#### Ambient44

##### Member

- Location
- Oak Lawn, IL

Thank you,

Tom

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- Thread starter Ambient44
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- Location
- Oak Lawn, IL

Thank you,

Tom

- Location
- London, England

As an approximation though, one HP=746 watts.

Therefore a 25 HP motor will have an output of 18.65 KW

The input will be more than this because the motor is not 100% efficient, we dont know what the losses are, but 10% might be a fair estimate.

That would give an input of about 20KW.

Multiply the prevailing cost per KWH in cents by 20, and that is the APPROXIMATE running cost in cents/hour, presuming that the motor is fully loaded.

It would be more accurate to fit a KWH meter and bill for the actual energy used.

- Occupation
- Electrician

Thank you,

Tom

Not taking into account in rush current or duty cycle, name tag apparent power would be 11.5 kVA.

That will have to be adjusted for power factor (for example a pf of .7 would mean that only about 8 kW is being used), duty cycle (what percent of the day will be pump actually be running) and peak power during start up that may be in excess of the name plate rating.

At 20 cents per kW, the pump costs approx. 40 cents of electricity to run per hour at a 100 percent duty cycle.

40 cents per hour may not seem like much, but it works out to 288 dollars per month.

- Occupation
- Electrician

As an approximation though, one HP=746 watts.

Therefore a 25 HP motor will have an output of 18.65 KW

The input will be more than this because the motor is not 100% efficient, we dont know what the losses are, but 10% might be a fair estimate.

That would give an input of about 20KW.

Multiply the prevailing cost per KWH in cents by 20, and that is the APPROXIMATE running cost in cents/hour, presuming that the motor is fully loaded.

It would be more accurate to fit a KWH meter and bill for the actual energy used.

That is not legal in most states as the sale of electricity must be overseen by the state's Public Service Commission. That is why landlords can't charge their tenants per kwh as well.

I doubt anyone would balk at paying a buck an hour and that would certainly be enough to compensate for the amount of additional electricity being used.

- Location
- Ann Arbor, Michigan

Your calculation of 11.5 KVA is not correct.

Convert the 460 to the equivalent line to neutral which is 266.

Multiply 25 * 266 = 6.65 KVA.

Then by 3 because there are three phases which produces 3 * 6.65 = 19.95 KVA.

I agree with your comment on selling electricity. Instead sell the services of the pump and embed the electric cost in the services. And a dollar per hour seems like a good add on.

.

.

Thank you,

Tom

Wattage?

Assume you mean power.

Are you supplying the power? If not, what would your basis for charging for that?

Anyway, 20hp is about 15kW. Losses might be around 10% so power supplied

By the hour that would be 16.5kWh.

Obviously.

- Location
- Oak Lawn, IL

Thanks

Thank you for the help in this matter, I assumed there was a power factor involved I was asking how to translate it into hours of use. I believe I got the answer I was seeking. For calculating three phase kilowatts what is the watts X 1.732 divided by 1000 then?

- Location
- Ann Arbor, Michigan

Ambient44:

Following is an intuitive way to think thru some 3 phase problems.

If you have a Y source with a Y resistive load, then it is obvious that the current in any phase is in phase with the voltage from line to neutral. Why? because the line to neutral voltage is the voltage across the resistor and the current thru a resistor is in phase with the voltage across the resistor.

So now if you know the line to neutral voltage and the current in the line, then you can calculated the VA to the load in that one phase. For a resistive load the VA and watts will be identical.

Assuming everything is balanced and you know the power in one phase load, then 3 times this is the total power in the 3 phase circuit.

You also know that line to neutral voltage is less than line to line. Thus, the way to use 1.732 (George Washington's birth year) is to divide the line to line voltage by 1.732 to get the line to neutral voltage. 1.732 is derived from drawing a triangle whose base is 1/2 of the line to line voltage, the hypotenuse is the line to neutral voltage, and cos 30 deg is the relationship between base and hypotenuse. cos 30 is (1/2)* sq-root of 3. Sq-root of 3 is 1.732 .

.

- Location
- Clarkesville, Georgia

- Occupation
- electrician

That is not legal in most states as the sale of electricity must be overseen by the state's Public Service Commission. That is why landlords can't charge their tenants per kwh as well.

