Chapter-1 Hydro Power Plant

1.1 Energy Conversion from hydraulic to Electrical:

Energy contained in the flowing water and storage water are form of hydraulic (mechanical)
energy. It exist as Kinetic Energy in flowing water and Potential Energy in the stored water at some
elevation with respect to a lower reference datum level.

Bernoulli’s equation of fluid gives:

1
 .g.h   .V2  Constant at any position (Or point) of flowing fluid.
2
Where, ρ = density of the fluid
h = Height of that point from reference datum level
V = Velocity of the fluid at that point
g = Acceleration due to gravity

Let us consider a water flowing system through a closed tube as shown in Fig1.1.

Fig.1.1 Illustration of Bernoulli’s theory

Let us consider a particular volume of water (∆V) at an elevation of h2 (i.e. at point – 2).
Let P2 = pressure of the water at point -2
V2 = velocity of water at point-2.
∆S2 = distance moved by that particular volume of water in a small time ∆t.

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Work done by that particular volume of water while moving from position-1 to position-2 is given
by:
∆W = P1A1∆S1 – P2A2∆S2 = (P1 –P2) ∆V ………. (1.1) Note: P1 > P2

Net change in kinetic energy while moving from position-1 to position-2 is given by:
1
K   .V (V 2  V2 ) ......... (1.2) Note : V  V
2 2 1 2 1
Net change in potential energy while moving from position-1 to position-2 is given by:
U   .V (h  h ). g ......... (1.3)
2 1
According to work-energy theory:
∆W = ∆K + ∆u
1
Or ( P - P ) V   .V (V 2  V2 )   .g.( h - h )V
1 2 2 2 1 2 1
1
Or ( P - P )   .(V2  V2 )   .g.( h - h ) ......... (1.4)
1 2 2 2 1 2 1
1 1
Or P   .V2   .g. h  P   .V2   .g. h  Constant ......... (1.5)
2 2 2 2 1 2 1 1
1
In general, - P   .V2   .g. h  Constant ......... (1.6)
2
Therefore, a mass of water at certain position has total energy = KE +PE and it remains constant at
any position. When height decreases, PE decreases and KE increases, but the total energy remains
constant. This energy possess by a water can be converted into Electrical Energy with the help of
some energy conversion devices such as Turbine – Generator coupled system as shown in Fig.1.2.

Fig.1.2 Schematic layout of hydro power plant

The power that can be generated by such plant is given by:

P = 9.81wQHƞ kW ……. (1.7)

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Where, w = Specific weight of water (kg/m3)
Q = Discharge of water through the penstock pipe (m3/sec)
H= Gross head available (m)
ƞ = Overall efficiency of the system

1.2 Water Hammer and Surge Tank:

At steady state operation, the water discharge (Q) through the penstock pipe is smooth and just
sufficient to produce a power equal to the power consumed by the load. When the electrical load
decreases, Q shall be decreased accordingly to reduce the input power to the generator so that Input
Power to Generator = Power Consumed by the load and the water flow is again smooth. For this
purpose, an automatic control mechanism (Governor) is used in hydro power plant.

A step large reduction in electrical load on the generator causes the governor to close the turbine
gate and thus create an increased pressure in the penstock pipe. This sudden increase in pressure
gives an intensive thrust to the penstock pipe. This effect is known as Positive Water Hammer. If
the strength of the penstock pipe is not sufficient to resist this sudden thrust, it may brust. The surge
tank helps to reduce this positive water hammer effect. The surge tank gives space for rejected
volume of water thus by reducing the sudden pressure in the penstock pipe.

An increase in electrical load by a large amount in step causes the governor to open the turbine gate
suddenly to allow more water flow through penstock pipe thus by increasing input power to the
generator. While doing so there is a tendency to create a vacuum (or negative pressure) in the
penstock pipe. Such a vacuum in the penstock pipe will obstruct the smooth flow of water through
the penstock pipe. This effect is known as negative water hammer. Surge tank also helps to reduce
the negative water hammer effect. During such condition of negative water hammer, surge tank
supplies additional amount of water to the penstock pipe thus by nullifying the vacuum.

1.3 Steady State operation of Hydro Power Plant:

Fig.1.3 shows the block diagram representation of a hydro power plant without speed governor. It
is simply turbine- generator coupled system and the generator supplying electric power to the load.

Fig.1.3 Block diagram representation of a hydro power plant without speed governor.

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During steady state operation, there is exact balance between mechanical power in put to generator
(Pm) and Electrical power output from the generator ( Pe).
i.e. Pm (input) = Pe (output) + Power loss in the generator
In such a case, the speed of the system remains constant and accordingly frequency of the generated
ac voltage remains constant.
For ideal case (where there is no power loss ):
Pm (input) = Pe (output), then speed remains constant
If Pm (input) > Pe (output), then speed increases
If Pm (input) < Pe (output), then speed decreases
In order to keep the speed constant at variable load conditions, Speed governor is used in the hydro
power plant.

1.4 Hydraulic Speed Governor:

It is an automatic control system, which adjust the amount of water flow into the turbine to keep the
speed constant at variable load conditions. Fig.1.4 shows the schematic diagram of speed governor.

Fig.1.4 Schematic diagram of speed governor.

The main functional part is the fly ball assembly whose shaft is driven by turbine shaft. When the
electric load on the generator decreases, the speed of the turbine increases and accordingly speed of
the shaft of the fly ball assembly increases. Then the rotating fly balls moves away due to
centrifugal force. This will cause the up-ward movement of the end-A of the floating lever and

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down ward movement of the end-B of the floating lever. This will move the shaft of control valve
piston down ward thus by opening the path-Aof the control valve. The under high pressure will flow
through the path-A and reaches to upper chamber of the hydraulic amplifier piston. The pressure in
the upper chamber increases and the piston move down ward and spear valve also moves down
ward thus be reducing the water inlet from the penstock to turbine. Hence, the speed decreases and
come back to pre set value. Just reverse process happens when the electric load on the generator
increases.

1.5 Transient in Turbine-Generator coupled system with Governor:

Fig.1.5 shows the control block diagram of turbine-generator coupled system with governor with
closed loop control system.

Fig.1.5 Control block diagram of turbine-generator coupled system with governor

The word “transient” refers to sudden change of some parameters of a system, which in turn,
disturbs the steady state condition of the system. During the transient period, the system behavior
will be abnormal with compare to the behavior in steady state condition. In case of turbine-
generator coupled system shown in Fig.1.5, the sudden change could be – sudden change in
electrical load by a large amount, short circuit across the generator terminals for few milliseconds
etc. In such a case, the frequency of generated voltage, magnitude of generated voltage, speed of the
system will deviate from the normal steady state values and may come back to original or new
steady state values after some time under the action of closed loop controlled system.

Fig.1.6 shows the responses of speed and mechanical power input to the generator during the
transient period due to the sudden step increase in electrical load.

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 Fig.1.6 Transient responses of turbine-generator coupled system

The system was operating in steady state operation from 0 to t1 second. During this period, the
electrical power output (Pe), speed (ω) and mechanical power input (Pm) are constant and equal to
Peo, ωo and Pmo respectively.
A step load of ∆Pe is switched ON at t = t1. This is the starting of transient period.
Now, Pe becomes > Pmo. Therefore, the speed will decrease as shown in Fig.1.6(b). At the same
time, the closed loop control system sense the speed and compare with the reference speed. The
error signal so obtained drives the controller, which in turn gives command signal to the gate valve
to increase the valve opening amount so that more amount of water flows into the turbine. Now
mechanical power output of the turbine increases as shown in Fig.1.6(C). However, due to inertia of
the rotating mass, speed can not increases suddenly. The speed is decreasing up to t = t2 at which the
Pm has increased to Pmo +∆Pe. At t > t2, Pm becomes > Peo +∆Pe, Therefore speed increases from
this point. At t=t3, speed has come back to initial value of ωo and Pm(new) becomes equal to Pe(new).
Therefore, speed again remains constant at ωo after t = t3. Therefore, t = t3 is the end of transient
period.

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Power Plant Equipment

Lecture No.2

1
2.1 Active Power(P) / Frequency and Reactive Powe (Q ) / Volatge magnitude tie-up in Power system

Change in active power in a power system mainly affects the frequency of the system, it does
not affect the magnitude of system voltage. Whereas, change in reactive power mainly affects
the magnitude of system voltage, it does not affect the system frequency.
The above facts can justify as follow:

Fig.2.1 P-f and Q-V control loops of a generating unit

Fig.2.1 shows a generating unit with synchronous generator driven by water turbine equipped
with speed governor (P-f loop) and voltage controller AVR (Q-V loop). The generator is
supplying power to the load which is mixture of active power and reactive power. The load
current lags the terminal voltage (V) by an angle of ϕ as shown in Fig.2.2

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 Fig.2.2 Phasor diagram and waveform of V and IL for inductive load.

