Understanding the nameplate
Understanding the nameplate
(OP)
Hi there,
I have tried to figure this one out buy searching, but without any results.
My case: On the nameplate for my motor it says:
Star: 3x400V / 2.2kW / 6.0A
Delta: 3x230V / 2.2kW / 10.4A
I am not sure about the efficiency but why is the motor power 2.2kW?!?
400 x 6 x 1.73 = 4.15kVA
230 x 10.4 x 1.73 = 4.14kVA
How do I know what the power factor is? Let us assume 0.85:
4.15kVA x 0.85 = 3.53kW
If this is the truth, then the efficiency is:
2.2 / 3.53 = 0.62
What am I missing?
I have tried to figure this one out buy searching, but without any results.
My case: On the nameplate for my motor it says:
Star: 3x400V / 2.2kW / 6.0A
Delta: 3x230V / 2.2kW / 10.4A
I am not sure about the efficiency but why is the motor power 2.2kW?!?
400 x 6 x 1.73 = 4.15kVA
230 x 10.4 x 1.73 = 4.14kVA
How do I know what the power factor is? Let us assume 0.85:
4.15kVA x 0.85 = 3.53kW
If this is the truth, then the efficiency is:
2.2 / 3.53 = 0.62
What am I missing?





RE: Understanding the nameplate
Power = E x I x 1.73 x P.F. x Eff.
where: P.F. = Power Factor
RE: Understanding the nameplate
What you don't say is what is the RPM on the nameplate.
Looking in an ABB motor catalog ..... here's some data on a
400v 2.2kW 1000 RPM motor.
Full Load Eff. 80.1 %
Power Factor .74
I 5.4 amps
Another motor I looked at of similar design and rating had these numbers
Eff. 83 %
P.F. .73
I 5.24 amps
When we look at an 750 RPM (8-pole) motor, we find this data ...
Eff. 80.5 %
P.F. .67
I 5.9 amps
From this I would conclude that your motor has a low efficiency and a low power factor.
RE: Understanding the nameplate
smaller motors tend to have lower pf*eff than larger.
slower motors tend to have lower pf*eff than faster.
older motors tend to have lower pf*eff than newer.
there are plenty of exceptions.
Manufacturer's info is the best available data for your motor (except perhaps actual measurements taken yourself).
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(2B)+(2B)' ?
RE: Understanding the nameplate
RE: Understanding the nameplate
It is 6 poles and 930 rpm - so I guess the effency is "low" due to a slow engine.
We am trying to size a UPS towards worst case - and even though full load is the best efficiency it will be still be "worst case" right?
Meaning: 400 x 6 x 1.73 = 4.15kVA.
4.15k x 0.83 = 3.44 kW
Just an info; it will be started up by a soft starter limited to 6.0 A.
RE: Understanding the nameplate
Limiting the starting current to 6 amps on a motor rated at 6 amps may inhibit the motors ability to start somewhat.
Check the specs on the inverter. It may be current or KVA rated rather than kW rated. You may have better luck with a VFD instead of a soft start in that size. Then you may be able to go with a single phase UPS.
Do you want the motor to be able to drive the occasional overload?
Many designers consider it good practice to oversize motor supply components by 25%. In many instances the codes mandate 125% capacity for some motor supply components.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding the nameplate
You know the product of efficiency times power factor is 2.2 / 4.15 = 0.53
It could be that the product pf*eff = 0.53 consists of eff = 0.83 and pf = 0.64.
But is it conservative? What if pf is a little higher and eff a little lower.
If you wanted to be really conservative (with respect to these particular factors), you might choose pf = 1, eff = 0.53, which leads to UPS requirement of 4.15kw. (I am assuming it is the real power drawn from stored energy of battery that is of interest)
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(2B)+(2B)' ?
RE: Understanding the nameplate
If you are looking at total power requirement (such as for battery amp-hour sizing), then you need to make an assumption about the split as I discussed above.
