## Relay Curves GEC CDG-23

## Relay Curves GEC CDG-23

(OP)

Is there, on the internet, manuals for older electromechanical relay, GEC type CDG-23, in order to make the settings. Also, if approximate equations have been made, I would like to find them.

## RE: Relay Curves GEC CDG-23

The GEC Measurements Protective Relays Application Guide, 3rd edition, gives the formula for the CDG13 very inverse relay as follows -

t=13.5/(I-1)

Where:

t= operating time (at 1.0 time dial)

I= current (multiple of plug setting)

## RE: Relay Curves GEC CDG-23

www.tde.alstom.com/p-c/en/contcent.htm

## RE: Relay Curves GEC CDG-23

Regards

Simon Morgan

Nilsen Electric

Engineering Services Queensland Australia

State Manager

Ph +61 (07) 38998866

Fax +61 (07) 38998766

Email: simonmorgan@nilsen.com.au

## RE: Relay Curves GEC CDG-23

1) CDG 12 (Inverse)

t = 111.82 i ^ -0.796

2 <= i <= 20 , i is multiple of the setting.

2) CDG 14 (Extremely Inverse)

t = 73.82 i ^ -2.229

2 <= i <= 6

For CDG 14, the characteristic curves go upto 40 times the setting, but the equations hold good only for the initial portion of the curve, upto 6 times the setting. Curve Fitting techniques failed to find an equation for the remaining part. But the equation shown above is very accurate upto 6 times setting

Once again, I admit that I don't have the data for CDG 23. If some one can send me the curves, I will try to find an equation.

Expecting comments,

Sajith Kumar

sajkumar@yahoo.com

## RE: Relay Curves GEC CDG-23

The formulas given in the GEC PRAG are as follows (they are also given by ABB, for the DPU2000R relay with IEC curves, and for the Alstom KCGG relays - see appropriate websites for relay manuals) -

Standard Inverse: t=0.14/((I^0.02)-1)

Very Inverse: t=13.5/(I-1)

Extremely inverse: t=80/((I^2)-1)

Long time inverse: t=120/(I-1)

You will notice the differences from your derived formulas- I would regard these as being definitive.

## RE: Relay Curves GEC CDG-23

So much for theory. What is implemented in our system is as follows. We have a 1.6 MVA, 11 kV/440 V Dy11 transformer that is protected with REF on the secy side only (star solid earthed) and no differential protection. We use 30% of the Full load current as the setting for the relay. We use a CAG 14 relay with range from 20%-80% of FLC).

The above referred book mentions that "for solidly earthed star-connected transformer windings, an effective setting of greater than 50% but less than 100% of rated current is usually specified. For transformer delta windings connected to resistance earthed systems, the setting specified should lie between 20 and 25% of the neutral resistor current setting."

If any further clarification or diagram is required, please feel free to contact me at sajkumar@yahoo.com

Sajith

## RE: Relay Curves GEC CDG-23

Setting the scheme requires knowledge of the maximum & minimum fault levels, the CT ratio, kneepoint voltage and secondary resistance, secondary wiring loop resistance. With a current based (CAG type) scheme, the pickup current is set as a percentage of the circuit rating (as noted by Sajith), but not more than say 30% of the minimum fault level. The scheme operating voltage is set by the stabilizing resistance to the value of the maximum voltage developed at the relay location due to the maximum external fault (Vs=Isec*(Rct+Rwiring). This value should not exceed 50% of the CT kneepoint voltage.

See the relay instruction manual for details.