Rural Electrification
Rural Electrification
Rural Electrification
إعداد:
ياســـــــر محمد علـــي الشـرجبي 2005/136 صالح أحمد عبدالوهاب الجنيد 2005/185
صالح محمد عبد الجبار عبد الحميد 2005/27 محمــد مسعــد قــايد الفتينــي 2005/257
حميــر قاســم ناجــي المرنعــي 2005/77
إشراف:
األستاذ الدكتور /توفيق محسن سفيـــــــــــــــــــــان
مشروع مقدم لقسم الهندسة الكهربائية وهو جزء من متطلبات نيل درجة البكالوريوس
في الهندسة الكهربائية تخصص القوى واآلت الكهربائية
2009 صنعاء
Scope
We chose this research because it is the main focus in our specialization, it is considered
as a review and covers most subjects that we studied during the five years in the collage;
the other reasons are summarized as follow:
There are more than 73.5% of the population of Yemen living in rural areas, and
70% of this population are without electricity. [Central Statistical organization,
Yemen,2003]
There are no existing studies for the alternative options for electrifying rural areas.
Rural areas need to be focused on from the point of view of electrification.
This project is practical and realistic and can be implemented.
Abstract
In this research we chose a district that consists of five villages in Dhamar governorate,
Alhada because we found information for this area, this information includes; weather,
terrain, climate and the information of the available alternatives. After that we studied
the four available alternatives (solar energy (PV), wind energy, fossil energy “Diesel”
and the extension with the national grid) in general form. Then from the information of
the available alternatives we excluded the wind energy for the following reasons:
The direction of wind speed is not stable during the seasons of the year.
Transportation process of wind farms components is very difficult such as wind
turbines due to unavailability of the roads required for this purpose.
Lack of technical or commercial skills and information: in Yemen markets, skilled
personnel who can install, operate, and maintain wind energy.
After that we estimated the approximate electrical loads in the five villages for next 20
years and with percentage development of 3% we found the average load for the five
villages to be (234.3, 83.8, 50.4, 77.8, 76.3 kVA). After load estimation we designed the
remaining three supply systems for each village.
We began in the design with the photovoltaic system by selecting the power ratings and
voltages for all required equipments and their quantities in each village. The second
design was the selection of the diesel generators where we chose the ratings of these
generators according to the available ratings in the markets and the estimated loads. The
third design was the extension of the national grid, in this design we chose five
transformers with the available ratings and according to the loads for each village and
also we chose the remaining equipments and it’s ratings after we got a dimensions map
for these villages from the Rural Electricity Sector. All designs were done at full load
condition.
After all calculations for the three alternatives, we estimated all costs for each supply
system which include the capital cost, the running cost and the cost of losses (only for
the extension of the national grid). We then calculated the total annual cost for each
system from which we estimated the cost of the energy for each system to be as follows:
Solar PV supply system (0.3535 $/kWh)
Diesel generator supply system (0.3296 $/kWh)
Extension with the national grid (0.0419 $/kWh)
After that we chose the best supply system in accordance to its reliability, its
environmental impacts and its cost factors. However, due to the loads not needing much
reliability, reliability factor was excluded from comparison as well as environmental
impacts. We found that the extension with the National Grid is the best economical
supply system to feed these villages.
Contents
Chapter 1 Introduction to Rural Electrification
2.1 Introduction 3
2.2 Importance of load estimation 3
2.3 The consideration of electrical load estimation 3
2.4 Load Estimate versus Load Calculation 4
2.5 Type of loads 4
2.6 Load curves and load duration curves 5
2.7 Installed power 5
2.8 Installed apparent power 5
2.9 Methods of estimation 5
2.10 Future expansion 10
2.11 Losses 11
2.12 References 12
3.2.1 Introduction 33
3.2.2 Wind Turbines 33
3.2.3 Wind farms 38
3.2.4 System situation 38
3.2.5 Reliability 39
3.2.6 Environmental impact 40
3.2.7 Field of applications 40
3.2.8 Advantages and Disadvantages 40
3.3 Diesel power plant 42
3.3.1 Introduction 42
3.3.2 Fuel characteristics 42
3.3.3 Fuel transportation 42
3.3.4 Principle of operation 42
3.3.5 Main Components of diesel generator 43
3.3.6 Performance of diesel generator 46
3.3.7 Diesel generator system Type 47
3.3.8 Accessories 47
3.3.9 General layout of diesel generator 48
3.3.10 System situation 48
3.3.11 Reliability 49
3.3.12 Environmental impact 49
3.3.13 Maintenance 50
3.3.14 Field of applications 50
3.3.15 Advantages and disadvantages 50
3.4.1 Introduction 51
3.4.2 Network conditions 51
3.4.3 System configurations 51
3.4.4 Type of extension 52
3.4.5 Main components of Overhead lines 52
3.4.6 Overhead Construction 59
3.4.7 Calculations 62
3.4.8 Protection 69
3.4.9 Reliability 68
3.4.10 Field of applications 68
3.4.11 Advantages and disadvantages 69
3.5 References 69
4.1 Introduction 71
4.2 Project position 71
4.3 Environment description 72
4.4 Information of the available alternatives 72
4.5 Importance of electricity for the project 76
4.6 Maximum Power demand estimation 76
4.7 Maximum Energy demand estimation 80
4.8 References 82
Chapter 5 supply systems design
5.1.1 Introduction 83
5.1.2 Sizing 83
5.4 References 97
6.1 Introduction 98
6.2 Photo voltaic supply system 98
6.3 Diesel generator supply system 102
6.4 Extend with network supply system 104
6.5 comparison 106
6.6 More economical supply system layout 108
6.7 Conclusion 109
6.8 Difficulties faced us 109
6.9 Recommendations 109
6.10 References 109
Appendix (A)
Appendix (B)
Appendix (C)
Appendix (D)
Appendix (E)
Appendix (F)
Appendix (G)
Appendix (H)
Chapter 1
Introduction To
Rural Electrification
Chapter 1 Introduction to Rural Electrification with the Best Economic Way
Energy is essential and without it societies can neither function socially nor
commercially. Without sufficient and adequate energy resources, developing countries
will not be able to foster the social and economic developments that are crucial for
growth especially in third world countries such as in our country (Yemen) where the rural
consumption does not exceed 3% [Central Statistical organization, Yemen,2003], which
is very little.
Where the provision of electricity supply to rural areas requires either the extension of the
distribution network or the establishment of standalone, decentralized networks not
connected to the national grid.