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(2B)+(2B)' ?
RE: Understanding the nameplate
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(2B)+(2B)' ?
RE: Understanding the nameplate
General purpose aluminum motors
Technical data for totally enclosed squirrel cage three phase motors
IP 55, IC 411; Insulation class F, temperature rise class B
2.2 kw 400 V 50 Hz 940 rpm
pf=0.71 eff.=0.77
Irat=5.9 A
RE: Understanding the nameplate
As Bill pointed out, limiting the starting current to 100% of FLC will result in a stall and an unsuccessful application. A soft starter can only limit current by limiting voltage, which limits torque. To attain a Current Limit at 100% of FLC when it wants to be 200% of FLC (the minimum required for even the lightest of loads) the soft starter will have to phase back the voltage to approximately 50% of RMS, so the torque will be approx. 25% of BDT, which would have been maybe 220% of FLT, so you will have roughly only 55% of FLT at the motor shaft at 100% Current Limit. I have never seen that work. Typical Current Limit for something easy like a pump or fan is 250-300% FLC, anything heavy may be 350-450%. In addition, Soft Starters need relatively clean power on the input so they can sense when to fire the SCRs. If the output of your UPS is PWM, chances are the Soft Starter will not work properly anyway.
Aside from that, your math is flawed as well. 2.2kW is 2.2kW, it doesn't matter how you get there. But in a motor, that 2.2kW rating is the MECHANICAL output of the motor, not the electrical input. The electrical input is the output (2.2kW) divided by the efficiency, which you do not know. But ASSuming (from the other posts) a 71% eff, the the electrical input would be 2.2kW/.71 = 3.1kW. But that would ONLY be for running load capability. Assuming the Soft Starter will not work (as I said above), your choices are DOL or VFD.
A VFD would be OK with PWM on the input because all it is doing is converting back into DC anyway, and because it controls frequency AND voltage, you will likely be able to start that motor with 3.1kW from the UPS, as long as starting time is irrelevant to you.
If you go with a DOL start, you will likely need a UPS that is capable of 18kW because unlike other power sources, a UPS is typically not built with any "reserve capacity" in the design. They are generally NOT intended to be used on induction motors.
The only other viable reduced voltage starting method for you would be an Autotransformer starter, but nobody makes one that small, you would have to roll your own from scratch.
Do NOT, I repeat, NOT attempt to use Star-Delta on this, the result will be a dead UPS and/or motor.
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RE: Understanding the nameplate
If Istart=6*5=30 A [D.O.L] a UPS of 20 A 400 V will be fair.
RE: Understanding the nameplate
Worst case could be kVA equal kW if we are looking at the UPS. We have a lot of equipment connected to this UPS (cameras, computers, IFF, radio, radar etc.). Which typically uses a controlled and know amount of power. But for the radar antenna running out in the open by a 3 phase engine - it gets a bit more complicated.
We have been looking at ABB motor controller (ACS550), and I got two choices. Either I will go for a three phase or one phase.
It looks like the 4.0 kW threes phase could do the job, but if I have a 4 kW ABB single phase, will this be enough? I am thinking current limitation from this one.
I have attached the name plate for the motor (not sure how it is shown since it is my first time to attached a link for a picture).
RE: Understanding the nameplate
You can often feed small a VFD with single phase or three phase and drive a three phase motor.
Worst case, KW=KVA, NO, when a motor is starting the KVA may be several times the KW.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding the nameplate
What to expect from a ABB during start up? I know there is many ways to set up the ABB for starting up the motor, but the time is not an issue, an example could be 15 sec. But can I expect 2 x kVA during start up? I know the UPS needs to be oversized but how much extra is needed for a motor like this?
RE: Understanding the nameplate
Pass the buck and try to get the UPS vender to take responsibility for sizing the UPS. That will be a lot safer, career wise.
Bill
--------------------
"Why not the best?"
Jimmy Carter
RE: Understanding the nameplate
RE: Understanding the nameplate