# 3-Phase Motor Power

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#### Jay Kirsch

I have been working on a project to calculate the enegy cost savings that would result by turning off large 3-phase pumps during peak energy cost
hours of the day.

How is the power drawn by a 3-phase motor calculated ? Is the equation, 3^0.5 * V * A, correct and independent of the load on the motor ? Is it more accurate than using motor horse-power and efficiency to calculate power draw ? What formula do the electric utilities use for their load shedding calculations ?

Jay Kirsch
Macro Automatics
2985 E. Hillcrest Drive, Ste 101
Thousand Oaks, CA 91362
[email protected]

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#### Kirk S. Hegwood

I don't know if AB still has it, but at one time they had a neat little program that would calculate energy savings using VFDs in place of
contactors. You could input kW costs, etc. I can't find my shareware version.

Kirk S. Hegwood
President
Signing for Hegwood Electric Service, Inc.
[email protected]

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#### Daniel Boudreault

Hello Jay,

> How is the power drawn by a 3-phase motor calculated ?

The cost of electricity depends on your supplier.
Around here, they charge by the formula that you mentioned.
"3^0.5 * V * A"
So, you would want to maintain your powerfactor to one (1).
The power utility company claims that they have losses on their power lines...so they need to charge for it.

> Is the equation, 3^0.5 * V * A, correct and independent of the load on the motor ?

Yes.

Dan Boudreault

A

#### Anthony Kerstens

The formula is actually:
KVA = 3^0.5 * V * A * PF / 1000

No, it's not the same because it's not usually the only thing the utilities use to calculate billing.

You should call your local utility rep. since one may charge differently than another. That said, there are several things to ask them about:
- peak demand charges
- peak kVA charges
- power factor penalty
- discounts if you own your substation
- services such as daily email of consumption and PF data.

Anthony Kerstens P.Eng.

#### PhilCorso

Responding to Jay Kirsch's query. Determining the power draw of a 3-phase motor can be done as follows:

1) Input kVAa = SQRT(3) x Ia x Va, where Ia and Va = actual operating values.

2) Rated kVAd = SQRT(3) x Id x Vd, where Id and Vd are taken from the nameplate.

3) Determine the motor's operating kVAas as a percent of its design rating, % = 100 x kVAa / kVAd

4) Ask motor manufacture to provide the design characteristic curves which should include power factor (PF) curves.

5) Input kW = kVAa x PFa, where PFa is the motor's power factor at the operating point.

6) If PF data are unavailable from manufacturer and you are unable to obtain them from reference books, then contact me. I can provide typical data.

7) Alternatively, a simpler, more accurate approach can be used if the plant's input power consumption kWhr meter is accessible to you. Take a reading with pump(s) in operation for a time interval, t. Then take a reading with the pumps out of service for the same time interval, t.

a) The motor(s) contribution in kwhr = [kwhr(w/motors) - kwhr(w/o motors)]. b) The motor(s) demand in kW, is kwhr / t. c) If the meter is not accessible to you, then the power company should be able to help.

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

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#### Dittrich, Rodolfo

Jay

The equation to calculate de power is P= sqrt(3) * V * I * cos (phi)

where V: voltage applied to the motor
I: current
cos (phi): angle between voltage and current, generally this value is between 0.8 and 0.85

Regards

R

#### Robert R. Stephens Pennzoil Products

3 phase power is 1.732*E*I* Cos(angle) but their are other factors to consider when calculating power savings, e.g. power demand charges and stair
stepped usage rates. In other words you have to know how you power bill is calculated. Most industrial facilities are charged from $4.5 to$10 per kilowatt of demand. So if you reduce the demand of a motor i.e. inrush current then your savings is more pronounced. Most usage charges (kilowatt-hours) are stair stepped starting with one charge, the highest for the base KWH and less as you use more. When you turn off a load the usage charge comes from the highest tier or cheapest usage rate. Also consider that some utilities pass through the fuel cost of electricity. At the industrial facility I work at half our costs are fuel adjustment costs. So it is not so simple as just adding up the usage savings.

