TESTING OF DC MACHINES

PROF. SWAPNIL JANI

INSTITUTE OF TECHNOLOGY

NIRMA UNIVERSITY

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Reference Book:

B.L. Theraja

ELECTRICAL TECHNOLOGY

PART-2 : AC, DC MACHINES

Some Basic

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EFFICIENCY OF DC MACHINE

• Efficiency can be determine by finding losses in the equipment, inputgiven and the output using below equation:

𝐸𝑓𝑓𝑖𝑐𝑖𝑒𝑛𝑐𝑦 𝑖𝑛 % =𝑜𝑢𝑡𝑝𝑢𝑡

𝑜𝑢𝑡𝑝𝑢𝑡 + 𝑙𝑜𝑠𝑠𝑒𝑠∗ 100

• The efficiency is calculated at 25%, 50%, 75%, 100% and 125% ratedload.

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Losses in the Equipment

• Friction and Windage Loss

• Iron Loss

• Armature Copper Loss

• Series Winding Copper Loss

• Inter-pole Winding Copper Loss

• Brush Contact Loss

• Shunt Field Copper Loss

• Stray Load Loss

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• Friction and Winding Loss: Separated by light load test

• Iron Loss: By separation of iron and friction loss by light load test

• Armature Copper Loss:

- Measure armature resistance

- Convert that to at 75˚C

- Find armature current at the load required

- Calculate 𝐼2 ∗ 𝑟𝑎

• Series Winding Copper Loss:

- Measure series winding resistance

- Calculate 𝐼𝑠𝑒2 ∗ 𝑅𝑠𝑒 at 75˚C and at the load required

• Inter-pole Winding Copper Loss: Same procedure to calculate 𝐼𝑐𝑝2 ∗ 𝑟𝑐𝑝

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• Brush Contact Loss:

- 1 V per brush or a total 2 V voltage drop

- (2 V * Current) = Watts in brush contact loss

• Shunt Field Copper Loss:

- Measure shunt field resistance

- Convert it to 75˚C and find normal field current at the loading condition

- Calculate 𝐼𝑓2𝑅𝑓

OR shunt field copper loss = (voltage across shunt winding * field current)

• Stray Load Loss: It can’t be measured accurately, approximately it is taken 1%of output

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Brake Test

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Brake Test

• Direct Method

• In which applying brake to water cooled pulleymounted on the motor shaft as shown in fig.. Thismethod is mainly used for small DC motor.

• In which one end is connected with earth via springbalance and another with the suspended weight.And the load over the motor adjusted till it carriesthe full load current.

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Torque developed Tsh = (W1 – W2) * R kg-m or 9.81(W1-W2)R N-m

Output Power = Tsh * 2πN or Tsh * ω watt

Input Power = V * I watt

Efficiency =61.68 𝑁 𝑊1−𝑊2 𝑅

𝑉 𝐼

Where, W1 = Suspended weight in kg

W2 = Reading on spring balance in kg

R = Radius of the pulley in meter

N = motor or pulley speed in RPS || ω = Angular velocity

V = Supply Voltage || I = Full load current taken by motor

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• Previous brake test is applied onlyfor small DC motor as there isproblem of dissipation of largeamount of heat generated at thebrake for large DC motor.

• For that another method is usedwhich is called Pony Brake. Inwhich rope is wound round alongthe pulley and both ends areconnected to individual springbalance (S1 and S2).

Efficiency = 61.68 𝑁 𝑆1−𝑆2 𝑅

𝑉 𝐼

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• Here a rope is wound around the pulley and its two ends are attachedwith the spring balance 𝑆1 and 𝑆2.

• Tension on this rope can be adjusted by means of adjustment ofswivels. And the force acting tangentially on the pulley is equal to thedifference between the readings of spring balance.

• Force acting on the pulley

= Difference between reading of spring balance =

= reading of 𝑆1 - reading of 𝑆2

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Swinburne’s Test

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Swinburne’s Test

• It is also know as No load test or Losses Method.

• It is simple method in which losses are measured separately and fromtheir knowledge efficiency can be measured at any desired load.Separately test required is only running no load test.

• This test is only required where the flux in the stator is remainingpractically constant.

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• At no load, the speed is beingregulated to its rated speed byuse of shunt regulator.

