Tony S
Senior Member
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- Resting under the Major Oak UK
Redesign the whole thing to run at the supply frequency. It would be cheaper in the long run.
Frequency conversion
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Redesign the whole thing to run at the supply frequency. It would be cheaper in the long run.
That's why I was suggesting just changing the motors. Then you probably wouldn't need a converter.
You mean something along the lines of what I suggested eaarlier?
And I thought you said costs didn't matter in Russia.
Tony S
Senior Member
- Location
- Resting under the Major Oak UK
You asked what I would do.
My colleague had a Russian open cheque book, something we all dream about.
I’ve always had financial constrains on projects. I’ll admit to some “creative accountancy” to pull a project in just about on budget. Three times over on one. Once it’s working the accountants can’t snatch the money back, they complain but have to get the cheque book out.
My colleague had a Russian open cheque book, something we all dream about.
I’ve always had financial constrains on projects. I’ll admit to some “creative accountancy” to pull a project in just about on budget. Three times over on one. Once it’s working the accountants can’t snatch the money back, they complain but have to get the cheque book out.
Yes.
Pretty what I suggested and for the same reasons.
i would try to run it at 50
you need a stepdown xfmr mv-lv anyways
get one that can be tapped 400-480 or as close as possible
set at 400 V so 400/50 = 480/60
Retap or replace cntl xfmrs if required
see what happens
if we knew more we could makevan informed decision
what type of machine, system, device is it?
what is the load profile
how many motors and what hp
how loaded are they? 50%, 75%, 100?
any existing vfd's
you need a stepdown xfmr mv-lv anyways
get one that can be tapped 400-480 or as close as possible
set at 400 V so 400/50 = 480/60
Retap or replace cntl xfmrs if required
see what happens
if we knew more we could makevan informed decision
what type of machine, system, device is it?
what is the load profile
how many motors and what hp
how loaded are they? 50%, 75%, 100?
any existing vfd's
It's usually the case in my experience.
Major condition met. Other conditions, motor cooling at 50hz, load requirement at 50 are also to be checked.
On the cooling side, what we usually do is change the ventilating fans-->solved!:blink:
Cooling would be an issue
current remains ~ the same v goes down 16%, power down about 20%
but torque should remain ~ the same???
assume 100 hp at 1760 rpm T = 300 lb ft
if we drop power to 80 hp and speed to 1470 so T = 285 lb ft, 5% less
makes sense since T ~ i ~ field strength
an issue may be controls
assuming 480:120
400/4 = 100 vac, marginal
current remains ~ the same v goes down 16%, power down about 20%
but torque should remain ~ the same???
assume 100 hp at 1760 rpm T = 300 lb ft
if we drop power to 80 hp and speed to 1470 so T = 285 lb ft, 5% less
makes sense since T ~ i ~ field strength
an issue may be controls
assuming 480:120
400/4 = 100 vac, marginal
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- Electrical Engineer
Torque remains the same since the flux strength is the same. Flux strength is determined by the V/Hz ratio and 400/50 is the same as 480/60. So if torque remains the same but speed drops by 20%, that's the only reason the power rating drops by 20%. For most machines, it's torque that we need, speed is secondary. It might still be important, yes, but if it were me, I'd try it. I've done this sort of thing dozens of times and generally speed is relatively easy to deal with.
You're right about the controls however. I've found that to be trickier than the speed issues a couple of times. Separating everything out to find the frequency sensitive devices was harder than just converting it all to 24VDC and using power supplies that don't give a darn about input frequency.
Do you have numbers to show that it's gonna be a problem?
More like 40%.
Are you assuming output proportional to speed? If no, why?
AFAIK, when your speed changes in a fan/pump, the afffinity laws are:
1) Volume output/discharge: Q1/Q2 = N1/N2, in this case : Q2 =5/6 x Q1 = 83%Q1 (17% decrease)
2) Change in head/pressure: p1/p2 = (N1/N2)^2: p1 = 25/36 x p1 = 69%p1 (31% decrease)
Lastly:
3) Power drawn: kW1/kW2 = (N1/N2)^3; kW2 = 58%kW1 (42% decrease)
Centrifugal is cube law.
A fan/pump output is measured in volume per unit time and in power.:roll:
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