Basically, the concept of rural electrification refers to the electricity supply to areas
outside of cities or rural electrification is the process of bringing electricity to rural
communities.
Rural electrification projects bring a range of benefits to the intended communities it’s
including:
Access to lighting
The ability to use household appliances such as televisions, fans, and refrigerators
Improved access to information
The ability to use labor-saving devices for agriculture and others activities.
Improved public services such as schools, community halls and health facilities
through such benefits, electrification projects help to improve the overall quality of
life including hygiene And leisure, in addition to rural productivity.
This research consist of six chapters covered main parts for generation of power sources
and complete details about design of some power sources and describes the most
important subject that it is the rural electrification process and its benefits by using
different sources of supply that includes solar energy (photo voltaic), wind energy, fossil
fuel energy (diesel generators) or extension with national grid. And this research
determine the most economic way for generate the required power from the studied
different available alternatives and this research depend on the international codes and
standards (IEC) because that IEC used in most world countries and it is especially for 50
Hz power system countries as in our country.
The second chapter covers the electrical load estimation, which consists of importance of
loads estimation, loads estimation considerations, types of loads, load and load duration
curves, installed power and installed apparent power (KVA), the different methods for
load estimation, future expansion, and the final title is power system losses.
The third chapter covers the supply systems, which shows the whole details about four
types of energy sources they are photovoltaic system (solar energy), wind energy, diesel
energy (isolated generator) and extension with national grid as an alternative of other
alternatives.
1
Chapter 1 Introduction to Rural Electrification with the Best Economic Way
Section one describes the PV system that includes solar cells, modules, arrays and its
design, photovoltaic orientations design ,solar PV systems, solar system situation,
reliability of PV system, environmental impacts ,maintenance, field applications of PV,
and advantages and disadvantage of this type of supply system.
Section two describes wind developing, wind turbines and wind parks (wind farms),
system situations, reliability, environmental impacts, field of applications, and
advantages and disadvantages of wind energy.
Section three explains diesel generators that includes fuel characteristics, fuel
transportation, principle of operation ,main components of diesel generators, performance
of diesel generators, diesel generator system types, accessories of diesel power station,
system situations, reliability, environmental impacts, maintenance ,field of applications,
and advantages and disadvantages of diesel energy.
Section four describes the extension with the national grid and includes network
conditions, system configurations, type of extension, main components of over head
lines, over head construction, calculations(voltage selection, voltage regulation, voltage
drop, load to be supplied, fault analysis, losses),protection, reliability, operating and
maintenance, field applications, and advantages and disadvantages.
The fourth chapter covers the required information for the intended villages, which
includes project position, environment description, information of the available
alternatives, and importance of electricity for the project, maximum power demand
estimation, and maximum energy estimation.
The fifth chapter contains all supply systems design:
Part one solar PV system that includes sizing (PV arrays, Inverters, batteries, load
growth), Part two diesel power plant that includes sizing, Fuel consumption, lubricating
oil, Accessories, Part three extension with the national grid that covers sizing (Voltage
selection, choosing transformers, choosing of fuses, Choosing Conductors, Choosing
poles, Choosing insulators, Choosing of surge arresters, Choosing accessories),spacing
(clearance between poles),and calculation (voltage drop, power losses). All supply
system designs are designed under full load condition that's mean after 20 years.
The sixth chapter shows the economic evaluation for the different designs and includes
capital cost, running cost, annual cost, cost of electricity, cost of losses for the network,
and the final comparison between all alternatives.
2
Chapter 2
Electrical
Load Estimation
Chapter (2) Electrical load estimation
2.1 Introduction
In order to design any installation, the actual maximum load demand likely to be
imposed on the power-supply system must be assessed. To base the design simply on
the arithmetic sum of all the loads existing in the installation would be extravagantly
uneconomical, and bad engineering practice. The aim of this chapter is to show how
some factors taking into account, the diversity (non simultaneous operation of all
appliances of a given group) and utilization (e.g. an electric motor is not generally
operated at its full-load capability, etc.) of all existing and projected loads can be
assessed. The values given are based on experience and on records taken from actual
installations. In addition to providing basic installation-design data on individual
circuits, the results will provide a global value for the installation, from which the
requirements of a supply system (distribution network, MV/LV transformer, or
generating set) can be specified.
If all the lights, irons, refrigerators, televisions, motors, and other equipments in all
the consumers‟ premises were to be turned on at one time, it would be impossible for
the utility company to supply the energy necessary to run all these appliances. For a
company to install generators, transformers, transmission and distribution lines, and
other equipment to provide for this total connected load would be extremely
uneconomical since all this equipment will not be operating at the same time.
The electrical design professional should determine a building‟s electrical load
characteristics early in the preliminary design stage of the building to select the
proper power supply system and equipments having adequate power capacity with
proper voltage levels, and sufficient space and ventilation to maintain proper
ambient. The most feasible method of supplying and distributing electrical power will
be determined by first quantifying the electrical power requirements (or maximum
demand load) for the installation. In the early design stages, this demand should be
based on area or population; in later design stages, summation of individual building
connected loads modified by suitable demand and diversity factors will be used.
3
Chapter (2) Electrical load estimation
4
Chapter (2) Electrical load estimation
5
Chapter (2) Electrical load estimation
This method is the wide spread method in load estimation, because it‟s easy to change
its tables to gives similarity with weather conditions and it gives more accuracy than
other methods for calculating load demand, this method is used to estimate the
maximum power demand for un similar loads, it includes some factors as following:
Demand factor (kd)
In normal operating conditions the power consumption of a load is sometimes less
than that indicated as its nominal power rating, where this factor must be applied to
each individual load, with particular attention to electric motors, which are very rarely
operated at full load. In an industrial installation this factor may be estimated on an
average at 0.75 for motors.
For incandescent-lighting loads, the factor always equals one. For socket-outlet
circuits, the factors depend entirely on the type of appliances being supplied from the
sockets concerned, the tables below show the demand factor of some domestic load
appliances and general lighting of some costumer loads.
For example
The maximum kW demand of Customer at some moment was found to be 6.18 kW,
in order to determine the demand factor; the total connected load of the customer
needs to be known. The total connected load will be the sum of the ratings of all of
the electrical devices at the customer‟s location. Assume that this total comes to 35
kW; then, the demand factor is computed to be
Maximum demand
Demand factor (2-1)
Total connected load
6
Chapter (2) Electrical load estimation
The simultaneous operation of all installed loads of a given installation never occurs
in practice, the factor (Ks) is applied to each group of loads. The determination of this
factor is the responsibility of the designer, since it requires a detailed knowledge of
the installation and the conditions in which the individual circuits are to be exploited.
