eee lab manual

8/13/2019 eee lab manual

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UNIT - 4TESTING OF DC MACHINES

Testing of DC machines can be broadly classified as

i) Direct method of Testingii) Indirect method of testing

DIRECT METHOD OF TESTING:

In this method, the DC machine is loaded directly by means of a brake applied to a water cooled

pulley coupled to the shaft of the machine. The input and output are measured and efficiency is

determined by =

. It is not practically possible to arrange loads for machines of large

capacity.

INDIRECT METHOD OF TESTING:

In this method, the losses are determined without actual loading the machine. If the losses are

known, then efficiency can be determined. Swinburnes test and Hopkinsons test are commonly

used on shunt motors. But, as series motor cannot be started on No-load,these tests cannot be

conducted on DC series motor.

(i) BRAKE TEST: is a direct method of testing.In this method of testing motor shaft is coupled to a

Water cooled pulley which is loaded by means of weight

as shown in figure4.1.

= suspended weight in kg

=Reading in spring balance in kg

R = radius of pulley

N = speed in rps

V = Supply voltage

I = Full Load Current

Net pull due to friction = ( kg

= 9.81 ( Newton . 1

Shaft torque =


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= 9.81 ( 2

Motor output power = Watt

= watts . 3

Or 9.81 ( watt.

= 61.68 N ( .. 4

Input power = VI watts . 5

Therefore efficiency =

=

6

This method of testing can be used for small motors only because for a large motor it is difficult to

arrange for dissipation of heat generated at the brake.

(ii)Swinburnes Test:

This test is a no load test and hence cannot be performed on series motor. The circuit connection is

shown in Figure 4.2. The machine is run on no load at rated speed which is adjusted by the shunt

field resistance.

ADVANTAGES

1. Economical, because no load input power is sufficient to perform the test2. Efficiency can be pre-determined3. As it is a no load test, it cannot be done on a dc series motor


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DISADVANTAGES

1. Change in iron loss from no load to full load is not taken into account. (Because of armaturereaction, flux is distorted which increases iron losses).

2. Stray load loss cannot be determined by this test and hence efficiency is over estimated.3. Temperature rise of the machine cannot be determined.4. The test does not indicate whether commutation would be satisfactory when the machine is

loaded.

IO= No load current; Ish= shunt field current

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

V= Supply Voltage

No load input =VIowatts.

No load power input supplies

(i) Iron losses in the core(ii) Friction and windings loss and(iii) Armature copper loss.

Let I = load current at which efficiency is required

Ia= I Ishif machine is motoring; I + Ishif machine is generating

Efficiency as a motor:

Input = VI; Ia2ra= (I- Ish)

2ra

Constant losses Wc= VIo (Io Ish)2ra 7

Total losses = (I- Ish)2ra+ Wc

Therefore efficiency of motor =

; =

. 8

EFFICIENCY OF A GENERATOR:

Output = VI

Ia2ra= (I+Ish)

2ra

Total losses = Wc+(I+Ish)2ra.. 9

Efficiency of generator =

=

10


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1. A 220 V DC shunt motor at No-load takes a current of 2.5 A. the resistance of the armatureand shunt field are 0.8and 200respectively. Estimate the efficiency of the motor when the

input current is 32 A.

SOLUTION:No-load input = 220 x 2.5 = 500 W

= 1.1 A; 2 = ;=

X 0.8 = 1.6 W

Constant losses = 550 1.6 = 548.4 W

When input current is 32 A, Ia= 32 1.1 = 30.9 A

=

8 = ; Total losses = 764 + 548.4 = 1315 W

Input = 220 X 32

Therefore output = (220 X 32) 1315 W

Therefore =

= 81.3 %

2. A 440V D.C Shunt motor takes no load current of 25A. The resistance of shunt field andarmature are550 & 1.2 respectively. The full load line current is 32A. Determine the full

load output &efficiency of the motor.

LI:LI:LI:LI:input on no load= V=4405=00W= =

= 08A; = , therefore, = = 508=7A

= = (7=348W = = 00348=0953W=3A, = 308=3A; Armature copper loss = (3 = 88 Total losses = 88 0953 =4W = =

(( 00 = 839 %

2. A 500 V DC shunt motor when running on No-load takes 5 A. Armature resistance is 0.5and

shunt field resistance is 250. Find the output in kW and of the motor when running on full

load and taking a current of 50 A.


