Masinde Muliro University of Science
and Technology
SCHOOL OF ENGINEERING AND BUILT ENVIRONMENT
DEPARTMENT OF ELECTRICAL AND COMMUNICATION ENGINEERING
Power System I
Lecture Notes
Subject Code: TEE 212
Lecturer: E. Kinoti
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Introduction
1. Sources of Energy
Figure 1. Energy Sources.
Choice of size and number of Generating Units
1. Economy: large size units have

low capital cost/kW,

need less land area

requires less operating labor
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
have better efficiency

system strength
2. Power plant capacity: neither small nor large number of units
3. Transmission facility
4. Reserve requirements
5. Status of technology
6. For hydro stations unit size depends on nature of flow, availability of head and water to
generate maximum power possible.
Types of Load

Domestic Load

Industrial Load

Commercial Load

Agricultural load
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Figure 2: Electrical Loads.
Important Terms
(i) Demand Factor = Maximum Demand / Connected Load
Connected Load: sum of continuous ratings of all outlets in a distribution circuit
Maximum Demand: maximum power that the distribution circuit is likely to draw at
any time
(ii) Group Diversity Factor = Sum of individual maximum demands / Maximum demand
of the group. GDF is always greater than unity
(iii)
Peak Diversity Factor = Sum of maximum demand of a consumer group /
demand of the consumer group at the time of maximum demand
(iv) Load Factor =Average Load / Peak Load
(v) Capacity Factor = Average Demand / Installed Capacity
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(vi) Utilization Factor = Maximum Load / Rated Plant Capacity
(vii)
Load Curve: It is the curve between load (MW) versus time.
(viii)
Load Duration Curve: It is the rearrangement of all the load elements of a load
curve in a descending order plotted as a function of time.
Figure 3: Load Curve.
(ix)
Energy Load Curve: It plots the cumulative integration of area under the load
curve.
(x) Mass Curve It gives the total energy used by the load up to each hour of the day.
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Figure 2: Load Duration Curve
Operating Reserves

How much generating capacity should be committed and how much should be left for
future expansion?

In electricity networks, the “operating reserve” is the generating capacity available to
system operator within a short interval of time to meet the changing demand or in case
a generator is out of service.
Table 1: Operating Reserves
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Electricity Tariffs
Definition: The amount of money frame by the supplier for the supply of electrical energy to
various types of consumers in known as an electricity tariff. In other words, the tariff is the
methods of charging a consumer for consuming electric power. The tariff covers the total cost
of producing and supplying electric energy plus a reasonable cost.
Objectives

Capital recovery

Operational cost of distribution utility

Cost of metering, billing and collection

Simple and comprehensible to general public

Uniform for a large population

Should provide incentives for using power in off-peak hours

Should have a provision of penalty for low power factor.
The actual tariffs that the customer pay depends on the consumption of the electricity. The
consumer bill varies according to their requirements. The industrial consumers pay more
tariffs because they use more power for long times than the domestic consumers. The
electricity tariffs depends on the following factors

Type of load

Time at which load is required.

The power factor of the load.

The amount of energy used.
The total bill of the consumer has three parts, namely, fixed charge D, semi-fixed charge
Ax and running charge by.
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Where,
C – Total charge for a period (say one month)
x – maximum demand during the period (kW or kVA)
y – Total energy consumed during the period (kW or kVA)
A – cost per kW or kVa of maximum demand.
B – Cost per kWh of energy consumed.
D – Fixed charge during each billing period.
This is known as three-part electricity tariff, and it is mainly applied to the big consumer.
Factors Affecting the Electricity Tariffs
The following factors are taken into accounts to decide the electricity tariff:

Types of Load – The load is mainly classified into three types, i.e., domestic, commercial,
or industrial. The industrial consumers use more energy for a longer time than
domestic consumers, and hence the tariff for the industrial consumers is more than the
domestic consumers. The tariff of the electric energy varies according to their
requirement.

Maximum demand – The cost of the electrical energy supplied by a generating station
depends on the installed capacity of the plant and kWh generated. Increased in
maximum capacity increased the installed capacity of the generating station.

The time at which load is required – The time at which the maximum load required is
also essential for the electricity tariff. If the maximum demand coincides with the
maximum demand of the consumer, then the additional plant is required. And if the
maximum demand of the consumers occurs during off-peak hours, the load factor is
improved, and no extra plant capacity is needed. Thus, the overall cost per kWh
generated is reduced.
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
The power factor of the load – The power factor plays a major role in the plant
economics. The low power factor increases the load current which increases the losses
in the system. Thus, the regulation becomes poor. For improving the power factor, the
power factor correction equipment is installed at the generating station. Thus, the cost
of the generation increases.

The amount of energy used – The cost of electrical energy is reduced by using large
amounts of energy for longer periods.
Types of Electricity Tariff
Some of the most important types of tariff are as follows;
1. Flat Demand Rate tariff
2. Straight-line Meter rate tariff
3. Block meter Rate tariff
4. Two-part tariff
5. Power factor tariff
6. Seasonal rate tariff
7. Peak load tariff
8. Three-part tariff
The different types of tariffs are explained below in details
1. Flat demand rate tariff – The flat demand rate tariff is expressed by the equation C = Ax.
In this type of tariff, the bill of the power consumption depends only on the maximum
demand of the load. The generation of the bill is independent of the normal energy
consumption. This type of tariff is used on the street light, sign lighting, irrigation, etc.,
where the working hours of the equipment are unknown. The metering system is not used
for calculating such type of tariffs.
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2. Straight-line meter rate tariff – This type of tariff is given by the equation C = By. The
generation of the bills depends on the energy consumption of the load. Thus, different types
of bills are generated by the consumers.
The charges for different types of consumption depends on the load and diversity factors of
the load. For example, the tariff for small devices is less as compared to the power loads.
Hence different meters are used for measuring the power consumption
3. Block meter rate tariff – In this type of tariff, the energy consumption is distinguished
into blocks. The per unit tariff of the individual block is fixed. The price of the block is
arranged in the decreasing order. The first block has the highest cost, and it goes on
decreasing accordingly.
The price and the energy consumption are divided into three blocks. The first few units of
energy at a certain rate, the next at a slightly lower rate and the remaining unit at a very
lower rate.
4. Two-part tariff – In such type of tariff, the total bill is divided into two parts. The first
one is the fixed charge and the second is the running charge. The fixed charge is because
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of the maximum demand and the second charge depends on the energy consumption by
the load.
The factor A and B may be constant and vary according to some sliding.
5. Power factor tariff – The tariff, which depends on the power factor of the load, is known
as the power factor tariff. The power factor tariff is mainly classified into two types.
a. kVA maximum demand tariff – This is also a two-part tariff.
The low power factor increases the KVA rating of the load.
b. kWh and kVarh tariff – The bill is calculated by the sum of the kVarh and Kwh rating of
the load.
The kVarh is inversely proportional to the power factor of the load.
c. Sliding Scale or Average power factor tariff – In Average power factor tariff, the
particular value of the power factor is taken as reference. If the power factor at the
consumer end is low, then the consumer has to pay the additional charges. Similarly, if the
power factor of the load is above from the reference value, then the discount will be given
to the consumer.
6. Seasonal rate tariff – Such type of tariff measures the high price in kWh used by the
consumer in one complete year. It is also known as the on peak season tariff. If the low
consumption occurs in the year, then it called the off-peak season tariffs.
7. Peak-load tariff – Such type of tariff is similar to peak load tariffs. The only difference is
that the seasonal tariff measures the peak hour of the year and the peak tariff calculates it
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for the day. If the power consumption is high, then it is known as the on-peak tariff, and
for low power consumption, it is called off-peak load tariffs.
The peak load and seasonal tariffs both are used for reducing the idle or standby capacity
of the load.
8. Three-part tariff – The three-part tariff is in the form of,
applied to the big consumer.
General Tariff Form
A = c*x + d*y + f
Where
A= total amount of bill for a certain period
x=maximum demand during a period (kW or kVA)
y= total energy consumed during the period in kWh
c=unit charge for maximum demand (Kshs/kWh or Kshs/kVA)
d=unit cost of energy (Kshs./kWh)
f= constant charge
HYDRO POWER PLANTS
Hydro Power Potential
P = g*ρ*Q*H
Where
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and it is
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.
Hydrology
• First requirement – Q (discharge)
• Hydrology deals with occurrence and distribution of water over and under earth’s surface.
– Surface Water Hydrology
– Ground Water Hydrology
• Watershed, catchment area or drainage area: length of the river, size and shape of the area it
affects, tributaries, lakes, reservoirs etc.
• Investigation of run-off for past few years is required for power potential studies of a HPP.
Objectives of Hydrology
• To obtain data regarding the stream flow of water that would be available,
• To predict the yearly possible flow
• To calculate the mean annual rainfall in the area under consideration from a record of the
annual rainfall for a number of years, say 25 to 30
• To note the frequency of dry years
• To find maximum rainfall and flood frequency
Various terms related to Hydrology
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• Rainfall is also known as precipitation and can be measured by rain gauges.
• Some part of precipitation is lost due to evaporation, interception and transpiration.
Transpiration: Plants absorbing moisture and giving it off to the atmosphere
• Stream flow = precipitation – losses
• Stream flow = surface flow + percolation to ground
• Surface flow is also known as run-off.
Hydrograph: shows the variation of stream flow in m3/s with time for a particular river site.
The time may be hour, week, month or a year. The area under hydrograph gives the total
volume of flow
Flow duration curve: shows the percentage of time during the period when the flow was equal
to greater than the given flow. The area under FDC gives the total quantity of run-off during a
period
Mass curve: Indicates the total volume of run-off in cubic meters up to a certain time. The
slope of the curve at any point shows the rate of flow at that time. Used for estimating the
capacity of storage reservoir
Storage: to ensure water availability during deficient flow and thus increasing the firm
capacity. Storage also results in more energy production
Pondage: Storing water in small ponds near the power plant as the storage reservoir is away
from plant. To meet the power demand fluctuations over a short period of time e.g. 24 hours
• Primary Power: power that will be available 90 % of the time
• Secondary Power: power that will be available 75 % of the time
• Dump Power: power that will be available 50 % of the time.
Maximum flow estimation: gives estimation of floods and helps in design of dam and spillway.
Site Selection for Hydropower Plants
• Availability of Water: Run-off data for many years available
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• 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
Classification of Hydropower Plants
 According to water flow regulation:

