## What is Circle Diagram :

####
Circle diagram is a graphical method to analyse the performance of a 3-phase induction motor. We can also compute the operation of induction motor using equivalent circuit, but it is easy and simple by using the circle diagram. If we neglect the parallel branch of the equivalent circuit of an induction motor, it reduces into a simple series circuit having a constant voltage supply. A diagram giving the locus of the current drawn by an induction motor is referred to as a ' Circle Diagram '.

The circle diagram can be drawn by using no-load and blocked-rotor test data. From the circle diagram it is possible to on obtain graphically a considerable range of information like full-load current and power factor, maximum power output, pull-out torque, full-load efficiency, etc.

Circle diagram is a graphical method to analyse the performance of a 3-phase induction motor. We can also compute the operation of induction motor using equivalent circuit, but it is easy and simple by using the circle diagram. If we neglect the parallel branch of the equivalent circuit of an induction motor, it reduces into a simple series circuit having a constant voltage supply. A diagram giving the locus of the current drawn by an induction motor is referred to as a ' Circle Diagram '.

The circle diagram can be drawn by using no-load and blocked-rotor test data. From the circle diagram it is possible to on obtain graphically a considerable range of information like full-load current and power factor, maximum power output, pull-out torque, full-load efficiency, etc.

## Theory :

####
This method aims at plotting the locus of input stator current I_{1 }for various values of slip, I_{1 }is the phasor sum of I'_{2 }( i.e., rotor current as referred to stator ) and I_{o}.

####
The phasor E_{2 }/ X_{2 }denotes a constant current. As seen from the above equation current lagging E_{2 }by 90°, is composed of two components viz. I_{2 } and - J
R_{2 }I_{2} / s X_{2}. These two components are at right angles to each other. Therefore the extremity of I_{2 }lies on the circumference of a circle is shown in figure below.

####
Since I_{2 }& I'_{2 }are related by a constant, the locus of I'_{2 }is also a circle. Since I_{1 }is the phasor sum of I'_{2 }and I_{o}, we can displace this circle of I'_{2 }from origin by the magnitude of I_{o}. Then the extremity of I_{1 }will lie on the circumference of the circle is shown in figure below.

## Construction :

####
The data available from the No. Load and Blocked Rotor Tests is used for drawing the circle diagram. All quantities are per phase values.

####
The No-load current I_{o }and its power factor angle Φ_{o }can be obtained from the No-Load test. The S.C. (blocked rotor) current I_{sc }for voltage V_{sc }and its power factor angle Q_{sc }can be found out from the blocked rotor test. This current must be converted corresponding to the rated voltage, V_{1 }of the motor.

The data available from the No. Load and Blocked Rotor Tests is used for drawing the circle diagram. All quantities are per phase values.

####
The No-load current I_{o }and its power factor angle Φ_{o }can be obtained from the No-Load test. The S.C. (blocked rotor) current I_{sc }for voltage V_{sc }and its power factor angle Q_{sc }can be found out from the blocked rotor test. This current must be converted corresponding to the rated voltage, V_{1 }of the motor.

#### S.C. current for normal voltage,

####
I_{SN} = I_{sc} ( V_{1} / V_{sc}
)

#### Knowing these data the circle diagram can be drawn as follows.

####
1. Draw a vertical line representing the rated voltage V_{1 }( reference phasor ).

####
2. Select a suitable scale for current. Draw line 'OA' lagging V_{1 }by an angle Φ_{o }equal to No-Load current I_{o}.

####
3. Draw a line 'OB' lagging V_{1 }by an angle Φ_{sc }equal to S.C. current for normal voltage.

####
Vector 'OB' represents the rotor current I'_{2 }as referred to stator.

#### 4. Join 'AB' which represents the output line of the motor.

#### 5. A perpendicular bisector of the line 'AB' is drawn cutting the horizontal line 'AD' ( drawn from A ) at the point 'C', which is the centre of the circle. Then with 'C' as centre and 'CA' as radius, draw the semi circle 'ABD'.

#### 6. Draw a vertical line from the point 'B', so as to meet the line 'AD' at the point 'F' Divide 'BF' in the ratio of rotor copper loss to stator copper loss at the point 'E ' i.e.,

#### Then 'AE' represent the torque line.

## Computation of Performance :

####
As the applied voltage V_{1 }is drawn vertically, all vertical distances represent the power or energy components of the currents. The vertical distance AQ ( = I_{o }CosΦ_{o }) represents the no-load input which supplies core loss, mechanical loss and a small stator copper loss.

