Demonstrative Video

Theoretical Explanations

Experiment

OBJECTIVE:

To conduct Hopkinson’s test on the given pair of two identical machines and obtain performance characteristics for both motoring and generating operations.

Name plate details of DC shunt motor and generator:

H.P./

kW rating

Voltage

Current

Speed (RPM)

Field Current

Winding type

Motor

3.5 KW

200 V

18.6 A

1500

0.95 A

Shunt

$R_a = 0.8~\Omega$

$R_{sh} = 172~\Omega$

Generator

3.5

200 V

16 A

1500

0.9 A

Shunt

$R_a = 1.5~\Omega $

$R_{sh} =181~\Omega $

APPARATUS REQUIRED

S.No.	Name of the Equipment	Range	Quantity	Type
1.	Voltmeter	300 V	2 Nos	Digital
2.	Ammeter	20 A	2 Nos	Digital
3.	Rheostat	360Ω/1.6A	2 Nos	Coil
4.	SPST (Switch)		1 No
5.	Tachometer	2000 Rpm	1 No	Digital

CIRCUIT DIAGRAM :

INTRODUCTION :

The Hopkinson’s test is performed on two DC machines connected in parallel across the supply.
By adjusting their excitations both of them can be simultaneously loaded (to any extent) where one (motoring) feeds mechanical power to the other machine (generating), while the generating machine feeds electrical power to the motoring machine.
The only power drawn from the mains is the losses of the both machines. Load test and heat-run test could thus be conducted with very little energy consumption while the machine carries full load current at rated voltage.

THEORY:

This test is called regenerative test or back to back test which can be carried out on two identical DC machines mechanically coupled to each other and simultaneously tested.
Thus the full load test can be carried out on two identical shunt machines without wasting their outputs.
One of the machines is made to act as a motor while the other as a generator.
The mechanical output obtained from the motor drives the generator whose electrical output supplies the greater part of input to the motor.
The motor is connected to the supply mains only to compensate for losses. since in absence of losses, the motor-generator set would have run without any external power supply.
But due to losses, the generator output is not sufficient to drive the motor. Thus motor takes current from the supply to account for losses.
The losses in a dc machine comprise of,

(a) Resistance losses in the armature and field circuits,

(b) Hysteresis losses in armature iron,

(c) Eddy current losses in the armature iron and pole faces and

(d) Mechanical losses due to friction and windage.

In the circuit diagram for Hopkinson’s test, the two shunt machines are connected in parallel.
The SPST switch S is kept open.
The 2nd machine which is coupled to the first will act as load on the first machine which is acting as a motor. Thus second machine will act as a generator.
The speed of the motor is adjusted to normal value with the help of the field rheostat. The voltmeter reading V2 is observed.
The voltage of the generator is adjusted by its field rheostat so that voltmeter reading is zero. This will indicate that the generator voltage is having the same magnitude and polarity as that of the supply voltage. This will prevent heavy circulating current flowing in the local loop of armature on closing the switch.
Now switch S is closed.
The two machines can be made to take up any load by adjusting their field rheostats.
The generator current I3 can be adjusted to any value by increasing the excitation of the generator or by reducing the excitation of the motor.

LAB SETUP :

CIRCUIT CONNECTIONS :

Supply Positive to L (3 Point Starter)
Supply Positive to Voltmeter (1) Terminal (Red)
Point Starter F to Rheostat (R1) Terminal (Red)
Point Starter F to Ammeter (4) Terminal (Red)
Ammeter (4) Terminal (Black) to Rheostat (R2) Terminal (Red)
Point Starter A to Ammeter (2) Terminal (Red)
Point Starter A to SPST Switch Terminal (Red)
SPST Switch Terminal (Black) to Ammeter (3) Terminal (Red)
SPST Switch Terminal (Red) to Voltmeter (2) Terminal (Red)
SPST Switch Terminal (Black) to Voltmeter (2) Terminal (Black)
Ammeter (1) Terminal (Black) to Motor Filed Terminal Z (Red)
Ammeter (2) Terminal (Black) to Motor Armature Terminal A (Red)
Ammeter (3) Terminal (Black) to Generator Terminal A (Red)
Rheostat (R2) Terminal (Black) to Generator Filed Terminal (Red)
Supply Negative Terminal (Black) to Voltmeter (1) Terminal (Black)
Supply Negative Terminal (Black) to Motor Field Terminal ZZ (Black)
Motor Field Terminal ZZ (Black) to Motor Armature Terminal AA(Black)
Motor Armature Terminal AA (B) to Generator Armature Terminal AA(B)
Generator Armature Terminal AA (B) to Generator Filed Terminal ZZ (Black)

PROCEDURE :

Connect the circuit.
The field rheostat of the motor should be kept at minimum position and field rheostat of the Generator should be kept at maximum position.
Switch ON the MCB, start the Motor-Generator set by slowly pulling the handle of 3-point starter.
The motor–generator set is brought to rated speed i.e. 1500 rpm by adjusting the field rheostat of the motor.
Vary the excitation of the generator by varying its field rheostat until the voltmeter across the switch reads zero; then close the switch.
If the voltmeter reading had been large before closing the switch, that indicates unsatisfactory condition for parallel operation. In such case reverse the generator field terminals to arrive at satisfactory condition and repeat step 5.
Now, adjust the generator and or motor field rheostat for obtaining different values of currents and note down the meter readings until the generator is fully loaded. i.e: Now increase I_gf and decrease I_mf in steps so as to vary the motor armature current to a maximum of 20% overload.
During all this time the motor’s speed should be maintained at its rated value and any change due to the above variations should be dully nullified by adjusting the external field resistance. Take about 8 to 10 readings for the motor armature currents between its minimum and maximum permissible values and record the same in the observation table.
While switching off the system, it is a healthy practice to go in the reverse order so that when you start the system there is no chance of the armature resistance being at its minimum i.e. exactly reverse the previous steps.
Measure the resistance of DC motor by V-I method.

