Circle Diagram of 3phase Induction MotorM.pptx

The circle diagram is a graphical representation of the performances of an induction motor. It is very useful to study the performance of an induction motor under all operating conditions. he construction of the circle diagram of an induction motor is based on the equivalent circuit of the motor which is shown in Figure-1.

By applying KCL in the equivalent circuit, we can write,

I1=I0+I'2…(1)

If the phase voltage V1 is taken along the vertical axis of OY as shown in Figure-2. Then, the no-load current (I0) lags behind the voltage V1 by an angle ϕ0. Where, ϕ0 is the no-load power factor angle and is of the order of 60°-80° because to produce the required flux per pole in the air gap of the machine, a large magnetising current is needed.

Now, at no-load, the slip s = 0, then

R2s=∞i.e.opencircuitedatno−load

∴I0=V1ZNL…(2)

Where,

ZNL=(R0||jX0)=Noloadimpedance

Here, all the rotational losses are taken into account by R0. Then, the no-load losses are given by,>

P0=V1I0cosφ0…(3)

The rotor current referred to the stator side is given by,

I'2=V1Z'e1=V1(R1+R'2s)2+(X1+X'2)2−−−−−−−−−−−−−−−−−−−−−√…(4)

This rotor current I′2 lags behind the voltage V1 by the impedance angle ϕ.

Refer Figure-3,

sinφ=X1+X'2(R1+R'2s)2+(X1+X'2)2−−−−−−−−−−−−−−−−−−−−−√…(5)

Hence, by combining the eqns. (4) & (5), we get,

I'2=V1X1+X'2sinφ;…(6)

Eqn. (6) is in the form of 𝑟 = 𝑎 sin θ, which represents a circle with diameter a. Therefore, the locus of current (𝐼′2) is a circle whose diameter is

diameter=V1X1+X'2

It is shown in Figure-4.

The radius of the circle is,

O'C=V12(X1+X'2)

From eqn. (1), it is clear that the stator current I1 is found by combining the results shown in Figures-2, 3, and 4. Then, the resulting figure obtained is shown in Figure-5 which is known as the circle diagram of the induction motor.

1. Tests on Induct

> No Load Test

2

5/24/2021

3 Phase
Supply at
rated
voltage and
frequency Motor
motor

Mechanical
Load

NMAM INSTITUTE

2. Tests on Induction

> No Load Test

> As the motor is running at no load, the total input power is equal to the constant iron loss,
friction and windage losses of the motor.

> Vi, is the input line voltage

> P,, is the total three-phase input power at the no load
> 1, is the input line current

> Vip is the input phase voltage

> Pon = V3Vinilo COS Po

> l= rl = Ip cos Po

> Ry =, X =~

| WN NITTE | SE CROSS

5/24/2001 DEPART

3. ction M

> No Load Test: Separation of Losses

> Friction and windage loss can be separated from the no-load loss Py. At no load various
readings of the No load loss are taken at the different stator applied voltages. The readings
are taken from rated to the breakdown value at rated frequency.

Po

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4. Tests on Induc

> Blocked Rotor (Locked Rotor) Test

Motor

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5. Tests on I ction Mo

> Blocked Rotor (Locked Rotor) Test:

> V., is the short circuit line voltage, I,, is the short circuit line current

> Poe = V3Vscilsct COS Psc

_ Psep _ Vsep = 2 2
Rey = le = Xe = ZU Ri,

> The slip of the induction motor varies between 2 to 4 percent, and the resulting rotor
frequency is in the range of 1 to 2 hertz for the stator frequency of 50 hertz at the normal
conditions.

> In order to obtain the accurate results, the Blocked Rotor Test is performed at a
frequency 25 percent or less than the rated frequency. The leakage reactances at the rated
frequency are obtained by considering that the reactance is proportional to the frequency.

> However, for the motor less than the 20-kilowatt rating, the effects of the frequency are
negligible, and the blocked rotor test can be performed directly at the rated frequency.

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6. Construc;

of the Circle Diagram

> Circle diagram of an induction motor can be drawn by using the data obtained from
(1) No-load test

(2) Short-circuit test or Blocked rotor test

(3) Stator Resistance Test

Current-

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7. Step No.1
v + From no-load test, lo and do can be calculated. Hence, as shown in fig, vector for lo
can be laid off lagging do behind the applied voltage V.

W y= V3 V, 1,cos 0,

LA

cos =
VF Y, Lo

where Y, = line voltage and 17, is no-load stator input.

> X
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8

8. , Step No. 2

V| + From blocked rotor test or short-circuit test, short circuit current Isw
corresponding to normal voltage and sare found.

+ The vector OA represents Isn

+ Isn=lsx V/Vs
+ where Is = short-circuit current obtainable with normal voltage V
* Is= short-circuit current with voltage Vs where 77, = total power input on short-circuit
+ Power factor on short-circuit is found from Vs, = line voltage on short-circuit
Ws = WV sp lp cosQ5: + COSO5=W5/ (V3 V 5,15) ls, = line current on short-circuit

* Vector O'A represents rotor current I2' as referred to stator.

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9. 10

5/24/2021

* For finding the centre C of this circle,
+ Drawaline perpendicular to voltage vector from O'

+ chord O'A is bisected at right angles-its bisector giving point C.

