Power output +Power loss
EXPERIMENTAL EXERCISE 1.1 Load Test on a Single-Phase Transformer
OBJECTIVES:
1.
To determine the polarity of the primary and secondary windings.2.
To find the turns-ratio.3.
To determine the efficiency at different loads.4.
To determine the voltage regulation.APPARATUS:Single-phase ac power supply 230 V; One single-phase transformer 2 kVA, 220-V/110-V; One Variac 0-270 V, 15 A; One wattmeter 10 A, 230 V; Two voltmeters (MI type) 0-230 V; Two ammeters (MI type) 0-10 A and 0-20 A; One resistive load.
CIRCUIT DIAGRAMS: The circuit diagrams are shown in Fig. 1.27.
(a) Circuit for polarity test and voltage ratio test.
(b) Circuit for determining the efficiency and the voltage regulation.
Fig. 1.27 Circuit diagrams for load test on a single-phase transformer.
BRIEF THEORY: (i) When a transformer is used, a terminal of primary (as well as of secondary) winding is alternately positive and negative with respect to other terminal. It becomes important to know the relative polarities of the primary and secondary terminals in situations such as follows:
(1) When two single-phase transformers are connected in parallel so as to share the total load on the system.
(2) When three single-phase transformers are connected to make a three-phase transformer.
The relative polarities can be determined by shorting one of the terminals of the primary and secondary windings (say, P2 and S2 in Fig. 1.27a), and then measuring the voltage across the other terminals of the primary and secondary windings (say, P1 and S1
in Fig. 1.27a).
If P1 and S1 have same polarity (say, positive at an instant), the situation is as shown in Fig. 1.28a. Obviously, in such a case, the voltage V3 = V1 –V2. On the other hand, if P1 and S1 have opposite polarity (say, positive and negative, respectively, at an instant), the situation becomes as shown in Fig. 1.28b. Obviously, in such a case, the voltage V3 = V1 +V2.
(a) P1 and S1 of same polarity. (b) P1 and S1 of opposite polarity.
Fig. 1.28 Reading of V3 depends on the relative polarity of primary and secondary.
(ii) The induced emf in winding is proportional to the number of turns on it. Therefore, the turns-ratio is given as
2 2 2
1 1 1
N E V
K = N = E V
(iii) If Pin is the input power and Po is the output power of a transformer, its efficiency is given as
in
o o
o l
P P
P P P
η = = +
where, Pl represents copper loss and iron loss in the transformer. The copper loss in the windings is proportional to the square of the current. However, the iron loss (hysteresis and eddy-current loss in the core) remains constant. As the load is increased, the efficiency also increases. It becomes maximum when the variable copper-loss becomes same as the fixed iron-loss. On further increasing the load, the efficiency starts decreasing (Fig. 1.29a).
(a) Efficiency versus load-current. (b) Output voltage versus load-current.
Fig. 1.29 Both efficiency and output voltage of the transformer vary with load current.
(iv) If V2(0) is the output no-load voltage and V2(FL) is the output full-load voltage, the percentage voltage-regulation is defined as
2(0) 2(FL) 2(FL)
% Regulation V V 100 % V
= − ×
PROCEDURE:
(i) For Polarity and Turns-Ratio Test:
1. Make connections as given in Fig. 1.27a.
2. Switch on the ac supply and adjust the variac so that voltmeter V1 reads about 100 V.
3. Record the reading of voltmeter V3.
4. If V3 < V1, the terminals P1 and S1 have same polarity.
5. If V3 > V1, the terminals P1 and S1 have opposite polarity.
6. Using variac, vary the input voltage.
7. For various values of the input voltage V1, note down the readings of the output voltage V2.
8. Calculate the ratio V2/V1. 9. Switch off the supply.
(ii) For Efficiency and Regulation Test:
1. Make connections as given in Fig. 1.27b.
2. Disconnect the load.
3. Switch on the ac supply and adjust the variac so that voltmeter V1 reads 220 V, the rated value for the given transformer.
