Electrical Machines Test 5

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Electrical Machines Test 5

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Weekly Quiz Competition

Question 1
1 / -0

A 400 V shunt motor has armature resistance 0.5 Ω and field resistance of 100 Ω. If the motor is delivering 10 kW at 90%. Efficiency then the value of induced emf will be – (in V)

A
387-389

B
3874-3891

C
3877-3896

D
3879-3892

Solution

Efficiency, \({\rm{\eta }} = \frac{{{P_{oput}}}}{{{P_{in}}}}\)
\( \Rightarrow 0.9 = \frac{{10 \times {{10}^3}}}{{{{\rm{P}}_{{\rm{in}}}}}}\)
⇒ P_in = 11.11 kW
Motor Line current, \({I_L} = \frac{{{P_{in}}}}{V} = \frac{{11.11 \times {{10}^3}}}{{400}} = 27.78A\)
\({I_{sh}} = \frac{V}{{{R_{Sh}}}} = \frac{{400}}{{100}} = 4A\)
Armature current, I_a = I_L - I_sh = 27.78 - 4 = 23.78A
Back emf, E_b = V - I_aR_a = 400 - (23.78) (0.5) = 388.11 V
Question 2
1 / -0

A 220 V shunt motor with armature resistance of 0.5 Ω, runs at 500 rpm and draws 30 A at full load. The shunt field is excited to give constant main field. What will be the speed at full load if a 1 ohm resistor is placed in series with the armature circuit?

A
427 rpm

B
453 rpm

C
557 rpm

D
401 rpm

Solution

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
\(\Rightarrow \frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{{N_1}}}{{{N_2}}}\)
\(\Rightarrow {N_2} = \frac{{175}}{{205}} \times 500 = 426.829 \approx 427\;rpm\)
Question 3
1 / -0

A DC machine has total armature ampere conductors of 4500 and total flux in the machine is 0.14 wb. The torque developed in the machine is _________ (in N-m/rad)

A
99-101

B
994-1011

C
997-1016

D
999-1012

Solution

Developed torque \(\left( T \right)=\left( \frac{PZ}{2\pi A} \right)\phi {{I}_{a}}\)
P is number of poles
Z is number of conductors
ϕ is flux per pole
I_a is armature current
A is number of parallel paths
Total flux = 0.14 wb
⇒ Pϕ = 0.14 wb
Current in each conductor \(=\frac{{{I}_{a}}}{A}\)
Armature ampere conductors \(=Z\left( \frac{{{I}_{a}}}{A} \right)\)
= 4500
\(\Rightarrow T=\frac{0.14\times 4500}{2\pi }=100.26~N-m/r\)
Question 4
1 / -0

A 4-pole dc motor is lap wound with 480 conductors. The pole shoe is 20 cm long and the average flux density over one-pole pitch is 0.4T, the armature diameter being 40 cm.
The torque developed when the motor is drawing 30 A and running at 1500 rpm is ______ (in N-m)

A
56.5-58.5

B
56.54-58.51

C
56.57-58.56

D
56.59-58.52

Solution

Diameter (d) = 40 cm = 40 × 10^-2 m
Length of pole shoe (l) = 20 cm = 20 × 10^-2 cm
Flux density (B) = 0.4 T
Flux/pole (ϕ) = B A
\(= 20 \times {10^{ - 2}} \times 0.4 \times \frac{{\pi\; \times \;\left( {40\; \times \;{{10}^{ - 2}}} \right)}}{4}\)
= 0.025
Number of conductors (Z) = 480
Number of Poles (P) = 4
Number of Parallel Paths (A) = P = 4
Speed (N) = 1500
\(Emf = \frac{{\phi ZNP}}{{60\;A}} = \frac{{0.025\; \times \;1500\; \times \;480 \times 4}}{{60\; \times \;4}} = 300\;V\)
Gross mechanical power developed (D) = E_b I_a
P = 300 × 30 = 9 kW
Torque developed. \(\left( T \right) = \frac{P}{{\left( {\frac{{2\pi N}}{{60}}} \right)}}\)
\(= \frac{{9\; \times \;{{10}^3}}}{{\left( {\frac{{2\pi\; \times \;1500}}{{60}}} \right)}} = 57.3\;Nm\)
Question 5
1 / -0

A 20 hp, 230 V, 1150 rpm dc shunt motor has 4 poles four parallel armature paths and 882 armature conductors. The armature resistance is 0.188 Ω. At rated speed and rated output, the armature current is 73 A and the field current is 1.6 A.
The electromagnetic torque is ____________ (N-m/r)

A
131-132

B
1314-1321

C
1317-1326

D
1319-1322

Solution

Number of poles (P) = 4
Number of parallel paths (A) = 4
Armature resistance (R_a) = 0.188Ω
Armature current (I_a) = 73 A
Number of conductors (Z) = 882
Field current (I_sh) = 1.6 A
Speed (N) = 1150 rpm
Voltage (V) = 230 V
Back emf (E_b) = V – I_aR_a
= 230 – 73(0.188)
= 216.276 V
\({{E}_{b}}=\frac{\phi ZNP}{60~A}\)
\(\Rightarrow 216.276=\frac{\phi \times 882\times 1150\times 4}{60\times 4}\)
⇒ ϕ = 12.8 mwb
\(T=\left( \frac{PZ}{2\pi A} \right)\phi {{I}_{a}}\)
\(=\frac{4\times 882}{2\pi \times 4}\times 12.8\times {{10}^{-3}}\times 73=131.39~N-m/r\)
Question 6
1 / -0

A 250 V shunt motor on no-load runs at 1000 rpm and takes 5 A. The total armature and shunt field resistances are respectively 0.2 Ω and 250 Ω. Calculate the speed (in rpm) when loaded and taking a current of 50 A, if armature reaction weakens the field by 3%

