PF based on capacitive load. 4

[aolalde]

Somebody asked about "getting the PF adding capacitors and measuring the current with a clamp-on instrument".

It will be possible for Inductive loads.
Get the “V-curve” of the load.
Record the total current It (Amperes). Then gradually add capacitive load.
The current will drop to a minimum, from then as capacitance is added the current will increase again.
You want to match the capacitive load that produced the minimum total current.
Now read the total minimum current ( Itm).

The power factor will be:

PF= cos(phi) = Itm / It

For instance; It= 12.5 amperes, the minimum total current Itm = 10 Amps after capacitive load is added and before "It" starts to increase again.

PF= 10/12.5 = 0.80

This is simple for single phase circuits. Balanced phase currents should be handled if the test is for polyphase circuits. WARNING: Comply with safety local and federal codes. Handling electricity by inexpert people could be lethal.

 

[justinn4]

Why?
is not PF=kW/kVA ?

 

[TomG33]

Exactly,
and you would measure KW and KVA by taking voltage and current measurements. Thus, Aololde's original post stands, (assuming the voltage remains constant).

Tom

 

[UKpete]

justinn4, the kVA is easy to find (it's just measured volts x measured amps), but kW is not so easy unless you have a power meter. Aolalde's method allows you to find the power factor of a load by artificially "tuning" out the load inductance by adding capacitance. (If you think in terms of radio ciruits, when you add the right amount of capacitance in parallel you have a parallel tuned ciruit of infinite impedance and minimum overall current).

The equation PF = Itm/I quoted in the original post can be proved by drawing a phasor diagram (I just tried it).

 

[digitrex]

I know it is theoretically correct to estimate PF in this way. But is it practically being done by anybody? Do you increase the capacitance in steps(by adding piece by piece) or does a variable capacitor exist?

 

[tinfoil]

I suspect the vast majority of 'real world' applications measure the kW and kVA 'directly' with an appropriate meter (fixed or portable) that can directly measure current, voltage, and phase angle.

Ignoring accuracy issues, Capacitors are fairly bulky things, whereas meters and clamp-on ammeters are not so bad.

 

[topire]

this is the power meter i use

http://www.duncaninstr.com/nanovip.htm

 

[Marke]

Hi

This method of determining the optimum correction to be applied needs to be used with caution.

If you apply this method to a motor, and at each iteration you leave the capaciotrs connected to the motor terminals when you switch the motor OFF (as is common practice with static correction), you could end up with a lot of damage.

It is very important with static correction (capacitors connected to the motor terminals or output of motor starter) that you only correct to 80% of the optimum correction.
This is because the capacitors form a resonant circuit with the motor inductance. At critical or optimum correction (pf = 1) the motor/capacitors are resonant at the line frequency. If you over correct, the circuit is resonant below line frequency.
If the motor and capacitors are disconnected from the supply and the voltage generated by the motor as it spins down passes through resonance, very high voltages and currents can be generated causing insulation damage. Additionally, as there can be electrical breakdowns, there can be severe torque transients as well.

I would strongly recommend using a separate control means to connect the capacitors when you are experimenting.

Best regards,

Mark Empson

http://www.lmphotonics.com

 

[jghrist]

You could determine the power factor from measurements of voltage, current without capacitors, and current after connecting a fixed amount of capacitors. It wouldn't have to minimize the current.

pf = sqrt(2·I2²·V²·VAR²-I2⁴·V⁴+2·I2²·V⁴·I1²-I1⁴·V⁴+2·I1²·V²·VAR²-VAR⁴)/(2·VAR·V·I1)

Where I1 and I2 are currents before and after switching in the capacitors, V is the voltage to neutral, and VAR is the rating of the capacitor per phase. This assumes that the system is stiff enough that the voltage does not increase significantly when the capacitors are switched in.

For example, if I1=210A, I2=200A, V=120V, VAR=2000, the pf of the load (without capacitors) is 78%.

 

[busbar]

 
Killer equation, jghrist! Where's my slide rule? Good work. Dare I ask where you found it?

 

[jghrist]

busbar,

Just a little symbolic manipulation with Mathcad, starting with:

I1²=IR²+IX²
I2²=IR²+(IX-IC)²
IC=VAR/V
pf=IR/I1