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(iii) Hopkinsons Or Regenerative Or Back To Back Test:

This is a regenerative test in which two identical DC shunt machines are coupled mechanically and

tested simultaneously. One of the machines is run as a generator while the other as motor supplied by

the generator. The set therefore draws only losses in the machines. The circuit connection is shown

in Figure 4.3. The machine is started as motor and its shunt field resistance is varied to run the motor

at its rated speed. The voltage of the generator is made equal to supply voltage by varying the shunt

field resistance of the generator which is indicated by the zero reading of the voltmeter connected

across the switch. By adjusting the field currents of the machines, the machines can be made to

operate at any desired load with in the rated capacity of the machines

ADVANTAGES:

i. The two machines are tested under loaded conditions so that stray load losses areaccounted for.

ii. Power required for the test is small as compared to the full load powers of the twomachines. Therefore economical for long duration tests like Heat run tests.

iii. Temperature rise and commutation qualities can be observed.iv. By merely adjusting the field currents of the two machines the two machines can be

loaded easily and the load test can be conducted over the complete load range in a short

time.

DISADVANTAGES:

i. Availability of two identical machinesii. Both machines are not loaded equally and this is crucial in smaller machines.iii. There is no way of separating iron losses of the two machines which are different because

of different excitations.

iv. Since field currents are varied widely to get full load, the set speed will be greater thanrated values.

The efficiency can be determined as follows:


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CIRCUIT DIAGRAM

V= supply voltage

Motor input = V(I1+I2)

Generator output = VI1 ----------------- (a)

If we assume both machines have the same efficiency , then,

Output of motor = x input = x V (I1+I2) = input to generator

Output of generator = x input = x V (I1+I2) = 2V(I1+I2)-----(b)

Equating (a) and (b),

VI1= 2V(I1+I2)

Therefore, = 11

Armature copper loss in motor = (I1+ I2 I4)2ra

Shunt field copper loss in motor = VI4

Armature copper loss in generator = ( Shunt field copper loss in generator = VI3

Power drawn from supply = VI2

Therefore stray losses = VI2- ( ( = W (say) 12


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Stray losses/motor = 13

Therefore for generator

Total losses = ( = Wg(say) . 14

Output = VI1, therefore generator=

=

.. 15

For motor,

Total losses =( = (Input to motor = V( Therefore motor=

(( 16


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ALTERNATIVE CONNECTION:

The Figure 4.4 shows an alternate circuit connection for this test. In this connection the shunt field

windings are directly connected across the lines. Hence the input current is excluding the field

currents. The efficiency is determined as follows:

Motor armature copper loss =(Generator armature copper loss = Power drawn from supply = VI1

Stray losses = (= W(say) .. 17Stray loss/motor =

. 18

MOTOR EFFICIENCY: motor input = armature input + shunt field input

= V( Motor loss = Armature copper loss + Shunt copper loss + stray losses = ( .. 19

Therefore motor= . 20

Generator efficiency : Generator output = VI2

Generator losses = 21

generator=

. 22


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1. The Hopkinsons test on two similar shunt machines gave the following Full load data.Linevoltage = 110 V; Line current = 48 A; Motor armature current = 230 A; Field currents are 3 A and

3.5 A; Armature resistanceof each machine is 0.035 ; brush drop of 1V/brush; Calculate the

efficiency of each machine.

SOLUTION:Motor:Armature copper loss = ( = 1851.5 WBrush contact loss = 230 X 2 = 460 W

Total armature copper loss = 1851.5 + 460 = 2312 W

Shunt field copper loss = 110 X 3 = 330 W

Total copper loss = 2312 + 330 = 2642 W

Generator: Generator armature copper loss = (188.5)2x 0.035 = 1244 W

Brush contact loss = 188.5 X 2 = 377 W

Total armature copper loss = 1244 + 377 = 1621 W.

Shunt field copper loss = 110 X 3.5 = 385 W

Therefore total copper loss = 1621 + 385 = 2006 W

Total copper loss for set = 2642 + 2006 = 4648 W

Total input = 110 X 48 = 5280 W

Therefore stray losses = 5280 4648 = 632 W

Stray losses/machine =

Total losses = 2312 + 330 + 316 = 2958 W

Motor input = 110 X 233 = 25630 W. Motor output = 22672 W

motor=

= 88.45%

generator: total losses = 1621 + 385 +316 = 2322

Output of generator = 110 X 185 = 20350 W.

