EEPP 522/521
POWER PLANT ENGINEERING
Course Outline:
I. Power Plant Prime Movers
II. Types of Power Plant
 principles of operation of each type of power plant
 comparison of power plants
III. Parallel operation of Alternators
IV. Load graph significance
 Power plant factors
 Application to performance calculations
V. Performance calculations of power plants
VI. Power plant economics & electric rates
VII. Power system Stability
VIII. Power Plant Layout
References:
1.
2.
3.
4.
5.
Principle of Power System by V. K. Metha 1997 ed. S Chand
Worked Examples in Electrical Power by A. P. Gupta, 4th ed 1995 SSMB
Elements of Power System Analysis by Stevenson, 4th ed.
A Textbook of Electrical Technology by B. L. Theraja 22nd ed. 1997 S Chand
Electrical Power Equipment by Tarboux 3rd ed. 1946 JMC
Engr. NELSON S. ANDRES P.E.E.
Instructor
INTRODUCTION
Power Plant Engineering
An art of designing and installing generating plant that will result in
maximum return on investment over the expected life of equipment and also,
operating this equipment to achieve reliable, continuous and cheap power
service.
1.
2.
3.
4.
Major Factors Affecting the Cost of Energy
Operating labor & superintendence
Maintenance
Fuel cost
Fixed charges such as taxes, interest & depreciation
Sources of Energy
1. Sun - By means of reflector, heat generated can be used to raise the steam
temperature and pressure and electrical energy can be produced.
2. Wind - wind energy (wind mill) is used to drive generator. For continuous
supply, a charged battery is utilized.
3. Water - Power is generated by falling water that drives the turbine.
4. Fuel - either solid, liquid or gas energy is generated by means of steam
engine, steam turbine or internal combustion engine.
5. Nuclear - heat is produced by nuclear fission.
1 kg of nuclear fuel can produce
the same amount of energy as 5000 tons of coal.
Comparison of energy Sources
Particular
Initial cost
Running cost
Reserves
Cleanliness
Simplicity
Reliability
Water Power
High
Less
Permanent
Cleanest
Simplest
More reliable
Fuel
Low
High
Exhaustible
Dirtiest
Complex
Less reliable
Nuclear
Highest
Least
Inexhaustible
Clean
Most complex
More reliable
Power Plant - a station or establishment which houses the prime mover, electric
generators and auxiliaries for conversion of mechanical energy, chemical energy
and or nuclear energy in electrical energy.
Different types of Prime Movers
1. Steam Turbine
2. Hydraulic Drive
a. Impulse wheel
b. Reaction wheel
c. Propeller wheel
3. Internal Combustion
Choice of Water Turbine
Types of Turbine
Propeller or Kaplan Turbine
Reaction or Francis Turbine
Impulse or Pelton wheel
Head of Water for which
is used
Up to 150 ft (45 m)
Up to 500 ft (150 m)
500-1000 ft (150-300 m)
Range of Specific
Speed (British Unit)
200 to 120
120 to 20
20 to 10
Specific Speed, Ns
is defined as the speed of a geometrically similar turbine working
under unit head and delivering unit power, the specific speed in
British unit is given by:
Ns  N
HP
5
H4
Where:
N= Rotational Speed in rpm.
H= Head of water in ft.
HP = Horse Power of the wheel.
For geometrically similar turbine, the following equation together with specific
speed equation are used to compare the turbine performance.
Q1
Q2
 constant
3 
N1D1 N2 D 32
H1
H2
 constant
2 2 
N1 D1 N22D 22
Problems (Energy Sources)
1. Coal reserves in the country are estimated to contain 2250 quads of energy. If
the energy content of this coal is 11, 500 btu per lb, determine the weight of
the coal reserve.
1015 btu
Energy  2250 quads x
 2250 x1015 btu
quad
2250x10 15 btu
Weight of Reserve Energy 
 1.957 x1010 lbs
btu
11,500
lb
2. In year 2000, in a certain country the consumption of energy (in quads) from
various sources was as follows; coal 16.1, Oil 32.1, natural gas 20.2, hydro
2.9 and nuclear 2.9 . Calculate the GWH of total electric energy that could be
produced from these sources assuming that the average power plant
conversion efficiency of 10%.
1015 btu
Fuel Energy  16.1  32.1  20.2  2.9  2.9  74.2 quads x
quad
 kwh 
12
6
 74.2 x1015 btu 
  2.17 x10 kwhr  2.17 x10 GWH
3413
btu


Where:
3. Suppose that the consumption of energy in certain country has a growth rate
of 4% per year. In how many years will the energy consumption be tripled?
E  E O e at
Where: Eo- initial energy consumption
E- energy consumption after time in years
a- energy consumption growth rate in decimal
t- time in years
3Eo  Eo e 0.04 t
Ln3
t
 23.1 years
0.04
4. The Natural gas reserves in a certain country are estimated at 100 x 10 9 ft3,
with an energy content of 0.025 Watt. yr/ft3. If the present peak power
demand is 0.5GW, the power demand growth rate is 5%., and all the energy
is to be supplied by natural gas, approximately how long will the reserve last?
Formula:

1  a x QT
t  Ln
 1
a  PO

QT  (100 x10 9 ft 3 )(0.025
watt  year
)  2.5 x10 9 watt  year
3
ft
 0.05 x 2.5 x10 9

1
t
Ln 

1
  4.46 yrs.
0.05 
0.5 x10 9

5. In a power station, 4 x 10 4GWh of energy is to be produced in 1 yr. To be
supplied either Coal or Natural gas, the energy content of coal is 900 W .
yrs/ton, and that of natural gas is 0.03W . yr/ft3.
a. How much fuel is required if the energy is to be supplied by coal
4x1013 watt  hr
Weight of Coal 
 5.07 x106 tons
watt - year 8760hrs
900
x
ton
yr
b. How much fuel is required if the energy is to be supplied by natural
gas.
4x1013 watt  hr
Volume of Natural Gas 
 1.522 x10 9 ft 3
watt - year 8760hrs
0.03
x
yr
ft 3
6. What is the average output of a wind turbine in kw having a blade diameter of
35 ft if the wind velocity ranges from 10 to 30 mi/hr.
Formula: P  2.46x10 -3 D 2V 3 watts
Where:
P-wind power developed in watts
D- windmill diameter in ft
V- wind velocity in miles per hour
Pmin  2.46x10-3 (35)2 (10) 3  3kw
Pmax  2.46x10-3 (35)2 (30) 3  81kw
3  81
Average Power 
= 42 kw
2
7. Calculate the power that can be generated from a tidal power station with a
tidal bay area of 128.6 sq.km and the tidal head available is 6m.
P  0.219AH 2 MW
Formula:
Where: P-tidal power generated in MW
A-tidal bay area in sq.km
H- tidal head in meter
P  0.219(126.8)(6)2  1000 MW
8. A hydraulic drive works under a head of 24 ft. and produces 2400 bhp at 250
rpm. Calculate
a. Rpm under a unit head for a unit power.
Specific Speed , N S  N
HP
H 1.25
2400
 231 rpm
241.25
b. Hp under a unit head at a unit rpm.
HP1=2400 hp
HP2=?
H1=24 ft
H2= 1 ft
N1=250 rpm
N2=1 rpm
For geometrically similar turbine,
HP
HP
N S  N1 1.251  N 2 1.252
H1
H2
N S  250
(unit head)
(unit rpm)
HP2
11.25
HP2=53,361 hp
231  1
c. Hp & Rpm under a unit head.
HP1=2400 hp
HP3=?
H1=24 ft
H3= 1 ft
(unit head)
N1=250 rpm
N3=?
For geometrically similar turbine, the following relation can be applied
H1
H
H
 22 2  23 2
2
2
N1 D1
N 2 D2
N 3 D3
1
24
 2 2
assume D1=D3
2
2
250 D1
N 3 D3
N3  51 rpm
then N S  N 3
HP3
 231 rpm
1.25
H3
HP
51 1.25 3  231 rpm
1
HP3=20.5 hp
9. The turbine installed in a powerhouse develops 2400 hp under a unit head of
400 ft. Find the specific speed of the turbine if it has to run at a speed of 500
rpm. If a similar turbine with runner diameter half in size of the original runs
under a head of 60 ft, find the speed of rotation in rpm.
2400
N S  500
 13.7 rpm
4001.25
HP1=2400 hp
HP2=?
H1=400 ft
H2= 60 ft
N1=500 rpm
N2=?
For geometrically similar turbine,
H1
H
 22 2
2
2
N1 D1
N 2 D2
60
400
 2
2
2
500 D1
N 2 (0.5D1 ) 2
N 2  387 rpm
D2=0.5D1
EXERCISES
1. In a certain region the growth rate of energy consumption is 6%. In how many
years will the energy consumption be quadrupled?
2. A certain amount of fuel contains 15 x 1010 Btu of energy is converted into
electric energy in a power station having a 12% overall efficiency. The
average demand on the station over a 24-h period is 5MW. In how many days
will the fuel be totally consumed?
3. A certain amount of fuel can produce 10 quads of energy. In how many days
will the fuel be totally consumed if it is used to satisfy a demand of
1013Btu/day at a power plant with an overall efficiency of 20%?
4. In a power station, 4 x 10 4GWh of energy is to be produced in 1 yr, one third
from coal and two thirds from natural gas. The energy content of coal is 900
W . yrs/ton, and that of natural gas is 0.03W . yr/ft3.
A. How much coal will be required?
B. How much natural gas will be required?
5. In a certain country the equivalent fuel reserve for power generation is 3 x 106
MW . yrs. The present peak power demand is 200GW, and the expected
power consumption growth rate is 2.1%. How long will the fuel reserve last?
6. A model turbine has a size 1/8th of the actual turbine. It is tested under a head
of 10 m and develops 40 hp at 600 rpm. If the actual turbine operates at a
head of 45 m, what will be its speed and hp?
7. A wind mill has a blade diameter of 50 ft. Calculate the power that can be
delivered for a constant wind velocity of 20 mph and overall efficiency of
80%.
8. A tidal power station has its tidal bay area of 300 sq.km and an expected tidal
head of 5.6 m. Calculate the power that can be developed. Also, calculate to
power that can be generated if the overall efficiency is 85%.
TYPES OF POWER PLANT (AS TO SOURCE)
1. Thermal Power Plant
a. Oil-fired Thermal Plant - Makes use of heavy fuel or bunker oil.
b. Coal-Fired Thermal Plant - makes use of pulverized coal as fuel.
c. Dendro-Thermal Plant - makes use of wood (ipil-ipil)
d. Nuclear Power Plant - makes use of steam generated in a reactor by
heat from the fission process of nuclear fuel. (Uranium 235 & 238)
e. Gas Turbine Plant - makes used of combustible gases.
f. Geothermal Power Plant - makes use of heat generated from
inherent steam from earth’s magma.
g. Solar steam Plant - makes use of steam generated from solar
radiation.
2. Hydro-Electric Power Plant
a. Run of River Plant - using pondage of steam flow as it occurs.
b. Plant with storage Capacity - associated with large reservoir. This
permits regulated supply of water for constant power output.
c. Pumped-Storage Plant - Energy is generated during periods of high
system demand using water which has been pumped into a reservoir
usually during periods of low system demand.
3. Diesel Power Plant - a plant of internal combustion engine prime mover
using diesel as fuel in producing energy.
4. Wind Mill Plant - using series of windmills as prime movers.
5. Sea waves/Ocean Tides Plant - makes use of natural rising of tides to
simulate flowing water.
6. Solar or Photo-voltaic Plant - Chemical conversion of radiant energy of the
sun to electric (dc) energy.
Types of Power Plant (as to use)
1. Base load Plant - Plant that is normally assumed load requirements under
normal condition.
2. Peaking Plant - Plant that is normally operated to provide power only during
peak periods.
3. Regulating Power Plants - Plants capable of carrying load for time interval
during off-peak or peak periods and usually responds to changes in system
frequency
4. Reserve or Standby Plant - for peak or system deficiency.
Types of Plant Reserve
1. Cold Reserve - a portion of the installed reserved, kept in operable condition
and available for service but not for immediate loading.
2. Hot reserve - refers to the units available, maintained at operating temp. &
ready for service although not in actual operation.
3. Spinning Reserve - Generating unit connected to the bus and ready to take
load.
Comparison of Hydro & Steam Plant
Steam plant - can be built very near the load center.
Hydro Plant -must be built at a considerable distance away hence high voltage
transmission line must be used to connect hydro plant to load center.
Most Important Factors in the comparison of Hydro versus Steam
1. Cost of energy at load center.
2. Continuity of reserve.
3. Reliability of service.
Choice of Proper Location of any Plant
Important factors to consider:
1. Steam Plant station
a. accessibility
b. coal and ash handling
c. water supply
d. stability of foundation
e. facility of extension
f. cost of real estate
g. restriction due to surrounding
2. Hydro-Electric Plant
a. water privileges
b. required fall
c. water supply
d. facility of extension
e. accessibility
f. stability of foundation
Hydro Electric Power Plant
Principles of Operation:
Hydro electric power stations are generally located at hilly areas where
dams can be built conveniently and large water reservoir can be obtained. In a
hydro-electric power station, water head created by constructing dams across a
river or lake. From the dam, water is led to a water turbine. The water turbine
captures the energy in the falling water and changes the hydraulic energy into
mechanical energy at the turbine shaft. The turbine drives the alternator which
converts mechanical energy into electrical energy. The hydro-electric power
station are becoming very popular because the reserves of fuel are depleting day
by day, they have the added importance for flood control, storage of water for
irrigation and water for drinking purposes.
Hydraulic Structures:
1. Dam - a barrier which stores water and created water head.
2. Spillway - Discharges the Surplus water from storage reservoir into the
river on the down stream side of the dam.
3. Head works - diverts flooding debris and sediments. Also it has valve for
controlling the flow of water into the turbine.
4. Surge Tank - is a small reservoir or tank in which water level rises or falls to
reduce pressure swings in the conduit. It is located near the beginning of the
conduit.
5. Penstock - an open or closed conduits which carry water to the turbine. It is
made of reinforced concrete or steel.
Advantages of H.E.P.P.
1. It requires no fuel as water is used for the generation of electrical
energy.
2. It is quite neat & clean as no smoke or ash produced.
3. It requires very small running charges because water available is free
of cost.
4. It is comparatively simple in construction and requires less
maintenance.
5. It does not require very long starting time as steam power station. In
fact, such plant can be put into service instantly.
6. Such plant serves many purposes. In addition to generation of electric
energy, they also help in irrigation and controlling floods.
7. Although such plant requires the attention of highly skilled person at
the time of construction, yet for operation a few experienced person
may do the job.
Disadvantages of H.E.P.P.
1. It involves high capital cost due to construction of dam.
2. There is uncertainty about the availability of huge amount of water.
3. Skilled and experienced hands are required to build the plant.
4. It requires high cost of transmission lines and the plant is located in
hilly areas which are quite away from the consumer.
Problem1 (Hydro):
At a potential hydro- electric site the average elevation of the head of water is
estimated at 600 m. The tail water elevation is estimated at 495 m. The average
annual flow of water was determined to be equal to that volume flowing through a
rectangular channel 9.1 m wide with a depth of 1.2 m and an average velocity of
6.1 m/sec. Find the annual energy that the site will produce with a water turbine
efficiency of 85%, a generator efficiency of 95% and loss in the head equal to 3%
of the available head.
Difference in Elevation  600 - 495  105m
Head loss  3%(105m)
Effective Head, H  105 - 0.03(105)  101.85m
Discharge, Q  ChannelArea x Ave Velocity  9.1x1.2x6.1  66.612m3 / sec
Generator Power Output, Po  Qwh T g  (66.612m3 / s)(9810 N / m3 )(101.85m)(0.85)(0.95)
PO  53,743,389 watts  53.7 MW
Annual Energy  Ave Po (8760hrs / yr )  53.7(8760)  470,792 MWHr
Problem 2 (Hydro):
Calculate the average power in kW and the annual energy that can be
generated in a hydro electric project from the following data:
Catchment Area , A = 150 sq. km.
Annual rainfall,
F = 1.25 m
Effective Head,
H = 300 m
Yield factor , K= 50%
Efficiency of the plant, 80%
If the load factor is 34%, what should be the rating of the generator installed?
Effective Water Level  FxK  1.25(0.5)  0.625m
1000 m 2 

Volume of Water for Generation  A * h  150km 2 (
) 0.625m   93.75 x106 m3
km


Annual Energy available in kwh 
Volh 93.75 x106 m3 (9810 N / m3 )(300m)(0.8)

N m
3.6x106
3.6 x106
kwhr
 61,312,500kwhr
Annual Energy 61,312,500kwhr

 7000kw
time
8760hrs
Average Power
Annual Average Power 
Load Factor 
Maximun Demand
7000KW
 20,588kw
0.34
Therefore , Generator Rating  20,000kw
Maximum Demand 
Problem 3 (Hydro):
A factory located near the water fall, the usable head for power generation
is 25m. The factory requires continuous power of 400 kW throughout the year.
The river flow in a year is
10 cu. m per sec. for 4 months
6 cu. m per sec. for 2 months
1.5 cu.m per sec. for 6 months
a. If the site is developed as run-of- river type of plant without storage, determine
the standby capacity to be provided. Assume that overall efficiency of the plant is
80 %.
Power Output, Po  Qh
When dischage is 10 m3 / s
Po  1962kw
When dischage is 6 m3 / s
Po  1177kw ok since greater than 400kw
ok since greater than 400kw
When dischage is 1.5 m / s Po  294kw Not ok since less than 400kw
Therefore, the required capacity of standby unit is (400-294) =106KW.
3
b. If a reservoir is arranged upstream, will any standby unit be necessary? What
will be the excess power available.
As plant with storage Capacity, the annual water discharge canbe regulated as
10(4)  6(2)  1.5(6)
m3
Q ave 
 5.083
12
sec
Ave. annual Output Power  (5.083m 3 / s(9810 N / m 3 )(25m)(0.8)  997 kw
Therefore, the excess Power  997  400  597 kw
Problem 4.
A hydro-power station has a reservoir of area 1.1 sq. mile and a capacity
of 180 million cu. ft. The net head of water at the turbine is 200ft. If the efficiency
of the turbine be 85% and generator be 90%, calculate the total energy in kwh
that can be generated from this power station.
Area  1.1sq. mile  2847769 .1 m2
Volume of water  180x106ft 3  5100948.9m3
Effective Head. H  200ft  60.98m
Volh
Annual Energy available in kwh 
3.6x106
5100948 .9m3 (9810 N / m3 )(60.98m)(0.85 x0.9)

 648,388kwhr
N m
3.6 x106
kwhr
If a load of 25 MW is being supplied for 6 hours, how much feet the level
of water in the reservoir falling?
Vol
5100948.9
Original Water Level, h 

 1.8m  5.88 ft
Area
2847769.1
Energy Output is proportion al to water level h
Energy Available power x time

h
h
648,388kwh 25,000kw(6hrs )