For this reason, it is not possible to give precise values for general application.
Factor of simultaneity for an apartment block
Some typical values for this case are given in table (2.3), they are applicable to
domestic consumers supplied at 230/400 V (3-phase 4-wires). In the case of
consumers using electrical heat-storage units for space heating, a factor of 0.8 is
recommended regardless of the number of consumers.
For example
Assume 5 stories (36, 24, 30, 36, 24) kVA apartment building with (25) consumers,
each having (6 kVA) of installed load.
The total installed load for the building is: 36 + 24 + 30 + 36 + 24 = 150 kVA
From table (2.2), for (25) consumers (KS = 0.46), so the apparent-power supply
required for the buildings is: 150 × 0.46 = 69 kVA
Factor of simultaneity for distribution boards
Table (2.3) shows hypothetical values of (Ks) for a distribution board supplying a
number of circuits for which there is no indication of the manner in which the total
load divides between them. If the circuits are mainly for lighting loads, it is prudent to
adopt (Ks) values close to unity. [2]
7
Chapter (2) Electrical load estimation
8
Chapter (2) Electrical load estimation
Average demand
Load factor (2-4)
Max. kW demand
The above formula is used to find the load factor.
Sometimes utility companies will encourage industrial customers to improve their
load factors. One method of encouragement is to penalize the customer on the electric
bill for having a low load factor.[1]
9
Chapter (2) Electrical load estimation
Where:
E is the energy consumption
N is the number of houses, mosques, schools or healthy centers
m is the number of the loads
2.9.3 Using power/area method
In this method we use standard tables to give how much watt per square meter are
used. First step is to determine the layouts of the location which we estimate its load,
then use the standard tables to find the load for this location, such as houses, hospitals
or schools. It can be used for specific circuit such as lighting. [5]
For example
The room area is (25m2) and from the tables the lighting is need (6W/m2), then the
room needs total power (25×6) = (150W).
There are general tables for all circuits such as lighting, air conditioning, pumping,
heating, etc. These tables give the total power which the consumers need.
For example
The house area is (400m2) and from tables the consumer needs (12.5W/m2) for all
circuits, then the house needs total power (400×12.5) = (5000W).
Because the table was designed for urban loads and for specified weather conditions,
this method is not used in rural design.
2.10 Future expansion
The estimated load by using the above methods is not the final load which we design
the generations or substations according to it, because of the population growth and
10
Chapter (2) Electrical load estimation
load expansion. For this reason the designers must be taking into account the future
expansion and the period of the project in years.
Future expansion is given by a percentage of country development, and then this
percentage is multiply by the first load and too sum with first load.
For example
If the total load of the consumers (50kVA), future expansion percentage (4%) and the
period of the project 20 years, then the installed power is equal to (50×0.04×20+50)
= (90kVA) by Ignoring distribution and transmission losses.
2.11 Losses
2.11.1 Loss Allocation Process
IEC 61968 standard calculates supplier hourly loads, including distribution and
transmission losses, in two step process. First, the hourly loads of each retail
customer are estimated at the customer‟s meter and multiplied by one of three
distribution loss factors, depending upon the customer‟s rate class. Customer hourly
loads including these distribution loss factors are then summed by supplier and the
total is compared to Distribution Company “Delivered load.” Delivered load is
defined as load measured at substation and tie-line metering points inside the IEC
61968 standard. Delivered load includes all distribution line losses but no
transmission losses. Differences between the sum of estimated customer hourly
loads, including distribution loss factors, and actual delivered loads are allocated to
suppliers based on their share in total estimated load. These differences can be
positive or negative and vary by hour, but their expected value is zero.
Second, IEC 61968 standard calculates transmission losses as the difference between
System load and delivered load. [6]
11
Chapter (2) Electrical load estimation
2.12 References
1. Nodal Load Estimation for Electric Power Distribution Systems-July 2003
2. Electric load forecasting: literature survey and classi®cation of methods, Hesham
K. Alfares* and Mohammad Nazeeruddin; International Journal of Systems
Science, 2002, volume 33, number 1, pages 23±34 Calculating Total power
requirements for data center- By Richard Sawyer
3. Electrical installed guide according; Dr. Khaled Y. Al-Khalaf Vice Chairman,
Board of Directors and Director General, SASO.
4. Electrical engineering portable handbook; ROBERT B. HICKEY, P.E.Electrical
Engineer, second edition2002.
5. ELECTRIC POWER application, engine& GENERATOR SIZING 2004.
6. Supplier load estimation allocation of losses to suppliers;Scovild;2003
12
Chapter 3
Supply Systems
Chapter 3 supply systems
13
Chapter 3 supply systems
When they are joined, the junction approaches a new thermodynamic equilibrium.
Such equilibrium can be achieved only when the Fermi level is equal in the two
materials. This occurs by the flow of electrons from one material to the other until a
voltage difference is established between the two materials which have the potential
just equal to the initial difference of the Fermi level. This potential drives the
photocurrent.
14
Chapter 3 supply systems
the current delivered to the external load equals the current I L generated by the
illumination, less the diode current ID and the ground-shunt current Ish. [7]
The open circuit voltage Voc of the cell is obtained when the load current is zero, i.e.,
when I = 0, and is given by the following equation:
IL Rs
I
Id Ish
Rsh V
Equivalent electrical circuit of PV module, showing the diode and ground leakage
currents, the diode current is given by the classical diode current expression:
AQKVocT
I d I D e
1 (3-2)
Where:
ID is the saturation current of the diode
Q is the electron charge = 1.6 · 10–19 Coulombs
A is the curve fitting constant
K is Boltzmann constant = 1.38 · 10–23 Joule/Ko
T is the temperature on absolute scale Ko
The load current is therefore given by the expression:
QVoc V
I I L I D e A K T 1 oc (3-3)
Rsh
Ish is the ground-leakage current, in practical cells is small compared to IL and ID, and
can be ignored. The diode-saturation current can, therefore, be determined
experimentally by applying voltage Voc in the dark and measuring the current going
into the cell or exit in manufactures data. This current is often called the dark current
or the reverse diode-saturation current.[5]
Open Circuit Voltage and Short Circuit Current:
The two most important parameters widely used for describing the cell electrical
performance is the open-circuit voltage Voc and the short circuit current Isc. The
short-circuit current is measured by shorting the output terminals, and measuring
15
Chapter 3 supply systems
the terminal current under full illumination. Ignoring the small diode and the ground-
leakage currents under zero-terminal voltage, the short-circuit current under this
condition is the photocurrent IL. The maximum photo voltage is produced under the
open-circuit voltage. Again, by ignoring the ground-leakage current, equation (3-3)
with I = 0 gives the open-circuit voltage as the following:
A K T IL
Voc Log n
I 1
(3-4)
Q D
The constant (K×T/Q) is the absolute temperature expressed in voltage (300K o =
0.026 volt).
