3 Phase Power Consumption

Thank You "JRAEF" and "BIGG"

So if I had 10 of these pumps in my excavation and I had to use a generator to power the pumps, then the total Hp would be 5hp x 10 pumps= 50 hp and the total KW would KW = 50 x .746 = 37.3 KW. Right?

If so, what size generator would I use? i.e. how much over sizing above 37.3 KW.

I am still confused if I should size the generator for the name plate amps, or the NEC amps or the above calculations. Should I size the generator for the KW calculated from the name plate amps (or NEC amps) and figure the electrical consumption from the known HP of the pumps?

Also, If electric was available from the electric utility company and it cost .10 per KWhr then would the maximum consumption cost of these 10 pumps for one year be calculated as follows

37.3 KW x 365 day x 24 hr/day x .10 $/KWHr = $32674

If this is correct and pumps worked full load for 2 months and 3/4 load for the next 10 months then the electric consumption cost would be less than $32,674 because of less mechanical work?

Also, Does NEC publish power factors? I just guessed at my .9 ASSumption.

Once again Thanks

I'm sure the sparky's will jump in here, but I'd have

10 x 9A x 480V x 1.73/1000 = 75 KVA max running

plus 36A x 480 x 1.73/1000 = 30 KVA to make sure I could start the last pump w/9 running

total 105 KVA (probably 100 KW generator if it can deliver 105 KVA), and manual starters so I don't get the 2 AM call that 'the generator went down and we can't get it going' because all the pumps are trying to start at one time. And yes, if this is a de-watering job, you'll probably have a good deal of time when the load is much less.

Guess I didn't reply to the other 2 parts.

Your KW calculation would need to include eff. So 5 x 10 x .746 / .9 = 41.5KW Your billing charge calc should be ok, unless the utiliity has charges for demand and low PF.

My statement regarding PF was based on Googling 'full load power factor for 5 hp motor', which led me to this gubment document:

http://www1.eere.energy.gov/manufacturing/tech_deployment/pdfs/10097517.pdf

It has nice curves for efficiency and PF vs load for various HP ranges. I didn't take the time to determine if they represent standard, high eff, or some kind of average motor.

Why are you adding 36 amps? Are you saying starting amps are 4 time running amps?

Actually allowed for 5 times, but that's a SWAG

Install a power analyser relay or datalogger and record the power consumption over a period of time.

Pretty good SWAG actually.

Generator sizing is more of an art than a science, mostly because not all generators are created equal. Some have more capacity for brief starting current surges than others, generally following the cost of the generator. You really need to have this conversation with your generator suppliers, your results will vary among them yet most of them will not be wrong for THEIR generator.

But if you want a quick and dirty "rule of thumb", use 3.5X the motor HP as the generator kW if you are starting them all at once. If you can ENSURE that they will start sequentially, as bigg suggested by using manual controls that require human intervention, then you can do 3X largest motor HP + HP total of remaining motors as kW. So in your case, 17.5kW + 20kW = 37.5kW, AS LONG AS THEY DON'T ALL TRY TO START AT ONCE. If there is a chance of that, then 87.5kW minimum (which will likely be 100kW).

If you use VFDs or Soft Starters, you can reduce the size of the generator a little. 2-2.5X for soft starters, 1.5X for VFDs.

I got lost. He has 10 - 5 HP motors.

(3 x 5 for the largest one) + (9 x 5 for the rest) = 60 KW?

I used 9 FLA (per his nameplate) instead of the more common 7.5 FLA. Don't know if these are high service factor motors or maybe the 5 HP is hydraulic horsepower, or ?. Anyway, if we upsize. 9 / 7.5 * 60 KW = 72 KW

72 KW / 0.75 PF = 96 KVA vs my 105 KVA we're in the same ballpark.

Still Confused, Sorry

The pumps are electrical submersible and yes they can be started one at a time

I did ask the generator supplier but he is a salesman from a rental company that just uses rules of thumb. He told me 60 KW would work. He used 1.2 times the total HP.

I am trying to get the theoretical answer then i will get practical (upsizing).

This is just one example I have 5 excavations, just used a simple example.

So far there have been 3 way to calculate full load amps

1. using HP to get amps (6 amps in this case)

.746HP=KW=I x E x 1.73 x pf/1000

2. using NEC of 7.5 amps for 5 hp motor

3. using nameplate amps (9 amps in this case)

Are we agreeing we should use nameplate amps and 3.5 x name plate amps for start up amps then

KW = 9 x 480 x 1.73 x .75/1000 = 5.61 for 9 pumps

The 10th Pump would be 5.61 *3.5 = 19.64 (using 3.5 for start up amps)

Total is 5.61 *9 + 19.64 = 70.13 (note please don't round in your responses so i can follow the logic)

Now do you divide this 70.13 by the power factor used in the KW calculation

So 70.13/.75 = 93.50 Kw

Power Factor still is confusing to me You multiply the PowerFactor to the KVA (I x E x 1.73) and then divide by the power factor to size the generator???

