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Electrical Machines-I

ECE-2107

Induction Motor-SL3

Fariya Tabassum
Assistant Professor, Dept. of Electrical & Computer Engineering
Rajshahi University of Engineering & Technology, Rajshahi-6204

“Allah does not burden a soul beyond that it can bear”.

[Sura Baqarah]

Rotor Torque

The torque T developed by the rotor is directly proportional to

Rotor current
Rotor e.m.f
Power factor of the rotor circuit

$∴ T \propto E_{2} I_{2} cos φ_{2}$

$Or T = K E_{2} I_{2} cos φ_{2}$

Where,
$I_{2} =$ Rotor current at standstill
$E_{2} =$ Rotor e.m.f at standstill
$cos φ_{2} =$ Rotor power factor at standstill

Starting Torque

The torque developed by the motor at the time of starting is known as starting torque of the motor.

Let, $E_{2} =$ rotor e.m.f. per phase at standstill
$R_{2} =$ rotor resistance per phase
$X_{2} =$ rotor reactance per phase at standstill

$Z_{2} = (R_{2}^{2} + X_{2}^{2}) = rotor impedance per phase at standstill$

$Then, rotor current I_{2} = \frac{E _{2}}{Z _{2}} = \frac{E _{2}}{( R _{2} ^{2} + X _{2} ^{2} )}$

$power factor cos φ_{2} = \frac{R _{2}}{Z _{2}} = \frac{R _{2}}{( R _{2} ^{2} + X _{2} ^{2} )}$

$Now, the starting torque T_{s t} = K_{1} E_{2} I_{2} cos φ_{2}$

$or T_{s t} = K_{1} E_{2} \frac{E _{2}}{( R _{2} ^{2} + X _{2} ^{2} )} \frac{R _{2}}{( R _{2} ^{2} + X _{2} ^{2} )} = K_{1} \frac{E _{2} ^{2} R _{2}}{R _{2} ^{2} + X _{2} ^{2}}$

Starting Torque

If the supply voltage is constant, then the flux $φ$ and hence $E_{2}$ both are constant.

$∴ T_{s t} = K_{2} \frac{R _{2}}{R _{2} ^{2} + X _{2} ^{2}} = \frac{K _{2} R _{2}}{Z _{2} ^{2}}$

$It can be shown that, K_{1} = \frac{3}{2 π N _{s}}$

$∴ T_{s t} = \frac{3}{2 π N _{s}} \frac{E _{2} ^{2} R _{2}}{R _{2} ^{2} + X _{2} ^{2}}$

Where, $N_{s} =$ Synchronous speed in r.p.s

For preparing your answer you can go through the article 34.13 of the book written by “B. L. Theraza”

Condition for Maximum Starting Torque

It can be proved that starting torque will be maximum when rotor resistance/phase is equal to standstill rotor reactance/phase.

$We know the staring torque, T_{s t} = K_{2} \frac{R _{2}}{R _{2} ^{2} + X _{2} ^{2}} (i)$

Differentiating equation (i) w.r.t $R_{2}$ and equating the result to zero we get,

$\frac{d T _{s t}}{d R _{2}} = K_{2} [\frac{1}{R _{2} ^{2} + X _{2} ^{2}} - \frac{R _{2} ( 2 R _{2} )}{( R _{2} ^{2} + X _{2} ^{2} ) ^{2}}] = 0$

$Or R_{2}^{2} + X_{2}^{2} = 2 R_{2}^{2}$

$∴ R_{2} = X_{2}$

Problems

Practice example 34.6, 34.7, 34.8, 34.9 and 34.11 of B. L. Theraza and also the related tutorial problem.

