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The house we bought is on well water, but no history on the well pump was transferred to us in the sale...such as horsepower or other specs. I only know it's 100 feet down to my static water level because I tied a fishing weight on fishing line and put it down the well casing until I heard it hit water.

On line I see that average residential well pumps are about 1- 1.5 horsepower.
With an amp meter I'm getting 12 running amps after an initial 25 amp surge. At 240v that equals 2,880 watts (or about 3.85 horsepower -at 746w per hp). That seems a little high. Does anyone know if 12 amps is about normal for these pumps? I'm wondering if I have a tired pump or acceptable numbers. thanks
 

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The largest well pumps I have seen are 2 hp in my area however given that your water table is 100 feet down I would have to assume that the pump is 150-300 feet down and according to what I've found online a 1.5 hp pump will pump to 275 feet ( this is for a utilitech pump on lowes site) and draw 11.5 amps if you have a 1.5 horse pump I'd say you're good for reference a 1 hp pump will draw about 9.8 amps I'm sure the numbers are different between manufacturers but hope this helped you may also be able to find info on the well at your county records office most counties keep the info on file at least in my neck of the woods
 

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Discussion Starter #4 (Edited)
thanks for the reply. I guess it's okay amp range. I'll have to check those records you mentioned and see what they have on file. strange why the math for back calculating the horsepower doesn't come out quite right. given there's 746 watts per horsepower..a pump drawing 12 amps should be: 12amps x 240v = 2880watts / 746w = 3.85hp. I guess these pumps are operating at higher horsepower than their rating.. unless the 1.5hp is just a static rating,, as in pumping no water
 

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More than likely the rating depends on the pump manufacturer for instance a more expensive Wayne pump could be closer to two hp plus has a better warranty and longer life span needed a utilitech pump from the big box stores is rated closer to its actual power but has a shorter warranty and lifespan comparing the specs for same hp pimps from both manufacturers there is a big difference in the lift capability between the two
 

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thanks for the reply. I guess it's okay amp range. I'll have to check those records you mentioned and see what they have on file. strange why the math for back calculating the horsepower doesn't come out quite right. given there's 746 watts per horsepower..a pump drawing 12 amps should be: 12amps x 240v = 2880watts / 746w = 3.85hp. I guess these pumps are operating at higher horsepower than their rating.. unless the 1.5hp is just a static rating,, as in pumping no water
Your math is wrong do to not knowing the power factor the motor is operating at. Also you are not taking your reading at the motor so the wiring heat loss is a factor as well. The third an most important item is what is the discharge pressure an flow GPM at the time of amperage an voltage measurement?
 

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Hi,

Even if it a is an old thread, it in important to note the Kre's comment: The power factor is a very important value to know the real power taken by a load with windings inside. These windings have a characteristic named inductance that modifies the simple formula used by Kanoa50. To correct this formula we need to use a factor named Power Factor.

When we use a DC supply, in a passive circuit (no transistors or electronics), the current is stable in a very short time. With no voltage variations, no current variations.

When an alternating current is used, there is a continuos variation in the voltage applied (60 times per second) and of course in the current (Amps) circulating. In the case of a pure resistive load, the voltage variations are followed by proportional current variations at exactly the same time. Voltage and current are in phase.

When the load contains coils or windings (as in motors or transformers) and are fed with AC, these coils react sending a voltage in the opposite direction, reducing the effective voltage applied. This reaction takes a time so the current increases a little after the voltage raises. It is the current lag with respect to the voltage and this value in a trigonometric representation forms an angle whose cosine is named cos φ (cos Phi)or power factor.
This PF varies in a motor or transformer according to the load applied. The cosine function varies from 0 to 1, but in the common electrical loads it is normal to found values between 0.4 to 0.95 (+or-) and is the factor that reduces the Volts x Amps product. At less load, less power factor.

Volts x Amps = Apparent power in VA or KVA
Volts x Amps x cos φ = real power Watts or KW.

In this case, assuming a PF = 0.8

Apparent power: 240 V x 12 Amp = 2880 VA or 2.8 KVA
Real power: 240 V x 12 Amp x 0.8 = 2304 Watts or 2.304 KW

But...This is the real power absorbed by the motor, but this motor is not perfect, it dissipates heat, electromagnetic waves, mechanical vibrations, noise etc etc. All this energy waste comes from our incoming 2304 Watts, so the mechanical output in the form of a rotational force applied to the pump's shaft can't be the same as the input power. The relationship between the output power and the input power is the efficiency and it will never reach the unit (100%).

In this case assuming an 80% efficiency :

Input power x Eff. = Power in the shaft
2304 Watts x 0.8 = 1843 Watts
1843 Watts / 746 W/Hp= 2.47 HP

This is only an approximation, assuming PF and Eff. This values change ,as Kre explains, with the flow of water (GPM) and the pressure (PSI). Every pump has its own set of curves that determine most of the variables found on it.

Regards
 
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