Voltage drop and current drawn

[Surefire]

Gentlemen

I have a question for your perusal. A 3kW 230volt submersible pump is connected to a long extension cable (60 metres). Approximate current drawn is I=3000watts/230volts = 13amps (ignoring PF and efficiency). Cable is rated at 0.03 volts/amp/metre therefore the volt drop is 0.03x13x60 = 23volts.

If the current drawn is recalculated subtracting the voltage dropped we have: I=3000/207 =14.5A

So if the pump continued to do 3KW’s worth of work it would now require 14.5amps due to the voltage dropped by the cable.

My question is this: would this happen in reality or would the added resistance of the cable just reduce the current to the pump?

Thank you for your attention.

 

[telectrix]

your pump would theoretically have a reduced power 207 x 13 = 2.7kW, if it still pulled 13A, but the most likely effect would be that the pump motor would be stalled and burn out, due to the undervoltage.

 

[Marvo]

It's unusual to get a 3kW pump in single phase. The kilowatt rating of the motor refers to the mechanical power output at the shaft. You can't work out the run current from the kW rating unless you know all the variables such as PF, efficiency etc and I wouldn't be surprised if a 3Kw pump would have an FLA of around 20A and not the 13A you calculated. I deal with 2.2 kW single phase pumps with a higher FLA than 13A but you really need to refer to the manufacturers data.

Voltdrop or low voltage at the pump won't necessarily cause higher run current although it may do. Again there's other variables at play, the efficiency of the motor plummets with low voltage which means less shaft power is produced (water volume at a given pressure will reduce) but more heat is produced due to greater internal losses.

Bottom line is that manufacturers data is the key. They will give you the run current and they will give you the minimum on-line voltage. You need to run the pump under the worst conditions it's likely to encounter and make the on-load tests to see if everything is within the specified limits.

 

[Knobhead]

OP, you're like a stuck record. Constantly repeating yourself.

 

[dingledong]

He's only said it once ! !

 

[Knobhead]

I've just opened you post.

Look at his past posts!

Your back on the ignore list.

 

[Lucien Nunes]

You need to read up on how motors work*. The full load current is exactly that, the current at full load. If you put in series resistance, the motor won't be able to achieve full load as the voltage will fall, so the full load current rating is now invalid.

To predict the behaviour of a loaded induction motor, you need to model it as an induction motor connected to a load with an appropriate curve, not a constant power load as you attempted.
With an induction motor decreasing the terminal volts will increase the load angle as the magnetising current reduces. This will affect both the real and reactive power and the power factor will vary. Therefore, your simplistic analysis ignoring the power factor will also fail to give the correct results.

*PS not you Tony LOL!