Option 3 : 76.4 N-m

**Concept:**

Speed of the motor is given as

\({ω} = \frac{{{N} \times 2\pi }}{{60}}\)

Torque developed by the motor is

T = P / ω

where,

Power developed P = E I_{a}

\(E = \frac{ϕ ZNp}{60A}\)

∴ \(T= \frac{\frac{ϕ ZNp}{60A}I_a}{\frac{{{N_s} \times 2\pi }}{{60}}}=\frac{ϕ ZpI_a}{2\pi A}\)

**Calculation:**

Given,

Armature current I_{a} = 50 A

Number of poles p = 4

Lap wound A = p = 4

Number of conductors Z = 480

The flux per pole ϕ = 20 mWb

\(T=\frac{ϕ ZpI_a}{2\pi A}=\frac{20\times 10^{-3}\times 480\times 50 \times4}{4\times 2\pi }\)

**T = 76.4 N-m**

Option 2 : 79.6 N-m

__Concept:__

**Torque:**

- When armature conductors of a DC motor carry current in the presence of stator field flux, a mechanical torque is developed between the armature and the stator.
- Torque is given by the product of the force and the radius at which this force acts.

**Power:**

- Mechanical power developed by the motor is given by P
_{m}= E_{b}× I_{a} - E
_{b }is EMF developed and I_{a }is armature current - In terms of torque and speed mechanical power developed is given by P
_{m}= torque × speed. - Speed is in radians per second (ω).
- \({\rm{\omega }} = 2\;\pi \frac{N}{{60}}\) radians per sec.

__Calculations:__

**\({\rm{\omega }} = \frac{{2\; \times \;{\rm{\pi }}\; \times \;1500}}{{60}}\)**

= 157.08 radians/sec.

Power = generated emf × armature current

= 250 × 50

= 12500 watts.

Torque = power/speed.

= \(\frac{{12500}}{{157.08}}\)

= **79.577 Nm.**

Option 1 : 44%

__Concept:__

For DC Machine, torque is directly proportional to armature current and flux.

**T ∝ ϕ Ia**

In dc shunt motors, flux is constant (not dependent on armature current). Hence torque is proportional to armature current.

**T ∝ I _{a}**

I**n dc series motors, flux is directly proportional to armature current. Hence torque is directly proportional to square of the armature current.**

**Ta ∝ Ia2 **(∵ Flux saturation is neglected)

In dc series motor at **saturation condition**, flux is constant. Hence torque is proportional to armature current.

**T ∝ Ia**

Where, T = torque, Ia = armature current, ϕ = field flux

__Calculation:__

Let at I = 10 A, the torque is T1 and at I = 12 A torque is T2

\(\frac{{{T_1}}}{{{T_2}}} = {\left( {\frac{{10}}{{12}}} \right)^2}\)

T2 = 1.44 T1

i.e. T2 increases by 44%

Option 2 : 202.5 N-m

**Concept:**

\(T = \frac{ZPϕ(I_a)}{A60}\)

Where,

T is the torque of the DC motor

Z is the number of conductors

A is the number of parallel paths

P is the number of poles of the motor

ϕ is the field flux of the motor

I_{a} is armature current

For any DC machine, \((\frac{ZP}{A60})\) remains constant

∴ T ∝ ϕ Ia

**Calculation:**

Given,

Torque T_{1} = 150 N-m

There is 10% reduction in field flux

⇒ ϕ_{2} = 90% ϕ_{1} = 0.9 ϕ_{1}

Also armature current increased by 50%

⇒ I_{a2} = 150% I_{a1} = 1.5 I_{a1}

As T ∝ ϕ I_{a}

\(\frac{{{T_2}}}{{{T_1}}} = \frac{{{\phi _2}}}{{{\phi _1}}} \times \frac{{{I_{a2}}}}{{{I_{a1}}}}\)

\(\frac{{{T_2}}}{{150}} = \frac{{0.9{\phi _2}}}{{{\phi _1}}} \times \frac{{1.5{I_{a2}}}}{{{I_{a1}}}} = 1.35\)

T_{2} = 202.5 N-m

**Therefore the new value of torque is 202.5 N-m**

Option 3 : Changing flux and armature current both

In a DC machine, torque (T) is given as

\(T = \frac{{{P_{out}}}}{ω } \) .......(1)