No load current = 𝐼0(measured by ammeter 𝐴1)

Shunt Field Current = 𝐼𝑠ℎ(measured by ammeter 𝐴2)

No load armature current =𝐼0 - 𝐼𝑠ℎ= 𝐼𝑎0

If supply voltage=V then i/p=V ∗ 𝐼0Power i/p to armature = V * (𝐼0 - 𝐼𝑠ℎ)

Power i/p to shunt = V * 𝐼𝑠ℎ

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• No load power i/p to armature suppliesthe following:

1. Iron losses in core2. Friction Loss3. Windage Loss4. Armature Cu loss = (𝐼0 − 𝐼𝑠ℎ)2*𝑅𝑎 = 𝐼𝑎0

2*𝑅𝑎• While calculating the Cu loss ‘hot’

resistance need to be considered. Whichcan be find by measuring resistance atroom temperature and then calculateresistance for some temp. rise.

• Ex.: Room Temp. = 15 ˚C , ∝0= (1 234.5) and Temp. Rise = 50 ˚C

So, 𝑅15 = 𝑅0 ∗ 1 + 15 ∝0 and 𝑅65 = 𝑅0 ∗ 1 + 65 ∝0

𝑅65 = 1.2 * 𝑅15

(Here, armature current will be found decrease with increase on armatureresistance as shown in fig. due to brush contact resistance as it is inverselyproposal to armature current.)

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• Now subtracting the armature Cu loss it can be easily found out theconstant losses.

𝑊𝑐 = 𝑉 𝐼0 − 𝐼0 − 𝐼𝑠ℎ2 ∗ 𝑅𝑎

• After which at any load point efficiency can be find out.

For which, armature current 𝐼𝑎 = I − 𝐼𝑠ℎ(for motoring action)

= I + 𝐼𝑠ℎ (for generating action)

• Efficiency as a motor = Ƞ𝑚 =𝑖𝑛𝑝𝑢𝑡 −𝑙𝑜𝑠𝑠𝑒𝑠

𝑖𝑛𝑝𝑢𝑡=𝑉 𝐼 −( 𝐼 − 𝐼𝑠ℎ

2∗𝑅𝑎 ) −𝑊𝑐

𝑉 𝐼

• Efficiency as a generator = Ƞ𝑔 =𝑜𝑢𝑡𝑝𝑢𝑡

𝑜𝑢𝑡𝑝𝑢𝑡+𝑙𝑜𝑠𝑠𝑒𝑠=

𝑉 𝐼

𝑉 𝐼 + 𝐼+ 𝐼𝑠ℎ2∗𝑅𝑎 +𝑊𝑐

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Advantage

• This method is convenient andeconomical as it needs smallpower equal to no load inputpower.

• Efficiency can be predeterminedat any load condition asconstant losses are known.

Disadvantage

• No account of change in iron lossis taken into consideration fromno load to full load. At a fullload, due to armature reactionflux is being distorted whichincreases the iron losses in somecase it is being gone up to 50%.

• As in this test only no load test isdone so no data of satisfactorycommutation and temp. with inspecific limit can be found.

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Regenerative or Hopkinson’s Test

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Regenerative or Hopkinson’s Test (Back to Back Test)

• In this method full load test is being carried out. Between twoidentical ones shunt machines.

• Without wasting each others energy by using mechanicalcoupling between each other and electrical coupling is such thatone is being used as motor and another one as generator.

• So by which mech. o/p of motor is being use to drive thegenerator and electrical o/p of generator is being use to drive themotor.

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• Ideally each can fulfil demand of other but as due to losses there is need ofexternal force.

• Which can be fulfil by mechanically by attaching extra motor to beltconnection or by electrically supply from the mains.

• Procedure:o Machine M start from the supply mains and main switch S of the another machine is open.

o Speed is adjusted to normal value by means of field regulator.

o Machine M drives machine G as generator and its voltage can be measured from voltmeter 𝑉1.

o Voltage of machine G is adjusted by field regulator until voltmeter 𝑉1 reads zero, thereby voltage wll be same inmagnitude and polarity as that of the mains supply.

o Thereafter S is closed to parallel the machines.

o By adjusting the respective field regulators any load can be thrown on the machines.

o Gen. current 𝐼1 can be adjusted to any desired value by increasing the excitation of G or by reducing the excitationof M.

o The electrical o/p of the Gen. + the small power taken from the supply by the motor = mech. power aftersupplying the motor losses.

