To conduct field’s test on a given two identical dc series machines and to determine the efficiency
S.no | Name | Range | Type | Quantity |
1 | Voltmeter | (0-300)V | MC | 2nos |
2 | Ammeter | (0-20)A | MC | 1nos |
3 | Ammeter | (0-2)A | MC | 1no |
4 | Techometer | (0-10,000) RPM | Digital | 1nos |
5 | Rheostat | 400Ω, 1.7A | Wire Wound | 1no |
6 | Load box | 230V,5KW/20A | Resistive | 1no |
7 | Connecting Wires | (0-20)A | *********** | Required |
Name plate details:-
Motor Generator
Voltage - 220V Voltage - 220V
Current -13.6A Current -13.6A
Speed - 1500 rpm Speed - 1300 rpm
Excitation type –series Excitation type - series
Circuit Diagram:
To determine armature resistance:
To determine series field winding resistance:
Open circuit test:
To determine armature resistance:-
To determine series field winding resistance:
Table1:
s.no | Input Voltage V(volts) | Armature current of motor I_{1}(amps) | Armature current of generator I_{2}(amps) | Motor voltage V_{1}(volts) | Generator voltage V_{2}(volts) | Input power Wi = V*I_{1}(watts) | Armature cu losses of motor Wcu,m = (I_{1}) ^{2}*Ra (watts) | Armature cu losses of generato r Wcu,g = (I_{1}) ^{2}*Ra (watts) | Motor field cu losses Wse,m = (I_{1})^{2}*Rse (watts ) |
1 | 220 | 3.5 | 4 | 220 | 90 | 770 | 23.37 | 30.4 | 22.05 |
2 | 220 | 4 | 4.5 | 220 | 80 | 880 | 30.4 | 38.475 | 28.8 |
3 | 220 | 4 | 5 | 220 | 80 | 880 | 30.4 | 47.5 | 28.8 |
4 | 220 | 4.5 | 5.5 | 220 | 80 | 990 | 38.4 | 57.47 | 36.45 |
5 | 220 | 5 | 6 | 220 | 75 | 1100 | 47.5 | 68.4 | 45 |
Generator field cu losses Wse,g =( I_{1}) ^{2}*Rse (watts) | Stray losses W _{s} = W_{i} -(W _{cu,m}+ W _{cu,g}+ W _{se,g}+ W _{se,m}) (Watts) | Total losses of motor W_{m} = W _{cu,m}+ W _{se,m} + W _{s}/2 (Watts) | Total losses of generator W _{g}= W _{cu,g}+ W _{se,g}+ Ws/2 ( Watts) | % efficiency of motor %ηm= (V_{1} I_{1}- W _{m}) *100 (V_{1} I_{1}) | % efficiency of generator %ηg = (V_{2}I_{2}) *100 (V_{2} I_{2}+ W_{g}) |
22.05 | 672.225 | 381.44 | 388.56 | 50.46 | 46.09 |
28.8 | 753.2 | 435.96 | 444.04 | 50.45 | 44.8 |
28.8 | 744.5 | 431.45 | 448.55 | 50.97 | 47.14 |
36.45 | 821.5 | 504.5 | 504.5 | 49.04 | 46.6 |
45 | 894.1 | 521.97 | 560.45 | 52.55 | 44.53 |
Table:3
V | I | R _{a}=(V/I) |
Table:2
V | I | R _{a}=(V/I) |
Model Graph
Calculations:
Input voltage V =___ volts
Voltage across series field and armature winding V_{1}=____ volts
Armature current of dc series motor, I_{1} =____amps
Armature current of dc series generator I_{2} =____amps
Terminal voltage of dc series generator V_{2} =____ volts
Input power to the set Wi =V* I_{1} watts
Armature copper losses of motor Wcu,m =(I_{2}) ^{2}*Ra watts
Wcu,g =______watts
Generator field copper losses Wse,g = (I_{1}) ^{2}*Rse watts
Wse,g = ______ watts
Motor field copper losses Wse,m = (I_{1}) ^{2}*Rse watts
Wse,m = ______ watts
Stray losses Ws = Wi- {W_{cu,m}+ W_{cu,g} + W_{se,g} + W_{se,m}}Watts
Total losses of motor Wm= W_{cu,m} + W_{se,m} + W _{s}/2 Watts
Total losses of generator W _{g} = W _{cu,g}+ W _{se,g}+ W _{s}/2 Watts
Percentage efficiency of motor %ηm= ((V_{1} I_{1} - Wm)/ V_{1} I_{1})*100
Percentage efficiency of generator %ηg= (V_{2}I_{2})/ (V_{2} I_{2}+ W_{g})*100
Result: Field test is conducted on a given dc series machine and hence efficiency is calculated for motor and generator .
Precautions:
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