Figure (3.3) Current versus voltage (I-V) characteristics of the PV module in sunlight
and in dark [3]
This is the current we would measure with the output terminals shorted (zero voltage).
The bottom right of the curve at zero current is called the open-circuit voltage. This is
the voltage we would measure with the output terminals open (zero current). In the
left shaded region, the cell works like a constant current source, generating voltage to
match with the load resistance. In the shaded region on the right, the current drops
rapidly with a small rise in voltage. In this region, the cell works like a constant
voltage source with an internal resistance. Somewhere in the middle of the two shaded
regions, the curve has a knee point. If the voltage is externally applied in the reverse
direction, say during a system fault transient, the current remains flat and the power is
absorbed by the cell. However, beyond a certain negative voltage, the junction breaks
down as in a diode and the current rises to a high value. In the dark, the current is zero
for voltage up to the breakdown voltage which is the same as in the illuminated
condition. [3]
3.1.3 Module
The solar cell described above is the basic building block of the PV power system.
Typically, it is a few square inches in size and produces little over 0.4 Vdc in bright
16
Chapter 3 supply systems
sunshine and about one watt of power. For obtaining higher voltages and high
power, numerous such cells are connected in series and parallel circuits on a panel
(module) area of several square feet. The actual construction parts of a module are:
17
Chapter 3 supply systems
I sc I ( 1 α ΔT) (3-5)
P Po 1 (α β) ΔT (3-8)
The three different solar modules available on the market each have different
temperature Coefficients these are:
1) Monocrystalline Modules: Monocrystalline Modules typically have temperature
coefficient (0.45% /Co). That is, for every degree above 25Co the output power is
derated by 0.45%
2) Multicrystalline Modules: Polycrystalline Modules typically have temperature
coefficient (0.5% /Co).
3) Thin Film Modules: Consult the manufacturer’s data.
The de-rating of the array due to temperature will be dependent on the type of module
installed and the average ambient maximum temperature for the location.
The temperature de-rating factor is calculated as follows:
F Temp 1 T (3-9)
Where:
FTemp = temperature de-rating factor, (dimensionless).
(3-10)
Power temperature coefficient per degree Celsius (see above)
18
Chapter 3 supply systems
3.1.4 Array
The solar array or panel is defined as a group of several modules electrically
connected in series-parallel combinations to generate the required current and voltage.
The (I-V) curve at 1000 W/m2 is for the module that faces the sun directly.
1000 W/m2
1.0
0.8
Current (A)
0.6
500 W/m2
0.4
0.2
100 W/m2
0 4 8 12 16 20
Voltage (V)
19
Chapter 3 supply systems
Figure (3.4) the effect that light intensity has on the I-V curve for a module [3]
Therefore the current from a module fixed in one position varies through the day;
even when the weather is clear with no clouds. The charging current rises during
the morning, reaches its highest level at midday, and then fall during the
afternoon. Most charging of a battery happens over a few hours in the middle of
the day.
To extend the period of charging, the angle of module can be changed during the
day so that it faces the sun more closely for longer. Changing the angle is called
tracking.
When the sun is covered by clouds, light still reaches the ground and some
current is generated by the module. With a thin layer of clouds, the irradiance
reaching the module from all part of the sky might be as much as 300 W/m2.
With thicker cloud cover, the irradiance could be reduced to100 W/m2 or less.
The (I-V) for 100 W/m2 in figure (3.4) is very low. This shows that with thick
cloud cover, the current at 16 V is almost zero and there is very little charging.
Module mismatch and wiring losses: The maximum power output of the total
PV array is always less than the sum of the maximum output of the individual
modules. This difference is a result of slight inconsistencies in performance from
one module to the next and is called module mismatch and amounts to at least a
2% loss in system power. Power is also lost to resistance in the system wiring.
These losses should be kept to a minimum but it is difficult to keep these losses
below 3% for the system. A reasonable reduction factor for these losses is 95%.
The operating temperature: This factor as mentioned above in module performance.
Effect of Climate: On a partly cloudy day, the PV module can produce up to 80
percent of their full sun power. Even on an extremely overcast day, it can
produce about 30 percent power. Snow does not usually collect on the modules,
because they are angled to catch the sun. If snow does collect, it quickly melts
mechanically.[1]
E PV
P PV (3-13)
H K1 K 2
20
Chapter 3 supply systems
PPV N
NM , N MP M (3-17)
PMax N MS
If we have gotten non integer number we round it to integer number and calculate
again the new value of the modules we have to purchase them after that we calculate
the exact new array size by using the following formulas:
N MN N MS N MP
21
Chapter 3 supply systems
22
Chapter 3 supply systems
roof to roof. If economic, installing a separate inverter for each section of the
array which has the same orientation and angle will maximize the output the
total array.
Multiple inverters allow a portion of the system to continue to operate if one
inverter fails.
Allows the system to be modular, so that increasing the system involves the
adding a predetermined number of modules with one inverter (allowing the
system to growth). The unique disadvantage of multiple inverters is that in
general, the cost of a number of inverters with lower power ratings is generally
more expensive then one single inverter with a higher power rating.
3.1.6.5 Batteries
One of the main components in PV system is the energy storage facility. The need for
storage arises when electricity is required during night time and on cloudy days. the
basic element for electrical energy storage in the PV system is the battery. The types
and characteristics of batteries that can be used will be discussed below. Most
professional photovoltaic system incorporates stationary application lead-acid
batteries (sealed battery construction).
There are many Types of Batteries:
a) Nickel-Cadmium
b) Lithium-Ion
c) Lead-acid
Lead-acid battery
There are two main types of lead acid batteries. The first type is the flooded/wet cell battery.
The second type of lead acid battery is the Value Regulated Lead Acid battery
(VRLA). This designation stands for maintenance-free lead-acid batteries. Two types
of VRLA batteries are the Absorbed Glass Mat (AGM) and Gel cell batteries. The gel
cell batteries are considered the most commonly used type.
A gel battery is another type of VRLA battery with a jellified electrolyte. Unlike a
traditional wet-cell lead-acid battery, these batteries do not need to be kept upright. In
addition, gel batteries virtually eliminate the electrolyte evaporation, spillage common
to the wet-cell battery, and boast greater resistance to extreme temperatures, shock,
and vibration. As a result, they are primarily used in automobiles, boats, aircraft, and
other motorized vehicles. These batteries are often referred to as sealed lead-acid
(SLA) batteries due to their non-leaking containers; chemically they are the same as
wet batteries except the lead plates are replaced by calcium. This preserves the
mechanical characteristics but helps ensure that gas does not escape from the battery.
The sealed battery construction despite its high cost is now becoming more widely
used especially in PV system due to the advantages of low maintenance as it requires
no replenishment of electrolyte. It is easier to install as the stationary electrolyte
eliminates the possibility of acid leakage and has high efficiency. [10]
The Battery Bank
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Chapter 3 supply systems
The basic building block of a lead-acid battery is a 2-volt cell. A battery bank is a
collection of connected 2, 6, or 12-volt batteries that supply power to the load in case
of outages or low production from renewable energy sources. The batteries are wired
together in series to produce 12, 24, or 48-volt strings. These strings are then
connected together in parallel to make up the entire battery bank. The battery bank
supplies DC power to an inverter, which produces AC power that can be used to run
appliances. The decision to select a 12, 24, or 48-volt battery bank will be determined
by the inverter‘s input, the type of battery you select, and the amount of energy
storage you require.
Size of batteries
To determine the number of batteries you need, you must first determine how much
energy storage you need in kilowatt-hours (kWh). If you are connected to the utility
grid, you can use your monthly utility bill to calculate past energy usage for your
household. A second way to determine your required kWh of energy storage is to
multiply the wattage of your appliances by the number of hours you use them in a
day.
Power for batteries that have to be stored in them for whether conditions:
E B (Wh/days)
C Ah (3-20)
V Bb CC B DOD
And the number of required batteries can be calculated using the following equations:
C Ah
N BP (3-21)
B Ah
V Bb
N BS (3-22)
VB
N B N BS N BP (3-23)
Batteries items
The maximum depth of discharge (DOD): refers to the lowest point to which the
battery can be discharged before having to be recharged; where DOD often
ranges from 20 to 80 %.
Days of autonomy (DOA): the number of days of those batteries will supply the
loads instead of PV in cloudy days and this may be safely taken to be 2 to 3 days
per week this depends on where the system is to be located especially in remote
areas.
Self discharge: it is a process that occurs in every battery and is the gradual loss
of available energy even when the battery is not connected to a load. The self
discharge may be irreversible (i.e. the capacity of the battery cannot be
recovered) or reversible (i.e. when it is recharged the battery will have the same
24
Chapter 3 supply systems
25
Chapter 3 supply systems
state that the voltage drop should not exceed 2.5% of the nominal voltage when the
conductors are carrying the full load current. The 2.5% limit is 6v in a mains voltage
of 240v. For solar systems the voltage drop is not set by any statutory regulations but
depends on the operating range of the battery supply and the appliances. Considering
the variation in voltage of the supply, lead acid batteries with nominal voltage of 12v
actually supply between 11.5 and 12.5v depending on the state of charge and how
much current is being drawn. The upper limit rises to 15v when they are being charge
during the day. The maximum acceptable voltage drop a long a cable is the difference
between the minimum supply voltage and the minimum voltage that appliances need.
But in order to ensure high reliability and effective operation of all appliances the
voltage drop that should be used is 5% which corresponds to 0.6v for 12v systems and
1.2v for 24v systems and it is considered the international standard.
Cable size from solar modules to batteries: the cable carrying the charging current
from the solar modules to the batteries must also be selected according to voltage drop
that has to be kept to 5%.
For one module, the short circuit current under STC (Standard Testing Conditions)
should be used for the maximum current in the below formula. When there is an array
of modules with a single cable leading to the batteries the short circuit current of each
module is combined using the following formula:
I max N pv I S.C.pv (3-24)
And the cross section area is calculated by the following formula:
3.1.7 Protection
3.1.7.1 Overcurrent Protection of PV Systems
Fuses and a circuit breaker are used to provide protection for the PV modules, charge
controller, inverter, batteries and appliances against system faults. Fuse sizes are
usually determined on the basis of fault currents which exceed rated currents by (20 –
56%). Necessary fuses or circuit breakers must be properly sized and specified for
each circuit. The IEC standard defines the maximum circuit current as 125% of the
short-circuit current of the PV module (Isc). The conductors and the over current
protective device are then sized at 125% of the maximum circuit current. [5]
Selecting Fuse Links for PV String Protection
26
Chapter 3 supply systems
Depending on the desired capacity of the PV system, there may be several PV strings
connected in parallel to achieve higher currents and subsequently more power.
Procedures of selecting of string fuse:
1. Define solar panel specifications: Ipm , Isc, max system voltage
2. Define conductor/fuse size per string 1.25 × Isc = I conductor/fuse rating
withstand.
3. Define number of parallel solar panels/strings N = parallel panels/string.
4. PV String Fuse In= 1.25 × Isc select equal to or next higher standard fuse rating
PV String Fuse Vn => max system voltage
Battery over current protection
It must have a sufficient voltage and ampere-interrupt rating (AIR) to withstand the
operating conditions of the battery system.
Inverter Output Circuit over current protection
It should be sized and protected according the manufacturer's directions. The circuit
and corresponding over current protection should be sized at a 125% of the maximum
continuous output of the inverter [Over current for Feeder Circuits].
27
Chapter 3 supply systems
PV array wiring should be done with minimum lengths of wire, tucked into the metal
framework, then run through metal conduit. Positive and negative wires should be
run together wherever possible, rather than being some distance apart. This will
minimize induction of lightning surges. Bury long outdoor wire runs instead of
running them overhead. Place them in grounded metal conduit if you feel you need
maximum protection.
Surge protection devices bypass the high voltages induced by lightning. They are
recommended for additional protection in lightning-prone areas or where good
grounding is not feasible (such as on a dry rocky mountain top), where Surge
protectors must be special for low voltage systems.
