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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 MARKS NAME : M. S. D. PERERA. REGNO : 311089590 CENTER : COLOMBO. TOTAL DATE OF SUBMISSION: 07/11/2012 %
  2. 2. Experiment 1: Open circuit and short circuit tests of a single-phase transformer. 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. Theory Working with a ideal transformer is easy. But life get complicated when it comes to theoretical electrical engineering to real world electrical engineering. Ideal model no longer useful in industrial electrical engineering applications. So we have to come up with a model that represent an normal non-ideal industry transformer. The real transformer have following things to be included and modeled when in when 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 resistor because it’s proportional to the number of turns in the winding. And magnetizing flux could also represent as a parallel component as well as it’s also proportional to the number of windings. We represent copper loss as a serial resistive component , because it’s just equal to a pure resistor passing current through it and disparaging energy. And also we represent leakage flux as also a serial inductive component, we could imagine it as a series inductor outside the transformer which is blocking some potential difference across it so it will reduce the gross potential difference among ideal transformer terminals. Bellow figure depicts this model diagrammatically. Since these two windings are magnetically coupled, we could get it’s thevean equivalent circuit as we seen from the primary. (This could be done to the secondary 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 to multiply each inductive/resistive component by square of turns ration. Which means, When you transforming primary into the secondary side, you have to divide it by 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 the voltage 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 2 Calculations: Since there are no power desperation on the secondary side we have only power desperations on the primary side. They are sum of copper loss+ core loss. But in here, since we have very little current flowing through primary winding, we could ignore copper loss and assuming that reading in the wattmeter is equal to core losses. So through that we could find two variables.
  5. 5. Graphs and Characteristics: Part B: Short Circuit Test Now we are going to short circuit the secondary side. We need to take caution here, because there is a potential to burn the fuses in the learning panel if we won’t be careful. So we keep the variac device at it’s lower position and powering up the switches. Here we are getting a one reading only. It’s at wattmeter and ammeter readings while variac kept at 9V. We use such a very small (9V) potential thus because this is a short circuit test 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 just because Rc and Xm are very large values, we could ignore them. So it’s safe to assume that all the power desperation is now equal to the copper loss. So, Observations, 9Ammeter Voltmeter Wattmeter 2.17 9 32.5 So we get, Discussion: Why HV side is open circuited and LV side is short-circuited when performing the practical? Well as it term derives it’s meaning that HV side will generate high voltages. So that will ramp up the short circuit current to a very large value just because there are no any resistance to limit current flow on secondary side. So that’s why we need to use LV side to be short circuited as well as we should use very low voltage ( like 9V in our experiment ) to avoid damaging or frying transformers, fuses or breakers. Experiment 2 : Load Test Of Transformers Apparatus 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. Leads Theory Voltage regulation is a principle to keep voltage value independent of the load. When it comes to voltage regulation we have to consider bellow facts into 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 not possible to get a ideal constant voltage, it will vary at least by a fraction of a million when it’s load current changes. By the way, we should have some standard index to measure how much bad or good 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 load Here we can’t assume that power factor is 1, we have to calculate it. Since ,
  9. 9. Graphs:

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