A
991-997

B
9914-9971

C
9917-9976

D
9919-9972

Solution

No load:
V = 250 V
N₁ = 1000 rpm
I₁ = 5 A
\({I_{sh}} = \frac{V}{{{R_{sh}}}} = \frac{{250}}{{250}} = 1\;A\)
I_sh = I_L1 - I_sh = 5 – 1 = 4 A
E_b1 = V – I_a1 R_a = 250 – (4) (0.2) = 249.2 V
Loaded condition:
I_L2 = 50 A
I_a2 = I_L2 - I_sh = 50 – 1 = 49 A
E_b2 = V – I_a2 R_a = 250 – (49) (0.2) = 240.2 V
ϕ₂ = 0.97 ϕ₁
E_b ∝ N ϕ
\(\Rightarrow \frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{{N_1}}}{{{N_2}}} \times \frac{{{\phi _1}}}{{{\phi _2}}}\)
\(\Rightarrow \frac{{249.2}}{{240.2}} = \frac{{1000}}{{{N_2}}} \times \frac{{{\phi _1}}}{{0.97\;{\phi _1}}}\)
⇒ N₂ = 993.69
Question 7
1 / -0

A 400 V DC shunt motor draws 30 amperes while supplying the rated load at a speed of 120 rad/sec. The armature resistance is 1 ohm and the field winding resistance is 250 ohms. Find the external resistance (in ohms) is inserted in series so that, the armature current should not exceed 150% of its rated value, when the motor is plugged.

A
17-17.2

B
174-17.21

C
177-17.26

D
179-17.22

Solution

Voltage (V) = 400 V
Load current (I_L) = 30
Armature resistance (R_a) = 1Ω
Shunt resistance (R_sh) = 250Ω
Field current \(\left( {{I}_{f}} \right)=\frac{V}{{{R}_{sh}}}=\frac{400}{250}=1.6~A\)
Armature current (I_a) = I_L – I_sh
= 30 – 1.6 = 28.4 A
Back emf (E_b) = V – I_aR_a
= 400 – (28.4) (1)
= 371.6 V
In plugging mode,
\(V+{{E}_{b}}={{I}_{ab}}\left( {{R}_{a}}+{{R}_{b}} \right)\)
\({{R}_{a}}+{{R}_{b}}=\frac{400+371.6}{1.5\times {{I}_{a}}}\)
\(\Rightarrow {{R}_{b}}=\frac{771.6}{1.5\times 28.4}-1=17.11~\text{ }\!\!\Omega\!\!\text{ }\)
Question 8
1 / -0

A 250 V shunt motor is driving a load at 600 rpm and the armature draws a current of 20 A. What resistance should be inserted in series to the shunt field if the speed is to be raised from 600 rpm to 800 rpm? The armature resistance and shunt field resistance are 0.5 ohm and 250 ohm respectively.

A
88 Ω

B
112 Ω

C
63 Ω

D
42 Ω

Solution

V = 250 V
R_a = 0.5 Ω, R_sh = 250 Ω
Case 1:
\({I_{sh1}} = \frac{V}{{{R_{sh}}}} = \frac{{250}}{{250}} = 1\)
I_a1 = 20 A
E_b1 = 250 – (20) (0.5) = 240 V
Case 2:
E_b2 = 250 – I_a2 (0.5) = 250 – 0.5 I_a2
Torque is constant.
T ∝ ϕ I_a
⇒ ϕ I_a = constant
ϕ ∝ I_sh
⇒ I_sh I_a = constant
⇒ I_sh1 I_a1 = I_sh2 I_a2
⇒ I_sh2 I_a2 = 20
I_b ∝ N ϕ ∝ N I_sh
\(\Rightarrow \frac{{{E_{b1}}}}{{{E_{b2}}}} = \frac{{{N_1}}}{{{N_2}}} \times \frac{{{I_{sh1}}}}{{{I_{sh2}}}}\)
\(\Rightarrow \frac{{240}}{{250 - 0.5\;{I_{a2}}}} = \frac{{600}}{{800}} \times \frac{1}{{{I_{sh2}}}}\)
\(\Rightarrow \frac{{240}}{{250 - 0.5\left( {\frac{{20}}{{{I_{sh2}}}}} \right)}} = \frac{{600}}{{800}} \times \frac{1}{{{I_{sh2}}}}\)
\(\Rightarrow 320\;\;{I_{sh2}} = 250 - \frac{{10}}{{{I_{sh2}}}}\)
\(\Rightarrow 320\;I_{sh2}^2 - 25\;{I_{sh2}} + 10 = 0\)
⇒ I_sh2 = 0.7389 A
Shunt resistance corresponding to I_sh2 is
\({R_{sh2}} = \frac{V}{{{I_{sh2}}}} = \frac{{250}}{{0.7389}} = 338\;{\rm{\Omega }}\)
Resistance to be added = 338 – 250 = 88 Ω

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Re-Attempt Weekly Quiz Competition

Electrical Machines Test 5

Result

Electrical Machines Test 5

Score

-

Rank

-

SHARING IS CARING

Question 1

387-389

3874-3891

3877-3896

3879-3892

Solution

Question 2

427 rpm

453 rpm

557 rpm

401 rpm

Solution

Question 3

99-101

994-1011

997-1016

999-1012

Solution

Question 4

56.5-58.5

56.54-58.51

56.57-58.56

56.59-58.52

Solution

Question 5

131-132

1314-1321

1317-1326

1319-1322

Solution

Question 6

991-997

9914-9971

9917-9976

9919-9972

Solution

Question 7

17-17.2

174-17.21

177-17.26

179-17.22

Solution

Question 8

88 Ω

112 Ω

63 Ω

42 Ω

Solution

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