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oSupply voltage = V

oGenerator output = V 𝐼1oMotor input = V (𝐼1 + 𝐼2)

(Here 𝐼2 is current taken from the supply)

Assumption: Both machine have same efficiency (Ƞ)

oOutput of motor = Ƞ * i/p = Ƞ * V (𝐼1 + 𝐼2)

oOutput of generator = Ƞ * i/p = Ƞ * Ƞ * V (𝐼1 + 𝐼2) = Ƞ2* V (𝐼1 + 𝐼2)= Ƞ * V 𝐼1

Ƞ = 𝐼1

𝐼1+ 𝐼2

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However there can not be equal efficiency of both machine as thearmature current and excitation in both machines are different.

Let,𝑅𝑎 = 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑚𝑎𝑐ℎ𝑖𝑛𝑒𝐼3 = 𝑒𝑥𝑐𝑖𝑡𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑔𝑒𝑛𝑒𝑟𝑎𝑡𝑜𝑟𝐼4 = 𝑒𝑥𝑐𝑖𝑡𝑖𝑛𝑔 𝑐𝑢𝑟𝑟𝑒𝑛𝑡 𝑜𝑓 𝑡ℎ𝑒 𝑚𝑜𝑡𝑜𝑟

Armature Cu loss in generator = (𝐼1 + 𝐼3)2 ∗ 𝑅𝑎

Armature Cu loss in motor = (𝐼1 + 𝐼2 − 𝐼4)2 ∗ 𝑅𝑎

Shunt Cu loss in Generator = V * 𝐼3

Shunt Cu loss in Motor = V * 𝐼4

As Total generator and motor losses are equal to the power supplied by the mains.

Power drawn from supply = V * 𝐼2

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Stray Losses for the set:

= V * 𝐼2 − [ 𝐼1 + 𝐼32 ∗ 𝑅𝑎 + 𝐼1 + 𝐼2 − 𝐼4

2 ∗ 𝑅𝑎 + 𝑉 ∗ 𝐼3 +𝑉 ∗ 𝐼4

= 𝑊𝑠𝑡𝑟𝑎𝑦

Assumption: Stray loss is equally distributed between two machines.

So, stray losses per machine = 𝑊𝑠𝑡𝑟𝑎𝑦/2

For Gen.:

Total losses 𝑊𝑔 = 𝐼1 + 𝐼32 ∗ 𝑅𝑎 + 𝑉 ∗ 𝐼3 + (𝑊𝑠𝑡𝑟𝑎𝑦/2) Ƞ𝒈 =

V ∗ 𝑰𝟏

V ∗ 𝑰𝟏+𝑾𝒈

Output = V * 𝐼1

For Motor:

Total losses 𝑊𝑚 = 𝐼1 + 𝐼2 − 𝐼42 ∗ 𝑅𝑎+ 𝑉 ∗ 𝐼4 + (𝑊𝑠𝑡𝑟𝑎𝑦/2) Ƞ𝒎 =

V ∗ (𝑰𝟏+𝑰𝟐) −𝑾𝒎

V ∗ (𝑰𝟏+𝑰𝟐)Input = V * (𝐼1+𝐼2)

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Advantage

• Power required for to do thistest is small as compared to fullload test of the two machines.

• As in this test machines areworking on full load, temp. riseand commutation qualities canbe observed.

• Any change in iron loss due toflux distortion at the full load isbeing taken into account.

Disadvantage

• Two identical machines arerequired which is the maindisadvantage or demerit of thistest.

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Field’s Test

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Field’s Test

• Applicable only for two similar seriesmotors.

• Two machines are coupled mechanically,one machine runs normally as a motor anddrives a generator whose o/p is wasted invariable resistor.

• Here the iron and friction loss of twomachines made equal.

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• How to make iron and friction loss of the two machines equal forfield’s test?

oBy joining the series field winding of the generator in the motorarmature circuit so that both machines are equally excited.

oBy running them at equal speed.

• Adjust the value resistor R such that the reading of 𝐴1notes full loadcurrent.

• And note the value at various voltmeter and ammeter.