Run- off river plants without pondage

Run- off river plants with pondage

Hydroelectric plants with storage reservoir
 According to Load:

Base load plants

Peak load plants

Pumped storage plants
 According to head:

High head plants (>100m)

Medium head plants (30-100 m)

Low head plants (<30 m)
Components of a HPP
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Figure 3: Schematic of a Hydropower 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
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
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.
Power House consists of:

Hydraulic turbines

Electric generators

Governors

Gate valves

Relief valves

Water circulation pumps

Air ducts

Switch board and instruments

Storage batteries
Hydraulic Turbines
Types of Hydraulic Turbines
1. According to the head and quantity of water available
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
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
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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.
Hydro Power Plant Auxiliaries
1. Governing oils systems
2. Lubricating oil pumps
3. Coolant oil pumps
4. Drainage pumps
5. Pipes, fans, ventilation
6. Air compressor
7. Cooling oil pumps for transformers
8. Head gates
9. Drain valves
10. Gantry cranes
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11. Station batteries
12. Instrumentation system
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.
Thermal Power Plants
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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.
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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
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
Super heater and Reheater

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 capacity uses coal of calorific value 6400 kCal/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%.
Nuclear Power Plant
A nuclear power plant is a thermal power station in which the heat source is one or more
nuclear reactors. As in a conventional thermal power station the heat is used to generate steam
which drives a steam turbine connected to a generator which produces electricity.
Nuclear power plants are usually considered to be base load stations, which are best suited to
constant power output.
Figure 4: Schematic of a Nuclear Power Plant
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Factors for Site Selection of NPPs
1. Availability of Water: working fluid
2. Distance from Populated Area: danger of radioactivity
3. Nearness to the load centre: reduction in transmission cost
4. Disposal of Waste: radioactive waste
5. Accessibility by Rail and Road: transport of heavy equipment
Advantages of NPPs
1. Reduces demand for fossil fuels
2. Quantity of nuclear fuel is much less: thus reducing transport and resulting costs
3. Area of land required is less: compared to a conventional plant of similar capacity
4. Production of fissile material
5. Location independent of geographical factors: except water requirement
Disadvantages of NPPs
1. Not available for variable loads (load factor-0.8): as the reactors cannot be controlled to
respond quickly
2. Economical reason should be substantial
3. Risk of leakage of radioactive material
4. Further investigation on life cycle assessment and reliability needs to be done
5. Perception problems
Numerical Problems on Nuclear Power Plant
1. Calculate the amount of coal containing the same energy as in 1 kg of Natural Uranium
under the following assumptions. Also calculate the number of fissions per second to produce
1 watt power.

Energy release from one fission of U235 = 200 MeV

Atoms in one gram pure U235 = 25.64 x 1020
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
Calorific value of coal = 6000 kcal/kg

