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To determine the efficiency of a dc motor by loss summation method(Swinburne's test)

                                                            Swinburne test

                    Loss summation method in DC shunt machines



1.  Introduction :


Swinburne‟s test is the most commonly used method for testing of D.C machines. It is an indirect method of testing D.C machines. In this method, the no load losses of the machine are determined experimentally and the additional losses on load are estimated from the known data of the machine, and with the help of the losses and input power the efficiency at any desired load is predetermined.



2.  Losses :


We are using the term machine in the discussion of power losses owing to the fact that no distinction need be made between the losses in the D.C generator and the motor. The law of conservation of energy dictates that the input power must always be equal to the output power plus the losses in the machine. There are three major categories of losses:


1)  Mechanical losses


2)  Magnetic losses


3)  Copper losses


2.1) Mechanical losses


Mechanical losses are the result of :


(a)    The friction between the bearings and the shaft,


(b)    The friction between the brushes and the commutator, and


(c)    The drag on the armature caused by air enveloping the armature (windage loss).


The bearing-friction loss depends upon the diameter of the shaft at the bearing, the shaft‟s peripheral speed, and the coefficient of friction between the shaft and the bearing. To reduce the coefficient of friction, the bearing are usually lubricated.


The brush-friction loss depends upon the peripheral speed of the commutator, the brush pressure, and the coefficient of friction between the brush and the commutator. The graphite is the brush helps provide lubrication to lessen the coefficient of friction.


The windage loss depends upon the peripheral speed of the armature, the number of slots on its periphery, and its length.


Mechanical losses due to friction and windage Pfw can be determined by rotating the armature of an unexcited machine at its rated speed by coupling it to a calibrated motor. Because there is no power output, the power supplied to the armature is the mechanical loss.



2.2) Magnetic loss


Since the induce E.M.F. in the conductors of the armature alternates with a frequency determined by the speed of rotation and the number of poles, a magnetic loss Pm (hysteresis and eddy-current) exists in the armature.


The hysteresis loss depends on the frequency of the induced E.M.F., the area of the hysteresis loop, the magnetic flux density, and the volume of the magnetic material. The area of the hysteresis loop is smaller for soft magnetic materials. This is one of the reasons why soft magnetic materials are used for electrical machines.


Although the armature is build using thin laminations, the eddy currents do appear in each lamination and produce eddy-current loss. The eddy-current loss depends upon the thickness of the lamination, the magnetic flux density, the frequency of the induced E.M.F. and the volume of the magnetic material.


A considerable reduction in the magnetic loss can be obtained by operating the machine in the linear region at a low flux density at the expense of its size and initial cost.



Rotational lossesIn the analysis of a D.C. machine, it is a common practice to lump the mechanical loss and the magnetic loss together. The sum of the two losses is called the rotational loss, Pr.

i.e. Pr  = Pfw  + Pm


The rotational loss of a D.C. machine can be determined by running the machine as a separately excited motor under no load. The armature winding voltage be so adjusted that the induced E.M.F. in the armature winding is equal to its rated value, Ea . If Vt is the terminal voltage and Ra is the armature-winding resistance, then the voltage that must be applied to the armature terminals is should


Va = Vt - IaRa For the motor.


Apply Va across the armature terminals and adjust the field excitation until the machine rotates at its rated speed. Then measure the armature current. Because the armature current under no load is small fraction of its rated value and the armature winding resistance is usually very small, we can neglect the power loss in the armature winding. As there is no power output, the power supplied to the motor, VaIa, must be equal to the rotational loss in the machine. By subtracting the mechanical loss, we can determine the magnetic loss in the machine.



2.3) Copper losses


Whenever a current flows in a wire, a copper loss, Pcu, is associated with it. The copper losses, also known as electrical or I2R losses, can be segregated as follows:


1.  Armature-winding loss


2.  Shunt field-winding loss


3.  Series field- winding loss


4.  Interpole field- winding loss


5.  Compensating field-winding loss

Stray-load loss


A machine always has some losses that cannot be easily accounted for; they are termed as stray-load losses. It is suspected that the stray-load losses in the D.C. machine s are the result of (a) the distorted flux due to armature reaction and (b) short-circuit currents in the coils undergoing commutation. As a rule of thumb, the stray-load loss is assumed to be 1% of the power output in large machines (above 100 horse power) and can be neglected in small machines.


3.Power flow diagram:


4. Circuit diagram of Swinburne test:








   We should basically find out the constant losses as follows -

   V = D.C supply voltage

    Io = no-load line current

    Ish = shunt field winding current

    Iao = no-load armature current = (Io-Ish)

    Pao = no-load armature copper losses

    No-load input power =VIo

    Constant losses (Pc) =Ps+Pf
     Pc = VIo – Pf – Pao + Pf


   Constant losses (Pc) = VIo - Pao


6. Calculation of total losses:

Let us consider,

we have to find out the efficiency of the D.C. shunt motor at the load

current of „I‟ amperes

then, Armature copper losses = I2aRa = (I – Is)2Ra

Constant losses = Pc (we found them above)

Therefore, total losses = Armature copper losses + constant losses

Total losses = (I - Ish)2Ra + Pc


7. Efficiency:

Efficiency = (output / input )*100

Input to the Motor (Total power) = VI

Motor efficiency

Ƞm = 1 – losses / input


8. Advantages of Swinburne’s test:

The main advantages of Swinburne‟s test are:

1. It is a convenient and economical metod of testing D.C. machines since power required to test a large machine is small.

2. The efficiency can be predetermined at any load because constant losses are known.


9. Main Disadvantages:

1. No account is taken of the change in iron loss from no-load to full load. At full load, due to armature reaction, flux is distorted which increases the iron losses.

2. As the test is on no load, it does not indicate whether the commutation on full load is satisfactory and whether the temperature rise would be within specified limits.


10. Limitations:

1. Swinburne‟s test is applicable to those machines in which the flux is practically constant, that is shunt machines and level compound generators.

2. Series machines cannot be tested by this metod as they cannot be run on light loads and secondly speed and flux vary greatly.


11. References:

1. Dr. Bhag S.Guru, “Electric Machinery and transformers”, oxford university  press, 3rd edition.

2. D.R. Kohli-S.K Jain, “A laboratory course in ELECTRICAL MACHINES” Nem Chand & Bros., 2000.




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