power design matlab

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1. Introduction...........................................................................................3

2. Case Study Analysis...........................................................................3

i) Draw the system diagram indicating all the impedances and the fault bus..........................................................................................................3

ii) Use MATLAB to get the bus admittance and impedance matrix.....4

iii) Use MATLAB to get the fault in per unit, and kA...........................5

iv) Calculate the rated short-circuit current and the rated momentary current....................................................................................................6

V) Using inverse IDMT characteristics to compute the value of time delay and plot the curve.........................................................................7

Vi) Justify the multiplying factor (k) is 1 when X/R ratio is less than 15............................................................................................................9

3.Conclusion......................................................................................11

4.References...................................................................................12

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1. IntroductionA power system consists generation, transmission and distribution. None of the power system in the world is perfect. A fault issue may occur in any power system. If the fault is not solved, it may cause problem to the company and the public who are making use of electricity from the power system. Hence, it is important to analyze the fault issue and solve it as soon as possible. A case study will be analyzed and a plan will be devised to prevent the fault from damaging the power system and cause disturbances.

2. Case Study AnalysisConsider a 3 bus system. In Bus 1 and 2, two generators of size 100 MVA, with transient reactance 10%, are connected via two transformers of size 100MVA, 11/110kV, and 5% reactance. All the three buses are interconnected with transmission lines, with reactance of 10%, with a base 100 MVA, 110kV. There is a three phase solid short-circuit fault in bus 3

i) Draw the system diagram indicating all the impedances and the fault bus

Figure 1: System Diagram

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ii) Use MATLAB to get the bus admittance and impedance matrix The next step is to form to a 3×3 admittance matrix.(Ybus). The impendence matrix of the 3-bus system is just the inverse of its admittance matrix. ([Z]=[Y]-1 )

Figure 2: MATLAB Program

Figure 3: Admittance and Impedance Result

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iii) Use MATLAB to get the fault in per unit, and kACalculate the equivalent impedance from the bus 3 and assume the voltage thevenin from bus 3 is 1pu. From these two values, fault current can be calculated in per unit value and kA.

Figure 4: Fault current in per unit and kA in MATLAB

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iv) Calculate the rated short-circuit current and the rated momentary currentGiven the R/L ratio is 7, for a 3 cycle(50 Hz) breaker, rated short-circuit current and rated momentary current able to obtained from MATLAB program.

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The momentum current is the fully offset RMS fault current capacity that allows the circuit breaker to withstand it without being damaged when the circuit breaker is closed and latched the contacts. In addition, the momentum current is not a function of the actual voltage of application .Normally fuses are used as a fast acting, protective devices that operate in the first cycle of fault which are rated at momentum current value. While the rated shorted current is the maximum rms current that the circuit breaker can disconnect at the rated voltage. It is lower than momentum current because it takes into account the short-circuit decrement with respect to time while the circuit breaker is opening.

Figure 5: Rated Short-circuit current and Rated momentary current in MATLAB

V) Using inverse IDMT characteristics to compute the value of time delay and plot the curveIn the Inverse IDMT characteristics, K=1 is the slowest time multiplier. Hence, time delay is able to compute in the MATLAB given Alpha and Beta. Given I base =524.8639A, Set I(pickup)=1.2 of I base.

Figure 6: Time delay of the current ratio in MATLAB

Figure 7: Plotting Curve Command in MATLAB

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Figure 8: Very Inverse IDMT Curve in MATLAB

Figure 9: Time Multiplier Computation for Curve Extremely Inverse

Time multiplier is rounded down to 0.1. If it is rounded up to 0.2, the circuit breaker will not trip and it will continue to wait. Hence, it will damage the circuit breaker.

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Vi) Justify the multiplying factor (k) is 1 when X/R ratio is less than 15 .

Figure 10: K Factor when f=50Hz, Cycle=3 and variable X/R ratio

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(Eq.1)

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Does System frequency has any impact?

There is no impact in the system frequency. According to the derivation of the multiplying factor (K) which is also known as asymmetry factor, the system frequency is cancelled out in the equation and it solely depends on the time constant (L/R) and the cycle (t). Hence, changing multiplying factor has no impact on the system frequency

3.ConclusionFrom the case study shown above, applying the fault knowledge enable electrical engineers to design an adequate protection to the power system in advance. Without setting up protection to the power system, the power system would not last for several years when fault occurred. In additional, and it is not wise to keep splurging money to build new power system when it is damaged since the cost is high. Hence, setting protection devices in power system helps to prolong the power system lifespan and continues its productivity to the consumers.

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4.References[1] EE4503 Lecture & Practice Notes (Academic year 2015/2016), Power Engineering

Design (Unpublished)[ 2 ]Power systemanalysis , Arthur R .Bergen ,Vijay Vittal ,Prentice Hall ,

Inc . ,2000.

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power design matlab

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