## 1. Torque line :

####
Under blocked rotor conditions at rated voltage, the power input is 'BG'. This represents core loss and copper losses The intercept 'FG' the ( = AG ) is approximately equal to core loss because through the mechanical losses are absent slightly increased under blocked rotor conditions.

Hence the intercept 'BF' represents the sum of stator and rotor copper losses.

The line 'AE' which separates the stator and rotor copper loss as known as torque line or rotor input line.

Torque is proportional to power input to rotor. The rotor copper loss at stand still is 'BE'.

Therefore Starting torque = BE

Under blocked rotor conditions at rated voltage, the power input is 'BG'. This represents core loss and copper losses The intercept 'FG' the ( = AG ) is approximately equal to core loss because through the mechanical losses are absent slightly increased under blocked rotor conditions.

Hence the intercept 'BF' represents the sum of stator and rotor copper losses.

The line 'AE' which separates the stator and rotor copper loss as known as torque line or rotor input line.

Torque is proportional to power input to rotor. The rotor copper loss at stand still is 'BE'.

Therefore Starting torque = BE

## 2. Location of Point 'E' :

## Cage Rotor :

####
The input power under blocked rotor conditions is approximately equal to copper losses because the iron losses are very small and neglected. Let R_{1} = stator resistance / phase ( found from stator resistance test )

####
Stator copper loss = 3 I_{s}^{2}
R_{1}

#### Therefore,

####
Rotor copper loss = S. C. input power - 3 I_{s}^{2} R_{1}

## Wound rotor :

#### Let,

####
R_{1 }and R_{2 }= stator and rotor resistance / phase respectively

####
I_{1 }and I_{2 = }stator and rotor resistance / phase respectively

## 3. Operating Point or Full Load Point 'P' :

#### Draw a vertical line 'BL' represents the full load output of the motor. From the point 'L' draw a line parallel to the output line and it cuts the circle at two points 'U' and 'P'. Since the normal operating slip is small, the point 'P' will be the operating point.

## 4. Full Load Current and Power Factor :

####
Join 'OP' which represents the full load current of the motor. Corresponding power factor angle φ_{1}_{ } ( = ∟_{
}V_{1 }O P ) gives the power factor at which the motor is operating.

## 5. Full Load Efficiency, Slip & Speed :

#### Draw a vertical line from the point 'P'. Then 'PM' & 'PK' respectively represent the output and input of the motor. 'JK' represents fixed losses. 'MN' & 'NJ' represent the rotor and stator copper losses.

#### Efficiency,

#### Full load slip,

####
Full load rotor speed, N = N_{s }( 1 - s )

## 6. Maximum Output :

#### Draw a line parallel to the output line 'AB' tangent to the semicircle. The actual point of contact 'H' is obtained by drawing a perpendicular to the tangent drawn from the centre 'C'. Now 'HI' is the maximum output.

## 7. Maximum torque :

#### It is also called as stalling or pull out torque. Draw a line the torque line 'AE ' tangent to the semi-circle. The actual point of contact 'R ' by drawing a perpendicular to the tangent drawn from the centre 'C'. Now 'RS' is the maximum torque.

#### The slip at maximum torque,

####
s_{m}
=
TS / RS

## 8. Maximum Input :

#### It occurs at highest point of the circle i.e., at point 'V' where the tangent to the circle is horizontal. Now 'VW' is the maximum input.

## From the circle diagram, it can be concluded that

#### Full current = OP

####

Full-Load power factor angle,

####
**φ**_{1} = ∟V_{1}0P or Cos
φ_{1} = PK / OP

_{1}= ∟V

_{1}0P or Cos φ

_{1}= PK / OP

#### Full-Speed,

####
Full-Load speed, N = N_{s }( 1 - s )

#### Maximum output = HI

#### Maximum torque = RS

#### Full-load torque = Full-load rotor input = PN

#### Starting torque = Rotor copper loss at Standstill = BE

#### Slip at maximum torque,

####
**s**_{m}
=
TS / RS

_{m}= TS / RS

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