Voltage Up, Current down......but not always?

Status
Not open for further replies.
L

larrybartowski

Member
I've always been taught this to be the case, but have recently been told otherwise. Can someone enlighten me as to why a large load bank would drawing less current at 208v than it would at 240v? I'm thinking it has something to do with resistive vs reactive loads, but that's about as much as I know.

Thanks for the help. :?
 
larrybartowski said:
I've always been taught this to be the case, but have recently been told otherwise. Can someone enlighten me as to why a large load bank would drawing less current at 208v than it would at 240v? I'm thinking it has something to do with resistive vs reactive loads, but that's about as much as I know.

Thanks for the help. :?
Click to expand...

Well, assuming you had a purely resistive load and you changed the supply voltage from 240 to 208 the current would drop about 20%. Of course to the power consumed would also be 20% less.
 
larrybartowski said:
I've always been taught this to be the case, but have recently been told otherwise.
Click to expand...

It has to do with what values are changing and why.

Taking some latitude via simplification.
If you take the traditional formula P=IE, if you keep the P value constant, such as when you are 'drawing power' to turn motor, then what you learned is true. As the voltage goes down, the current goes up.

However, you also need to remember the formula V=IxR, in this case if R is constant, like it is with your load bank resistor, then as the voltage goes down, the current goes down also. By the way, in this example the P value also changes.
 
Last edited:
jim dungar said:
It has to do with what values are changing and why.

Taking some latitude via simplification.
If you take the traditional formula P=IE, if you keep the P value constant, such as when you are 'drawing power' to turn motor, then what you learned is true. As the voltage goes down, the current goes up.

However, you also need to remember the formula V=IxR, in this case if R is constant, like it is with your load bank resistor, then as the voltage goes down, the current goes up. By the way, in this example the P value also changes.
Click to expand...
Umm... I believe you got that backwards.

Equivalent P is voltage down and current up.

Equivalent R is voltage down, current down.
 
texie said:
Well, assuming you had a purely resistive load and you changed the supply voltage from 240 to 208 the current would drop about 20%. Of course to the power consumed would also be 20% less.
Click to expand...
Current would drop ~13%. Power would drop ~25%.
 
Smart $ said:
Current would drop ~13%. Power would drop ~25%.
Click to expand...

Of course you are correct. That's what I get for shooting from the hip without doing the math before posting. I guess both Jim and I get lashes this morning.:)
I hope we don't confuse the OP. Now if the OP comes back and says this is all motor load this will get a little messy.:)
 
Last edited:
Voltage Up, Amps Down

Voltage Up, Amps Down

LarryBartowski...

As was mentioned in earlier posts for a predominantly resistive-load (eg, heater) decreasing the source voltage will result in a reduction of load-current, and power decreases!

Conversely a predominantly inductive-load (eg, motor) decreasing the source voltage will result in an increase of load-current if the driven-load is constant, i.e., essentially independent of relatively small changes in motor speed!

If the load is a combination of the two, and load-power is constant, and its power factor is lagging, then the load impedance can be expressed as a constant impedance, thus:

ZL = [VL? x Cos(Theta-L) / PL] / [Cos(Theta-L)+ jSin(Theta-L)], where:

ZL = Load Impedance (Ohms); PL = Load Power (Watts); Theta-L= Load Power-Factor angle (degrees)

Do you you need additional information?

Regards, Phil Corso
 
Last edited:
Phil Corso said:
Conversely a predominantly inductive-load (eg, motor) decreasing the source voltage will result in an increase of load-current if the driven-load is constant, i.e., essentially independent of relatively small changes in motor speed!

Regards, Phil Corso[/FONT][/COLOR][/SIZE]
Click to expand...
I find this potentially misleading wording.
The reason motors are different is not because they are inductive loads. An inductor will exhibit exactly the same proportionality between voltage and current as a resistor or a capacitor. An inductive load such as an unloaded transformer will show a decrease in idling current with applied voltage.

The thing that makes motors different is the fact that the current is determined not solely by winding inductance and resistance but by the back EMF generated by the moving magnetic fields. This in turn is dependent on the slip angle and thus indirectly the power being pulled from the motor.


Also, for a motor, the reason that the current increases with decreasing voltage is exactly that there was a small change in motor speed, reducing the back EMF. The reason that the power delivered is constant is however that the decrease in motor speed is small.
 
Phil Corso said:
Goldigger (one of the more predominant of the PTP's)...

Does the equation represent that of a motor's running-impedance or doesn't it?

Phil
Click to expand...
Yes and no.
I find it problematic that you state that it expresses the motor's constant running impedance when the voltage and load power are held constant and yet it contains both power and voltage as factors. As such, it does not really adequately address the issue we are discussing, namely what happens to a particular motor when you make an incremental change in the voltage.
If you make a small change in the motor speed relative to synchronous speed, you will be changing the power factor angle by a large amount, among other things, so again it is helpful but not necessarily the best way to understand what is happening.

Not all true equations are helpful equations, depending on what problem you are trying to solve.
 