Torque Under Running Conditions

Let the rotor at standstill have per phase induced e.m.f. $E_{2}$ , reactance $X_{2}$ and resistance $R_{2}$ . Then under running conditions at slip s

$Rotor e.m.f per phase E_{r} = s E_{2}$

$Rotor reactance per phase X_{r} = s X_{2}$

$Rotor impedance per phase Z_{r} = (R_{2}^{2} + (s X_{2})^{2})$

$Rotor current per phase I_{r} = \frac{E _{r}}{Z _{r}} = \frac{s E _{2}}{( R _{2} ^{2} + ( s X _{2} ) ^{2} )}$

$power factor cos φ_{r} = \frac{R _{2}}{Z _{r}} = \frac{R _{2}}{( R _{2} ^{2} + ( s X _{2} ) ^{2} )}$

Torque Under Running Conditions

$Then, running torque T_{r} \propto E_{r} I_{r} cos φ_{r}$

$T_{r} \propto φ I_{r} cos φ_{r}$

$\propto φ \frac{s E _{2}}{( R _{2} ^{2} + ( s X _{2} ) ^{2} )} \frac{R _{2}}{( R _{2} ^{2} + ( s X _{2} ) ^{2} )}$

$\propto φ \frac{s E _{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

$= K φ \frac{s E _{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

$= \frac{K _{1} s E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

If the stator supply voltage is constant, then stator flux and hence $E_{2}$ will be constant.

Torque Under Running Conditions

$∴ T_{r} = \frac{K _{2} s R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

Where, $K_{2}$ is another constant.

It may be seen that running torque is

Directly proportional to the slip i.e., if slip increases (motor speed decreases), the torque will increase and vice-versa.
Directly proportional to square of supply voltage (*)**

Torque Under Running Conditions

$It can be shown that, K_{1} = \frac{3}{2 π N _{s}}$

Where, $N_{s} =$ Synchronous speed in r.p.s

$∴ T_{r} = \frac{3}{2 π N _{s}} \frac{s E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}} = \frac{3}{2 π N _{s}} \frac{s E _{2} ^{2} R _{2}}{Z _{r} ^{2}}$

$At standstill, when s=1, T_{r} = \frac{K _{1} E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}} (or = \frac{3}{2 π N _{s}} \frac{E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}})$

Same as the starting torque.

Condition for Maximum Torque Under Running Conditions

The torque of a rotor under running condition is

$T_{r} = K φ \frac{s E _{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}} = \frac{K _{1} s E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

The condition for maximum torque may be obtained by differentiating the above expression w.r.t to slip, s and then putting it equal to zero. However, it is simpler to put $Y = \frac{1}{T _{r}}$ and then differentiate it.

$∴ Y = \frac{R _{2} ^{2} + ( s X _{2} ) ^{2}}{K φ s E _{2} R _{2}} = \frac{R _{2}}{K φ s E _{2}} + \frac{s X _{2} ^{2}}{K φ E _{2} R _{2}}$

$∴ \frac{d Y}{d s} = \frac{- R _{2}}{K φ s ^{2} E _{2}} + \frac{X _{2} ^{2}}{K φ E _{2} R _{2}} = 0$

$∴ \frac{R _{2}}{K φ s ^{2} E _{2}} = \frac{X _{2} ^{2}}{K φ E _{2} R _{2}} or R_{2}^{2} = s^{2} X_{2}^{2} or R_{2} = s X_{2}$

Condition for Maximum Torque Under Running Conditions

Hence torque under running condition is maximum at that slip which makes rotor reactance per phase equal to rotor resistance per phase. This slip is sometimes written as $s_{b}$ and the maximum torque as $T_{b}$ .

$The slip corresponding to maximum torque is s =^{R_{2}} /_{X_{2}}$

After putting $R_{2} = s X_{2}$ in the expression of torque, we get

$T_{ma x} = \frac{K φ s ^{2} E _{2} X _{2}}{2 s ^{2} X _{2} ^{2}} (or \frac{K φ s E _{2} R _{2}}{2 R _{2} ^{2}})$

$Or T_{ma x} = \frac{K φ E _{2}}{2 X _{2}} (or \frac{K φ s E _{2}}{2 R _{2} ^{2}})$

Condition for Maximum Torque Under Running Conditions

$Substituting value of s =^{R_{2}} /_{X_{2}} in another expression of torque, we get$

$T_{ma x} = K_{1} \frac{( R _{2} / X _{2} ) E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( R _{2} / X _{2} ) ^{2} X _{2} ^{2}} = K_{1} \frac{E _{2} ^{2}}{2 X _{2}}$

$Since K_{1} = \frac{3}{2 π N _{s}}, we have T_{ma x} = \frac{3}{2 π N _{s}} \frac{E _{2} ^{2}}{2 X _{2}} N-m$

It is evident from the above expression that,

The value of rotor resistance does not alter the value of the maximum torque but only the value of slip at which it occurs.
The maximum torque varies inversely as the standstill reactance. Therefore it should be kept as small as possible.
The maximum torque varies directly with the square of the applied voltage .
To obtain maximum torque at starting, the rotor resistance must be made equal to rotor reactance at standstill.

