Starting Current wth Soft Starter 8

[jwilson3]

I'm in a disagreement with an Engineer with a major electrical equipment manufacturer over the basics of a thyristor controlled soft starter. I am trying to calculate the voltage drop profile on the supply system during starting.

The Engineer's analysis shows the initial starting voltage at 30% and the motor current at 330A with 20%PF and ramps up to 598A as the voltage is raised. That's the range of current values I say needs to be used in the voltage drop calc. The Engineer says the current to use is the current on the source side of the starter, which would be 30% of the current on the motor side at start and would rise to equal the motor side current as the voltage is ramped up.

I say he's wrong, that the current is virtually the same on either side of the starter.

 

[davidbeach]

Power is virtually the same on either side of the starter. There is no law of conservation of current across any device that can change voltage.

 

[rbulsara]

The engineer and davidbeach are correct.

Think of the reduced voltage starter as a varibale voltage transformer. Input V*I = Output V*I

The source side voltage is not changing only output(motor) side is, so obviously the current will be different on the two sides when voltages are different to deliver a given power.

 

[jraef]

Huh???

There is NO transformer action on a solid state starter! Maybe you were thinking of an Auto-Transformer starter. In a solid state starter, voltage is not transformed, it is reduced by using partial waveforms. So really it can be considered line potential for smaller time slices, hence a lower RMS voltage, which in combination with the motor impendence, reduces the current consumption.

Line current and motor current are the SAME in a solid state soft starter!

JRaef.com
"Engineers like to solve problems. If there are no problems handily available, they will create their own problems." Scott Adams
For the best use of Eng-Tips, please click here -> faq731-376

 

[davidbeach]

jraef, I wasn't trying to imply a transformer action, and maybe only the phase angle of the current changes, but if you have a different rms voltage on each side of the starter, and power is conserved (accounting for starter losses), there has to be some difference in the currents. If that difference is only phase angle, it might be hard to detect with a ammeter, and the magnitude might be the same, but it isn't the same current.

 

[jwilson3]

I agree there might be some slight difference in the line and motor, but that difference is not defined by Input V*I = Output V*I.

It's easy to show how a conversion device like a transformer changes voltage and current on either side of the device and that Input V*I = Output V*I, but a solid state soft starter is not a conversion device, so I'd like Mr davidbeach, or Mr rbulsara to explain to me how a soft starter changes the current as suggested in their comments.

 

[davidbeach]

If output V (phasor) differs from input V, such as being at a lower rms value then output I will have to differ from input I.

Assume for the moment that the starter is lossless. One method that can be used is to turn on the thyristors for a portion of each half cycle. V in is a sinusoid, V out is non-sinusoidal and has a much lower rms value. Power out equals power in, therefore since V in and V out are not equal, it is not possible for I in and I out to be equal. It is entirely possible for I (scalar quantity, magnitude only) in to equal I out.

 

[rbulsara]

I am open to learning.

I would be interested to know if the voltage does not change how does the motor see "reduced" voltage? and if does, then why V.I in will not be V.I out? Less the losses in the electronics?

You possibly can't have 480V in and 144V (30%) out and not have different currents. If so then the solid state starter will have losses equal to a primary resistor or reactor starter. Is that true?

 

[itsmoked]

jwilson3;

The Engineer says the current to use is the current on the source side of the starter, which would be 30% of the current on the motor side at start...

But of course, since right at start the motor is not being sent full voltage. The motor side will have the power consumption defined by that (partial) starting voltage and the current. The same amount of 'power' will be drawn by the starter but at V_source. The result is you will need less current on the start side than the current of the motor side. 30% in this case.

The key is "constant power" as there is no energy "storage" or consumption in the starter. (except losses)

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[cbarn24050]

There seems to be some confusion here. A conventional solid state starter has three thyristor switches, one in each line. The switch has only 2 possible states, closed or open. When its closed the line current is the same as the motor current and when its open the line and motor current are zero.

 

[jraef]

First let me start off by saying that after 15 years of providing technical support for a soft starter manufacturers (3 of them), I no longer surprised by the number of very knowledgeable and intelligent engineers out there who have no experience with them and therefore don't understand how they work. So please do not construe any of this as condescending, I am just trying to illuminate the problems of misconception out there. Before I took that first job, I too was confused. I had to see it (and deconstruct it) to truly understand, and I accept that not everyone has that opportunity.

And for the moment, trust me when I say that if you put a CT on the input side or the output side of an RVSS starter, the current measures almost exactly the same (1.5W or heat per amp per phase is the only power loss).

Stop thinking of power "conversion", be it by rectification, as in a VFD, or by induction via transformation or by restriction via resistance/impedance. An RVSS is "restricting", but not with resistance/impedance. It plays with time, not magnetism or chemical interaction, to NOT ALLOW full power through the devices. To understand what I mean, go back to basics for a moment. What is a Volt? The potential that makes 1 Amp flow through a conductor and dissipate 1 watt. What is an Amp? 1 coulomb of charge per second through a conductor. What an RVSS is doing is altering the time domain of that basic equation. By only allowing power flow for a restricted time, the current, and by definition the voltage, is lowered. So even though you can measure the RMS voltage on the load side as being lower than the line side, it is different by virtue of a reduced net amount of current flowing through the circuit. But from an Amperage standpoint, the current is the current, regardless of which side of the SCR you measure it on.

JRaef.com
"Engineers like to solve problems. If there are no problems handily available, they will create their own problems." Scott Adams
For the best use of Eng-Tips, please click here -> faq731-376

 

[jraef]

And since I didn't expressly say it jwilson3, the "engineer" in your original post is wrong. You were right to question him on that minor point, not that it ultimately matters in the application you were considering..

JRaef.com
"Engineers like to solve problems. If there are no problems handily available, they will create their own problems." Scott Adams
For the best use of Eng-Tips, please click here -> faq731-376

 

[aolalde]

I think we are missing the fact that the waves are no longer 100% sinusoidal.
The current "in" must be the same as the current "out". . However the incoming current is zero for an extended time until the SCRS fire on. On the output the current is the same but the voltage is chopped (not reduced as for a transformer). The product “VxI” is the same on both sides.
See figure below.

 

[jwilson3]

Thanks to JRaef, for a fundamental explanation. I was beginning to doubt my own understanding of electrical theory,since so many knowledgeable people were disagreeing with me.

The hardest thing to overcome is the concept that full voltage on the line side times the current passing thru to the motor side doesn't equal the motor side voltage times the current. If you consider that the voltage on the line side is only "driving" current during the part of the cycle that the thyristors are firing, and you disregard the voltage of the non-conducting part of the cycle, then the applicable line side and the motor side RMS voltages are the same. Then the fundamental equations that we grew up on remain valid. Everyone can then be happy.

 

[rbulsara]

aoldale/Jraef:

Good picture. OK. I buy that. Learn a thing everyday.

 

[cbarn24050]

Well it would be a good picture if the current waveform was correct.

 

[aolalde]

Thank you for the advice cbarn24050. If you have a picture from an oscilloscope I will appreciate it. I kind of guess-deducted the current feeding an inductive load, just for a quick illustration.

 

[cbarn24050]

Oh you drew it yourself, I just assumed you got it from somewhere. It shouldn't be too hard to find it.

 

[itsmoked]

http://www.saftronics.com/LITERATURE/TN_VFD_GEN_016-G.pdf

Keith Cress
Flamin Systems, Inc.-

http://www.flaminsystems.com

 

[aolalde]

Hi Itsmoked. I think we are talking of a "soft starter" not a "VFD".