UNIT - V

Testing of D.C. Machines

 Testing of D.C. Machines
Syllabus:
• Losses: Constant & Variable losses
• calculation of efficiency – condition for maximum
efficiency.
• Methods of Testing – direct, indirect and
regenerative testing
• Brake test
• Swinburne’s test
• Hopkinson’s test
• Field’s test
• separation of stray losses in a D.C. motor test.
 Losses in a D.C. Machine
➢ The losses in a d.c. machine (generator or
motor) may be divided into three classes viz.,
(i) Copper losses
(ii) Iron or Core losses and
(iii)Mechanical losses.
➢ All these losses appear as heat and thus raise
the temperature of the machine.
➢ They also lower the efficiency of the machine
Losses in a D.C. Machine
 Losses in a D.C. Machine
1. Copper losses
These losses occur due to currents in the various
windings of the machine.
Armature copper loss = Ia2Ra
Shunt field copper loss = Ish2Rsh
Series field copper loss = Ise2Rse
Note: There is also brush contact loss due to brush
contact resistance (i.e., resistance between the surface
of brush and surface of commutator).
This loss is generally included in armature copper
loss.
 Losses in a D.C. Machine
2. Iron or Core losses:
These losses occur in the armature of a d.c.
machine and are due to the rotation of
armature in the magnetic field of the poles.
They are of two types viz.,
(i) Hysteresis loss
(ii) Eddy current loss.
 Losses in a D.C. Machine
2. Iron or Core losses:
i) Hysteresis loss:
➢ Hysteresis loss occurs in the armature of the d.c.
machine since any given part of the armature is
subjected to magnetic field reversals
as it passes under successive poles.
➢ The fig. shows an armature rotating in two-pole
machine.
 Losses in a D.C. Machine
2. Iron or Core losses:
i) Hysteresis loss:
➢ Consider a small piece a-b of the armature.
➢ When the piece a-b is under N-pole, the magnetic
lines pass from a to b.
➢ Half a revolution later, the same piece of iron is
under S-pole and magnetic lines pass from b to a so
that magnetism in the iron is reversed.
➢ In order to reverse continuously the molecular
magnets in the armature core, some amount of
power has to be spent which is called hysteresis loss.
 Losses in a D.C. Machine
2. Iron or Core losses:
i) Hysteresis loss:
It is given by Steinmetz formula. This formula is
Hysteresis loss, Ph = η Bmax1.6f V watts
where Bmax = Maximum flux density in armature
f = Frequency of magnetic reversals
= NP/120 where N is in r.p.m.
V = Volume of armature in m3
η = Steinmetz hysteresis co-efficient
In order to reduce this loss in a d.c. machine, armature
core is made of such materials which have a low value of
Steinmetz hysteresis co-efficient e.g., silicon steel.
 Losses in a D.C. Machine
2. Iron or Core losses:
ii) Eddy current loss :
➢ In addition to the voltages
induced in the armature
conductors, there are also
voltages induced in the armature core.
➢ These voltages produce circulating currents in the
armature core as shown in fig..
➢ These are called eddy currents and power loss due to
their flow is called eddy current loss.
➢ The eddy current loss appears as heat which raises the
temperature of the machine and lowers its efficiency.
 Losses in a D.C. Machine
2. Iron or Core losses:
ii) Eddy current loss :
Eddy current loss is given by,
Pe = KeBmax2f 2 t 2 V watts
Where, Ke = Constant depending upon the electrical
resistance of core and system of units used
Bmax = Maximum flux density in Wb/m2
f = Frequency of magnetic reversals in Hz
t = Thickness of lamination in m
V = Volume of core in m3
Note: Eddy current loss depends upon the square of
thickness. To reduce Eddy current losses, lamination of
core is done as shown..
 Losses in a D.C. Machine
3. Mechanical losses:
These losses are due to friction and windage.
➢ Friction loss e.g., bearing friction, brush friction
etc.
➢ Windage loss i.e., air friction of rotating
armature.
• These losses depend upon the speed of the
machine.
• But for a given speed, they are practically
constant.
 Losses in a D.C. Machine
Constant and Variable Losses:
Constant losses:
• Those losses in a d.c. machine which remain constant at all loads
are known as constant losses.
• The constant losses in a d.c. generator are:
• Iron losses
• Mechanical losses
• Shunt field losses
Variable losses
• Those losses in a d.c. machine which vary with load are called
variable losses. The variable losses in a d.c. machine are:
» A) Copper loss in armature winding ( I2R ) a

» B)Copper loss in series field winding ( Ise2 R ) se

• Total losses = Constant losses + Variable losses

• Note. Field Cu loss is constant for shunt and compound generators.