4. Note down the reading of the ammeter A1. This gives no-load primary current.
5. Note down the reading of the voltmeter V2. This gives no-load output voltage V(0). 6. Note down the reading of the wattmeter W. This gives the iron loss Pi in the transformer.
7. Switch on the load in parts till the full-load current [
I
2(FL)= (2 kVA/110 V)=18.18 A
] is reached. For each load, note down the readings of wattmeter W, voltmeter V2 and ammeter A2.8. Calculate the efficiency for each value of load and then plot the curve of efficiency versus the load current.
9. Calculate the percentage regulation.
10. Switch off the supply.
OBSERVATIONS AND CALCULATIONS: (i) Polarity Test:
Reading of the voltmeter V1 = Reading of the voltmeter V2 = Reading of the voltmeter V3 =
If V3 < V1, the terminals P1 and S1 have same polarity, and if V3 > V1, the terminals P1 and S1 have opposite polarity.
(ii) Table for Turns-Ratio Test:
Sr. No. V1 V2 Turns-Ratio = V2/V1 1
2 3
(iii) Table for Efficiency Test:
Sr. No. W V1 V2 Output = V2I2 Efficiency = V2I1
/
WPlot the graph of efficiency versus load current (as in Fig. 1.29a).
(iv) Regulation Test:
Secondary voltage on no-load, V2(0) = Secondary voltage on full-load, V2(FL) =
2(0) 2( )
2( )
% Regulation
FL100 %
FL
V V
V
∴ = − × =
RESULTS:
1. The polarities of the primary and secondary windings have been marked.
2. The turns-ratio has been found almost same as the specified value.
3. The efficiency versus load-current curve has been plotted. The maximum efficiency occurs at a load current of _____
amperes.
PRECAUTIONS:
1. Before switching on the supply, the zero reading of the wattmeter, voltmeters and ammeters should be checked.
2. The meters of proper range should be selected.
3. While determining efficiency, use an ammeter in series with the wattmeter so as to keep a check that the current through the wattmeter does not exceed its rating.
VIVA-VOCE:
1. Can you name some applications of transformers?
Ans.: To step-up generated voltage before transmission and to step-down at consumer end. Current and voltage transformers are used in instrumentation. Small size transformers are used in electronics and communication circuits, e.g., radio, TV, etc.
2. Can you name the material normally used for making the core of transformers?
Ans.: Cold rolled grain oriented (CRGO) silicon steel.
3. What is the purpose of adding silicon to steel?
Ans.: By adding silicon (about 4 %) to the steel, its resistance increase. Hence the eddy-current loss is reduced.
4. Any harm if you add more than 4 % silicon?
Ans.: By adding more silicon, the resistance increases further. But, the steel become very brittle.
5. Why do we use laminations for making the core of a transformer?
Ans.: To decrease the eddy-current loss.
6. Usually, how much is the thickness of laminations used.
Ans.: It is about 0.35 mm.
7. A transformer is rated as 0.5 kVA, 220 V/110 V, 50 Hz. Which winding, primary or secondary, is made of thinner wire?
Ans.: Since the primary carries less current, it is made of thinner wire.
8. If I apply 220 V at 40 Hz, how much voltage will you get across the secondary?
Ans.: Sincevoltage transformation ratio does not depend on the frequency of the supply, we will still get 110 V across the secondary.
9. If I apply 220 V at 0 Hz, will I still get 110 V across the secondary?
Ans.: No, we will get 0 V, as the flux is not varying at all for zero frequency (it is a dc source). Moreover, the transformer will get burnt due to flow of large current, as there will be no back emf induced in the primary.
10. There are 780 turns on the primary and 240 turns on the secondary winding of a transformer. The primary winding is connected to an ac supply and secondary is connected to a load such that the primary current is 10 A. Now, suppose 100 turns of the secondary winding suddenly get short-circuited. Will the primary current increase or decrease?
Ans.: The short-circuited 100 turns of the secondary will draw a large current. Hence the primary current will increase by large amount.