5.88 ft
h
h  1.36 ft
Coal-Fired Thermal Power Plant
A generating station which converts coal combustion into electrical energy
Principles of Operation
A steam power station basically works on the Rankine process. Steam is
produced in the boiler by utilizing the heat of the coal combustion. The steam
expanded in the prime mover and is condensed in the condenser to be fed in the
boiler again. The steam turbine drives the alternator which converts mechanical
energy of the turbine into electrical energy. This type of station is suitable where
coal and water are available and large amount of power is to be generated.
Stages of Coal fired thermal Plant
1. Coal & ash handling arrangement
2. Steam Generating plant
a. Boiler
b. Super heater
c. Economizer
d. Air pre-heater
3. Steam Turbine
4. Alternator
5. Feed Water
6. Cooling Arrangements
Advantages of Coal-fired
1. fuel is quite cheap
2. Less initial cost
3. It can be installed at any place irrespective of existence of coal. Coal is
transported by rail or road.
4. Requires less space as compared with hydro-electric plant.
5. Cost of generation is lesser than diesel Plant.
Disadvantages of coal fired:
1. It pollutes the atmosphere due to production of large amount of fumes and
smokes.
2. Its running cost is higher than Hydro-electric plant.
Overall Thermal Efficiency
The heat of combustion of the coal fired in the boiler is not wholly converted into
electrical energy as there are heat losses in the boiler and turbine and also
mechanical and electrical losses in the alternator. The overall efficiency of the
station is depends upon the efficiencies of boiler, turbine and alternator.
Overall thermal Efficiency=
Heat...equivalent... per...kwh
Coal..consumed.. per...kwh.x..Calorific...Value...ofCoal
In modern steam power Station, the overall thermal efficiency is about 30%
Heat equivalent per kwh (MKS units)
1 kwh=3,600X1000 watt-sec
1 Calorie = 4.18 joules
1 kwh = 860 kcal
Heat equivalent per kwh (English units)
1 Btu = 252 calories
1 kwh = 3413 btu
Heat or Calorific Value (HHV)
Is the amount of heat produced by complete combustion of a unit weight
of coal and is expressed in btu per pound of coal or kcal per kg. of coal.
Range:
11,500 – 13,000 btu/lb.
Station Heat Rate- defined as the heat energy in fuel devided by the station
output in kwh and is expressed in btu per kwh or kcal per kwh. The station heat
rate gives us an idea of the performance of the power station related to its overall
thermal efficiency by the following formula:
Station...Heat...Rate.. 
3413btu / kwh
Overall...thermal...Efficiency
Station...Heat...Rate.. 
860..Kcal / kwh
Overall...thermal...Efficiency
Diesel Power Station
In a diesel power station, diesel engine is used as the prime mover. The
diesel burns inside the engine and the product of this combustion acts as
“ working fluid” to produce mechanical energy. The diesel engine drives the
alternator which converts mechanical energy into electrical energy. As the
generation cost is considerable due to high price of diesel, therefore, such power
station are only used to produce small power. The plant is also used as standby
sets for continuity of supply such as in hospitals, radio stations, cinema and
telephone exchanges.
Advantages:
1. The design and layout of the plant is quite simple.
2. It occupies less space as the number and size of auxiliaries are small.
3. It can be located at any place.
4. It can be started quickly and can pick-up load in short time.
5. There are no standby losses.
6. It requires less quantity of water for cooling.
7. The overall cost is much less than steam of the same capacity.
8. Thermal efficiency is higher than steam power station.
9. It requires less operating staff.
Disadvantages:
1. High running charges as the fuel used is costly.
2. Does not work satisfactorily under over load.
3. Generate small power.
4. Cost of lubrication is generally high.
5. The maintenance charges are generally high.
Auxiliaries:
1. Fuel supply system
 storage tank
 strainer
 fuel transfer pump
 fuel tank
2. Air intake system
 Pipes for the supply of fresh air to the engine manifold
 Filter to remove dust particle.
3. Exhaust system
 A silencer usually incorporated to reduce noise level.
4. Cooling system
 water source
 pump
 cooling tower
5. Engine starting systems
 manual for small set
 compressed air for large unit
Problem 1
A 55 MW coal fired thermal power plant working at a load factor of 80%
has a boiler, turbine and alternator efficiency of 35%, 86% and 93% respectively.
Coal with heating value of 13,000 btu/lb. cost P1.50 / lb .
a. What is the overall efficiency?
Overall Efficiency,  bTg  ((0.35)(0.86)(0.93)  0.28
b. What is the coal consumption in tons per day ?
Energy balance:
Input Heat to the Plant=Energy Released by Fuel
KWo (3413btu / kwh )
 m f xHHV
overall
Where: KWo=average power output
mf= mass of fuel intake to the boiler in pounds/hr
HHV-high heating value of fuel in btu/pound
Ave output=(Load Factor)(Maximum demand)
Substituting the values,
55,000(0.8) KW (3413btu / kwh )
 m f x13,000btu / lb.
0.28
m f  41,256lbs. / hr  450tons / day
c. What is the fuel cost in producing one kWh?
Fuel Cost per kwhr=P1.50/lb(consumption in lb. per kwhr)
m
(3413btu / kwh )
3413btu / kwhr
Fuel Consumptio n per kwhr  f 

kwo
( ) HHV
(0.28)13,000btu / lb.
 0.94lb. / kwhr
Therefore;
Fuel cost Component=1.5(0.94)=P1.41/kwhr
Problem 2
A 100 MW power plant with a heating value of 13,000 btu per lb. has a
heat rate of 2.88x106 Calories per kWh. it is a base load plant and runs at full
load 24 hrs a day.
a. How many tons of coal is needed per day ?
KWo (3413btu / kwh )
 m f xHHV
overall
Station Heat Rate 
3413btu / kwhr
overall
KWo ( Heat Rate)  m f xHHV
As base load plant, at full load : KW o 100,000
100,000 KW (2.88 x106 cal / kwhr )  m f (13,000btu / lb)(252cal / btu )(2200lb / ton)
m f  40tons / hr  960tons / day
b. calculate the cost of producing one kwh if the coal cost P2.50/lb
( Heat Rate )
m
2.88 x106 cal / kwhr
Fuel Consumptio n per kwhr  f 

kwo
HHV
13,000btu / lb.(252cal / btu )
 0.879 lb. / kwhr
Therefore;
Fuel cost Component=2.5(0.879)=P2.2/kwhr
Problem 3
A 55 MW steam power plant uses coal of heating value of 12,000 btu/lb. The fuel
consumption when supplying full rated load is 32,550 lbs/hr.
a. What is the overall efficiency of the station?
KWo (3413btu / kwh )
 m f xHHV
overall
55,000 KW (3413btu / kwh )
overall
overall  48%
 (32,550lbs / hr )(12,000btu / lb)
b. What is the annual cost of coal if coal cost P2.50 per lb.
Annual fuel consumption  32550lbs. / hr(8760hrs)  285,138,000lbs
Annual fuel cost  P2.50 / lb(285,138,000lbs)  P712,845,000
Problem 4(assignment)
A 75 MW steam power station working at 80% load factor has an overall thermal
efficiency of 30%, the coal burnt has a calorific value of 6950 kcal/kg. Calculate
a. Overall heat rate of the plant in kcal/kwh
b. the coal consumption in kg per kwh generated,
c. the fuel cost of producing one kwh if the coal cost P2500/ton.
d. the coal consumption in tons per day.
Problem 5
A diesel generator set burns with heating value of 19,500 btu/lb. The diesel
engine has an efficiency of 30% and alternator efficiency of 90%. The specific
weight of fuel is 800 kg/m3
a. Determine the fuel cost component of producing one kWh if diesel cost
P28/liter.
KWo (3413btu / kwh )
 m f xHHV
overall
Fuel Consumptio n per kwhr 
mf
(3413btu / kwh )
3413btu / kwhr


kwo ( g )(eng ) HHV (0.3)(0.9)19,500btu / lb.
3
 kg  m  1000liters 


 0.648 lb. / kwhr 

  0.368liters / kwhr
m3
 2.204lb  800kg 

Therefore;
Fuel cost Component=P28/liter(0.268liters/kwhr)=P10.29/kwhr
b. if the generator delivers 100 kW, calculate the volume of fuel in liters used per
day.
Volume of fuel used / day  0.368liters/ kwhr(100kw)(24hrs / day)  883.2liters
Gas Turbine Plant
In a gas turbine plant, air is used as working fluid. The air is compressed
by the compressor and is led into a combustion chamber where heat is added to
air, thus raising the temperature. Heat is added to the compressed air either by
burning fuel in the chamber
or by the use of air heater. The hot and high
pressure air from their combustion chamber is then passed to the gas turbine
where it expands and does mechanical work. Since the compressor, gas turbine
and alternator are mounted on the same shaft, the plant is being used as standby
plant for hydro electric power stations, as starting plant for driving auxiliaries in
power plant etc.
Main Components:
1. Compressor
2. Regenerator
3. Combustion chamber
4. Gas Turbine
5. Alternator
6. Starting Motor
Advantages of Gas Turbine Plant
1. It is simple to design as compared with steam power station since no
boiler & auxiliaries required.
2. It is much smaller in size as compared with steam power plant of the
same capacity.
3. The initial and operating cost are much lower than equivalent steam
power plant.
4. It requires comparatively less water and no condenser used.
5. The maintenance charges are very small.
6. Gas turbine is simpler in construction and operation than steam
turbine.
7. It can be started quickly from cold conditions.
8. There is no standby loss.
Disadvantages
1. There is problem in starting the unit because compressor has to be
operated for which external power is required. But when it get started,
external supply is not needed.
2. The net output is low since greater power developed by the turbine is
used in driving the compressor.
3. The overall efficiency is low because the exhaust gases from the
turbine contains sufficient heat.
4. The temperature of combustion chamber is quite high(3,000°C) so that
the life is comparatively low.
Nuclear Power Plant
A generating station in which the energy is converted into electrical
energy. Heavy element such as uranium U235  or Thorium  Th232  are subjected
to nuclear fission is utilized in raising the steam at high temperature and
pressure. The steam runs the steam turbine which convert energy into
mechanical energy.
Nuclear reactor - an apparatus which nuclear fuel is subjected to nuclear
fission. It controls the chain reaction that start once the fission is done.
Fission - the breaking up of nuclei of heavy atoms into two equal parts
with release of huge amount of energy.
Chain Reaction - The nuclear fission is done by bombarding uranium
nuclei with slow moving neutrons. This split uranium nuclei which release huge
amount energy and emission of neutrons. These fission cause for that fission. If
this power continuous, then is a very short time huge amount of energy will be
released and cause explosion. But in a reactor, controlled chain reaction is
allowed. This is done by, something symmetrically removing the fission neutrons
from the reactor. The greater the number of fission neutrons remove, the lesser
the intensity of energy is released.
Parts of a nuclear reactor
1. Fuel rod - uranium
2. Moderator - slows down neutron before they bombard
3. Control rods - are of cadmium, as string neutron absorber and regulates the
supply neutron for fission.
Types of a nuclear reactor
1. Boiling Water reactor (BWR)- simplest type
2. Pressurized Water Reactor (PWR)
3. CANDU – Canadian advance
4. The Breeder Type – in this reactor, the fissionable type of fuel is produced
more than fuel consumed.
Advantages of Nuclear Power station
1. The amount of fuel is quite small. Therefore savings in the cost of fuel
transportation
2. Requires less space.
3. Low running charges
4. Economical for producing bulk electric power
5. Can be located near the load centers, therefore the cost of primary
distribution is reduced.
6. Large deposit of nuclear fuel is available all over the world. therefore,
such plant ensures continuous supply of energy for thousands of
years.
7. It ensures reliability of operation.
Disadvantages
1. The fuel used is expensive and difficult to recover
2. The fuel cost is very high
3. The construction and commissioning of the plant requires greater
technical know-how.
4. The fission by-products are generally radioactive.
5. Maintenance charges are high due to lack of standardization. High
salaries of trained personnel, etc.
6. Nuclear power plant are not suitable for varying loads as the reactor
does not respond to the load fluctuation efficiently.
7. Disposal of radioactive products is the big problem.
GEOTHERMAL POWER PLANT
Basic Component:
1.
2.
3.
4.
Steam wells
Steam pipes
Steam separator
Steam conditioning
General Evaluation of Power Plants
1. Site
Anywhere: Diesel
2. Initial Cost
Highest: Nuclear
Lowest: Diesel
3. Running Cost
Highest: Diesel
Lowest: Hydro
4. Limitation of source
HEPP – dependable
Diesel - limited reserve
Coal - Limited reserve
Nuclear – Sufficient
5. Fuel Cost
Lowest: HEPP
Highest: Steam
6. Cleanliness & Simplicity
Simple and clean –HEPP
Least Clean – Steam
7. Overall efficient
Most efficient – HEPP
Less efficient – Steam
8. Starting
Longest - Steam
Shortest – Diesel
Easy – Nuclear PP
9. Space required
Less space – Nuclear
Large Space – HEPP
10. Maintenance Cost
Highest: Nuclear
Lowest: Diesel
11. Transmission and Distribution Cost
Least: Diesel
Highest: HEPP
12. Standby losses
Lowest: HEPP
Highest: Steam
Load Graph and Significance
Load Graph & Load Curve - is a graphic record showing the power demands for
every instant during a certain time interval.
Load graph can be obtained by:
a. Use of recording graphic meter.
b. Manually plotting values of power indicated by wattmeter.
Importance:
1. The load curve shows variation of load on the power station during
different hours.
2. The area under a load gives the number of kWh generated a day.
3. The highest point in the load curve represents maximum demand on
the station on that day.
4. The load curve helps in selecting the size and number of generating
units.
5. The load curve helps in preparing operating scheduling of the station.
Load Duration Curve – when the load elements of a load curve are arranged in
the order of descending magnitudes, the curve thus obtained is called a load
duration curve.
Important Terms and Factors
1. Connected Load (CL) - the sum of continuous rating of the equipment
connected to the supply system.
2. Maximum Demand (MD) - it is the greatest demand of the load on the power
station during a given period. It helps in determining the installed capacity of
the station.
3. Demand Factor (DF) - the ratio of the maximum demand of the station to its
connected load.
MD
DF 
 1.0
CL
It is vital in determining the capacity of the plant equipment.
4. Average load or average Demand(AL) - the average load occurring on the
power station in a given period(day/month/year).
No. of kwh Generated/day
Daily Ave. Load =
periodic Hrs.in a Day
No. of Kwh Generated/ mo.
Monthly Ave. Load =
No. of Hrs. in a Month
No. of Kwh Generated/ yr
8,760 hr
yr
5. Load Factor (LF)- is the ratio of the average load to the maximum demand
during a given period. It plays key role in determining overall cost/KWh. The
higher the load factor of the station, the lesser will be the cost per kwh
generated.
Yearly Ave. Load =
LF 
Total Energy Produced (TEP)
MD  Service Hour.
Note: If the Service hour is equal to periodic hour
LF 
Ave.Load
MD
5. Diversity Factor (Div. F) - the ratio of the sum of individual maximum demand to
the simultaneous maximum demand on the power station.
Div.F 
 MD's
> 1.0
TMD
The greater Div. F, the lesser the cost of generation of power.
6. Coincident factor (Co. F) - the reciprocal of diversity factor
Co.F 
TMD
 MD's
7. Capacity Factor (CF) - is the ratio of the actual energy produced to the maximum
possible energy to have been produced during a given period.
CF 
TEP
Ave Load

IC  Pr
IC
where: IC - installed Capacity
Pr - periodic hr.
It is an indication of the reserve capacity of the plant
Reserve Capacity = Plant Capacity - MD
8. Plant Use Factor (PUF) - is the ratio of energy generated to the product of plant
capacity and the number of hours for which plant was in operation.
PUF 
TEP
IC  Sr
where: Sr – actual number of service hours
9. Utilization Factor (UF) - the ratio of maximum generator demand to the generator
capacity.
UF 
MD
IC
Note: If the plant is in operation, UF 
Station Output Kwh
 Capacity  Hrs. of Use
Problem 1:
A power station has a daily load cycle as under
275 MW for 6 hrs
250 MW for 8 hrs
200 MW for 4 hrs
150 MW for 6 hrs
If the power station is equipped with 4 sets of 75 MW each, calculate:
a. TEP per day of the plant
TEP  Area under load curve
TEP  275MW (6hrs )  250 MW (8hrs )  200 MW (4hrs )  150 MW (6hrs )  5350 MWHr
b. Daily load factor
Total Energy Produced (TEP)
5350MWHrs
Daily LF 

 81%
MD  Service Hour.
275MW (24Hrs)
c. Capacity factor
TEP
5350MWHrs
d. dail Daily CF 

 74.3% y requirement if the
IC  Pr 300MW (24Hrs)
heating value of oil used were 19,500 btu/lb. and average heat rate of the
station were 2650 btu/kwh..
Problem 2:
The daily demands of four consumers are given below.
Time
Consumer 1
Consumer 2
Consumer 3
12 MN-8 AM
1,400 w
1,500 w
1,500 w
8 AM-4 PM
2,500 w
1,200 w
2,800 w
4 PM-10 PM
1,500 w
1,900 w
1,500 w
10 PM-12 MN
4,000 w
900 w
1,200 w
Consumer 4
3,500 w
1,500 w
2,500 w
1,800 w
Plot the load curve and find a) Maximum demand of each consumer; b) Daily
energy consumption of each consumer; c)Load factor of each consumer; d) Diversity
factor of the system; e) Station Load factor.
a. Maximum demand of all individual consumers
MD1=4000W
MD2=1900W
MD3=2800W
MD4=3500W
b. Daily Energy consumption of individual consumer, E
E1=1400W(8hrs)+2500W(8hrs)+1500W(6hrs)+4000W(2hrs)=48,200Whrs
E2=1500W(8hrs)+1200W(8hrs)+1900W(6hrs)+900W(2hrs)=34,800Whrs
E3=1500W(8hrs)+2800W(8hrs)+1500W(6hrs)+1200W(2hrs)=45,800Whrs
E4=3500W(8hrs)+1500W(8hrs)+2500W(6hrs)+1800W(2hrs)=58,600Whrs
c. Load factor of individual consumers
Energy 1
48200 Hrs
LF1 

 50.2%
MD1  Sr. 4000W (24 Hrs)
Energy 2
34800 Hrs
LF2 

 76.3%
MD2  Sr. 1900W (24 Hrs )
Energy 3
45800WHrs
LF3 

 68.2%
MD3  Sr. 2800W (24 Hrs )
Energy 4
58600WHrs
LF4 

 69.8%
MD 4  Sr. 3500W (24 Hrs )
d. Diversity factor
Overall Demand for each period are:
Time
Horizontal 
12 MN-8 AM
7900 w
8 AM-4 PM
8000 w
4 PM-10 PM
7400 w
10 PM-12 MN
7900 w
MD' s 7900  8000  7400  7900
System Div.F  

 1.925
TMD
8000
e. Load factor of the station
System LF 
Total Energy
(48200  34800  45800  58600)WHrs

 97.6%
TMD  Sr.
8000W (24 Hrs )
Problem 3:
A power station has to meet the following demand
Group A: 200 MW between 8 AM and 6PM
Group B: 300 MW between 6 AM and 10 AM
Group C: 150 MW between 10 PM and 4 AM
Group D: 175 MW between 10 AM and 6 PM & 75MW between 6 PM and 6 AM
Plot the load curve and determine a) Diversity Factor; b) Energy Generated per day; c)
Load Factor.
Solutions:
a. Diversity Factor
Total Maximum demand is 300+200MW occurs at time 8:00am-10:00am
MD' s 200  300  150  175
Div.F  

 1.25
TMD
500
b. Energy generated per day
E=200MW(10hrs)+300MW(4hrs)+150MW(6hrs)+175MW(8hrs)+75MW(
12hrs)=6050MWhrs
c. Load Factor
Daily LF 
Total Energy
6050 MWHrs

 50.46%
TMD  Sr.
500 MW (24 Hrs )
Problem 4:
The yearly duration curve of a certain plant be considered as a straight line from
40 MW to 20 MW for 5840 hours and 20 MW to 16 Mw for the remaining hours. To
meet this load, three (3) turbo-generators of two rated 20 MW and one at 10 MW are
installed. Determine a) Installed capacity; b) TEP of the Plant; c) Plant Factor.
Solutions:
a. The installed capacity
IC=20+20+10=50MW
b. TEP of the plant.
TEP=Area under load curve
1
1
= 40  20(5840hrs)  20  16(2920hrs)  227,760MWH rs
2
2
c. Plant Factor.
Total Energy
227,760 MWHrs
Annual CF 

 52%
IC  Pr.
50MW (8760 Hrs )
Problem 5:
The maximum demand on a power station is 100 MW. If the annual load factor is 50%
and the capacity factor is 40%, calculate a) Annual average load in KW, b) Annual
energy generated in KWh, c) Plant Capacity or Installed Capacity, d) Reserve capacity
Solutions:
a. Annual average load in KW
Ave.Load
Ave.Load
LF 
0.5 
Ave Load=50MW
MD
100MW
b. Annual energy generated in KWh
Under this condition, Sr=Pr=8760hrs
TEP
TEP
LF 
0.5 
TEP=438,000MWHrs
MD  Sr.
100MW  8760hr.
c. Plant Capacity or Installed Capacity
50MW
TEP
CF 
0.4 
IC  Pr.
IC
IC=125MW
d. Reserve capacity
Reserve Capacity=IC-MD=125MW-100MW=25MW
Problem 6:
A generating station has a maximum demand of 100 MW, annual load factor of
70%, annual plant capacity factor of 52.5% and a plant use factor of 60% Calculate:
a. the number of hours not in service per year
b. the annual energy produced
c. the reserve capacity of the plant
Since PUF is not equal to CF, then Sr is not equal to Pr. Then use energy equations
TEP
TEP
LF 
0.7 
-------- 1
MD  Sr.
100  Sr
TEP
TEP
0.525 
CF 
--------2
IC  Pr.
ICx8760
TEP
TEP
PUF 
0.6 
--------3
IC  Sr.
ICxSr
a. Solving for Sr, equate 2 and 3, thus
0.525(8760)  0.6( Sr )
Sr=7665hrs
Then, Hours Not in Service=8760-Sr=8760-7665=1095hrs
b. Solving for TEP, substitute Sr value to equation 1. Thus,
TEP=0.7(100MW)(7665hrs)=536,550MWHrs
c. Solving for IC, substitute TEP value to equation 2. Thus,
536,550
IC=
 116MW
0.525(8760)
Therefore, Reserve Capacity=IC-MD=117-100=17MW
Problem 7:
The annual peak load on a 15,000 kW power plant is 10,500 kW. Two substations are
supplied by this plant. The annual energy dispatched through substation A is 27,500,000
kWh with a peak at 8,900 kW while 16,500,000 kWh are sent through b with peak at
6,650 kW . Calculate:
a. Diversity Factor
b. Capacity Factor
c. Utilization Factor
MD' s 8900  6650

 1.48
a. Div.F  
TMD
10500
b. Annual CF 
c. UF 
Total Energy
(27,500,000  16,500,000) MWHrs

