Kilo Watts estimated

Although you don't define the voltage, 15 Amps is about the full load current for a 115V single phase 1 Hp motor.

Assuming that is true, the formula for single phase watts is: volts x amps x power factor. The power factor may be listed on the motor nameplate but, assuming .75 as typical for this type of motor results in watts = 1294 or 1.294 kW x 40 = 51.76 or about 52 kW-hrs.

That should be a reasonable ballpark figure but you can probably calculate a more accurate number from the actual motor characteristics.

For a 2-speed motor, is the horsepower rating related to the higher speed, or the lower speed, or to both? It may be significant, as a fan is a load for which the power requirement rises strongly with RPM. Some clarification here would be very helpful. Thanks.

I don't know. This is all the information I have.

Thank you .

netmaker

Blue,

So what you're saying is at the higher rpms, this blower will run approx. 52 kw in 40 hours or in one hour?

Thank you.

netmaker

Net,

In 40 hours. In one hour, it would consume about 1.3 kW; or (to put that in perspective) about the same as one 1300 W (or 13 x 100 W ) incandescent light bulbs.

OK.

I am looking at a condenser type cooler system that they guarantee will work in my location.

It has a blower motor with those specs.

In the shop I have an open 15A circuit which the manufacturer says should hold it.

I just wanted to see what this thing will look like in terms of wattage used.

Do you think a 15A breaker will be enough?

Thank you.

netmaker

Blue,

This guy also claims a 55 mph velocity wind out of the vent.

Do you think this is possible?

netmaker

The minimum size breaker for that motor is a 25 amp (or 30 amp) single pole breaker if it rated at 120 volts. I would recommend wiring the motor with #10 wire unless it has thermal overload heaters installed. Then you can probably get away with #12 wire. Don't forget the equipment ground wire!

Please do not mix up!<52 kw in 40 hours >

KW is power at any moment

KW-HR is Energy over 40 Ho.urs that you get billed for.

I guessed(in an earlier post) that you would be billed no more than 20KW-HRS for running 40 Hrs at the lower speed of this pole-changing motor.

At the lower speed you do less work and hence draw less power from line.

By Definition 1 electric horsepower = 746 Watts. Reference: "Guide for the Use of the International System of Units" published by the US National Institute of Standards and Technology.

That's 0.746 kW

Therefore running this motor for 40 hours should add almost 30 kilowatt hours to your electric meter reading.

One more thing since the price of electricity in Louisiana is about 9.36 cents per kilowatthour then your electric bill will increase by $2.81 for each 40 hours that you run the motor.

If you want the cost per hour it is about 7 cents per hour.

H Net Maker,

Don't let them confuse you with the facts...

VxAxPF=Watts Where V=Voltage, A is the current in Amps., PF= power factor

CB (circuit breaker) should be at least FLA (Full Load Amperes)X1.5 or little higher ,say FLAx1.75 if single phase motor is used. Use relatively slow acting Magnetic circuit breakers. Or better use "Motor Guard" of some kind. Wire (individual) should be size 14AWG if no longer than 15' or 12AWG for up to 50' and 10AWG if longer. Jacketed cable, always one size bigger.

The KWh is in simple terms the price you are going to pay for using the blower. Looking at the power factor, The closer it is to 1 the better. The utility company may require you to use PF correctors which means more $$$. Assuming that your line voltage is 115V, single Phase, or?. Wind at 55mph, To be able to really know the wind speed and CFM you need to know the RPM, blades diameter, and pitch, and for wind speed you will need to know if in restricted area(duct for example) or free air. Your best bet is to use a portable anemometer, and see for yourself. Otherwise you will be ...(Go back to the first line above.)

Wangito.

Hi Netmaker,

A very interesting thread. It starts with your question that looks like you're asking how much power the unit will draw (in units of Kilowatts) but with the question about 40 hours, the perception changes to one that wants to answer the energy used (in Kilowatt-Hours) query. After a lot of pretty sensible chatter about kWh, $ it will cost, etc. you clarify the original request with the question about circuit capacity or current in Amperes...

The reason I said 'pretty sensible' is that most seemed to ignore your original specification of 1140 RPM (unless you added that later).

Anyway, I'm going to take a shot and say the current draw, power used, etc. is going to run proportional to the square-root of the blower speed.