I may be wrong but I believe that in Georgia it is perfectly legal for a landlord to charge tenants per kwh, as long as landlord doesn't mark up price and sells at or below cost?

So, you can't buy some kWh meter off the internet and use that for the purpose of billing.

- Location
- NE Nebraska

Wattage?

Assume you mean power.

Are you supplying the power? If not, what would your basis for charging for that?

Anyway, 20hp is about 15kW. Losses might be around 10% so power suppliedifthe motor is running at rated load would be about 16.5kW.

By the hour that would be 16.5kWh.

Obviously.

If energy cost is 10 cents per kWh one hour costs 1.65. Those that suggested a buck an hour are losing money.

- Location
- NE Nebraska

Thank you for the help in this matter, I assumed there was a power factor involved I was asking how to translate it into hours of use. I believe I got the answer I was seeking. For calculating three phase kilowatts what is the watts X 1.732 divided by 1000 then?

Watts is energy, and is not dependent on voltage, current or number of phases. Think of it as how much work is being done not how many amps are being drawn. Kilowatts is just 1000 watts.

Change voltage or current or number of phases and you will have different levels of the others if the same wattage is being consumed.

Two identical prime movers one supplying single phase generator the other supplying three phase generator - each supplying loads that provide the same amount of work should use the same amount of fuel to get the work done assuming all efficiencies are also equal.

Watts is energy, and is not dependent on voltage, current or number of phases. Think of it as how much work is being done not how many amps are being drawn. Kilowatts is just 1000 watts.

Watts are a measure of power, not energy. Energy is billed in kW

- Location
- NE Nebraska

Watts are a measure of power, not energy. Energy is billed in kWh, not W or kW.

I knew that, I was focusing more on the fact that he was trying to multiply watts by 1.73 to convert to three phase values.

A better statement is that power and/or energy is same for a given amount of work (assuming same efficiencies of equipment) no matter what voltage or number of phases is or even if AC vs DC.

power and/or energy is same

Energy is power integrated over time.

- Location
- Ann Arbor, Michigan

I made a mistake in supporting the $1 per hour charge. In my mind I may have been thinking of $1 per KWH. In any event the real cost of electricity for the area or conditions (gasoline powered generator) needs to be used to arrive at a price.

kwired:

i somewhat support your loose use of power relative to what you were trying to explain. I think in loose conversation it is easier to say power than energy. We usually speak of power companies, but the use of energy companies is becoming more common.

.

Energy and work are equivalent quantities.I knew that, I was focusing more on the fact that he was trying to multiply watts by 1.73 to convert to three phase values.

A better statement is that power and/or energy is same for a given amount of work (assuming same efficiencies of equipment)

Power is not the same as work.

Power is the

In SI (metric) units, the unit of work (and energy) is the Joule.

One Joule is a force of one Newton moved through a distance of one metre.

Do that in a time of one second and that gives you the

For Imperial, if you move a force of one lb(f) through a distance of 550 feet or 550lbf through a distance of one foot, you will you will have done 550 ft-lbf of work. Do either in a second and your rate of work will be one horsepower. Or about 746 Watts.

- Location
- San Francisco Bay Area, CA, USA

- Occupation
- Electrical Engineer

1) Motor electrical power consumption = 746W per HP, plus losses,

2) The only information you have about the motor is the full load HP, you don't know what the actual load is on the motor.

But it's a relatively safe bet it's probably around 80-90% loaded, that's a good design practice. So since you don't know how loaded the motor is, but you also don't know the efficiency (losses), you can pretty much trade the unknown loading for the unknown losses and call it a wash.

So for NON-REGULATED energy charges, I would just go with the 15kW and if you want to charge for 1 hour of work, that's 15kWh. Keep it simple.

If you want to be gnat's ass accurate, buy a kWh meter.

But....I think that before getting too far off on tangents, just stick to the facts

is not, in fact, a fact.power and/or energy is same for a given amount of work

Just thought I'd clear that one up.

- Location
- Massachusetts

That is not legal in most states as the sale of electricity must be overseen by the state's Public Service Commission.

You can still install a meter to monitor kwh and in most areas you can charge what you paid for the electricity.

In other words the illegal part would be reselling the power for profit.

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