 33.5%
IC  Pr.
15,000 MW (8760 Hrs )
MD 10,500MW

 70%
IC 15,000MW
Problem 8:
At the end of a three phase 240-volts power distribution system a certain feeder
supplies group of customers whose connected are as follows:
Feeder
Connected load
Demand Factor
Diversity of group
F-1
85 kw
0.6
1.4
F-2
125 kw
0.7
2.5
F-3
75 kw
0.8
1.6
If the diversity factor among feeder is 1.5, load factor is 60% and the power factor is
80%. Find the following:
a. Maximum demand kw at each feeder.
b. Minimum kva capacity of the transformer
c. Annual energy supplied by the substation
Solutions:
a. Div.F 
MD' s  (CL * DF )
TMD
TMD
85kw(0.6)
 36.4 KW
MDF-1=
1.4
125kw(0.7)
 35 KW
MDF-2=
2.5
75kw(0.8)
 37.5KW
MDF-3=
1.6
b.
TMD 
 MD' s  MD
Div. F
F-1
 MD F-2  MD F-3
36.4  35  37.5

 72.6 KW
Div. F
1.5
72.6
KW
 91 KVA
TMD=
0.8
Pf
Use 100 KVA transformer
KVA 
c.
Annual Energy supplied= MD*LF*Pr=72.6KW(0.6)(8760hrs)=381,586 KWHrs
LOAD CURVES & SELECTION OF GENERATING UNITS
The load on the Power Station varies from time to time. A single generating unit
will not be economical because it will have very poor efficiency during periods of light
load on Power Station. Therefore, in actual practice, the number of generating units of
different sizes are installed in the Power Station. The selection of the number and sizes
of the units is decided from the annual load curve of the station. The number and size of
the units are selected in such a way that they correctly fit the station load curve.
Important Points in the Selection of units
1. The number of units should be so selected that they approximately fit the annual
load curve of the station.
2. The units should be preferably of different capacities to meet the load
requirements, although the use of identical units ensure savings in cost, they often
do not meet load requirement.
3. The capacity of the plant should be 15% to 20% more than the maximum demand
to meet the future load requirements.
4. There should be a spare generating unit so that repairs and overhauling of
working units can be carried out.
5. The tendency to select the large number of units of small capacity in order to fit
the load accurately should be avoided. It is because the investment cost per kw
capacity increases as the size of unit decreases.
BASE LOAD AND PEAK LOAD ON THE POWER STATION
Base load - unvarying load which occurs almost the whole day on the station.
Peak load - Various peak demand of load over and above the load of the station.
Methods of Meeting the load
To achieve overall economy, the best method to meet the load is the interconnect
two different Power Stations. The most efficient plant can be used as base load station
while less efficient plant used to supply peak loads.
Interconnected Grid systems
 the connection of several generating station in parallel
Advantages of interconnected systems
1. Exchange of peak load – If the load curve on the power station shows a peak
demand that is greater than the rated capacity of the plant, then the excess load can be
shared by other stations interconnected with it.
2. Use of older plants - older and less efficient plant can carry short peaks of load
when interconnected with modern power plants.
3. Ensures economical operation – sharing among the stations is arranged in such a
way that more efficient station work continuously throughout the year at a high load
factor and the less efficient plants work for peak load hours only.
4. Increase diversity factor - load curves of different interconnected station are
generally different, reduced total maximum demand improves the diversity factor and
increases the capacity of the system.
5. Reduces the plant reserve capacity – when several units are in parallel, the reserve
capacity of the system is much reduce, this increases the efficiency of the system.
6. Increase reliability of the supply – If a major breakdown occurs in one station,
continuity of supply can be maintained by other healthy stations.
Conditions Involved in Determining minimum cost/kWh
1. To deliver a given peak load, the total installed capacity of a plant increases as the
number of units is decreased if a reverse unit is provided in each case.
2. For a given installed capacity of plant, fixed charges increases with the increased
number of units.
3. For a given installed capacity of plant, maintenance charges increases with a number
of units.
4. As general rule, large units have better efficiency than small units.
5. For a given total capacity of standby equipment, the standby losses increase with the
number of units, particularly in steam plant.
6. For a given installed capacity of plant, the operating cost increase with the increase
number of units.
7. The total costs per kWh decreases as the load factor of the load increases.
Problem 1:
A proposed station has the following annual load cycle.
Month
Jan Feb Mar Apr May June July Aug Sept Oct Nov Dec
Load in MW 20 40 50
35 70
40
20
50 70
80 90 100
Draw the load curve and select suitable generator units from 10, 20, 25, 30 MW. Prepare
the operation schedules for the machines selected.
Problem 2:
The annual load duration curve of a typical heavy load shown below is being
serve by steam station, Diesel station and hydroelectric station and the kWh supplied by
the station are as follows:
Steam : Hydro : Diesel
7
:
4
: 1
The Steam station is capable of generating power continuously and work as base
load plant. The diesel station work as a peak load plant. Determine:
a. Maximum demand of each station.
b. Load factor of each station.
Solution:
MW
hrs
Annual load Graph
TEP of the system= Area under the load curve
1
= 200  100(8760hrs )  1.314,000 MWHrs
2
Energy Supplied by each Station based on the energy supplied Ratio
TEP of Steam Station=(7/12)(1,314,000MWHrs)=766,500MWHrs
TEP of Hydro Station=(4/12)(1,314,000MWHrs)=438,000MWHrs
TEP of Diesel Station=(1/12)(1,314,000MWHrs)=109,500MWHrs
a. Solving for MD of each Station
MD of Steam Station
Area under the curve TEP of Steam 766,500 MWHrs


 87.5MW
=
base
Pr
8760 Hrs
MD of Diesel Station, y
1
Triangular Area= TEP of Diesel=109,500= xy
2
y 100
x  87.6 y
By similar triangle, 
x 8760
Combining the equations
1
109,500  (87.6 y 2 )
y=50
2
MD of Diesel=y=50MW
Also Sr of Diesel=x=4380hrs
MD of Hydro= 200-MD of Steam- MD of Diesel=200-87.5-50=62.5 MW
b. Solving For LF of each Station
766500MWHrs
 100%
(87.5)(8760hrs)
438000MWHrs
LF of Hydro=
 80%
(62.5)(8760hrs)
109,500MWHrs
LF of Diesel=
 50%
(50)(4380hrs)
LF of Steam=
ECONOMIC DISPATCH OF GENERATORS
PARALLEL OPERATION AND LOAD SHARING OF ALTERNATORS
Since a large number of generating stations are linked together in any
given power system and in each of the station the number of generating units
could be more than one, the problem of parallel operation of alternators, their
sharing of loads and speed regulation are of considerable interest of power
engineers. The operation of alternators is known as “synchronizing” and certain
conditions have to be fulfilled before this can be accomplished.
CONDITIONS OF PARALLEL OPERATION OF TWO ALTERNATORS
1. The terminal voltage of the incoming machine must be equal to busbar
voltage.
2. The phase sequence of the machines must be the same as that of the
busbars.
3. The frequency of the machine voltages must be equal to that of the busbar
voltage.
STEPS TO SYNCHRONIZE GENERATORS
1. Make sure that the breaker of generator 2 is open
2. Bolt the disconnect links of generator 2 to the connecting bars Switch
the voltage regulator to automatic
3. Start the engine or turbine of the generator 2 and bring up to speed
4. Using the governor control switch, adjust the frequency of generator 2
to approximately 1/10 cycles higher than the bus frequency
5. Using field regulator, Adjust the voltage of generator 2 to approximately
equal or slightly higher than the bus voltage
6. Switch the synchroscope to generator 2 and adjust the frequency until
pointer revolves slowly in fast direction
7. Close the circuit breaker of generator 2 at the instant the synchroscope
pointer passes into zero position
8. Turn off the synchroscope
9. Adjust the governor switch of generator 2 switch to “raise” and
generator 1 to “lower” until the desired KW load for each machine is
obtained
10. Adjust the field regulator switch of generator 2 switch to “raise” and
generator 1 to “lower” until the desired KVAR load or power factor for
each machine is obtained
LOAD SHARING BETWEEN TWO ALTERNATORS
Effect of change of excitation
When two alternators are operating in parallel the amount of kw load taken by
each alternator is governed by the fuel input i. e. the steam supply to its prime
mover. If the excitation of one of the alternators is changed it will only change
the reactive power taken by the alternator, its Kw power output remaining the
same. As regards the reactive component of the entire load, the strongly excited
machine takes a larger portion of it.
Effect of change in steam supply
If we increase supply to the prime mover of one of the alternators then the
machine can be made to take up an increased share of the load. In that case
the load on the other alternators should be decreased by the amount by a
adjusting its steam supply.
MACHINE CONNECTED TO INFINITE BUSBARS
Consider an a.c. generator (cylindrical rotor machine) connected to
infinite busbar (constant voltage and constant frequency busbar) and let the
excitation voltage of the generator remain constant as E 1 and the infinite busbar
voltage be E2. Then if the steam supply to the prime mover of the machine is
increased the machine will be loaded. With E1 and E2 remaining constant, there
is a definite limit of power which the generator can supply to the load, the
expression for which can be derived.
SAMPLE PROBLEMS
1. Two 2500 kVA ac generators running in parallel supplies the total loads of
2,500 kW at 0.80 p.f. lagging . One machine is loaded to 1000 kW at 0.7
p.f. lagging. Calculate the kw load and operating power factor of the other
machine.
1000
S1 
 -arcos0.7=1428.7   45.57 0
0.7
2500
SL 
  arcos0.8  3125  36.87 0
0.8
S2  SL  S1  3125  36.870  1428.7  45.570  1500  j854.76  1726 - 29.7o kVA
P2  1500 KW

 854.76 
Pf 2  cos arctan
  0.87 lagging
 1500 

2. Two identical three-phase generators operating in parallel share, equally a
load of 700 Kw at 6,000 volts and a power factor of 0.8 lagging.
The excitation of the machine is changed and its current adjusted to 40
amps at a lagging power factor: find (a) the current delivered by the
second machine and (b)the power factor of each machine.
700,000
IL 
  ar cos(0.8)  84.2  36.87 0
3 (6000)(0.8)
When sharing equally,
1
I1  I 2  (I L )  42.1  36.87 0  33.68  j 25.26
2
When the excitation is changed, the current of unit 1 is adjusted to 40Amp,
while no change in energy input
 33.68 
0
I1 '  40  ar cos
  40  32.648
 40 
Pf1 '  cos(32.648 0 )  0.842 lagging
I 2 '  I L  I1 '  84.2  36.870  40  32.6480  33.68  j 28.94
I 2 '  44.4  40.67 0
Pf 2 '  cos(40.67 0 )  0.76 lagging
3. Two identical alternators each rated 2500 KW are connected to the same
busbars in a power station. The governor droop of one machine is such
that the speed drops uniformly from 60 c/s on no load to 59.2 c/s on full
load. The corresponding uniform speed drop for the second machine is
from 60 to 58.8 c/s. Find how the two machine will share a load (a)
4000 Kw (b) What would be the maximum load, that can be delivered
without overloading either machine.
a) P1  P2  4000 - - - - - -1
by similar triangle,
60  59.2 60  f
then, P1  3125(60  f ) -------2

2500
P1
60  58.8 60  f
then, P2  2083.33(60  f ) -------3

2500
P2
Substitute equation 2 and 3 to equation 1,
3125(60  f )  2083.33(60  f )  4000
f  59.23 Hz
P1  2400 KW
P2  1600 KW
b) For maximum load without overloading either machine, the system
frequency must be 59.2 Hz.
Pmax=3125(60-59.2)+2083.33(60-59.2)=4167kw
4. The speed regulation of two 1000 Kw alternators A and B running is
parallel are 4% & 3% respectively. How will the two alternators share a
load of 1750 Kw?
No load Frequency- Full Load Frequency
Speed/Frequency Regulation 
x100%
Full load Frequency
assume full load frequency=60 Hz
f NL1  60(1.03)  61.8Hz
f NL 2  60(1.04)  62.4 Hz
To share the load,
By similar triangle,
61.8  60 61.8  f

1000
PB
62.4  60 62.4  f

1000
PA
PA  PB  1750 ---------------------1
then, PB  555.5(61.8  f ) -------2
then, PA  416.67(62.4  f ) -----3
Substitute equation 2 and 3 to equation 1
555.55(61.8  f )  416.67(62.4  f )  1750
f  60.26Hz
PB  856 KW
PA  892 KW
5. A three phase turbo-alternator has a reactance of 10 ohm and negligible
resistance. The machine delivers an armature current of 150-amps at 0.8
p.f. lagging when running on 11,000 volts infinite busbars:
(a) If the steam admission is constant but the e.m.f.(excitation) is reduced
by 5% calculate the value of the armature current and p.f.
(b) If the excitation is held constant but the steam admission is reduced by
5% calculate the value of the armature current and p.f.
(c) If both steam supply and excitation are reduced by 5% simultaneously.
Solution
When supplying 150 Amp at 0.8 power factor lagging,
I   150  36.870  120  j90 , the excitation emf is
11,000 0
E   V  I  Z S 
0  150  36.870 (0  j10)  7349.59.4 0
3
a) When excitation is reduced by 5%, while steam admission is
held constant
E  '  0.95(7349.5)  6982 volts
V '  V  6350.85 volts
since infinite bus
I  '  I e ' jI q '  120  jIq'
Ie' 120A
Using the voltage equation,
E  '  V 'I  ' Z S
since constant steam admission
6982  6350.85 + (120  jIq' )(0  j10)
6982  (6350.85  10Iq' )  j1200
(6982) 2  (6350.85  10Iq' ) 2  (1200) 2
Iq'  52.7
The new current is, I  '  120  j52.7  131  23.7 0 Amp
Pf’=cos 23.7o=0.92 lagging
b) If the steam admission is reduced by 5% while excitation is held
constant.
since constant excitation
E  ' '  E   7349.5 volts
V ' '  V  6350.85 volts
since infinite bus
I  ' '  I e ' ' jI q ' '  114  jIq' '
Ie' '  0.95(Ie)  0.95(120)  114A
Using the voltage equation,
E  ' '  V ' 'I  ' ' Z S
7349.5  6350.85 + (114  jIq' )(0  j10)
(7349.5) 2  (6350.85  10Iq' ) 2  (1140) 2
Iq'  90.97A
The new current is, I  ' '  114  j90.97  145.8  38.6 0 Amp
Pf’’=cos 38.6o=0.781 lagging
c) If the excitation and steam admission is reduced by 5% simultaneously.
E  ' ' '  6982volts
since infinite bus
V ' ' '  V  6350.85 volts
I  ' ' '  I e ' ' ' jI q ' ' '  114  jIq' ' '
Ie' '  0.95(Ie)  0.95(120)  114A
Using the voltage equation,
E  ' ' '  V ' ' 'I  ' ' ' Z S
6982  6350.85 + (114  jIq' )(0  j10)
(6982) 2  (6350.85  10Iq' ) 2  (1140) 2
Iq'  53.7A
The new current is, I  '  114  j53.7  126  25.2 0 Amp
Pf’=cos 25.2o=0.91 lagging
Exercises:
1. Two 2500 kVA ac generators running in parallel supplies the total loads of
4000 kW at 0.80 p.f. lagging . One machine is loaded to 2000 kW at 0.7
p.f. lagging. Calculate the kw load and operating power factor of the other
machine.
2. Two identical alternators each rated 2500 KW are connected to the same
busbars in a power station. The governor droop of one machine is such
that the speed drops uniformly from 60 c/s on no load to 59.6 c/s on full
load. The corresponding uniform speed drop for the second machine is
from 60 to 59.2 c/s. Find how the two machine will share a load (a)
4000 Kw (b) 5000 kw, (c) What would be the maximum load, that can be
delivered without overloading either machine.
3. Two alternators are driven by shunt motors. The shunt motors have
speed-load droop characteristics of 3 % and 4 % respectively. The
alternators are in parallel and each carrying 50 kW. There is no automatic
speed-load control. An additional 50 kW load is switched on. What are the
resulting loads of the alternators assuming that the speed-load control of
each is not adjusted?
4. A 3 ,Y-connected, 6.6 kV, 1000 kVA alternator has a reactance of 20%
but negligible resistance. When supplying full load at 0.8 pf lagging to a
large power system, If the excitation is reduced by 20%, determine the
new current and power factor
ECONOMICAL LOADING OF GENERATING UNITS
In order to determine the optimum load distribution between various
generating units in a steam power station, it is necessary to plot the
input/output curve of each generating unit. The curve is plotted in terms of
fuel input in kcal/hr versus output in Mw. The slope of the input/output curve
gives the incremental fuel rate expressed in kcal/kwh. The cost of fuel in Peso
per kcal is determined to convert the input/output curve in Peso per hour
versus output in Mw. The curve may be expressed mathematically as:
F = (C0+C1P+C2P2) Peso/hour
Where: P – is the power output in MW
F – the cost in pesos
Co,C1 & C2 - constants
The slope of this curve gives the incremental fuel cost for the unit in
peso per Mwh.
For Optimum load distribution between two generating units A & B, the
incremental fuel cost of Unit A must be equal to the incremental fuel cost of
unit B. Thus,
dFA dFB

dPA dPB
EXAMPLE
The incremental fuel costs for two units of a generating plant are:
dF1
 0.1P1  20
dP1
and
dF2
 0.15P2  15
dP2
where F1 and F2 are in pesos per hour and P is in Megawatts. What is the
most economical manner in which a total power of 100 Mw can be allotted
between these two units? Determine also the corresponding incremental
cost.
SOLUTION
For economical load division between units 1 and 2 incremental
fuel costs must be equal.
0.1 P1 + 20 = 0.15 P2 +15
or
P1-1.5 P2 + 50= 0
…(i)
If the total load is 100 Mw
Then
P1 + P2 = 100
…(ii)
Subtracting (1) from (2) we get,
2.5 P2 = 150
or
P2 = 60 MW
&
P1 = 40 MW
Thus unit 1 and 2 should take 40 MW and 60 MW power respectively.
dF1
Incremental cost,
= 0.1 P1 + 20=0.1 X 40 + 20
dP1
= Rs. 24/MWHR
If the load is shared equally between each plant, what will be the loss per
day compared with the economical sharing of the loads.
2
P1
P
F1   (0.1P1  20)dP1   0.1
 20P1 0
2
When P1=40MW, F1=P880/hr
When P1=50MW, F1=P1125/hr
Loss=880-1125=P245/hr
F2   (0.15P2  15)dP2   0.15
2
P2
 15P2
2
When P2=60MW, F2=P1170/hr
When P2=50MW, F2=P937.5/hr
Gain=1170-937.5=P232.5/hr
0P
Combined Operation Cost=F1+F2= -245+232.5=P-12.5/hr
Loss per day for operating at 50% sharing is,
Loss per day=(P12.5/hr)(24hrs/day)=P300/day
ECONOMICS OF POWER GENERATION
The art of determining the per unit (one kWh) cost of production of electrical
energy.
A power station is required to deliver power to a large number of consumer to
meet their requirements. While designing and building a power station, efforts should be
made to achieve overall economy so that the per unit cost of production is as low as
possible. This will enable the electric supply company to sell electrical energy at a profit
and ensure reliable service.
Terms used in the economics of power generation:
1. Interest - the cost of used money
2. Depreciation - the decrease in value of the power plant equipment and building due to
constant use.
Cost Of Electrical energy
1. Fixed cost - It is the cost which is independent of maximum demand and units
generated. The fixed cost is due to the following:
a. annual cost of central organization,
b. interest on capital cost of land and
c. salaries of high ranking officials.
2. Semi-fixed Cost - it is the cost which depends upon the maximum demand but it is
independent of units (kWh) generated. The semi-fixed cost is directly proportional to
the maximum demand on power station and is an account of
a. annual interest and depreciation on the capital investment of building equipment,
b. taxes and salaries of management and clerical staff.
3. Running or Operating Cost - it is the cost which depends only upon the number of
units generated. The running cost is an account of annual cost of fuel, lubricating oil,
maintenance, repairs and salaries of operating staff.
Expression for Cost of Electrical energy.
1. Three part Form
Total Annual Cost = Fixed cost(constant) + Semi-fixed cost (Proportional to
MD) +Running cost (Proportional to KWh)

= P a  bKwMD   cKwhr 

2. Two Part Form
Total Annual Cost = Fixed Sum per KWMD + Running Charge per KWHR
= PB  Kw MD  C  Kwhr 
Where:
a – a fixed cost representing the annual cost due to cost of central organization interest on
capital cost of land and salaries of high ranking officials.
b - annual interest and depreciation on the capital investment of building equipment,
taxes and salaries of management and clerical staff express in P/KWMD
c - annual cost of fuel, lubricating oil, maintenance, repairs and salaries of operating staff.
express in P/KWH
Problem 1:
The annual working cost of a power station is represented by the formula
P a  b  Kw MD  c  kwh  where the various terms has their usual meaning. Determine
the values of a, b, and c for a 50 MW station working at annual LF of 40% from the
following data:
a. Capital cost of building and equipment is P 50M
b. Annual cost of fuel, oil taxes and wages of operating staff is P15M.
c. Interest and depreciation on building and equipment are 10% per annum.
d. Annual cost of central organization is P 50,000.00
Solution:
Fixed Cost, a=P50,000
Semi-fixed cost=annual interest and depreciation
= 10%(Capital cost) = 10%(P50M)= P5M
Semi - fixed cost
P5,000,000
Semi - Fixed Charge 

