Calculating Electric motor consumption when a VSD is fitted

Calahonda52

Want to calculated the power consumed by an electric motor with a VSD, where the motor RPM is not available:

Assumptions:
1: Motor efficiency: 85%
motor name plate output power, o/p, divided by this to give i/p: input power

2: Load Factor: 70%
i/p multiplied by this

3: VSD efficiency: 95%
Value from 2 above divide by this

Then:

If
RPM = N = (Hz*(2/No of poles)*60
And
P1/P2 = (N1/N2)*3 cubed where P is the power consumed.
Then
P1/P2 = ((Hz1*(2/No of poles)*60)/ ((Hz2*(2/No of poles)*60))*3 cubed
P1/P2 =(Hz1/Hz2)*3 cubed

So P2= P1*(Hz2/Hz1)*3 cubed.
Eg
P at 20 Hz = (P at 50 Hz) * (20/50)*3 cubed
= 3.47*(0.4*0.4*0.4)= 0.22 kW

This make sense?

ted1

Calahonda52 wrote: »

Want to calculated the power consumed by an electric motor with a VSD, where the motor RPM is not available:

Assumptions:
1: Motor efficiency: 85%
motor name plate output power, o/p, divided by this to give i/p: input power

2: Load Factor: 70%
i/p multiplied by this

3: VSD efficiency: 95%
Value from 2 above divide by this

Then:

If
RPM = N = (Hz*(2/No of poles)*60
And
P1/P2 = (N1/N2)*3 cubed where P is the power consumed.
Then
P1/P2 = ((Hz1*(2/No of poles)*60)/ ((Hz2*(2/No of poles)*60))*3 cubed
P1/P2 =(Hz1/Hz2)*3 cubed

So P2= P1*(Hz2/Hz1)*3 cubed.
Eg
P at 20 Hz = (P at 50 Hz) * (20/50)*3 cubed
= 3.47*(0.4*0.4*0.4)= 0.22 kW

This make sense?

Unless your doing an academic excercise it makes much more sense to sub Meter the load.

Calahonda52

ted1 wrote: »

Unless your doing an academic excercise it makes much more sense to sub Meter the load.

This is much more than an academic exercise: I have 146 extract fans and 54 AHUs, all on VSDs

Nukem

If there is a VSD you should be able to see the amps (L1,L2,L3) & voltage onthe HMI (toggle down), so potentially do the V*I calc? Check the VSD spec for the likes of PF (e.g. >.98) and if the motor was matched correctly and programmed right, there should have the motor efficiency entered and accounted for

Just check for 3ph as you may need to multiply by 3*SQRT (1.732)

Calahonda52

Nukem wrote: »

If there is a VSD you should be able to see the amps (L1,L2,L3) & voltage onthe HMI (toggle down), so potentially do the V*I calc? Check the VSD spec for the likes of PF (e.g. >.98) and if the motor was matched correctly and programmed right, there should have the motor efficiency entered and accounted for

Just check for 3ph as you may need to multiply by 3*SQRT (1.732)

Thanks for this.
I have done that V*I math but I need to be able to calculate the power requirement as the VSDs on the AHUs ramp up when the extract fans come on so as to maintain + pressure and meet the ACH requirements

2011

Calahonda52 wrote: »

Thanks for this.
I have done that V*I math but I need to be able to calculate the power requirement as the VSDs on the AHUs ramp up when the extract fans come on so as to maintain + pressure and meet the ACH requirements

Well that is easy then, simply run the motor at different speeds and using a grip on ammeter measure the current drawn on the supply side of the VSD. You already know the line voltage and the power factor is going to be a constant so a power calculation is straight forward.
It would then be possible to put a table together of RPM V power consumed. If you wanted to get really fancy you could graph it.

Calahonda52

Thanks for all the insights.
As there are c 200 fan motors in this complex, ranging from (0.75 to 25) kW, I need to know is my math correct.

2011

Calahonda52 wrote: »

Thanks for all the insights.
As there are c 200 fan motors in this complex, ranging from (0.75 to 25) kW, I need to know is my math correct.

That’s easy.
Power consumed = Root 3 x Line voltage x line current x power factor

Now all you need to do is measure the current.

Calahonda52

Calahonda52 wrote: »

Want to calculated the power consumed by an electric motor with a VSD, where the motor RPM is not available:

Assumptions:
1: Motor efficiency: 85%
motor name plate output power, o/p, divided by this to give i/p: input power

2: Load Factor: 70%
i/p multiplied by this

3: VSD efficiency: 95%
Value from 2 above divide by this

Then:

If
RPM = N = (Hz*(2/No of poles)*60
And
P1/P2 = (N1/N2)*3 cubed where P is the power consumed.
Then
P1/P2 = ((Hz1*(2/No of poles)*60)/ ((Hz2*(2/No of poles)*60))*3 cubed
P1/P2 =(Hz1/Hz2)*3 cubed

So P2= P1*(Hz2/Hz1)*3 cubed.

P1=P2*(Hz1/Hz2)*3 cubed
Eg
P at 20 Hz = (P at 50 Hz) * (20/50)*3 cubed
= 3.47*(0.4*0.4*0.4)= 0.22 kW

Error corrected