B747-400 Electrical Generators

[Zulu]

Hi,
How much power does each electrical generator on the B747-400 produce?
I believe they're rated at 90 kVA; is that equivalent to 90 kW?
Z.

[catchup]

Don't know about the 747, but kVA and kW are not the same. kVA is always higher due to cos Phi (0,8)

so, kW= kVA x 0,8

regards

[VR-HDB]

I'm far from being an expert on electricity - in fact, I'm just a pilot... ;)


But this is the way I understood it:

Watts=(Volt x Amps) x power factor


REAL Power: Watts
APPARANT Power: VA

A three-phase generator has a "Power factor" (cos phi) of 0.8 - so as "catchup" said the (k)VA is always higher.

Hopefully someone can explain it better!

[catchup]

Yeah, it has to do with the "phase-shift" of a three-phase generator.

Good enough for us pilots.;)

regards

Yeah, 90 kVA is correct. Six of them....

[Zulu]

...so with all engines, and both APU generators running, that would be...

(6 x 90 kVA x 0.8) W = 432 kW?

[gas path]

Er not quite! the main gennies and apu gennies cannot be pararrelled. Indeed, even the apu gennies cannot be pararrelled as the split system breaker will open.
So it would be 90KVA *4 (main gennies all pararrelled) or apu 90KVA *2 (split between busses 1+2 and 3+4)
Mind you, if you were pulling that much load.....there might be a problem :ooh:

[GearDoor]

An AC circuit has three types of loads: Pure resistance, inductive and capacitive. True power (P) is the power consumed by the resistance elements only, and as someone pointed out P=IR. That would be exactly the same a calculating the power of a DC circuit. However, the apparent power (U) takes into account all resistances plus the inductvive and/or capacitive reactances. (reactance is kinda the same as resistance) A circuit can be predominantly inductive or predominantly capacitive. When you sum all the inductive and capacitive reactances, you end up with either a capacitive reactive power or an inductive reactive power.

To arrive at the apparent power (U), you must vectorially add the true power (P) and the reactive power (Q) (inductive or capacitive). Apparent power is always greater than or equal to reactive power. If you can imagine a graph, reactive power increases along the +ve x-axis, and inductive power goes up along the +ve y-axis and capacitive power goes down along the -ve y-axis. Apparent power points either up and to the right (inductive) or down and to the right (capacitive).

True power (P) is measured in watts (W). Apparent power (U) is measured in volt-amps (VA) and reactive power (Q) is measured in volt-amps-reactive (VAR).

The power factor (PF) is the ratio of true power to apparent power. PF=P/U. It is basically the circuits efficiency. It is not a costant value (like 0.8) but depends on the phase angle. PF=cosine (phase angle). PF=1 for purely resistive circuits, and PF=0 for purely reactive circuits. You might be thinking of the RMS voltage/current which is 0.707 times the maximum voltage/current.

[Milt]

Do the constant speed drives (Sunstrand I presume) enable full electrical power at engine idle ?

[cirrus01]

Milt......

Yes, Engine at Gnd Idle the AC Generators provide full power.

BTW there is no seperate CSD unit, as IDGs ( Intergrated Drive Generator ) are fitted.

Normal AC generators fitted on the APU.


:ok: