TEE 212: Power Systems I.
Sources of Electrical Energy
Fig. 1: Sources of Energy
1. HYDRO POWER PLANTS
Hydro Power Potential
P = g*ρ*Q*H
Where
P = Power available in water
g = 9.81 m/s2
Q = flow or discharge (m3/s)
H = Height of fall of water or head (m)
P = 9.81*1000*Q*H*10-3 kW = 9.81 QH kW
P= 9.81 QHη kW where η = efficiency of the turbine-generator assembly

Rain falling on earth’s surface has potential energy relative to oceans.

This energy is converted to shaft work when the water falls through a vertical distance.

This shaft work is used to drive water turbines to generate electricity.
Site Selection for Hydropower Plants
• Availability of Water: Run-off data for many years available
• Water Storage: for water availability throughout the year
• Head of Water: most economic head, possibility of constructing a dam to get required head
• Geological Investigations: strong foundation, earthquake frequency is less
• Water Pollution: excessive corrosion and damage to metallic structures
• Sedimentation: capacity reduces due to gradual deposition of silt
• Social and Environmental Effects: submergence of areas, effect on biodiversity, cultural and
historic aspects
• Access to Site: for transportation of construction material and heavy machinery new railway
lines or roads may be needed
• Multipurpose: power generation, irrigation, flood control, navigation, recreation; because
initial cost of power plant is high because of civil engineering construction work.
Components of a HPP
Schematic Diagram of Hydroelectric power Plant
The various components of HPP are as follows:
1. Catchment area
2. Reservoir
3. Dam
4. Spillways
5. Conduits
6. Surge tanks
7. Draft tubes
8. Power house
9. Switchyard for power evacuation
Dam

Develops a reservoir to store water

Builds up head for power generation
Spillway

To safeguard the dam when water level in the reservoir rises
Intake

Contains trash racks to filter out debris which may damage the turbine
Conduits

Headrace is a channel which lead the water to the turbine

Tailrace is a channel which carries water from the turbine
Penstocks are closed conduits for supplying water “under pressure” from head pond to the
turbines.
Surge Tank

A surge tank is a small reservoir in which the water level rises or falls to reduce the
pressure swings so that they are not transmitted to the penstock.
Hydraulic Turbines
Types of Hydraulic Turbines
1. According to the head and quantity of water available

Low head (2-15m)

Medium head (16-70m)

High head (71-500m)

Very high head (>500m)
2. According to the name of the originator

Francis

Kaplan

Pelton
3. According to the nature of working of water on blades
4. According to the direction of flow of water

Radial

Axial

Tangential (Deriaz)
5. According to the axis of the turbine shaft: vertical, horizontal
Comparison of Turbines
Table 2: Comparison of Turbines
Specific Speed (Ns): It is defined as the speed of a geometrically similar turbine to produce 1
kW of power under 1 m head. Its units are ‘rpm in (m-kW)’ or ‘rpm in (m-mhp)’.
Where N = rotational speed of the turbine in rpm
P = Power output of the turbine in kW or mhp
H = Head of the turbine in meters

Specific speed is the basis of comparison of the characteristics of hydraulic turbines.
Higher the specific speed for a given head and power output, the lower the cost of
installation as a whole.
Example 1:
Find out the specific speed of a turbine of 10 MW capacity working under a head of 500m and
having the normal working speed of 300 RPM.
Solution:
Ns = 300x Sqrt (10000) / 500^ (1.25) = 12 rpm in (m-kW)
Runaway Speed
It is the maximum speed at which a turbine would run under the worst conditions of operation
i.e. with all gates open so as to allow all possible water inflow under maximum head and
corresponding to the condition of the load being suddenly thrown off from the generator.
8. Head gates
Numerical Problems on Hydro Power Plants
1. A hydro plant operates under an effective head of 100 m and a discharge of 200 m3/s.
If the efficiency of the turbine alternator set is 0.9, find the power developed. (Ans.
176.52 MW)
2. A hydro-electric station has an average available head of 100 meters and reservoir
capacity of 50 million cubic meters. Calculate the total energy in kWh that can be
generated, assuming hydraulic efficiency of 85 % and electrical efficiency of 90%. (Ans.
10.423 x 106 kWh).
3. One million cubic meters of water is stored in a reservoir feeding a water turbine. The
density of water is 993 kg/m3. If the centre of mass of water is 50m above the turbine
and the losses are negligible, what will be the energy in MWh produced by that volume
of water?
4. The utilizable water from a catchment is 60x106 cu m annually and the hydro station
has a head of 40 m. Assuming ideal generator and turbine, find power that can be
theoretically generated? (Ans. 250/300/500/750 kW)
5. A hydroelectric station is designed to operate at a mean head of 205 m and fed by a
reservoir having a catchment area of 1000 km2 with an annual rainfall of 125 m of
which 80% is available for power generation. The expected load factor is 75%.
Allowing a head loss of 5 m and assuming efficiency of turbine and generator to be 0.9
and 0.95 calculate suitable MW rating of the power station. Comment on the type of
turbine to be used.
2. Thermal Power Plant
Selection of site for thermal power plant

Transportation network: Easy and enough access to transportation network is required
in both power plant construction and operation periods.

Power transmission network: To transfer the generated electricity to the consumers, the
plant should be connected to electrical transmission system. Therefore the nearness to
the electric network can play a roll.

Geology and soil type: The power plant should be built in an area with soil and rock
layers that could stand the weight and vibrations of the power plant.

Earthquake and geological faults: Even weak and small earthquakes can damage many
parts of a power plant intensively. Therefore the site should be away enough from the
faults and previous earthquake areas.

Topography: It is proved that high elevation has a negative effect on production
efficiency of gas turbines. In addition, changing of a sloping area into a flat site for the
construction of the power plant needs extra budget. Therefore, the parameters of
elevation and slope should be considered.

Water resources: For the construction and operating of power plant different volumes
of water are required. This could be supplied from either rivers or underground water
resources. Therefore having enough water supplies in defined vicinity can be a factor in
the selection of the site.

Environmental resources: Operation of a power plant has important impacts on
environment. Therefore, priority will be given to the locations that are far enough from
national parks, wildlife, protected areas, etc.

Need for power: In general, the site should be near the areas that there is more need for
generation capacity, to decrease the amount of power loss and transmission expenses.
Schematic of a Thermal Power Plant
Major Components of a Thermal Power Plant

Coal Handling Plant

Pulverizing Plant

Draft or Draught fan

Boiler

Ash Handling Plant

Turbine and Generator

Condenser

Cooling Tower And Ponds

Feed Water Heater

Economizer

Super heater and Re-heater

Air pre heater

Alternator with Exciter

Protection and control equipment

Instrumentation
Numerical Problems on Thermal Power Plant
1. A steam power station of 100 MW capacities uses coal of calorific value 6400 k
Cal/kg. The thermal efficiency of the station is 30% and electrical generation efficiency
is 92%. Find the coal requirement per hour when the plant is working on full load.
2. Assuming efficiency of 33%, how much coal is needed to be burnt to supply energy for
average household in a year. Given connected load: 1 kW, load Factor: 60%.
3. Diesel Power Plant.
General Layout
Essential elements of Diesel Power Plant

Engine System. This is the main component of the plant which develops required
power. The engine is generally directly coupled to the generator. Generally classified as
two stroke engine and four stroke engines.