U235 content in Natural Uranium = 0.7 %

Fission efficiency = 50 %

One Joule = 0.239 cal
2. Find the power produced by fissioning 5 grams of U235 per day. Number of atoms in one
gram of U235 =2.563 x 1021
3. Find the U235 fuel used in one year in a 235 MW pressurized water reactor. Assume overall
plant efficiency of 33 % and 100% load factor throughout the year. Number of fissions
required for 1 watt-sec= 3.1x1010. Number of atoms in one gram of U235 = 2.563 x 1021
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Power Supply System
The electrical energy is produced at generating stations, and through the transmission
network, it is transmitted to the consumers. Between the generating stations and the
distribution stations, three different levels of voltage (transmission, sub-transmission and
distribution level of voltage) are used.
The high voltage is required for long distance transmission and, the low voltage is required for
utility purposes. The voltage level is going on decreasing from the transmission system to the
distribution system. The electrical energy is generated by the three-phase synchronous
generator (alternators) as shown in the figure below. The generation voltage is usually 11kV
and 33 KV.
Power supply Systems
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This voltage is too low for transmission over long distance. It is, therefore, stepped up to 132,
220, 400 KV, or more by step-up transformers. At that voltage, the electrical energy is
transmitted to the bulk power substation where energy is supplied from several power
substations.
The voltage at these substations is stepped down to 66KV and fed to the sub-transmission
system for onward transmission to the distribution sub-stations. These substations are located
in the region of the load centers.
The voltage is further stepped down to 33KV and 11KV. The large industrial consumers are
supplied at the primary distribution level of 33KV while the smaller industrial consumer is
supplied at 11KV.
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The voltage is stepped down further by a distribution transformer located in the residential and
commercial area, where it is supplied to these consumers at the secondary distribution level of
400V three phase and 230V single phase.
Power system symbols
Power flow Single Line Diagram Equivalent Circuit
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Advantage of Interconnection of Generating Stations
The power system consists two or more generating stations which are connected by tie lines.
Interconnection of generating stations has the following important advantages.
1. It enables the mutual transfer of energy from surplus zone to deficit zone economically.
2. Lesser overall installed capacity to meet the peak demand.
3. Lesser standby reserve generating capacity is required.
4. It permits the generation of energy at the most efficient and cheapest station at every
time.
5. It reduces the capital cost, operating cost and cost of energy generated.
6. If there is a major breakdown of a generating system unit in an interconnected system,
then there is no interruption of power supply.
The interconnection provides the best use of power resources and greater security of supply. It
enables overall economic generation by optimum use of the high capacity economical
generating plant. The interconnection between networks is done either by HVAC (high voltage
alternating current) links or through HVDC (High Voltage Direct Current) links.
Active, Reactive and Apparent Power
1. Active Power
Definition: The power which is actually consumed or utilized in an AC Circuit is called True
power or Active power or Real power. It is measured in kilowatt (kW) or MW. It is the actual
outcomes of the electrical system which runs the electric circuits or load.
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2. Reactive Power
Definition: The power which flows back and forth that means it moves in both the directions in
the circuit or reacts upon itself, is called Reactive Power. The reactive power is measured in
kilo volt-ampere reactive (kVAR) or MVAR.
3. Apparent Power
Definition: The product of root mean square (RMS) value of voltage and current is known
as Apparent Power. This power is measured in kVA or MVA.
It has been seen that power is consumed only in resistance. A pure inductor and a pure
capacitor do not consume any power since in a half cycle whatever power is received from the
source by these components, the same power is returned to the source. This power which
returns and flows in both the direction in the circuit, is called Reactive power. This reactive
power does not perform any useful work in the circuit.
In a purely resistive circuit, the current is in phase with the applied voltage, whereas in a
purely inductive and capacitive circuit the current is 90 degrees out of phase, i.e., if the
inductive load is connected in the circuit the current lags voltage by 90 degrees and if the
capacitive load is connected the current leads the voltage by 90 degrees.
Hence, from all the above discussion, it is concluded that the current in phase with the voltage
produces true or active power, whereas, the current 90 degrees out of phase with the voltage
contributes to reactive power in the circuit.
Therefore,

True power = voltage x current in phase with the voltage
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
Reactive power = voltage x current out of phase with the voltage
The phasor diagram for an inductive circuit is shown below:
Taking voltage V as reference, the current I lags behind the voltage V by an angle ϕ. The
current I is divided into two components:

I Cos ϕ in phase with the voltage V

I Sin ϕ which is 90 degrees out of phase with the voltage V
Therefore, the following expression shown below gives the active, reactive and apparent power
respectively.

Active power P = V x I cosϕ = V I cosϕ

Reactive power Pr or Q = V x I sinϕ = V I sinϕ

Apparent power Pa or S = V x I = VI
Active component of the current
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The current component, which is in phase with the circuit voltage and contributes to the active
or true power of the circuit, is called an active component or watt-full component or in-phase
component of the current.
Reactive component of the current
The current component, which is in quadrature or 90 degrees out of phase to the circuit
voltage and contributes to the reactive power of the circuit, is called a reactive component of
the current.
Power Triangle
Power Triangle is the representation of a right angle triangle showing the relation between active
power, reactive power and apparent power.
When each component of the current that is the active component (I cosϕ) or the reactive component (I
sinϕ) is multiplied by the voltage V, a power triangle is obtained shown in the figure below:
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The power which is actually consumed or utilized in an AC Circuit is called True power or Active
Power or real power. It is measured in kilowatt (kW) or MW.
The power which flows back and forth that means it moves in both the direction in the circuit or reacts
upon it, is called Reactive Power. The reactive power is measured in kilovolt-ampere reactive (kVAR) or
MVAR.
The product of root mean square (RMS) value of voltage and current is known as Apparent Power. This
power is measured in KVA or MVA.
The following point shows the relationship between the following quantities and is explained
by graphical representation called Power Triangle shown above.

When an active component of current is multiplied by the circuit voltage V, it results in
active power.it is this power which produces torque in the motor, heat in the heater,
etc. This power is measured by the wattmeter.

When the reactive component of the current is multiplied by the circuit voltage, it gives
reactive power. This power determines the power factor, and it flows back and forth in
the circuit.

When the circuit current is multiplied by the circuit voltage, it results in apparent
power.

