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  • ot well pump voltage

    We have been trying to find an answer can we run a--- (single phase)230 volt submersible pump on (single phase) 208 volt. Every person I ask says you can't run it on 208 3 phase. That is not what I asked them.
    SOL

  • #2
    I asked a similar question once and was told it's all the same.

    ....?

    If it's all the same then why not pick one and be done?

    Anyway, result is the pump works just fine.
    Len

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    • #3
      Short answer is yes, it will run. If the nametag does not specifically state it is a dual voltage motor(208/230 V), chances are it will burn out when you try to actually pump water with it.
      Location: Saskatoon, Saskatchewan, Canada

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      • #4
        Every person I ask says you can't run it on 208 3 phase. That is not what I asked them.
        Actually, that is what you asked. Single phase 208 is what you get from line to line on a three phase service. To run properly the motor must be 200 volt rated. If it isn't it will draw excessive current and overheat.
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        • #5
          Evan is on the mark. Ohm's law states that for a given load (wattage) if the voltage goes down, the amperage goes up, and vise versa. The motor you have (240 volt) would be running with 208 volts at the best case. Factor in the footage that the circuit is, and there will be some voltage drop in addition to using 208V. The fact of the voltage being low would mean the windings would carry more current than it is designed for to perform the work, and would shorten the life of the motor considerably; if not burn it up in short order.

          Your options in this case would be to add a transformer to bring your voltage up to the 240V that you need, find a 208V rated motor, or hook it up and see how long it lasts (with the understanding that it may not live very long).
          Why buy it for $2 when you can make it for $20

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          • #6
            Then again, if you don't ask for max gpm at max lift, it may last a long time.

            Just idling at 208 won't hurt it at all, it is the extra current under power that kills it by heating.

            Cutting current in half reduces heating by 4X, so even a relatively small amount below full power would have significantly less heating and might make up the 10% difference in voltage.

            At 70% power, heating is reduced to half. At 90% power, heating is reduced 20%.
            1601

            Keep eye on ball.
            Hashim Khan

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            • #7
              LOL... When does a water pump idle?

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              • #8
                It idles in a shallow application like a spring as opposed to a deep well. It runs at reduced load when it is designed to run to a specific depth, say 200' and it is only 100' down the well.

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                • #9
                  Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.

                  Example: block the intake of your vacuum. The motor speeds up.
                  Last edited by Evan; 07-28-2006, 11:09 AM.
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                  • #10
                    The power requirements of a pump, whether loaded or unloaded will depend on the design of the pump.

                    Positive displacement pumps aside, rotary pumps can have several different designs, a turbine pump will behave quite differently than a centrifugal. A centrifugal with an open face impeller will behave differently than one with a closed face. Impellers can be designed to be overloading or non overloading, and can be trimmed to use only a certain amount of power. There are many more variations that will influence the performance of a pump.

                    Depending on the application and pump design, the motor may or may not hold up. The best source of information would be to contact the manufacturer or supplier. More information about the performance characteristics of the pump are needed to give good advice.

                    Generally speaking, running a motor that far below the nameplate voltage will result in it's destruction unless load is reduced. If a clamp on ammmeter is available, the load can be evaluated, and if actual amperage is less than the nameplate by a factor approaching or exceeding the difference in voltage, it will possibly survive.
                    Jim H.

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                    • #11
                      In this case we are talking about a submersible well pump. They are multi-stage centrifugal design and produce maximum flow at minimum back pressure. That means they operate at or near maximum current at all times they are running. As back pressure increases current reduces somewhat but not a great deal and any lessening in current is offset by additional heating caused by reduced water flow as the motors are water cooled.
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                      • #12
                        Originally posted by Evan
                        Nope. A pump runs at maximum load when the outlet is at minimum pressure. The lower the outlet pressure the higher the volume of water it moves and the heavier the load.
                        The lower the output pressure the lower the mass of water being moved. F=MA, Newton’s second law, the higher the mass the greater the force required to move it. Force in our case equals load. The greater the load the greater the current required to move that load.

                        Example: block the intake of your vacuum. The motor speeds up.

                        You’re confusing velocity with current load. A motor with no load will turn faster then one with a load. The loaded motor, spinning slower will have greater current required to turn the shaft regardless of how fast it spins. This is torque (newton/ meters). The greater the torque the higher the current or work required to do it.

                        The current load increases with respect to the amount of work it is doing. Not the speed it attains.

                        Joule = 1 watt/second.

                        Velocity does not enter into the equation.

                        The vacuum intake blocked creates a greater load and therefore a higher current draw. It does not speed up, it slows down regardless of the increased volume of the motor.

                        As for the well pump, I agree with JC.

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                        • #13
                          The lower the output pressure the lower the mass of water being moved. F=MA, Newton’s second law, the higher the mass the greater the force required to move it. Force in our case equals load. The greater the load the greater the current required to move that load.
                          I believe you are the person confused. The lower the output pressure the greater the flow and therefore the greater the mass moved.

                          The pressure of the outlet of a pump is determined by the degree of resistance to flow. As the resistance goes up the flow rate goes down as does the amount of mass moved while pressure goes up.

                          You’re confusing velocity with current load. A motor with no load will turn faster then one with a load.
                          Precisely. No confusion there either. The fact that the motor speeds up is an indication of reduced load which proves my point.

                          The vacuum intake blocked creates a greater load and therefore a higher current draw. It does not speed up, it slows down regardless of the increased volume of the motor.
                          No it doesn't. Try it.

                          BTW, that last quote of what you said doesn't really make sense.
                          Last edited by Evan; 07-28-2006, 01:13 PM.
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                          • #14
                            Originally posted by Evan
                            In this case we are talking about a submersible well pump. They are multi-stage centrifugal design and produce maximum flow at minimum back pressure. That means they operate at or near maximum current at all times they are running. As back pressure increases current reduces somewhat but not a great deal and any lessening in current is offset by additional heating caused by reduced water flow as the motors are water cooled.
                            Anything that does work must adhere to laws of physics. The greater the force the greater the work. The less the force the less work being done. Period.

                            A motor that is moving a specific volume of water does more work then a motor that moves less water. Period.

                            The more work a motor does the more current it requires to do that work. Period.

                            We can’t get something for nothing as much as we would like to!

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                            • #15
                              Anything that does work must adhere to laws of physics. The greater the force the greater the work. The less the force the less work being done. Period.

                              A motor that is moving a specific volume of water does more work then a motor that moves less water. Period.

                              The more work a motor does the more current it requires to do that work. Period.

                              We can’t get something for nothing as much as we would like to!
                              So? Pressure isn't work.

                              You don't know how pumps work. If a deep well pump is set deep enough that it is working to it's maximum pressure, say 200 psi, then no flow at all happens. All that happens is that it stirs the water. The only work it does is to heat the water it is stirring. The lower the head the more work it does and the higher the current.
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