Benha University
Faculty of Engineering- Shoubra
Electrical Engineering Department
Third Year Communication
Final Term Exam
Date: 30 / 5 / 2013
Power & Electrical Machines
Duration : 2 hours
 Answer all the following question
 Illustrate your answers with sketches when necessary.
 The exam. Consists of two pages
 No. of questions of 1st Page: 2
 Total Mark of the pages:70M
 The first Page 35 Marks
1.a) Discuss and draw in details each of the following:
1- The Phasor diagram for the secondary circuit under a load.
(2 Marks)
2- The Short Circuit Test.
(2 Marks)
3- The Current Transformer.
(2 Marks)
4- The Three-Phase Transformer Connections (Y – Y) and (Y – Δ).
(4Marks)
1.b) A 2400/480 volt transformer with Rp = 0.3 Ohm, Xp = 0.75 Ohm, Rs = 0.003 Ohm and
Xs= 0.006 Ohm is loaded by an effective resistive load R L of 0.65 Ohm.
Use the
approximated equivalent circuit of the transformer referred to the primary, determine:1- The equivalent impedance of the transformer referred to the primary, (2 Marks)
2- The primary current flows if the applied voltage = 2400 volt.
(2 Marks)
3- The total circuit copper losses (by two methods).
(4 Marks)
2.a) How can the power plants classified on the basis of Fuel used, Nature of Load and
Location?
(5 Marks)
2.b) What are the various factors to be kept in view while designing a power station?
(4 Marks)
2.c) The following loads: 18, 15, 11.5, 9, 3.5 Mw are supplied from a power station which
has the maximum demand of 30 Mw, the annual energy supplied of 150 x 106 Kwh, the
station has a capacity of 35 Mw. Find the percentage: a) Load Factor,
(2 Marks)
b) Demand Factor
(2 Marks)
c) Capacity Factor
(2 Marks)
d) Diversity factor
(2 Marks)
Good Luck &
Prof. Dr. Mohamed Moenes
Please Turn over the page
The Model Answer
1. The secondary circuit of the Transformer on load
The secondary winding current flows ( Is ).
Is will generate Φ2 in the transformer,
Ip will generate Φ1 in the transformer.
Φ1 = Φ2
Xp and Xs are the leakage reactances of the primary and secondary windings, and also for
the resistances Rp and Rs. Rs and Xs are connected in series with the secondary. The load
current causes a voltage drop across Rs and Xs. Similarly, Rp and Xp are connected in series
with the primary windings, and voltage drops are produced across them when a primary
current flows. The Complete equivalent circuit for a transformer is shown as
Φ
Es
Is Xs
Φs
Φo
Vo
Is Rs
Is
Phasor diagram for the secondary circuit of a transformer under load
Vo sin Φo
Φ
Is Xs
Es
Φs
Vo
) Φo
Is Rs
Is
Vo co
The secondary induced voltage Es is the phasor sum of Vo, Is Rs and Is Xs.
Es = √ ( Vo cos Φo + Is Rs )2 + ( Vo sin Φo + Is Xs )2
And,
( Vo sin Φo + Is Xs )
Φs = tan-1
( Vo cos Φo + Is Rs )
The core flux Φ leads Es by 90 degrees
2- The Short Circuit Test
It is performed with the secondary terminals short-circuited.
The primary current Ip is measured by the Ammeter, the primary voltage Ep is measured by
the Voltmeter and the input power is measured by the Wattmeter as shown in the figure.
The ammeter current must indicate the normal full-load primary current.
Short-circuit test on a transformer.
The voltage required to produce full load primary and secondary currents is around 3 % of
the normal input voltage level, the core losses are so small and can be neglected.
The windings are carrying normal full-load current, and thus the input is supplying the
normal full-load copper losses (the Wattmeter reading P ).
Then,
P
Req p =
Ip2
P is the true power (active power),
P = Ip2
Req p , The apparent power S = Ep Ip
Ep is the reading of the Voltmeter, and Ip is the reading of the Ammeter.
The Reactive power Q = √ S2 – P2 = √ ( Ep Ip )2 - P2
Also,
Q = Ip2
Xeq p
Q
Xeq p =
Ip2
Therefore,
Then, the resistive and reactive components of the full-load equivalent circuit referred to the
primary. Then, from the short-circuit test it can be obtained each of:- Req p
- Xeq p
3- The Current Transformer
A high-level alternating current can be most easily measured by accurately transforming the
current to a much lower level.
Conductor
Ip
Np 
Is
Ammeter
iron core
Secondary coil Ns
Current Transformer
Np
Is = Ip
Ns
The figure illustrates the principle of the current transformer. A conductor carrying a large
current passes through a circular laminated iron core. The conductor constitutes a one-turn
primary winding. The secondary winding consists of a number of turns of much finer wire
wrapped around the core as shown.
4- The Three-Phase Transformer Connections:
1 - Star to star connection ( Y – Y ):VLp
The secondary line voltage V L s =
, where a =
A
ILp
1
=
ILs
a
VLp
VLs
Y – Y Connection
The secondary line current
ILs=aILp
3 - Star to delta connection ( Y – Δ ):Addition factor of √ 3
that arises from the line –to- phase values.