 P100/kW
KW Maximum Demand 50,000 kw
Operating Cost= annual cost of fuel, oil, taxes etc = P15M
Operating or Running Charges, C 
Operating cost
15,000,000

 P0.085/kwhr
TEP per year 50,000kw (0.4LF)(8760hrs)
Therefore, the three part form of Annual Charges=P(50,000+100/kW+0.085/kWh)
Problem 2:
A certain generating station has the following data:
Plant Capacity
50MW
Annual load factor
40%
Capital cost
P3.6B
Annual cost of wages, taxation etc. (25% fixed)
P400T
Annual Cost of fuel, lubrication, maintenance etc.
P35M
Annual interest and depreciation
10%
a. Calculate the fixed charge per kw maximum demand and running charge per kwh
generated
b. Calculate the overall generation charge per kwh generated.
Solution:
TEP per year=MD(LF)(8760hrs)=50,000(0.4)(8760)= 175.2x106kwhr
Annual Depreciation  Annual fixed salaries, taxes etc
Fixed Charge 
KW Maximum Demand
Fixed Charge 
10%(P3,600,000,000) 25%(P400,000)
 P7,202/kW
50,000kw
Running Charge 
Annual Cost of Fuel etc  Anual wages etc of Operating Staff
Total energy produced/year
P35,000,000  75%(P400,000)
 P0.2/kWhr
175.2x10 6 kwhr
Overall Charges=P7202/kw+P0.2/kWhr
Running charge 
Overall Generation charge/kWhr
Overall Generation cost Fixed cost  Annual Wages  Annual cost of fuel


TEP per year
TEP
10%(P3,600,000,000)  P400,000  P35,000,000

 P2.26 / kwhr
175.2 x10 6 kwhr
Problem 3.
The capital cost of a hydro electric power station of 50 MW capacity is P1000 per
kw. The annual depreciation charges are 10% of the capital cost. A royalty of P1 .00 per
kw per year and P0.01 per kwh generated is to be paid for using the river water for
generation of power. The maximum demand on the power station is 40MW and annual
load factor is 60%. The annual cost of salaries, maintenance charges etc. is P700,000. If
20% of this expenses is also chargeable as fixed charges, calculate the generation cost in
two part form.
Solution
Capital cost=P1000/kw(50,000 kw)=P50,000,000
MD=40,000 kw
TEP/year=(40,000kw)(0.6LF)(8760 hrs)=210.24x106 kwhr
Annual Depreciation  Annual Wages, taxes etc.
 Annual Royalty per kw
KW Maximum Demand
10%(P50,000,000)  20%(P700,000)

 P1.00/kw  P129.50/kW
40,000kw
Annual Wages, taxes etc
Running Charge 
 Part of Annual Royalty per kwhr
TEP/year
80%(P700,000)

 P0.01/kwhr  P0.01267/kwhr
210.24x10 6 kwhr
Therefore, the two part form is
Annual Charges=P129.50/kW+P0.01267/kWhr
Fixed Sum, B 
CHOICE OF POWER PLANTS AND THEIR COMBINED GENERATING COST
When making a selection between steam and hydro-power plants a comparison
has to be made of the overall generating cost per unit for each type of power plant and
whichever is more economical to operate should be selected.
Important Factors Governing the choice of Power Plants (Thermal and Hydro)
1. Initial cost of hydro is quite heavy due to large amount of excavation, and earth work,
due to the land and water rights and cost of rehabilitation of population from areas that
may be inundated by pondage
2. Running or operating cost of steam power station due to high consumption of coal,
lubricants etc.
3. In hydro-station, another important factor is the cost of HV transmission line, there is
loss of energy which increases the cost per unit of power delivered.
Problem 1
A power station having a maximum demand of 75 Mw has a load factor of 60%. It is to
be supplied by either of the following schemes.
a. A steam power station in conjunction with HEPP. The latter is supplying
125  106 kwh per annum with maximum output of 30 Mw.
b. A steam station capable of supplying the whole load.
c. A hydro station capable of supplying the whole load.
The following data are as follows:
STEAM
HEPP
Capital cost/kw installed capacity
P 8,000
P 12,000
Interest/depreciation on capital cost
12%
10%
Operating cost/kWh
P 0.25
P0.025
Transmission cost/kWh
Negligible
P 0.006
Calculate the overall cost/kWh in each scheme (neglect spares)
Solutions:
Scheme A: Mixed capacity
Overall TEP per year=MD*LF*Pr=75,000kw(0.6)(8760 hrs)=394.2x106 kwhr
Annual Generation Cost=Running Cost + Fixed Cost
FOR HYDRO:
Generation cost =(P0.025+0.006/kwhr)(125x106kwhr)+10%(P12,000/kwx30,000kw)
=P39.875x106
FOR STEAM
MD=75MW-30MW=45MW
TEP per year=394.2-125)x106=269.2x106 kwhr
Generation cost =P0.25/kwhr(269.2x106 kwhr+12%(P8,000/kw)(45,000kw)
=P110.5x106
COMBINED GENERATION COST=P39.875M+P110.5M=P150.375M
Combined Generation cost
P150,375,000
Overall Cost per kwhr 

 P0.38/kwhr
Oveall TEP per year
394.2x106 kwhr
Scheme B: Steam Station operating alone
Annual Interest & Depreciation
Oveall TEP per year
12%(P8000/kw * 75,000kw)
 P0.25 / kwhr 
 P0.43/kwhr
394.2x10 6 kwhr
Cost per kwhr  Operating Cost per kwhr 
Scheme C: Hydro Station operating alone
Annual Interest & Depreciation
Oveall TEP per year
10%(P12000/kw * 75,000kw)
 P0.025 / kwhr  P0.006/kwhr 
 P0.259/kwhr
394.2x10 6 kwhr
Cost per kwhr  Operating Cost per kwhr 
Problem 2:
A load having a maximum demand of 150,000 kW can be supplied by either
HEPP or Steam Power Plant. The cost are as follows:
Steam
HEPP
Capital Cost
P 7,000/kw
P16,000/kw
Operating Cost
P0.13/Kwh
P0.016/kwh
Interest & Dep.
7%
7%
a. Calculate the overall generation cost/kWh for each at 50% load factor.
b. Calculate the minimum LF above which the HEPP will be more economical.
Solution
Overall Generation Cost per kwhr=Running Charges + Fixed Charges
Annual Interest & Depreciation
Overall Cost per kwhr  Operating Cost per kwhr 
Oveall TEP per year
FOR STEAM STATION:
7%(P7000/kwx150,000kw)
 P0.13/kwhr 
 P0.242/kwh
150,000kw(0.5LF)(8760kwhr)
FOR HYDRO STATION:
7%(P16000/kwx150,000kw)
 P0.016/kwhr 
 P0.272/kwh
150,000kw(0.5LF)(8760kwhr)
Therefore, at 50% load factor Steam Station is cheaper than Hydro
FOR MINIMUM LOAD FACTOR @ which Hydro is economical, annual Cost of
steam must be equal to annual cost of hydro, thus
Let X- be the minimum LF
7%(P16000/kwx150,000kw)
7%(7000kwx150,000kw)
P0.016/kwhr 
 P0.13wh 
150,000kw(X)(8760hr)
150,000KW(X)8760hr
Minimum LF=X=63.1%
Problem 3. (Assignment)
A load having a maximum demand of 100,000 kW can be supplied by either
HEPP or Steam Power Plant. The cost are as follows:
Steam
HEPP
Capital Cost
P 8,000/kw
P16,000/kw
Operating Cost
P0.25/Kwh
P0.026/kwh
Transmission cost/kWh
Negligible
P 0.006
Interest & Dep.
12%
12%
1. Calculate the overall generation cost/kWh for each at 38% load factor.
2. Calculate the overall generation cost/kWh for each at 55% load factor.
3. Calculate the minimum LF above which the HEPP will be more economical.
TARIFF
The rate at which electrical energy is supplied to the consumer
Objectives:
1. Recovery of cost of producing electrical energy at the power station
2. Recovery of cost on the capital investment in transmission and distribution
systems.
3. Recovery of cost of operation and maintenance of supply of electrical
energy e.g. metering equipment, billing etc.
4. A suitable profit on the capital investment
Desirable Characteristic of a Tariff
1. Proper return
2. Fairness
3. Simplicity
4. Reasonable profit
5. Attractive
Types of Tariff
1. Simple tariff – there is a fixed rate per unit energy consumed
Disadvantages
a. no discrimination between different types of consumer since every consumer
has to pay equitably for the fixed charges.
b. The cost per unit delivered is high.
2. Flat rate – different types of consumers are charged at different uniform per unit rates.
Disadvantages:
a. Since the flat rate varies according to the way the supply is used, separate
meters are required for lighting load, power load etc.. This makes the
application of such tariff expensive and complicated.
b. A particular class of consumer is charged at the same rate irrespective of
magnitude of energy consumed. However a big consumer should be charged
at a lower rate as his case the fixed per unit are reduced.
3. Block rate Tariff – a given block of energy is charged at specified rate and the
succeeding blocks of energy are charged at progressively reduced rates
4. Two Part Tariff – the rate of electrical energy is charged on the basis of maximum
demand of the consumer and the units consumed.
Advantages:
a. it is easily understood by the consumer
b. it recovers fixed charges which depends upon the maximum demand of the
consumer but independent of units consumed.
Disadvantages:
a. The consumer has to pay the fixed charges irrespective of the fact whether he
as consumed or not consumed the electrical energy.
b. There is always error in assessing the maximum demand of the consumer.
5. Power factor tariff – the tariff in which the power factor of the consumer is taken into
consideration. A low power factor increases the rating of equipment and line losses,
therefore a consumer having low power factor must be penalized.
6. Three Part Tariff - the total charge to be made form the consumer is split into three
parts namely fixed charge, semi-fixed charge and running charge. This is generally
applied to big consumers.
Problem1:
Calculate the annual bill of a consumer whose maximum demand is 250 kW, pf = 80%
lagging and load factor = 60%. The tariff used is P 1500/Kva of maximum demand plus
P1.00/kwh consumed.
Solution:
Annual Energy consumed=250Kw(0.6LF)(8760 hrs)=1,314,000 kwhr
Maximum Dekand KW 200
Maximum Demand KVA 

