KVA AND KW Difference

[Toros]

Hi,
What is the difference between KVA and Kw or VA and W

 

[Sahib]

KVA is the maximum power that a power system can supply and KW is the actual power it supplies to the load connected.

 

[m sleem]

Hi,
What is the difference between KVA and Kw or VA and W

KVA = Apparent Power
KW = Actual Power
KVAR = Reactive Power

KVA*= KW*+KVAR* or KVA^2=KW^2+KVAR^2
* angular value

KW=KVA COS$
KVAR=KVA SIN$

COS$=Power Factor (cosine the angle between volt & ampere)

Why we use sometimes KW & KVA??

We use KW (when we know the power factor which will be drawn by the actual load) & this applicable for the equipment at the end user.

We use KVA (when we don't know the power factor which will be drawn by the actual load) & this applicable for (transformer, generator, UPS)

 

[electrofelon]

KVA is the maximum power that a power system can supply and KW is the actual power it supplies to the load connected.

personally I find the first part of that kind of a strange way to put it and perhaps not so helpful to the OP. VA is simply volts X amps. Volts X amps is also watts IF the voltage and current curves (assuming a graphical representation here) are lined up - reaching their peaks at the same time. With AC, you can have inductive and capacitive loads that make voltage lead or lag the current and in these cases you will not get as much power as if the curves were peaking at the same time. We need to be able to quantify this leading and lagging of current and we call in power factor. So watts is technically volts X amps X power factor.

 

[Sahib]

VA is simply volts X amps.

By stating KVA as the maximum power that a power system is able to supply, and KW is the actual power supplied to the load, the necessity of introducing power factor into the explanation is avoided. The power factor concept may be introduced later after the OP has understood the first concept.

 

[GoldDigger]

Which will be confusing if the questioner encounters a generator or fuel cell or other power plant for which the kW limit is the actual power that can be delivered, because of the limitations of the energy source rather than the current source.
And the kVA figure can only be delivered because not all of it represents real power.

My personal feeling is that it is appropriate to introduce the concepts of phase angle and power factor very soon after making the transition from DC concepts to AC concepts.
It is no more fundamentally difficult than RMS values in the first place, IMHO.


Tapatalk...

 

[Sahib]

Which will be confusing if the questioner encounters a generator or fuel cell or other power plant for which the kW limit is the actual power that can be delivered, because of the limitations of the energy source rather than the current source.
And the kVA figure can only be delivered because not all of it represents real power.

It would be confusing to the OP when he encounters such equipment, if he leaves this thread with the understanding of KVA and KW but without understanding the power factor, the connecting link between the two.

 

[mivey]

VA is just volts times amps. One reason it is used because system losses are a function of (VA)^2. Another is because we have to consider both iron losses (related to voltage) and copper losses (related to current) and our equipment has to be able to handle thermal loading from both. Another is because it shows the relation between real power (W) and reactive power (var).

VA and W are the same only when we have a pure resistance (not usually the case for AC). The relationship is that W = VA * power factor. Power factor is cosine of the phase angle between voltage and current (phase angle difference because the voltage and current waves reach maximum peaks at different times).

W is the real work that is done. var is power that is a temporarily exchanged from source and load but does no net work. VA^2 = W^2 + var^2.

One reason VA is used to size equipment is to make sure we can handle the thermal load associated with the W and var exchange that will be going through our equipment.

 

[fmtjfw]

Another is because we have to consider both iron losses (related to voltage) and copper losses (related to current) and our equipment has to be able to handle thermal loading from both.

I understand that an unloaded transformer has losses due to the impedance of the primary and due to heating the iron core. But are there additional iron losses as the transformed due to secondary loading?

Just trying to understand.

Thanks

 

[JoeStillman]

I understand that an unloaded transformer has losses due to the impedance of the primary and due to heating the iron core. But are there additional iron losses as the transformed due to secondary loading?

Just trying to understand.

Thanks

Yes, transformer losses are proportional to the load.

 

[mivey]

I understand that an unloaded transformer has losses due to the impedance of the primary and due to heating the iron core. But are there additional iron losses as the transformed due to secondary loading?

Just trying to understand.

Thanks

To put it simply, the iron losses are related to the voltage (eddy currents, etc.) and the copper losses are related to the load current. Transformers are thermally loaded just from being energized even if there is no load attached to the secondary (iron losses). The secondary load adds additional thermal load (copper losses).

 

[kwired]

It would be confusing to the OP when he encounters such equipment, if he leaves this thread with the understanding of KVA and KW but without understanding the power factor, the connecting link between the two.

And that connecting link is a very important part that was left out of what you said in post #2, OP asked what the difference was, your answer was not wrong, but came short of explaining the difference.

 

[Besoeker]

By stating KVA as the maximum power that a power system is able to supply, and KW is the actual power supplied to the load, the necessity of introducing power factor into the explanation is avoided.

OK. I have a 3-phase 110kW, 400V motor. The efficiency is 93%.
Explain how you would calculate the supply current at full load.

 

[Besoeker]

Hi,
What is the difference between KVA and Kw or VA and W

For DC, 1Vx1A is 1W.
With AC, it gets a little more complicated.
The voltage and current vary over a cycle, 360 degrees. If you multiply them at any point over that cycle you get Watts at that point. Given that volts and amps vary over the cycle, so do the watts.

Here is one example of this:



It's how 1V and 1A (rms) can appear over a cycle. (I've slightly adjusted the volts an amps so that they can they don't obscure each other.)
They are exactly in phase. That is they both cross zero going in the same direction at the same time. Multiply the values at all instants in time and you get watts at those instants. The average value is 1W. The 1V and 1A also give you 1VA

Another example.



It's still 1V and 1A so still 1VA.
But now they don't cross zero at the same time. There are places were the current is negative and the voltage is positive. The power at any point is still the voltage times the current at that point. But with negative current and positive voltage, you get negative power as the chart shows. And the maximum values for current and voltage don't occur at the same so the peak of the power is lower.

The effect is that, over a complete cycle, the power has a lower average value than in the first example. In this case, 0.707W.
Thus the ratio of W/VA is 0.707. The power factor. PF.
W = VAxPF

Numerically, the PF is the cosine of the angle between the zero crossing points of voltage and current. AKA phi. In the example above that is 45deg. Cos 45 is 0.707.
So we get P (power) = VIcos phi.
VA is just VI and does not change with PF.

 

[GoldDigger]

FWIW, your statement is only strictly true for constant voltage (flat line ) DC. For pulsed

Tapatalk...

 

[Jraef]

Yes, transformer losses are proportional to the load.

Well, there are both. There are fixed no-load losses, called the magnetic losses or "iron losses" because they are mostly associated with issues related to magnetizing the core such as hysteresis and eddy currents. Then there are the losses associated with winding resistance, the "copper losses" or "I^2R losses" that are proportional to current, thereby load. No load losses are always much smaller when looking at the full load capacity of the transformer, but when a transformer is lightly loaded or even unloaded, they make up a greater and greater percentage of your losses.

 

[GoldDigger]

They may be small in comparison, but there will also be copper losses associated with the no-load magnetizing current.


Tapatalk...

 

[Besoeker]

FWIW, your statement is only strictly true for constant voltage (flat line ) DC. For pulsed

Tapatalk...

Different topic.