VFD and Motor Power Calculation

I am not a VFD expert but your calculation looks ok (except the use of 0.9173 ? in the formula). Yes, it is true that the actual % value of power is showing less. The VFD display shows the actual power in percentage of the Motor Nominal Power (parameter 99.09). I am in doubt that the value of the parameter 99.09 (parameter group # 99, parameter # 09) is not set correctly. Please verify the value of this parameter. Its value should be 7 kW (motor nameplate kW). You can use the document from the below link and follow the instruction in page 28 to view the parameter value.

http://www05.abb.com/global/scot/scot201.nsf/veritydisplay/4b7e4f55e27c97fcc12577150030cfde/$File/EN_800_IPC_FW_C.pdf

- MS

91.73 is the efficiency of the motor. Now I came to realize that the power shown by the drive Display is the Electrical input Power i-e P = 1.732 *V *I*P.F

Actually the original nameplate is like that

VEM :11/ 7 Kw, 400 V, 50 Hz, 21/15.5 A, P.F 0.85/ 0.75, 1450/1465, 40/52 C

(Normally temperature in our region remains hot (45 C) during these days therefore I assume that it will be 7 KW like many other motors where we put de-rated values in Simocodes but I will check it tomorrow the parameter U have mentioned).

but how these values are wright. Let say 11 KW case for first measurement i-e

(25.57 %, 13.24 A, 28.49 Hz, 857 RPM). The Power will be P = 1.732 *228*13.24*.85 = 4.444 KW which is 40 % of 11 KW instead of 25.57 %

(34.88 %, 13.20 A, 39.25 Hz, 1160.0 RPM). P= 1.732*314*13.20*0.85 = 6.10 KW which is 55.47 % of 11 KW instead of 34.88 %.

Actually where the problem lies i-e Value of the P.F or the value of the Voltage applied at the terminals.

( I tried to measure the voltage at the terminals of the motor but Digital meter shows wrong values, for one moment it shows KHZ, other moment resistance, then a Buzzer sound and so on etc)

Check parameters for voltage and torque and review your calculation,check your motor parameter settings,moreover,your motor shows different kw11 and 7 at the same voltage is this correct or there are different terminal connections.

You are calculating power based on what it COULD be in the drive, but the drive is measuring what the power ACTUALLY IS. Assuming you have a centrifugal (quadratic) load such as a pump or fan, the power required by the load varies at the cube of the speed reduction. So at 857RPM, that is (857/1465=) 58% speed. So the LOAD will require only .58³ or roughly 20% power, plus whatever losses exist in the VFD, motor and power train components. The VFD is measuring this and displaying it; their formula is simply Power In - Power Out.

The Load is not Variable torque Load Like Fan or Pump. The load is a conveyor belt.It is a constant Torque Load. This conveyor belt take material of 180 Ton/ hr to the Cement mill. Fluctuation will be from 170 to 190 t.If it is a constant Load. As there should not be so huge difference in the calculated values vs the actual values. There is something wrong here may be V/f ratio or something else

Now take another example of a VFD. There is a 2000 KW, 690V, 46.4 Hz, 1985 A, P.F 0.87, 923 RPM motor attached to a fan.A reading Taken from the VFD

812.2 RPM, Frequency 40.6 Hz, Current 1520 A, Torque 74 %, Power 1374 KW

Now V/F = 690/46.4 = 14.87 So at 40.6 Hz V= 604 V

P = 1.732 *604*1520*0.87 =1383.4 KW

Now in our calculated values and values given by the VFD we have only a minor difference i-e 0.07 %

Hello

have you checked if the drive is woking in V/F mode or DTC mode?

I work on ABB ACS800 multidrive inverters with ambient temp above 50C but with Aircon.

have some problems in calculated torque.

Yes the Drive is in DTC mode. Motor Power is 11 KW and and drive rating is 12.5 KW

Percentage power,current frequency and Rpm are insufficient parameters,we require voltage and Torque parameters as well to calculate the exact power to the motor

I feel there may be a parameter setting problem in the motor map,kindly check motor parameters..

I am a newbie; sorry for a kiddy question.

I came across ACS800 multi drive system.

In Drive Window;

______________________________________

P01.06: Motor Current is 1150A

P01.11: Motor Voltage is 695.0V

P01.09: Power Percentage is 35.98

Nominal Power entered is 2990kW

cos(phi) = 0.85

____________________________________

I calculated Power as Root(3) V I Cos(phi) = 1176.8 kW which is 39.35% of 2990kW

On other hand parameter P01.09 says 35.98%..

Am I doing something wrong in the calculation ? Or how drive is calculating the percentage of power?

"Lost in ACS800"

for ABB drive Par 01.06 Motor Shaft Power: P = T*w - losses (in case of 200kW losses are about 0 - 4kW) where (w = omega)
Open loop Torque accuracy ± 4% of motor nominal torque
Open loop Power accuracy ± 4% of motor nominal power
Using formula P=√3 *V*I*PF*Eff is quite unreliable because there is too many changing factors.
For example V, PF
Using this formula in all situations is very tricky, as you would need data table of correct values in different situations.
PF ratio is only around half in 1/4 load but around 80% in full load. You can find the details in motor data sheet. Use that number in calculation and you will get calculated result same as actual signal.

Why is any of this a problem?