The three phase currents IL(abc)

produces rotating magnetic poles in the
stator core. The active components (IX)
of IL(abc) produces south-pole of the
rotating magnetic field in the stator
core gets magnetically lock-up
(Magnetic Tie-UP) with the north-pole
of the rotor as shown in Fig.2.3. Both
of these poles are rotating at
synchronous speed. This magnetic tie-
up acts as opposing force to the turbine
and turbine produces driving torque
against this opposing force and energy
conversion take place.
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When active power of the load increases, the active component of IL= IX = ILcosϕ increases.
Therefore, magnetic tie-up between stator pole and rotor pole increases and speed decreases.
The speed is sensed and compared with the ref-speed, the error so obtained is passed through
the controller. The controller gives command to the valve and water flow increases and speed
comes back to pre-set value. It will take some time to come back the speed to pre-set value.
This time is known as transient period.

Fig.2.4 shows the speed response of a

generating unit with isochronous speed
governor. Up to t1, the generating unit is
running at constant speed and giving
constant frequency (f0) of generated
voltage. At t1, a step load is increased by ∆P
and frequency (speed) decreases and comes
backs to f0 at t2 again due to the action of
speed governor. If the governor has PI Fig.2.4 frequency response of generating unit with change in load
controller, the speed comes to f0 with zero
steady state error. During the transient
period, the magnitude of voltage also
decreases due to the decrease in speed.
However after transient period, the voltage
magnitude also comes back to normal
value.
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When reactive power of the load increases, the reactive component of IL = IL sinϕ = IY
increases. The rotating magnetic poles produced by 900 lagging currents (IY(abc)) lags by 900
with compared to rotating pole produced by in-phase currents (IX(abc)) as shown in Fig.2.5.

Fig.2.5 Magnetic tie-up between stator pole and rotor pole due 90 0 lagging current IY

Since, stator pole and rotor poles are 900 apart, there will be no magnetic tie-up. Therefore there
will be no stress to turbine due to increase in 900 lagging current. Hence, speed does not
decreases due to the increase in reactive load. However, stator current magnitude increases and
it will cause more voltage drop in the stator winding impedance and terminal voltage decreases.
The voltage is sensed and compared with the ref-voltage, the error so obtained is passed through
the controller (AVR). The controller increases the excitation current If and emf induce increases
and terminal voltage increases to pre-set value.

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 In case of transmission system also, the voltage drop due to increase in active power is
negligible and voltage drop due to the increase in reactive power is very significant. This
facts can be illustrates as follow:

Fig.2.6 Transmission line model Fig.2.7 Phasor diagram

Fig.2.6 shows the circuit model of a transmission line. Here, the shunt capacitance has been
neglected to simplify the illustration. Fig.2.7 shows the phasor diagram of the transmission
line with inductive load at receiving end.
Here, |VR| = OA
|VS| = OE, Neglecting a small length of DE,
|VS| ≈ OD = OA + AC + CD OR |VS| = |VR| + AC + CD
OR |VS| - |VR| = AC + CD OR Magnitude of voltage drop in the line = AC +GH

From the right angle triangle ∆ ACG, AC = (IR R) Cosϕ

From the right angle triangle ∆ GHF, GH = (IR X) Sinϕ

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Therefore, Magnitude of voltage drop in the line
= (IR R) Cosϕ + (IR X) Sinϕ
(V . I Cos )R  (V .I Sin ) X 
(P.R )

(Q.X )
 R R R R
V V
V R R
R
= Voltage drop due to Active power + Voltage drop due to reactive power

For a practical transmission line, R << X

Therefore, Voltage drop due to Active power << Voltage drop due to reactive power
Hence, Total voltage drop in the line is mainly affected by Reactive power.

(Q.X )
Hence, |VR| ≈ | VS| -
V
R

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2.2 Pump Storage Hydro Power Plant
This is an unique type of peak load power plant , whose turbine –generator set will be used as
pump during the light load period to pump back the water collected at tail race dam to upper
storage pond so that it can generate energy during peak load period. In fact, it is just like a
storage of energy during light load period and releasing out during peak load period. Fig.2.7
shows the schematic layout of such power plant.

During the light load period when there is

excess power in the grid, the un-utilized
energy of the grid is used to drive the turbine-
generator set as pump in reverse direction to
pump back the water at lower storage pond to
upper storage pump. During this period, valve
V1 is open and valve V2 is closed. Francis
turbine is used in such a plant as it can used a
pump in reverse direction and synchronous
generator is used in such a plant as it can be
used as motor to drive the pump. Fig.2.7 Schematic layout of pump storage hydropower plant.

During peak load period, the water stored in the upper storage pump is allowed to flow down to
operate the plant in generating mode. Thus the same water is used again and again and some
extra water is required only to take care of evaporation and seepage. Hence, such plant even
can be installed at load centre, where there is no natural resources of water.

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The efficiency of such power plant is only 60% to 70%. Therefore such power plant is only
economical only if there is un-utilized energy available during light load period and not enough
energy during peak load period. Such power plant is economically feasible when integrated
with thermal power plant ( such as steam power plant , nuclear power plant) dominated grid.
The steam and nuclear power plants operates with better efficiency when they are operated at
full load. If such plant is closed during light load period , it takes longer time to re-start it again
and energy loss in starting process will be high. Therefore the thermal power plant can be
operated nearly at full load even during light load period by using the excess of energy to pump
back the water in pump storage power plant. It has been estimated that the a pump storage
power plant can be set up at a half cost of a nuclear power plant of same capacity.

The followings are some of the distinct advantages of the pump storage hydro power plant:

•Peak load can be supplied at a lower cost with compare to that supplied by thermal power
plant.
•The steam and nuclear power plants can be operated at almost unity load factor which ensures
the better efficiency of the thermal power plant.
•Pump storage plant need very less starting time. It needs only 2-3 seconds to start and can be
fully loaded in about 15 seconds. This will reduce spinning reserve requirement in the system.
•In the event of extra demand coming up suddenly on the system, such plant can be
immediately switched on to meet the load demand.

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Fig.2.8 illustrate the example of utilization of pump storage hydro power plant.

Let us consider a power system in which daily load demand

varies from 100 MW to 160 MW as shown in Fig.2.8(a)

One of the option to meet this load demand is to have a

base load plant(steam power plant) of 100MW and 60 MW
peaking plant (Diesel generator) as shown in Fig.2.8(b). The
hatched area represents the energy supplied by the peaking
plant and the blank area below the hatch area represents
energy supplied by base load plant (i.e. steam power plant).

Another option is to install little larger base load plant

(Steam power plant) of 130 MW and smaller 30 MW of
pump storage power plant as shown in Fig.2.8(c). During
the light load periods ( t0 to t1, t2 to t3 and t4 to t5), the load
demand is less than the capacity of steam power plant. The
surplus power from steam power plant can be utilized to
pump up the water in pump storage plant. The lower
hatched area in Fig.2.8(c) represents the energy consumed
the pump storage plant from the grid. During the peak load
periods ( t1 to t2, and t3 to t4), the load demand is more than
the capacity of steam power plant. Pump storage plant can
generate power during these period to make power balance.
The upper hatched area in Fig.2.8(c) represents the energy
supplied by the pump storage plant to the grid.
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The above scheme with pump storage plant as peak load plant has following advantages:
•The base power plant (Steam power plant ) is larger and consequently it is more efficient.
•The peak load plant (diesel generator) in Fig.2.8(b) has higher operating cost. The pump
storage plant is smaller than diesel power plant and its operating cost is less.

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Power Plant Equipment

Lecture No.3

1
3.1 Modeling of Speed Governor
The basic concept of speed governor can be illustrated by considering an isolated
generating unit supplying a local load as shown in Fig.3.1.

Fig.3.1 Basic concept of speed governing system

When there is imbalance between mechanical power input (Pm) and electrical power
output (Pe), speed increases or decreases. The speed is sensed and the governor
produces command signal for gate valve to adjust the water flow to match with the
electrical load.
Let Pm = Mechanical power input to the generator
Pe = Electrical power output from the generator
Tm = Mechanical torque input to the generator shaft
Te = Counter electrical torque developed by the generator Counter Electrical torque

2
When generator supplies electric current to the load, the three phase
stator current produces rotating magnetic field in the stator core. The
N-pole of the rotating magnetic poles in the stator core gets
magnetically tie-up with the S-pole of the rotor, which behaves as
counter electrical torque in the opposite direction of mechanical input
torque and energy conversion takes place successfully.

In the absence of governor, how above system responses with change in electrical load can be
explained as follow. Then it will be easier to develop the mathematical model of the system
with governor.

