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1 REGULATION OF 3 PHASE ALTERNATOR BY EMF AND MMF METHOD EMF METHOD MMF METHOD

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Page 1: EMF METHOD - files.teceee-in.webnode.infiles.teceee-in.webnode.in/200000001-752dd76282/Circuit daigrams...ย ยท 1 regulation of 3 phase alternator by emf and mmf method emf method mmf

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REGULATION OF 3 PHASE ALTERNATOR BY EMF AND MMF METHOD

EMF METHOD

MMF METHOD

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FORMULA:

Synchronous Impedance: ๐‘ง๐‘  =๐ธ1

๐ผ๐‘ ๐‘ ohms

Where E1 = Open circuit voltage for a certain field current

Isc = Short circuit current for the same field current.

Synchronous Impedance: ๐‘‹๐‘  = โˆš๐‘๐‘ 2 โˆ’ ๐‘…๐‘Ž

2 Ohms

Lagging power factor: ๐ธ0 = โˆš(๐‘‰ cos โˆ… + ๐ผ๐‘…๐‘Ž)2 + (๐‘‰ sin โˆ… + ๐ผ๐‘‹๐‘ )2

Leading Power factor: ๐ธ0 = โˆš(๐‘‰ cos โˆ… + ๐ผ๐‘…๐‘Ž)2 + (๐‘‰ sin โˆ… โˆ’ ๐ผ๐‘‹๐‘ )2

Unity power factor: ๐ธ0 = โˆš(๐‘‰ + ๐ผ๐‘…๐‘Ž)2 + (๐ผ๐‘‹๐‘ )2

Where V= terminal voltage; I-Full load current

Regulation: %๐‘… =๐ธ0โˆ’๐‘‰

๐‘‰ร— 100

GRAPH:

The regulation curve is drawn between regulation and power factor

VIVA QUESTIONS:

l. What is an alternator?

2. What are the types of alternatorโ€™?

3. Define voltage regulation of an alternator.

4. Mention the methods by which voltage regulation can be determined.

5. Which method gives the result nearer to the actual value?

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V AND INVERTED V CURVES OF SYNCHRONOUS MOTOR

GRAPH:

The curve is drawn between

o Armature current VS excitation current

o Power factor VS Excitation current

FORMULA:

tan โˆ… = โˆš3[๐‘Š2โˆ’๐‘Š1

๐‘Š2+๐‘Š1]

โˆ… = tanโˆ’1[โˆš3 [๐‘Š2โˆ’๐‘Š1

๐‘Š2+๐‘Š1]]

Power factor: ๐œ๐จ๐ฌ โˆ…

VIVA QUESTIONS:

1. With what condition synchronous motor can be used as a synchronous condenser.

2. What are the special applications of an over excited synchronous motor.

3. Explain the effect of change of excitation of a synchronous motor on its armature current.

4. Explain the effect of change of excitation of a synchronous motor on its power factor.

5. With the given excitation a synchronous motor draws a unity PF current. If the mechanical

Load is increased what will be the power factor and current for the same excitation.

6. Why V curve shift upwards and inverted V curve shift right as the load increases.

7. Explain the effect of change of excitation of a synchronous generator on its armature current.

8. Explain the effect of change of excitation of a synchronous generator on its power factor.

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LOAD TEST ON 3 PHASE INDUCTION MOTOR

MODEL GRAPH

FORMULA:

๐‘บ = ๐‘บ๐Ÿ โˆ’ ๐‘บ๐Ÿ Kg ; ๐‘ป๐’๐’“๐’’๐’–๐’† = ๐’” โˆ— ๐Ÿ—. ๐Ÿ–๐Ÿ โˆ— ๐‘น Nm [R-radius of drum; S1, S2- Spring balance]

Input = ๐’˜๐Ÿ + ๐’˜๐Ÿ ;

Output power = ๐Ÿ๐…๐‘ต๐‘ป

๐Ÿ”๐ŸŽ watts; [T- Torque in Nm; N- Speed in rpm]

Power Factor = ๐‘พ

โˆš๐Ÿ‘๐‘ฝ๐‘ณ๐‘ฐ๐‘ณ

Efficiency =

๐’

๐’‘ ๐’‘๐’๐’˜๐’†๐’“

๐’Š

๐’‘ ๐’‘๐’๐’˜๐’†๐’“

ร— ๐Ÿ๐ŸŽ๐ŸŽ

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GRAPH:

The performance characteristic curves are drawn as

1) Output Power vs. current

2) Output Power vs. Torque

3) Output Power vs. Speed

4) Output Power vs. Efficiency,

Taking Output Power along X-axis and current, torque, speed and efficiency

Along Y-axis.