Torque-slip characteristics

If a curve is drawn between the torque and slip for a particular value of rotor resistance $R_{2}$ , the graph thus obtained is called torque-slip characteristic.

It shows a family of torque-slip characteristics for a slip-range from s = 0 to s = 1 for various values of rotor resistance.

$We know that torque, T = \frac{K φ s E _{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

The following points may be noted carefully

At s = 0, T = 0 so that torque-slip curve starts from the origin.
At normal speed, slip is small so that $s X_{2}$ is negligible as compared to $R_{2}$ .

$∴ T \propto s / R_{2}$ $\propto s [as R_{2} is constant]$

Hence torque slip curve is a straight line from zero slip to a slip that corresponds to full-load.

Family of torque-slip curves for different values of rotor resistance R2 up to maximum torque Tm

Torque-slip characteristics

As slip increases beyond full-load slip, the torque increases and becomes maximum at $s = R_{2} / X_{2}$ . This maximum torque in an induction motor is called pull-out torque or break down torque.

As the slip further increases (i.e. motor speed falls) with further increase in motor load, then $R_{2}$ becomes negligible as compared to $s X_{2}$ Therefore, for large values of slip

$T \propto \frac{s}{( s X _{2} ) ^{2}} \propto \frac{1}{s}$

Thus the torque is now inversely proportional to slip. Hence torque-slip curve is a rectangular hyperbola.

The maximum torque remains the same and is independent of the value of rotor resistance. Therefore, the addition of resistance to the rotor circuit does not change the value of maximum torque but it only changes the value of slip at which maximum torque occurs.

Torque-speed characteristics

Torque-speed characteristics of induction motor for resistances R, 4R, 6R showing variation of slip and speed percentage of Ns

SlidesByMaam/L-04_ECE-2107.md

9.2 KB · commit d2f3b6e

Electrical Machines-I

ECE-2107

Induction Motor-SL3

Fariya Tabassum
Assistant Professor, Dept. of Electrical & Computer Engineering
Rajshahi University of Engineering & Technology, Rajshahi-6204

“Allah does not burden a soul beyond that it can bear”.

[Sura Baqarah]

Rotor Torque

The torque T developed by the rotor is directly proportional to

Rotor current
Rotor e.m.f
Power factor of the rotor circuit

$∴ T \propto E_{2} I_{2} cos φ_{2}$

$Or T = K E_{2} I_{2} cos φ_{2}$

Where,
$I_{2} =$ Rotor current at standstill
$E_{2} =$ Rotor e.m.f at standstill
$cos φ_{2} =$ Rotor power factor at standstill

Starting Torque

The torque developed by the motor at the time of starting is known as starting torque of the motor.

Let, $E_{2} =$ rotor e.m.f. per phase at standstill
$R_{2} =$ rotor resistance per phase
$X_{2} =$ rotor reactance per phase at standstill

$Z_{2} = (R_{2}^{2} + X_{2}^{2}) = rotor impedance per phase at standstill$

$Then, rotor current I_{2} = \frac{E _{2}}{Z _{2}} = \frac{E _{2}}{( R _{2} ^{2} + X _{2} ^{2} )}$

$power factor cos φ_{2} = \frac{R _{2}}{Z _{2}} = \frac{R _{2}}{( R _{2} ^{2} + X _{2} ^{2} )}$

$Now, the starting torque T_{s t} = K_{1} E_{2} I_{2} cos φ_{2}$

$or T_{s t} = K_{1} E_{2} \frac{E _{2}}{( R _{2} ^{2} + X _{2} ^{2} )} \frac{R _{2}}{( R _{2} ^{2} + X _{2} ^{2} )} = K_{1} \frac{E _{2} ^{2} R _{2}}{R _{2} ^{2} + X _{2} ^{2}}$

Starting Torque

If the supply voltage is constant, then the flux $φ$ and hence $E_{2}$ both are constant.

