I’m trying to understand motors and the amount of current to turn them. Getting new toys sometimes causes more questions. I recently picked up a low price a/c amperage gauge. It is designed to clamp over one of the power legs of a devise and read the current draw. My well motor by the paperwork shows it to be ½ hp. The small panel at the well has labeling from the manufacturer showing it to be ½ hp. When I measure the amperage I get 6.34 amps. The well runs on 220 volt. From this I calculate about 1.5 to 1.6 hp. I understand there are correction factors I maybe missing, but would they account for more then a hp? Would this indicate that my well motor may have been replaced at sometime or should be?

Thanks,

Eric D

Eric,

Please post you HP formula to this thread so I know exactly how you are arriving at your final amount?

Unfortunately for complex AC loads, ohms law of P=VI doesn't work. For sinusoidal voltages and sinusoidal currents, the formula is P = VI*Cos(theta). The cos(theta) is also known as the power factor and is the difference in phase between voltage and current.

AC Meters to measure voltage and current also get expensive. The voltage measured should be the DC voltage with the same "heating value". e.g. An AC voltage of 120 V should generate the same heat as a DC voltage of 120 V when applied to a resistive element.

Most meters measure the average voltage and multiply it by a constant, such that sine waves read the correct voltage. Other meters are TRMS or True RMS meters.

The nameplate of the motor mayhave a VA rating. Even though V*A is watts, it may give you a more correct value of A.

Quote:

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Originally Posted by donf

Eric,

Please post you HP formula to this thread so I know exactly how you are arriving at your final amount?

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Hi Don,

Here is what I tried to use:


E x I x %Eff x Pf = hp for single phase
746

In my case:

230 x 6.34 x .86 x .91 x 1.15 = 1.76 hp
746

Thanks,

Eric D

Quote:

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Originally Posted by KeepItSimpleStupid

Unfortunately for complex AC loads, ohms law of P=VI doesn't work. For sinusoidal voltages and sinusoidal currents, the formula is P = VI*Cos(theta). The cos(theta) is also known as the power factor and is the difference in phase between voltage and current.

AC Meters to measure voltage and current also get expensive. The voltage measured should be the DC voltage with the same "heating value". e.g. An AC voltage of 120 V should generate the same heat as a DC voltage of 120 V when applied to a resistive element.

Most meters measure the average voltage and multiply it by a constant, such that sine waves read the correct voltage. Other meters are TRMS or True RMS meters.

The nameplate of the motor mayhave a VA rating. Even though V*A is watts, it may give you a more correct value of A.

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Do you know if an a/c current probe can be fooled by the power factor to read higher amperage then what is truly being used? I can't find any data except for the control box that is wired to the submerged well pump. It states 1/2 hp but no amperage or VA number.

Thanks,

Eric D

Do me a favor, check the accuracy of your probe. Use a toaster as the load. Make sure it's warm. Check the wattage and see if you can measure it on the current scale of the DVM. That might be 10 amps or so. Then try to measure it with your clamp on meter.

You can buy a gismo just for measuring current with a clamp meter or you can take a socket and plug and wire it so the conductors are available for the clamp meter. The A202 on this page. Digital Multimeter and Clamp-on Accessories

Remember to put the single cable in the middle of the clamp and parallel to the jaws. SOme clamps need to be zeroed.

Your 1/2 HP motor, you may be able to do the same.

Eric,

With my compliments to Google please visit this link. Great information.

[PDF] The Numbers Game: A Primer on Single-Phase A.C. Electric Motor... File Format: PDF/Adobe Acrobat - View as HTML
Often stated as "develops 5 HP." The peak power of. A motor in this... use the amperage as a rule of thumb. Motor efficiency is still the wildcard,.
users.goldengate.net/~kbrady/motors.pdf

Don,

That is a very good pdf. Thanks for looking it up and posting it. It backs up my thoughts that either my submerged well pump isn't ½ hp or I'm leaking current somewhere. I have tested my current probe to a high dollar unit at work and it matched to within 2%. I'll try some resistive load like you suggested at my home to see if there is something different happen there compared to my work place.

One of the guys at work suggested I check both legs of the power to make sure they are the same. I'll make that test a little later this evening.

Thanks again,

Eric D

Don, nice white paper, reads well, very clear and informative.

Eric, do the amp test on both legs, I suspect your will see a difference, may not be much thou, as most motors, esp ,wet ones, will have ground fault currents. If the current imbalance is more than 10% between legs, then concern is in order.

Let us know what the results are.

Judging by the real amp reading and the chart, you probably do have a half horse motor.

