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Old January 19th, 2005, 09:11 PM   #1
Combo
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Connecting a motor in triangle with inverters, Torque untill 87Hz

My boss has an idea:

We have a project with 2 drives en 2 motors. 2 MM440's from Siemens. He tells me to connect the Motors in Triangle...

The Motors are 2,2 kW and the drives 4 kW; the drives have a greater power, because, the motors will take more current (triangle).

We have a max torque untill 87Hz


Can someone explain me what's the benefit of this ?

Maybe a calculation of the Torque and current as an example,


Ow, yes: the machine = 2 rolheads mechanical coupled.

tnx
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Old January 19th, 2005, 09:30 PM   #2
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From reading your other posts-
I think you need a lot of help. Please call a
Siemens engineer to your plant to help you.
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Old January 19th, 2005, 11:02 PM   #3
leitmotif
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Triangular motor hookup ????????

OK here is what little I understand.

1. YOu want to use two motors driving in tandem
2. You want to hook the two motors onto the same variable freq power supply

So far so good. What I want to know is how to hook motors in triangle.
I have only seen hooking motors in parallel or in some cases series then parallel or vice versa. I have connected motor windings delta (triangle ??) or wye (aka star).

I want to see a wiring diagram. I may need to learn this cause I surely have not seen it yet.

Dan Bentler
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Old January 19th, 2005, 11:06 PM   #4
LadderLogic
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I am sure Combo means delta connection.

I just happen to know that in some languages, what we call "wye" and "delta", is literally called "star" and "triangle". Combo is just not fluent in English techno slang, that's all.
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Old January 19th, 2005, 11:12 PM   #5
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Combo,

1. As Guest said, you need to talk to an experienced drives expert.

2. If you need help here, explain what it is that you want to achieve - then people can assist you how to achieve it.
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Old January 20th, 2005, 02:33 AM   #6
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Always read the motor plate.

A small motor like that probably says 380/400V in star, 230V in delta ("traingle"). A 4kW inverter probably has a 400V 3-phase input and output, so the motor must be connected in star. Connecting a 4kW inverter to a 2.2kW motor does NOT make a 4kW motor.
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Old January 20th, 2005, 05:09 AM   #7
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I think they figured they need higher current....ie 4KW because they were connecting in triangle (delta).

Combo if you have a 2.2KW motor at 460vac it will draw 2.9 amps
At 230vac it will draw 5.8 amps....appproximate
You obtain KW by multiplying Current and Voltage (approximate)
This page has the formulas: http://www.patchn.com/motorformula.htm

I agree, call your Siemens rep and get a drive specialist in to assist.

I am still confused on mechanucally coupling 2 motors then running at different speeds.
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Last edited by rsdoran; January 20th, 2005 at 05:53 AM.
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Old January 20th, 2005, 06:05 AM   #8
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Dividing kW by voltage is not a good way to get motor current, because they are not resistive loads. You are better off using tables, so a 2.2kW squirrel cage motor with a 400V 3-phase supply (most European countries use 230V L-N, 400V L-L 50Hz) will take about 4.8A full load. Even better, like I say, is read the plate!
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Old January 20th, 2005, 06:21 AM   #9
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I stated...approximate.
How do you think tables are created?
The exact formula, if KW is known and amps are needed:
Code:
(Kw x 1000) divided by( 1.73 x E x pf)
e is efficiency
E is voltage
pf is power factor

Noone bothers to look at the links and get the details involved with the response.
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Last edited by rsdoran; January 20th, 2005 at 06:28 AM.
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Old January 20th, 2005, 07:09 AM   #10
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Well, that's all true of course.
I assumed the tables came from the motor manufacturers, since I might know kW, and I might know V, but only they know efficiency and pf. Unless I read it from the motor pla... sorry, I already said that didn't I.
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Old January 20th, 2005, 09:34 AM   #11
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Without going into needless detail, Ron is correct in his assumption for determining power electrically, but there is a bit more to the problem.

There are several things happening at the same time to determine shaft power. You have the input electrical current, and if 3 phase, that will be V line-line times the square root of three ("root 3"). So for a 480V system, that's about 830 VA per amp. VA times power factor is Watts (power).

