DC Motor Speed Regulation MCQ Quiz in தமிழ் - Objective Question with Answer for DC Motor Speed Regulation - இலவச PDF ஐப் பதிவிறக்கவும்

P_out = 60 kW

\({P_{in}} = \frac{{60\;kW}}{\eta } = \frac{{60}}{{0.85}} = 70.59\;kW\)

P_in = VI

⇒ 70.59 = 600 × I

⇒ I = 0.117 kA = 117.65 A

\({I_f} = \frac{{600}}{{100}} = 6A\)

I_a = I - I_f = 117.65 – 6 = 111.65 A

V = I_aR_a + E

⇒ 600 = (111.65 × 0.16) + E

E = 582.14 V

E₁ = 582.14 V, N₁ = 900 rpm

At no load: E ≈ V = 600 V

⇒ E₀ = 600 V

We know that E ∝ N

For shunt motor, ϕ = constant

\(\frac{{{E_0}}}{{{E_1}}} = \frac{{{N_0}}}{{{N_1}}}\)

\(\Rightarrow {N_0} = \frac{{{E_0}\; \times \;{N_1}}}{{{E_1}}} = \frac{{600\; \times \;900}}{{582.14}} = 927.6\;rpm\)

Speed regulation \(= \frac{{{N_0} - {N_1}}}{{{N_1}}} \times 100\)

\(= \frac{{927.6 - 900}}{{900}} \times 100 = 3.1\%\)

Line current (I_L) = 40 A

For series motor, I_a = I_L = 40 A

E_b1 = V - I_a (R_a + R_se)

= 230 – 40 (0.2 + 0.1) = 218 V

ϕ₂ = 0.6 ϕ₁

For I_L = 20 A,

E_b2 = 230 – 20 (0.2 + 0.1) = 224 V

\(\frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{{N_1}{\phi _1}}}{{{N_2}{\phi _2}}}\)

\(\Rightarrow \frac{{218}}{{224}} = \frac{{\left( {1000} \right)\left( {{\phi _1}} \right)}}{{{N_2}\left( {0.6\;{\phi _1}} \right)}}\)

⇒ N₂ = 1712.54 rpm

Voltage (V) = 120 V

shunt field resistance (R_sh) = 240 Ω

Shunt field current \(\left( {{{\rm{I}}_{{\rm{sh}}}}} \right) = \frac{{120}}{{240}} = 0.5{\rm{\;A}}\)

Armature resistance (R_a) = 0.8 Ω

No load:

Load current (I_LO) = 2 A

Armature current (I_ao) = I_Lo – I_sh

= 2 – 0.5 = 1.5 A

Let No load speed is N_o

Back emf (E_bo) = V – I_ao R_a = 120 – (1.5) (0.8) = 118.8 V

Full load:

Load current (I_Lf) = 7 A

Armature current (I_af) = I_Lf – I_sh

= 7 – 0.5 = 6.5 A

Back emf (E_bf) – V - I_af R_a

= 120 – (6.5) (0.8)

= 114.8 V

Full load speed (N_f) = 1200 rpm

In a DC shunt motor

E_b ∝ N

\(\Rightarrow {{\rm{N}}_{\rm{o}}} = \frac{{{{\rm{E}}_{{\rm{bo}}}}}}{{{{\rm{E}}_{{\rm{bf}}}}}} \times {{\rm{N}}_{\rm{f}}} = \frac{{118.8}}{{114.8}} \times 1200 = 1241.8{\rm{\;rpm}}\)

V = 200 V, R₁ = R_a + R_se = 1 ohm, I_a = 15 A

N₁ = 600 rpm, N₂ = 475 rpm

In a DC series motor, E_b = V – I_aR

E_b ∝ Nϕ and ϕ ∝ I_a

\(\frac{{{E_1}}}{{{E_2}}} = \frac{{{I_{a1}}}}{{{I_{a2}}}} \times \frac{{{N_1}}}{{{N_2}}}\)

E₁ = 200 – 15 × 1 = 185 V

Given that I_a1 = I_a2

\( \Rightarrow \frac{{185}}{{{E_2}}} = \frac{{800}}{{475}}\)

Therefore, E₂ = 110 V

110 = 200 – 15 × R₂

⇒ R₂ = 6 Ω

Therefore, Extra Resistance to be added in Series:

R_ex = R₂ – R₁

= 6 – 1 = 5 Ω

At rated conditions,

E_b1 = V – I_a1 (R_a + R_s)

= 120 – (58) (0.2 + 0.16) = 99.12 V

At armature current of 35 A,

E_b2 = V – I_a2 (R_a + R_s)