Checking both legs shows they are the same or very close to the same amperage, 6.34. I also found more information about my well. Here are the details:

Franklin Electric
½ hp 230 volts, s.f. 1.6, Motor model #2145059004S, 300 trust, 19#wt., Control box #2801054915

Single phase three-wire submersible motors require the use of control boxes. Operation of motors without control boxes or with incorrect boxes can result in motor failure and voids warranty. Control boxes contain starting capacitors and a starting relay.

Ok, could the capacitors be causing the amperage values to look strange? What does the s.f. value mean?

Thanks,

Eric D

s.f. - service factor. This page is specific for submersible pumps.

Franklin AID (Application Installation Data) Vol. 15 No. 3 May/June 1997

Quote:

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Originally Posted by KeepItSimpleStupid

s.f. - service factor. This page is specific for submersible pumps.

Franklin AID (Application Installation Data) Vol. 15 No. 3 May/June 1997

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Excellent find! Thank You! This helps with a number of questions I have. I wish I could find some amperage numbers like the example they have for the 10 hp pump, but for the 1/2 hp. If I calculate the amperage I get a maximum of 2.59 amps, based on .8 hp max on the service factor, equals 596.8 watts, divided by the 230 volts gives me 2.59 amps. I only have to account for the 3.75 amps over the 2.59! I’m still reading a total on each leg of 6.34 amps.

This submerged pump is a three wire from the manufacturer’s paperwork. Does this mean it’s a three phase motor running on single phase and the box in my basement that feeds it is a 1 to 3 phase converter? Or, is the third wire just used for starting?

By my calculations the current draw fits if the pump is a 1.5 hp. I know my pump sets at about the 60 foot level in my 168 foot well and will flow 25 gpm @35 psi. I ran this test before purchasing my geothermal unit using a 2500 gallon storage tank to calculate the flow rate. The geothermal draw is about 7 gpm and the well is working well to supply it.

Thanks again for your input, it’s been a great help.

Eric D

It is probably your 2 hots and a ground(the 3 wires). If ground is separate, then the 3rd wire could be for 2 speed operation(not likely)

This article should help you understand measurements made with a current clamp. http://www.clsfacilityservices.com/p...eRMSMeters.pdf

You did say the pump had a controller, but we have no idea if the "controller" modifies the AC waveform at all.

Gentlemen,

I humbly accept all koo dos for the in-depth work I did on the previously mentioned.

Unfortunately I can accept none of them. All I did was use the facilities of Google to find a read-able article on horse power calcs. Please extend the praise to the correct folks.

But donf:

You know what terms to use. There is some story where an engineer charges an absurd amount of money and uses an inexpensive part to solve a lant problem. Part $1.00, where to put it $24,999 or something like that.

Search terms can be difficult to come up with at times especially when words have different meanings. Rarely do I have to use ask.com or dogpile.com, but sometime you do.

I got "Where did you find that?" all the time.

KeepItSimpleStupid,

Interesting paper on using true rms. Thanks! From what I read, it seems this normally causes a low read. My problem is a high reading. It does mention switching load causing a square wave and harmonics. Tonight I’ll hookup my oscilloscope and measure voltage and take a look at the waveform off the pump lines. Maybe this will give some clues. I can also check to see if it has any harmonics other then the normal 60Hz. My scope also has a simple FFT function.

Don,

You still helped with finding the info, even if it was from Google!:cool:

Eric D

We never know the experience level of the user we are talking too. I think we assume a DIY'er with minimal tools. For instance tkrussell assumes that DIY's don't own a meter that can read mV AC.

Ok, for those of you still reading this overly long thread I'll add the final touch to close it out. I learned with everyone's help what I was seeking to find. The major question I had and now have the answer, my pump is drawing the current level it should be. Below is the specification for my submerged well pump.

Submersible Pump Motor, Capacitor-Start, Totally Enclosed Nonventilated, 1/2 HP, 3450 RPM, 230 Volts, 1 Phase, Service Factor 1.60, Full Load Amps 5.0/3.5, Max SF Amps 6.0/4.3, Thrust Rating 300 Lb, Number of Wires 3, Thermal Protection Auto, Mounting NEMA, Diameter 4 In, Control Box Required 1LZV8, Amps for CSIR/CSCR Control Boxes 6.0/4.3

Note the S.F. amperage rating, 6.0/4.3. This confused me at first why there were two numbers. Long story short I ended up talking with a Franklin pump engineer. All of these pumps have a matching control box that is installed at the same time as the pump and they are a matched set. In the case of my pump they offer two types. One uses capacitor start, induction run, and the other capacitor start, capacitor run. The induction run control draws 6.0 amps, this is the one I have, and the capacitor run control box draws 4.0 amps, the one I want! I'll be placing an order for the control that uses CSCR to replace my current CSIR box. According to the Franklin engineer there is no loss of pump performance changing to CSCR. Changing to the lower amperage controller will have a payback in less then a year in electricity usage.

Thanks again for everyone's input,

Eric D:)