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V = Supply Voltage 𝑉2=Terminal potential drop of Generator

𝐼1= Motor Current 𝐼2 = Load Current

Intake of the whole set = 𝑉 ∗ 𝐼1 Output = 𝑉2 ∗ 𝐼2

Total losses in the set, 𝑊𝑡 = 𝑉 ∗ 𝐼1 − 𝑉2 ∗ 𝐼2

Armature and Field Cu losses, 𝑊𝐶𝑢 = 𝑅𝑎 + 2 𝑅𝑠𝑒 𝐼12 + 𝐼2

2𝑅𝑎

𝑅𝑎 = 𝐻𝑜𝑡 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑚𝑎𝑐ℎ𝑖𝑛𝑒

𝑅𝑠𝑒 = 𝐻𝑜𝑡 𝑠𝑒𝑟𝑖𝑒𝑠 𝑓𝑖𝑒𝑙𝑑 𝑟𝑒𝑠𝑖𝑠𝑡𝑎𝑛𝑐𝑒 𝑜𝑓 𝑒𝑎𝑐ℎ 𝑚𝑎𝑐ℎ𝑖𝑛𝑒

Stray losses for the set = 𝑊𝑡 − 𝑊𝐶𝑢

Stray losses/ machine, 𝑊𝑠 =𝑊𝑡 −𝑊𝐶𝑢

2(Assumption: Stray losses are equally divided)

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• Motor EfficiencyMotor input = 𝑉1𝐼1

Motor losses = armature + field Cu loss + Stray loss = (𝑅𝑎 + 𝑅𝑠𝑒)𝐼12 +𝑊𝑠 = 𝑊𝑚

Ƞ𝒎 =𝑽𝟏𝑰𝟏 − 𝑾𝒎

𝑽𝟏𝑰𝟏

• Generator EfficiencyGenerator output = 𝑉2𝐼2

Field Cu loss = 𝐼12𝑅𝑠𝑒 Armature Cu loss = 𝐼2

2𝑅𝑎Stray Losses = 𝑊𝑠

Total Losses = 𝐼12𝑅𝑠𝑒 + 𝐼2

2𝑅𝑎 +𝑊𝑠 = 𝑊𝑔

Ƞ𝒈 =𝑽𝟐𝑰𝟐

𝑽𝟐𝑰𝟐 + 𝑾𝒈

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Heat Run Test

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Heat Run Test:

• Back to back test is done for the temperature run also.

• Normal run for time for temperature run test for large machine is 6hours.

• As the time span to reach the steady state value of temp. is very longsome short time span test is being done in between and ultimatetemp. rise of the machine determined.

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Temp. Rise = 𝜃𝑡𝜃𝑡 = 𝜃𝑚 1 − 𝑒 −𝑡

𝜏

Let, 𝑒 −1𝜏 = 𝐾 Then, 𝜽𝒕 = 𝜽𝒎 𝟏 − 𝑲𝒕

When, t=1 𝜃1 = 𝜃𝑚 1 − 𝐾

t=2 𝜃2 = 𝜃𝑚 1 − 𝐾2

𝜃2

𝜃1= 1 + K → 𝐾 =

𝜃2−𝜃1

𝜃1

Max. Temp. Rise = 𝜃𝑚

𝜽𝒎 =𝜽𝟏

𝟏−𝑲

If initial temp. rise 𝜃0 = 0

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Successive interval between tworeadings is 30 min. or even smaller.

K is the ratio of increase in temp. rise during two successive interval of time

Cooling Curve

• Cooling curve is shown with heatingcurve, mirror image of the other,displaced by the 𝜃𝑚𝑎𝑥.

• Eq. for Temp. for cooling curve

𝜃𝑡 = 𝜃𝑚𝑎𝑥𝑒 −𝑡𝜏 = 𝜃𝑚𝑎𝑥𝐾

𝑡

• Constant will have the same constantvalue as of heating curve.

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• Graphical Method

• It is convenience to use graphical methodto predict the temp. rise from readingstaken from the short time test. In fig. Aand B are the points on temp. rise curvefor 𝜃1 and 𝜃2 at time t = 1 and 2 , where tis any convenient time interval.

• Construct QL and NC line intercept atpoint Q such that

𝑄𝐿 =𝐴𝑁

1 − 𝐾= 𝜃𝑚𝑎𝑥

• Simple in construction can also beapplied to any three consecutivereadings at equal time interval. No needof initial reading.

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Retardation Test

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Retardation Test or Running Down Test

• It is used for finding out straylosses of shunt wound DC motor.

• In this method, machine under testis being speed up slightly morethan normal speed and thensupply to the armature is cut off.