Solar array
Charger Power
Controller inverter
Load
Batteries
Busbar
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Chapter 3 supply systems
with a plan known as Net Metering. At times when you are not using all of the
electricity produced by your system, your meter will spin backwards selling the
electricity back to the grid at retail rate. These systems do not include a battery. Power
is obtained from the utility grid when the system is not producing electricity.
Solar array
Grid
Charger Power
Controller inverter
Load
Batteries
Busbar
3.1.9 Reliability
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Chapter 3 supply systems
Modern solar electric systems have been shown to be very reliable. With no moving
parts, the reliability depend on warranties of key components and some others factors
that affect the operation of these components.
Most solar photovoltaic modules have warranties of 25 years and there are various
factors effects solar modules performance and reduce its efficiency, they previously
mentioned in de-rating module performance in section (3.1.3). Modern inverters also
have minimum warranties of 5 years with some manufacturers increasingly offering
10 year warranties. Realistically, there will be some solar modules that fail "out of the
box", but these are spotted immediately by the installer. Some inverters have also
failed quickly, but are also quickly replaced. And the whole Solar photovoltaic
systems are given service lifetimes between 30-50 years.
The reliability of the PV system is also determined by the design of arrays, inverters
and the batteries. For more reliability the arrays are divided into various arrays where
if one of the existing arrays has a failure however the remaining arrays continue
supplying the load without any barriers and the same thing has done for the inverters
as mentioned in section (3.1.6.3) and the batteries, But we can‘t forget the high cost
for this design but it is necessary for more reliable so the PV system is considered
from the most reliable systems.
3.1.11 Maintenance
A solar electric system requires a regular maintenance to ensure proper operation and
the full life of the components.[9]
30
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31
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Disadvantages
The biggest barriers to increasing solar power generation are the cost, the
amount of land required for large-scale electricity production, and the
intermittent nature of the energy source.
In terms of the latter, thermal systems do not work at night or in inclement
weather. Storage of hot water for domestic or commercial use is simple,
needing only insulated tanks, but storage of the higher-temperature liquids
needed to generate electricity on a large scale or storage of the electricity itself
requires further technological development.
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Pt 0.5 CP A V 3 (3-26)
33
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Where is the air density (1.225 kg/m3), A is the swept area of the turbine and V
is the wind velocity, The Cp parameter is called the power coefficient It has a
theoretical maximum value of 0.593 (the Betz limit) and rather lower peak values are
achieved in practice and is dependent on the ratio between the linear velocity of
the blade tip (r t ) and the wind velocity (V ) . This ratio, known as the tip-speed
ratio, is defined as
(r Wt )
(3-27)
V
Where 'r' is the radius of the turbine. [1]
Gear-box
transmission
High-speed
Housing shaft
Blades Tower
Rotor
The portion of the wind turbine that collects energy from the wind is called the rotor.
The rotor usually consists of two or more wooden, fiberglass or metal blades which
rotate about an axis at a rate determined by the wind speed and the shape of the
blades. The blades are attached to the hub, which in turn is attached to the main shaft.
The rotor is converting the wind energy to mechanical energy.
Drag Design: Blade designs operate on either the principle of drag or lift. For the
drag design, the wind literally pushes the blades out of the way. Drag powered wind
34
Chapter 3 supply systems
turbines are characterized by slower rotational speeds and high torque capabilities.
They are useful for the pumping, or grinding work, For example, a farm-type
windmill must develop high torque at start-up in order to lift water from a deep well.
Lift Design: The lift blade design employs the same principle that enables airplanes,
and birds to fly. The blade is essentially an airfoil, or wing. When air flows past the
blade, a wind speed and pressure differential is created between the upper and
lower blade surfaces. The pressure at the lower surface is greater and thus acts to
"lift" the blade. When blades are attached to a central axis, like a wind turbine rotor,
the lift is translated into rotational motion. Lift-powered wind turbines have much
higher rotational speeds than drag types and therefore well suited for electricity
generation.
Generators
The generator is what converts the mechanical energy of a wind turbine's blades into
electricity. The generator's rating, or size, is dependent on the length of the wind
turbine's blades because more energy is captured by longer blades. An important
step for installation of wind energy system is to select the turbine rating, the
generator, and the distribution system. In general, the output characteristics of the
wind turbine power do not follow exactly those of the generator power; so they
have to be matched in the most reasonable way possible. Based on the maximum
speed expected for the turbine and taking into account the cubic relationship
between the wind speed and the generated power, the designer must select the
generator and the gearbox so as to match these limits. The most sensitive point here
is the correct selection of the rated speed for the generator. If it is too low, the high
speed of the primary source wind will be wasted; if it is too high, the power factor
will be harmed. The characteristics of the commercially available turbines and
generators must be matched to the requirements of the project with regard to cost,
efficiency, and maximum generated power in an iterative design process. Several
types of generators can be coupled to the rotating wind power turbines: dc and ac
types, parallel and compound dc generators, with permanent magnets or electrical
field excitation, synchronous or non-synchronous, and, especially, induction
generators. The dc machines are not usually employed because of their high cost,
bulky size, and maintenance needs. The right choice of generator depends on a wide
range of factors related to the primary source, the type of load, and the speed of the
turbine. Besides, systems differ with respect to their applications, whether they are
stand-alone or connected to the grid or batteries, their degree of interrupt ability,
and the quality and cost of their output. Because of the way it works as a motor or
generator, the possibility of variable speed operation, and its low cost compared to
other generators. [2]
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Chapter 3 supply systems
The induction machine offers advantages for rotating power plants, like the wind
power, in both standalone and interconnected applications, Because they are widely
and commercially available and very inexpensive. It is also very easy to operate them
in parallel with large power systems, because the utility grid controls voltage and
frequency while static and reactive compensating capacitors can be used for
correction of the power factor and harmonic reduction. Although the induction
generator is mostly suitable for hydro and wind power plants, it can be efficiently
used with prime movers driven by diesel, biogas, natural gas, gasoline and alcohol
motors. Induction generators have outstanding operation as either motor or
generator; they have very robust construction features, providing natural protection
against short-circuits, and have the lowest cost with respect to other generators.
Abrupt speed changes due to load or primary source changes, as usually expected in
small power plants, are easily absorbed by its solid rotor, and any current surge is
damped by the magnetization path of its iron core without fear of demagnetization,
as opposed to permanent magnet based generators. [4]
Transmission
The number of revolutions per minute (rpm) of a wind turbine rotor can range
between 40 rpm and 400 rpm, depending on the model and the wind speed.