During steady state operation, the electric power output will be exactly equal to mechanical
power input and speed remains constant. When the electric load changes, it is reflected
instantly as a change in electrical torque (Te). This will cause mis-match between the
mechanical torque input (Tm) and the counter electrical torque (Te), which in turn produces
speed variation as determined by the equation of motion (swing equation) which given by:
d 1
 (T -T ) .......... .. (3.1)
2H m e
dt
Where, H = Inertia of the rotating mass of the system
d 1
Replacing by Laplace operator ' s' , s  (T - T )
2H m e
dt

3
 1
OR   (T - T ) ......... (3.2)
2Hs m e
Equation (3.2) can be represented by transfer function block diagram shown in Fig.3.2

Fig.3.2 Transfer function relating speed deviation with accelerating torque

In power system load –frequency studies, it is preferable to express the above relationship in
terms of power rather than torque. The relationship between power and torque of a rotating
system is given by:
P = ωT ……. (3.3)
Considering small deviation (denoted by prefex ∆) from initial steady state values (denoted by
subscript 0), it can be written as:

P = P0 + ∆P
T = T0 + ∆T ………. (3.4)
ω = ω0 + ∆ ω
Substituting these values in eqn(3.3) gives:

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P0 + ∆P = (ω 0 + ∆ ω).( T0 + ∆T)
OR P0 + ∆P = (ω 0 T0+ ω 0 ∆ T +∆ ω T0 +∆ ω ∆T)
Assuming that ∆ ω ∆T ≈ 0 (product of small numbers) and P0 = ω 0 T0, the above
eqn can be written as:
∆P = ω 0 ∆ T + T0 ∆ ω ……. (3.5)
Expressing the above equation in terms of differences between mechanical and electrical quantities:

∆Pm - ∆Pe = ω 0 (∆ Tm -∆Te) + (Tm0 – Te0 )∆ ω …….. (3.6)

Tm0 – Te0 = 0 (in steady state) and Tm0 – Te0 ≈ 0 (for small change in elect load), and
expressing the above equation in p.u. (i.e ω 0 = 1 pu):
P - P   T -  T .......... .. (3.7) in pu
m e m e
Hence, Fig.3.2 can be written as follow in pu system:

Fig.3.3 Transfer function relating speed deviation with accelerating power in pu

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In general, power system loads are composite in nature, which the mixture of frequency
independent restive load and frequency dependent inductive load. The overall frequency
dependent characteristics of a composite load can be expressed as follow:

∆Pe = ∆PL + D. ∆ω ……… (3.8)

Where, ∆PL = Change in frequency independent load
D. ∆ω = Change in frequency dependent load
D = Load damping constant
Load damping constant is expressed as a percentage change in in load due to to 1% change
in frequency. Typical values of D may vary from 1 to 2.
A value of D = 2 means that 1% change if frequency will cause 2 % change in load.
Considering the effect of frequency dependent load as per eqn (3.8), the transfer function
model shown in Fog.3.3 can be written as follow:

Fig.3.4 Transfer function model with effect of frequency dependent load

It is now like a feed back control loop with feedback path gain = D and forward path gain
= 1/MS.

6
 Fig.3.5 Overall transfer function of feedback control system

According to control system theory, overall transfer function of feedback control system
shown in Fig.3.5 is given by :
G(S)
Overall T.F. 
1  G(S). H(S)
Hence, the overall transfer function of the system shown in Fig.3.4 is given by:
1
MS 1
Overall T.F.   .......... (3.9)
D MS  D
1
1
MS
Hence the system shown in Fig.3.4 is equivalent to system shown in Fig.3.6.

Fig.3.6 Transfer function model with effect of frequency dependent load in the absence of governor

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In the absence of speed governor, the system response to a change in load is determined by
the inertia constant and load damping constant. If the change in load is small, the inertia
constant and load damping constant could be sufficient to keep the speed in acceptable
value. If the change in load is large, then inertia constant and load damping constant will
not be sufficient to keep the speed in acceptable value. In such a case, speed governor is
required to keep the speed within acceptable values for large change in load.

Basically, there are two types of speed governors:

•Isochronous governor
•Governor with speed-droop characteristic

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 3.1 Isochronous Governor
The word Isochronus means constant speed. An Isochronous governor adjust the turbine gate-
valve opening to bring the speed and frequency back to the normal pre-set value. It uses P-I
controller and ensures zero steady state error. Fig.3.7 shows the control system (in block
diagram) of an Isochronous governor.

Fig.3.7 Control system of hydro generating unit with Isochronous governor.

When the electric load increases, speed decreases. The speed is sensed and compared with the
desired reference speed (ω0). The error signal (∆ω) is amplified by –K and integrated to
produce a control signal (∆Y) which actuates the gate-valve of the turbine to increase the gate
opening thus by increasing the water flow into the turbine and speed increases. When there is
decrease in speed, the proportional gain shall be such that it shall increase Pm. Therefore,
negative sign is assigned with the gain K. Since the controller is PI type, the control signal
(∆Y) reaches to a new steady state value only when the speed error is zero. Therefore, this
type of Isochronous governor ensures to bring back the speed to the preset value with zero
steady state error.

9
Fig.3.8 shows the time response of a hydro generating unit with an Isochronous Governor with
increase in electric load.

Fig.3.8 Time response of a hydro generating unit with an Isochronous Governor.

When the electric load increases, speed decreases at a rate determined by the inertia and load
damping. As the speed decreases below the ref speed, the governor comes into action to increase
the water flow into the turbine thus by increasing the mechanical power output of the turbine.
This in turn causes a reduction in the rate of decrease of speed and then an increase in speed
when the mechanical power (Pm) is greater than the electric load power (i.e. beyond the point ‘P’
in the Pm response curve. Finally the speed will come back to its preset reference value and
steady state turbine power increases by an amount equal to increase in electrical load.

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Isochronous governor is only suitable for isolated single generating unit. If two generating units
with their respective isochronous governor are operated in parallel as shown in Fig.3.9, it will not
give stable operation with proper load sharing by each generating units.

Fig.3.9 Parallel operation of two generating units with their respective Isochronous Governor
Once these two generating units are synchronized to operate in parallel, they are bound to
operated at a common frequency. When a step load of ∆PL is switched on, the frequency decreases
to a common value or speed of both generating units decreases by a common value of ∆ω. Now
each generator will understand that there is increase in load by ∆PL each governor will act
independently to increase the Pm by ∆PL. So there will be total increase of mechanical power of
2∆PL. Now Pm(new) becomes > PL(new) by an amount of ∆PL . Therefore speed increase beyond ω0
and both governors come into action again to reduce the Pm each by ∆PL. In this way both
governors fights with each other and try to compensate the total change in load by alone itself.
Hence they will not results a stable operation. This is because there were no coordinated control
between two generating units.
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3.2 Governor with speed-droop characteristics:
When two or more generating units are to be operated in parallel, this type of governors are
required. This type of governors can be tuned to have coordinated operation among each other so
that it give stable parallel operation at a common frequency and shares the load in proportional to
their capacities. The speed-droop characteristics can be obtained by adding a steady state feed
back loop around the PI controller of isochronous governor as shown in Fig.3.10.

Fig.3.10 Block diagram of generating unit with governor having speed-droop characteristic

The feedback path has gain of R. This is knows as speed regulation droop. By setting the
values of R properly for two generating units, a coordinated operation can be achieved to share
the load in proportional to their respective capacities.

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The closed loop transfer function between point A to B can be obtained as follow:
Forward path gain G(S) = K/S G(S)
Closed loop T.F. 
Feedback path gain H(S) = R 1  G(S). H(S)

K K
S S K S
OR closed lopop T.F. between A and B    
K S  KR S S  KR
1 R
S S

K K/ (KR)
OR closed lopop T.F. between A and B  
S  KR 1  S/KR
1/ R 1
OR closed lopop T.F. between A and B  .......... (3.10) Where, T 
1  S.T G KR
G

Hence, the closed loop control system shown in Fig.3.10 can be written as shown in Fig.3.11.

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 1/ R
OR closed lopop T.F. between A and B 
1  S.T
G

Fig.3.11 Reduced block diagram of governor with speed-droop characteristic

When the electric load on the generator increases, the speed decreases against the inertia and
load damping and the decreasing rate depends on the value of R in the feedback path. Fig.3.12
shows the steady state speed versus load characteristic curve.

Fig.3.11 Steady state speed versus load characteristic curve.

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The slope of this curve represent the gain of the feedback path R. Hence, R is defined as the ratio
of frequency (OR speed) deviation (∆f) to the change in gate-valve opening (∆Y) OR change in
electric load (∆P).

pu change in frequency f 
Therefore, R    ......... (3.11)
pu change in electric power P P

For example, a 5% droop (R) means that a 5% change in frequency deviation causes 100%
change in output power.

15
Load sharing between parallel generating units :
If two or more generating units with speed-droop characteristics are connected in parallel to a
common bus, there will be an unique frequency at which they will share the load in proportional
to their capacities provided the values of their droop regulations (R) are properly tuned. In a
commercial speed governor there will be a hardware knob to set the value of R to a desired
value. Fig.3.12 shows two generating units connected in parallel to a common bus to share a
common load (PL) connected to that bus.

Fig.3.12(a) Fig.3.12(b) Fig.3.12 (c)

Fig.3.12 Two generating units connected in parallel to a common bus

Both generating units have their respective speed governor with speed-droop characteristics. Let
us assume that the capacity of generating unit-1 is greater than that of the generating unit-2
Let R1 = droop regulation of generating unit-1
And R2 = droop regulation of generating unit-2

16
The slope of droop regulation of smaller generating unit-1 (R1) shall be higher than the slope of
droop regulation of smaller generating unit-2 (R2) so that the smaller generating unit shares less
load with compare to that shared by larger generating unit. In order to make the load sharing in
proportional to their capacities, the values of R1 and R2 shall be tuned appropriately.