 Efficiency of a D.C. Machine
• The power that a d.c. machine receives is called the input and the power it
gives out is called the output.
• Therefore, the efficiency of a d.c. machine, like that of any energy-
transferring device, is given by;

Efficiency = Output
Input

• Output = Input - Losses and Input = Output + Losses

• Therefore, the efficiency of a d.c. machine can also be expressed in the

following forms:

Efficiency = Input - Losses

Input

Efficiency = Output + Losses

Output
 Efficiency of a D.C. Machine
Power Stages:
➢The various power stages in a d.c. generator
are represented diagrammatically in Fig..
➢ A - B = Iron and friction losses
➢ B - C = Copper losses
 Efficiency of a D.C. Machine
Power Stages:
➢Overall efficiency = C/A
➢Mechanical efficiency = B/A
➢Electrical efficiency = C/B
Condition for Maximum Efficiency
 Methods of Testing
➢Direct Testing : Brake Test
➢Indirect Testing : Swinburne’s test
➢Regenerative testing : Hopkinson’s test
 Brake Test
Direct Testing
➢In this method, a brake is applied to a water-
cooled pulley mounted on the motor shaft as
shown in Fig.
➢One end of the rope is fixed to the floor via a
spring balance S and a known mass is
suspended at the other end.
➢If the spring balance reading
is S kg-Wt and the suspended
mass has a weight of W kg-Wt,
then,
 Brake Test
Net pull on the rope = (W - S) kg-Wt = (W - S) x 9.81 newtons
then the shaft torque Tsh developed by the motor is

Tsh = (W - S) x 9.81 x r N-m

If the speed of the pulley is N r.p.m., then,

Output power = 2Π N Tsh = 2 Π N x (W - S) x 9.81x r watts

60 60

Let V = Supply voltage in volts

I = Current taken by the motor in amperes
Input to motor = V I watts
Efficiency = 2 Π N(W - S) x r x 9.81
60 x VI
 Swinburne’s Test
In-Direct Testing:
➢In this method, the d.c. machine (generator or
motor) is run as a motor at no- load and
losses of the machine are determined.
➢Once the losses of the machine are known, its
efficiency at any desired load can be
determined in advance.
➢It may be noted that this method is applicable
to those machines in which flux is practically
constant at all loads e.g., shunt and compound
machines.
 Swinburne’s Testing
Step-1: Determination of hot resistances of windings
➢ The armature resistance and shunt field resistance
are measured using a battery, voltmeter and
ammeter.
➢ These resistances are measured when the machine is
cold, they must be converted to values
corresponding to the temperature at which the
machine would work on full-load.
➢ Generally 40°C or above the room temperature.
 Swinburne’s Testing
Step-2: Determination of constant losses
➢The machine is run as a motor on no-load with
supply voltage adjusted to the rated voltage
i.e. voltage stamped on the nameplate.
➢The speed of the motor is adjusted to the
rated speed with the
help of a field
regulator R as
shown is Fig
 Swinburne’s Testing
Step-2: Determination of constant losses
➢ Let V = Supply voltage
➢ I0 = No-load current read by ammeter A1
➢ Ish = Shunt-field current read by ammeter A2
➢ No-load armature current, Ia0 = I0 - Ish
➢ No-load input power to motor = V I0
➢ No-load power input to armature = V Ia0
= V(I0 - Ish)
➢ Since the output of the motor is zero, the no-load input
power to the armature supplies
(a) iron losses in the core
(b) friction loss
(c) windage loss
(d) armature Cu loss
 Swinburne’s Testing
Step-2: Determination of constant losses
➢ Constant losses, WC = Input to motor - Armature Cu loss
or
WC = V I0 - (I0 - Ish )2 Ra

➢ Since constant losses are known, the efficiency of the

machine at any other load can be determined.
➢ Suppose it is desired to determine the efficiency of the
machine at load current I.
➢ Then,
Armature current, Ia = I – Ish ... For Motor
= I + Ish ... For Generator
 Swinburne’s Testing
Step-3: Determination of Efficiency
Efficiency when running as a motor
Input power to motor = VI
Armature Cu loss = Ia2Ra = (I - Ish)2 Ra
Constant losses = WC ….. found above
Total losses = (I - Ish )2 Ra + WC
Motor efficiency, η m = Input - Losses = VI - (I - Ish )2 Ra - WC
Input VI
Efficiency when running as a generator
Output of generator = VI
Armature Cu loss = (I + Ish )2 Ra
Constant losses = WC …… found above
Total losses = (I + Ish )2 Ra + WC