(1140)^2/((1725)^2)=.44 or 44% thereabouts. So if the motor is 15A at 1725, it ought to be 15*.44= about 6.6A; lets throw in a little and say its 8 Amps at 115V at the slower speed. From where I sit, the 15A circuit is good to go. One caveat: the installation would wire the motor permanently to this speed!

A clamp-on ammeter would confirm or refute all this....

Ask your local electrician or electrical inspector's office for guidance at the end of the day, not anyone else.

Hugh

PS all the other #'s for cost and such should scale with .44 or .5 for conversation purposes.

As coment # 16 states. There are .754 KW per horspower. This is correect. For your motor this would be at 100 % efficiency, but you have to consider the losses, and a good efficiency estimate for this size mortor is about 80 %.

One hour's use of your motor will use about .746/0.8 = 0.9325 watts. (almost 1 KWH )

The power factor, speed, Volts, current, amps are not included it the formula and have no bearing on this calculation other than perhaps small effects on the efficiency.

The motor design determines the efficiency.

The power quantity you pay for (in America) is the KWH's (look on your domestic power bill)

Horspower is energy (work) 32,548 Ft lbs / minute

0.746 Watts is also equal to 32,548 Ft lbs / minute

Note: There is no power factor or other terms in the equation.

It is amazing the number people that do not understand power factor.

Information Source. 35 years in the electric meter department of a Northern power utility. Also Member IEEE.

There are 0.754 KW per horsepower.

The correct is :

HP (550 ft-lb_f /sec) = 0.7457 KW

HP (metric) = 0.735499 KW

HP (electric) = 0.746 KW

Abdel..

You are absouletly correct. I transposed some numbers.

Snakers.

You are correct, power factor is only important from the supply (utility) side. Just curious, does your utility monitor load power factor and impose a surcharge if the PF routinely drops below some level over some time period?

Any idea at what level they install PF correction capacitors?

Thanks.

IN 40 HOURS the fan could not have consumed more than 20Kw-Hr energy!

That means the Electric Supply Company would have billed you for 20 UNITS

Am I missing something here, or can't you just multiply volts times amps to get watts?

Replying to Comment by Guest: (Use Copy & Paste or drag text to quote the original text.)

Am I missing something here, or can't you just multiply volts times amps to get watts?

Yes you can IF IT IS DIRECT CURRENT........

However if it is Alternating current the formula becomes. Volts times Amperes times the Cosine of the phase angle between the voltage and the current. If the angle is zero the the cosine of zero = one. Motors ( unless specially designed) have a lagging phase angle so the cosine is less than one.

Volts time amps alone does not give the correct answer.

Cosines go from 0 to 1

Hello ALL,

Let me clarify this situation and recap what I think you all are telling me;

1. This unit is self conatained and has a blower fan (turbine type) as opposed to a blade fan (like propellors).

2. It will be placed 20' from the circuit breaker box.

3. It will be no further than 10 ' from a receptacle.

The specs given, are all I have:

Fan Motor 1 Hp / Two Speed / 15Amp / 60Hz
Motor R.P.M. 1725 / 1140 R.P.M.

What size breaker should I be using?

What sized wire should I run?

As for the kw used, i understand from your many posts, this unit will run less than my current 2- 36" circular fans and the four 20" oscillating fans.

Thank you one and all.

netmaker

Breaker Rating = 0>10Amps settable or 15 Amp@5 Seconds

Wires 2.5mm²Copper/Aluminium would do.

Wires 2.5mm²Copper/Aluminium would do.

Could you tell me this in US sizes

No.10, No.12, No.14, etc.

Thank you.

netmaker

12 ga wire, 20 A breaker

Hello again, NetMaker.

All the answers to all of your questions are given in my previous post. Don't worry about PF, just install the damn blower and go to work... Turbine or blades, they are the same. measure it yourself. And if it does the work, what else is important?

What is important, pay attention to the wiring. Most probably there will be a different wiring for each RPM, and sometimes there will be a different running/starting capacitor. (rarely so).

Wangito.

It sounds like you are using a 120 voltage input, multiply the voltage by the amperage and divide the product by 1000 to obtaion the KW input:

120V x 15A = 1800W x 1KW / 1000 W = 1.8 KW