Nozzle energy equation and nozzle efficiency

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

 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: $p_{02} < p_{01}$.