Motor current

grousedogtom

Hi, can someone explain to me why a submersible motor at 220 volts would draw 4.2amps and at 230 volts 4.3 amps, should the current drawn be lower with the higher voltage, please see attached picture, thanks

grousedogtom

Here's picture

tomdempsey200

no i would expect it to rise

TheBody

grousedogtom wrote: »

Hi, can someone explain to me why a submersible motor at 220 volts would draw 4.2amps and at 230 volts 4.3 amps, should the current drawn be lower with the higher voltage, please see attached picture, thanks

Voltage=current[latex]\times[/latex]Resistance

So if the resistance stays the same then if the voltage goes up, the current must increase too.

grousedogtom

Power =voltage X current, so if my power was 550 watts and voltage 230v =2.39 amps , same power at 550 watts and voltage 250 = current 2.2 amps, does this make sense?

TheBody

grousedogtom wrote: »

Power =voltage X current, so if my power was 550 watts and voltage 230v =2.39 amps , same power at 550 watts and voltage 250 = current 2.2 amps, does this make sense?

Yes, that's it.

grousedogtom

grousedogtom wrote: »

Power =voltage X current, so if my power was 550 watts and voltage 230v =2.39 amps , same power at 550 watts and voltage 250 = current 2.2 amps, does this make sense?

The above shows higher voltage lower current and vice versa.?

TheBody

grousedogtom wrote: »

The above shows higher voltage lower current and vice versa.?

Correct. Provided the power stays the same.

coylemj

But the power consumption will go up with the voltage because as pointed out, the resistance will be the same. If the device contains sensitive components then it will usually incorporate some level of intelligence to moderate the current based on the voltage in order to keep the power consumption constant, otherwise the current will rise in direct proportion to the voltage.

If you have a dimmer switch for a filament bulb then as you adjust the dimmer, the voltage at the bulb is raised and lowered so the light gets brighter (more power) or lower (less power). Something as simple as a motor will behave the same, it will consume more power as you raise the voltage because the current is going up with the higher voltage.

Bruthal

grousedogtom wrote: »

The above shows higher voltage lower current and vice versa.?

TheBody wrote: »

Correct. Provided the power stays the same.

But power does not stay the same unless the item has a switching regulator of some sort, which would not be the norm for most items.

A 2kw 230v item takes 8.7 amps. If we forget about power factor etc, the 2kw item will have an impedance of 26 ohms.

It will only be a 2kw item when the voltage is 230v.

People assume voltage x current = kw, so increasing the voltage must mean current reduces. It doesnt. If you double the voltage to a 2kw immersion element, it will become an 8kw element.

In the case of motors, they can actually show a current reduction when voltage is increased, such as when the motor has a mechanical load, and is being supplied with too low a voltage to get up to proper running speed. But thats a different topic than the one here.

Bruthal

grousedogtom wrote: »

Power =voltage X current, so if my power was 550 watts and voltage 230v =2.39 amps , same power at 550 watts and voltage 250 = current 2.2 amps, does this make sense?

The 550 wat 230v item will only be 550 watts when powered by 230v.


As a crude comparison. A car crash test.

Speed x weight = impact force (in a general way)

If you use the same methodology as for the electrical load, then for a given impact force, the faster the car is going when it crashed, the lower the weight must be.

But that does not mean that the faster a given car is crashed, that the lower its weight will become, which is simply because its weight is fixed. It just means a different lighter car will have a higher impact speed to produce the same impact force. Just like a 230v 2kw load will not be a 2kw load at 330v, which is simply because its resistance is fixed.

grousedogtom

Thanks for all the previous replies, a. 55kw motor at 230 volts should draw 2.39 amps yet the motor plates states @ 230 volt current equal 4.3 amps, is this the ampage when motor is fully loaded?

Bruthal

grousedogtom wrote: »

Thanks for all the previous replies, a. 55kw motor at 230 volts should draw 2.39 amps yet the motor plates states @ 230 volt current equal 4.3 amps, is this the ampage when motor is fully loaded?

