--- title: "Nozzle energy equation and nozzle efficiency" subject: "Turbo Machines" chapter: 4 key: "semester_7/turbo_machines:nozzle-energy-equation-efficiency" source: https://www.bctengineer.com/q/semester_7/turbo_machines/nozzle-energy-equation-efficiency site: BCT Engineering --- # Nozzle energy equation and nozzle efficiency Turbo Machines · Chapter 4: Gas Turbine Nozzles - Asked in: year not given, 6 marks - Answer checked by Claude Opus 5.5 (high effort) on 2026-10-14 - Page: https://www.bctengineer.com/q/semester_7/turbo_machines/nozzle-energy-equation-efficiency - Whole chapter: https://www.bctengineer.com/md/semester_7/turbo_machines/chapter-4 ## Question Write the energy equation for a gas nozzle and derive an expression for the exit velocity. Define nozzle efficiency, velocity coefficient and discharge coefficient and show how the loss appears on a T-s diagram. ## Short answer Steady adiabatic nozzle energy equation: h01 = h2 + C2^2/2, giving C2 = sqrt(2(h01 - h2)). Nozzle efficiency eta_n = (h01 - h2)/(h01 - h2s) = C2^2/C2s^2. Velocity coefficient C_v = C2/C2s = sqrt(eta_n); discharge coefficient C_d = m_dot_actual / m_dot_ideal. Loss appears as higher exit temperature T2 > T2s and lower total pressure p02 < p01 on the T-s diagram. ## Answer #### Energy equation For steady adiabatic flow with no shaft work, per unit mass (inlet 1, exit 2): $$ h_1 + \frac{C_1^2}{2} = h_2 + \frac{C_2^2}{2} $$ so the exit velocity is $$ C_2 = \sqrt{2(h_1 - h_2) + C_1^2} = \sqrt{2(h_{01} - h_2)} $$ For a perfect gas with negligible inlet velocity and an isentropic process 1-2s: $$ C_{2s} = \sqrt{2c_p T_1\left[1 - \left(\frac{p_2}{p_1}\right)^{(\gamma-1)/\gamma}\right]} $$ #### T-s diagram ```text T | 1 ------ p1 | |\ | | \ | | \ p2 | 2s 2 <- 2 is at higher T than 2s | +--------------- s ``` Friction heats the gas, so the actual exit state 2 has a higher enthalpy than 2s and less kinetic energy is produced. #### Efficiency and coefficients - **Nozzle efficiency:** ratio of actual kinetic-energy gain to the isentropic gain between the same pressures. $$ \eta_n = \frac{h_{01} - h_2}{h_{01} - h_{2s}} = \frac{C_2^2/2}{C_{2s}^2/2} \approx \frac{T_{01} - T_2}{T_{01} - T_{2s}} $$ - **Velocity coefficient:** $C_v = \dfrac{C_2}{C_{2s}} = \sqrt{\eta_n}$. Typical values 0.95 to 0.99 give $\eta_n \approx 0.90$ to 0.98. - **Coefficient of discharge:** $C_d = \dfrac{\dot m_{actual}}{\dot m_{ideal}}$, usually 0.95 to 0.99 for convergent nozzles, reduced by boundary layer blockage. - **Nozzle loss coefficient:** $\zeta = \dfrac{h_2 - h_{2s}}{C_2^2/2}$, where $h_2 - h_{2s} = (1 - \eta_n)(h_{01} - h_{2s})$. The nozzle losses come from skin friction, mixing and separation, and, in a supersonic nozzle, from shock waves and over- or under-expansion. The total temperature is unchanged in the actual process, but total pressure falls: $p_{02} < p_{01}$. ## Check yourself The answer's key points as multiple-choice questions, with the correct option and why. 1. For a steady adiabatic nozzle with no shaft work, the exit velocity is A. C2 = sqrt(2(h2 - h1)) B. C2 = sqrt(2(h1 - h2) + C1^2) C. C2 = 2(h1 - h2) D. C2 = sqrt(h1 - h2) Answer: B. From h1 + C1^2/2 = h2 + C2^2/2, C2 = sqrt(2(h1 - h2) + C1^2) = sqrt(2(h01 - h2)). (Key point: Exit velocity from energy equation) 2. Nozzle efficiency is A. actual mass flow / ideal mass flow B. actual exit pressure / inlet pressure C. isentropic temperature drop / actual drop D. actual kinetic energy gain / isentropic kinetic energy gain between the same pressures Answer: D. eta_n = (h01 - h2)/(h01 - h2s) = C2^2/C2s^2 for the same pressure ratio. (Key point: Nozzle efficiency definition) 3. How is the velocity coefficient related to nozzle efficiency? A. Cv = sqrt(eta_n) B. Cv = eta_n C. Cv = eta_n^2 D. Cv = 1 - eta_n Answer: A. Cv = C2/C2s and eta_n = C2^2/C2s^2, so Cv = sqrt(eta_n); Cv of 0.95 to 0.99 gives eta_n about 0.90 to 0.98. (Key point: Velocity coefficient) 4. On the T-s diagram, where does the actual exit state 2 lie relative to the isentropic state 2s? A. At a lower temperature than 2s B. At the same state as 2s C. At a higher temperature (enthalpy) than 2s because friction heats the gas D. At a higher pressure than 2s Answer: C. Friction adds entropy, so h2 > h2s and less kinetic energy is produced. (Key point: Loss on T-s diagram) 5. Which statement about the actual nozzle process is correct? A. Total temperature falls and total pressure is unchanged B. Total temperature is unchanged but total pressure falls, p02 < p01 C. Both totals are unchanged D. Cd is the ratio of actual to ideal exit velocity Answer: B. The process is adiabatic so T0 is constant, but losses reduce p0. Cd is the ratio of actual to ideal mass flow. 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