Let us assume that the parallel units are operating at a common frequency of f1 and supplying
power of PL to a common load. As shown in Fig.3.12(a), Fig.3.12(b) and Fig.3.12(c),
generating unit-1 is supplying a power of P1 and generating unit-2 is supplying a power of P2 so
that P1 + P2 = PL. As the slope R1 is greater than slope R2, the generating unit-1 is supplying
less power than that supplied by the generating unit-2. The load shared by each units can be
made in proportional their capacities by selecting the proper values of R1 and R2.

17
 From no-load to load PL, change in power in G1 = ∆P1 = P1-0 = P1
And change in power in G2 = ∆P2 = P2-0 = P2
Let f0 = common frequency at no-load
When the system operates at a common frequency of f1, G1 supplies P1 and G2 supplies P2.
And P1 + P2 = PL ……… (3.12)

The proportions of P1 and P2 can be calculated as follow:

According to definition of droop regulation,

f f -f f f -f
R   0 1 And R   0 1 .
1 P P 2 P P
1 1 2 2

18
 R
OR R . P  f And R P  f . That means R . P  R P Or P  1 P
1 1 2 2 1 1 2 2 2 R 1
2
R R R R
And P  P  P Or P  1 P  P OR P (1  1 )  P Or P ( 2 1 )P
1 2 L 1 R 1 L 1 R L 1 R L
2 2 2
R
Therefore, P  ( 2 )P ..... (3.13)
1 R R L
1 2
R
Similarly, P  ( 1 )P ..... (3.14)
2 R R L
1 2
According to eqn(3.13) and eqn (3.14), it is clear that selecting the appropriate values of R1 and
R2 , P1 and P2 can be made in proportional to the capacities of G1 and G2.

Let us assume that the system is operating at a common frequency of f1. G1 supplies P1 and G2
supplied P2 as shown below:

19
 If the load is increased by ∆PL, the system frequency decreases to a common frequency of f2 . At
this reduced frequency, generating unit-1 generates total power of P1(new) and generating unit-2
generates total power of P2(new)
Then, ∆f = f1 - f2
Let ∆P1 = load shared by generating unit-1
And ∆P2 = load shared by generating unit-2
∆P = ∆P1 + ∆P2 …… (3.15)
f
R  = Speed droop regulation of generating unit-1
1 P
1
f
R  = Speed droop regulation of generating unit-2
2 P
2

20
 R
OR R . P  f And R P  f . That means R . P  R P Or P  1 P
1 1 2 2 1 1 2 2 2 R 1
2
R R R R
And P  P  P Or P  1 P  P OR P (1  1 )  P Or P ( 2 1 )  P
1 2 L 1 R 1 L 1 R L 1 R L
2 2 2

R
Therefore, P  ( 2 ) P ..... (3.16)
1 R R L
1 2
R
Similarly, P  ( 1 ) P ..... (3.14)
2 R R L
1 2

21
An Illustrative Numerical Example:
QN1 Two generators of capacities 600MW and 400MW are connected in parallel to supply a
common load of 600 MW. Their droop regulations are 4% and 5 % respectively w.r.t. their
respective ratings. At no-load, they operates at a common frequency of 51 Hz. How they will
share the common load of 600MW? When the load is increased by 100 MW, at which
frequency they will operate and calculate the power supplied by each generator.
Solution:

Let Pbase = 1000 MW, fbase = 50 Hz

Pbase(new) 1000
R R   0.04   0.066 pu
1(new) 1(old) Pbase(old ) 600
Pbase(new) 1000
R R   0.05   0.125 pu
2(new) 2(old) Pbase(old ) 400
fNL (pu) = 51 Hz / fbase = 51 / 50 = 1.02 pu

22
 Case-1: When a common load PL = 600 W = 600/1000 = 0.6 pu
Let us assume that generators are operated at no-load with fNL = 51 Hz = 1.02 pu
And when loaded up to PL = 600 MW, let frequency decreases to f1.
Then change in frequency ∆f1 = fNL – f1 and G1 supplies P1 and G2 supplies P2.
Therefore from no-load to PL = 600MW
∆P1 = P1 and ∆P2 = P2 as shown in Figure.

As per definition of droop regulation,

f
R OR f  R  P
P
Therefore, ∆f1 = R1 ×∆P1 for G1
and for G2 ∆f1 = R2 ×∆P2
As ∆f1 is common for both generators, ∆f1 = R1 ×∆P1 = R2 ×∆P2
P . R
OR P  2 2 (eqn 1) And ∆P1 + ∆P2 = PL = 0.6 pu (eqn 2)
1 R
1

23
Case-2: When a load ∆PL = 100 W = 100/1000 = 0.1 pu is switched ON:
Let new frequency = f2 as shown in Fig below:

Here, ∆P1 + ∆P2 = 0.1 pu (eqn-3) and ∆f2 = R1 ×∆P1 = R2 ×∆P2

24
∆P2 = 0.0345 pu×Pbase = 0.0345 × 1000 MW = 34.5 MW
∆f2 =R1×∆P1 = 0.066×0.0655= 0.00423 pu = 0.00423×fbase = 0.00423 ×50Hz =0.216 Hz
Therefore, f2 = f1 – ∆f1 = 49.29 Hz – 0.216 Hz = 49.07 Hz

Power supplied by each generator:

Power supplied by G1 = P1 + ∆P1 = 393 MW + 65.5 MW = 458.5 MW
Power supplied by G2 = P2 + ∆P2 = 207 MW + 34.5 MW = 241.5 MW
_______

25
Power Plant Equipment

Lecture No.4
(Numericals)

1
An Illustrative Numerical Examples:
QN1 Two generators of capacities 600MW and 400MW are connected in parallel to supply a
common load of 600 MW. Their droop regulations are 4% and 5 % respectively w.r.t. their
respective ratings. At no-load, they operates at a common frequency of 51 Hz. How they will
share the common load of 600MW? When the load is increased by 100 MW, at which
frequency they will operate and calculate the power supplied by each generator.
Solution:

Let Pbase = 1000 MW, fbase = 50 Hz

Pbase(new) 1000
R R   0.04   0.066 pu
1(new) 1(old) Pbase(old ) 600
Pbase(new) 1000
R R   0.05   0.125 pu
2(new) 2(old) Pbase(old ) 400
fNL (pu) = 51 Hz / fbase = 51 / 50 = 1.02 pu

2
 Case-1: When a common load PL = 600 MW = 600/1000 = 0.6 pu
Let us assume that generators are operated at no-load with fNL = 51 Hz = 1.02 pu
And when loaded up to PL = 600 MW, let frequency decreases to f1.
Then change in frequency ∆f1 = fNL – f1 and G1 supplies P1 and G2 supplies P2.
Therefore from no-load to PL = 600MW
∆P1 = P1 and ∆P2 = P2 as shown in Figure.

As per definition of droop regulation,

f
R OR f  R  P
P
Therefore, ∆f1 = R1 ×∆P1 for G1
and for G2 ∆f1 = R2 ×∆P2
As ∆f1 is common for both generators, ∆f1 = R1 ×∆P1 = R2 ×∆P2
P . R
OR P  2 2 (eqn 1) And ∆P1 + ∆P2 = PL = 0.6 pu (eqn 2)
1 R
1

3
Case-2: When a load ∆PL = 100 W = 100/1000 = 0.1 pu is switched ON:
Let new frequency = f2 as shown in Fig below:

Here, ∆P1 + ∆P2 = 0.1 pu (eqn-3) and ∆f2 = R1 ×∆P1 = R2 ×∆P2

4
∆P2 = 0.0345 pu×Pbase = 0.0345 × 1000 MW = 34.5 MW
∆f2 =R1×∆P1 = 0.066×0.0655= 0.00423 pu = 0.00423×fbase = 0.00423 ×50Hz =0.216 Hz
Therefore, f2 = f1 – ∆f1 = 49.29 Hz – 0.216 Hz = 49.07 Hz

Power supplied by each generator:

Power supplied by G1 = P1 + ∆P1 = 393 MW + 65.5 MW = 458.5 MW
Power supplied by G2 = P2 + ∆P2 = 207 MW + 34.5 MW = 241.5 MW
_______

5
QN2 Figure below shows two generators operating in parallel and supplying a load of 800
MW. G1 is rated as 800 MW and G2 is rated as 300MW. G1 supplies 600 MW and G2 supplies
200 MW and system frequency is 50 Hz. At no-load, they operate at a common frequency of 51
Hz. Calculate droop regulations R1 and R2 of G1 and G2 with respect to their ratings. Assume
base power = 1000 MW. When the load is decreased, the frequency increases to 50.2 Hz.
Calculate the total load in MW and load shared by each generator at 50.02Hz.

