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Home Electrical and Electronics Electrical Machines 1 To conduct the Hopkinton’s test on a given dc-motor and to determine the efficiency it’s efficiency when acting as a generator and motor
Electrical Machines 1 Lab Experiments

To conduct the Hopkinton’s test on a given dc-motor and to determine the efficiency it’s efficiency when acting as a generator and motor

Sep 24 2023


Aim

To conduct the Hopkinton’s test on a given dc-motor and to determine the efficiency.

  1. It’s efficiency when acting as a generator
  2. It’s efficiency when acting as a motor.

Apparatus Required:

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:

circuit-diagram

To measure armature resistance:

diagram

Procedure:-

  1. Connect the circuit as per the circuit diagram.
  2. Ensure that the motor field rheostat should be in minimum position and generator field rheostat should be in minimum output voltage position and also The SPST switch should be in open position at starting.
  3. Give dc supply to the dc machine by closing the DPST switch and start the motor with the help of 3-point starter.
  4. Adjust the motor field rheostat till the rated speed is obtained.
  5. By varying the generator field rheostat note down the readings of terminal voltage, generator field current, motor field current, output current of generator and total input current from the supply.
  6. Repeat step 6,till the rated current of generator is obtained.
  7. Switch off the dc supply by opening the DPST switch.

To determine armature resistance:

  1. Connect the circuit as per the circuit diagram
  2. Switch on dc supply
  3. Increase the load and note down the voltage and current.

Armature Resistance

V I RaV/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– I2)- Wm *100 V(I1+I2- 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 2watts

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=(I1+ I 3) 2*R awatts

W cu,g =______watts

Motor field copper losses W sh,m = V* I 4watts

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/( (V1)+ Wg)

Result:Hopkinson’s test is conducted on a given DC shunt machines and efficiency is calculated for both motor and generator

Viva voce

  1. What is the main advantage of the Hopikinson’s test?
  2. What is the main disadvantage of the Hopikinson’s test?
  3. What is the need of the voltmeter which connected between motor and generator?








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