For simple minds like mine, I think the way to look at is that if power is constant, change in current is inversely proportional to change in voltage, but if resistance is constant, change in current is directly proportional to change in voltage.
 
Goldigger...

Haven't you applied the equation in stability studies? It is used determine an induction motor, or group of induction motors, impact on the power source following a voltage-dip, or even an interruption of power for several seconds? It's also used for reacceleration studies of industrial plants?

Reread my post. I never said "constant voltage!' I said constant power! And the change in PF angle is brought about by slip change, hence resulting in a VAr or kVAr change!

Phil
 
Last edited:
Phil Corso said:
Goldigger...

Haven't you applied the equation in stability studies? It is used determine an induction motor, or group of induction motors, impact on the power source following a voltage-dip, or even an interruption of power for several seconds? It's also used for reacceleration studies of industrial plants?

Reread my post. I never said "constant voltage!' I said constant power! And the change in PF angle is brought about by slip change, hence resulting in a VAr or kVAr change!

Phil
Click to expand...
I have not done stability studies, but I agree that it is useful for that purpose.
And if you go far enough away from the intended voltage operating point, the slip rate required for a given output power will be different and since the winding DC resistance is constant, the power factor will also change. I agree that for a first approximation, for small changes in anything but power, the watts and the VA will both tend to change very slowly and therefore can be considered to first order as constant.
To apply that knowledge to a practical situation (short of a stability study where the exact impedance is important), you do not really need any more than that. :)
 
Phil Corso said:
Goldigger...

VA, if you mean "Apparent Power", changes because VAr changes! Furthermore, the winding DC resistance has little to do with "Real Power" used in the equation!

Phil
Click to expand...
It does when you are starting or approaching burning out the motor.....
The majority of the real power is indeed going to the load.
 
Phil Corso said:
GoldDigger...

Another PtP-applied diversionary manuver! My original question was, "Does the equation represent that of a motor's running-impedance or doesn't it?"

Phil
Click to expand...
And I replied yes.
Speaking of diversionary, my two objections to your post were not related to the equation but to other wording in your post surrounding it. So why did you ask me if the equation was correct? :)
 
larrybartowski said:
I've always been taught this to be the case, but have recently been told otherwise. Can someone enlighten me as to why a large load bank would drawing less current at 208v than it would at 240v? I'm thinking it has something to do with resistive vs reactive loads, but that's about as much as I know.

Thanks for the help. :?
Click to expand...

A little simpler form of understanding the above is to just look at ohms law.

A load bank is nothing but a pure form of a resistor, lets say it is a 100 amp bank at 240 volts, using ohms law this would give you 2.4 ohms of resistance, now at 120 volts we divide 120/2.4 and we find it only pulls 50 amps, or if we go up to 480 volts we find it pulls 200 amps.

The above is only if we keep the resistance the same and change the voltage, but what happens if we keep the power the same and change the voltage, if we take the same above example of 100 amps at 240 volts we fine it will use 24kva or 24000 watts of energy, now if we use a 24kva rated lets say heater element at 120 volts we will find that now it is drawing 200 amps this is because the resistance now has to be .6 ohms to keep the same wattage output or power the same, so in simple form its all about what changes and what doesn't.

In most cases when we apply a different voltage to an appliance or load, we make changes to its resistance to keep the power the same by changing taps or re-arranging the elements in series or parallel to allow them to be run on different voltages, we also do the same for motors and transformers but we always keep the output power the same, with the load bank this doesn't always happen if it is a fixed resistance load bank as the resistance stays the same but the voltage changes, so you kind of have to get your head wrapped around ohms law as well as the power formula to understand what happens.

As for what happens when the voltage changes on motors when the voltage changes but the motor is kept wire for the same voltage, it was covered pretty well in the previous post above.
 
A simplified guide as to what happens to the current draw if the voltage is altered is as follows.

For a simple resistance, such as most heating elements, then a 10% increase in voltage will result in 10% more current being used, and wattage being increased by about 21%

For tungsten lamps, a 10% increase in voltage will increase the current by about 5%, the wattage will be about 116% of that used originaly.

For induction motors, a 10% increase in voltage will result in about 10% LESS current, and about the same horsepower, PROVIDED THAT THE NEW VOLTAGE IS STILL WITHIN THE NORMAL OPERATING RANGE for which the motor was designed. If the increased voltage is in excess of that for which the motor was designed, then EXTRA current will be drawn, and the motor liable to overheating.

For switched mode power supplies (including variable frequency drives and electronic lamp ballasts) a 10% increase in voltage will result in about 10% less current being drawn, and the wattage input and wattage output being virtually unchanged.

For thermostaticly controlled heating elements, a 10% increase in voltage will result in 10% more current, as above, whilst the element is "on"
However the heat output is increased by 21%, with the result that the average current goes DOWN ! (10% more current, but for about 21% LESS minutes in each hour, results in an AVERAGE current that is reduced by about 10%)
 
Status
Not open for further replies.
Top