Generators typically require rpm's of 1,200 to 1,800. As a result, most wind turbines
require a gear-box transmission to increase the rotation of the generator to the
speeds necessary for efficient electricity production.[3]
Towers
The tower on which a wind turbine is mounted is not just a support structure; it also
raises the wind turbine so that its blades safely clear the ground and so it can reach
the stronger winds at higher elevations. Larger wind turbines are usually mounted
on towers ranging from 40 to70 meters tall. [3]
Tip Speed Ratio
The tip-speed is the ratio of the rotational speed of the blade to the wind speed. The larger this
ratio, the faster the rotation of the wind turbine rotor at a given wind speed. [3]
36
Chapter 3 supply systems
case, a tail vane is usually used to keep the blades facing into the wind. Other
designs operate in a downwind mode so that the wind passes the tower before
striking the blades. Without a tail vane, the machine rotor naturally tracks the wind
in a downwind mode. Some very large wind turbines use a motor-driven mechanism
that turns the machine in response to a wind direction sensor mounted on the
tower.
Figure (3.9) downwind turbine Figure (3.10) upwind turbine Figure (3.11)
Vertical Axis
Although vertical axis wind turbines have existed for centuries, they are not as
common as their horizontal counterparts. The main reason for this is that they do
not take advantage of the higher wind speeds at higher elevations above the ground
as well as horizontal axis turbines.
37
Chapter 3 supply systems
Figure (3.12)
Betz Limit
It is the flow of air over the blades and through the rotor area that makes a wind
turbine function. The wind turbine extracts energy by slowing the wind down. The
theoretical maximum amount of energy in the wind that can be collected by a wind
turbine's rotor is approximately 60%. This value is known as the Betz limit. In
practice, the collection efficiency of a rotor is not as high as 59%. A more typical
efficiency is 35% to 45%. A complete wind energy system, including rotor,
transmission, generator, storage and other devices, which all have less than perfect
efficiencies, will (depending on the model) deliver between 10% and 30% of the
original energy available in the wind. [1]
Dcw
Dr
Ddw
38
Chapter 3 supply systems
CB
T
Utility
Turbine
Figure (3.14) 0ne line diagram of wind farm electrical network [2]
Batteries
Figure (3.15) Load
39
Chapter 3 supply systems
from the grid to meet the demand. However, in times of good wind speed and low
power
demand, it is possible for additional generation from the turbine to be fed back into
the electricity grid. [2]
Grid
Synchronizing
breaker
Rectifier Filter Inverter
Figure (3.16)
Load
3.2.5 Reliability
The availability of a wind farm, defined as the percentage of time it is able to produce
electricity, is function of the reliability, maintainability and serviceability of the hard
and software used in the whole system.
The required reliability levels to achieve sufficient availability have been investigated
in Figure (3.17). It shows for various levels of reliability of the technology how the
wind farm availability varies as a function of the accessibility (percentage of the time
in which the plant is available for service operations, depending on weather
circumstances).
Today onshore wind turbines reliability levels result in an estimated availability of
only due to the limited access. Improved technology will lead to availability of 88 %.
In order to increase availability to 98 %, the technology should be highly improved
technology, see figure (3.17). Wind turbines only produce energy when the wind is
blowing, and energy production varies with each gust of wind. The variable forces
acting on a wind turbine throughout its expected lifetime of 120,000 operating hours
could be expected to exert significant.
High quality modern wind turbines have an availability factor above 98 percent, i.e.
the turbines are on average operational and ready to run during more than 98 percent
of the hours of the year. This availability factor is beyond any other electricity
generating technology.
40
Chapter 3 supply systems
Figure (3.17)
Figure (3.17) show the relationships between availability and accessibility for
different grades of wind turbine reliability [5]
41
Chapter 3 supply systems
42
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43
Chapter 3 supply systems
Principle of operation:
The diesel internal combustion
engines use a higher compression of
the air to ignite the fuel. In the
diesel engine, only air is introduced
into the combustion chamber. The
air is then compressed with a
compression ratio typically between
15 and 22 resulting into a 40 bar
pressure. This high compression
heats the air to 550 °C. At about this
moment (the exact moment is
determined by the fuel injection Figure (3.19)
timing of the fuel system), fuel is injected directly into the compressed air in the
combustion chamber. The fuel injector ensures that the fuel is broken down into small
droplets, and that the fuel is distributed as evenly as possible. The heat of the
compressed air vaporizes fuel from the surface of the droplets. The vapour is then
ignited by the heat from the compressed air in the combustion chamber, the droplets
continue to vaporize from their surfaces and burn, getting smaller, until all the fuel in
the droplets has been burnt. The start of vaporization causes a delay period during
ignition, and the characteristic diesel knocking sound as the vapour reaches ignition
temperature and causes an abrupt increase in pressure above the piston. The rapid
expansion of combustion gases then drives the piston downward, supplying power to
the crankshaft. As well as the high level of compression allowing combustion to take
place without a separate ignition system, a high compression ratio greatly increases
the engine's efficiency. [2]
44
Chapter 3 supply systems
Electrical generator is used to convert the mechanical power on the shaft to electrical
power. Electric generators may be classified many ways, but the following are
deemed as fully representative:
By application domain
By principle
The application domain implies the power level. The
classifications by principle unfolded here include commercial
(widely used) types together with new configurations, still in
the laboratory (although advanced) stages.
By principle, there are three main types of electric generators:
Synchronous
Induction
Figure (3.20)
Parametric, with magnetic anisotropy and permanent magnets
Parametric generators have in most configurations doubly salient magnetic
circuit structures, so they may be called also doubly salient electric generators.
In diesel generator, synchronous generators are used.
Larger industrial generators can range from 8kVA - 30kVA for homes, small shops
and offices up to 2000kVA used for large office complexes, factories. [1]
Generator Ratings and Sizing
Rating: Generators must be capable of delivering the power required for the hours per
day anticipated by the designer to allow reliable operation and prevent damage.
Typically a given set can deliver more power for fewer hours per day, or less power
continuously. That is a standby set is only expected to give its peak output for a few
hours per day, whereas a continuously running set, would be expected to give a
somewhat lower output. To meet the above criteria manufactures give each set a
rating based on internationally agreed definitions. These standard rating definitions
are designed to allow correct machine selection and valid comparisons between
manufacturers to prevent them from misstating the performance of their machines,
and to guide designers.