Therefore =

= 89.4%


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2. In a Hopkinsons test on a pair of 500 V, 100 kW shunt generator. The following data was

obtained:Auxiliary supply 30 A at 500 V; Generator output current 200 A; Field current 3.5 A and

1.8 A; ra = 0.075 for each machine; voltage drop at brushes = 2 V/machine; calculate the

efficiency of the machine as a generator.

SOLUTION:

Motor armature copper loss = (230)2X 0.075 + 230 X 2 = 4428 W

Motor field copper loss = 500 X 1.8 = 900 W

Generator armature copper loss = (200)2X 0.075 + 200 X 2 = 3400 W

Generator field copper loss = 500 X 3.5 = 1750 W.

Total copper loss for 2 machines = 4428 + 900 + 3400 + 1750 = 10478 W

Power drawn = 500 X 30 = 15000 W

Therefore stray loss for the two machines = 15000 10478 = 4522 W.

Stray loss / machine =

= 2261 W

Therefore total losses in generator = 3400 + 1750 + 2261 = 7411 W

Generator output = 500 X 200 = 100000 W

Therefore generator=

= 93.09 %

3. In a Hopkinson test on 250 V machine, the line current was 50 A and the motor current is 400 A

not including the field currents of 6 and 5 A. the armature resistance of each machine was 0.015.

Calculate the efficiency of each machine.


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SOLUTION:

Motor armature copper loss = (400)2X 0.015 = 2400 W

Generator armature copper loss = (350)2X 0.015 = 1838 W

Input power = 250 X 50 = 12500 W

Ws= stray losses = 12500 (2400 + 1838) = 8262 W; W sper machine =

Motor efficiency:armature copper loss of motor = 2400 W;

Motor field copper loss = 250 X 5 = 1250 W; Total motor losses = 2400 + 1250 + 4130 = 7780 W

Motor input = 250 X 400 + 250 X 5 = 101250 W.

Therefore =

= 92.3 %

Generator efficiency:

Generator armature copper loss = 1838 W;Generator field copper loss = 250 X 6 = 1500 W

Total losses of Generator = 1828 + 1500 + 4130 = 7468 W

Generator Output = 250 X 350 = 87500 W

Efficiency of Generator =

4. The following test results were obtained while Hopkinsons test was performed on two similar DC

shunt machine. Supply voltage = 250 V; Motor Field Current = 2 A; Generator Field Current = 2.5A;

Generator Armature Current = 60 A; Current taken by the two armature from supply = 15 A;

Armature Resistance/Machine = 0.2; Calculate the efficiency of motor and generator under these

conditions.


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(iv) Field test for series motor:

Figure 4.8 shows the circuit for fields test. This test is applicable to two similar series motor. One of

the machine runs as a motor and drives a generator whose output is wasted in a variable load R.

Both machine field coils are in series and both run at same speed so that iron and friction losses are

made equal.

Load resistance R is varied till the motor current reaches its full load value.

V = Supply voltage

I1 = Motor current

V2= Generator terminal voltage

I2= Load current

Input = VI1and output = V2I2

Raand Rse= hot resistances.

Total losses in the set Wt= VI1 V2I2.... 23

Armature and Field copper losses Wc= (Ra+ 2 rse) I12+ Ia

2Ra 24

Stray losses forthe set = Wt Wc..25


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SOLUTION:

Total input = 630 X 56 = 35280 W

Output = 400 X 44 = 17600 W

Total losses = 35280 17600 = 17680 W

Rse= = 0.714

Total copper loss = (0.3 + 2 X 0.714) 562+ 44

2X 0.3 = 6006 W

Therefore stray losses = 17600 6006 = 11674 W

Stray losses per motor =

= 5837 W

Motor efficiency: motor input = 590 X 56 = 33040 W

Armature circuit copper loss = (0.714 + 0.3) 562= 3180 W

Stray losses = 5837 W

Total losses = 3180 + 5837 = 9017 W

Therefore efficiency =

00 = 77%Generator efficiency: Armature copper loss = 0.3 X 44

2= 581 W

Series field copper loss = 40 X 56 = 2240 W.

Stray losses = 5837 W

Therefore total losses = 581 + 2240 + 5837 = 8658 W

Output = 400 X 44 = 17600 W

Therefore generator=

00 = 7%3 A fields test on two similar series machines gave the following data; Motor: Armature current:60A; Voltage across armature: 500V; Voltage across field: 40V; Generator: Terminal voltage: 450V;

Output current 46A; Voltage across field:40V; Armature resistance of each machine is 0.25.