Starting System. The function of this system is to start the engine from cold by
supplying compressed air at about 17 bar supplied from an air tank. Fuel is admitted to
the remaining cylinders and ignited in the normal way causing the engine to start.
 Lubrication System. It includes the oil pumps, oil tanks, filters, coolers and connecting
pipes. The purpose of the lubrication system is to reduce the wear of the engine moving
parts. Part of the cylinder such as piston, shafts, and valves must be lubricated.
Lubrication also helps to cool the engine
 Fuel System. It includes the storage tank, fuel pump, fuel transfer pump, strainers and
heater. Pump draws diesel from storage tank to day tank through the filter. Diesel is
filtered before being injected into the engine by the fuel injection pump.
 Cooling System. The temperature of the hot gases inside the cylinder may be as high as
2750 c. If there is no external cooling, the cylinder walls and piston will tend to assume
the average temp. of the gases. Cooling is necessary because:
i. To avoid deterioration or burning of lubricating oil.
ii.
The strength of the materials used for various engine parts decreases with
increase in temperature. Local thermal stress can develop due to uneven
expansion of various parts.
iii. Due to high cylinder head temp., the efficiency and hence power O/P of the
engine is reduced.
 Exhaust System. This includes the silencers and connecting ducts. The exhaust gases
coming out of the engine is very noisy. silencer (muffler) is provide to reduce the noise
ADVANTAGES
 Simple design & layout of plant
 Occupies less space & is compact
 can be started quickly and picks up load in a short time
 requires less water for cooling
 Thermal efficiency better that of Steam Power Plant of same size
 No ash handling problem
 Less operating and supervising work is required
DISADVANTAGES:
 High running charges due to costly price of Diesel
 Generates small amount of power
 Cost of lubrication very high
 Maintenance charges are generally high
 Noise problem
 Capacity is restricted. Cannot be of very big size.
Applications of diesel power plant

They are used as central station for small or medium power supplies.