 250
Power Factor
0.8
Annual Bill=Demand Charge(MD)+Energy Charge(Energy Consumed)
= P150/kva(250 kva) + P1.00/kwhr(1,314,000 kwhr)=P1,351,500
Problem 2:
A consumer requires 0ne million kwh per year and his annual load factor is 50%. The
tariff enforce is P120 per kw maximum demand plus P0.05 per kwh consumed. Estimate
the savings in energy charges if he improves his load factor is raised to 100%.
Solution:
Annual Energy Consumed
LF 
MDx8760hrs
At 50% Load Factor:
1,000,000kwhr
MD 
 228.31 kw
0.5(8760hrs)
Annual Bill=P120/kw(228.31 kw) +P0.05/kwhr(1,000,000 kwhr)=P77,397.20
At 100% Load Factor:
1,000,000kwhr
MD 
 114.15 kw
1.0(8760hrs)
Annual Bill=P120/kw(114.15 kw) +P0.05/kwhr(1,000,000 kwhr)=P63,698.00
Annual Savings=77,397.20-63698=P14,000
Problem 3.
The monthly reading of consumers meter are as follows:
MD = 75 kW; Energy Consumed = 36,000 kWh; Reactive Energy Consumed = 23,400
Kvar-hr. If the tariff is P 100/kw of MD plus P 0.8/kwh +P0.005 per kwh for each 1% of
pf above or below 86%. Calculate the monthly bill of consumer.
Solution:
kvarhr
 23400 
 Arctan 
 33 o

kwhr
 36,000 
0
Power Factor= cos  =cos 33 =83.9%
Monthly Charges=P100/kw+P0.8/kwhr+P0.005/kwh(Reference P.F.- Actual P.F.)
Monthly Bill=P100/kw(75kw)+P0.8/kwhr(36,000 kwhr + P0.05/kwh(86-83.9) (36,000
kwhr)=P7500+28,800+3780=P40,080.00
Power Factor Angle  Arctan
Problem 4:
The following two tariff are offered:
Scheme No. 1 = P (1000 + 1.00/KWH) per month
Scheme No. 2 = flat rate at P 3.00/KWH per month
a. Which scheme is economical? if the monthly consumption is 400 kwh,
b. At what consumption in kwh is the first tariff Economical ?
Solution:
a. At 400 kwhr consumption:
Scheme No. 1 Monthly Bill=1000+1.00(400)=P1,500
Scheme No.2 Monthly Bill=3.00(400)=P1,200
Scheme No. 2 is economical
b. For Minimum Consumption at which Scheme No. 1 is economical, equate the cost.
Let X- be the minimum consumption
1000+1.00(x)=3.00(x)
X=500 kwhr
Problem 5:
A medium size industrial plant has a maximum demand of 500 KVA at 70.7% pf lagging
and yearly consumption of 1  106 kwh . You have to improve the pf to 90% lagging.
Using shunt capacitor. The first cost of capacitor is P2,500/KVA and the annual fixed
charges is 10%.
Electric rates are given as:
Demand Charge = P 1,500/KVA max.-demand/yr.
Energy Charge = P 4.90/kwh
Calculate:
a. The annual electric bill before and after pf correction.
b. The net savings/yr. after putting capacitor.
c. The economical operating pf that will result to maximum savings.
Solution:
Annual Charges=P1,500/kva+P4.90/kwhr
Before Pf correction:
Annual Bill=P1500/kva(500 kva)+P4.90/kwhr(1,000,000 kwhr)=5,650,000
After Pf correction:
New Demand kva=500(0.707/0.9)=392.8 kva
Annual Bill=P1500/kva(392.8 kva)+P4.90/kwhr(1,000,000 kwhr)=5,489,200
Gross Saving by Pf correction=5650,000-5,489,200=P160,800
Net Savings = Annual Gross savings – Annual Interest & Depreciation
P = 500(0.707)=353.5 kw
 =arcos 0.707=450
 ’=arcos 0.9=25.840
CKVAR of required Capacitor=P(tan  -tan  ’)=353.5(tan450-tan25.840)
= 182.3 kvar
Capital Cost of required Capacitor=P2500/kvar(182.3kvar)=P455,767.15
Annual Interest & Depreciation= 10%(P455,767.15)=45,476.70
Therefore, Net Savings per year = P160,800-45,576.70=P115,233.30
Economical Operating pf that will result to maximum Savings

 i 
 capacitor cost/kva  100  


Economical operating pf  1  
Demand Charge






2
2

 10%  
 2500 100%  

  =98.6%
Economical operating pf  1  
1500






Problem 4. (Assignment)
A power plant is working at its maximum kva capacity with a lagging power factor of
0.70. It is now required to increase its kw capacity to meet the demand of additional load.
This can be done by raising the power factor to 85% by correction apparatus or by
installing extra generating plant which cost P8000/kva. Find the minimum cost per kva of
the power factor correction apparatus to make its use economical that the additional
generating plant.
STABILITY OF POWER SYSTEMS
The term “STABILITY” denotes a condition in which the various synchronous machines
of the system remain in synchronism. The system is said to be “UNSTABLE” when the machine
cease to operate in parallel and violent fluctuation of voltage and frequency occurs on the system.
The stability limit may be determined both under steady state as well as under transient condition.
1. Steady State Stability Limit – is the maximum flow of power through a
particular point of power system without loss of stability when the power is
increased gradually.
2. Transient Stability Limit – is the maximum flow of power through a point
without loss of stability when a sudden disturbance occurs such as sudden
change of load, short circuit or fault on the system.
1. Calculation of Steady State Stability Limit
Power Limit of Transmission System
VS = AVR + BIR
VS/ = A/ VR/0
+ B/ IR
SR = PR – j QR = VR*IR
P R – j QR
2

VR VS    AVR    
=

B
B
Equating real Parts of both sides
PR
2

VR VS cos(    ) AVR  cos(    )
=

B
B
 is variable
At maximum power,  = 
PR MAX =
VR VS   AVR 2 cos(   )
B
B
For negligible shunt admittance, A=1,  = 0
PR MAX =
VR VS   AVR 2 cos 
B
B
For negligible Shunt Admittance and Resistance,  = 0 &  = 900
PR MAX =
VR VS 
X
Power Limit of Two Machine System
Eg = Em + IZ
Eg/ = Em/0+ I(Z/
Power delivered to the motor
Sm = Pm – j Qm = Em*I
Pm – j Qm =
EmEg    Eg2 
Z
Z
Equating real Parts of both sides
Pm =
EgEmcos(    )  Em2 cos 
Z
 is variable
At maximum power,  = 
Pm MAX =
Z
EgEm  Em2 cos 
Z
Z
For negligible shunt admittance,  = 0
Pm MAX =
EgEm  Em2 cos 
Z
Z
For negligible Shunt Admittance and Resistance,  = 0 &  = 900
Pm MAX =
EgEm
X
2. Calculation of Transient Stability
Consider a synchronous generator connected to an infinite busbar, the swing equation of
the system is given by
M
d 2
dt 2
 Ps  Pe
where Ps – the shaft power
Pe – electrical power developed
Eg – the voltage of the infinite busbar
Em the voltage behind the transient reactance of the motor
X- transient reactance of the system(sum of reactances of motor,
lines &transformer)
Pe =
EgEm sin = P
X
max
 - the torque angle
sin
Effect of Sudden Increase of load
The torque angle corresponding to Po
 Po 
o = sin-1 
 P max 
The torque angle corresponding to Ps
 Ps 
s= sin-1 
 P max 
On application of sudden load Ps the torque angle  will begin to increase and swing
between o to m around the equilibrium position s.
Foe remissible maximum load which can be allowed to be taken by the motor without loosing
stability, the relation be
(-s-o)sins = coss + coso
from the equation above,s can be calculated and the load P can be found which is the maximum
permissible load which the motor can take without losing stability.
Methods of Improving Stability Limit
1. For Steady State Stability – the stability can be improved by using additional
machines in parallel or two lines in parallel so that total reactance of the
system is reduced. The other method adopted are using higher excitation
voltages and quick response excitation system.
2. For transient Stability – the stability limit can be improved by using quick
acting and auto-closing circuit breakers and relays having minimum time
lags. The other alternative method consist in using machines of high inertia
and high speed governors. Also, the use of quick acting voltage regulators
helps the system stability
Problem 1
The generalized circuit constant of a nominal-pi circuit representing a three phase
transmission line are:
A=D= 0.9/0.3o
B= 82.5 /76
C= 0.0005/90o
a. Find the steady state stability of the system if the V S and VR are held constant at
110 kv.
b. What is the steady state stability if shunt admittance is assumed to be zero
c. What is the steady state stability limit if the shunt admittance is assumed to be
zero and the series resistance is neglected.
Solution:
a. Pmax 
VS . VR
B

A.V2R
B
cos(    )
Pmax 
110 .110
82.5

0.9 .110 2
cos(76  0.3)  114 MW
82.5
b. If shunt admittance is zero, then A=1.0/0o
Pmax 
Pmax 
c.
If R=0,
Pmax
VS . VR
B
110 .110
82.5

A . VR

2
cos
B
1.0 .110 2
82.5
cos76 0  111MW
X=82.5 sin76o=80
and =0, then
110.110
VS . VR
=

 151MW
80
B sin 
Problem 2
A 3-phase transmission line has a resistance and reactance of 10 ohms and 30 ohms per
phase respectively. The sending and receiving end voltages are both maintained at 132
kv. Determine the maximum steady state power limit and calculate the maximum
additional load which could be suddenly switched on the line without loss of stability if
the line were already carrying 80 MW.
Solution:
ZL = 10 + j 30
then
=71.6o
Z = 31.6
Pmax 
VS . VR
Z

. VR 2
Z
cos =
132 .132
31.6
For Negligible resistance, Pmax 

VS . VR
.132 2
31.6

cos71.6 0 =378 MW
132.132
 581MW
30
X
The maximum Additional Load suddenly switched on to the line is
Pmax 
VS . VR
X
sin   Pmax sin 
Let the initial power carried by the line be Po and angle between voltages be
 0  arcsin
0
Po
 7.9o =0.138 rad
Pmax
consider the condition:
(-s-o)sins = coss + coso
(3.14-s-0.138)sins = coss + cos7.9o
Substituting the value of o, by approximation s will give 48o
The increase in load could suddenly be applied without loss of stability is
= Pmax sins - Po
= 581 sin 48o - 80 = 350 MW
END
Electric power Plant Circuit layout
Important Factors in planning Power Plant layout
1.
2.
3.
4.
5.
6.
7.
Flexibility
Adequacy
Reliability
Simplicity
Safety
Space
Cost
Typical Power Plant circuit layout
1. Single bus bar, single breaker - used for small stations where simplicity and
economy is the primary requisites.
2. Double bus, single Breaker System - This arrangement greatly increases the
chances for continuity of service, one bus bar may be used as an auxiliary
only or one may used as lighting load while the other feed a power load.
3. Double Bus, Double Breaker System- This arrangement gives the same
advantage double bus, single breaker system, with double assurance against
shut down to circuit breaker.
4. Ring Bus System, Bus Sectionalization - Suitable for station of medium size
were great flexibility and maximum economy in cost is desired. The
arrangement requires small amount of copper in the bus bar.