Question 6.13: Problem: Argon enters an adiabatic nozzle at 4 bar and 850 °...
Problem: Argon enters an adiabatic nozzle at 4 bar and 850 °C and exits at 1 bar. If the isentropic efficiency of the nozzle is 90%, find the exit velocity and temperature of the gas.
Find: Exit velocity \pmb{V} _2 and temperature T_2 of argon.
Known: Inlet temperature T_1 = 850 °C = 1123.15 K, inlet pressure P_1 = 4 bar, exit pressure P_2 = 1 bar, isentropic efficiency η_n = 0.9.
Assumptions: Argon is an ideal gas with constant specific heats.
Governing Equations:
Isentropic nozzle efficiency \eta _{\text{nozzle}}=\frac{\pmb{V}_2^2}{\pmb{V}_{2s}^2}
Nozzle exit velocity \pmb{V}_2=\sqrt{2(h_2 – h_1)} =\sqrt{2c_p(T_2 – T_1)}
Isentropic process (ideal gas, \frac{T_2}{T_1} =\left\lgroup\frac{P_2}{P_1} \right\rgroup ^{(\gamma -1)/\gamma }
constant specific heats)
Properties: Argon at 25 °C (approximation) has specific heat c_p = 0.520 kJ / kgK (Appendix 1), specific heat ratio of argon at 25 °C (approximation) \gamma = 1.667 (Appendix 1).
Assuming isentropic flow, the exit temperature is
T_{2s}=T_1\left\lgroup\frac{P_2}{P_1} \right\rgroup ^{(\gamma -1)/\gamma }= 1123 \ K \left\lgroup\frac{1 \ bar}{4 \ bar} \right\rgroup ^{\frac{1.667-1}{1.667} }=644.973 \ K
The exit velocity assuming isentropic flow is
\pmb{V}_{2s}=\sqrt{2c_p(T_1 -T_2)} =\sqrt{2\times 0.520 \times 10^3 \ kJ/kg (1123.15 \ K -644.973 \ K) } =705.198 \ m/s.
The real velocity is then
\pmb{V}_2=\sqrt{\eta _{\text{nozzle}}\pmb{V}_{2s}^2} =\sqrt{0.9(705.198 \ m/s)^2} =669.009 \ m/s.
Then the real outlet temperature can be calculated:
\pmb{V}_2=\sqrt{2c_p(T_1-T_2)}
669.009 \ m/s=\sqrt{2 \times 0.520 \times 10^3 \ kJ/kg (1123.15 \ K-T_2)}
T_2=692.791 \ K
Answer: The exit velocity is 669.0 m / s and the exit temperature is 693 K.