They can be used as stand-by plants to hydro-electric power plants and steam power
plants for emergency services.
 They can be used as peak load plants in combinations with thermal or hydro-plants.
 They are quite suitable for mobile power generation and are widely used in
transportation systems such as automobiles, railways, air planes and ships.
Economics of power generation
The following are the commonly used methods for determining the annual depreciation
charge:
(i) Straight line method
(ii) Diminishing value method
(iii) Sinking fund method
(i) Straight line method: In this method, a constant depreciation charge is made every year
on the basis of total depreciation and the useful life of the property. Obviously, annual
depreciation charge will be equal to the total depreciation divided by the useful life of the
property. Thus, if the initial cost of equipment is Kshs. 100,000 and its scrap value is Kshs
10,000 after a useful life of 20 years, then,
In general, the annual depreciation charge on the straight line method may be expressed
as:
Where
P = Initial cost of equipment
n = Useful life of equipment in years
S = Scrap or salvage value after the useful life of the plant.
The straight line method is extremely simple and is easy to apply as the annual depreciation
charge can be readily calculated from the total depreciation and useful life of the equipment.
The figure below shows the graphical representation of the method. It is clear that initial value
P of the equipment reduces uniformly, through depreciation, to the scrap value S in the useful
life of the equipment.
The depreciation curve (PA) follows a straight line path, indicating constant annual
depreciation charge. However, this method suffers from two defects. Firstly, the assumption of
constant depreciation charge every year is not correct. Secondly, it does not account for the
interest which may be drawn during accumulation.
(ii) Diminishing value method: In this method, depreciation charge is made every year at a
fixed rate on the diminished value of the equipment. In other words, depreciation charge is
first applied to the initial cost of equipment and then to its diminished value. As an example,
suppose the initial cost of equipment is Rs.10, 000 and its scrap value after the useful life is
zero. If the annual rate of depreciation is 10%, then depreciation charge for the first year will
be 0·1 × 10,000 = 1,000.
The value of the equipment is diminished by Rs 1,000 and becomes Rs 9,000.For the second
year, the depreciation charge will be made on the diminished value (i.e. Rs 9,000) and
becomes 0·1 × 9,000 = Rs 900.The value of the equipment now becomes 9000 − 900 = Rs
8100.For the third year, the depreciation charge will be 0·1 × 8100 = Rs 810 and so on.
Mathematical treatment:
Let
P = Capital cost of equipment
n = Useful life of equipment in years
S = Scrap value after useful life
Suppose the annual unit depreciation is x. It is desired to find the value of x in terms of P, n
and S.
But the value of equipment after n years (i.e., useful life) is equal to the scrap value S.
Similarly, annual depreciation charge for the subsequent years can be calculated. This method
is more rational than the straight line method. The figure below shows the graphical
representation of diminishing value method. The initial value P of the equipment
reduces, through depreciation, to the scrap value S over the useful life of the equipment. The
depreciation curve follows the path PA.
It is clear from the curve that depreciation charges are heavy in the early years but decrease to
a low value in the later years. This method has two drawbacks. Firstly, low depreciation
charges are made in the late years when the maintenance and repair charges are quite heavy.
Secondly, the depreciation charge is independent of the rate of interest which it may draw
during accumulation. Such interest money, if earned, are to be treated as income.
(iii) Sinking fund method: In this method, a fixed depreciation charge is made every year
and interest compounded on it annually. The constant depreciation charge is such that total of
annual instalments plus the interest accumulations equal to the cost of replacement of
equipment after its useful life.
Electricity tariff – Types of Tariff
Let
P = Initial value of equipment
n = Useful life of equipment in years
S = Scrap value after useful life
r = Annual rate of interest expressed as a decimal
Cost of replacement = P − S
This total fund must be equal to the cost of replacement of equipment P − S.
The value of ‘q’ gives the uniform annual depreciation charge. The parenthetical term in eq.
(i) is frequently referred to as the “sinking fund factor”.
Though this method does not find very frequent application in practical depreciation
accounting, it is the fundamental method in making economy studies and most
effective method of determining depreciation.
Load Curve in a Power Plant – Daily, Monthly and Annual
A load curve is a plot of the load demand (on the y-axis) versus the time (on the x-axis) in the
chronological order. It is a curve showing the variation of load on the power station with
respect to (w.r.t) time.
Daily Load Curve in a Power Plant
From out of the load connected, a consumer uses different fractions of the total load at various
times of the day as per his/her requirements.
Since a power system has to supply load to all such consumers, the load to be supplied varies
continuously with time and does not remain constant. If the load is measured (in units of
power) at regular intervals of time, say, once in an hour (or half-an-hour) and recorded, we
can draw a curve known as the load curve.
Daily Load Curve
A time period of only 24 hours is considered, and the resulting load curve, which is called
a ‘Daily load curve‘, is shown in the figure.
Monthly Load Curve
The monthly load curve can be obtained from the daily load curves of that month. For
this purpose, average values of power over a month at different times of the day are
calculated and then plotted on the graph.
The monthly load curve is generally used to fix the rates of energy.
Yearly/Annual Load Curve
The yearly load curve is obtained by considering the monthly load curves of that particular
year.
The yearly load curve is generally used to determine the annual load factor.
However, to predict the annual requirements of energy, the occurrence of the load at different
hours and days in a year and in the power supply economics, ‘Annual load curves’ are used.
An annual load curve is a plot of the load demand of the consumer against time in hours of the
year (1 year: 8,760 hours).
Daily Load Curve in which Load is measured in an hour
The significance of Load Curve
From the daily load curve shown above, the following information can be obtained.

Observe the variation of load on the power system during different hours of the day.

The area under this curve gives the number of units generated in a day.

The highest point on that curve indicates the maximum demand on the power station
on that day.

The area of this curve divided by 24 hours gives the average load on the power station
in the day.

It helps in the selection of the rating and number of generating units required.
Importance of Daily Load Curve
The daily load curves have gained great importance in the generation industry due to the fact
that they supply the following information.
Daily Load curve on Power Plant
1. On the daily load curve, you can see how the load on the power station varies during
each hour of the day.
2. The area under the daily load curve gives the number of units generated on that day.
 Units generated/day = Area (in kWh) under daily load curve.
3. The peak on the daily load curve indicates the maximum demand on the station on that
day.
4. By dividing the area under the daily load curve by the total number of hours, the daily
average load is obtained.