VIVA QUESTIONS:

1. What are the two types of induction motors?

2. What are the methods of starting of an induction motor?

3. How can the direction of rotation of the motor be reversed?

4. In which induction motor, can you arid external resistance to the rotor?

5. Whether a single phase induction motor is self-starting?

LOAD TEST ON 1 PHASE INDUCTION MOTOR

FORMULA:

๐‘บ = ๐‘บ๐Ÿ โˆ’ ๐‘บ๐Ÿ Kg ; ๐‘ป๐’๐’“๐’’๐’–๐’† = ๐’” โˆ— ๐Ÿ—. ๐Ÿ–๐Ÿ โˆ— ๐‘น Nm [R-radius of drum; S1, S2- Spring balance]

Input = ๐’˜๐Ÿ + ๐’˜๐Ÿ ;

Output power = ๐Ÿ๐…๐‘ต๐‘ป

๐Ÿ”๐ŸŽ watts; [T- Torque in Nm; N- Speed in rpm]

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Power Factor = ๐‘พ

โˆš๐Ÿ‘๐‘ฝ๐‘ณ๐‘ฐ๐‘ณ

Efficiency =

๐’

๐’‘ ๐’‘๐’๐’˜๐’†๐’“

๐’Š

๐’‘ ๐’‘๐’๐’˜๐’†๐’“

ร— ๐Ÿ๐ŸŽ๐ŸŽ

Percentage of slip = ๐‘ต๐’”โˆ’๐‘ต

๐‘ต๐’” ร— ๐Ÿ๐ŸŽ๐ŸŽ

Power factor = Input power/V1I1

GRAPH:

The performance characteristic curves are drawn as

1) Output Power vs. current

2) Output Power vs. Torque

3) Output Power vs. Speed

4) Output Power vs. Efficiency,

Taking Output Power along X-axis and current, torque, speed and efficiency Along Y-axis.

VIVA QUESTIONS:

l. What are the two types of induction motors?

2. What are the methods of starting of an induction motor?

3. How can the direction of rotation of the motor be reversed?

4. Whether a single phase induction motor is self-starting?

REGULATION OF 3 PHASE ALTERNATOR BY SALIENT POLE ALTERNATOR BY

SLIP TEST

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FORMULA:

Direct Axis Synchronous Reactance: Xd = ๐ผ๐‘“๐‘ ๐‘

๐ผ0 ๐‘๐‘’๐‘Ÿ ๐‘ข๐‘›๐‘–๐‘ก

Quadrature axis Synchronous Reactance Xq = ๐‘‹๐‘ž

๐‘‹๐‘‘๐‘  ๐‘‹๐‘‘ ๐‘๐‘’๐‘Ÿ ๐‘ข๐‘›๐‘–๐‘ก

MODEL CALCULATION:

๐‘ฟ๐’… =๐‘ฌ๐’Ž๐’‚๐’™

๐‘ฐ๐’Ž๐’Š๐’ ; ๐‘ฟ๐’’ =

๐‘ฌ๐’Ž๐’Š๐’

๐‘ฐ๐’Ž๐’‚๐’™

Where , Emax, Emin ,Imax ,Imin where taken in the Reading

Now calculate for Power factor

cos โˆ… = 0.8 , So find the value of ใ€–โˆ… ใ€—

Therefore, tan ๐œ‘ = ๐‘‰ sin ๐œ‘+๐ผ๐‘‹๐‘ž

๐‘‰ cos ๐œ‘+๐ผ๐‘…๐‘Ž from this equation find ๐œ‘

Where, ๐œ• = ๐œ‘ โˆ’ โˆ… calculate the value of โˆ‚

Now to find the regulation:

We know that, ๐ธ0 = ๐‘‰ cos ๐œ• + ๐ผ๐‘ž๐‘…๐‘Ž + ๐ผ๐‘‘๐‘‹๐‘‘

Where ๐ผ๐‘‘ = ๐ผ๐‘Ž sin ๐œ‘ ๐ผ๐‘ž = ๐ผ๐‘Ž cos ๐œ‘

Now find the Regulation with above equations by subtitling proper values in below equation we

get,

%๐‘…๐‘’๐‘”๐‘ข๐‘™๐‘Ž๐‘ก๐‘–๐‘œ๐‘› =๐ธ0 โˆ’ ๐‘ฃ

๐‘ฃ ร— 100

Calculate for other power factor values with the help of above equations

GRAPH:

The graph is drawn between % regulations vs. Power factor

VIVA QUESTIONS:

1. What is the purpose of slip test on 3 phase alternator?

2. What is meant by direct axis reactance?

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3. What is meant by quadrature axis reactance?