From the power triangle shown above the power, the factor may be determined by
taking
the
ratio
of
true
power
P a g e | 32
to
the
apparent
power.
As we know simply power means the product of voltage and current but in AC circuit except
for pure resistive circuit there is usually a phase difference between voltage and current and
thus VI does not give real or true power in the circuit.
Power Factor Correction
Power Factor Correction uses parallel connected capacitors to oppose the effects of inductive
elements and reduce the phase shift between the voltage and current.
Power Factor Correction is a technique which uses capacitors to reduce the reactive power
component of an AC circuit in order to improve its efficiency and reduce current.
P a g e | 33
When dealing with direct current (DC) circuits, the power dissipated by the connected load is
simply calculated as the product of the DC voltage times the DC current, that is V*I, given in
watts (W). For a fixed resistive load, current is proportional to the applied voltage so the
electrical power dissipated by the resistive load will be linear. But in an alternating current
(AC) circuit the situation is slightly different as reactance affects the behavior of the circuit.
For an AC circuit, the power dissipated in watts at any instant in time is equal to the product of
the volts and amperes at that exact same instant, this is because an AC voltage (and current) is
sinusoidal so changes continuously in both magnitude and direction with time at a rate
determined by the source frequency.
In a DC circuit the average power is simply V*I, but the average power of an AC circuit is not
the same value as many AC loads have inductive elements, such as coils, windings,
transformers, etc. where the current is out of phase with the voltage by some degrees resulting
in the actual power dissipated in watts being less than the product of the voltage and current.
This is because in circuits containing both resistance and reactance, the phase angle (Θ)
between them must also be taken into account.
We saw in the tutorial about Sinusoidal Waveforms that the phase angle (∠Θ) is the angle in
electrical degrees by which the current lags behind the voltage. For a purely resistive load,
voltage and current are “in-phase” since there is no reactance.
However, for an AC circuit containing an inductor, coil, or solenoid or some other form of
inductive load, its inductive reactance (XL) creates a phase angle with the current lagging
behind the voltage by 90o. Therefore there is both resistance (R) and inductive reactance (XL)
both given in Ohms, with the combined effect called Impedance. Thus impedance, represented
P a g e | 34
by the capital letter Z, is the resulting value given in Ohms due to the combined effect of a
circuit’s resistance and reactance.
Consider the RL series circuit below.
RL Series Circuit
Since it is a series circuit, the current must therefore be common to both the resistor and the
inductor so the voltage dropped across the resistor, VR is “in-phase” with the series current
while the voltage drop across the inductor, VL “leads” the current by 90o (ELI). As a result the
voltage dropped across the resistor is placed on the current vector because both vectors are inphase, while the voltage developed across the inductor coil is drawn in a vertical direction due
to the voltage leading the current by 90o.
Thus the vector diagram drawn for each component will have the current vector as its
reference with the two voltage vectors being plotted with respect to their position as shown.
R and L Vector Diagrams
P a g e | 35
The resistor voltage VR is plotted along the horizontal or “real axis” and the inductor voltage
VL is plotted on the vertical or “imaginary axis”. In order to find the resulting voltage
VS developed across the series connected circuit we must combine together the two individual
vectors using the current as our reference. The resulting vectorial voltage can easily be found
using Pythagoras’ theorem as the combination of VR and VL forms a right angled triangle as
shown below.
Phasor Diagram for the Series RL Circuit
P a g e | 36
Power Factor Correction Example No1
An RL series circuit consists of a resistance of 15Ω and an inductor which has an inductive
reactance of 26Ω. If a current of 5 amperes flows around the circuit, calculate:
1) The supply voltage.
2) The phase angle between the supply voltage and circuit current.
3) Draw the resulting phasor diagram.
1). the supply voltage VS
We can double check this answer of 150Vrms using the impedances of the circuit as
follows:
P a g e | 37
2). the phase angle Θ using Trigonometry functions is:
3). The resulting phasor diagram showing VS
The calculated voltage dropped across the resistor (the real component) was 75 volts while the
voltage generated across the inductor (the imaginary component) was 130 volts. Clearly the
sum of 75 volts plus 130 volts equals 205 volts which is far greater than the calculated 150
volts. This is due to the fact that the value of 150V represents the phasor sum. Knowing the
individual voltage drops and impedances we can convert these values into values that
represent the power consumed, either real or imaginary in the circuit.
P a g e | 38
Power in a RL Series Circuit
In a circuit containing reactance, the current, i will either lead or lag the voltage by some
amount depending on whether the reactance is capacitive or inductive. The power consumed
by the resistor in watts is called the “real power” so is given the symbol “P” (or W). Watts can
also be calculated as I2R, where R is the total resistance of the circuit. However, to calculate the
value of the real power in terms of the rms voltage and rms current (Vrms*Irms), we must also
multiply these values by the cosine of the phase angle, cosϕ giving:
Real Power, P = V*I cos (ϕ)
Since, as we have seen above, the voltage and current are “in-phase” for a resistance, the phase
angle is therefore zero (0), thus giving us cos(ϕ)= 1. Multiplying V*I*1 will therefore give us
the same real power value as does using I2R. Then using our coil example above, the power
dissipated by the 15Ω resistor is:
PR = I2R = 52x 15 = 375 watts
Equivalent to:
PR = VR*I cos (ϕ) = 75 x 5 x cos (ϕ) = 375 watts
When the voltage and current are “out-of-phase” with each other because the circuit contains
reactance, the product of V*I is called the “apparent power”, given the units of volt-amperes
(VA) instead of watts. Volt-amperes have the symbol “S“. For a purely inductive circuit the
current lags the voltage by 90o so the reactive power for an inductive load is given as: V*I cos
(+90o) which becomes: V*I*0. Clearly then there is no power consumed by an inductance so
there is no power loss, thus PL = 0 watts. However to show that this wattless power exists in an
P a g e | 39
AC circuit, it is called volt-amperes reactive (VAR) and is given the symbol “Q“. So the voltamperes reactive or simply “reactive power” for an inductive circuit uses the symbol QL.
Similarly, for a purely capacitive circuit the current leads the voltage by 90o, the reactive
power for a capacitive load is given as: V*I cos (-90o) which again becomes: V*I*0. Clearly then
and as before, there is no power consumed by a capacitance so there is no power loss as PC = 0
watts. So to show that this wattless power exists in a capacitive circuit, it is called volt-amperes
reactive capacitive and is given the symbol QC. Note here that the reactive power of a
capacitance is defined as being negative, resulting in -QC.
So again using our example above, the reactive power flowing in and out of the inductor at a
rate determined by the frequency is given as:
QL = I2XL = 52x 26 = 650 VAR
As there is a 90o phase difference between the voltage and the current waveforms in a pure
reactance (either inductive or capacitive) we multiply V*I by sin(Θ) to give the vertical
component that is 90o out-of-phase. However, the sine of the angle (sin 90o) gives the result as
“1” so we can find the reactive power by simply multiplying the rms voltage and current
values as shown.
QL = I2XL = V*I*sin (ϕ) = 130*5*sin (90o) = 130*5*1 = 650 VAR
Then we can see that the volts-amperes reactive or VAR part has a magnitude (the same as for
the real power) but no phase angle associated with it. That is reactive power is always on the
90o vertical axis. So if we know that:
PR = I2R = 375 Watts
and
P a g e | 40
QL = I2XL = 650 VAR (ind.)
We can construct a power triangle to show the relationship between P, Q, and S as shown.
Inductive Power Triangle
Capacitive Power Triangle
Again we can use the previous Pythagoras’ Theorem and the Trigonometry functions of Sine,
Cosine and Tangent to define a power triangle.
P a g e | 41
Power Triangle Equations
Power Factor Correction Example No2
A coil has a resistance of 10Ω and an inductance of 46mH. If it draws a current of 5 Amperes
when connected to a 100Vrms, 60Hz supply, calculate:
P a g e | 42
1) The voltages across the components.
2) The phase angle of the circuit.
3) The different powers consumed by the series RL circuit.
First find the impedances
1). the voltages across the resistor, VR and inductor, VL
P a g e | 43
2). the phase angle of the circuit
3). the circuit power
We can confirm that the circuit draws 500VA of complex power from the supply as S = I2Z, so
52 x 20 = 500VA and the construction of a power triangle would also confirm this as being
correct.
However, this complex or apparent power being consumed by the series RL circuit is large
because the phase angle (ϕ) by which the voltage leads the current (ELI) is also large resulting
in a poor power factor of 0.5 (cos60o) lagging. So we need to cancel some of this inductive
reactive power being consumed (433 VAR) by the coil used to sustain the coils magnetic field
by adding some more reactance to it but of the opposite type to the circuit. Should we be
concerned about the coils low power factor? Well yes as power factor is the ratio of the coils
real power to its apparent power (Watts/Volt-Amperes), it gives an indication of how
effectively the electrical power being supplied is being used. Thus a low power factor means
P a g e | 44
that the electrical power being supplied is not fully utilized as in our example coil above, at
50% power factor (W/VA = 250/500) it takes 500VA to produce just 250W of real power. If
the coil has inductive reactance which is positive, then we must add some capacitive reactance
which is negative to cancel it out and improve the coils overall power factor value. Adding
capacitors to reduce a circuits phase angle and reactive power consumption is referred to
as power factor correction which allows us to reduce a circuit’s power factor to nearer 1,
unity.
Power Factor Correction
Power Factor Correction improves the phase angle between the supply voltage and current
while the real power consumption in watts remains the same, because as we have seen a pure
reactance does not consume any real power. Adding an impedance in the form of capacitive
reactance in parallel with the coil above will decrease Θ and thus increases the power factor
which in turn reduces the circuits rms current drawn from the supply.
The power factor of an AC circuit can vary from between 0 and 1 depending on the strength of
the inductive load but in reality it can never be less than about 0.2 for the heaviest of inductive
loads. As we have seen above, a power factor of less than 1 means that there is reactive power
consumption which increases the closer it gets to 0 (fully inductive). Clearly then a power
factor of exactly “1” means the circuit consumes zero reactive power (fully resistive) resulting
in a power factor angle of 0o. This is referred to as “unity power factor”.
Adding a capacitor in parallel with the coil will not only reduce this unwanted reactive power,
but will also reduce the total amount of current taken from the source supply. In theory
capacitors could provide 100% of compensated reactive power required in a circuit, but in
practice a power factor correction of between 95% and 98% (0.95 to 0.98) is usually
P a g e | 45
sufficient. So using our coil from example no2 above, what value of capacitor is required to
improve the power factor from 0.5 to 0.95.
A power factor of 0.95 is equal to a phase angle of: cos (0.95) = 18.2o thus the amount of VAR
required is:
Therefore for a phase angle of 18.2o we need a reactive power value of 82.2VAR. If the original
uncorrected VAR value was 433VAR and the new calculated value is 82.2VAR, we need a
reduction of 433 – 82.2 = 350.8 VAR (capacitive). Therefore:
The capacitor required to reduce the reactive power to 82.2VAR must have a capacitive
reactance of 28.5Ω at the rated supply frequency. Therefore the capacitance of the capacitor is
calculated as:
P a g e | 46
So to improve the power factor of the coil in example no2 from 0.5 to 0.95 requires a parallel
connected capacitor of 93uF. Using the values from above we can now calculate the amount of
real power supplied by the source after the power factor correction has been applied.
New Volt-Amperes Value
We can also construct a power triangle to show the before and after values for VA (S) and VAR
(Q) as shown.
Power Triangle
P a g e | 47
If the circuit’s apparent power has been reduced from 500VA to just 263VA, we can calculate
the rms current supplied as:
S = V*I, therefore: I = S/V = 263/100 = 2.63 Amperes
So just by connecting a capacitor across the coil not only improves its overall power factor
from 0.5 to 0.95, but reduces the supply current from 5 amperes to 2.63 amperes, a reduction
