Power Conversion

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Zyb

Zyb

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Maine
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Design Engineer
Hi,

I'm trying to estimate the starting power of sump pump.
Motor: 1/2 HP
Voltage: 115V
Amps: 4.7 - 9.4 A

If I convert HP to W then it is equal to 368W
If I multiply V x A, it is equal to (4.7 x 115) = 541A or 9.4A x 115V = 1,081W

If I want to estimate the starting power which is more accurate using HP or VxA

assuming 60% efficient and starting power is x 3
(368÷0.60) x 3 = 1,840W

(1,081÷0.60) x 3 = 5,405W

Thank you
 
Dont forget PF

do an educational experiment for yourself.
Scope and motor with load.
Measure voltage and current on 2 different channels, Trigger scope on single sweep.
Now you have all the data you need.
Super easy to plot all values if your scope has a math function.
 
Do you want starting power or starting current?

As a rough approximation, use running current * 6 for starting current.

The 4.7-9.4A value looks suspiciously like values for 2 different operating voltages, not an operating current range.

-Jon
 
Zyb said:
Hi,

I'm trying to estimate the starting power of sump pump.
Motor: 1/2 HP
Voltage: 115V
Amps: 4.7 - 9.4 A

If I convert HP to W then it is equal to 368W
If I multiply V x A, it is equal to (4.7 x 115) = 541A or 9.4A x 115V = 1,081W

If I want to estimate the starting power which is more accurate using HP or VxA

assuming 60% efficient and starting power is x 3
(368÷0.60) x 3 = 1,840W

(1,081÷0.60) x 3 = 5,405W

Thank you
Click to expand...
Typically the fewer conversions the better.
 
Keep in mind centrifugal pumps are typically low inertia loads. They really don't mechanically load the motor much until they get up to speed. So your starting currents will be relatively low and of a very brief duration.
 
Zyb said:
Hi,

I'm trying to estimate the starting power of sump pump.
Motor: 1/2 HP
Voltage: 115V
Amps: 4.7 - 9.4 A

If I convert HP to W then it is equal to 368W
If I multiply V x A, it is equal to (4.7 x 115) = 541A or 9.4A x 115V = 1,081W

If I want to estimate the starting power which is more accurate using HP or VxA

assuming 60% efficient and starting power is x 3
(368÷0.60) x 3 = 1,840W

(1,081÷0.60) x 3 = 5,405W

Thank you
Click to expand...
Please search for the locked-rotor kVA CODE letter from the motor specs. Refer to the locked-rotor kVA per HP corresponding to that letter code. Then you calculate the starting current of your motor.
Say you read Code = J. The kVA per HP of a code J is 7.1 to 7.99 kVA per HP. Let's choose 7.5 (mid-value). The starting kVA will then be =1/2HP X 7.5 kVA/HP = 3.75 kVA, with the starting current = 3,750/115= 32.6A, The starting current will be smaller if the code letter is B (3.15 to 3.54 kVA/HP); starting current = 1/2 X 3.3 (1000)/115= 14.3 A!
 
As was said, 6x nameplate amps is a general rule of thumb for locked rotor current allowance. As was also said, measuring actual draw is where the rubber really meets the road. If you want to engineer it down to a gnat's left wing, you will need empirical data & that means taking measurements.

That being said, centrifugal pumps are a funny thing. The torque load varies with RPM quite a bit. That throws off the usual 6x rule. Viscosity of the medium being pumped can have a noteworthy effect. Loose mud is harder to pump compared to clean water & therefore draws more current. Head pressure also matters. If you are pumping the water very high above the pump base, you will draw more current than you would for a shorter rise. All this adds up to my answer being (as it often is) - it depends.

That being said, every 1/2 hp sump pump I have ever seen has run without issues on a standard 15 amp 120vac circuit, unless a lot of that capacity was already being drawn down by other equipment.

If you need to minimize circuit loading & inrush current, you can use a VFD to ramp up the pump slowly and/or run at reduced speed, but single phase output VFDs are not the most common. You need to do a little digging to find them. They do exist, but they probably represent less than 5% of the market from what I have seen & a lot of suppliers don't know that they exist.

What's the actual application? Do you need to run a pump off of a circuit that is already heavily loaded?
 
Every pump varies a bit, but the basic rule of thumb for a centrifugal pump is power required increases by the square of the speed.

So if this pump requires 1/2 HP at 3600 RPM, it will require 1/8 HP at 1800 RPM, 1/32 hp at 900 RPM, 1/128 HP at 450 RPM,,,,,,,
 
Joethemechanic said:
Every pump varies a bit, but the basic rule of thumb for a centrifugal pump is power required increases by the square of the speed.

So if this pump requires 1/2 HP at 3600 RPM, it will require 1/8 HP at 1800 RPM, 1/32 hp at 900 RPM, 1/128 HP at 450 RPM,,,,,,,
Click to expand...

I believe the affinity law for pumps states ideally the power input varies as the CUBE of rpm... so the 1/2 HP @ 3600 would require 1/16 HP @ 1800 rpm etc. The head or pressure follows the square rule.
 
dkarst said:
I believe the affinity law for pumps states ideally the power input varies as the CUBE of rpm... so the 1/2 HP @ 3600 would require 1/16 HP @ 1800 rpm etc. The head or pressure follows the square rule.
Click to expand...
Yeah, you're right. I forgot to account for the reduction in volume that is being pumped
 
winnie said:
Do you want starting power or starting current?

As a rough approximation, use running current * 6 for starting current.

The 4.7-9.4A value looks suspiciously like values for 2 different operating voltages, not an operating current range.

-Jon
Click to expand...
Yes, and is probably a 115/230 volt motor. The 4.7 figure would be for 230 volts, which is in the range of motor current tables for a motor this size.
 
ggunn said:
If the voltage is constant, what's the difference?
Click to expand...
System capacity. If the application is for running this pump from let’s say a battery backup system through an inverter, the inverter will need to be able to provide all of the starting current, ie the kVA/HP as topgone said earlier.
 
Jraef said:
System capacity. If the application is for running this pump from let’s say a battery backup system through an inverter, the inverter will need to be able to provide all of the starting current, ie the kVA/HP as topgone said earlier.
Click to expand...
What I meant to ask was that if the voltage is constant the % change in current and the % change in power are the same; what's the difference which number you use?
 
ggunn said:
What I meant to ask was that if the voltage is constant the % change in current and the % change in power are the same; what's the difference which number you use?
Click to expand...
If the voltage really is constant, it would make no difference which % change you'd use. But, more current will mean more voltage drop, so holding voltage constant is only an approximation to reality.

From what Dr Campbell said earlier, it sounds like % change in current is the more important number for this application.
 
ggunn said:
What I meant to ask was that if the voltage is constant the % change in current and the % change in power are the same; what's the difference which number you use?
Click to expand...
The voltage may well not be constant when you power from and inverter or generator.
So, particularly for a generator, delivering a surge beyond rated current is much easier than delivering a surge beyond rated power (limited by the prime mover.)
 
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