Calculate the efficiency of both the machines.SOLUTIONSOLUTIONSOLUTIONSOLUTION::::

Total input = 58060=34800W

Output =45046=20700W

Therefore total losses in 2machines = 34800-20700=14100W


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Total cu loss = (0.25+ 20.67)(60)2+ (46)

2=6253W

Stray losses of the set = 14100-6253= 7847W

Stray loss per machine =

=3923.5W

Efficiency of Motor;

Motor input= motor armature voltage armature current =540 60 =32400W

Armature cu loss = (0.25 + 0.67) (60)2= 3312W

Total losses = 3312 + 3923.5 =7235.5W

ofmotor= 324007235532400

7767%

Efficiency of Generator;generator output= 450 46 =20700W

Armature cu loss = (0.25) (46)2= 529 W

Series field cu loss = 40 60 =2400W

Total losses = 2400 + 529 + 3923.5 =6852.5W

ofgenerator20700

2070068525 100 751%


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(V) Retardation or running down test:

This method is applicable to shunt motors and generators and is used for finding the stray losses. If

armature and shunt copper losses are known for a given load, efficiency can be calculated. The

circuit is shown in figure 4.11.

Mach pdd p h bd rad pd ad h pp c r harar h p h d xcd Arar d ad c r dd raa rc ad da

Kc r h arar = I = M ra h arar

= Aar cRaa ;N = Ra KERa Kc r W =

I X W 30T qa d b (((( M Ira IM Ira IM Ira IM Ira I(((( rrrr ( bca( bca( bca( bca (i) FdFdFdFd :The voltmeter V in the circuit shown in Figure 4.11 is used as speed indicator

by suitably graduating it because E when the supply is cut off, the armature speed and


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W W1 = ( IN 36; dN a

=ra cha pd h crca ad=ra cha pd h crca ad

=

37 r

=

r W= W1

r W= W1

. 38

1 A rarda ad a para xcd DC ach a a r h dcda a r 0V 5V 5 c p h arar crc & c dd cha h arar cc r pp a ad rac a 10A(ara Fd h cc h ach h r a a r& a a crr 5A a pp 50V Th rac ar 0 & ha d d 50SOLUTION:SOLUTION:SOLUTION:SOLUTION:

Ara a acr ad = 05 = 5Ia= 10APr abrbd = = 5 X 10 = 51=0c & = c W = ra Thrr W = W X ; hrr W = 5 X

= 1

Ip crr = 5A; Ih= = 1A hrr Ia= 5 1= AArar Crr = (X 0 = 0

Sh crr = 50 X 1 = 50Thrr a = 1 0 50 = 115 Ip = 5 X 5 = 50 p = 50 115 = 505Thrr cc = = 080 r 80 I a rarda a para r h dcd h arar a r 0V 190V 0cd dcc h arar r h pp Th a a a pac 0cd da ar dcc arar ccd a rac hcha 10A (ara dr h a d h ra rSOLUTION:SOLUTION:SOLUTION:SOLUTION:W = ra Ara a acr rac = = 195Ara crr = 10A; Pr abrbd W = 1950=

; W = 1950

= 900


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(A rarda carrd a 1000rp DC ach Th a r h pd a r 100 rp 90rp (a c h xca (b 15c h xca& (c 9 cd h xca & h arar pp a xra ad 10A a 19VCaca (1 Th ra h arar ( Ir ad (Thchaca a h a pd 1000rp

SOLUTION:SOLUTION:SOLUTION:SOLUTION:( Wh h ar d h xca c r d r cchaca ( x Ir

( Wh xca Kc Er d pp chaca & r ra ( I M Ira h ra r WaMchaca W=

I N dN = 10090 = 0rp d = cd N = 1000rp

W=

I Na (1

Sar W= I N a (

W= X = 19 X 10 X

= 85 a

U qa ( 85 = 1000 ; I = 5

Dd (1 b ( =

W= 85 X = 19 a

& r =WW= 8519 = 191a I a rarda a DC r h d ra xcd h pd r 155 15 rp 5cd Wh a ara ad 10 ppd b h arar h apd drp ccrrd 0cd Fd h M Ira h ra par SOLUTION:SOLUTION:SOLUTION:SOLUTION:W = I N

& W =W X

W= 1 = 1000a & 1= 5 cd& = 0cdW = 1000 X = 000N = 1500rp (ara pd; dN = 15515 = 50rp

d = 5 ; 000 = 1500 X

I = 118

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