To conduct the Hopkinton’s test on a given dc-motor and to determine the efficiency.
S.no | Name | Range | Type | Quantity |
1 | Voltmete | (0-300)V | MC | 1nos |
2 | Voltmete | (0-500)V | MC | 1nos |
Ammeter | (0-25)A | MC | 1 no | |
3 | Ammeter | (0-2)A | MC | 1nos |
4 | Tachometer | (0-10,000)rpm | Digital | 1no |
5 | Rheostat | 250 Ω/ 1.7A | Wire wound | 2nos |
6 | Connecting wires | (0-20)A | - | Required |
7 | SPST switch | (0-20)A | Open | 1 nos |
Name plate details:-
Motor Generator
Voltage - 220V Voltage - 220V
Current -19A Current -13.6A
Speed - 1500 Speed - 1500 rpm
Excitation type - shunt Excitation type - shunt
Circuit diagram:
To measure armature resistance:
To determine armature resistance:
Armature Resistance
V | I | R_{a}V/I |
Tabular column1:
S.No | Input Voltage V (volts) | Total input current I _{1}(amps) | Genera tor field current I _{4}(amps) | Motor Field current I _{2}(amps) | Generator Output current I _{3}(amps) | Input power Wi = V*I _{1}(watts) | Armature cu losses of motor Wcu,m = ( I _{1}+I _{3}– I _{2}) ^{2}*R _{a}(watts) | Armatur e cu losses of generato r Wcu,g = ( I _{4}+ I _{3}) ^{2} *R _{a} | Motor field cu losses Wsh m = V* I _{2}(watts) |
1 | 230 | 3 | 0.5 | 0.8 | 1 | 690 | 19.45 | 4.29 | 184 |
2 | 230 | 3 | 0.55 | 0.8 | 1.5 | 690 | 26 | 7.98 | 184 |
3 | 230 | 3 | 0.57 | 0.8 | 2.5 | 690 | 41.97 | 17.9 | 184 |
4 | 230 | 3 | 0.58 | 0.8 | 3 | 690 | 51.37 | 24.35 | 184 |
5 | 230 | 3 | 0.6 | 0.8 | 3.5 | 690 | 61.73 | 31.94 | 184 |
Tabular column1:
Generator field cu losses W _{sh,g} = V*I _{4}(watts) | Stray losses Ws = Wi -(W _{cu,m}+W _{cu,g}+ W _{sh,g }+ W _{sh,m}) (Watts) | Total losses of motor Wm = W _{cu,m} + W _{sh,m}+ Ws/2 (Watts) | Total losses of generator Wg = W _{cu,g} + W _{sh,g} + Ws/2 ( Watts) | % efficiency of Generator %ηg= [(VI3)/((V I3)+ Wg)]*100 | % efficiency of motor %ηm= V(I _{1}+I _{3}– I_{2})- Wm *100 V(I_{1}+I_{2}- I _{4}) |
115 | 367.2 | 387.1 | 302.9 | 43.16 | 57.9 |
126.5 | 345.5 | 382.7 | 307.2 | 52.9 | 63.02 |
131.1 | 315.03 | 383.4 | 306.5 | 65.23 | 69.9 |
133.4 | 296.87 | 383.8 | 306.6 | 69.26 | 72.2 |
138 | 276.38 | 383.9 | 304.07 | 73.74 | 74.32 |
Calculation:
Input voltage V=___ volts
Total input current from the supply I _{2} =____amps
Generator field current I _{4}=____amps
Motor field current I _{3}=____amps
Generator output current I _{1} =____amps
Input power to the set Wi = V* I _{2}watts
Armature copper losses of motor W _{cu,m}=( I _{1}+I _{2}- I _{4}) ^{2} *R _{a} watts
W _{cu,m} = ______ watts
Armature copper losses of generator W _{} c _{u,g}=(I_{1}+ I _{3}) ^{2}*R _{a}watts
W _{cu,g} =______watts
Motor field copper losses W _{sh,m} = V* I _{4}watts
W _{sh,m}= ______ watts
Stray losses Ws = Wi - {W _{cu,m}+ W _{cu,g} + W _{sh,g}+ W _{sh,m}}Watts
Ws =_________Watts
Total losses of motor Wm = W _{cu,m}+ W _{sh,m}+ Ws/2 Watts
Total losses of generator Wg = W _{cu,g} + W _{sh,g}+ Ws/2 Watts
Total losses of generator Wg = _________ Watts
Percentage efficiency of motor %ηm = V (I _{1}+I _{2}- I _{4})- Wm *100 / V(I _{1} +I _{2}- I _{4})
Percentage efficiency of generator %ηg = (V I _{1}) *100/( (V_{1})+ Wg)
Result:Hopkinson’s test is conducted on a given DC shunt machines and efficiency is calculated for both motor and generator
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