Where,

P_{out} = Output power in kW

ω = Angular speed in radian per second = 2π N

E_{b} = Back emf

I_{a} = Armature current

N = Speed in rpm

Z = Number of conductors

A = Number of parallel paths

P = Number of poles

Output power of DC motor is given by

Pout = Eb Ia ........(2)

Back EMF of DC motor is given by

\(E_b = \frac{{{ZP\phi N}}}{60A } \) .......(3)

From equations (1), (2) and (3), we get

\(T = \frac{E_bI_a}{2π N} = \frac{ZP\phi(I_a)}{2π \times60A}\)

For any DC machine, \((\frac{ZP}{2π \times60A})\) remains constant

∴ **T ∝ ϕ Ia**

**So that for DC Machine, torque is directly proportional to armature current and flux.**

**Hence torque of a given motor can be varied by changing flux and armature current both .**

__Note:__

In DC shunt motors, flux is constant (not dependent on armature current). Hence torque is proportional to armature current.

T ∝ Ia

In DC series motors, flux is directly proportional to armature current. Hence torque is directly proportional to square of the armature current.

Ta ∝ Ia2

In dc series motor at saturation condition, flux is constant. Hence torque is proportional to armature current.

T ∝ Ia

In a DC shunt motor, the torque produced is proportional to:

Option 1 : Armature current

For DC Machine, torque is directly proportional to armature current and flux.

T ∝ ϕ Ia

**In dc shunt motors, flux is constant (not dependent on armature current). Hence torque is proportional to armature current.**

**T ∝ Ia**

In dc series motors, flux is directly proportional to armature current. Hence torque is directly proportional to square of the armature current.

Ta ∝ Ia2

In dc series motor at saturation condition, flux is constant. Hence torque is proportional to armature current.

T ∝ Ia

Where, T = torque, Ia = armature current, ϕ = field flux

Option 2 : 202.5 N-m

Given ϕ_{2} = 0.9ϕ_{1} and I_{a2 }= 1.5 I_{a1}

T ∝ ϕ I_{a}

\(\frac{{{T_1}}}{{{T_2}}} = \frac{{{\phi _1}}}{{{\phi _2}}} \times \frac{{{I_{a1}}}}{{{I_{a2}}}}\)

\({T_2} = 150 \times \frac{{0.9{\phi _1}}}{{{\phi _1}}} \times \frac{{1.5{I_{a1}}}}{{{I_{a1}}}} = 202.5\;Nm\)Option 4 : 143.2 N-m

__Torque of D.C. Motor:__

Torque is meant the turning or twisting moment of a force about an axis.

Torque is the turning moment of a force about an axis and is measured by the product of force (F) and radius (r) at a right angle to which the force acts i.e.

D.C. Motors

In a DC Motor, each conductor is acted upon by a circumferential force F at a distance r, the radius of the armature as shown below.

T = F × r.

Let in a DC Motor

r = average radius of armature in m

\(l\) = effective length of each conductor in m

Z = total number of armature conductors

A = number of parallel paths

i = current in each conductor = Ia/A

B = average flux density in Wb/m2

ϕ = flux per pole in Wb

P = number of poles

Force on each conductor, (F) = \(\large{B\ i\ l}\) newtons

Torque due to one conductor = F × r newton- meter

Total armature torque, (Ta) = Z F r newton- meter

∴ \(\large{T_a=ZBilr}\) newton- meter ... (1)

Now,

i = Ia/A, (current in each conductor)

B = ϕ/a; where a is the x-sectional area of flux path per pole a radius r.

And, \(\large{a=\frac{2π rl}{P}}\)

Substitute these values in equation (1),

\(\large{T_a=Z× \frac{ϕ}{a}× \frac{I_a}{A}× l× r}\)

\(\large{T_a=Z× \frac{ϕ}{2π rl/P}× \frac{I_a}{A}× l× r}\)

\(\large{T_a=0.159Zϕ I_a×(\frac{P}{A})}\)

Since, number of Poles (P) and number of the parallel paths (A) is constant,

∴ Ta ∝ ϕ Ia

(i) For a shunt motor, flux ϕ is practically constant

∴ Ta ∝ ϕ Ia

(ii) For a series motor, flux ϕ is directly proportional to armature current Ia provided magnetic saturation does not take place.

