Gas Turbine Nozzles

A nozzle is a passage of varying cross-section that converts the enthalpy (pressure energy) of a fluid into kinetic energy. From the steady flow energy equation for an adiabatic nozzle:

$$h_0 = h + \frac{C^2}{2} \;\Rightarrow\; C = \sqrt{2(h_0 - h)}$$

and for isentropic flow of a perfect gas the area-velocity relation is

$$\frac{dA}{A} = \frac{dC}{C}\left(Ma^2 - 1\right)$$

Area-velocity relation

Types

 Convergent      Convergent-divergent    Divergent (supersonic)
 ---\            ---\       /---         |\
 ====>   ==>     ===> throat ====>       | \===>
 ---/            ---/       \---         |/

1. Convergent nozzle: area decreases; exit velocity at most sonic. If the pressure ratio is above the critical value ($p_b/p_0 < 0.528$ for air) the nozzle is choked: exit pressure stays at $p^*$ and the extra expansion occurs outside as shock and expansion waves. Used on subsonic and low-supersonic aircraft engines, and as the final nozzle of turbojets.
2. Convergent-divergent (de Laval) nozzle: subsonic in the convergent part, $Ma = 1$ at the throat, supersonic in the divergent part, where pressure continues to fall. Used when the pressure ratio is high: rocket engines, afterburning and supersonic jet engines.
3. Divergent nozzle: a divergent passage with supersonic entry (the divergent part alone). With subsonic entry it acts as a diffuser, used in engine inlets and compressor diffusers.

In gas turbines the turbine nozzle guide vanes form a ring of convergent passages that accelerate the gas and turn it onto the rotor blades; the propelling nozzle produces the jet thrust.

The stagnation (total) state is the state reached when a flowing fluid is brought to rest adiabatically (and isentropically for the pressure). The static properties $T$, $p$ are those measured moving with the fluid.

Stagnation temperature

From the steady flow energy equation for adiabatic deceleration to rest:

$$h_0 = h + \frac{C^2}{2} \;\Rightarrow\; c_p T_0 = c_p T + \frac{C^2}{2}$$ $$T_0 = T + \frac{C^2}{2c_p}$$

Using $c_p = \dfrac{\gamma R}{\gamma - 1}$ and $a^2 = \gamma R T$, so $C^2 = Ma^2 \gamma R T$:

$$\boxed{\frac{T_0}{T} = 1 + \frac{\gamma - 1}{2}Ma^2}$$

Stagnation pressure

For isentropic deceleration, $p_0/p = (T_0/T)^{\gamma/(\gamma-1)}$:

$$\boxed{\frac{p_0}{p} = \left(1 + \frac{\gamma - 1}{2}Ma^2\right)^{\gamma/(\gamma-1)}}$$

At $Ma = 1$ (critical state): $T^*/T_0 = 2/(\gamma + 1) = 0.833$ and $p^*/p_0 = 0.528$ for air ($\gamma = 1.4$).

Points to note

• In adiabatic flow without work, $T_0$ stays constant along a nozzle even with friction; $p_0$ falls because of friction (entropy rise).
• In a compressor or turbine, $T_0$ changes by the work done, so $w = c_p \Delta T_0$.
• For small Mach number, $p_0 \approx p + \tfrac12\rho C^2$.

Why used

1. The kinetic energy is large in gas turbines ($Ma$ of 0.3 to 1 or more), so static and total values differ greatly.
2. A thermocouple or pitot tube measures nearly total values, and performance (efficiency, pressure ratio) is defined on a total basis, e.g. $\eta_{tt}$.
3. Total conditions are independent of the flow velocity, so they give a common reference at each engine station (inlet, compressor exit, turbine exit).