Also the NEC sugestion is 7.5 amps @ 460 (not 480)

Thanks

Proceeding at great risk (I've adjusted my #'s slightly from earlier posts),

here's a link to a diesel generator in the general size range you'll need http://www.industrialdieselgenerators.com/_dgcAdmin/files/pro_docs/Item%20No.%20200%20(J80UC3II).pdf

It is rated at 2 conditions: 1) Emergency Standby (ESP), 2) Prime Power (PRP) ESP is maximum for a 1 hour period, PRP is maximum continuous. You can see it is rated at 120 stand-by amps. The generator is driven by a diesel engine rated at 126 BHP (94 KW) maximum stand-by power (at rated generator speed)

• ..ESP. . . .PRP
• KW/KVA KW/KVA
• .80/100. .73/91

The KW's (real power) are 80% of the KVA's (apparent power) because the generator is designed for 0.8 (80%) PF. Let's check the stand-by KVA:

120A x 480V x 1.73 / 1000 = 99.6KVA ≈ 100KVA ESP. Note that if the load had PF of 1 that would be 100KW and would exceed the engine BHP of 94KW.

Now for your running load: 10 x 5HP x 0.746 / 0.9(eff. assumed) = 41.4KW which is less than 73KW PRP available.

Check the maximum apparent load: (9 x 9A x 480V x 1.73 / 1000) + (1 x 9A x 3.5 x 480V x 1.73 / 1000) = 67.3KVA (9 pumps running at FLA) + 26.2KVA (10th pump starting) = 93.5KVA which is less than 100KVA ESP available.

Seems like a good starting point to me. If you see the actual load is less, you can always down-size, at least the additional systems.

The big inconsistency I see is the 9FLA for 5HP seems high.

Thanks Bigg. Your explanation is perfect.

Maybe, If the motor does pull 9 FLA then maybe the motor produces more than 5 HP.

Can you provide a link to the pumps you intend to use?

Sorry, for some reason I saw 5 pumps, not 10. Oops.

You got it right except that you have not put efficiency factor in secodn calculations.i..e. 0.9 x 6.73 = 6.057. That shows your pf is not 0.9 but is aroudn .7 to 0.7 to 0.8. also efficiency figures need to check.

I disagree. The 9FLA represents real + imaginary power. Whether the 'real' power is useful or wasted, it has to be provided. Therefore, efficiency doesn't matter. I agree with your PF estimate.

I not trying to disagree (because i am trying to learn) but the published formula I found is KW= I x E x 1.73 x pf /1000

The other formula I used is I=HP x 746 / (1.73 x E x % eff x pf)

Substituting Formula 2 into 1 you get KW = HP x .746/% eff.

note: when you do this the pf canceled which seems odd to me but makes me happy since i never know pf.

Some other observations (or confusion) below:

If the power factor is 70% then is that like saying the output is 30% less than the input? (i.e. 30% is lost in motor windings, heat, etc). If this is the case and I was hooked up to the electric grid then why wouldn't I still pay for this total input power. (i.e. why is the pf above the division line in the equation KW=IxEx1.73xpf/1000).

So a well made 5 hp pump might have a lower full load amps and a higher power factor and a poorly made 5 hp pump might have a higher full load amps and a lower power factor?

note: my first statement was in the above response was to Shivaganti comment.

Your equations are ok.

The losses, due primarily to heating, however, are accounted for by efficiency, not power factor.

Power factor is a little trickier to understand. You understand that power input is I x E (let's stay with single phase to simplify this). KW = I x E / 1000 This is right for a resistive load such as a heater or ordinary light bulb, because the current and voltage are in phase with each other - the peaks and valleys in waveform happen at the same instant in time. However, for a load with a high inductive component such as a solenoid or motor, the inductance causes the current peak to lag behind the voltage peak. In order to provide the same power (KW), more current must flow (voltage being constant), because (in layman's terms) it is doing it's work at a lower point in the voltage waveform. Then, the KW = I x E x PF / 1000, the higher current being reflected on the nameplate rating. The wire, switches, breakers, transformers, etc must be sized for this higher current, although it doesn't produce useful power. That's why this extra current is related to what's sometimes called 'imaginary power'.

Apparent power (KVA) = Real power (KW) + Imaginary power (KVAR, KVA reactive)

Can a power factor be calculated?

Say it is known that a 5 hp pump with %eff=.9 and Nameplate Full Load Amps is 9.0

Then KVA = 9 x 480 x 1.73/1000 = 7.47

KW = .746*5/.9 = 4.14

PF=KW/KVA = 4.14/7.47 = .554 (maybe this is considered estimated and not calculated)

(No one has every answered do you use 480 volts or 460 volts in a calculation) I beginning to realize it really doesn't matter but, Does NEC give a recommendation to use 460?

There is no role of PF in calculating efficiency. KVA x PF = KWs. Now KWs X effiency is the BHP received for Pump/ Motor.