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E

R

Z X

If R = 0 then sin = 1 i.e., = 900

I

If R = then sin = 0 i.e., = 00

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Current (I)

V

R1 < R2 < R3 < R4

0 I1

I1= V/X at R = 0, the angle between V & I will be 900

I2

I2= V/Z is the current at R = R1

I3

I3= V/Z is the current at R = R2

I4

I4= V/Z is the current at R = R3

I5

I5= V/Z is the current at R = R4

If we vary the value ofRfrom 0 to then the current (I)

is varying from it to zero (0)

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Steps to construct the Circle DiagramRequirements:

OC Test (Or) No Load Test SC Test (Or) Blocked Rotor Test

Step 1: From No load test calculate the angle 0

Step 2:

From SC test calculate the angle SC

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Let 1cm = 5 Amps (Current Scale)

Let I0 = 8 A then I0 in current scale = 8/5 = 1.6 cms

Draw the line of I0

=1.6cms with an angle of0

Let ISN = 100 A then ISN in current scale = 100/5 = 20 cms

Draw the line of ISN=20cms with an angle ofSC

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The line is at an

angle of 450

with3.5 cm Length

Draw a line is with and angle of 450 with 3.5 cm Length

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Current (I)

V

I0

0

sc

001

A

CGF

ERotor

Cu Losses

Stator

Cu Losses

Fixed

Losses

Draw the OO1 line with No-load current I0 with an angle 0

from O.C Test. Take 1cm = 5 Amps

Join O

1

A line that is Output line

XM

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Draw the OA line with short circuit current ISN when

Normal voltage is applied with an angle sc from Blocked

Rotor Test. ISN = V(Isc/Vsc), Convert the ISN into current

scale

Draw the line joining O1 & E that is the Torque line

Draw the perpendicular line from A to O1X1 & OX

at F & G

Draw the line parallel to OX to O1

Mark the point E at the center of the AF

Draw the circle taking C as Center joining O1 , A

and cuts the line O1X1 line at M

Cut the O1A (Output line) at center by using

Compass taking greater than half of the line, then it

cuts the O1X1 line at C

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Power Scale: Calculate the length ofAG with the help of Scale. (Let AG = 8.1cm)

Calculate the WSN from the SC test values (Let WSN = 28000 watts)

Where WSC is the SC test Power VSC is the SC test Voltage

V is the applied voltage (normal Voltage)

Power Scale is WSN / AG ( = 28000 / 8.1) = 3456.7

i.e., 1 cm = 3456.7 watts is the Power Scale

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In the Problem it will be given full load motor output

Let Motor output =14.9 Kw

Convert this Motor output into Power scale

= 14900/ Power scale

= 14900/ 3456.7

= 4.31 cms

Draw a line extending the line AG to S

such that SA = 4.31 cm

Draw a parallel line to output line which meets

the Point S and cuts the circle at L

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Current (I)

V

I0

0

sc

001

A

CG

F

E

Rot

Cu L

St

Cu

Fixed Lo

1

I1

L

S

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V

I0

0

sc

0

01

A

C

G

F

E

RCu

S

Cu

Fixed L

1

I1

Fixed Losses

Rotor

Output

L

M

N

J

K

Power Factor = Cos

= LK / OL

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Efficiency = Output / Input= ML / LK

1 Slip = rotor output / rotor input = N / NS= LM / LN

Slip = Cu Losses / P2

= MN / (LM + MN) = MN / LNRotor Input (P2) = Rotor output + Rotor cu Losses

= LM + MN