MEE115-INTRODUCTION TO ELECTRICAL MACHINE & INSTALLATION OSUNTECHCOLLEGE 2017/2018 FIRST SEMESTER EXAMINATION

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ANWSER

Question 1:

(a) Self-induction: Self-induction is the phenomenon where a change in current in a coil induces an electromotive force (e.m.f) in the same coil due to the changing magnetic field.

Mutual Induction: Mutual induction occurs when a changing current in one coil induces an e.m.f in a neighboring coil due to the magnetic field interaction between them.

(b) Effects of Self-Inductance in Electrical Circuits:

In AC circuits, self-inductance opposes the change in current, leading to energy storage in the magnetic field.
In DC circuits, self-inductance affects the switching of currents, leading to voltage spikes.
Practical Demonstration: A coil connected to a DC source with a switch and a bulb. When the switch is turned off, a high voltage spike may cause a spark across the switch due to self-induction.

(i) Flux produced by a current of 5A
Flux linkage $λ=L×I\lambda = L \times I$
$λ=10H×5A=50\lambda = 10H \times 5A = 50$ Weber-turns.
Since $λ=Nϕ\lambda = N\phi$ ,
$ϕ=λN=50300=0.1667\phi = \frac{\lambda}{N} = \frac{50}{300} = 0.1667$ Weber

(ii) Average value of induced e.m.f when the current reverses in 8ms
Using the formula:
$e=LdIdte = L \frac{dI}{dt}$
Here, $dI=10AdI = 10A$ (as the current reverses from +5A to -5A), and $dt=8ms=0.008sdt = 8ms = 0.008s$

$e=10×100.008=12,500Ve = 10 \times \frac{10}{0.008} = 12,500V$

Question 2:

(a) Proof of e.m.f formula $e=BLUe = B L U$
From Faraday’s Law, induced e.m.f is:
$e=dΦdte = \frac{d\Phi}{dt}$
Magnetic flux $Φ=B×A\Phi = B \times A$ , where $A=L×U×dtA = L \times U \times dt$ ,
$e=B×L×Ue = B \times L \times U$

(b) Flux calculations for rectangular coil
Given:

$200mm×100mm200mm \times 100mm$ coil
Rotates about midpoint of 100mm side
Magnetic field $B=0.05TB = 0.05T$

(i) Maximum flux
$Φmax⁡=B×A=0.05×(0.2×0.1)\Phi_{\max} = B \times A = 0.05 \times (0.2 \times 0.1)$
$Φmax⁡=0.001\Phi_{\max} = 0.001$ Weber

(ii) Flux when inclined at 45°
$Φ=Φmax⁡×cos⁡45∘\Phi = \Phi_{\max} \times \cos 45^\circ$
$Φ=0.001×0.707=0.000707\Phi = 0.001 \times 0.707 = 0.000707$ Weber

(c) Fleming’s Right-Hand Rule
It determines the direction of induced current when a conductor moves in a magnetic field.

Thumb = Motion
Forefinger = Field
Middle finger = Induced current

Question 3:

(a) Transformer Definition:
A transformer is a static electrical device that transfers electrical energy between two or more circuits through electromagnetic induction.

(b) Working Principle:
A transformer operates on the principle of mutual induction. When an alternating current passes through the primary coil, it produces a changing magnetic field, which induces an e.m.f in the secondary coil.

(c) RMS value of induced e.m.f per turn:
The formula is:
$E=4.44fΦmE = 4.44 f \Phi_m$
where:
$ff$ = Frequency,
$Φm\Phi_m$ = Maximum magnetic flux

Question 4:

(a) Given a 250kVA, 11000V/400V, 50Hz transformer with 80 turns on the secondary, calculate:

(i) Approximate primary and secondary current
Using: $S=V×IS = V \times I$
Primary current:
$Ip=SVp=250,00011,000=22.73AI_p = \frac{S}{V_p} = \frac{250,000}{11,000} = 22.73 A$
Secondary current:
$Is=SVs=250,000400=625AI_s = \frac{S}{V_s} = \frac{250,000}{400} = 625 A$

(ii) Number of primary turns
Using: $NpNs=VpVs\frac{N_p}{N_s} = \frac{V_p}{V_s}$
$Np=11,000400×80=2,200 turnsN_p = \frac{11,000}{400} \times 80 = 2,200 \text{ turns}$

(iii) Maximum flux
Using $E=4.44fΦmNpE = 4.44 f \Phi_m N_p$
Rearrange for $Φm\Phi_m$ :
$Φm=E4.44fNp=11,0004.44×50×2,200\Phi_m = \frac{E}{4.44 f N_p} = \frac{11,000}{4.44 \times 50 \times 2,200}$
$Φm=0.0227 Weber\Phi_m = 0.0227 \text{ Weber}$

(b) Transformer Losses:

Copper loss: Due to winding resistance
Iron loss: Hysteresis and eddy current losses
Stray losses: Leakage flux effects
Dielectric loss: Insulation failure

Question 5:

(a) Phasor Diagram of Transformer on No-load:

Voltage and flux are in phase.
Magnetizing current lags behind the applied voltage by 90°.
Core loss component is in phase with voltage.

(b) No-load current and Power Factor:

No-load current consists of magnetizing and loss components.
Power factor is low, mainly due to magnetizing current.

Effective cross-sectional area: $150cm2=0.015m2150 cm^2 = 0.015m^2$
Low-voltage winding turns: $8080$

(i) Maximum Flux Density $Bmax⁡B_{\max}$ :
$Bmax⁡=V4.44fANB_{\max} = \frac{V}{4.44 f A N}$
$Bmax⁡=30004.44×50×0.015×NpB_{\max} = \frac{3000}{4.44 \times 50 \times 0.015 \times N_p}$
For primary turns:
$Np=3000200×80=1,200N_p = \frac{3000}{200} \times 80 = 1,200$
Now,
$Bmax⁡=30004.44×50×0.015×1200B_{\max} = \frac{3000}{4.44 \times 50 \times 0.015 \times 1200}$
$Bmax⁡=0.075 TB_{\max} = 0.075 \text{ T}$

(ii) Number of turns in high voltage winding:
Using $NpNs=VpVs\frac{N_p}{N_s} = \frac{V_p}{V_s}$ ,
$Np=3000200×80=1200N_p = \frac{3000}{200} \times 80 = 1200$ turns.