Rocket engine thrust and specific impulse calculation

Data: $\dot m_p = 120$ kg/s, $C_e = 2800$ m/s, $A_e = 0.5$ m², $p_e = 80$ kPa.

(a) Sea-level thrust

$$F = \dot m_p C_e + (p_e - p_a)A_e = 120 \times 2800 + (80000 - 101300) \times 0.5$$ $$F = 336000 - 10650 = 325350\ \text{N} = 325.4\ \text{kN}$$

The nozzle is over-expanded ($p_e < p_a$), so the pressure term is negative.

$$C_{eff} = \frac{F}{\dot m_p} = \frac{325350}{120} = 2711.2\ \text{m/s}$$

(b) Specific impulse

$$I_{sp,SL} = \frac{C_{eff}}{g_0} = \frac{2711.2}{9.81} = 276.4\ \text{s}$$

In vacuum $p_a = 0$:

$$F_{vac} = 336000 + 80000 \times 0.5 = 376000\ \text{N}, \qquad C_{eff,vac} = 3133.3\ \text{m/s}, \qquad I_{sp,vac} = \frac{3133.3}{9.81} = 319.4\ \text{s}$$

(c) Propulsive efficiency and thrust power

$$\frac{v}{C_{eff}} = \frac{1500}{2711.2} = 0.5533$$ $$\eta_p = \frac{2(v/C_{eff})}{1 + (v/C_{eff})^2} = \frac{2 \times 0.5533}{1 + 0.3061} = 0.847$$ $$P_T = F v = 325350 \times 1500 = 488025\ \text{kW}$$

Answer: $F = 325.4$ kN, $C_{eff} = 2711.2$ m/s; $I_{sp} = 276.4$ s (sea level), 319.4 s (vacuum); $\eta_p = 0.847$; $P_T = 488025$ kW.