∴ Ta ∝ (Ia)2

**Calculation:**

Given,

P = 4

ϕ = 25 mwb

I_{L} = Ia = 30 A (Since it is DC series motor)

Z = 600

For Wave Winding, A = 2

From the above concept,

\(\large{T_a=0.159Zϕ I_a×(\frac{P}{A})}\)

or, \(\large{T_a=0.159× 600× 25× 10^{-3}× 30×(\frac{4}{2})}\)

∴ Gross Torque = 143.2 Nm

Option 2 : 10 N-m

**Concept:**

For separately excited dc generator armature current and load current is the same.

The output power of generator = Terminal voltage × Armature current.

Torque exerted by the armature T = Power output / ω

where ω = (2 π N) / 60

N = speed of the armature in rpm.

**Calculation:**

Given sped of the generator N = 3000 rpm

Terminal voltage V_{t} = 157 V

Delivered load current I_{L} = 20 A

Power developed by generator **P = V _{t} × I_{L}**

P = 157 × 20 = 3140 W

Torque exerted by armature **T = P / ω **

where **ω = (2 π N) / 60**

ω = (2 π × 3000) / 60 = 314 rad / sec

∴ **T = 3140 / 314 = 10 N-m**

**Concept:**

The torque of the motor is, \(T = \frac{P}{\omega } = \frac{{V{I_a}}}{\omega }\)

Where \(\omega = \frac{{2\pi {N_s}}}{{60}}\)

I_{a} = Armature current, V = Excitation voltage

**Calculation:**

No load speed, N = 1400 rpm

I_{a} = 8A, V = 230 V

\(T = \frac{{V{I_a}}}{\omega } = \frac{{V{I_a}}}{{\frac{{2\pi\; \times \;1400}}{{60}}}} = \frac{{230\; \times\; 8}}{{\frac{{2\; \times \;\pi \; \times \;1400}}{{60}}}}\)

\(T = \frac{{9.55\; \times \;230\; \times\; 8}}{{1400}}\)

T = 12.554 Nm

Option 2 : 30 A

Voltage (V) = 220 V

Torque (τ) = 70 N-m

Speed (N) = 900 rpm

Armature resistance (R_{a}) = 0.02 Ω

Power (P) = 10 kW

⇒ E_{b }I_{a} = 10 × 10^{3}

⇒ (V - I_{a }R_{a}) I_{a} = 10,000

⇒ [220 – I_{a}(0.02)] I_{a} = 10,000

⇒ I_{a} = 45.75 A

P = τω

\(\Rightarrow 10 \times {10^3} = {\rm{\tau }}\left( {\frac{{2{\rm{\pi N}}}}{{60}}} \right)\)

\(\Rightarrow {\rm{\tau }} = \frac{{10 \times {{10}^3} \times 60}}{{2{\rm{\pi }} \times 900}} = 106.1{\rm{\;N}} - {\rm{m}}\)

Torque α ϕ I_{a}

For separately excited motor

T α I_{a}

T_{1} = 106.1 N-m, I_{a1} = 45.75 A

T_{2} = 70 N-m

\({{\rm{I}}_{{{\rm{a}}_2}}} = \frac{{{{\rm{T}}_2} \times {{\rm{I}}_{{{\rm{a}}_1}}}}}{{{{\rm{T}}_1}}} = \frac{{70 \times 45.75}}{{106.1}} = 30.18{\rm{\;A}}\)

An Electric motor, developing a starting torque of 15 Nm, starts with a load torque of 7 Nm on its shaft. If the acceleration at start is 2 rad/s^{2}, the moment of inertia of the system is (neglecting viscous and Coulomb friction)

Option 3 : 4 kg m2

**Concept:**

Accelerating torque in an electric motor is given as

**T _{a} = T_{m} - T_{e }**

**T _{a} = J α **

Where, T_{m} = Mechanical torque or starting torque

T_{e} = Electrical torque or load torque

J = Moment of inertia

α = Angular acceleration

**Calculation:**

Given-

T_{m }= 15 Nm, T_{e} = 7 Nm,_{ }α = 2 rad /sec^{2 }

J = (15 - 7) / 2

**J = 4 Nm ^{2}**

__Concept__

Under steady state load torque will be equal to The developed torque by the motor.

