Calculating Power Consumption

There are meters that clip on the circuit and measure the PF. Google PF measurement.

http://cr4.globalspec.com/thread/25669#comment270757

This helped me--Amp reading times 1.732.

Mark 684 received a good answer and I believe that North of 60 also deserves one--Read up and cross reference them and vote if it helps you.---GOOD STUFF!

Hai, The 3 - phase power can be measured by following expression: P=Root(3) x VL X IL X COS(p.f) Where VL - Line voltage of the supply IL - Line current Cos(p.f) - Power factor The power factor measurement can be done by 2-Wattmeter method (care should be taken for using LPF or UPF meters) Thanks, Vinod

Without suitable power metering for exact measurement, an estimate could be made by taking the full load power factor on the nameplate to estimate no load current and factoring in efficiency or by motor speed.

i.e. Io = Iflc * (sq root (1- (cosphi)2))

e.g. 5kW 90% eff 10A flc motor cos phi (nameplate) = 0.8 squared = 0.64. Then 1-0.64 = 0.36. Then sq root of 0.36 = 0.6. Io = 10A 8 0.6 = 6A.

If this motor drew 8A, then at half load approx 2.5kW, at 90% eff then consumption roughly 2.75kW.

Motor load can also be assessed by motor speed if you have a reliable tacho.

e.g. 5kW motor nameplate = 1430 Motor speed = 1465 therfore half load again etc. In my experience, true full load speed is normally more than nameplate to the order of 80% of slip. So I would say if 1430, real full load speed could be 1444rpm (80% of stated slip)

Any input from motor designers??

No, not without at least one other value; either phase angle or watts, however you could ratio your measured values to the stated nameplate values for this motor and come close. [(V_n*A_n/v*a)*Hp_n]/746 = watts(approx).

p.f.=watts/va

The relation between values is the following:

S = U * I * pf * eff * sqrt(3) = P * pf

where

S = apparent power [kVA] (input power)

U = voltage (line-to-line) [V]

I = current drawn [A]

pf = power factor [-]

eff = efficiency [-]

P= active(real) power [kW] (input power)

The input power includes output mechanical power at the shaft (torque x rotating speed) + all losses in the machine. The losses are included in the efficiency ( eff = output power /(ouput power + losses).

You measured U and I. In order to be able to determine the power factor you need additional data: torque + speed + efficiency.

If the measurements were made at full load (rated, nominal) , some recommended values are given below. At lower load, the values are lower.

It the list below, the rating is given in HP ( 1 HP = 746 W for USA, 1 HP = 736 W international) (output mechanical power) power factor and efficiency are indicated in [%]. In the formula indicated earlier, values should be entered as decimal.

The values are those recommended as minimal in the US for a "Energy Star" qualification.

The theory is good but beware of using assumptions of pf when at part load.

The pf decreases dramatically with load e.g. 0.9 at full load, 0.7 at half load.(c. 5kW)

Many suppliers do not list pf below half load as the pf is so poor.

So, if the motor is not near fully loaded and known to be, be aware the calulated result may have much more error in overestimating power consumption which is, in effect, apparent only.

You are absolutely right. The pf is the ratio of real (ohmic) power vs apparent power. the reactive power requirement is basically the power required for magnetization and is nearly constant along the load spectrum ( from no-load to full load) while the real power has to cover the output power (dependant on load) + losses ( some dependent on load ( losses in windings) and other nearly constant (mechanical).

The numbers given were for the full load situation.

It is very easy to determine the Power, Current, Voltage, Power Factor for this Formula.

Power (KW) =1.732*Voltage*Current*Power factor.