Auxiliary Power Unit (APU) generators seem to be rated in kilo-volts-amperes (kVA) and not kilo-watts (kW). Are kVA not the same as kW? Is there a reason why the generator is rated in terms of kVA instead of kW?
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1$\begingroup$ A Volt-Ampere is the same thing as a Watt, just a different way to express the quantity. $\endgroup$– J WJun 12, 2016 at 12:04
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8$\begingroup$ @JonathanWalters No, they're not. VA is apparent power, watts are real power. $\endgroup$– SomeoneSomewhereSupportsMonicaJun 12, 2016 at 12:51
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$\begingroup$ @SomeoneSomewhere Let me clarify that for DC circuits the product of Volts times Amperes is the same as the real power in Watts. However, if the generator is rated in KVA, that may be an indication that it is an AC generator, not DC, in which case you are correct that KVA would not be strictly equivalent to KW. $\endgroup$– J WJun 12, 2016 at 13:14
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$\begingroup$ @SomeoneSomewhere what is the difference between the two? $\endgroup$– ant BldelJun 12, 2016 at 14:39
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3$\begingroup$ @antBldel (This is an oversimplification.) When you supply AC to a motor, power flows to the windings of the motor to build a magnetic field, but then when the magnetic field collapses, "unusued" power is returned to the source. Watts measure just the power used. kVA measures the flow. A motor that consumes 10kW might need a supply that can handle 15kW (with 5kW "ping ponging" between the supply and the load). So a 10kW generator may not be able to run a 10kW motor. $\endgroup$– David SchwartzJun 13, 2016 at 8:31
2 Answers
Current and voltage don't always vary at the same time, one usually lags, this lag is the root of evils found in electrical engineering.
Any device is a mix of resistive load (no lag), inductive load (current lags) and capacitive load (voltage lags).
The lag (phase shift angle) is denoted with Greek letter φ, and usually expressed as cos φ, or power factor. Power factor $=100 \cos φ$. Each device has its own individual φ, which may vary with conditions.
Lag effect is only visible when voltage or current changes. This happens transiently in DC when the circuit is closed or opened (effect used to create high DC voltage from low DC voltage in a contact breaker), but constantly in AC (not mentioning other induced variations, e.g. changes in mechanical demand on an electric engine will change the impedance of the coil, so the current). The effect of the phase angle increases with circuit frequency, and is significant in aircraft circuitry, with major loads working at 400 Hz.
With this lag, the voltage and current mean (RMS) values are not sufficient to size electrical circuits. A phase shift introduces additional power in the circuit. This reactive component is not described by the RMS values, it's only visible when looking at instantaneous values, for example on this animation of the addition of two phasors, where the fixed circle radii indicate the mean values, and the moving dots show the instantaneous values. In simple cases one can determine the power required using the triangle of powers:
(Triangle of powers, source)
The apparent power is the sum of the real part of the power (resistive load in W) and the imaginary part (reactive load in VAr).
Work (Joule) is only produced by active power (the reactive power produces no actual work) and the active power rating of the generator must be higher or equal to the active power rating of the load. However the reactive power produces a reaction on the generator (by phase shifting the current) and must be accounted for when sizing the generator. The generator must have an apparent power rating higher or equal to the apparent power rating of the load, at the working phase angle.
A higher apparent power, regardless of the active power, requires a more robust generator and wiring to prevent overload.
One way to simplify the required power calculation is to just add the apparent power value of all devices (in VA). The result in VA or a multiple like kVA is conservative. This allows to select the generator required, provided we know its VA rating too. This is why generator manufacturers provide this value.
Notes:
All units, VA, VAr, W, are of the same kind (dimension of power, W). VA and VAr denote the fact that voltage, current and power are not varying in-phase in AC circuits. These units describe the kind of "power" we are taking about.
When the apparent power rating (VA) is the only rating known for a device, the active power rating (W) can be assumed to be no more than 60%, however the actual value depends on multiple factors and this is only a quick approximation. A generator must be provided with a chart showing the relationship between VA and W under different conditions of use.
A generator cannot work under any power factor, and cannot maintain the same efficiency across the valid power factor range. It is provided with a table or graph showing the acceptable conditions. For example, for this generator, the power factor must be kept in the green area (inductive load, between 80 and 100%), and will start being unreliable at some power factor, depending on the active power delivered and the type of reactance (red area).
(Source)The explanations above are really simplifications, because in AC, the phase difference is only one side of the problem of reactive power. The other side is the presence of harmonics which are per definition out of phase.
The general principle in electricity is that voltage is set by the generator, and current is set by the load. The load can set a current which has the same frequency than the voltage, this is what we assumed for the answer. However most of the time the load sets a current variation which doesn't coincide with the voltage frequency and/or shape.
This non-linearity between current and voltage creates 'harmonics' and unbalances phases (so a current start flowing in the neutral, which is not something we want).
A little story about that relates to first datacenters in years 1990. Standby generators (diesel) not well sized were unable to sustain computers loads, due to a capacitive reactive power created by computer switching power supplies (non linearity and a cos φ as low as 0.6) Fires ensued, it was time to understand how to deal with apparent power.The $φ$ angle just comes from the general definition of a sine wave current. Current at time $t$ is defined by $a = a _{peak} \space sin \space (\omega t \space + \varphi)$ where $a _{peak}$ is the maximum current, $ω$ the angular frequency (314 rd/s for the EU grid) and $φ$ the phase shift angle.
Generators generally produce Alternating Current (AC).
Considering that the voltage is regulated, the demand on the generator determines the current, and the heat produced in the generator is directly related to the current, therefore the higher the current the higher the internal heating effect.
From the user's point of view, if the load is purely inductive (for example a solenoid) there is no consumed power, because the instantaneous taken energy is restored back to the supply immediately after, so the net consumed energy is zero while the generator is developing internal heat resulting from the current.
If the load is purely resistive, the delivered current is only producing heat and the produced kVA, as a calculated number, is equal to the delivered kW, as a number.
Since the generator is limited by the current it can deliver at a regulated voltage, it is normal to give the generator capacity in KVA.
To make the best use of the delivered KVA it is up to the user to reduce the induction in the load, for instance by addition of capacitors.
Of course this applies only to alternating current generation. For filtered DC generation, kVA or kW would mean the same, because there is no way in DC generation for the user to restore back the instantaneously used energy.
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$\begingroup$ I thought generators produce DC, alternators produce AC. $\endgroup$ Jun 19, 2019 at 10:42
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1$\begingroup$ @Juan Jimenez, Generator is more general than alternator, but in APUs generators are likely alternators. More precisely: In electricity generation, a generator is a device that converts motive power (mechanical energy) into electrical power for use in an external circuit. $\endgroup$ Jun 19, 2019 at 16:25
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2$\begingroup$ @JuanJimenez A dynamo produces DC, an alternator AC, and both are generators. $\endgroup$ Jun 19, 2019 at 18:30