# Need help with calculation of engine airflow and jet velocity and efficiency

This is an image of a set of questions that are relevant to an Airbus A380 type of aircraft (4 engines):

As the answers are incorrect, I'd like to understand how these need to be calculated.

Q1) I used the equation q = v x A. The velocity being 250.02 (486 knts) and the area being pi * 1.48 squared. This gave the answer of 1720.474

Q2) With the airflow answer, I assumed it was the mass flow. So T = m(Vj - V0) so 400,000 = 1720.474 * (Vj - 250.02) I assumed that the air speed of the air going into the engine was the same as the airspeed. This gave the exhaust velocity being 482.514 m/s or 248.23 knots

Q3) Since jet efficiency = 2 / (1 + (Vj / V0)) I did 2 / (1 + (482.514 / 250.02)) which I got 68.26 %

I would like to know the correct answers (and what I did wrong).

• What are the units your are assumed to use?
– mins
Jul 12, 2021 at 15:01
• I used si units, so velocity is meters per second, area is meter squared, thrust is newtons and efficency is percentage Jul 12, 2021 at 15:13
• General remarks, 1) how many engines are there, 2) don't mix up units. For Q1) you might need to express this in kg/s. For Q2) think about a single engine, and the amount of thrust for that engine. Currently the jet has a lower velocity... Jul 13, 2021 at 13:52
• There's 4 engines, my bad Jul 13, 2021 at 20:12

One engine has a diameter of 2.96 m, the area is 6.88 m2. Airplane speed of 486 kts is 250 m/s. The density at FL280 is 0.493070 kg/m3.

@ Q1) In case there are 2 engines (not specified), having each an intake flow of (assuming that the column of air is sucked into the engine without inlet spillage (drag)): $$\dot m = {\rho\times \dot V = \rho\times A\times v = [kg/m^3]\times[m^2]\times[m/s]=[kg/s]}$$ $$\dot m = 0.493070\times6.88\times250=848.1\:[kg/s]$$

@ Q2) For one engine the thrust is 200 kN (under the assumption that 2 engines are on the aircraft and split the load evenly; from a later comment it became clear that there are 4 engines, the exercise continues for 2 engines; for four engines, the thrust value needs to be halved = 100 kN)

(This is an approximation as we are assuming that the inlet mass flow equals the outlet mass flow, no added fuel, and no pressure difference between the inlet and the outlet.) $$T=\dot m \times(v_j - v_0)=[kg\;m/s^2] = [kg/s]\times[m/s] = 200,000 = 848.1\times(v_j-250)$$ $$=> v_j = 486\;[m/s]$$

@ Q3) $$\eta_j = \frac{2}{1+\frac{v_j}{v_0}}= \frac{2}{1+\frac{486}{250}}=0.68=68\;\%$$

I would like to know the correct answers

For four engines you will get 368 m/s (average) jet velocity leading to a 81 % jet efficiency; it is left to the reader to calculate this.

(and what I did wrong).

What went wrong is that the thrust is specified for all engines as a total, you need to work out the thrust per engine. Note that the exercise is a simplified version of reality, assumptions are stated in the answer, but for proper calculation should be considered; however, this would require more input.

• Hi oscar, thank you for the answer. The engines on the plane is 4 as it is an A380, how would the answers vary? Jul 13, 2021 at 20:11
• @UmerRiaz You can do the math for four, but, note the A380 uses turbofan engines, not straight jets. You can calculate a mean velocity increase as there are 2 streams coming out of the engine. I thought the mass flow was high, so edit the question to include the amount of engines. Jul 13, 2021 at 20:49
• So would I just have to times the airflow answer by 2? Jul 13, 2021 at 20:56
• And what would Q2's answer be assuming that the 4 engines is there? Sorry I'm asking alot of questions xD im taking this course online and I havent even started A levels Jul 13, 2021 at 20:58
• @UmerRiaz If 2 engines are 200 kN each to reach 400 kN, how much would one engine have is there are four to reach 400 kN? Q1 is "per engine" so I don't see why you want to multiply that by 2, it is just the column of air in front of the engine, to be precise, each engine sees such a column, but that is not asked in Q1 (Calculate the air flow per engine at this alltitude). Jul 14, 2021 at 7:51