> X
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10. u

5/24/2021

{+ With centre C and radius = CO’, the circle can be drawn.
+ The diameter O'D is drawn perpendicular to the voltage vector

* Scale of current vectors should be so chosen such that the diameter is more than 20 cm

|
D

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The line O'A is known as output line.

It should be noted that as the voltage vector is drawn vertically, all vertical distances
represent the active or power or energy components of the currents.

the vertical component O’P of no-load current OO’ represents the no-load input, which

supplies core loss, friction and windage loss and a negligibly small amount of stator copper
loss

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13. 14

5/24/2021

Step No. 3

Torque line: This is the line which separates the stator and the rotor copper losses

When the rotor is locked, then all the power supplied to the motor goes to meet
core losses and Cu losses in the stator and rotor windings.

The vertical component AG of short-circuit current OA is proportional to the motor
input on short circuit.

Out of this, FG (= O'P) represents fixed losses i.e. stator core loss and friction and
windage losses.

x
P Ca" fixed losses
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14. 4 + AF is proportional to the sum of the stator and rotor Cu losses.
+ The point E is such that

AE _ rotor Cu loss

EF stator Cu loss

+ line O’E is known as torque line.

+ This is the line which separates the stator and the rotor copper losses.

NT Rotor Cu losses

E
wai Stator Cu losses

o € Fir D

x

E “Ea
fixed losses;
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15. How to locate point E ?
(i) Squirrel-cage Rotor. Stator resistance/phase i.e. Ris found from stator-resistance test.
Now, the short-circuit motor input Ws is approximately equal to motor Cu losses (neglecting iron losses)
Stator Cu loss = 3/,°R,
, rotor Cu loss = W,—3 1,” Ry

AE MR
EF 3RR,

(ii) Wound Rotor. In this case, rotor and stator resistances per phase r, and r, can be easily computed.

For any values of stator and rotor currents I, and I, respectively, we can write

IBA LAGE oo
Æ = 22-22 Now, = transformation ratio
EF Rn All
AE _ M, 1 _n/K? _n/ _ equivalent rotor resistance per phase
EF je Ke if stator resistance per phase

Value of K may be found from short-circuit test itself by using two ammeters, both in stator and rotor circuits.
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1/2021 Y NTE OF TECHNOLOGY

16. w

5/24/2021

Let us assume that the motor is running under certain load and taking a current OL at pf of cos,
Draw a line from point L vertically downwards till it intersects horizontal axis.

Then,

the perpendicular JK represents fixed losses,
JN is stator Cu loss,

NL is the rotor input,

NM is rotor Cu loss,

ML is rotor output

LK is the total motor input.
i

Rotor Cu losses

Stator Cu losses

x

fixed losses
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17. 5/24/2021

V3.V,.LK = motor input V3.V,.JK = fixed losses

V3 .V,.JN = stator copperloss 3.7,.MN = rotor copper loss
= total loss V3 .V,.ML = mechanical output
= rotor input = torque

ML/LK = output/input = efficiency

MN/NL = (rotor Cu loss)/(rotor input) = slip. s.

ML _ rotor output _ | _¿= N - actual speed
NL rotorinput N, _ synchronous speed
4. IK = power factor

Rotor Cu losses

Stator Cu losses

BD Xx

fixed losses NMAM INSTITUTE
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18. Maximum Quantities

(i) Maximum Output
* Itoccurs at point where the tangent is parallel to output line O'A.

(ii) Maximum Torque or Rotor Input
+ It occurs at point where the tangent is parallel to torque line O’E.

(ii) Maximum Input Power
+ Itoccurs at the highest point of the circle i.e, at a point where the tangent to the circle
is horizontal.

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19. Maximum Quantities
Vv (i) Maximum Output
* Itoccurs at point S where the tangent is parallel to output line OA.
+ Or Point S may be located by drawing a line CS from point C such that it is perpendicular to
the output line O'A.

+ Maximum output is represented by the vertical SS’, (From point S, draw a line vertically
downwards till it intersects output line i.e. point 5’)

Na Rotor Cu losses
\
E \
wal Stator Cu losses
|
|
D
» Si P is E fixed losses :
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5/24/2021 pa rta Mr data ira

20. 4 Maximum Quantities

M (i) Maximum Torque or Rotor Input
«It occurs at point T where the tangent is parallel to torque line O'E.

or
* point T may be found by drawing CT perpendicular to the torque line.

+ Its value is represented by TT’. (From point T, draw a line vertically downwards till it
intersects torque line i.e. point T’)

+ Maximum torque is also known as stalling or pull-out torque.

Rotor Cu losses

e

wal Stator Cu losses

N D x
6 a
fixed losses
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21. 5/24/2021

4 Maximum Quantities

M (iii) Maximum Input Power

* It occurs at the highest point of the circle i.e. at point R where the tangent to the circle is horizontal.

* Itis proportional to RR’. (From point R, draw a line vertically downwards till it intersects horizontal axis
ie. point R’)

* Asthe point Ris beyond the point of maximum torque, the induction motor will be unstable here.
However, the maximum input is a measure of the size of the circle and is an indication of the ability of
the motor to carry short time over-loads. Generally, RR’ is twice or thrice the motor input at rated

load.
—> Max. Input (RR’)

<A

Na Rotor Cu losses

\

E
7 Stator Cu losses
LS |
IS D

x
- fixed losses (N) NITE | NMANINSETUTE