4. How is the regulation of alternator predetermined by slip test?

5. What is the difference between salient pole alternator and cylindrical rotor type alternator?

REGULATION OF 3 PHASE ALTERNATOR BY ZPF AND ASA METHOD

GRAPH:

1) A graph is drawn b/w If and V which is known as OC curve, by taking If on X-axis and V on

Y-axis.

2) A graph is drawn b/w If and ISC which is known as SC curve, by Taking If on X-axis and

ISCV on Y-axis.

DRAWING ZPF CURVE

1. OCC is drawn.

2. Point A is located such that OA gives If corresponding to Irated. Under short circuit test.

3. Point B is located such that it gives If to voltage from ZPF test.

4. Points A and B joined by curve parallel to OC called ZPF curve.

5. From the curve, ZPF curve is extended.

6. From H, HD is drawn parallel to OCC line.

7. From B, BH is drawn parallel and equal to OA.

8. Point D is point to B and BHD is tangent is obtained.

9. From D, perpendicular to BH at E is drawn.

10. DE gives Ia XL. BE gives If necessary to overcome demagnetizing effect of armature

resistance. EH gives If necessary for balancing armature leakage reactance drop DE.

11. Internal emf, E1 is calculated as

E1 = โˆš((Vph cosะค + Ia Ra)2 + (Vph sinะค + IaXL)2)

โ€˜+โ€™ โ€“ for lagging pf and

โ€˜-โ€˜ โ€“ for leading pf.

12. Find If1 corresponding E1 from OCC.

13. If2 is field current, required to overcome armature reaction (BE)

14. If = โˆš(If12+If2

2-2If1If2 cos(90ยฑะค)) โ€˜+โ€™ โ€“ for lagging pf and

โ€˜-โ€˜ โ€“ for leading pf.

15. From internal emf E1, a horizontal line is drawn cutting the OCC.

16. The regulation is calculated as

% regulation = ((Eo โ€“ Vph)/ Vph)x100.

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NO LOAD AND BLOCKED ROTOR TEST ON SINGLE PHASE INDUCTION MOTOR

FORMULA:

For NO load test:

cos ๐œ‘0 =๐‘Š0

๐‘‰0๐ผ0

๐ผ๐‘Š = ๐ผ0 cos ๐œ‘0 ๐ด

๐ผ๐œ‡ = ๐ผ0 sin ๐œ‘0 ๐ด

๐‘…0 =๐‘‰

๐ผ๐‘Š ๐‘œโ„Ž๐‘š

๐‘‹0 =๐‘‰

๐ผ๐œ‡ ๐‘œโ„Ž๐‘š

For blocked rotor test:

.

cos โˆ…๐‘ ๐‘ =๐‘Š๐‘ ๐‘

๐‘‰๐‘ ๐‘๐ผ๐‘ ๐‘

๐ผ๐‘Š = ๐ผ0 cos ๐œ‘0 ๐ด

๐ผ๐œ‡ = ๐ผ0 sin ๐œ‘0 ๐ด

๐‘…01 =๐‘Š๐‘ ๐‘

๐ผ๐‘ ๐‘2 ๐‘œโ„Ž๐‘š

๐‘01 =๐‘‰๐‘ ๐‘

๐ผ๐‘ ๐‘ ๐‘œโ„Ž๐‘š

๐‘‹01 = โˆš๐‘012 โˆ’ ๐‘…01

2 ๐‘œโ„Ž๐‘š

.