0.55 kw I assume you mean.

Yea 550 watts should take 2.39 amps. However, with motors there is power factor. Due to the characteristics of motor windings, only some of the current drawn Is doing any work. In this case the 2.39 amps is doing the work, the rest is gone in the power factor, which is another technical topic.

Also the rating plate is probably the motor output. So actual amps drawn will be higher than output power due to motor inefficiency.

Steve

Bruthal wrote: »

However, with motors there is power factor.

Yeah, that's it - look at the pic posted, PF=0.98@220V, 0.94@230V. The increase in voltage is just creating more wattless current.

To explain to those who don't understand AC, normally I=P/V, this only works for DC or for resistive AC loads. If there is inductance or capacitance involved then the out of phase current doesn't count as far as real power is concerned.

ted1

grousedogtom wrote: »

Hi, can someone explain to me why a submersible motor at 220 volts would draw 4.2amps and at 230 volts 4.3 amps, should the current drawn be lower with the higher voltage, please see attached picture, thanks

V=IR
As R remains constant constant and V increases then I must also I increase

Cerco

ted1 wrote: »

V=IR
As R remains constant constant and V increases then I must also I increase

Your formula is correct but your assumption is wrong in this case. As pointed out above by Steve and others your are dealing with impedance when a motor is analysed. It is therefore necessary to include the reactance of the coil (windings). Hence the power factor comes into play... see post from Bruthal.

TheChizler

You possibly thinking of transformers OP? A more generalised formula would be V = IZ, Z being impedance. If you hold the frequency (& dynamics) constant, the current will always rise proportionally to the voltage.

brightspark

ted1 wrote: »

V=IR
As R remains constant constant and V increases then I must also I increase

One thing you can be certain of is that current flow in a motor cannot be determined by V=IR

Factors to be considered are

Power factor, Motor slip, Speed, Load, etc.

A motor has impedance, not just simple resistance, if you measure the resistance at standstill you will find it is much lower than V/I, as the motor increases in speed it creates a back EMF that limits the current flow.

Steve

In the case of motors (or any non resistive load) P=I*V*PF. That's for consumed electrical power, not output power. For output power you also have to know the motor efficiency.

grousedogtom

2.39 amps versus 4.3 amps, would it be normal for current to almost double considering power factor and efficiency?

TheChizler

grousedogtom wrote: »

2.39 amps versus 4.3 amps, would it be normal for current to almost double considering power factor and efficiency?

Can you explain where you're getting these numbers OP?

grousedogtom

TheChizler wrote: »

Can you explain where you're getting these numbers OP?

550 watts divide by 230 volts =2.39 amps and 4.3 amps as motor nameplate as previous picture

TheChizler

grousedogtom wrote: »

550 watts divide by 230 volts =2.39 amps and 4.3 amps as motor nameplate as previous picture

Oh yeah, sorry, no sleep last night...

brightspark

The 550 watts is the output power of the motor, not the input power.

Bruthal

brightspark wrote: »

One thing you can be certain of is that current flow in a motor cannot be determined by V=IR

Factors to be considered are

Power factor, Motor slip,

If it was a universal motor, there would be no slip as such. The equivalent to induction motor slip is generated back emf, which reduces as the universal motor is slowed by loading, the same way slip increases by mechanical loading on induction motors. The effect in both cases is higher current drawn. The generated back emf can't reach supply voltage just like the slip can't be zero in induction motors. If they did, there would be no driving force.

Stoner

I always think of the electrical slip from the delta frequency point of view between the two fields, speed and the chasing field, never catching because if it did the whole principle of sweeping the current carrying conductor with a magnetic filed is gone out the window.
I know its just a ratio but I remember the first time a mechanical engineer held the shaft of an induction motor and moved it backwards and forwards and found the "play" between the shaft engaging and showing that he had to double the distance from going from standstill to forward to forward to reverse, then he called it the slip of the motor due to hysteresis loss, since then i always check if its elec or mec hysteresis loss that they are talking about, not that I've ever had to do much checking.