Ct For Generator Protection-rev1

Project Name -Lodha IT Park Document No -

CT Sizing & Max Fault Level of Generator Diesel Generating set comprises of diesel engine as the prime mover. When a short circuit occurs on the system powered by a generator, the generator continues to produce voltage at the generator terminals as the field excitation is maintained and the prime mover drives the generator at normal speed. The generated voltage causes a large magnitude fault current flow from the generator to the short circuit. The flow of fault current is limited only by the generator impedance and the impedance of circuit between the generator and short circuit. In case of a short circuit at the generator terminals, the fault current is limited by generator impedance only

Objective – To calculate the fault current of the generator Generator Rating – 2000KVA Rated Voltage -415V Sub Transient Reactance Xd “ - 0.11 ( Reference Technical Data Sheet of Alternator ) Fault Current If = Fault MVA / √3 x kV Fault MVA = MVA / Xd” Fault MVA = 2/0.11 = 18.1818 Fault Current = 18.1818/ √3 x 0.415 = 25.295 kA Full Load Current of the Generator = MVA / √3 x kV =2.782kA or 2782 A This means that the fault current is (25.295/2782 ) times i.e approximately 9times the full load current . Due to this reason Class 5P10 CTs will suffice for Protection , as they will not saturate for upto 10times the rated current .

Further Calculations Let us calculate the Knee Point Voltage requirement of CT for differential protection . The following assumptions are made during such a calculation for stability under worst external fault .The settings are such that the relay shall not operate for a spill current due to CT saturation under worst external fault conditions

Consider a case where fault occurs outside the protected zone (i.e-external fault) . If= Fault Current RL= Lead Resistance RCT = CT Secondary resistance Vs= Voltage across the relay at the time of fault If

Neutral CT

Phase CT G If

R

Vs

Vs = If (RCT + 2RL) Ifp = 25.295 kA (Primary Fault Current ) RCT = 0.582 ohms RL =0.11ohms ( Considering Length between AMF panel to DG as 15metres )

If = (Ifp*5)/3200

=39.52A

Vs= 39.52(0.582+0.22)= 31.69V Voltage across the relay during worst fault = Vs =31.69V

Vk = Knee point Voltage of CT. Knee Point voltage of CT should be higher than this voltage across the relay

Let us calculate the knee point volatge of the CT under consideration (Class 5P10) Vk = (VA x ALF ) / CT secondary + (CT secondary x ALF x RCT @ 75deg C) ALF = 10 VA =15 RCT @75deg C = RCT X 1.15 Vk = (15 x 10) / 5

+

(5 x 10 x 0.582X1.15 ) = 30 + 33.465 = 63.465 V

Conclusion – The knee point Voltage Vk of the Class PS CT is much higher than the voltage across the relay Vs for a worst external fault .Therefore Class 5P10 CT will suffice as it will not saturate under this condition .