Case-1: When G1 supplies 600 MW and G2 supplies 200 MW and system frequency is 50 Hz. At no-load,
they operate at a common frequency of 51 Hz.
These facts are shown in Droop chac curves in Fig shown below:

Let Pbase = 1000 MW, fbase = 50 Hz

From No-load to 800 MW load:
P1 = ∆P1 = 600MW=600/1000 = 0.6 pu
P2 = ∆P2 = 200MW = 600/1000 = 0.2 pu

When the load increases from no-load to

800MW, the change in frequency is :
∆f = fNL – f1 = 51 Hz – 50Hz = 1 Hz
= 1 Hz/ fbase = 1/50 = 0.02 pu

6
7
Case-2: When the load decreases by some amount, the frequency increases to f2 = 50.2 Hz
P1(new) =? P2(new) = ?
This is shown in droop curves below:

The operating point for G1 is shifted to ‘A’ and power generation decreases to P1(new)
The operating point for G2 is shifted to ‘B’ and power generation decreases to P2(new)
Here, ∆f = f2 – f1 = 50.2 Hz – 50Hz = 0.2 Hz = 0.2 Hz/ fbase = 1/50 = 0.004 pu

8
QN3 Three generating units of 600MW, 400MW and 200 MW capacities respectively are
operating in parallel and supplying power to a common load at a common frequency of 50 Hz.
The first unit supplies 300MW, the second unit supplies 200MW and the third unit supplies
100MW. At no-load, they operates at a common frequency of 50.6 Hz. When the load is
increased by 200MW, at which frequency they will operate. Also calculate the power supplied
by each generator.

The figure shows the operation at initial loading condition and additional load of
∆P = 200 MW is switched ON later.

9
 Case-1 : No-load to initial load condition:

G1, G2 and G3 operates at initial operating points A1, B1 and C1 respectively and G1, G2 and G3 supplies
power of P1, P2 and P3 respectively.
Let Pbase = 1000 MW, fbase = 50 Hz
Then P1 = 300MW / Pbase = 300/1000 = 0.3 pu
P2 = 200MW / Pbase = 200/1000 = 0.2 pu
P3 = 100MW / Pbase = 100/1000 = 0.1 pu

From No-load to initial load, change in frequency is :

∆f1 = fNL –f1= 50.6 Hz – 50 Hz = 0.6 Hz = 0.6 Hz / fbase = 0.6 / 50 = 0.012 pu

10
 Based on Pbase = 1000MW

Case-2 : When the load is increased by : ∆P = 200 MW /1000 = 0.2 pu

Frequency decreases to f2 (say) as shown in Fig below:

Now the new operating points of G1, G2 and G3 are A2, B2 and C2 respectively and corresponding new
generations are P1(new), P2(new) and P3(new) respectively.
11
 Therefore, for load change from 600MW to 600MW+200MW corresponding increase in power generation
are: ∆P1 , ∆P2 and ∆P3 respectively.
f f f f
R  2  OR P  2 P  2 And P  3
1 P 1 R 2 R2 3 R
1 1 3
Also Total change in load is : ∆P = ∆P1 + ∆P2 + ∆P3
f f f    1 1 
OR P  2  2  2  f  1  1  1  OR P  f  
1
  f (49.99)
R R R 2R R R  2  0.04 0.06 0.12  2
1 2 3  1 2 3

P 0.2
Therefore, f    0.004 pu
2 49.99 49.99

In absolute value: f  0.004 pu  f  0.004  50  0.2 Hz

2 base
12
 f0.004
Then, P  2   0.1 pu
1 R 0.04
1
f 0.004
P  2   0.0666 pu
2 R 0.06
2
f 0.004
P  2   0.0333pu
3 R 0.12
3
In absolute values:
P  0.1 pu  P  0.1  1000MW  100MW
1 base
P  0.066 pu  P  0.0666  1000MW  66.6 MW
2 base
P  0.033 pu  P  0.0333  1000MW  33.3MW
3 base
New frequency f2 = f1 - ∆f2 = 50 Hz – 0.2 Hz = 49.8 Hz.

New generations:
Power supplied by G1: P1(new) =P1 + ∆P1 = 300 MW + 100 MW = 400 MW
Power supplied by G2: P2(new) =P2 + ∆P2 = 200 MW + 66.6 MW = 266.6 MW
Power supplied by G3: P3(new) =P3 + ∆P3 = 100 MW + 33.3 MW = 133.3 MW

13
 Excitation System

Excitation is the DC current supplied to the filed winding of the AC generator and
it helps to produce magnetic flux in the air gap between stator core and rotor
core. Excitation system is a closed loop control system to maintain the
magnitude of generated voltage constant within the acceptable range at varying
load condition. As explained in the previous section, the change in reactive power
load affects the magnitude of terminal voltage of the generator, the closed loop
control system for excitation system is designed by sensing the terminal voltage
of the generator as shown below in Fig.4.1.

Fig.4.1 Q-V loop for excitation control system

1
When the reactive power of the load increases, the magnitude of terminal
voltage decreases. The voltage sensor is used to sense the magnitude of
terminal voltage. The sensed voltage is compared with the reference value and
error signal so obtained is passed through a AVR with PI-controller. The AVR
increases the dc excitation current (If) so that the terminal voltage increases
until the error is significantly small. That means the magnitude of terminal
voltage increases back to pre-set value.

There are various types of excitation control system. Following two types
shall be studied in this course.
2
4.1) Conventional Excitation System with Slip ring arrangement:

Exciter

Main Generator

Fig.4.2 Conventional excitation control system

3
A dc excitor generator is fitted on the shaft of the main generator. The dc
current (Idc) generated by the armature of dc excitor is supplied to the field
winding of the main generator via slip ring and carbon brush arrangement.
The voltage sensor is used to sense the magnitude of terminal voltage. The
sensed voltage is compared with the reference value and error signal so
obtained drives the controller which generates a controlled Vdc. The controlled
Vdc signal is passed through the gate signal generating logic circuit, where it
is compared with triangular carrier single and produces gate pulses based on
intersection between Vdc signal and triangular carrier signal is AVR with PI-
controller. The gate pulses is passed to the gate terminal of the controlled
rectifier circuit. The controlled rectifier circuit supplies appropriate value of
excitor current (Ie) and the armature of the DC excitor supplies appropriate
value of DC excitation current (Idc) to the field winding of the main generator
and the magnitute of terminal voltage of the main generator comes back to
pre-set value.

The main dis-advantage of this type of excitation is wear and tear of carbon
brushes and slip ring. It produces carbon dust and needs very frequent
maintenance. To over come this disadvantage, brushless excitation system is
developed as described below.

4
4.2 Brush less Excitation System :
Stator

N
If (dc)
N
AC Exciter I ac Rotating Diode

N
I (e)
S

AVR To Load

Voltage
-
Command signal + Sensor
Vref

Fig.4.3 Brushless excitation control system

5
 This type of excitation system has AC excitor instead of DC excitor which do
not have carbon brushes and slip rings. The armature of ac excitor generates
ac current (Iac) which is rectified into If (dc) through the rotating diode rectifier
circuit mounted on the shaft main generator so that no slip rings arrangement
is required. The operating control logic is same as that explained in the
conventional excitation system.

4.3) Modeling of Excitation System:

In order to study the dynamic behavior of the excitation system, it necessary
to have the mathematical model with the transfer function. For the purpose of
developing the mathematical model, the excitation system in general can be
represented by the schematic diagram shown in Fig.4.4.

6
 Fig.4.4 Schematic diagram of excitation control system

The terminal voltage of the main generator is sensed and its magnitude is compared with the
reference voltage (Vref). The error signal is passed through a error amplifier. A stabilizing
transformer is connected between the error amplifier and the power amplifier. The stabilizing
transformer provides the derivative feedback to improve the dynamic response of the excitation
system. The power amplifier provides the correct value of VR (input voltage to field winding of
DC excitor) which in turn provides the correct value of correct value of excitation voltage (VE)
for the field winding of the main generator.

7
The transfer function of the DC exciter can be derived as follow:
KVL for field winding circuit of DC exciter gives:
dI
V  R .I  L e OR V (s)  R . I (s)  s.L . I (s)
R e e e dt R e e e e
The out put voltage (VE) from the armature of dc exciter is proportional to magnetic flux-
linkage of in the field winding of dc exciter.
Therefore, VE = K.N.ϕ = K.Le. Ie Or VE(s) = K.Le. Ie(s)
V (s) K.Le . Ie(s) K.Le . K. Le/Re Ke
Therefore, E    
V (s) Re. Ie(s)  S.Le. Ie(s) Re.  S.Le 1  S. Le/Re 1  s. Te
R

Where, Ke = K. Le/Re , Te = Le/Re = Time constant of dc exciter field winding

8
 The transfer function of the main generator can be derived as follow:
KVL for field winding circuit of main generator gives:
dI
V  R .I  L f OR V (s)  R . I (s)  s.L . I (s)
E f f f dt E f f f f
The out put voltage (Vt) from the main generator is proportional to magnetic flux-linkage of
in the field winding of main generator.
Therefore, Vt = K.N.ϕ = K.Lf. If Or Vt(s) = K.Lf. If(s)
V (s) K.L . I (s) K.L . K. L /R K
Therefore, t  f f  f  f e  G
V (s) R . I (s)  S.L . I (s) R .  S.L 1  S. L /R 1  s. T
E f f f f f f f e G
Where, KG = K. Lf/Rf
TG = Lf/Rf = Time constant of field winding of main generator
9
Stabilizing Transformer: Since Te and TG are large, the dynamic response of the
excitation system could be slow. That means, when a disturbance occurs in the system,
it will take longer time to settle down to a new steady state condition. It is well known
that the dynamic response of a control system can be made faster by internal
derivative feed back loop. This is provided by means of a stabilizing transformer excited
by the exciter output voltage E. The output of the stabilizing transformer is fed
negatively at input terminal of power amplifier.

