To perform open circuit and short circuit test on a single phase transformer and to pre-determine the efficiency, regulation and equivalent circuit of the transformer.
S.no | Name | Range | Type | Quantity |
1 | Voltmeter | (0-300)V/(0-150)V | MI/MI | 1nos/1nos |
2 | Ammeter | (0-2)A/(0-20)A | MI/MI | 1nos/1nos |
3 | Wattmeter | (0-150)V LPF/(0-2.5)A | Dynamo type | 1 no |
4 | Wattmeter | (0-150)V UPF/(0-10)A | Dynamo type | 1nos |
5 | Connecting Wires | (0-20)A | *********** | Required |
Transformer Specifications:
Transformer Rating :( in KVA) _________
Winding Details:
LV (in Volts): _______________________
LV side current:_____________________
HV (in Volts): _____________________
HV side Current:___________________
Type (Shell/Core):___________________
Auto transformer Specifications:
Input Voltage (in Volts):______________
Output Voltage (in Volts): ____________
frequency (in Hz):____________________
Current rating (in Amp):_____________
Circuit Diagram:
Open circuit test:
Short Circuit Test:
Observation:
(I) For OC test
Tabular columns:
S No. | Voltmeter Reading(Vo)in Volts | Ammeter Reading (Io) in Amps | Wattmeter Reading (Wo) in Watts | Ro (Ω) | Xo(Ω) | Cos ɸo |
1 | 115 | 0.75 | 44 | 302.63 | 178.29 | 0.51 Lag |
(II) For SC test
S No. | Voltmeter Reading ( VSC) in Volts | Ammeter Reading (ISC) in Amps | Wattmeter reading (WSC) in Watts | R_{ 01}(Ω) | X_{ 01}(Ω) | Z_{ 01}(Ω) |
1 | 20 | 6.52 | 110 | 2.6 | 1.64 | 3.07 |
Model Calculation:
R_{0}=V_{1}/I_{w},where I_{w} =I_{0}Cosɸo
X_{0}= V_{1}/I_{m} , where I_{m} =I _{0}Sin ɸ_{0}
R_{01}= W_{SC}/ I_{SC} ^{2},Z_{01}= V_{SC}/ I_{SC},X_{01}=√Z_{01}^{2}- R _{01} ^{2}
K=V_{2}/V_{1}, where K=Transformation Ratio
Find the equivalent circuit parameters R_{0}, X_{0}, R_{01}, R_{02}, X_{01} and X_{02}rom the O. C. and S.
C. test results and draw the equivalent circuit referred to L. V. side as well as H. V. side.
Let the transformer be the step-down transformer
Primary is H. V. side.
Calculations to find efficiency and regulation:
For example at ½ full load
Cupper losses = W_{SC} x (1/2) ^{2} watts, where W_{SC} = full – load cupper
losses Constant losses = W_{0} watts
Output = ½ KVA x cos Φ [cos Φ may be assumed] Input = output + Cu. Loss + constant loss_________
Efficiency =(output /input)*100
Efficiency at different loads and P.f’s
Regulation: From open circuit and Short circuit test
% Voltage Regulation=(I_{2rat}/V_{2})(R_{02}Cosϴ ± X_{02}Sinϴ)*100
‘+’ for lagging power factors
‘-‘ for leading power factor
Power factor | Lagging pf % Regulation | Leading pf % Regulation |
0 | 6.21 | -6.21 |
0.2 | 7.42 | -4.74 |
0.4 | 8.36 | -3.01 |
0.6 | 8.98 | -0.956 |
0.8 | 9.07 | 1.62 |
CosΦ = 1.0
S No. | Load fraction | Load current | Wcu (W) | O/P(W) | I/P (W) | η (%) |
1 | 1 | 1.3 | 100 | 1495 | 1637 | 91 |
1 | 1 | 1.3 | 100 | 1495 | 1637 | 91 |
2 | 3/4 | 0.75 | 56.2 | 1121.5 | 1219.5 | 91.9 |
3 | 1/2 | 6.5 | 25 | 707.2 | 814.5 | 91.7 |
4 | 1/4 | 3.25 | 6.25 | 373.7 | 422 | 88.5 |
5 | 1/8 | 1.625 | 1.5 | 186.8 | 230.4 | 81.09 |
CosΦ = 1.0
S No. | Load | Wcu (W) | O/P(W) | I/P (W) | η (%) | |
1. | - | - | - | - | - | - |
Graphs:Plots drawn between
1. % Efficiency Vs output
2. % Regulation Vs Power factor
Precautions:
Result:By conducting open circuit and short circuit test of a single phase transformer calculated the efficiency and regulation of the transformer.
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