same pressure ratio. Because the actual process still starts and ends at states connected by the same energy balance ($q=0$, no work), the "lost" kinetic energy reappears as extra internal energy, raising the actual exit temperature above the isentropic exit temperature at the same exit pressure.

Common pitfall. Applying $\eta_n$ to the exit temperature drop or exit velocity directly, rather than to the enthalpy drop specifically — because $c_p$ is not constant here, $\Delta T$ and $\Delta h$ are not simply proportional, so the efficiency ratio must be taken on enthalpy (energy), not temperature.

Source. Cengel & Boles, Thermodynamics: An Engineering Approach, 8th ed., Ch. 17 (Compressible Flow § Isentropic Flow through Nozzles).

Question 5 of 9: Adiabatic Nozzle, Variable Specific Heats

Question 5: Adiabatic Nozzle, Variable Specific Heats (15 marks)

Approach