What are the differences?

Status
Not open for further replies.
KVA (Kilivolt-ampere) is apparent (possible) power whereas KW (Kilowatt) is true (measureable) power.

Roger
 
KW is the actual power being used but sometimes is less than if you multiplied Volts times Amps which gives you kVA. If you have a 100% resistive load- strip heat, incandescent lamps, etc.- kW will equal kVA. If on the other hand, you are looking at an inductive load- motors, transformers, etc.- the consumption will be lower because the electricity needed to raise a magnetic field is recovered as the field collapses and returns elecytricity back into your system; you "only borrow" the electricity to make the fields. That is why you caculate amps for conductors; you want the maximum current the conductor will see- kVA.
It is also why power companies charge a penalty for low power factor- they must provide the maximum wire size the system will need but can only charge for the power consumed- kW.
 
A Fundamental Answer:

A Fundamental Answer:

kda3310 said:
What are the differences of KVA and KW? It seems to be the same thing. Why list it two differences ways?

Jason
Click to expand...

In AC circuits, the voltage and current are sometimes in phase, and sometimes not.

For example, an electric space heater may draw 10A @ 120V, and the current and voltage are in phase. Then apparent and real powers are numerically equal, but the units are different.

Pa = ExI = 120Vx10A = 1200VA

Pr = ExI = 120Vx10A = 1200W

Now consider a motor drawing the same current but with the current lagging the voltage by 30 degrees (1/12 of a period). The cosine of this angle is called the "Power Factor".

In this case, apparent power is,

Pa = 120Vx10A = 1200VA as before,

But real power is,

Pr = ExIxPF = 120Vx10Axcos(30) = 1039W

Watt-hour meters integrate real power and read out in KWH, not KVA as some would have you believe.
 
rattus said:
Watt-hour meters integrate real power and read out in KWH, not KVA as some would have you believe.
Click to expand...

Hi rattus,

So the power company takes the hit for a residence having a poor PF? It takes the same amount of generation to feed a 20-amp load at unity PF as it does to feed a 16-amp load at 80% PF?
 
hardworkingstiff said:
Hi rattus,

So the power company takes the hit for a residence having a poor PF? It takes the same amount of generation to feed a 20-amp load at unity PF as it does to feed a 16-amp load at 80% PF?
Click to expand...

The power company has generating units they can put online that generate electricity with a leading PF to counteract the effects of poor power factor, and supply no actual energy into the grid. Its like they switched in a giant capacitor bank.

The power company does have to pay for the I^2R losses in the distribution system, which is why they penalize large users with PF issues.
 
hardworkingstiff said:
Hi rattus,

So the power company takes the hit for a residence having a poor PF? It takes the same amount of generation to feed a 20-amp load at unity PF as it does to feed a 16-amp load at 80% PF?
Click to expand...

Stiff, it takes the same amount of coal to generate a KWH at any PF, (at the plant) but it takes more current to deliver that KWH to the customer with a PF less than 100%. This means bigger transformers and cables for the poco plus the heat losses. Add in these losses and the poco has to burn more coal to deliver that KWH to you at lower PFs. To recoup the extra expense, the poco slaps a demand charge on commercial customers. Residential rates have the PF built in so to speak.

Your example is backwards. Assume 20A @ 120V and 80%. The apparent power is,

Pa = 120V x 20A = 2400W

Real power is,

P = 120V x 20A x 0.8 = 1920W

Now assume 16A @ 120V and 100%,

Pa = 120V x 16A = 1920VA = 1.92KVA

Real power is,

P = 120V x 20A x 1.0 = 1920W = 1.92KW

Simply put, it costs the poco more to deliver electrical energy if the PF is less than 100%. Likewise the customer pays for the heat losses in his building, and he must provide bigger transformers, etc. as well as paying a demand charge.
 
petersonra said:
The power company has generating units they can put online that generate electricity with a leading PF to counteract the effects of poor power factor, and supply no actual energy into the grid. Its like they switched in a giant capacitor bank.

The power company does have to pay for the I^2R losses in the distribution system, which is why they penalize large users with PF issues.
Click to expand...

Not sure how this would work, but such generators would mean bigger capital and operating expenses. I would think they would have to be located at the load which sounds impractical to me. Capacitors would be cheaper.
 
the generators just up the p.f. in the generating plants whenever the load p.f. goes down. generators always get operate with a 0.8 p.f and lower / raise the kVARS if the load requires it.
 
Status
Not open for further replies.
Top