Hope this helps. Good luck.

O

#### O'Connor, Denis

>I have been working on a project to calculate the enegy cost savings that
>would result by turning off large 3-phase pumps during peak energy cost
>hours of the day.

In addition to using Anthony's formula and suggestions about understanding your electric bill, make sure you use actual measurements and not motor nameplate information. If you don't have power measurement equipment, you can also determine the motor load using a strobe light and comparing actual rpm to nameplate rpm.

When calculating your savings, don't make the mistake a lot of people do and add the total bill up and divide by the total kWh to get the cost per kWh. If you are going to take demand credits, make sure you understand when you set your demand and whether the equipment you are shutting off is running then.

Denis O'Connor
Manager of Operations
EUA Cogenex Corp.
Boot Mills South
100 Foot of John Street
Lowell, MA 01852

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#### Dobrowolski, Jacek

Hi Jay
There is only one way to find out how much power motor takes - measure it. To do it you need two wattmeters. Equation to calculate it looks like this: P = 1.73 * U * I * cos fi where: P - active
power U - voltage between phases I - current drawn by one phase cos fi - phase shift between voltage and current Equation to calculate reactive power equation looks the same but instead of cos fi there's sin fi. Both formulas are correct when you know U, I and cos fi. But I as well as cos fi depends on the load. Values
which can be read from the motor plate are measured for nominal torque. If those motors are driven by inverters, be aware that power you can read out from inverters is mostly apparent power or only estimated active power.

Regards,

Jacek Dobrowolski

Software Eng.
Electrical and Software Design Department
Secondary Division
International Tobacco Machinery Ltd. Poland

#### PhilCorso

Responding to Anthony's reply:

A slight Technical Correction... you gave the formula for kW, not kVA.

Yes, the power company's requirements are some of the ones you listed, but the most important are: energy (kwHr); and maximum demand (Peak kW).
The latter is usually over a period of time, say 15 minutes.

Their contract with your company should list the requirements.

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

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#### leeenglock

for large motors in the 2, 4 or (maybe) 6 pole range the power factor is around 0.9. when the poles go up the power factor drops. you can get to download motormaster software free from the doe website and this has thousands of motors in the database, details of the service factor, power factor, efficiency, etc are all in the software.

rgds
leelock

#### PhilCorso

Responding to Corso's 2-Aug-00, 11:09am response to Jay Kirsch's query:

P.E., huh. You are Technically Incorrect:

In order for the left-hand side of equations 1) and 2) to correspond to kilo-VA, then, the right-hand side must be divided by 1,000 ! ! !

PRACTICE WHAT YOU PREACH ! ! !

Regards,
Phil Corso, PE
Trip-A-Larm Corp
(Deerfield Beach, FL)

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#### Rimmele_Martin

I have the shareware that Kirk mencioned. I'll mail it to you if you want.( or to somebody else who want it.)
Regards,

Martin Rimmele
[email protected]

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#### Ken Codd

Martin,

Could you send me a copy of the shareware?

Thanks,
Ken Codd
[email protected]

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#### bong buerano

Can you give me a copy also of that Shareware? Thanks and regards. Bong Buerano E&I Engineer [email protected]

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#### Paul - University student

(considering a balanced load) with resistance R the instantaneous power delivered to a balanced three phase load is constant. power is determined by 3/2* V^2 /R The total average power delivered to the balanced Y-connected load and triangular connected load is given by P= 3*VP*IL*cos(pf), V and I are the effective values of the phase voltage and line current. It is easier to measure the line to line voltage and the line current of a circuit. Also recall that the line current = the phase current and that the phase voltage is VP = VL/1.732 Therefore P=1.732*VL*IL*cos(pf) = 3Vp*IL*cos(pf) L = load Vp = phase voltage This is the equation needed Paul Uni student

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#### Bud Graham

Please email a copy of the shareware mentioned.
Regards,

Bud Graham