Sizing well pump inverter (first post)

[jtbartlett]

1. Normal power is just 8.2 x 230. You are adding in an extra factor of two.

2. The starting amperage (except maybe for the first fraction of a cycle) will be less than or equal to the locked rotor amperage. Fortunately not all of that current will be in phase with the source voltage so the power required, in watts, will be somewhat lower than the product of volts times amps.

3. But the inverter or generator which has to start the motor will need to be able to supply at least half of that current to get the motor going.
With a generator you have the inertia of the rotating system to help handle the surge even though the windings of the generator may not be able to carry that load for more than a minute or two.
With an inverter, the batteries have to be able to supply the full surge current and the output circuit of the inverter has to be able to handle it for at least a few seconds without overheating.

So, my operating voltage is 8.2*230= 1886 and maximum (s.f.load) is 9.8*230= 2254. That sounds better. However, starting amperage will be >.5*(41.8*230)=4807. There will probably be occasional, concurrent house loads. SO... I am probably looking at a 6kw (+) inverter (?). This seems larger than I had originally thought but this is the one load that I cannot control - pump is down the hole at 480' and not coming out again until it dies. This is 48v, offgrid application.
I have been trying -unsuccessfully- to find a way to run this load with smaller equipment. Cistern pumping/ gravity feed is not possible due to topography and climate.

[inetdog]

My load is a 1hp, 230v 3 wire, deep well pump. (goulds 7gs10 to be exact). This will be my largest load so I am trying to calculate the size of the power inverter to match. From the spec sheet the following: Rated input, each leg 8.2 amps, 1200 watts. Maximum input, each leg 9.8 amps, 1600 watts. Locked rotor amps is 41.8. Am I correct in calculating the max running watts as (9.8*2)A*230V = 4508? I understand that there is a surge, or starting load, but I'm not sure how to find the values. Locked rotor amperage*voltage is 9614 which seems unlikely. Any thoughts on where my calculations go wrong? Thanks.

1. Normal power is just 8.2 x 230. You are adding in an extra factor of two.

2. The starting amperage (except maybe for the first fraction of a cycle) will be less than or equal to the locked rotor amperage. Fortunately not all of that current will be in phase with the source voltage so the power required, in watts, will be somewhat lower than the product of volts times amps.

3. But the inverter or generator which has to start the motor will need to be able to supply at least half of that current to get the motor going.
With a generator you have the inertia of the rotating system to help handle the surge even though the windings of the generator may not be able to carry that load for more than a minute or two.
With an inverter, the batteries have to be able to supply the full surge current and the output circuit of the inverter has to be able to handle it for at least a few seconds without overheating.

[jtbartlett]

My load is a 1hp, 230v 3 wire, deep well pump. (goulds 7gs10 to be exact). This will be my largest load so I am trying to calculate the size of the power inverter to match. From the spec sheet the following: Rated input, each leg 8.2 amps, 1200 watts. Maximum input, each leg 9.8 amps, 1600 watts. Locked rotor amps is 41.8. Am I correct in calculating the max running watts as (9.8*2)A*230V = 4508? I understand that there is a surge, or starting load, but I'm not sure how to find the values. Locked rotor amperage*voltage is 9614 which seems unlikely. Any thoughts on where my calculations go wrong? Thanks.