converting alternator 3/0 to single phase, capacity? 2

[brutus1955]

if i have a 12kva 3 phase alternator that is connected star/wye
and i want to run it as single phase, what is its new capacity?

for some reason i have something like 58% of 12kva, but can't remember where i get that except it is the reciprocal
of the 1.73 delta to star/wye voltage multiplier.

then there is the thought that it might be 2/3 capacity as single phase, converted from 3 phase or, 12kva now becomes 9kva?

which is it?

thanks

bob g

 

[waross]

The basic figure is 2/3.
But, if the set is rated at 80% power factor, the three phase kW will be 9.6 kW.
Single phase the KVA will be 8 KVA and the engine has enough power to easily generate 8 kW so the PF rises to 100% or unity.
These figures are based on the assumption that the set is rewired in double delta or zig-zag for true single phase output.
12 KVA is 4 KVA per phase, or 33.33 amps at 120 volts.
If you use two phases at 208 Volts, you only get 33.33A x 208V or 6.9 KVA. With all 120 Volt loads you can get 8 KVA
If you have a 10 lead machine you will have to use a bar-diamond connection to get full capacity on single phase.

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

 

[wayne440]

"waross" please excuse me for this slight diversion- Would the "bar-diamond" connection of a 10 lead unit yield the same voltages as low zig-zag in a similar 12 lead set? If so, can you point me to a diagram of same?

thanks,

Wayne

 

[waross]

I've never seen a diagram, I made that up. But it will work.
With zig-zag connection, the single phase connections are:
L1> 1,7
N> 4,5,10,11
Insulated connection> 2,6,8,12
L2> 3,9
With a 10 lead machine, build one zigzag thus;
L1> 7
N> 10
Insulated connection> 6,8
L2> 3
Now builds another, inverted, zig-zag thus;
L1> 1
N> 4,10
Insulated connection> 9,5
L2> 2
Now common 1 and 7 for L1
Common 4,10 for neutral
Common 3,2 for L2
There will be two insulated connections, 6,8 and 5,9

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

 

[wayne440]

I have thought about the connection you describe many times,and it "looks" like it should work to me. Coincidently, I have my eye on a nice 45kw set that is 10 lead, but I have yet to develop enough confidence in my own conclusion to buy it using my own money. It is reassuring to see you bring that connection scheme up.

 

[jtpasto]

I have a Caterpillar D311 Electric Set s/n 4V7151
It is a Caterpillar R-Series self regulating generator
It operates at 1800 rpm
It is a ten wire; wires numbered T1 thru T9 plus 00
The exciter is a Century Electric Company Model DG-204-BCA-6-212724-01, Spec No. 54784, volts 62.5, rpm 3600, S/N R6, winding SPSH

Can you tell me what single phase voltage I would achieve using the above referenced 10 wire zig-zag connection. The problem I am trying to resolve is my water well pump is a 3hp Gould submersible pump and I am told I need 230 or close to it and that 208 will probably be a problem when the pump starts.

Jerome

 

[waross]

It is probably a 120/208V, 240/416V set.
The bar-diamond or Collins connection will give you single phase, 120/240 volts.This is the recommended supply voltage for a 230 volt motor.
My post of 14 Aug 09 19:31 gives the connections.

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

 

[rmarotta]

Can any of these connections be used on a standard induction motor for use as a generator? (Perhaps using capacitors for excitation?)
I don't care about efficiency, but if it's possible to do this, what could one expect?
Regards,
Rmarotta

 

[waross]

That is an interesting question. I don't know. The voltage vectors say that it will work, but the current vectors don't agree. The current will be single phase and so will be out of phase with the voltage in 2/3 of the windings.
If you have the facilities to try this, please let us know what you find out.
A suggestion;
Try connecting the capacitors from line to neutral on one half of the windings first. (the #7, #8, #9 windings have an internal wye point that is not accessible.)
God luck.

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

 

[rmarotta]

Bill, Thanks for your interest!
As I'm not an engineer, I'm curious about why (if all 12 leads are brought out of the motor windings) would the #7, #8, and #9 connections be "inaccessible"?

Here's where I'm going with this...
As I understand from various reading, an induction motor can be made to generate power if oversped slightly.
(The amount of overspeed equal to synchronous speed, plus the difference between synch. and nameplate speed.)
Also, if there is no residual, it won't produce unless excited somehow.
So, I guess what I'm asking is this:

If I drive a 3ph motor connected for 1ph as above, at the correct speed, could I expect to get any usable power out?

I don't have any "facilities" here, but wish I did. Maybe a 3ph 3HP motor and a lawnmower engine?

Regards,
Ralph

 

[waross]

All twelve leads are seldom brought out of a motor. That makes #10, #11, and #12 inaccessible.

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

 

[waross]

Also, an induction generator will generate over a fairly wide speed range. It only needs to turn above synchronous speed if it is going to regenerate into another system.

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