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Thread: Why does higher voltage results in higher current drawn?

  1. #1
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    Why does higher voltage results in higher current drawn?

    In the factory test, the 460V, 60Hz, 3ph, FLA of 100A pump gives 300 litre/min. The current measured was 100A. When installed on site with a 480V, 60Hz, 3ph power supply, the pump now gives 330 litres/min and the current measured 110A.
    Gievn some voltage drop across the cables which I am sure it is within 3%, shouldn't a slightly higher voltage results in a lower current drawn? Why do I measure a higher current instead?
    Please let me know if you need more information. Thank you.

  2. #2
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    Quote Originally Posted by Crowbar View Post
    In the factory test, the 460V, 60Hz, 3ph, FLA of 100A pump gives 300 litre/min. The current measured was 100A. When installed on site with a 480V, 60Hz, 3ph power supply, the pump now gives 330 litres/min and the current measured 110A.
    Gievn some voltage drop across the cables which I am sure it is within 3%, shouldn't a slightly higher voltage results in a lower current drawn? Why do I measure a higher current instead?
    Please let me know if you need more information. Thank you.
    If after the voltage increase the output remained at 300 litre/min and efficiency and power factor also remained the same then current should drop.

    But you said output went up to 330 litres/min, that is an increase in load of 110%, and increase in voltage of about 104%. You need to look into differences in power factor and efficiency at both load levels to see the rest of the puzzle.

  3. #3
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    Quote Originally Posted by Crowbar View Post
    In the factory test, the 460V, 60Hz, 3ph, FLA of 100A pump gives 300 litre/min. The current measured was 100A. When installed on site with a 480V, 60Hz, 3ph power supply, the pump now gives 330 litres/min and the current measured 110A.
    Gievn some voltage drop across the cables which I am sure it is within 3%, shouldn't a slightly higher voltage results in a lower current drawn? Why do I measure a higher current instead?
    Please let me know if you need more information. Thank you.
    instead of an inductive load of a motor, think of a resistive load.
    if you increase the voltage, the current will also increase.

    in your case, voltage, current, and work done all increase with the voltage increase.
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  4. #4
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    10% is a lot to expect from a simple voltage difference. 1-2% maybe, 3% possibly, 10% unlikely.

    It's far far more likely that a slight difference in static head, slightly different pipe size, fewer turns and bends in the piping providing less fluid friction or some other aspect of the piping resulted in the increased flow, which then in a centrifugal pump results in an increase in current.
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  5. #5
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    The key factor is that when, for whatever reason, the flow rate of a centrifugal pump increases the power delivered by the motor increases and therefore the current drawn by the motor (for a fixed voltage) has to increase too.
    Since the current is above the nominal FLA of the motor, it is a very strong indication that the motor is mismatched to the combination of pump and load and is being used at higher than its nominal maximum mechanical power output. Unless there is a service factor on the motor, that is at least a technical abuse of the motor.

  6. #6
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    Post the pump curve
    Jraef imo hit the nail on the head
    the system curve is different, a bit less head than the test point

    throttle/close the discharge valve down a bit while measuring i and Q



  7. #7
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    Quote Originally Posted by Jraef View Post
    10% is a lot to expect from a simple voltage difference. 1-2% maybe, 3% possibly, 10% unlikely.

    It's far far more likely that a slight difference in static head, slightly different pipe size, fewer turns and bends in the piping providing less fluid friction or some other aspect of the piping resulted in the increased flow, which then in a centrifugal pump results in an increase in current.
    Didn't really think too much about that before, but yes. As long as frequency remained same in both situations, pump speed and would remain the same, and pump load would remain same if piping, head, etc remained the same, so very good chance the media being pumped has different characteristics in some way at the site application then it had in the factory test.

  8. #8
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    Quote Originally Posted by kwired View Post
    pump speed and would remain the same, and pump load would remain same if piping, head, etc remained the same, .
    Not quite due to slip ? Am assuming an induction motor?
    As Ign.. said, need the pump curve. Motor curve would be good also. Or at least type of motor, pump speed would stay the same if sycn motor, but then would not expect power draw to increase.

    Flow went up, assuming piping stayed the same, that means head increased slightly also.
    Power proportional to CUBE of pump speed.
    From pump curve, would be able to see how much slip of motor (hence slightly increased speed) decreased.
    OP said kVA went up 14%, from which one could crank numbers and infer that the increased flow velocity in the piping to increase head, etc.....
    Cube root of 14% is 4.4% increase in speed (approximately), say 3400 to 3550 RPM, which seems possible unless synchronous motor.

  9. #9
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    Quote Originally Posted by junkhound View Post
    Not quite due to slip ? Am assuming an induction motor?
    As Ign.. said, need the pump curve. Motor curve would be good also. Or at least type of motor, pump speed would stay the same if sycn motor, but then would not expect power draw to increase.

    Flow went up, assuming piping stayed the same, that means head increased slightly also.
    Power proportional to CUBE of pump speed.
    From pump curve, would be able to see how much slip of motor (hence slightly increased speed) decreased.
    OP said kVA went up 14%, from which one could crank numbers and infer that the increased flow velocity in the piping to increase head, etc.....
    Cube root of 14% is 4.4% increase in speed (approximately), say 3400 to 3550 RPM, which seems possible unless synchronous motor.
    I guess what I meant was synchronous speed would be the same if frequency was the same, which means motor will be aiming for same speed, it is voltage and load differences that will cause a change in slip. If pump is turning at the same speed in both locations and the pump, piping, and media characteristics are identical, then the driven load will be the same in both locations - then an increase in voltage should see a decrease in current if the same work is being done, since OP saw an increase in current, as well as an increase in flow rate, and presumably same reference frequency which should give same base motor speed, we do not have the same load conditions in both places.

  10. #10
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    Exactly. At 100A that 100HP motor was not fully loaded, even at 110A I doubt that it is. That to me removes the likelihood of this being related to a difference in slip speed. Yes, at a slightly higher voltage the torque will be slightly higher, but the motor is not very likely to be more than 3-5% slip under normal operating conditions, so going from 5% slip to 4% slip will not make that much difference, unless the motor was in high slip the fist time because of being overloaded, and the evidence at hand doesn't show that to be the case.

    This is my favorite graphic for cases like this, showing the effects of voltage variation on an AC induction motor. Notice that the curve labeled as speed doesn't vary enough to warrant this being relevant here.
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    I believe it is load related and the increase in current is the result of an increase in flow, but the increase in flow is not the result in that slight voltage variation. Remember, a motor labeled as 460V is DESIGNED to operate at 480V; 460V +-10%
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