kW is kilowatts - a measure of real power. 1 hp = 0.746 kW.
kVA is kilo Volt-amperes, which is just volts times amps, basically. If the voltage sine wave is exactly in phase with the current sine wave, the power factor = 1.0 and kVA = kW. Otherwise, the two are not the same.
Try looking up "power factor" on Wikipedia or Google.
Well to be more precise
P - Active power is measured in kW
Q - Reactive power is measured in kVAr
And
S - apparent power is mesured in kVA
S=sgrt(P^2 + Q^2)
Real power is S(underlined) =P+jQ = cos(f)S + jsin(f) in VA also. It is consisted of active and reactive part.
f- it is phase angle and its cosinus or sinus values are factors that dpc mentioned. You have that definition in any book of basis of electrotehnics.
If not, kW says how much work is done. In a resistor, all current is producing heat. kVA is simply volts times amps with the "k" meaning we are counting thousands ov VA. You can have 1 kVA and close to zero kW in a circuit. A high-quality capacitor or a low loss inductor can give you that. High current and high voltage, but no heat. Zero kW.
Visualize a small cart on a railway track.
The track is oriented north and south.
It takes about 100 lbs. pull on a rope to start the cart moving. If you stand to the north of the cart and pull the rope with 100 lbs. of force the cart will start moving.
If you stand to the east or west of the cart, no amount of force on the rope will cause the cart to move north.
If you move to the north-east or north-west of the cart, it will take about 141 lbs. pull on the rope to start the cart moving.
The same effect is seen in electrical circuit when the voltage and current are not at the same angle.
In the example above, if the cart is moved 1 foot, the actual work done is 100 ft.lbs.
If you are to the north, the work that you do is 100 ft.lbs.
If you are to the north-east, the apparent work done is 141 ft.lbs., but despite the increased pull on the rope, when the cart moves one foot, the actual work done is 100 ft.lbs.
The 141 lbs pull from the north-east is analogous to KVA and the 100 lbs pull from the north is analogous to KW.
respectfully
... and the reason they are different (or "Why would I stand to the NE or NW when I could just stand to the North?") is because of the nature of the load you are referring to. As others have said, with a purely resistive load such as a heater element there is no difference (analogous to standing to the North on Waross' example). When your load is a motor or other inductive load, the natural properties of the motor construction cause a delay between the voltage and the current, know as "power factor" which essentially describes the angle by which your real power (kW) consumption is different from the kVA. Hence, back to the train track analogy, just because your load is a motor you are automatically relegated to the NE or NW of the track.
Pay no attention to this (but I had to): The NW position would be a funny motor, I think. Since it would represent a leading phase angle. As I said, do not pay any attention to this, not MUCH attention, anyhow.
And the reactive power is analogous to the side thrust of the wheels on the rails when the pulling force is not straight north. This is a real, measurable, force but there is no movement to the east or west and so no real work is done by this component of the total force.
respectfully
Electical is mostly an abstract subject, analogy to "seen" things helps to non-electrical persons or new to electrical. Mechanical is much more easily understood. If it were so simple, the OP would not have asked to begin with.
Are we not engineers? Do we not know what a vector product is? I think we do know that. All engineering schools teach math - and vectors are part of that.
Then, why should we not accept the plain truth?
The plain truth is that voltage and current can be represented by two vectors, U and I (sorry can't put that vector bar on U and I).
All engineers know that the scalar product of two vectors U and I, with angle phi between them, is UxIxcos(phi)
Making people understand that should not be so difficult. Remember, these fora are for engineers.
In my industry and no doubt in many others there are a lot of staff who are either not engineers or at least not electrical engineers. The technicians who operate the plant are one example: they do not all have electrical backgrounds - some have mechanical backgrounds from working in steam raising plants - but they need to have some understanding of how the generator they are controlling behaves. We don't need electrical experts in that role but we need to the ops techs to be able to have a reasonable technical discussion with the engineers. Analogies are a useful starting point.
