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# Electrical power ecx3232 lab report

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### Electrical power ecx3232 lab report

1. 1. LAB REPORT: 01 OPEN CIRCUIT AND SHORT CIRCUIT TESTS OF A SINGLE-PHASE TRANSFORMER ECX 3232 ELECTRICAL POWER Q. NO MARKSNAME : M. S. D. PERERA.REGNO : 311089590CENTER : COLOMBO. TOTALDATE OF SUBMISSION: 07/11/2012 %
2. 2. Experiment 1: Open circuit and short circuit tests of a single-phasetransformer.Apparatus:  500VA, 230V/230V 1:1 transformer.  (0-250V,AC) voltmeter.  (0-150V,AC) voltmeter.  (0-1A,AC) Ammeter.  Wattmeter.  230/(0-250V) variac.  Leads.TheoryWorking with a ideal transformer is easy. But life get complicated when itcomes to theoretical electrical engineering to real world electricalengineering. Ideal model no longer useful in industrial electricalengineering applications. So we have to come up with a model that representan normal non-ideal industry transformer.The real transformer have following things to be included and modeled when inwhen we drawing it’s equivalent circuit.  Losses, There are two main losses related to a power transformer. And they are, * Core Losses * Hysteresis loss. * Eddy current loss. * Copper loss.  Leakage flux.  No load flux(also known as magnetizing flux).In an equivalent circuit we represent core losses as a parallel resistorbecause it’s proportional to the number of turns in the winding. Andmagnetizing flux could also represent as a parallel component as well as it’salso proportional to the number of windings.We represent copper loss as a serial resistive component , because it’s justequal to a pure resistor passing current through it and disparaging energy.And also we represent leakage flux as also a serial inductive component, wecould imagine it as a series inductor outside the transformer which isblocking some potential difference across it so it will reduce the grosspotential difference among ideal transformer terminals.Bellow figure depicts this model diagrammatically.Since these two windings are magnetically coupled, we could get it’s theveanequivalent circuit as we seen from the primary. (This could be done to thesecondary too).
3. 3. Bellow figure depicts how we see it from primary side.In the case of transforming secondary side to primary side we have tomultiply each inductive/resistive component by square of turns ration. Whichmeans,When you transforming primary into the secondary side, you have to divide itby square of turns ratio,EXPREMENT:PROCEDURE:Part A: Open Circuit Test. (a) The voltage ratings of the transformer is, 500VA, 230/230V 1:1 transformer. So KVA rating is ½ KVA. (b) Rated voltages,
4. 4. (c) :This is open circuit test. We log no load current and iron while changing thevoltage through variac device.This is the data we have collected. Impressed No Load Current Iron Loss(W) Voltage(V) (I/A) 230 0.6115 15 180 0.224 12 160 0.195 10 140 0.116 8 120 0.142 6 100 0.121 5 80 0.101 4 60 0.081 3 40 0.060 2Calculations:Since there are no power desperation on the secondary side we have onlypower desperations on the primary side. They are sum of copper loss+ coreloss. But in here, since we have very little current flowing through primarywinding, we could ignore copper loss and assuming that reading in thewattmeter is equal to core losses. So through that we could find twovariables.
5. 5. Graphs and Characteristics:Part B: Short Circuit TestNow we are going to short circuit the secondary side. We need to take cautionhere, because there is a potential to burn the fuses in the learning panel ifwe won’t be careful. So we keep the variac device at it’s lower position andpowering up the switches. Here we are getting a one reading only. It’s atwattmeter and ammeter readings while variac kept at 9V.We use such a very small (9V) potential thus because this is a short circuittest and we are not supposing to burn that expensive learning panels.Theory:
6. 6. In here there are no power desperation on the secondary side. And justbecause Rc and Xm are very large values, we could ignore them. So it’s safeto assume that all the power desperation is now equal to the copper loss.So,Observations, 9Ammeter Voltmeter Wattmeter 2.17 9 32.5So we get,Discussion:Why HV side is open circuited and LV side is short-circuited when performingthe practical?Well as it term derives it’s meaning that HV side will generate highvoltages. So that will ramp up the short circuit current to a very largevalue just because there are no any resistance to limit current flow onsecondary side. So that’s why we need to use LV side to be short circuited aswell as we should use very low voltage ( like 9V in our experiment ) to avoiddamaging or frying transformers, fuses or breakers.Experiment 2 : Load Test Of TransformersApparatus 1. 500VA, 230V/230V single phase transformer 2. 0-250V,AC voltmeter 3. 0-5A,AC ammeter 4. Wattmeter 5. 230/(0-250V) variac 6. Resistor bank 7. Capacitor bank 8. LeadsTheoryVoltage regulation is a principle to keep voltage value independent of theload. When it comes to voltage regulation we have to consider bellow factsinto consideration.  Different loads will take different currents at same voltage.  Different loads will have different leading/lagging reactive components.
7. 7.  A load may vary how much it will draw dynamically, take a washing machine for a example, When it washing clothes it will have drive motors and there will be a lagging current component and when it switched to drying clothes it will turn it’s motors off and turn on it’s heaters which will dynamically change gross load inductive load to a resistive load.Above facts are making voltage regulation a difficult subject. So it’s notpossible to get a ideal constant voltage, it will vary at least by a fractionof a million when it’s load current changes.By the way, we should have some standard index to measure how much bad orgood a particular device could regulate against varying load currents.In transformers we use ,And phasor diagram of a transformer when loaded with power factor load.And the efficacy of the transformer is given by,The first experiment is about resistive loads, so we could use as zero.Observations And Calculations:
8. 8. Graphs:For a capacitive loadHere we can’t assume that power factor is 1, we have to calculate it.Since ,
9. 9. Graphs: