Torque Slip Characteristics MCQ Quiz in मराठी - Objective Question with Answer for Torque Slip Characteristics - मोफत PDF डाउनलोड करा

Concept:

The alternator always run with synchronous speed which is given by:

\(N_s={120f\over P}\)

The magnetic field of the 3ϕ induction motor rotates with synchronous speed whereas the induction motor rotates with a speed slightly less than the synchronous speed. 

The rotor speed of an induction motor is given by:

\(N_r=(1-s)N_s\)

Calculation:

For alternator:

Given, P = 8

N_s = 750 rpm

\(750={120f\over 8}\)

f = 50 Hz

For induction motor:

Given, P = 6

\(N_s={120\times 50\over 6}\)

N_S = 1000 rpm

\(N_r=(1-0.03)\times 1000\)

N_r = 970 rpm

Form the below diagram we can observe that the maximum torque of an induction motor is independent of rotor resistance but slip at which maximum torque occurs depends on rotor resistance and they change on adding the additional resistance to the rotor circuit.

The maximum torque of an induction motor is given by

\({T_{max}} = \frac{{kE_{20}^2}}{{2{X_{20}}}}\)

Hence, the maximum torque is only dependent on the reactance of the rotor and is independent of the rotor resistance.

Sm the value of slip corresponding to the maximum torque is

\({S_m} = \frac{{{R_2}}}{{{X_{20}}}}\)

Therefore, slip at which maximum torque occurs is directly proportional to rotor resistance R2.

Hence, both statement (I) and statement (II) are individually true but statement (II) is not the correct explanation of statement (I)

Slip is given by the formula:-

 \(s = {{N_s - N} \over N_s}\)

Where N_s is the synchronous speed and N is the rotor speed.

• In generating mode, the induction machine acts as a generator, and for this more torque is required. Mechanical input needs to be supplied to overcome the negative torque developed inside the machine.
• An induction generator takes reactive powers from the mains supply and supplies active power to the mains
• For generating mode Ns \(s = {{N_s - N} \over N_s}\) < 0
• therefore, in generating mode, an induction machine operates as a generator with a shaft speed which is greater than the synchronous speed, if the slip is less than zero.

Concept:

For the induction machine, the value of slip is given by

\(s = \frac{{{N_s} - {N_r}}}{{{N_s}}}\)

Where,

NS is the synchronous speed of the induction machine

Nr is the rotor speed of the induction machine

Application:

In the induction generator operation, a prime mover drives the rotor at a speed greater than the synchronous speed. That is, Nr > Ns

Therefore, under such conditions, the value of slip s is negative for the induction generator.

In low slip region (stable region)

\(T = \frac{1}{{{\omega _s}}}.\frac{{V_s^2}}{{r_2^1}}\)

\(\frac{V}{f}is\;constnat\)

Slip speed = SN_s = N_s – Nr

Ns ∝ f

\(T \propto \frac{{{V^2}}}{{{N_s}}}.\frac{{S{N_s}}}{{{N_s}}} \propto \frac{{{V^2}}}{{N_s^2}}.S{N_s}\)

\({\left( {\frac{V}{{{N_s}}}} \right)^2} \to Constant\) 

Slip speed, S_Ns → Constant

Concept

The slip of an induction motor is given by:

\(s={N_s-N_r\over N_s}\)

where, s = Slip 

N_s = Synchronous speed

N_r = Motor speed

Calculation

Given, s = 100% = 1

\(1={N_s-N_r\over N_s}\)

\(N_s={N_s-N_r}\)

The above expression is possible only, when the value of N_r = 0 which means the motor is not rotating.

Hence, the correct answer is option 1.

Additional Information If the value of slip, s = 0

\(0={N_s-N_r\over N_s}\)

\(0={N_s-N_r}\)

\({N_s=N_r}\)

When the value of slip is zero, the motor runs with full load speed.

• In a three-phase induction motor, the torque reaches its maximum value at a specific value of slip (s), known as slip at maximum torque or breakdown torque.
• This slip is typically between 0 and 1 and is less than the full-load slip.
• Maximum Torque: The maximum torque occurs at a slip value between 0 and 1, depending on the motor design and load characteristics. This slip is where the motor produces the highest torque it can generate before the torque starts decreasing as slip increases further.
• Slip = 1: This corresponds to the starting condition (when the rotor is stationary), and while starting torque can be high, it is not the maximum torque for the motor's operational range.
• Slip = 0: This corresponds to synchronous speed, where the torque is zero because there is no relative motion between the stator magnetic field and the rotor.
   

Additional Information

Torque equation:

The torque equation of a three-phase induction motor is given by,

\(T = \frac{{180}}{{2\pi {N_s}}}\left( {\frac{{sV^2{R_2}}}{{\left( {R_2^2 + {s^2}X_2^2} \right)}}} \right)\)

• Where Ns is the synchronous speed
• V = supply voltage
• R2 = rotor resistance
• X2 = rotor reactance
• s is the slip
• From the above expression, we can say that the torque of an induction motor depends on rotor resistance and slip.
   

Conditions for maximum torque:

The condition to get the maximum torque at starting is,

\({s_m} = \frac{{{R_m}}}{{{X_m}}} = 1\)

Xm = R­m

Where,

Rm = Motor resistance per phase

Xm = Motor reactance per phase

At the starting of the three-phase slip ring induction motor

Slip (s) = 1 (At the starting Nr = 0)

Therefore, \({s_m} = \frac{{{R_m}}}{{{X_m}}} = 1\)

Rotor frequency corresponding to 50 Hz (f_r) = 2 Hz

Slip corresponding to 50 Hz = 2/50 = 0.04

Since stator parameters are not given, the stator impedance is neglected.

T_max = maximum torque per phase \(= \frac{{0.5E_2^2}}{{{\omega _s}{x_2}}}\)

\(\Rightarrow {T_{max}} \propto \frac{{E_2^2}}{{{f^2}}}\)

Slip corresponding to maximum torque at 50 Hz, s_max1 = 0.1

The slip corresponding to maximum torque is, \({s_{max}} = \frac{{{r_2}}}{{{x_2}}}\)

\( \Rightarrow {s_{max}} \propto \frac{1}{f}\)

Now, the slip corresponding to maximum torque at 40 Hz, \({s_{max2}} = {s_{max1}} \times \left( {\frac{{50}}{{40}}} \right) = 0.125\)

Let T_0.04 be the torque developed at a slip of 0.04

Again, it is given that the torque-slip characteristic is linear between zero torque and maximum torque in the working region.

\({T_{0.04}} = \frac{{{T_{max}}}}{{{s_{max}}}} \times 0.04 = \frac{{{{\left( {\frac{{400}}{{\sqrt 3 }}} \right)}^2}}}{{{{\left( {50} \right)}^2}}} \times \frac{{0.04}}{{0.1}}\)

For 340 V, 40 Hz system,

\({T_{0.04}} = \frac{{{T_{max}}}}{{{s_{max}}}} \times 0.04 = \frac{{{{\left( {\frac{{340}}{{\sqrt 3 }}} \right)}^2}}}{{{{\left( {40} \right)}^2}}} \times \frac{{{s_{340}}}}{{0.125}}\)

Where s₃₄₀ is the slip corresponding to 340 V supply.

It is given that the torque developed is the same for both the cases.

So, from the above two equations. We get

\(\frac{{{{\left( {\frac{{400}}{{\sqrt 3 }}} \right)}^2}}}{{{{\left( {50} \right)}^2}}} \times \frac{{0.04}}{{0.1}} = \frac{{{{\left( {\frac{{340}}{{\sqrt 3 }}} \right)}^2}}}{{{{\left( {40} \right)}^2}}} \times \frac{{{s_{340}}}}{{0.125}}\)

⇒ s₃₄₀ = 0.0446

\({N_S} = \frac{{120 \times 50}}{6} = 1000rpm\)

\(s = \frac{{1000 - 960}}{{1000}} = 0.04\)

Equivalent load resistance per phase, \({R_L} = {R_2}\left( {\frac{1}{S} - 1} \right)\)

\(= 1\left( {\frac{1}{{0.04}} - 1} \right)\)

= 24 Ω

Torque slip characteristic of 3-phase induction motor:

Torque-slip characteristics let us divide the slip range (s = 0 to s = 1) into two parts

1) Low slip region

2) High slip region

Low slip region: T ∝ s

• In low slip region torque is directly proportional to slip.
• The torque slip characteristics of induction motor is a straight line. So as load increases, speed decreases, increasing the slip.
• This increases the torque which satisfies the load demand. This region is called Stable region of operation.

High slip region: T ∝ 1/s

• At higher values of slip (i.e. the slip beyond that corresponding to maximum torque) torque is approximately inversely proportional to slip(s) and the torque slip characteristics of induction motor is rectangular hyperbola as shown in the figure.
• The region (extending from s = s_m to s = 1) is called unstable region. In this region with the increase in load, slip increases but torque decreases.
• The result is that the motor could not pick up the load and slows down and eventually stops. In the unstable region, the value of slip is large so this region is also called the high-slip region.

The given slip 0.8 p.u. is high so the motor has unstable torque