• Due to which armature slowsdown and its kinetic energy will beused for meeting rotational losses(Stray Loss).

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Kinetic energy of the armature =1

2𝐽 𝜔2

𝐽 = 𝑚𝑜𝑚𝑒𝑛𝑡 𝑜𝑓 𝑖𝑛𝑒𝑟𝑡𝑖𝑎 𝑜𝑓 𝑎𝑟𝑚𝑎𝑡𝑢𝑟𝑒

𝜔 = 𝑎𝑛𝑔𝑢𝑙𝑎𝑟 𝑠𝑝𝑒𝑒𝑑 𝑎𝑡 𝑡ℎ𝑒 𝑖𝑛𝑠𝑡𝑎𝑛𝑡

Rational losses, 𝑃𝑠 = 𝑟𝑎𝑡𝑒 𝑜𝑓 𝑐ℎ𝑎𝑛𝑔𝑒 𝑜𝑓 𝐾. 𝐸.=𝜕

𝜕𝑡

1

2𝐽 𝜔2 = 𝐽𝜔

𝜕𝜔

𝜕𝑡

𝜔 =2𝜋𝑁

60

Rotational Losses, 𝑃𝑠 = 𝐽2𝜋𝑁

60

𝜕

𝜕𝑡

2𝜋𝑁

60=

2𝜋

60

2𝐽𝑁

𝑑𝑁

𝑑𝑡

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N = Speed of armature in rpm at that instant

Determination of𝝏𝑵

𝝏𝒕

• Voltmeter connected across armature to find instantaneous back emf. Which isdirectly proposal to speed N.

• Voltmeter is calibrated such that it shows speed.

• When supply to armature is cut off speed of the motor decreases.

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• Speed or readings of the voltmeter is noted at differentinterval and draw graph of speed and time as shown infig.

• Draw tangent of any point lying on the curve (P) whichmeet X and Y axis at point A and B.

𝜕𝑁

𝜕𝑡=

𝑂𝐵 𝑖𝑛 𝑅𝑃𝑀

𝑂𝐴 𝑖𝑛 𝑠𝑒𝑐

Determination of Moment of Inertia

• First draw speed-time curve for which moment of inertia need tofound. Then key a flywheel with the shaft which has known momentof inertia 𝐽1 and again draw speed-time curve.

• From the above two curves determine𝜕𝑁

𝜕𝑡1and

𝜕𝑁

𝜕𝑡2.

• As due to flywheel in second case moment of inertia will be morethan in first case and losses will be same.

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Rotational losses in first case, 𝑃𝑠 =2𝜋

60

2JN

𝜕𝑁

𝜕𝑡1

Rotational losses in second case, 𝑃𝑠 =2𝜋

60

2(J + 𝐽1)N

𝜕𝑁

𝜕𝑡2

As the losses are same,

2𝜋

60

2JN

𝜕𝑁

𝜕𝑡1=

2𝜋

60

2(J + 𝐽1)N

𝜕𝑁

𝜕𝑡2

By simplifying above equation, 𝐽 =𝐽1

𝜕𝑁

𝜕𝑡1𝜕𝑁

𝜕𝑡1−

𝜕𝑁

𝜕𝑡2

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Second Method to determine of Moment of Inertia

• In this method, time is being noted down for slowing down of speed by 5%. Take it as 𝑡1sec.

• Now add an additional load known as retarding torque, mechanical or electrical is applied and same way note the time taken for slowing down by 5% (𝑡1sec).

• As shown in fig. of retardation test double throw switch is closed over terminal 1’ and 2’ to connect additional load just after disconnection of supply to armature.

• Which will make K.E. energy of armature to supply the power to non inductive resistor R (P’) along with the rotational loss.

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𝑃𝑠 =2𝜋

60

2

𝐽𝑁𝜕𝑁

𝜕𝑡1

So, 𝑃𝑠 + 𝑃𝑠′ =2𝜋

60

2𝐽𝑁

𝜕𝑁

𝜕𝑡2

By taking ration of above equation,

𝑃𝑠 + 𝑃𝑠′

𝑃𝑠=

𝜕𝑁𝜕𝑡2𝜕𝑁𝜕𝑡1

𝑃𝑠 = 𝑃𝑠′ ∗

𝜕𝑁𝜕𝑡1

𝜕𝑁𝜕𝑡2

−𝜕𝑁𝜕𝑡1

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