generator data 1

[tonyflair]

Not familar with this topic at all......do manufacturers such as CAT usually have the available short circuit current of their generators ? If not, I would like to know how to calculate it. What I do know about the generator in question is,

1.) 60 Hz, 1200 rpm
2.) 650 kW
3.) 600 V
4.) X''d = 0.16096, X''q = 0.17763, X'd = 0.25258
Xd = 1.73750, Xq = 0.93658, X2 = 0.16929, X0 = 0.09050
where X2=negative sequence and X0=zero sequence
**all reactances are in p.u.

Any help would be greatly appreciated

 

[cuky2000]

The SC at the generator connector could be estimated as follow:

1- Determine the rated Amperes:

I_rated = Gen Rated kVA/(1.73*Gen rated kV).
Gen. Rated kVA = Gen. Rated kW/PF
- PF = Power Factor (0.85 Assumed)
- Gen rated kV = 0.6kV (600 V)
_{NOTE: double check the rated operating voltage. This value typically could be 480 V or similar)}

2- Short Circuit at Generator:

SC_Gen=I_rated/X_d"

X_d" is in pu value[/sub]

 

[Jerry123]

You should be able to call you local CAT rep. and they can help you out with the information you need.

 

[Rodmcm]

As a guick rule of thumb the short circuit level is the kVA divided by the sub transient reactance x"d, hence a 600kW genset at o.8pF say is 750kVA so fault is 750/0.16096 = 4659kvA or 4483 Amps at 600V

 

[alehman]

If the generator neutral is solidly grounded you will need to evaluate the line-to-ground fault current as well, as it may be higher than 3-phase. It is roughly

3V / (2X"d + X0)

 

[jbartos]

Suggestion: Approximately calculated:

Isc,pu=Vpu/Xd"pu=1.0/0.16096=6.2127, pu

Irated,amps=(kW/PF) x 1000/(sqrt3 x E)=
=(650kW/0.8assumed) x 1000/(sqrt3 x 600V)=
=1354.17 amps

Short Circuit Current in amps:
Isc,amps=Isc,pu x Irated,amps=
=6.2127 pu x 1354.17 amps = 8413 amps.

General Reference: IEEE Std 141 Red Book Chapter 4 Shor Circuit Current Calculations

 

[Rodmcm]

Previous message
I calculate the rated current at 721Amps, I think your 1354 is a tad high!

 

[ScottyUK]

jB -

Your calculations appear to have an arithmetic error and have lost a "sqrt 3" factor which you included in your equation. With corrections, the calculation gives a full-load current of 781A and an s/c current of 4854A.

Rodmcm,

Your rule-of-thumb when recalculated based on a 650kW set as noted by the OP rather than 600kW gives an s/c fault of 5044kVA which again equates to 4854A.


The two methods are exactly the same, but jB has broken the calculation down further and it therefore looks more complex than the way Rodmcm has expressed it. Alehman's comment is well worth noting.






-----------------------------------

"Never look down any at anybody, unless you are helping them up."

Jesse Jackson.

 

[Rodmcm]

woops, must get my glasses fixed