ASK THE EXPERTS: Running a Well Pump Off-Grid

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Do I need to install a bigger inverter...or a smaller pump?

Hello! I’m attempting to run a 240 V well pump on a Xantrex SW4048 inverter via a 120/240 V, 4 KVA step-up transformer. The pump will start and run if no other load is present, but with even a few lights on, the inverter overloads and trips out. I called the pump motor manufacturer and they confirmed that the two-wire, 1 hp pump pulls 48 amps at 240 V for about 0.2 seconds during startup. However, since the inverter is supplying 120 V, the actual current the inverter has to supply is 96 amps! The surge ratings for the Xantrex SW4048 from the literature I have says 110 amps for 0.01 second, 77 amps for 0.1 second, and 73 amps (resistive load) for 5 seconds.

I also reviewed the manual for our newly installed submersible well pump (Goulds), and they have a section on sizing a generator to run the pump. For a 1 hp, three-wire pump, they recommend a minimum of 4,000 watts. For a two-wire pump, they recommend increasing the generator size by 50 percent.

I’m not sure why the 4,000 watt inverter isn’t having problems starting the pump without any other loads. The calculated pump motor inrush current of 96 amps for 0.2 seconds surely seems like it would exceed the short time ratings of the inverter. Maybe the motor manufacturer was being conservative with the 0.2 second duration for the inrush current. I think even if it works, it’s on the hairy edge of overload. Given the pump that’s been installed, I think I need a bigger inverter.

The 1 hp pump seems rather large for a residential well, especially a cabin in the woods. It’s more common to find a 3/4 hp unit. Is a pump that big really necessary, or was the well guy just covering himself? The well is 146 feet deep. My conclusion for fixing the problem is to install a bigger inverter or a smaller pump. Am I missing something? Help! Best Regards,

Kirk Wishowski • via email

 

Hello Kirk, You have correctly diagnosed the problem. This problem is compounded by the fact that the pump is not a resistive load, it is inductive. The power factor of the pump is probably in the range of 0.75 to 0.80 and this means that the inverter must produce roughly 25 percent more current than the pump’s rating.

This is a common problem. Well pump installers are not used to powering pumps with inverters, and routinely install huge pumps, since they assume that the grid is present. The best solution is to replace the pump. Here’s an example from our own system here on Agate Flat. Our well is 300 feet deep and the pump is located 260 feet down the well. We use a 1/3 hp Franklin motor, running on 120 VAC, driving a standard Goulds rotary pump. This unit can easily be started and powered by a 1 KW sine wave inverter. This pumping setup will deliver 5 gallons per minute to our water storage tanks.

Consider having the pump hauled out of the well and replacing its motor with a 1/2 to 1 hp motor running on 120 VAC. If the wire run to the well is very long, consider powering the pump on 240 VAC since you already have the step-up transformer. But running it on 120 VAC will be more efficient since you will not have the transformer’s loss.

Richard Perez • Home Power

Comments (1)

David Butler's picture

Kirk, not sure if this blog has automatic reply notification, but I came across this and wanted to correct something that both you and Richard said (or implied).

Neither the well depth nor the depth of the pump have anything to do with the required pumping power. The determining factor is the water table (and desired flow rate). 1/3 HP is typically fine for wells with a shallow water table, for example, 30 or 40 feet. If you really do need a 1 HP pump, then your water table must be pretty deep.

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