Convergent-divergent nozzle with efficiency

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.