Turbojet Thrust and Duct Areas: 04-BS-7, December 2015, Q7

Question 7: Turbojet Thrust and Duct Areas (5 marks)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given.

Quantity	Value
Ambient pressure p₀	100 kPa
Inlet air temperature	20°C = 293.15 K
Exhaust gas temperature	700°C = 973.15 K
Exhaust gas velocity V_jet	900 m/s
Aircraft velocity V_a	900 km/h = 250 m/s
Exhaust flow area A_e	0.3 m²
Gas constant R (treat as air)	287 J/kg·K

Find. Thrust F; inlet flow area A_in; exhaust nozzle area (check).

Fig. Q7 — control volume around the engine: air enters at aircraft speed through A_in, hot exhaust gas leaves through the given A_e at V_jet; since p_exit=p₀, thrust is pure momentum flux.

Approach. Because the exhaust and inlet static pressures are both atmospheric, the pressure-thrust term vanishes and thrust reduces to the momentum-flux form F = ṁ(V_jet−V_a); get ṁ from the exhaust conditions (ideal gas law for density) then use continuity to size the inlet.

Exhaust gas density (ideal gas at ambient pressure and exhaust temperature): $$\rho_{exhaust} = \frac{p_0}{R\,T_{exhaust}} = \frac{100\,000}{287(973.15)} = 0.3580\ \text{kg/m}^3$$
Mass flow rate (fuel mass neglected, so inlet and exit mass flow are equal): $$\dot{m} = \rho_{exhaust}\,A_e\,V_{jet} = 0.3580(0.3)(900) = \boxed{96.7\ \text{kg/s}}$$
Thrust (pressure term is zero since $p_{exhaust}=p_0$): $$F_{thrust} = \dot{m}(V_{jet}-V_a) = 96.7(900-250) = \boxed{62.8\ \text{kN}}$$
Inlet flow area (inlet air density from the ideal gas law at inlet temperature; note it reproduces the paper's own CONSTANTS-page value of 1.19 kg/m³ for air at 20°C, a good self-check): $$\rho_{inlet} = \frac{p_0}{R\,T_{inlet}} = \frac{100\,000}{287(293.15)} = 1.19\ \text{kg/m}^3$$ $$A_{in} = \frac{\dot{m}}{\rho_{inlet}V_a} = \frac{96.7}{1.19(250)} = \boxed{0.325\ \text{m}^2}$$
Exhaust nozzle area, expressed the same way (self-consistency check against the given 0.3 m²): $$A_e = \frac{\dot{m}}{\rho_{exhaust}V_{jet}} = \frac{96.7}{0.3580(900)} = \boxed{0.300\ \text{m}^2}\ \checkmark$$

Quantity	Result
Mass flow rate	96.7 kg/s
Thrust	62.8 kN
Inlet flow area	0.325 m²
Exhaust nozzle area (check)	0.300 m² (matches given)

Topic: Turbojet propulsion — momentum-flux thrust and continuity sizing.

Key relations: F = ṁ(V_jet−V_a) + A_e(p_e−p₀); ideal gas ρ=p/RT; continuity ṁ=ρAV.

Why this works. A control volume drawn around the whole engine converts the propulsion problem into a momentum-balance bookkeeping exercise: since the exhaust is stated to exit at ambient pressure, the only net force on the air is the change in its momentum flux between inlet and exit.

Common pitfall. Using the ambient (inlet) air density for the exhaust mass-flow calculation instead of the much hotter, less dense exhaust gas would understate ṁ and every downstream result — always match the density to the temperature/station at which it is evaluated.

Source: White, Fluid Mechanics, Ch. 3 (control-volume momentum and propulsion).

Question 7 of 13: Turbojet Thrust and Duct Areas

Question 7: Turbojet Thrust and Duct Areas (5 marks)