of some 47%. The final circuit will look like this.
Why improve power factor? The benefits that can be achieved by applying the correct power
factor correction are:
 Environmental benefit. Reduction of power consumption due to improved energy
efficiency. Reduced power consumption means less greenhouse gas emissions and fossil
fuel depletion by power stations.
 Reduction of electricity bills Extra kVA available from the existing supply
 Reduction of I2R losses in transformers and distribution equipment
 Reductions of voltage drop in long cables.
P a g e | 48
 Extended equipment life – Reduced electrical burden on cables and electrical
components.
Star – Delta configuration
P a g e | 49
P a g e | 50
Electrical Transformer
Definition: The transformer is the static device which works on the principle of
electromagnetic induction. It is used for transferring the electrical power from one circuit to
another without any variation in their frequency. In electromagnetic induction, the transfer of
energy from one circuit to another takes places by the help of the mutual induction. I.e. the
flux induced in the primary winding is linked with the secondary winding.
Construction of an Electrical Transformer
The primary winding, secondary winding and the magnetic core are the three important of the
transformer. These coils are insulated from each other. The main flux is induced in the
primary winding of the transformer. This flux passes through the low reluctance path of the
magnetic
core
and
linked
with
the
secondary
Transformer Working Principle
P a g e | 51
winding
of
the
transformer.
Consider the T1 and T2 are the numbers of the turn on the primary and the secondary winding
of the transformer shown in the figure above. The voltage is applied to the primary winding of
the transformer because of which the current is induced in it. The current causes the magnetic
flux which is represented by the dotted line in the above figure.
The flux induces in the primary winding because of self-induction. This flux is linked with the
secondary winding because of mutual induction. Thus, the emf is induced in the secondary
winding of the transformer. The power is transferred from the primary winding to the
secondary winding. The frequency of the transferred energy also remains same.
EMF Equation of Electrical Transformer.
The emf induced in each winding of the transformer can be calculated from its emf equation.
The linking of the flux is represented by the faraday law of electromagnetic induction. It is
expressed as,
The above equation may be written as,
P a g e | 52
Where Em = 4.44ωΦm = maximum value of e. For a sine wave, the r.m.s value of e.m.f is given
by:
The emf induced in their primary and secondary winding is expressed as,
The secondary RMS voltage is;
Where φm is the maximum value of flux in Weber (Wb), f is the frequency in hertz (Hz) and
E1 and E2 in volts.
If, Bm = maximum flux density in the magnetic circuit in Tesla (T)
A = area of cross-section of the core in square meter (m2)
P a g e | 53
The winding which has the higher number of voltage has high voltage while the primary
winding has low voltage.
Voltage Ratio and Turns Ratio
The ratio of E/T is called volts per turn. The primary and secondary volts per turns is given by
the formula
The equation (1) and (2) shows that the voltage per turn in both the winding is same, i.e.
The ratio T1/T2 is called the turn ratio. The turn ratio is expressed as:
The ratio of primary to secondary turn which equals to primary to secondary induced
voltage indicates how much the primary voltage lowered or raised. The turn ratio or induced
P a g e | 54
voltage ratio is called the transformation ratio, and it is denoted by the symbol a. Thus,
The any desired voltage ratio can be obtained by shifting the number of turns.
Ideal Transformer
Definition: The transformer which is free from all types of losses is known as an ideal
transformer. It is an imaginary transformer that has no core loss, no ohmic resistance, and no
leakage flux. The ideal transformer has the following important characteristic.
1. The resistance of their primary and secondary winding becomes zero.
2. The core of the ideal transformer has infinite permeability. The infinite permeable
means less magnetizing current requires for magnetizing their core.
3. The leakage flux of the transformer becomes zero, i.e. the whole of the flux induces in
the core of the transformer links with their primary and secondary winding.
4. The ideal transformer has 100 percent efficiency, i.e., the transformer is free from
hysteresis and eddy current loss.
The above mention properties are not possible in the practical transformer. In an ideal
transformer, there is no power loss. Therefore, the output power is equal to the input power.
P a g e | 55
Since El ∞ N2 and E1 ∞ N1, also E1 is similar to V1 and E2 is similar to V2 Therefore, the
transformation ratio will be given by the equation shown below:
The primary and the secondary currents are inversely proportional to their respective turns.
Behavior of Ideal Transformer
Consider the ideal transformer shown in the figure below:
P a g e | 56
The voltage source V1 is applied across the primary winding of the transformer. Their
secondary winding is kept open. The N1 and N2 are the numbers of turns of their primary and
secondary winding.
The current Im is the magnetizing current flows through the primary winding of the
transformer. The magnetizing current produces the flux φm in the core of the transformer.
As the permeability of the core is infinite the flux of the core link with both the primary and
secondary winding of the transformer.
The flux link with the primary winding induces the emf E1 because of self-induction. The
direction of the induced emf is inversely proportional to the applied voltage V1. The emf
E2 induces in the secondary winding of the transformer because of mutual induction.
Phasor Diagram of Ideal Transformer
The phasor diagram of the ideal transformer is shown in the figure below. As the coil of the
primary transformer is purely inductive the magnetizing current induces in the transformer
lag 90º by the input voltage V1.
P a g e | 57
The E1 and E2 are the emf induced in the primary and secondary winding of the transformer.
The direction of the induced emf is inversely proportional to the applied voltage.
Phasor Diagram of an Ideal Transformer.
Transformer Loses and Efficiency.
An ideal transformer is loss free, but in the practical transformer there are following losses
taking place:
P a g e | 58
Fig. Transformer Losses
Losses in transformer are explained below (I) Core Losses or Iron Losses
Eddy current loss and hysteresis loss depend upon the magnetic properties of the material used
for the construction of core. Hence these losses are also known as core losses or iron losses.

