Thrust and Flow Areas of a Turbojet Engine: 04-BS-7, December 2013, Q7

Question 7: Thrust and Flow Areas of a Turbojet Engine (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 / exhaust pressure, p	100 kPa
Inlet air temperature, T_in	20°C = 293.15 K
Exhaust gas temperature, T_ex	700°C = 973.15 K
Exhaust gas velocity, V_ex	900 m/s
Aircraft (=inlet air) velocity, V_ac	900 km/hr
Exhaust flow area, A_ex	0.3 m²

Turbojet control volume: inlet air enters at the aircraft velocity, exhaust gas leaves at the stated exit velocity; fuel mass flow neglected.

Find. Engine thrust; the inlet flow area giving inlet velocity = aircraft velocity; the exhaust nozzle area giving the stated exit velocity.

Approach. Treat inlet air and exhaust gas as ideal gases at the stated pressure and temperature to get their densities, use the exhaust conditions to get the (fuel-neglected) mass flow rate, apply the momentum (thrust) equation, then use the same mass flow with the inlet density to size the inlet area.

Exhaust gas density and mass flow. $\rho_{ex}=p/(RT_{ex})=100{,}000/(287\times973.15)=0.3580\ \text{kg/m}^3$, so $$\dot m = \rho_{ex}A_{ex}V_{ex} = 0.3580\times0.3\times900 = \boxed{96.7\ \text{kg/s}}$$
Thrust. Converting aircraft speed, $V_{ac}=900/3.6=250.0\ \text{m/s}$, and applying the momentum-thrust relation (equal inlet/exhaust pressure, fuel mass flow neglected): $$F_{thrust} = \dot m(V_{ex}-V_{ac}) = 96.7\times(900-250) = \boxed{62{,}840\ \text{N} = 62.8\ \text{kN}}$$
Inlet air density and inlet area. $\rho_{in}=p/(RT_{in})=100{,}000/(287\times293.15)=1.189\ \text{kg/m}^3$; by mass continuity, the SAME mass flow must pass through the inlet at the aircraft's own velocity, so $$A_{in} = \frac{\dot m}{\rho_{in}V_{ac}} = \frac{96.7}{1.189\times250.0} = \boxed{0.325\ \text{m}^2}$$
Exhaust nozzle area check. The exhaust area needed to deliver the stated 900 m/s exit velocity at this same mass flow is, by definition, the value already used in Step 1 — $A_{ex}=0.3\ \text{m}^2$ — confirming the given exhaust area is self-consistent with the required exit velocity.

Quantity	Value
Mass flow rate	96.7 kg/s
Thrust	62.8 kN
Required inlet area	0.325 m²
Required exhaust area (confirms given value)	0.300 m²

Topic: Jet-propulsion thrust from the linear momentum equation.

Key relations: ideal gas $\rho=p/RT$; continuity $\dot m=\rho AV$; thrust $F=\dot m(V_{ex}-V_{ac})$ when inlet/exhaust static pressures match ambient.

Why this works: with equal inlet and exhaust pressures the pressure-area terms in the full momentum equation cancel, leaving thrust as simply the rate of momentum change of the working fluid — the same mass flow rate must be conserved from inlet to exhaust, which is what links the inlet-area calculation back to the exhaust-derived mass flow.

Common pitfall: using the exhaust density (hot, low ρ) when sizing the inlet, or vice versa — each area must use the gas density AT THAT STATION, not a single averaged value.

Source: White, Fluid Mechanics, Ch. 3 (linear momentum equation, jet/rocket propulsion).

Question 7 of 13: Thrust and Flow Areas of a Turbojet Engine

Question 7: Thrust and Flow Areas of a Turbojet Engine (5 marks)