๐‘‹๐‘  = โˆš๐‘๐‘ 2 โˆ’ ๐‘…๐‘ 

2 ๐‘œโ„Ž๐‘š

๐‘…๐ฟ = ๐‘…2 [1

๐‘ โˆ’ 1] ๐‘œโ„Ž๐‘š

๐‘…2 = [๐‘…01 โˆ’๐‘…๐‘š

๐‘…๐‘ ] ๐‘œโ„Ž๐‘š

๐‘2 = [๐‘01 โˆ’๐‘๐‘Š

๐‘๐‘ ] ๐‘œโ„Ž๐‘š

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Equivalent circuit diagram:

Calculate and determine of RM and RS

Calculate the determine of ZM and ZC and ZS

VIVA QUESTION:

1. What is a 1-phase induction motor?

2. Write the classification of 1-phase induction motor?

3. Why do we draw the equivalent circuit of 1-phase induction motor?

4. What is double-field revolving theory?

5. Why 1-phase induction, motor is not self-starting?

NO LOAD AND BLOCKED ROTOR TEST ON THREE PHASE INDUCTION MOTOR

FORMULA:

For NO load test:

cos ๐œ‘0 =๐‘Š0

โˆš3๐‘‰0๐ผ0

๐ผ๐‘Š = ๐ผ0 cos ๐œ‘0 ๐ด

๐ผ๐œ‡ = ๐ผ0 sin ๐œ‘0 ๐ด

๐‘…0 =๐‘‰

๐ผ๐‘Š ๐‘œโ„Ž๐‘š

๐‘‹0 =๐‘‰

๐ผ๐œ‡ ๐‘œโ„Ž๐‘š

For blocked rotor test:

.

cos โˆ…๐‘ ๐‘ =๐‘Š๐‘ ๐‘

โˆš3๐‘‰๐‘ ๐‘๐ผ๐‘ ๐‘

๐ผ๐‘Š = ๐ผ0 cos ๐œ‘0 ๐ด

๐ผ๐œ‡ = ๐ผ0 sin ๐œ‘0 ๐ด

๐‘…01 =๐‘Š๐‘ ๐‘

๐ผ๐‘ ๐‘2 ๐‘œโ„Ž๐‘š

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๐‘01 =๐‘‰๐‘ ๐‘

๐ผ๐‘ ๐‘ ๐‘œโ„Ž๐‘š

๐‘‹01 = โˆš๐‘012 โˆ’ ๐‘…01

2 ๐‘œโ„Ž๐‘š

.

๐‘‹๐‘  = โˆš๐‘๐‘ 2 โˆ’ ๐‘…๐‘ 

2 ๐‘œโ„Ž๐‘š

๐‘…๐ฟ = ๐‘…2 [1

๐‘ โˆ’ 1] ๐‘œโ„Ž๐‘š

๐‘…2 = [๐‘…01 โˆ’ ๐‘…๐‘Ž๐‘]๐‘œโ„Ž๐‘š

๐‘…๐‘Ž๐‘ = [๐‘…๐‘‘๐‘ โˆ— 1.6]๐‘œโ„Ž๐‘š

Equivalent circuit diagram:

Draw the equivalent circuit with the help of above parameters

SEPARATION OF NO LOAD LOSSES IN 3 PHASE INDUCTION MOTOR

MODEL GRAPH:

FORMULA:

For no load test

cos ๐œ‘0 =๐‘Š0

โˆš3๐‘‰0๐ผ0

๐ผ๐‘Š = ๐ผ0 cos ๐œ‘0 ๐ด

๐ผ๐œ‡ = ๐ผ0 sin ๐œ‘0 ๐ด

๐‘…0 =๐‘‰

๐ผ๐‘Š ๐‘œโ„Ž๐‘š

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๐‘‹0 =๐‘‰

๐ผ๐œ‡ ๐‘œโ„Ž๐‘š

Loss calculation:

Stator Copper Loss: 3๐ผ02๐‘…

Constant loss =Input power โ€“Stator copper loss

Constant loss/phase =Constant loss / 3

GRAPH:

The graph drawn between constant losses (watts) and input voltage (volts)

MEASURMENTS OF NEGATIVE SEQUENCE AND ZERO SEQUENCE IMPEDANCE

OF ALTERNATORS

FORMULA:

For negative sequence:

๐‘ =๐‘ฃ

โˆš3๐ผ

๐‘‹2 =๐‘Š

โˆš3๐ผ2

๐‘…2 = โˆš(๐‘๐‘Ž2 โˆ’ ๐‘‹๐‘Ž

2

For Zero sequence:

๐‘0 =๐‘ฃ0

โˆš3๐ผ0

๐‘‹0 =๐‘Š

โˆš3๐ผ02

๐‘…2 = โˆš(๐‘02 โˆ’ ๐‘‹0

2