Estimation of the electrical efficiency of the engine in an aircraft such as A320 or B737

[adgirard]

Hi everybody,

I would like to have an estimation of the efficiency of producing electricity in a narrow bodies aircraft.
I am looking for a value which link the electrical watts consumed by a device on board and the kilograms of kerosene required for this.

A gross estimation will be far enough.

Thank you very much.

 

[rb1957]

i'd suggest asking an engine OEM (PW, GE, RR, ...), presumably with the eninge in cruise power setting, how does the fuel consumption change with adding an electrical load ?

compared to the power put out by the engine (lots of kW i suspect) i expect the effect of an electrical load of 1kW (?) would be very small

Quando Omni Flunkus Moritati

 

[debodine]

I can give you some ballpark figures that will give you part of the information you can use. A typical narrow body aircraft (twin engine) will have two engine driven electrical generators rated at 90 kVA (Kilo Volt Amps), usually derated to around 86 kVA. In cruise both generators will typically be required to provide about 70 kVA each, so you could assume 140 kVA total electrical energy is required for cruise flight. If you know the electrical load of the system you want to analyze, now you can determine what percentage of the total electrical load belongs to your system.

Also one typical Boeing 757 engine uses approximately 3,000 pounds of fuel per hour in cruise (or 6,000 pounds per hour total because it is a twin engine aircraft).

What I do NOT know that you still need to know (as rb1957 pointed out) is what percentage of the total energy output of each engine at cruise is required to drive each constant speed drive generator.

So let's run a guesstimate for ease of demonstration: Say your system uses 7 kVA in cruise flight. Thus ten percent of the total electrical outpput of one generator (producing 70 kVA) is used for your system. Then say the engine uses one percent of it's total energy output to drive the generator (the engine energy used to drive the generator is where I can only guess so do NOT trust this number), and it does this with a fuel flow of 3,000 pounds of fuel per hour. The fuel used to drive your 7 kVA system would be 3,000 pounds per hour X .01 (to derive the energy needed from the total engine energy for that one generator) X .1 (to derive the amount of the electrical output of the generator used to power your system).

 

[VEBill]

The P-3 has four turboprop engines, but engine No 1 does not have a generator. Perhaps you can find some info about fuel flow differences between No 1 and the other three. This would include all generator load (including parasitic losses), not just the delta per unit electrical power.

 

[adgirard]

Thank you very much.

I am looking for this very value debodine: the amount of fuel consumed by the engine to produce electricity. I have found something else in documentation. We have basically three devices involved in electricity production in an aircraft. Turbofans convert chemical energy in to mechanical one (thrust = rotating shaft, so it is ok ?), IDG which permits to purify the rotation of the shaft, then alternator which convert shaft rotation in to electricity.

Fuel -> Turbofans (efficiency: 0.6 - 0.4) -> IDG (0.70) -> Alternator (0.98) -> Electricity.

Then if we multiply all the efficiency involved we have 0.5 x 0.7 x 0.98 = 0.34
What do you think about this ?

Thank you.

 

[IRstuff]

The math is relatively simple, starting with gasoline containing about 48 MJ/kg

TTFN
faq731-376

Need help writing a question or understanding a reply? forum1529

 

[adgirard]

So you agree IRstuff for the schema of the calculation?

This is what I am asking for, the value of Jet Fuel A (aircraft kerosene) is not very hard to find, and by the way you made a mistake.
It is more 43 MJ/kg according to

http://en.wikipedia.org/wiki/Jet_fuel.

But what do you think about the method?

 

[debodine]

adgirard:

I see your method, but I have a difficult time agreeing that 34% of the energy generated by the engine is used to drive the generator/alternator. I think what is throwing me is that in your formula when you use .6 - .4 for the fuel conversion to energy efficiency, the rest of your numbers appear to assume that the energy is used 100% to drive the generator/alternator. I would guess that the great majority of the energy available from the .6 - .4 efficiency of the fuel conversion is used to spin the turbofan blades to produce thrust, not to spin the generator/alternator. So one more factor that you need in your formula chain is how much of the energy produced by the .6 - .4 efficient engine is actually used to drive the generator/alternator? And that number is something I have never worked with and don't even have a ball park guess.

I hope this helps.

 

[Compositepro]

The most efficient modern stationary power plants are about 50% efficient in converting thermal energy to electrical. These use multiple expansion cycles. The Carnot efficiency (maximum theoretical)of a heat engine is about 35%. So the efficiency of converting the combustion energy of fuel to electricity on a jet plane is probably in the range of 20-30%.

 

[adgirard]

debodine: Thank you and you are right. But I do not have this number neither, but actually I think I do not need it.
The thrust in the engine comes from the rotating motion of the shaft. Thus, I can, somehow, assume that the efficiency I give (0.6 - 0.4) is the efficiency to convert the chemical energy laying in the fuel in to the rotation of the shaft. I know that it is not perfectly true but it is close to the reality I guess. Then, the alternator will be driven by the rotation of the shaft too, so I can keep the same efficiency.

Then, I assume that the engine will sometimes consume fuel only to drive the alternator and rest of the time for the thrust. Which is mathematically equivalent to say that 1% of the the fuel consumed goes to the alternator and 99% goes to the thrust. Then, because I have the electrical power I need for my device, I can going backwards and estimate the fuel consumption for this device.

What do you think about my method ?

Thank you very much.

 

[debodine]

adgirard:

I think you and I have arrived at the same conclusion, which is that for the purpose of your analysis it is a reasonable assumption that 1% of the energy developed by the engine is used to drive the generator/alterator. I have zero data to back that up, but it appears to be a decent guess.

 

[GregLocock]

Incidentally I would not assume an alternator efficiency of 98%. Automotive alternators are about 80%, power stations 95%. I assume an aircraft generator will be somewhere between those two extremes.

Cheers

Greg Locock


New here? Try reading these, they might help FAQ731-376

http://eng-tips.com/market.cfm?