Power factor is derived from any of several methods, two of which I'll supply:

Watts divided by voltaamperes (VA)
real power divided by total power (previous statement re-worded)
Cosine of the phase angle difference between the current and voltage

The problem is that, typically, we only know the load requirements, which means the power required AFTER the motor. The formulas above don't account for that, so in addition to the above, you need to factor in losses in the motor and in the mechanical power transmission system.

That includes the further term "efficiency," which can pretty much be summed up as "heat losses." Not all motor power is applied to the final load. Anything giving off heat is a loss in the total system, and must be accounted for in the design.

Delta vs. Wye (triangle vs star) systems don't change the actual voltage felt by each winding. Stars (Wyes) require more phase voltage (L-L) as each winding feels V L-L divided by 1.732 (root 3) which is V L-N. Delta windings feel total L-L voltage. Therefore, you need 1.732 times more voltage in a star than in a delta for each winding's rating.

The interesting thing is that the winding currents don't change, but the phase currents do... by exactly the factor you use to convert the voltages. Total KVA required won't change. Total KW required also will not change.
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Old January 20th, 2005, 10:25 AM   #12
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Phase vs line current and voltage

It's a good thing Navy training drives things into your skull because they are still there after XXX years.

Power = I x E If you know two of the values you can always calculate the third. With motors and other non resistive loads you have to plug in Power factor - even guessing at that you will come close.

All these calculations are estimates and are not the truth. You get the truth when you connect the motor and take readings with a known load.

Line vs phase volt or current

Wye or star
is a series circuit.
SO with 480 line voltage phase volts are 277.
Phase current = line current.

Delta (and now triangle I guess)
is parallel.
SO 480 volts line = 480 phase
480 amps line = 277 amps phase

I would be following factory recommendations (nameplate connections) for how to connect (delta vs star).
FOR SURE I would take readings to ensure the motor is OK after reconnect (wye vs star) and keep a close eye on it.

Dan Bentler
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Old January 20th, 2005, 04:27 PM   #13
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Post okay

I am sure Combo means delta connection.

I just happen to know that in some languages, what we call "wye" and "delta", is literally called "star" and "triangle". Combo is just not fluent in English techno slang, that's all.


That's indeed what I mean, delta.

I have 2 motors, both:

delta/star
220V/380V xA/xA 2.2kW

What I wanna do is, connect the motors in Delta; and the connect tot the frequency drives. I know that I shood connect in Star normally. Shat am I gonna do:

When u connect in delta, then u will haven 400 Volts on each winding (max); It Wil take more current, and the Torque wil be great untill 87Hz. Yes 87Hz, comes from sqrt 3 x 50Hz
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Old January 20th, 2005, 04:45 PM   #14
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Ok I know there is a language issue but I would like some clarification if possible.

What application would allow 2 motors to be mechanically coupled but run at different speeds? If mechanically coupled then one will act as a driver and the other be driven...ie a motor and generator. One will be, more or less, full load all the time and the other with no load.

Am I missing something?

I really dont understand the torque issues, the MM420 is overall a decent drive, it should allow full torque up to approximately 90HZ. Which is basically what you stated.

My earlier reply was WRONG...I messed up the calculations.

The other issue is it doesnt matter which voltage you use...KW will remain the same. As was stated volts x amps (approximate) will give you KW...ie 460x4.8=2.2KW (approximate) 230x9.5=2.2KW. All this means is that on 230vac you will need larger wire for the supply and motor.

Use the highest voltage possible, follow (as was mentioned) the nameplate ratings and connections. Also see if you can have a drive specialist assist you.

Please explain the mechanical coupling.
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Last edited by rsdoran; January 20th, 2005 at 04:48 PM.
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Old January 20th, 2005, 04:45 PM   #15
PhilipW
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If the motor is rated delta for 220v 50Hz then if the VSD input supply is connected to 380v then:

380/220 * 50Hz = 89.36 Hz

The key to making this work is to set up the VSD so that:

Base Voltage = 380v

Base Freq = 87Hz

This is a very nice trick for getting MORE than the rated kw from a motor. ie if it is rated for 2.2kW at 50Hz it will produce:

87/50 * 2.2kW = 3.8kW

The current will remain constant. The torque will remain constant up to 87Hz.
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