Y – Δ Connection
( I Lp)
1
=
(I s ph)
a
(I Ls) = (I s ph) √ 3 = a (I Lp) √ 3
I Ls =
a √ 3
I Lp
And,
(V L p / √ 3)
V p ph
a=
=
V s ph
Then,
VLp
VLs =
a √ 3
VLs
1.b) A 2400/480 volt transformer with R p = 0.3 Ohm, Xp = 0.75 Ohm, Rs = 0.003 Ohm and
Xs= 0.006 Ohm is loaded by an effective resistive load R L of 0.65 Ohm.
Use the
approximated equivalent circuit of the transformer referred to the primary, determine:a- The equivalent impedance of the transformer referred to the primary, (2 Marks)
b- The primary current flows if the applied voltage = 2400 volt.
c- The total circuit copper losses (by two methods).
(2 Marks)
(4 Marks)
Solution:The transformation ratio a = Vp / Vs = 2400 / 480 = 5
The total equivalent resistance referred to the primary,
Req p = Rp + a2 Rs = 0.3 + (5)2 x 0.003 = 0.3 + 0.075 = 0.375 Ohm
The total equivalent reactance referred to the primary,
Xeq p = Xp + a2 Xs
Then:-
= 0.75 + (5)2 x 0.006 = 0.75 + 0.15 = 0.9 Ohm
a- The total internal impedance referred to the primary,
Zeq p = Req p + j Xeq p
= 0.375 + j 0.9 = 0.975
I 67.38°
Ohm
b- The equivalent load resistance referred to the primary,
RL' = a2 RL =
(5)2 (0.65) = 16.25 Ohm
The total equivalent circuit referred to the primary,
Zt = (0.375 + 16.25) + j 0.9 = 16.625 + j 0.9 = 16.649
I
3.099°
Ohm
Vp
Ip =
= 2400 / (16.649 I
3.099° ) = 144.153
I
- 3.099°
Amp.
C- The total circuit copper losses:
- The first method:
The total circuit copper losses = Ip2 (Rtp + RL') = (144.153)2 (16.625) = 345.469 Kwatt
- The Second method:
The secondary current Is can be calculated as :
Is = Ip (a) = 144.153
I
- 3.099° (5) = 720.764
The total circuit copper losses = Ip2 (Rp) + Is2 (Rs + RL) =
I
- 3.099°
A
(144.153)2 (0.3) + (720.764)2 (0.003 + 0.65) = 345.468 Kwatt
- The Third method:
The total resistance of the circuit referred to secondary Rts = Rs + RL + Rp'
= 0.003 + 0.65 + (0.3) (1/a)2 = 0.003 + 0.65 + (0.3) (1/5)2 = 0.665
Ohm
The total circuit copper losses = Is2 (Rts ) = (720.764)2 (0.665) = 345.468 Kwatt
---------------------------------------------2.a) How can the power plants classified on the basis of Fuel used, Nature of Load and
Location?
(5 Marks)
Types of Power Plants:
1- On the basis of fuel used:
a- Steam power plants.
c- Nuclear power plants.
b- Diesel power plants.
d- Hydro-electric power plants.
e- Gas turbine power plants.
2- On the basis of Nature of Load:
i - Base load plants.
ii - Peak load plants
3- On the basis of Location:
i - Central power stations
ii - Isolated power stations
2.b) What are the various factors to be kept in view while designing a power station?
(4 Marks)
Outline design
The various factors to be kept in view while designing a power station are as follows:1- Economy of expenditure.
2- Efficiency. 3- Safety of plant and personnel.
4- Reliability.
5- Easy of maintenance.
6- Good working conditions.
7- Useful life of a plant.
8- Prediction of loads.
2.c) The following loads: 18, 15, 11.5, 9, 3.5 Mw are supplied from a power station which
has the maximum demand of 30 Mw, the annual energy supplied of 150 x 106 Kwh, the
station has a capacity of 35 Mw. Find the percentage: (8 Marks)
a) The load Factor, b) The demand Factor c) The Capacity Factor d) The Diversity factor
Solution:
a) The Average power Pav = The Kwh supplied annually / 8760
6
= 150 x 10 x 1000 / 8760 = 17.123
Mw
The percentage Load Factor = 100 x Average power / Maximum Power
= 100 x 17.123 / 30 = 57.1 %
b) The Demand Factor = 100 x Maximum Demand / Sum of the connected load
= 100 x 30 / (18 + 15 + 11.5 + 9 + 3.5) = 100 x 30 / 57 = 52.6 %
c) The Capacity Factor = 100 x Average power / Capacity = 100 x 17.123 / 35 = 48.9 %
d) The diversity factor = 100 x (P1max + … + Pnmax) / Maximum of the total load
= 100 x (18 + 15 + 11.5 + 9 + 3.5) / 30 = 100 x 57 / 30 = 190 %