10
11
Combining the transfer functions of all the components together, the overall
mathematical model of the entire excitation system can be drawn as shown in fig.4.5.

Exciter Gen
Vref(s) E(s) V1(s) V2(s) Vt(s)
Ke Kg
+- Ka(s) +- 1+ Te S 1+ Tg S

Vt(s)
Vst(s)

S. Kst
1+ Tst S

Stabilizing
Transformer

Fig.4.5 Overall mathematical model of the excitation system

12
Power Plant Equipment

Lecture No.6
(Reactor)

1
6. Reactor
Reactor is the current limiting inductor (having very low resistance) connected in the power
system network to reduce the magnitude of fault current so that Circuit Breaker of lower
breaking capacity can be used. It also helps to reduce the voltage magnitude disturbance
during fault condition. A reactor can be installed at various locations and accordingly they are
classified as follow:
5.1 Generator Reactor
It is the reactor inserted between generator and generator bus as shown in Fig.6.1.

Fig.6.1 Illustration of Generator Reactor

Here, XG1 and XG2 are generator reactors.

2
 If a 3-phase to ground fault occurs on the one of the
out going feeder, the fault current supplied by G1
and G2 are Given by:
E E
I  And I 
f1 X"  X f2 X"  X
1 G1 2 G2

Where, X1” and X2” are sub-transient reactance of G1 and G2 respectively.

Total fault current on the outgoing feeder If = If1 + If2
It is very clear from the above equations that XG1 and XG2 helps to reduce fault currents If1 and
If2. Accordingly, the total fault current on faulty feeder also reduces.

During the fault, voltage drop take place in the both reactors XG1 and XG2 due to the fault
current If1 and If2. Therefore, fault on one feeder affects the voltage of other healthy feeder.

Modern generators are designed to have higher value of sub-transient reactance which is
sufficient to reduce the fault current due to three phase short circuit across the generator
terminal to a safe value. Hence, separate generator reactor may not required for modern
generator. However, when a new generator is added in parallel with an old generator, generator
reactor may have to be installed in the old generator.

3
6.2 Feeder Reactor
It is the reactor connected in series with the outgoing feeder as shown in Fig.6.2.

In the event of fault on any one feeder, voltage

drop due to fault current only take place in the
faulty feeder. Voltage of other feeder is not
affected much, hence the other feeder continue
to supply power. However, this scheme does
not provide protection to old generator (having
lower sub-transient reactance) against short
circuit across the generator terminals.

Fig.5.2 Illustration of Feeder Reactor

The total fault current supplied by G1 and G2 together is given by:

E E
I  
f X"  X X"  X
1 f1 2 f1

4
6.3 Bus Bar Reactor
In the above two schemes, the normal system current flows continuously through the reactor
during normal operation without fault. These normal current cause some voltage drop and
power loss continuously. This disadvantages can be eliminated by using Bus Bar Reactor
scheme. Followings are two types of bus bar reactor schemes.

6.3.1 Ring type bus bar reactor:

In this scheme, sections are made of generator and feeder and these sections are connected to
each other at common bus through a reactor as shown in Fig.6.3.1.

Fig.5.2 Illustration of ring type bus bar reactor

5
Here, XB1 and XB2 are ring type bus bar reactor. During normal operating condition, potential of
points ‘A’ , ‘B’ , ‘C’ and D will be nearly equal. Therefore nearly zero current flows through the
reactors XB1 and XB2. Hence, power loss and voltage drop in these reactors will be negligible
during normal operation.

In case of fault on the first feeder, only first generator supplies major portion of the total fault
current, while the fault current supplied by the other two generators will be limited by the bus
bar reactors. Hence, the reaming two feeders can continue to supply power, while there is fault
on first feeder.

6
The fault current supplied by G1, G2 and G3 are Given by:
E
I 
f1 X"  X
1 G1
E
I 
f2 X"  X
2 B1
E
I  Total fault current on the faulty feeder If = If1 + If2+ If3
f3 X"  X  X
3 B1 B2

7
6.3.2 Tie-bar bus bar reactor:
In this scheme, generators are connected to the common bus through the reactors, but the
feeders are connected to the generator side of the reactor as shown in Fig.6.3.2.

Fig.6.3.2 Illustration of tie-bar bus bar Reactor

If there is fault on feeder-1, the fault current supplied by G2 flows through X2 +X1, and fault
current supplied by G3 flows through X3 + X1. Therefore, voltage of feeder-2 and feeder-3
remains nearly constant to E. If the number of sections are increased, the fault current
through the faulty current feeder will not increase significantly. Therefore CB of existing
feeder need not be up-graded with higher breaking capacity.

8
Some Illustrative Numerical Examples:
QN1 For the system shown below, calculate the fault level in MVA at out going feeder for
a three phase to ground fault on this feeder. Calculate the value of reactance to be
connected in the feeder in order to reduce the fault level by 40%.

Fig.1 shows the diagram of given system

Let Pbase = 25 MVA and Vbase = 11 kV
V
P
base  25  1000kVA  1312A  1.312kA Z  base  11  1000V  4.84 
I  base 3I 3  1312A
base 3V 3  11kV base
base

9
Given that X1 = 0.15 pu based on 25MVA and 11kV
And X2 = 0.3 pu based on 25MVA and 11kV
Since the pu values both X1 and X2 are based on same base power and base voltage, they
need not required to convert on common base
Equivalent reactance up to fault point ‘F’ is given by parallel equivalent of X 1 and X2 and
its given by:
X X
X  1 2  0.15  0.3  0.1 pu
eq X  X 0.15  0.3
1 2
Then equivalent faulted circuit can be written as shown in Fig.2
Fault current on the outgoing feeder is given by:

E (pu) 1
I    10 pu
f X 0 .1
eq (pu)

Absolute value of fault current is given by: If = If(pu) × Ibase = 10 pu ×1.312 kA = 13.12 kA
1 1
The fault level at point ' F'   10 pu
X 0.1
eq (pu)
Absolute value of fault level (power) is given by:
Faulty power = Fault power (pu) × pbase = 10 pu ×25MVA = 250 MVA
10
Part-II: Fault level to be reduced by 40%
Therefore desired fault level = 10 pu – 40% of 10pu = 6 pu

In order to reduce the fault level to 6 pu, an reactance of Xadd shall be connected on the
feeder as shown in Fig.3 below and its equivalent circuit is shown in Fig.4

Now the total equivalent reactance up to the fault point is given by:
Xf = Xeq + Xadd
New fault level = 1/Xf (pu) = 6 pu OR Xf = 1/6 = 0.166 pu

We have, Xf = Xeq+ Xadd Or Xadd = Xf - Xeq = 0.166 – 0.1 = 0.066pu

Absolute Value of Xadd = 0.066 pu × Zbase

= 0.066pu × 4.84 Ω = 0.32 Ω

11
 QN2 Following figure shows two generator operating parallel

i) If a 3-phase to ground fault occurs on the out going feeder, calculate the fault current and
fault level on the feeder.
ii) If a 3rd generator (G3: 10MVA,11kV, X3= 0.2 pu) is added at point ‘C’, calculate the value
of reactor to be added on the outgoing feeder so that the fault level remains same as before.
Part-I:
Let Pbase = 20 MVA and Vbase = 11 kV

base  20  1000kVA  1049.75 A  1.049kA

P
Then I 
base 3V 3  11kV
base
base  11  1000V  6.05 
V
And Z 
base 3I 3  1049.75 A
base
12
X1 and X2 are in parallel, therefore equivalent
reactance up to the fault point is given by:
X X
X  1 2  0.05 pu
12 X  X
1 2
E (pu) 1
Therefore, Total fault current: I    20 pu
f X 0.05
12 (pu)

Absolute value of If = 20pu × Ibase = 20×1.049 kA = 20.98 kA

1 1
Fault Leve l (or fault power) in pu    20 pu
X 0.05
12 (pu)
Part-II:
When 3rd generator is added, Let Xadd be the
reactance to be added on the feeder as
shown in Fig below so that the fault level
remain same as before.
Here X3 = 0.2 pu based on 10MVA, 11kV
Value of X3 based on 20 MVA is given by:
P
base( new) 20
X  X   0.2   0.4 pu
3( new) 3(old) P 10
base(old )