__Explanation__

Given,

no load speed (ω0) = 200 rad/sec

Rated torque (T_{r}) = 500 Nm

Rated Speed (ωr) = 180 rad/sec

Load torque (TL) = 2.78 ω_{r}

using equation of straight line

Developed torque can be expressed as,

\(T=\frac{200-ω_r}{20}\times500\)

At steady state,

load torque = developed torque

⇒ T_{L} = T

⇒ 2.78 ω_{r} = \(\frac{200-ω_r}{20}\times500\)

⇒ ω_{r} = 180 rad/sec

Therefore, steady state speed is 180 rad/sec

__ Concept__:

__Back emf:__

- In dc motor also generator action takes place.
- Because of this generator action the rotating conductor's cuts the flux and emf induced in the conductors.
- This induced emf is called back emf. It always opposes the supply voltage.

The back emf induced in the dc motor can be given by,

Eb = (ϕZNP) / (60 A)

Where,

ϕ = flux/pole

Z = total number of conductors

A = number of parallel paths

N = speed in RPM

P = number of poles

__ Calculation__:

\(E = \frac{{\phi ZNp}}{{60A}} = \frac{{0.01 \times 666 \times 4 \times 1000}}{{60 \times 2}} = 222 V\)

I_{a} = (V - E) / R_{a} = (230 - 222) / 0.267

I_{a} = 29.96 A

Internal power ( P_{d}) \(= \;EI_a = 222\; \times \;29.96 = 6651.685 Watt\)

Shaft Power output,

P_{out} = P_{d }- Rotational losses

\(\begin{array}{l} {P_{out}} = 6651.685\;-\;600\; = 6051.685\\ T = \frac{{{P_{out}}}}{\omega } = \frac{{6051.685}}{{2\pi \times \frac{{1000}}{{60}}}} = 57.78\;Nm \end{array}\)

__Concept__:

Equivalent circuit of separately excited D.C motor,

V = E_{b} + I_{a}R_{a}

__ Explanation__:

Let,

Torque T_{1} at N_{1} = 1000 rpm

Torque T_{2 }= at N_{2} = 500 rpm

∴ T ∝ N^{2} {Given}

⇒ \(\frac{T_2}{T_1}=\left(\frac{N_2}{N_1}\right)^2\)

⇒ \(\frac{T_2}{T_1}=\left(\frac{500}{1000}\right)^2\)

⇒ \(\frac{T_2}{T_1}=\frac{1}{4}\)

Load torque can be expressed as,

T = k ϕ. I_{a}

T ∝ ϕ I_{a}

T ∝ I_{a} {∴ ϕ = constant}

⇒ \(\frac{T_2}{T_1}=\frac{Ia_2}{Ia_1}\)

⇒ \(\frac{Ia_2}{Ia_1}=\frac{1}{4}\)

⇒ \(Ia_2=\frac{10}{4}=2.5 \) A

Therefore,

\(E_{b_1}=V_1-I{a_1}R_a=100-10=90\) V

∴ \(E_b=\frac{NP\phi Z}{60\ A}\)

⇒ \(\frac{E_{b_1}}{E_{b_2}}=\frac{N_1}{N_2}\)

⇒ \(E_{b_2}=E_{b_1}\frac{N_2}{N_1}\)

⇒ \(E_{b_2}=\frac{90×500}{1000}\)

⇒ \(E_{b_2}=45\) V

Therefore,

\(V_2=E_{b_2}+I_{a_2}R_a\)

⇒ V_{2} = 45 + 2.5 × 1

⇒ V_{2} = 47.5 V

Hence, The armature voltage is 47.5 V

Option 1 : 881 rpm

**Concept:**

In motor back emf is given by

E_{b} = V - I_{a}r_{a}

And

E_{b} = K_{n}ϕN

Where N = speed in rpm

K_{n} in V/rpm

ϕ = flux per pole

Torque T = K_{a}ϕI_{a}

\({K_a} = \frac{{30}}{\pi }{K_n}\)

**Calculation:**

Given V = 250 volt

I_{a} = 5 A

\({I_f} = \frac{{250}}{{200}} = 1.25\;A\)

I_{a} = 5 – 1.25 = 3.75 A

E_{b} = 250 – 3.75 × 0.5 = 248.125 volt

\({K_n}\phi = \frac{{248.125}}{{1000}} = 0.248125\)

\(T = \frac{{30}}{\pi }{K_n}\phi {I_a}\)

At T = 150 Nm

\({I_a} = \frac{{150 \times \pi }}{{0.248 \times 30}} = 63.33\;A\)

E_{b} = 250 – 63.33 × 0.5 = 218.335

\(N = \frac{{218.335}}{{0.248125}} = 881rpm\)