EE4503 Power Engineering Design

Assignment 2

Name: Swagato Sen

Matric Number: U1122661G

Question i

Surge Arrester is used to protect electrical equipment from overvoltage transients caused by lightning or switching surges. Correct capacitance needs to be selected so that the clip voltage can withstand the impulse voltage.

Characteristics of ECI Waveshape

EC ( t )=2(1−e−tZC )EU ( t )−2(1−e

−(t−τ )ZC )EU ( t−τ )

Exponential Current Impulse Equation

For the 4/10 waveshape, the maximum point of the impulse wave occurs after the period of the unit step and it is 4s.

Using the ECI equation and t value of 4μs, we obtain the desired capacitance value.

EC ( t )=2(1−e−tZC )EU ( t )−2(1−e

−(t−4 μs)ZC )EU ( t−4 μs)

Using

t=4 μs Ec (4 μs )=10kV Zo=300Ω

EC (4 μs )=2 (1−e−4μsZC )EU (t )−2 (1−e

−(4 μs−4μs )ZC )EU (4 μs−4 μs)

10kV=2(1−e−4μs300∗C )∗15kV

−t300∗C

=ln(1− 10kV2∗15kV )

C=32.884nF

Using MATLAB to solve:

Plotting the ECI system with this value of capacitance

4/10 ECI Waveshape

It can be seen that the waveform shown is an approximate of the 4/10 waveshape with T1 = 4μs and T2 = 10 μs (approximate 50% of Ec(max)).

The minimum capacitance is 32.884nF

Question ii

IDMT relay is not preferred for busbar protection because a fault identified in one area will cause operation to stop at other areas due to time discrimination of the fault that occurred in the other area. Hence, an area without a fault will cease to operate even though a fault has not been identified there.

To determine which relay acts faster, we need to examine the underlying mechanism of the relays. The Bucholz relay uses oil and gas to trip the relay. When there is a fault, the transformer insulating oil will decompose producing gas and this accumulated gas closes a switch which results in the circuit breaker to trip. The operation of the differential relay on the other hand is based on Kirchoff’s Current Law (KCL). Based on the KCL, under normal operation the total currents entering the protection unit is equal to the total current leaving the unit. If there is any imbalance, it means there is a fault and detecting imbalance of current and causing the circuit breaker to trip is faster than the gas accumulating in the case of the Bucholz relay. Hence, the differential relay acts faster.

Question iii

To use an IDMT relay, we need to specify values for α, β, K and current ratio II set

. First we determine

the current ratio by figuring out the maximum current possible that can be flowing in the system.

It is given that impedence is 5% which means X pu=0.05 pu

Let Vpu = 1

Imax=V pu

X pu

= 10.05

Imax=20 pu

Hence, our current ratio setting should be 20.

The criteria for the relay is that we should get the fastest tripping time. The IDMT relay tripping time is given by this equation

t= K∗β

( II set

)α

−1

Hence, this means the value of the time multiplier K is 0.1 since t is proportional to K and the smaller K value is 0.1.

Next our job is to select suitable values of α and β by choosing one of the four sets of IDMT curve parameters as shown in the following table.

Since our current ratio has a high value of 20, we should use the Long time inverse settings because at high current ratio values, the response of the circuit breaker is faster than the other settings. A comparison can be seen from the IDMT curves for a chosen value of K.

The pink curve shows the time of operation for a IDMT relay with Extremely Inverse characteristics. For current ratio values higher than 5, we observe that the fastest response is given by the Extremely Inverse characteristics and hence our chosen value of α = 2.0 and β = 80.

Based on these settings our tripping time is

t= K∗β

( II set

)α

−1= 0.1∗80

(20 )2−1=0.02005 seconds

Fastest time is preferred because transformers are very expensive equipment. Slower tripping time would mean the overcurrent would pass through the transformer equipment for a longer duration and this would cause more damage as the transformer circuit is experiencing an overcurrent for a long duration. Hence, fastest tripping time would minimize this overcurrent exposure and minimize damage.

Long Time InverseNormal InverseVery InverseExtremely Inverse

Question iv

When a motor is turned on it requires a large amount of current. In this case it is 10 times the nominal current and this current is called the locked rotor current. In our case the locked rotor time is 0.6 seconds. During this period, the IDMT protection system should not mistake this overcurrent as a fault and trip the circuit breaker. Hence the response time of the IDMT protection should be more than 0.6 seconds for a current ratio value of 10.

The IDMT relay tripping time is given by this equation

t= K∗β

( II set

)α

−1

Our job is to determine the time multiplier for each type of IDMT curve.

By manipulation of the previous equation we obtain

K=t∗[( II set )

α

−1]β

We solve for K for each IDMT curve type by substituting the following values of α and β. The current ratio value is 10 and t is 0.6.

IDMT setting α β Calculated K K value to be usedNormal Inverse 0.02 0.14 0.202 0.3Very Inverse 1.0 13.5 0.4 0.5Extremely Inverse 2.0 80 0.7425 0.8Long Time Inverse 1.0 120 0.045 0.1

For each calculated value of K, we need to choose the next higher value of K available because t is proportional to K and we want our tripping time to faster than 0.6 seconds.

Question v

Voltage imbalance occurs when voltage magnitude of each phase differs or when phase angle is not the nominal value of ± 120 of each other⁰ . This imbalance causes winding temperature to increase and may lead to damage of insulation. To prevent excessive damage, the voltage imbalance factor should not be more than 6%.

Obtain the sequence voltage

(V p

V nV 0

)=( 13 )[1 R R2

1 R2 R1 1 1 ][V 1V 2V 3]

R=e− j 23π=1∠−120

For a given VUF value of 6%, calculate phase angle of V3.

VUF=100∗|V n|

|V p|

[V 1V 2V 3]=[ 1∠01∠1201∠∅ ]

Case 1 : V3 with phase shift of -130.29⁰

[V 1V 2V 3]=[ 230∠0230∠120

230∠−130.29]

Case 2 : V1 with phase angle shift of -10.29⁰

[V 1V 2V 3]=[230∠−10.29230∠120230∠−120 ]

Conclusion

After trying several values of the phase angle, I obtained that the phase angle imbalance has to be within ±10.29⁰ to ensure the VUF does not exceed 6%.

In the following case, we investigate what happens if we change the voltage magnitude equally in the 3 phases.

Case 3 : Equal Voltage Magnitude Change

[V 1V 2V 3]=[ 400∠0400∠120400∠−120]

Conclusion

From the pasted MATLAB code, we see the VUF value to be close to zero. Hence, equal change in voltage magnitude in all the phases does not affect the VUF.