SUBJECT- Energy Conversion Devices (BEEC2214)
LECTURER NO-4
Expression for EMF induced:-
Let Ø=flux/pole in webers…….
Z=total number of conductors= no. of slots x no. of conductors/slot
P= no. of generator poles….
A= no. of parallel armature paths
N= armature rotations in revolutions per minute (r.p.m)
E= Back Emf of motor
So, Eb=emf generated in any 1 of the parallel paths i.e. E
Average emf per conductor= dØ/dt volts (N=1)
Now, flux cut/conductor in 1 revolution= dØ=Øp Wb
No. of revolutions /second= N/60
Hence, time for 1 revolution, dt=60/N second
So, according to Faraday’s Laws of EMI,
EMF for 1 conductor= dØ/dt= Øpn/60
Therefore, emf for Z conductors for A parallel paths= ØNZP/60A
For a simplex wave winding, A= 2
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S.I.E.T,DHENKANAL

SUBJECT- Energy Conversion Devices (BEEC2214)
For a simplex lap winding, A=P
Voltage Regulation
The change in terminal voltage of a generator between full and no load (at constant speed) is called the voltage regulation, usually expressed as a percentage of the voltage at full-load.
% Voltage regulation= [ (V
NL
-V
FL
)/V
FL
] × 100 where V
NL
= Terminal voltage of generator at no load
V
FL =
Terminal voltage of generator at full load
Note that voltage regulation of a generator is determined with field circuit and speed held constant. If the voltage regulation of a generator is 10%, it means that terminal voltage increases 10% as the load is changed from full load to no load
Problems:-
Q-1) A 4 pole lap wound dc generator has 600 conductors on its armature. The flux per pole is 0.02 Wb. Calculate (i) the speed at which the generator must be run to generate
300 V. (ii) what should be the speed if the generator were wave wound.
Soln:- (i) Lap wound
E g
= ØNZP/60A
N= E
G x 60A/ PØZ = 300 X 60X 6/ (6 X0.02X 600) = 1500 RPM
(II) Wave wound
N= E
G x 60A/ PØZ = (300 x 60 x2)/( 6 X0.02X 600) = 500 rpm
Q-2) The armature of a 6 pole , 600 rpm lap wound generator has 900 slots. If each coil has 400nnturns, calculate the flux per pole required to generate an emf of 288 volt.
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SUBJECT- Energy Conversion Devices (BEEC2214)
Ans-
Z= 90 X 4X 2=720
E g
= ØNZP/60A
SO, Ø= E g X
60 X A/PZN = 288 X 60 X 6/(6 X 720 X 600) = 0.04Wb.
Q-3). Calculate the e.m.f generated by a 4 pole wave-wound generator having 65 slots with 12 conductors per slot when driving at 1200 r.p.m. The flux per pole is 0.02 Wb.
Sol. Eg = P flux Z N / 60 A
Here,
P=4, Flux=0.02 Wb , N=1200rpm , Z=12x65=780 , A=2
So, Eg= 4 x 0.02 x 780 x 1200 / 60 x 2
=624 volts
Q-4). A 6-pole lap wound DC generator has 600 conductors on its armature. The flux
per pole is 0.02wb.
(i)Calculate the speed at which the generator must be run to generate 300V.
(ii) What would be the speed if the generator were wave wound?
Sol. (i) Lap wound
Eg=P flux Z N/60 A
So, N=Eg 60 A/P flux Z
= 300 x 60 x 6 / 6 x 0.02 x 600
=1500 rpm
(ii) Wave wound
N= Eg x 60 A/ P flux Z
300 x 60 x 2 / 6 x 0.02 x 600=500 rpm
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S.I.E.T,DHENKANAL

SUBJECT- Energy Conversion Devices (BEEC2214)
Q-5). An 8-pole lap wound armature rotated at 350 rpm is required to generate 260 V.
The useful flux per pole is 0.05 Wb.If the armature has 120 slots, calculate the number of conductors per slot.
Sol. Eg=P flux Z N/60 A
Z= Eg x 60A / P flux N
=260 x 60 x 8 / 8 x 0.05 x 350
=890
Number of conductors per slot=890 / 120 = 7.41
Since the value must be even so conductor/slot = 8
Q-6). The armature of a 6-pole, 600 rpm lap wound generator has 90 slots. If each coil
has 4 turns, calculate the flux per pole required to generate an emf of 288 volts.
Sol. Each turn has two active conductors and 90 coils are required to fill 90 slots.
Z = 90 x 4 x 2 = 720
Eg = P flux Z N / 60 A
So flux = Eg x 60A / P Z N
=288 x 60 x 6 / 6 x 720 x 600
=0.04 wb.
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S.I.E.T,DHENKANAL

SUBJECT- Energy Conversion Devices (BEEC2214)
Q-7). The armature of a dc generator has 81 slots and the commutator has 243 segments. It is wound to give lap winding having 1 turn per coil. If the flux per pole is
30mWb, calculate the generated emf at speed of 1200 rpm. Number of pole is 6.
Sol. The number of coils is equal to the number of commutator segments. Each turn has 2 active conductors.
So , Z = 243 x 2 = 486
Flux = 0.03 Wb
Eg = P flux Z N / 60 A
= 6 x 0.03 x 486 x 1200 / 60 x 6
=291.6 volts
Q-8). The armature of a 2 pole, 200 v generator has 400 conductors and runs at 300 rpm. If the number of turns in each field coil is 1200, what is the average value of emf induced in each coil on breaking the field if the flux dies away completely in 0.1 s?
Sol. Here,
P / A = 1
Z=400, N=300rpm, Eg = 200V
So,
200 = 1 x flux x 400 x 300 / 60
Flux = 0.01 Wb
If n (=1200) is the number of turns of each field coil, then the magnitude of emf induced on breaking the field
= n d(flux)/dt
=1200 x 0.1/0.1
=1200 V
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