Average load = Area (in kWh) under daily load curve / 24 hours
5. Load factor is calculated from daily load curve by taking the ratio between the area
under the load curve and the area of the rectangle within which the load curve is
contained.

Load Factor = Average Load / Maximum demand (peak demand)
6. The load curve is used to select the size and number of generating units. It is chosen
based on the load curve which determines how many and what size of generating units
are needed. In this way, the generating units are operated at or near their maximum
efficiency.
7. When preparing the operating schedule for the power station, load curves are helpful.
A plant’s operating schedule is the timing and sequence in which various generating
units (such as alternators) will be put into operation.
Terms & Definition:
1. Connecting Load: It is the sum of ratings in kilowatts (kw) of equipment installed in the
consumers premises.
For example – if a consumer has connection for 4 lamps of 60w each and point of
500w and a radio consuming 60w, then the total connected load of the consumer =4 X
60 + 500 +60 = 800 w.
2. Demand factor: it is defined as the ratio of maximum demand to connecting load.
Demand factor = Maximum demand / Connecting Load.
3. Load Factor: It is defined as the ratio of average load to the maximum demand load
factor & demand factor always less than unity.
4. Plant capacity factor: It is defined as the ratio of actual energy that could have been
produced during the same period.
Plant capacity factor = E / (c x t)
Where,
E = Energy produced (kWh) in a given period.
C = Capacity of the plant in kW.
t = total number of hours in the given period.
5. Plant use Factor: It is defined as the ratio of actual energy product to the maximum
possible energy that could have been produced during the actual number of hours the
plant was in operation.
Plant use factor = E / (c x t1)
Where,
E = Energy produced (kWh) in a given period
C = Capacity of the plant in kW.
t1 = Actual number of hours the plant has been operation.
Problem: 1
A power station has a Maximum demand of 80×10 8 kW and daily load curve is defined as
follows:-
Determine e the load factor of power station.
Solution:
Given,
Maximum demand = 80 × 103 kW
To find,
Load factor =?
Here,
Energy generated
= (40×6) + (50×2) + (60×40) + (50×2) + (70×4) + (80×4) + (40×2)
= 1360 Mw-h
= 1360 X 103 kW-h
Average load = Energy generated /24
= (1360 X 103) / 24
= 56,666 kW
Load factor = Average load / Maximum demand
= 56,666 / 80,000
= 0.71
Problem: 2
A central power station has annual factor as follows-
Load factor
= 60%
Capacity factor
= 40%
Use factor
= 45%
Maximum demand = 15,000 kw
Determine(a) Annual energy production.
(b) Reserve capacity over and above peak load.
(c) Hours per year not in service.
Solution:
Given,
Load factor
= 60% = 0.6
Capacity factor = 40% = 0.4
Use factor
= 45% = 0.45
Maximum demand = 15,000 kW.
To find,
(a) Annual energy production =?
(b) Reserve capacity over and above peak load =?
(c) Hours per year not in service =?
Load factor = Average load / Maximum demand
=> Average load = load factor × Maximum demand
= 0.6 × 15,000
= 9000 kW
Annual energy produced, E = Average load × Total hours in one year (t)
= 900 × 8760
= 78.84 × 106 kWh
I.e. Annual energy production is 78.84×10 6 kw-h [Answer].
Capacity factor = E / (C x t)
=>
C = E / (Capacity factor x t)
= 78.84 x 106 / (0.4 x 8760)
= 22500 kW
Reserve capacity = capacity of the plant(c) – Maximum demand
= 22,500 - 15000
= 7500kw
I.e. Reserve capacity over and above peak load is 7500 Kw (answer)
Use factor = E / (C X t1)
=>
T1 = E / (Use factor x C)
= 78.84 x 106 / (0.45 x 22,500)
= 7786 hrs.
Where,
t1= Actual number of hours of the year for which the plant remains in operation.
Hours per year not in service = t - t1
= 8760 -7786
= 974 hours
i.e. 974 hours per year not in service [answer].