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

Critical pressure ratio for air:

$$\frac{p^*}{p_0} = \left(\frac{2}{\gamma + 1}\right)^{\gamma/(\gamma-1)} = 0.5283 \;\Rightarrow\; p^* = 0.5283 \times 500 = 264.1\ \text{kPa}$$

(a) Back pressure 350 kPa

Since $350 > 264.1$ kPa, the nozzle is not choked and the exit pressure equals the back pressure, $p_e = 350$ kPa.

$$T_e = T_0\left(\frac{p_e}{p_0}\right)^{0.2857} = 500 \times 0.7^{0.2857} = 451.6\ \text{K}$$ $$C_e = \sqrt{2c_p(T_0 - T_e)} = \sqrt{2 \times 1005 \times 48.4} = 312.0\ \text{m/s}$$ $$\rho_e = \frac{p_e}{RT_e} = \frac{350}{0.287 \times 451.6} = 2.701\ \text{kg/m}^3$$ $$\dot m = \rho_e C_e A_e = 2.701 \times 312.0 \times 0.0020 = 1.685\ \text{kg/s}$$

Exit Mach number $= 312.0/\sqrt{1.4 \times 287 \times 451.6} = 0.73$.

(b) Back pressure 200 kPa

Since $200 < 264.1$ kPa, the nozzle is choked. The exit pressure is $p^* = 264.1$ kPa (the remaining drop to 200 kPa takes place outside the nozzle) and $Ma_e = 1$.

$$T^* = \frac{2T_0}{\gamma + 1} = \frac{2 \times 500}{2.4} = 416.7\ \text{K}$$ $$C^* = \sqrt{\gamma R T^*} = \sqrt{1.4 \times 287 \times 416.7} = 409.2\ \text{m/s}$$ $$\rho^* = \frac{p^*}{RT^*} = \frac{264.1}{0.287 \times 416.7} = 2.209\ \text{kg/m}^3$$ $$\dot m_{max} = \rho^* C^* A = 2.209 \times 409.2 \times 0.0020 = 1.808\ \text{kg/s}$$

Lowering the back pressure further does not change the mass flow; it can only be raised by increasing $p_0$ or $A$.

Answer: (a) $C_e$ = 312.0 m/s, $\dot m$ = 1.685 kg/s; (b) choked, $p_e$ = 264.1 kPa, $C_e$ = 409.2 m/s, $\dot m$ = 1.808 kg/s.

Data: $p_0 = 800$ kPa, $T_0 = 600$ K, $p_e = 100$ kPa, $\eta_n = 0.92$, $\dot m = 2$ kg/s.

(a) Throat area

Pressure ratio $100/800 = 0.125 < 0.528$, so the nozzle is choked and a divergent part is needed.

$$\begin{aligned} p^* &= 0.5283 \times 800 = 422.6\ \text{kPa}, \quad T^* = \frac{2 \times 600}{2.4} = 500.0\ \text{K} \\ C^* &= \sqrt{1.4 \times 287 \times 500.0} = 448.2\ \text{m/s}, \quad \rho^* = \frac{422.6}{0.287 \times 500.0} = 2.945\ \text{kg/m}^3 \\ A^* &= \frac{\dot m}{\rho^* C^*} = \frac{2}{2.945 \times 448.2} = 15.15\ \text{cm}^2 \end{aligned}$$

(b) Exit temperature and velocity

Isentropic exit temperature:

$$T_{es} = 600\left(\frac{100}{800}\right)^{0.2857} = 331.2\ \text{K}, \qquad \Delta h_s = 1.005 \times (600 - 331.2) = 270.1\ \text{kJ/kg}$$

Actual enthalpy drop:

$$\Delta h = \eta_n \Delta h_s = 0.92 \times 270.1 = 248.5\ \text{kJ/kg}$$ $$T_e = 600 - \frac{248.5}{1.005} = 352.7\ \text{K}, \qquad C_e = \sqrt{2 \times 1000 \times 248.5} = 705.0\ \text{m/s}$$

(The isentropic velocity would be 735.0 m/s.)

(c) Exit Mach number

$$a_e = \sqrt{1.4 \times 287 \times 352.7} = 376.5\ \text{m/s}, \qquad Ma_e = \frac{705.0}{376.5} = 1.87$$

(d) Exit area

$$\rho_e = \frac{100}{0.287 \times 352.7} = 0.988\ \text{kg/m}^3, \qquad A_e = \frac{\dot m}{\rho_e C_e} = \frac{2}{0.988 \times 705.0} = 28.72\ \text{cm}^2$$

Answer: $A^* = 15.15$ cm²; $T_e = 352.7$ K, $C_e = 705.0$ m/s; $Ma_e = 1.87$; $A_e = 28.72$ cm².