$∴ T_{s t} = K_{2} \frac{R _{2}}{R _{2} ^{2} + X _{2} ^{2}} = \frac{K _{2} R _{2}}{Z _{2} ^{2}}$

$It can be shown that, K_{1} = \frac{3}{2 π N _{s}}$

$∴ T_{s t} = \frac{3}{2 π N _{s}} \frac{E _{2} ^{2} R _{2}}{R _{2} ^{2} + X _{2} ^{2}}$

Where, $N_{s} =$ Synchronous speed in r.p.s

For preparing your answer you can go through the article 34.13 of the book written by “B. L. Theraza”

Condition for Maximum Starting Torque

It can be proved that starting torque will be maximum when rotor resistance/phase is equal to standstill rotor reactance/phase.

$We know the staring torque, T_{s t} = K_{2} \frac{R _{2}}{R _{2} ^{2} + X _{2} ^{2}} (i)$

Differentiating equation (i) w.r.t $R_{2}$ and equating the result to zero we get,

$\frac{d T _{s t}}{d R _{2}} = K_{2} [\frac{1}{R _{2} ^{2} + X _{2} ^{2}} - \frac{R _{2} ( 2 R _{2} )}{( R _{2} ^{2} + X _{2} ^{2} ) ^{2}}] = 0$

$Or R_{2}^{2} + X_{2}^{2} = 2 R_{2}^{2}$

$∴ R_{2} = X_{2}$

Problems

Practice example 34.6, 34.7, 34.8, 34.9 and 34.11 of B. L. Theraza and also the related tutorial problem.

Torque Under Running Conditions

Let the rotor at standstill have per phase induced e.m.f. $E_{2}$ , reactance $X_{2}$ and resistance $R_{2}$ . Then under running conditions at slip s

$Rotor e.m.f per phase E_{r} = s E_{2}$

$Rotor reactance per phase X_{r} = s X_{2}$

$Rotor impedance per phase Z_{r} = (R_{2}^{2} + (s X_{2})^{2})$

$Rotor current per phase I_{r} = \frac{E _{r}}{Z _{r}} = \frac{s E _{2}}{( R _{2} ^{2} + ( s X _{2} ) ^{2} )}$

$power factor cos φ_{r} = \frac{R _{2}}{Z _{r}} = \frac{R _{2}}{( R _{2} ^{2} + ( s X _{2} ) ^{2} )}$

Torque Under Running Conditions

$Then, running torque T_{r} \propto E_{r} I_{r} cos φ_{r}$

$T_{r} \propto φ I_{r} cos φ_{r}$

$\propto φ \frac{s E _{2}}{( R _{2} ^{2} + ( s X _{2} ) ^{2} )} \frac{R _{2}}{( R _{2} ^{2} + ( s X _{2} ) ^{2} )}$

$\propto φ \frac{s E _{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

$= K φ \frac{s E _{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

$= \frac{K _{1} s E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

If the stator supply voltage is constant, then stator flux and hence $E_{2}$ will be constant.

Torque Under Running Conditions

$∴ T_{r} = \frac{K _{2} s R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

Where, $K_{2}$ is another constant.

It may be seen that running torque is

Directly proportional to the slip i.e., if slip increases (motor speed decreases), the torque will increase and vice-versa.
Directly proportional to square of supply voltage (*)**

Torque Under Running Conditions

$It can be shown that, K_{1} = \frac{3}{2 π N _{s}}$

Where, $N_{s} =$ Synchronous speed in r.p.s

$∴ T_{r} = \frac{3}{2 π N _{s}} \frac{s E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}} = \frac{3}{2 π N _{s}} \frac{s E _{2} ^{2} R _{2}}{Z _{r} ^{2}}$

$At standstill, when s=1, T_{r} = \frac{K _{1} E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}} (or = \frac{3}{2 π N _{s}} \frac{E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}})$

Same as the starting torque.

Condition for Maximum Torque Under Running Conditions

The torque of a rotor under running condition is

$T_{r} = K φ \frac{s E _{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}} = \frac{K _{1} s E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

$∴ Y = \frac{R _{2} ^{2} + ( s X _{2} ) ^{2}}{K φ s E _{2} R _{2}} = \frac{R _{2}}{K φ s E _{2}} + \frac{s X _{2} ^{2}}{K φ E _{2} R _{2}}$

$∴ \frac{d Y}{d s} = \frac{- R _{2}}{K φ s ^{2} E _{2}} + \frac{X _{2} ^{2}}{K φ E _{2} R _{2}} = 0$

$∴ \frac{R _{2}}{K φ s ^{2} E _{2}} = \frac{X _{2} ^{2}}{K φ E _{2} R _{2}} or R_{2}^{2} = s^{2} X_{2}^{2} or R_{2} = s X_{2}$

Condition for Maximum Torque Under Running Conditions

$The slip corresponding to maximum torque is s =^{R_{2}} /_{X_{2}}$

After putting $R_{2} = s X_{2}$ in the expression of torque, we get

$T_{ma x} = \frac{K φ s ^{2} E _{2} X _{2}}{2 s ^{2} X _{2} ^{2}} (or \frac{K φ s E _{2} R _{2}}{2 R _{2} ^{2}})$

$Or T_{ma x} = \frac{K φ E _{2}}{2 X _{2}} (or \frac{K φ s E _{2}}{2 R _{2} ^{2}})$

Condition for Maximum Torque Under Running Conditions

$Substituting value of s =^{R_{2}} /_{X_{2}} in another expression of torque, we get$

$T_{ma x} = K_{1} \frac{( R _{2} / X _{2} ) E _{2} ^{2} R _{2}}{R _{2} ^{2} + ( R _{2} / X _{2} ) ^{2} X _{2} ^{2}} = K_{1} \frac{E _{2} ^{2}}{2 X _{2}}$

$Since K_{1} = \frac{3}{2 π N _{s}}, we have T_{ma x} = \frac{3}{2 π N _{s}} \frac{E _{2} ^{2}}{2 X _{2}} N-m$

It is evident from the above expression that,

The value of rotor resistance does not alter the value of the maximum torque but only the value of slip at which it occurs.
The maximum torque varies inversely as the standstill reactance. Therefore it should be kept as small as possible.
The maximum torque varies directly with the square of the applied voltage .
To obtain maximum torque at starting, the rotor resistance must be made equal to rotor reactance at standstill.

Torque-slip characteristics

If a curve is drawn between the torque and slip for a particular value of rotor resistance $R_{2}$ , the graph thus obtained is called torque-slip characteristic.

It shows a family of torque-slip characteristics for a slip-range from s = 0 to s = 1 for various values of rotor resistance.

$We know that torque, T = \frac{K φ s E _{2} R _{2}}{R _{2} ^{2} + ( s X _{2} ) ^{2}}$

The following points may be noted carefully

At s = 0, T = 0 so that torque-slip curve starts from the origin.
At normal speed, slip is small so that $s X_{2}$ is negligible as compared to $R_{2}$ .

$∴ T \propto s / R_{2}$ $\propto s [as R_{2} is constant]$

Hence torque slip curve is a straight line from zero slip to a slip that corresponds to full-load.

Family of torque-slip curves for different values of rotor resistance R2 up to maximum torque Tm

Torque-slip characteristics

As the slip further increases (i.e. motor speed falls) with further increase in motor load, then $R_{2}$ becomes negligible as compared to $s X_{2}$ Therefore, for large values of slip

$T \propto \frac{s}{( s X _{2} ) ^{2}} \propto \frac{1}{s}$

Thus the torque is now inversely proportional to slip. Hence torque-slip curve is a rectangular hyperbola.

The maximum torque remains the same and is independent of the value of rotor resistance. Therefore, the addition of resistance to the rotor circuit does not change the value of maximum torque but it only changes the value of slip at which maximum torque occurs.

Torque-speed characteristics

Torque-speed characteristics of induction motor for resistances R, 4R, 6R showing variation of slip and speed percentage of Ns