Generator efficiency,
ηg = Output = VI
Output + Losses VI + (I + Ish )2 Ra - WC
 Swinburne’s Testing
Advantages:
➢ The power required to carry out the test is small because it is
a no-load test. Therefore, this method is quite economical.
➢ The efficiency can be determined at any load because
constant losses are known.
➢ This test is very convenient.
Disadvantages:
➢ It does not take into account the stray load losses that occur
when the machine is loaded.
➢ This test does not enable us to check the performance of the
machine on full-load. Like, whether commutation on full- load
is satisfactory and whether the temperature rise is within the
specified limits etc.,
➢ This test does not give quite accurate efficiency of the
machine. It is because iron losses under actual load are
greater than on No-load.
Regenerative or Hopkinson’s-Test
This method of determining the efficiency cf a d.c. machine saves power and gives more accurate results.
In order to carry out this test, we require two identical d.c. machines and a source of electrical power.

Circuit Connections:
➢ Two identical d.c. shunt machines are mechanically coupled and are connected in parallel across
the d.c. supply.
➢ By adjusting the field strengths of the two machines, the machine M is made to run as a motor
and machine G as a generator.
➢ The motor M draws current I1 from the generator G and current I2 from the d.c. supply so that
input current to motor M is (I1 + I2).
➢ Power taken from the d.c. supply is VI2 and is equal to the total motor and generator losses. The
field current of motor M is I4
and that of generator G is I3.
Regenerative or Hopkinson’s-Test
Operating Procedure:
➢ Two identical d.c. shunt machines are mechanically coupled and connected in parallel across the
d.c. supply.
➢ By adjusting the field excitations of the machines, one is run as a motor and the other as a
generator.
➢ The electric power from the generator and electrical power from the d.c. supply are fed to the
motor.
➢ The electric power given to the motor is mostly converted into mechanical power, the rest going
to the various motor losses.
➢ This mechanical power is given to the generator.
➢ The electrical power of the generator is given to the motor except that which is wasted as
generator losses.
➢ Thus the electrical power taken from the d.c. supply is the sum of motor and generator losses and
this can be measured directly by a voltmeter and an ammeter.
➢ Since the power input from the d.c. supply is equal to the power required to supply the losses of
the two machines, this test can be carried out with a small amount of power.
➢ By adjusting the field strengths of the machines, any load can be put on the machines.
➢ Therefore, we can measure the total loss of the machines at any load.
➢ Since the machines can be tested under full-load conditions (of course at the expense of power
equal to the losses in the two machines), the temperatures rise and commutation qualities of the
machines can be observed.
Regenerative or Hopkinson’s-Test
Alternate Connections for simplified calculations:
➢ The Fig. shows the alternative connections for Hopkinson’s test. The main difference is that now
the shunt field windings are directly connected across the lines.
➢ Therefore, the input line current is I1, excluding the field currents.
➢ The power VI1 drawn from the d.c. supply is equal to the total losses of the two machines except
the shunt field losses of the two machines
V I1= Total losses of two machines except shunt field losses of two machines
Regenerative or Hopkinson’s-Test
Calculations:
➢ Motor armature Cu loss = (I1 + I2 )2 Ra
Generator armature Cu loss = I22Ra
➢ Iron, friction and windage losses of the two machines are
➢ VI1 minus armature Cu losses of the two machines i.e..
➢ Iron, friction and windage losses of the two machines
➢ = VI1 - [(I1 + I2 )2 Ra + I22Ra ]= W
➢ Iron, friction and windage losses of each machine = W/2
Regenerative or Hopkinson’s-Test
Calculation of Efficiency:
Motor efficiency
Motor input, Pi = V(I1 + I2 + I3 )
Motor losses = (I1 + I2 )2 Ra + VI3 + W/2 = Wm (say)
Motor efficiency,
ηm = Motor input - Losses = Pi – Wm
Motor input Pi
Generator efficiency
Generator output = VI2
Generator losses = I22R + VI4 + W/2 = Wg
Generator efficiency,
ηg= VI2 /(VI2 + Wg )
Regenerative or Hopkinson’s-Test
Advantages of Hopkinson’s Test
➢ The total power required to test the two machines is small
compared with the full-load power of each machine.
➢ The machines can be tested under full-load conditions so that
commutation qualities and temperature rise can be checked.
➢ It is more accurate to measure the loss directly than to measure
it as the difference of the measured input and output.
➢ All the measurements are electrical which are simpler and more
accurate than mechanical measurements.

The main disadvantage is that two similar D.C. machines are

required.