Turbojet thrust, efficiencies and fuel consumption

Data: $V = 250$ m/s, $\dot m_a = 40$ kg/s, $f = 0.015$, $C_j = 600$ m/s, $p_j = p_0$, $LHV = 43000$ kJ/kg.

Fuel flow $\dot m_f = f\dot m_a = 0.015 \times 40 = 0.60$ kg/s; jet mass flow $\dot m_j = 40 + 0.60 = 40.60$ kg/s.

(a) Net thrust

$$F = \dot m_j C_j - \dot m_a V = 40.60 \times 600 - 40 \times 250 = 14360\ \text{N} = 14.36\ \text{kN}$$

(Specific thrust $F/\dot m_a = 359.0$ N s/kg.)

(b) Thrust (propulsive) power

$$P_T = F V = 14360 \times 250 = 3590\ \text{kW}$$

(c) Propulsive efficiency

Rate of increase of kinetic energy of the working fluid:

$$\Delta KE = \tfrac12 \dot m_j C_j^2 - \tfrac12 \dot m_a V^2 = \tfrac12 (40.60)(600^2) - \tfrac12 (40)(250^2) = 6058\ \text{kW}$$ $$\eta_p = \frac{P_T}{\Delta KE} = \frac{3590}{6058} = 0.593$$

(The simple formula $2V/(C_j + V) = 0.588$ ignores the fuel mass; the value above is the exact one.)

(d) Thermal and overall efficiency

Energy input $= \dot m_f \times LHV = 0.60 \times 43000 = 25800$ kW.

$$\eta_{th} = \frac{\Delta KE}{\dot m_f LHV} = \frac{6058}{25800} = 0.235, \qquad \eta_o = \eta_p\eta_{th} = \frac{P_T}{\dot m_f LHV} = 0.139$$

(e) Thrust specific fuel consumption

$$TSFC = \frac{\dot m_f}{F} = \frac{0.60 \times 3600}{14.36} = 150.4\ \text{kg/(kN h)}$$

Answer: $F = 14.36$ kN; $P_T = 3590$ kW; $\eta_p = 0.593$; $\eta_{th} = 0.235$; $\eta_o = 0.139$; TSFC = 150.4 kg/(kN h).