Electrical Machines Test 3

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Electrical Machines Test 3

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Weekly Quiz Competition

Question 1
1 / -0

A 6.6 kV/ 415 V delta/star connected distribution transformer has 4% resistance and 8% reactance. The voltage regulation of transformer at full load 0.6 Pf lead is ________ (in %)

A
-4.1--3.9

B
-4.14--3.91

C
-4.17--3.96

D
-4.19--3.92

Solution

Given that,
% R = 4%, % x = 8%
Power factor = 0.6 lead
⇒ cosϕ = 0.6, sin ϕ = 0.8
Regulation = % R cosϕ - % X sin ϕ
= (4 × 0.6) - (8 × 0.8)
= 2.4 - 6.4 = -4%
Question 2
1 / -0

Transformer connections of two three phase transformers are given in the options. Which of the pair is not suitable for the parallel operation.

A
∆/∆, Y/Y

B
∆/Zig Zag Y, Y/Y

C
∆/Y, Y/∆

D
Y/Zig Zag Y, Y/Y

Solution

For the parallel operation, phase displacement angle between secondaries of both transformers must be zero. That means both the transformers must belong to same phase group.
Transformer connections
Phasor displacement angle
∆/∆, Y/Y, ∆/Zig Zag Y
0° (or) 180°
∆/Y, Y/∆, Y/Zig Zag Y
± 30° (or) 180° ± 30°

In the given options,
Y/Zig Zag Y and Y/Y belongs to different phasor groups. Hence this pair is not suitable for parallel operation.
Question 3
1 / -0

A 3-phase, 50 Hz transformer has an iron cross section of 200 cm². If the flux density be limited to 1.5 wb/m². The voltage ratio is 4400/220 V, the high voltage side being connected in star and low voltage in delta. Stacking factor is 0.8. The sum of number of turns per phase on high and low voltage winding is

A
510-550

B
5104-5501

C
5107-5506

D
5109-5502

Solution

Given that, gross cross section = 200 cm²
Stacking factor = 0.8
A_i = 200 × 0.8 = 160 cm²
∅ = B_max × A_i = 1.5 × 160 × 10^-4
= 0.024 wb
HV side phase voltage, \({{\rm{E}}_{{\rm{P}}1}} = \frac{{4400}}{{\sqrt 3 }} = 2540.3{\rm{V}}\)
LV side phase voltage, E_P2=220V
Turns Per Phase on low voltage winding
\({{\rm{N}}_2} = \frac{{{{\rm{E}}_{{\rm{P}}2}}}}{{4.44{\emptyset _{max}}{\rm{f}}}} = \frac{{220}}{{4.44 \times 0.024 \times 50}} = 41.29 \approx 42\)
Turns Per Phase on high voltage winding,
\({N_1} = {N_2} \times \frac{{{E_{P1}}}}{{{E_{p2}}}} = \frac{{42 \times 2540.3}}{{220}} = 485\)
Sum = 42 + 485 = 527
Question 4
1 / -0

A 500 kVA, 13 kV/ 440V star-star connected transformer is reconnected as delta-star. The rating of the newly connected transformer is

A
\(500\sqrt 3 {\rm{\;kVA}},{\rm{\;}}13\sqrt 3 {\rm{\;kV}}/440{\rm{V}}\)

B
\(500{\rm{kVA}},{\rm{\;}}13{\rm{kV}}/\frac{{440}}{{\sqrt 3 }}{\rm{V}}\)

C
\(500{\rm{kVA}},{\rm{\;}}\frac{{13}}{{\sqrt 3 }}{\rm{kV}}/440{\rm{V}}\)

D
\(\frac{{500}}{{\sqrt 3 }}{\rm{kVA}},\frac{{13}}{{\sqrt 3 }}{\rm{kV}}/440{\rm{V}}\)

Solution

For star – star connected transformer,
Phase turns ratio\(= \frac{{13\;kV}}{{\sqrt 3 }}:\frac{{440}}{{\sqrt 3 }}\)
When transformer is reconnected as delta-star.
In delta connection, phase voltage = line voltage
In star connection, phase voltage\(= \frac{1}{{\sqrt 3 }} \times line\;voltage\)
\(\begin{array}{l}\frac{\Delta }{Y} = \frac{{13\;kV}}{{\sqrt 3 }}:\frac{{440}}{{\sqrt 3 }} \times \sqrt 3 \\ = \frac{{13kV}}{{\sqrt 3 }}:440\end{array}\)
KVA remains constant irrespective of connection.
Hence rating of newly connected transformer is \(500kVA,\frac{{13kV}}{{\sqrt 3 }}/440V\)
Question 5
1 / -0

If the OC test data on delta side of a three phase Y - Δ transformer is 66 kV, 11 A, 45 kW, the magnetizing component of phase current will be ________A.

A
6.28-6.35

B
6.284-6.351

C
6.287-6.356

D
6.289-6.352

Solution

We know that
\(P = \sqrt 3 {V_L}{I_L}\cos \phi\)
\(\cos {\phi _o} = \frac{{45 \times {{10}^3}}}{{\sqrt 3 \times 66 \times 11 \times {{10}^3}}} = 0.0358\)
sin ϕ_o = 0.99
Magnetizing component = I_m
No load component I_o = 11 A
\({I_m} = \frac{{{I_o}}}{{\sqrt 3 }}\sin {\phi _o} = \frac{{11}}{{\sqrt 3 }} \times 0.99 = 6.287A\)