Current discrepancy between VSD and Switchgear 1

[yk052052]

In our plant, we have 6kV compressor connected to VSD.

And I got some question regarding current discrepancy between VSD and Switchgear.

I found that during compressor operation, the current indicated in Switchgear (21A) is lower than in VSD(28A).

If the current in VSD is lower then I would understand since I would understand this due to current loses in VSD.

But I cannot understand that current in SWGR is lower than in VSD.

I asked VSD vendor and they answered that voltage is variable in VSD so current in VSD can be larger than in SWGR.
And they explained it with Ohm's law. For instance, 2V x 2A (SWGR) = 1V x 4A (VSD) like this.

But still I don't get it clear. By Kirchhoff's law, input current in VSD should be same as output,ideally. But considering loses in VSD, output could be less than input.

Which part am I wrong at?

Please give me explanation. Thanks.

 

[djs]

In most VSDs, the incoming voltage is rectified and then filtered by a capacitor bank and possibly an inductor. Due to this arrangement the VSD acts as a power converter, not a voltage converter. So power in will equal power out plus the efficiency losses in the VSD. You must also take into account the power factor both on the line side and also on the load side.

 

[waross]

You are possibly running at less than full load at less than full speed.
Try comparing kW in to kW out. That that will take into account the effective voltage that the VFD is outputting.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[ScottyUK]

Think of the VSD as a converter of power - it converts line voltage and current to output voltage and current. As djs has noted, the output power is always less than the input power because there are losses in the drive but at a (very) simplistic level we can assume that power in equals power out.

At frequencies below the motor base speed the drive output voltage has to be lower than the motor base voltage (for now it just does, OK ;-) ) and to a first approximation this is a linear relationship so at half the motor base frequency the output voltage will be half the base voltage. The corresponding input current therefore would be half of the output current to maintain the (power in) = (power out) balance.

 

[yk052052]

Thanks you all.

At first, I though of just simple circuit for understand. But considering VSD consists of capacitor bank, inductor, etc. which is not that simple as circuit and is considered as power converter, I realized that I shouldn't think just like (input current) = (output current).
There is still I need to look into about VSD though, I roughly got it.
Appreciate your replies.

 

[itsmoked]

Yes, sooooOOO very much more.

Keith Cress
kcress -

http://www.flaminsystems.com

 

[jraef]

Minor correction. Yes it is a power converter, but your question was why the CURRENT showing on the output of the VFD is higher than the current on the input (switchgear) side. The answer to THAT specific question is within what they said, but more explicitly it's because the VFD is correcting the motor's displacement power factor as read by the switchgear meter, whereas on the output side of the drive, the current being read is inclusive of the reactive current being delivered to the motor. So if you do the math on calculating Power Factor when watts, current and voltage is known (just use full voltage output on the VFD output side for this), you will calculate the input PF as being roughly .95 and the output PF will be something significantly lower, like a .70.


"You measure the size of the accomplishment by the obstacles you had to overcome to reach your goals" -- Booker T. Washington

 

[LionelHutz]

What the VSD manufacturer explained is wrong. It's more complex than just using input voltage X input current = output voltage X output current.

The VSD input only supplies the real power. The motor reactive power flows between the VSD and the motor. If you ran an unloaded motor at line frequency, the VSD input power is quite low since it's only the power required for the motor losses (copper, fan etc) and VSD losses. Yet, the motor current is mostly reactive and still a fairly high ratio of rated current (35% to 50% typically). The end result is that the VSD output current to the motor typically ends up being 10x to 20x the VSD line current. For example, the last motor I ran at 60Hz was drawing 230A from the VSD with 14A of line current. This demonstrates how bad using a simple input VxI = output VxI approximation can be.

As posted by waross, you can only compare input vs output power (kW).

 

[crshears]

So solving for "real power in plus reactive power in minus losses equals real power out plus reactive power out" should account for all variables, right?

CR

"As iron sharpens iron, so one person sharpens another." [Proverbs 27:17, NIV]

 

[LionelHutz]

No, that formula would not be correct.

 

[ScottyUK]

jraef, Lionel

Adding to what you said, if the motor is delivering somewhere near full torque at low rotational speed then the output current will be high to develop torque, but the power throughput will be low and the input current will be correspondingly low too.

 

[waross]

Yes. At full rated torque the current will be rated current. At low speed both the frequency and the voltage will be low. So a 1760 RPM motor running at about 410 RPM will have about 25% effective voltage applied to maintain the V/Hz ratio.
(1800RPM - 1760RPM = 40RPM slip; 1800RPM / 4 = 450RPM 450RPM - 40 RPM slip = 410RPM) There may be some non-linearities with regard to the magnetizing or reactive current complicating this but you get the idea.
Remember the slip frequency and the V/Hz ratio.

Bill
--------------------
"Why not the best?"
Jimmy Carter

 

[LionelHutz]

If you want a simple formula then the input power can be approximated by the speed ratio times the torque ratio.
As examples,
Motor is at 1/2 speed and producing rated torque - input power is 1/2 of rated motor power
Motor is at 1/2 speed and producing 1/2 of rated torque - input power is 1/4 of rated motor power
Motor is at 1/4 speed and producing 3/4 of rated torque - input power is 3/16 of rated motor power
Motor is at 3/4 speed and producing 1/2 of rated torque - input power is 3/8 of rated motor power

If you wanted to get more accurate then you'd have to account for the slip, motor losses and VSD losses, and the losses vary as the speed and load vary.