13
Equivalent of X1, X2 and X3 in parallel:

X X
X  12 3  0.05  0.4  0.0444 pu
123 X  X 0.05  0.4
12 3

After adding Xadd, the fault level on feeder shall remain same as before = 20 pu
That means : New equivalent reactance up to fault point shall remains same = 0.05 pu
Therefore X123 + Xadd = 0.05 pu
OR Xadd = 0.05 pu - X123 = 0.05 – 0.0444 = 0.0056pu
Absolute value of Xadd = 0.0056pu × Zbase = 0.0056 ×6.05 = 0.0333 Ω

14
QN4 The figure below shows four identical generators, each rated as 25MVA, 11 kV and each
having a sub-transient reactance of 16% on its ratings.

i) If there is 3-phase to ground fault on the outgoing feeder, calculate fault current and fault
level (power) to be carried by the circuit breaker and breaking capacity of CB required.
ii) If the fault level is to be limited to 500MVA or less, Calculate the value of bus bar reactor
(X) to be connected between point A and B.
Ans:
i) Let Pbase = 25 MVA and Vbase = 11 kV
P 25  1000kVA
I  base   1312A  1.312kA
base 3V 3  11kV
base
V
Z  base  11  1000V  4.84 
base 3I 3  1312A
base
15
 Since all 4 generators have same ratings and their pu reactance are already on the common
base quantities. Equivalent reactance of 4 machines in parallel is given by:
1  1 1 1 1 
    
X  X
eq  1
X
2
X
3
X 
4
1  4   X  0.16
Since all reactance are equal,    OR Xeq   1    0.04 pu
X  X 
eq  1   4
   4

E (pu) 1
Fault current on the outgoing feeder is given by: If    25 pu
X 0.04
eq (pu)
Absolute value of fault current is given by: If = If(pu) × Ibase = 25 pu ×1.312 kA = 32.8kA
1 1
Fault level on C.B. =   25 pu
X 0.04 Absolute value of fault level (power) is given by:
eq (pu)

Faulty power = Fault power (pu) × pbase = 25 pu ×25MVA = 625 MVA

The standard breaking capacities of commercial C.B are 500 MVA , 700 MVA
Therefore, C.B. with 700MVA breaking capacity shall be selected. 16
Part-II: Calculate the value of X to be connected between A and B to reduce the fault
level to 500MVA.

The fault power supplied by G1 and G2

remains same before, Therefore G1 and G2
together supplies fault power
= 625 MVA /2 = 312.5 MVA

Now by adding X between A and B, the fault power

supplied by G3 and G4 together shall be limited to
500MVA – 312.5 MVA = 187.5 MVA =
187.5/Pbase = 187.5 / 25 = 7.5 pu

Equivalent of X3 and X4 in parallel is given by X34 = 0.16 / 2 = 0.08 pu

And equivalent of X34 in series with X is given by:
1
Xt = X34 + X = 0.08 +X (eqn.1) Therefore, X  7.5 pu OR X  1  0.133 pu
t 7.5
t (pu)
From (eqn.1) Xt = 0.08 +X (eqn.1) Therefore X = Xt - 0.08 = 0.133 – 0.08 = 0.053 pu
Absolute value of X = 0.053 pu × Zbase = 0.053 pu × 4.84 Ω = 0.256 Ω

17
 Power Plant Equipment

Lecture No.7
(Ancillaries Services in Power Plant Station)

1
7.1 Fire Fighting System for Power station :

Carbon dioxide (CO2) fire extinguish system had been recommended by the code of practice
for electrical fire hazards. The other fire extinguishing system like- water jet, foam spraying are
not suitable as they may damage the electrical equipment.

CO2 is a non-combustible gas that can penetrate and spread to all parts of a fire, thus by
diluting the available oxygen to a low concentration which will not support combustion. CO2
does not conduct electricity and can be spray over the energized electrical equipment. However,
it is a dangerous gas to human life. Therefore, special precaution shall be taken by the fire
fighting personnel and normal occupants in the room shall go out as soon as possible.

Generally two types of CO2 fire extinguishing systems are in practice.

i) Hand hoses from portable CO2 storage cylinder
ii) Fixed piping system

i) Hand hoses from portable CO2 storage cylinder:

CO2 portable storage cylinder can be installed at appropriate locations where there are personal
attending all the time. In case of fire, a person can operate it to extinguish the fire by manual
operation.

ii) Fixed piping system:

Fixed piping system is suitable to cover large space where personnel attendance is not possible
for all 24 hours. Fig.7.1 shows a typical example of Fixed Pipe system.

2
Fig.7.1 A typical example of Fixed Pipe system.
3
Electronic type smoke detectors (S) are installed on the ceiling at an interval specified
by the manufacturer of the detector. In case of fire, smoke reaches on the ceiling and it
will activate the smoke detector. The smoke detector sends signal to Fire Alarm Control
Panel (FACP) installed at security room which is 24 hour attended by the shift security
personnel. FACP indicate the zone at which the fire had been taken place and at the
same sounders will produce alarm sound to alert the personnel inside the building and
all the persons inside the building will come out from the building. The response
indicator of a particular room having fire will indicate by flipping its LED indicator so
that the fire fighting person can easily identify the room having fire.

There is a CO2 pressurized tank at the convenient location from which metal pipes are run
on the ceiling around the rooms at an suitable interval. Nozzles are fitted on the bottom of
the metal pipes at an suitable interval. The bottom of the nozzle has a weak layer which
breaks at certain threshold value of temperature. When there is fire in particular spot of a
room, the temperature of that spot will rise above the threshold value of temperature and
the nozzles in that spot will break and releases the CO2 gas thus by extinguishing the fire.
The sounder produces alarm sound before the release of CO2 so that people inside the
building has enough time to escape out from the building before CO2 gas is released.

4
7.2 Power Line Communication:
It is a communication system between two electric sub-stations by using the power
transmission line conductor as medium to carry the communication high frequency signal
without disturbing the power transmission process. Fig.7.2 shows the schematic diagram of
such communication system.

Fig.7.2 shows the schematic diagram of such communication system.

Line traps shown in the figure are parallel resonant tuned circuit having negligible
impedance to power frequency of 50 Hz, but offers very high impedance to high frequency
communication signal. The coupling capacitor offers very low Impedance (1/2πfC) to high
frequency communication signal, but it offers very high impedance to low frequency power
signal (50Hz).
5
When a voltage of positive polarity is impressed on the control circuit of transmitter at one sub-
station, it generates a high frequency output voltage ( 30 kHz – 500 kHz) and consumes very
small power. This output voltage is applied between one phase conductor and earth. Now the
transmission line conductor carries the power frequency signal as well as high frequency
communication signal. The high frequency signal is blocked by the line trap at that sub-station.
When the transmitted signal reaches to S/S-2, line trap at that S/S blocks the high frequency
communication signal and only 50 Hz power signal flows to bus of S/S-2. Therefore the high
frequency communication signal passes thorough the coupling capacitor and reaches to receiver
at S/S-2 and ring tone will come at S/S-2 hand set. Now the person at S/S-2 can make
conversation with the person at S/S-1.

6
7.3 Power Transformer and its Components:
Power transformers are large capacity transformers used in generating sub-station to step up the
voltage to transmission level and transformer used in receiving end of the transmission line to
step down the voltage level. The various part of a power transformer are shown in Fig.7.3.

Fig.7.3 Various parts of a power transformer

7
Transformer Oil:
The core and winding of a power transformer is immersed in the transformer oil contained is an
enclosed tank. The transformer oil serves the double purpose of cooling the whole transformer
and insulating the winding from the tank. The transformer oil is bad conductor of electricity ,
but good conductor of heat. The heat generated by the winding and core of the transformer will
be dissipated to the atmosphere through the transformer oil. Transformer oil is the mineral oil
having following properties:

•Flash point temperature > 1600 C ( it is the temperature at which the oil gets vaporized and gets ignites)
•It shall not contain impurities like – sulphur and its compound which may cause corrosion on metallic
parts.

Explosion vent:
It is a vertical pipe fitted on the top of transformer tank. Top end of the pipe is bended and
sealed with some weak material. If the excessive pressure develops inside the tank due to
over-loading or short circuit, the weak material breaks and the release the pressure thus by
protecting the transformer from bursting.
Conservator Tank :
It is a cylindrical tank fitted above the transformer tank. The bottom end of the conservator
tank is connected to the transformer tank via Buchholz relay and upper end of the conservator
tank is connected to the breather as shown in the Fig. It provides space for expansion and
contraction of transformer oil due to heating and cooling.

8
Breather :
It is a small box with fine holes on the bottom and filled with silica gel. The breather is
connected to upper surface of the conservator tank through a pipe. It can be called as
breathing nose of the transformer. When the transformer is heated at full load condition, the
transformer oil expands and the oil level in the conservator tank rises up by displacing some
volume of air to the atmosphere through the breather. When the transformer gets cooled
during light load period, the oil level in the conservator tank falls down. While doing so the
atmospheric air enters into the conservator tank through the breather. The silica get in the
breather absorbs the moisture content in the air and moisture free air is passed to the
conservator tank.
Cooling Tubes:
These tubes helps to improve the cooling performance in the transformer. The heated oil
rises up and passes through the upper end of the cooling tubes. While the oil come down
through the tubes, it gets cooled due to contact with atmospheric air.