Hysteresis loss in transformer: Hysteresis loss is due to reversal of magnetization in the
transformer core. This loss depends upon the volume and grade of the iron, frequency
of magnetic reversals and value of flux density. It can be given by, Steinmetz formula:
Wh=ηBmax1.6fV (watts)
where, η = Steinmetz hysteresis constant
V = volume of the core in m3

Eddy current loss in transformer: In transformer, AC current is supplied to the primary
winding which sets up alternating magnetizing flux. When this flux links with
secondary winding, it produces induced emf in it. But some part of this flux also gets
linked with other conducting parts like steel core or iron body or the transformer,
which will result in induced emf in those parts, causing small circulating current in
P a g e | 59
them. This current is called as eddy current. Due to these eddy currents, some energy
will be dissipated in the form of heat.
(Ii) Copper Loss in Transformer
Copper loss is due to ohmic resistance of the transformer windings. Copper loss for the
primary winding is I12R1 and for secondary winding is I22R2. Where, I1 and I2 are current in
primary and secondary winding respectively, R1 and R2 are the resistances of primary and
secondary winding respectively. It is clear that Cu loss is proportional to square of the current,
and current depends on the load. Hence copper loss in transformer varies with the load.
Efficiency of Transformer
Just like any other electrical machine, efficiency of a transformer can be defined as the output
power divided by the input power. That is efficiency = output / input.
Transformers are the most highly efficient electrical devices. Most of the transformers have full
load efficiency between 95% to 98.5%. As a transformer being highly efficient, output and
input are having nearly same value, and hence it is impractical to measure the efficiency of
transformer by using output / input. A better method to find efficiency of a transformer is
using, efficiency = (input - losses) / input = 1 - (losses / input).
Condition for Maximum Efficiency
Let,
Copper loss = I12R1
Iron loss = Wi
P a g e | 60
Hence, efficiency of a transformer will be maximum when copper loss and iron losses are
equal.
That is Copper loss = Iron loss
Example 1.
An ideal transformer with a turns ratio of 2:7 is fed from a 240 V supply.
Determine its output voltage. A turns ratio of 2:7 means that the transformer has 2 turns on
P a g e | 61
the primary for every 7 turns on the secondary(i.e. a step-up transformer); thus (N1 /N2 ) =
(2/7). For an ideal transformer, (N1 /N2) = (V1 /V2 ) hence
(2/7) = (240/V2). Thus V2 = 240 * 7/2= 840 V.
P a g e | 62