The armature circuit consists of two resistances in series. They are armature winding resistance and resistance due to the brushes and the brush drop. The brush contact drop behaves like a non-linear resistance.
To separate the armature circuit resistance and brush resistance, a number of V-I readings are taken.
For large values of I, the equivalent armature resistance is taken to be V/I ohm.
If the value of brush drop V_b can be neglected then the armature resistance,
Or use a multimeter to get the value of armature resistance of the generator and motor.

OBSERVATION TABLE :

S.NO	1	2	3	4	5	6	7	8	9	10
V
I₁
I₂
I₃
I₄

Resistance of DC motor, R_am =
Resistance of DC generator, R_ag =

Expressions for Losses & Efficiencies :

$I_1 = I_{fm}=$ field current of motoring machine
$I_2 = I_{am}=$ armature current of motoring machine
$I_3 = I_{ag}=$ armature current of generating machine
$I_4 = I_{fg}=$ field current of generating machine
$V_1 = V$ supply voltage
$I_a = I_{am} - I_{ag}$
Motor armature copper losses = $I_{am}^2 \cdot R_{am}$
Generator armature copper losses = $I_{ag}^2 \cdot R_{ag}$
Total input to the armature = $P_{in (arm)}= V I_a = V (I_{am }-I_{ag})$
Total armature copper loss = $P_a = I_{am}^2 \cdot R_{am} + I_{ag}^2 \cdot R_{ag}$
Total stray loss (Total iron and mechanical losses) = $ P_{in (arm)} - P_{a}$
Total stray loss of each machine (assumed equal) = $P_s= 0.5 [P_{in (arm)} - P_a]$
Motor field copper loss = $V I_{fm}$
Generator field copper loss = $V I_{fg}$
Total Motor power loss = $P_{Lm} = P_s + V I_{fm} + I_{am}^2 \cdot R_{am}$
Total Generator power loss = $P_{Lg} = P_s + V I_{fg} + I_{ag}^2 \cdot R_{ag}$
Efficiency of Motor,
Efficiency of Generator,

Calculation of Stray Losses ($P_s$) :

V = ---------------------

S.No.	1	2	3	4	5	6	7	8	9	10
$$ \begin{aligned} P_{\mathrm{in(arm)}} &= V\cdot I_a \\ & = V \cdot (I_{am}-I_{ag}) \end{aligned} $$
$$ \begin{aligned} P_a & = I_{am}^2 \cdot R_{am} + I_{ag}^2 \cdot R_{ag} \end{aligned} $$
$$ \begin{aligned} & P_{\mathrm{in(arm)}} - P_a \end{aligned} $$
$$ \begin{aligned} P_s & = 0.5 \cdot \left(P_{\mathrm{in(arm)}} - P_a\right) \end{aligned} $$

Calculation of Efficiency :

S.No.	1	2	3	4	5	6	7	8	9	10
Input to motor $$ \begin{aligned} P_{\mathrm{in(motor)}} & = V \cdot (I_{am}+I_{fm}) \end{aligned} $$
Armature copper losses $$ \begin{aligned} & = I_{am}^2 \cdot R_{am} \end{aligned} $$
Field copper loss $$ \begin{aligned} & = V \cdot I_{fm} \end{aligned} $$
Total power loss$$ \begin{aligned} P_{Lm} & = P_s + V \cdot I_{fm} + I_{am}^2 \cdot R_{am} \end{aligned} $$
Motor output $$ \begin{aligned} P_{\text{out}} & = P_{\text{in}} - P_{Lm} \end{aligned} $$
Efficiency of the Motor $$ \begin{aligned} & P_{\text{out}}/ P_{\text{in}} \end{aligned} $$
Output of the Generator $$ \begin{aligned} P_{\text{out}} & = V \cdot (I_{ag}-I_{fg}) \end{aligned} $$
Generator field copper loss $$ \begin{aligned} & = V \cdot I_{fg} \end{aligned} $$
Total power loss $$ \begin{aligned} P_{Lg} & = P_s + V \cdot I_{fg} + I_{ag}^2 \cdot R_{ag} \end{aligned} $$
Input to the generator $$ \begin{aligned} & P_{\text{out (gen)}} - P_{Lg} \end{aligned} $$
Efficiency of the generator $$ \begin{aligned} & P_{\text{out}} / P_{\text{in}} \end{aligned} $$

Model Graphs :

GRAPHS TO BE PLOTTED

Plot the graph Efficiency of motor vs Line current (I am+ Ifm)
Plot the graph Efficiency of motor Vs Motor output (Pout m)
Plot Efficiency of generator Vs Load current of generator (Iag)
Plot Efficiency of generator Vs Generator output (Pout, g)

Note: Graphs 1 & 3 should be plotted in one graph sheet and graphs 2 & 4 should be plotted in one graph sheet

Results :

From the graphs infer the values:

Load	$\eta_m$	$\eta_g$
1/4
1/2
3/4
FULL

Questions? :

1.What is the difference between Swinburne’s and Hopkinson’s tests?

2.What are the advantages and disadvantages of Hopkinson’s test?

3.Can Hopkinson’s test be performed on any two DC machines? If so what should be the specifications of those machines?