9
Buchhloz Relay :
It is a protection device installed between transformer tank and the conservator tank. Fig.7.4
shows the details of Buchholz relay.

Fig.7.4(a) Fig.7.4(b)
Fig. 7.4 Details of Buchholz relay.

During various faults like – a broken joint in winding, earth fault, sudden short circuit etc, they
produce local heating and generate gases. This generation of gas is utilized to operate a relay
which in turn gives alarm and completes the relay trip circuit and opens the circuit breaker of
the transformer.

10
Fig.7.4(a) shows the cross-sectional view of Buchholz relay during normal condition. Vessel of the relay is full
of transformer oil at normal condition and the contacts ‘C1’ and ‘C2’ are open. The vessel contains two floats
‘b1’ and ‘b2’ which are hinged so as to pressed by their buoyancy against two strip. If the gas bubbles are
generated inside the transformer tank due to fault, the gas bubbles will rise up (as shown in Fig.7.4(b) ) and
passes through the pipe line toward the conservator tank and will be trapped in the upper part of the relay
vessel there by displacing the oil and lowering the floats ‘b1’ and closes the contact ‘C1’ there by closing the
circuit of alarm. If the gas generation is excessive due to longer fault , the float ‘b2’ will also lower down and
closes the contact ‘C2’ which will energize the electromagnet. The electromagnet pulls down the lever thus by
completing the circuit of trip coil of the relay of C.B of transformer and C.B opens.
11
SCADA in power system :
SCADA (Supervisory Control and Data Acquisition ) is a system developed to control and
monitor a power system (or any system) by collecting information from various points of the
system and transmitting them to a central operator, where the received information are
displayed and necessary corrective actions can be made automatically or manually. Fig.5.8
shows the schematic diagram of typical SCADA system.

Fig.7.5 Schematic diagram of typical SCADA system.

12
S/S-1 and S/S-2 are two particular points of a power system. CT, PT and temperature sensors
collects the information of current, voltage and temperature from the equipment. The Remote
Terminal Unit (RTU) collects the information from the sensors and performs some micro-
computations and signal conditioning task. It delivers the information (data) to the Central
Control Centre via radio frequency modium by wireless system. The input / output server at
control center receives the information. The pre-programmed software in the input/output
server will take automatic actions according to the information obtained. For example - it the
temperature of equipment at S/S-1 exceeds the pre-set value, the C.B. of the equipment at S/S-1
trips automatically. Manual command also can be made from Control Center to operate the
C.B.
13
 Power Plant Equipment

Lecture No.8
(Modeling of Water Turbine)

1
2
3
4
5
6
7
8
9
10
11
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13
 Power Plant Equipment
BEL 4th year part-I

QN1 Two generators of capacities 600MW and 400MW are connected in parallel to supply a
common load of 600 MW. Their droop regulations are 4% and 5 % respectively w.r.t.
their respective ratings. At no-load, they operates at a common frequency of 51 Hz. How
they will share the common load of 600MW? When the load is increased by 100 MW,
at which frequency they will operate and calculate the power supplied by each generator.
[Ans: Case-1: P1= 393 MW, P2 = 207 MW, f1 = 49.29 Hz.
Case-2: f2 = 49.07 Hz, P1(new) = 458.5 MW, P2(new) = 241.5 Hz]

QN2 Figure below shows two generators operating in parallel and supplying a load of 800
MW. G1 is rated as 800 MW and G2 is rated as 300MW. G1 supplies 600 MW and G2
supplies 200 MW and system frequency is 50 Hz. At no-load, they operate at a common
frequency of 51 Hz. Calculate droop regulations R1 and R2 of G1 and G2 with respect to
their ratings. Assume base power = 1000 MW. When the load is decreased, the frequency
increases to 50.2 Hz. Calculate the total load in MW and load shared by each generator at
50.2 Hz.
[Ans: Case-1: R1= 0.026 pu , R2 = 0.03 pu
Case-2: Total load = 640 MW, P1(new) = 460 MW, P2(new) = 180 Hz]

QN3 Three generating units of 600MW, 400MW and 200 MW capacities respectively are
operating in parallel and supplying power to a common load at a common frequency of
50 Hz. The first unit supplies 300MW, the second unit supplies 200MW and the third
unit supplies 100MW. At no-load, they operates at a common frequency of 50.6 Hz.
When the load is increased by 200MW, at which frequency they will operate. Also
calculate the power supplied by each generator.
[Ans:f(new) = 49.8 Hz, P1(new) =400MW, P2(new) = 266.6 MW, P3(new) =100.3 MW ]

QN4 Two generating units of 500MW and 250 MW capacities respectively are operating in
parallel and supplying power to a common load. When each generator is half loaded, they
operate at a common frequency of 50 Hz. If the droop regulation of 500MW generator is
set at 5% based on its rating, what must be the droop regulation of 250 MW generating
unit based on its own rating show that they share the change in load according to their
capacities. {Ans: 0.05 pu)

QN5 Two generators G1 and G2 rated as 100 MW and 500 MW respectively are operating in
parallel. At half load, they operate at a common frequency of 50 Hz. When the load is
increased by 100 MW, the frequency decreases to 49.5 Hz. What must be the individual
droop regulations (with respect to their own rating) so that two generators shares the
increased load in proportional to their capacities. { Ans: R1 = 0.06pu, R2 = 0.06 pu }

1
 Numericals on Reactor

QN1 For the system shown below, calculate the fault level in MVA at out going feeder for a
three phase to ground fault on this feeder. Calculate the value of reactance to be
connected in the feeder in order to reduce the fault level by 40%.
[Ans: 250 MVA, 0.32 ohm]

QN2 Following figure shows two generator operating parallel

i) If a 3-phase to ground fault occurs on the out going feeder, calculate the fault current and
fault level on the feeder.
ii) If a 3rd generator (G3: 10MVA,11kV,X3= 0.2 pu) is added at point ‘C’, calculate the
value of reactor to be added on the outgoing feeder so that the fault level remains
same as before.
[Ans: Part-1: 400 MVA, Part-2: 0.033 ohm]

2
QN3 Fig below show the single line diagram of parallel operated generators. Their ratings are:
G1: capacity = 50 MW, 11kV, X1= 0.16 pu based on its rating
G2: capacity = 40 MW, 11kV, X2= 0.08 pu based on its rating
XL= 2 ohms

Calculate the value of inductor (in mH) of the reactor to be connected in series with X1
so that both generators delivers equal amp of fault current during 3phase to ground fault
on the feeder. [Ans: 0.461 mH]

QN4 The figure below shows four identical generators, each rated as 25MVA, 11 kV and each
having a sub-transient reactance of 16% on its ratings.

i) If there is 3-phase to ground fault on the outgoing feeder, calculate fault current and fault
level (power) to be carried by the circuit breaker and breaking capacity of CB
required.
ii) If the fault level is to be limited to 500MVA or less, Calculate the value of bus bar
reactor (X) to be connected between point A and B.

[Ans: Part-1: 625 MVA, Part-2: 0.256 ohm]

3
QN5 Fig. below shows a bus-bar reactor scheme with four generators. If a 3phase to ground
fault occurs on the outgoing feeder, calculate the fault current (in kA) supplied by each
generator and fault MVA required for the circuit breaker (CB) on the feeder. Ratings of
generators are given as follow:
G1/G2 : 25MVA, 11kV, Xg1” = 0.2 pu
G3/G4: 50MVA, 11kV, Xg2” = 0.1 pu

[Ans: If1 = If2 = 8.4 kA , If3 = If4 = 12.6 kA, 800 MVA]

QN6 Fig. below shows a generator reactor scheme with four generators.
Given that:
Each generator is rated as: 20MVA, 11kV, Xg” = 0.2 pu
X1 = X2 = X3 =X4 = 0.5 ohms (generator reactors)
a) If a 3-phase to ground fault occurs on outgoing feeder, Calculate fault current
supplied by each generator and the fault level (MVA) on the feeder.
b) If bus bar rectors each of 1 ohm are added between A-B and C-D, calculate the fault
current supplied by each generator fault level (MVA) on the feeder.

[Ans: Part-a : 3.54 kA, 263 MVA, Part-b: If1= If4 = 2.34kA, If2= If3 = 3.67kA, 229.3 MVA]

4
QN7 Fig. below shows a generator reactor and bus reactor scheme with two generators.
Each generator is rated as: 20MVA, 11kV, Xg” = 0.2 pu
X1 = X2 = 0.5 ohms (generator reactors)
XB = 0.3 ohm (Bus bar rector)
a) If a 3-phase to ground fault occurs on outgoing feeder, Calculate fault current
supplied by each generator and the fault level (MVA) on the feeder.
b) Calculate the value of feeder reactor to be connected on the outgoing feeder so that
fault level on the feeder reduces by 25%.

[Ans: Part-a : Part-a: If1= 3.71 kA, If2 = 1.35 kA, 96.6 MVA, Part-b: 0.06 ohm ]