= 120 – 35 (0.2 + 0.16) = 107.4 V

E_b ∝ N ϕ and ϕ ∝ V

\( \Rightarrow \frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{{N_1}}}{{{N_2}}} \times \frac{{{V_1}}}{{{V_2}}}\) 

N₁ = Rated speed = 1050 rpm

N₂ = speed at armature current of 35 A

From the given magnetization curve,

V₁ = voltage at armature current of 58 A = 134 V

V₂ = Voltage at armature current of 35 A = 115 V

\( \Rightarrow {N_2} = \frac{{{E_{b2}}}}{{{E_{b1}}}} \times \frac{{{V_1}}}{{{V_2}}} \times {N_1}\) 

\( = \frac{{107.4}}{{99.12}} \times \frac{{134}}{{115}} \times 1050\) 

Generator:

P = 4000 W

\(I = \frac{P}{V} = \frac{{4000}}{{200}} = 20\;A\)

E_g = 200 + (0.4) (20) = 208 V

Motor:

I = 20 A

200 = I_a R_a + E_m

E_m = 200 – 8 = 192 V

We know that E ∝ ϕ N

\(\frac{{{E_g}}}{{{E_m}}} = \frac{{{\phi _g}\;{N_g}}}{{{\phi _m}\;{N_m}}}\)

\(\Rightarrow \frac{{208}}{{192}} = \frac{\phi }{{1.1\;\phi }} \cdot \frac{{{N_g}}}{{{N_m}}}\)

\(\Rightarrow \frac{{{N_g}}}{{{N_m}}} = \frac{{208\; \times \;1.1}}{{192}} = 1.19 \approx 1.2\)

Concept:

DC Shunt motor and its currents are shown below

In motor back emf is given by

E_b = V - I_ar_a

And

E_b = K_nϕN

Where N = speed in rpm

K_n in V/rpm

ϕ = flux per pole

Calculation:

Given V = 240 V

R_a = 0.1 Ω

Armature current I_a = 70 A

E_b = 240 – 70 × 0.1 = 233 volt

\({K_n}\phi = \frac{{233}}{{850}}\)

Now when it runs on 650 rpm

Back emf \({E_b} = \frac{{233}}{{850}} \times 650 = 178.17\;V\)

Current flows through armature, I_a = 50 A

\(R = \frac{{240 - 178.17}}{{50}} = 1.23\;{\rm{\Omega }}\)

\(E = \frac{{P\phi ZN}}{{60\;A}} = k\;\phi \omega\)

\(\omega = \frac{{2\pi N}}{{60}},k = \frac{{PZ}}{{2\pi A}}\;\)

\(T = \frac{{P\phi Z{I_a}}}{{2\pi A}} = k\;\phi {I_a}\)

At no load:

\({E_{b0}} = {V_s} = k\;\phi \omega\)

\(k\phi = \frac{V}{\omega } = \frac{{110}}{{2\pi\; \times \;\frac{{9000}}{{60}}}}\)

⇒ kϕ = 0.2626

At load:

τ = 0.6 Nm

⇒ 0.6 = kϕI_a

\({I_a} = \frac{{0.6}}{{k\phi }} = \frac{{0.6}}{{0.2626}} = 2.284\;A\)

At V₁ = 55V,

E₁ = V₁ - I_a R_a = 55 – 2.284 (1.5) = 51.574 V

E₁ = k ϕ ω₂

\({\omega _2} = \frac{{{E_1}}}{{k\phi }} = \frac{{51.574}}{{0.2626}} = 196.393\;rad/sec\)

\(N = \frac{{60\;\omega }}{{2\pi }} = 1875.4\;rpm\)

The circuit of the above question can be drawn as shown in fig (a)

Given that, R_a = 0.5 Ω

I_a = 2A, leakage inductance, L = 0.01 H

V = 300 V

E_{no load} = V - I_aR_a

= 300 – 2 (0.5)

= 300 – 1 = 299 V

When load is applied, we have

R_a = 0.5 Ω & I_a = 15 A

The circuit can be drawn is shown in fig (b)

Hence, E_load – V - I_aR_a

= 300 – 15 × 0.5 = 292.5 V

E ∝ N

\(\frac{{{E_{no\;load}}}}{{{E_{load}}}} = \frac{{{N_{no\;load}}}}{{{N_{load}}}} \Rightarrow \frac{{299}}{{292.5}} = \frac{{900}}{{{N_{load}}}}\) 

⇒ N_load = 880.434 rpm

Torque is constant

T ∝ ϕ I_a

In DC shunt motor, ϕ is constant

⇒ I_a = constant.

E_b1 = 220 – 30 (0.5) = 205 V

E_b1 = 220 – 30 (0.5 + 1) = 175 V

E_b ∝ N

\(\Rightarrow \frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{{N_1}}}{{{N_2}}}\)