Fault MVA Calculation

-by Tarun Goel-by Tarun Goel

Why Fault Analysis?

Fault Analysis to:� Calculate Power System Currents and Voltages

during Fault Conditions� Check that Breaking Capacity of Switchgear is not

exceeded� Determine the Quantities which can be used by � Determine the Quantities which can be used by

Relays to Distinguish between Healthy and Faulty Conditions

� Appreciate the effect of the method of earthing on the detection of Earth Faults

� Select the best relay Characteristics for Fault detection

� Conduct post fault analysis

Simple Generator Model

� Generator Model X will vary with time. Xd’’ – Xd’ – Xd

X

E

X

Parallel Generators

� If both generator EMFs are equal –they can be thought of as resulting from the same ideal source.

11 KV

20 MVA

20 MVA

X1=0.2pu

X1=0.2pu

j0.05

11 KV 11 KV

Parallel Generators (cont..)

j0.05 j0.05

1.0 1.0

j0.2j0.2

1.0

j0.2j0.2

Transformer Model

P S Zp = Primary Leakage Reactance

Zs = Secondary Leakage Reactance

Z = Magnetizing

P SZP ZS

ZM = Magnetizing Impedance = Large compared with Zp and Zs

ZM = Infinity = Open Circuit

ZT1 = Zp + Zs = positive sequence Impedance

ZM

N1

P SZT1 = Zp + Zs

N1

Motors

� Fault Current contribution decays with time

� Decay rate of the current depends on the system. (100-150ms)

� Typically modeled as a voltage behind as Impedance

M

Xd”

Induction Motors

� Small Motors

� Motor Load < 35kW – NEGLECT

� Motor Load > 35kW

� SCM = 4 X sum of FLCM

� Large Motors� Large Motors

� SCM = (motor FLC)/Xd”

� Approx. SCM = 5 X FLCM

Synchronous Motors

� Large Synchronous Motors

� SCM

: 6.7 X FLCM for 1200 rpm

: 5 X FLCM for 514 – 900 rpm

: 3.6 X FLCM for 450 rpm or less

Balanced 3-ph faults

� 3-ph faults may be represented by 1-ph circuit

� Valid because system is maintained in a balanced state during the faulta balanced state during the fault

� Volt equal and 120 apart

� Currents equal and 120 apart

� Power System plant symmetrical

� Phase Impedances equal

� Mutual Impedances equal

� Shunt Impedances equal

Balanced 3ph Faults

Generator TrafoLine X Line Y

LOADS

3 ph fault

EaEa

Ea

Ea

ZG ZT IaFZLX

IcF

IbF

ZLY

ZLOAD

Balanced 3ph faults(cont..)

Ea

Ea

Ea

ZG ZT IaFZLX

IcF

IbF

ZLY

ZLOAD

EaZG1 ZT1

Ia1 = IaF

ZLX ZLY

N1

F1

Positive Sequence (Single Phase) Circuit:-

ZLOAD

How to Analyze a different voltages

System?

� Per Unit System� Used to simplify calculations on systems with more than 2

voltages

pu value = (actual value)/(Base value in same units)

� Particularly useful when analyzing large systems with several voltage levels.

� All system parameters referred to common base quantities

� Base quantities fixed in one part of the system

� Base quantities at other parts at different voltage levels depend on ration of intervening transformers

Base quantities and Per Unit Values

� Base quantities normally used are:

� BASE MVA = MVAb = 3ph MVA

� Constant at all voltage levels

� Value – MVA rating of largest item of plant or 100MVA100MVA

� BASE VOLTAGE = KVb = ph-ph volt in KV

� Fixed in one part of the system

� This value is referred through transformers to obtain base voltages on other parts of system

� Base voltages on each part of transformer are in same ratio as voltage ratio.

� Other base quantities

� Base Imp = Zb = (kVb)X(kVb)/MVAb

� Base Current = I = MVA /kV (ph-e)� Base Current = Ib = MVAb/kVb(ph-e)

Example

Base voltage on each side of a transformer be in the same ratio as voltage ratio of transformer

11.8kV 11.8/141kV 132/11kV

Distribution System

Incorrect selection of kVb : 11.8kV / 132kV / 11kV

Correct selection of kVb : 11.05kV / 132kV / 11kV

: 11.8kV / 141kV / 11.75kV

Conversion of per unit values from one

Set of Quantities to Another

Zpu1 -> Zpu2

Actual Z = Za

Zpu1 = Za/Zb1

Zpu2 = Za/Zb2 = Zpu1 X Zb1/Zb2

= Z X (kV )X(kV )/MVA X MVA /(kV )(kV )= Zpu1 X (kVb1)X(kVb1)/MVAb1 X MVAb2/(kVb2)(kVb2)

Example

11kV20 MVA

11/132kV50 MVA

Line X Line Y

LOADS

3 ph Fault

132/33 kV50 MVA

0.3 pu

Line Y

LOADS

Fault

132/33 kV50 MVA

B

200/1

A

50/1

Relay settings depending on Fault MVA:

For Standard Inverse

Relay A: CTR = 50/1, Setting: 50A + TMS = 0.2

For fault current = 875A,

t(op) for A = 0.47sec

Grading required = 300 ms

Relay B: CTR = 200/1, Setting: 200A. TMS = ?

t(op) for B = 0.47 + 0.3 = 0.77sec

Hence TMS = 0.16

1

14.0

2−

×=

r

opI

TMSt

Fault

THANK YOU