Sizing: Typically it is the size of the maximum load that has to be connected and the
acceptable maximum voltage drop which determines the set size, not the maximum
load. If the set is required to start motors, then the set will have to be at least 3 times
the largest motor, which is normally started first. This means it will be unlikely to
operate at anywhere near the ratings of the chosen set.
3.3.5.3 Governor
The purpose of a governor is to control the speed of an engine. If an
engine is loaded beyond its rated capacity, it will slow down or may
45
Chapter 3 supply systems
even stop. Governors act through the fuel injection system to control the amount of
fuel delivered to the cylinders. The quantity of fuel delivered, in turn, governs the
power developed.
Description and operation
The principal parts of the governor are shown in the figure (3.23) when the engine is
running at the speed set on the governor; the land on the pilot valve plunger covers the
regulating port in the bushing. The plunger is held in this position by the flyweights.
However, if the engine loads decreases, the engine speeds up and the Figure (3.21)
additional
centrifugal force move the flyweights outward raising the pilot valve plunger. This
opens the regulating port of the bushing, and trapped oil from the power cylinder is
then allowed to flow through the pilot valve cylinder into a drainage passage to the oil
sump. As the trapped oil drains to the oil sump, the power spring forces the piston
down, actuating the linkage to the
fuel system controls, and the
supply of fuel to the engine is
diminished. As the engine speed
returns to the set rate, the
flyweights resume their original
position and the, pilot valve
plunger again covers the regulating
port. If the load increases, the
engine slows down, and the
flyweights move inward. This
lowers the pilot valve plunger,
allowing pressure oil to flow
through the pilot valve chamber to
the power cylinder. This oil
supplied by a pump is under a
pressure sufficient to overcome the
pressure of the power spring. The
power piston moves upward,
actuating the linkage to increase Figure (3.22)
the amount of fuel injected into the
engine cylinders. Once again, as the speed returns to the set rate, the flyweights
resume their central position. [1]
46
Chapter 3 supply systems
engine to determine its performance characteristics, such as: Indicated power (I.P),
Brake power (B.P), Frictional power (F.P), Mechanical efficiency (ηm), Indicated
thermal efficiency (ηi), fuel consumption and also specific fuel consumption etc.
Below, we shall discuss how these quantities are measured. [2]
Indicated mean effective pressure (IMEP)
In order to determine the power developed by the engine, the indicator diagram of
engine should be available. It is possible to find the average gas pressure that while
acting on piston throughout one stroke would account for the national grid done. This
pressure is called Indicated mean effective pressure (IMEP).
Indicated horse power (I.H.P)
The indicated horse power (I.H.P) of the engine can be calculated as follow:
Pm .L. A.N .n
I .H .P (3-28)
4500
Pm = indicated mean effective pressure.
L = length of stroke in metres.
A = piston area in cm2.
N = speed in ‗rpm‘.
n = number of cylinders.
К = 1 for two stroke engine and 2 for four stroke engine.
Brake horse power (B.H.P)
Brake horse power is defined as the net power available at the crankshaft. It is found
by measuring the output torque with dynamometer.
2NT
B.H .P (3-29)
4500
Where
T = torque in kg.m
N = speed in ‗rpm‘.
Frictional horse power
The difference of I.H.P and B.H.P is called F.H.P. it is utilized in overcoming
frictional resistance of rotating and sliding parts of the engine.
F.H .P I .H .P B.H .P (3-30)
Indicated thermal efficiency (ηi)
It is defined as the ratio of indicated work to thermal input.
I .H .P 4500
i (3-31)
W Cv J
Where
W = weight of fuel supplied in kg per second
Cv = calorific value of the fuel in kcal/kg.
J = Joules equivalent = 427.
47
Chapter 3 supply systems
3.3.8 Accessories
The accessories include the following:
Water storage
Is used for cooling the diesel generator, this tank must be opened from the above to
expose the water to the air. The capacity of the tank for small diesel generator less
than (0.5 MW) approximated to (10-15 barell) depending on the ambient temperature.
Diesel storage
Diesel storage is used for storage diesel for feed the diesel generator along months or
weeks or days depending on the capacity of the diesel generator and the fuel
consumption. We can calculate the capacity of diesel tank if we knew the period
which we need to storage it, the capacity of the generator and the fuel consumption of
the diesel generator as follows:
48
Chapter 3 supply systems
Internal
Mechanical
Gears box combustion
energy
chambers
Electrical Electrical
generator energy
Figure (3.23)
3.3.10 System situation
3.3.10.1 Stand alone system
This type of system is used where it is not necessary or practical to have a connection
to the mains electricity supply. The diesel generator in this situation is connected
directly to the loads. Typical installations would include power provision to remote
locations, where the cost of installing a mains connection is uneconomical or
impractical, isolated buildings and community projects, boats and caravans etc, the
connections of this system are shown below.
Diesel generator
Bus
Load
Figure (3.24)
3.3.10.2 Grid connected system
49
Chapter 3 supply systems
In this system the diesel generator is connected to the load and power grid. This
system is used as stand by system for critical loads and to meet peak demand in the
grid, the connections of this system are shown below.
Grid
Diesel generator
Bus
Load
Figure (3.25)
3.3.11 Reliability
Diesel-electric generating plants should be designed to maximize operating reliability
and ease of maintenance. Space must be provided around equipment and components
for easy access. Controls should be provided in multiple unit installations to prevent
maintenance activity taking place on one unit from interfering with operating units.
Spare diesel-engine generator sets are required for electric generating plants in
accordance with the applicable duty type criteria. Packaged electric-generating units
may be considered for stand-alone installations but they must comply with applicable
criteria.
3.3.13 Maintenance
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51
Chapter 3 supply systems
3.4.1 Introduction
In this supply the process involves taking power from the transmission system to
customers by using overhead lines or underground cables depend on the condition of
the site. And also we need geographical information about the site which we supply it
by electricity. In addition we need the distances between the loads and exists
transmission lines. There are two types of extension configurations, radial, which the
feeder is installed in one path to the load, loop, the feeder in this design is installed
from more paths to the load. In this part we will study many subjects; national grid
conditions, system configurations, calculations which is show (voltage selection,
voltage regulation, conductor selection, losses, fault analysis) and we will talking
about Reliability, Environmental impact, Field of applications and finally we will list
the advantages and disadvantages of extension with the national grid.
52
Chapter 3 supply systems
53
Chapter 3 supply systems
Copper Conductors
Copper has high conductivity, more Weights and it is more expensive. Copper is used
in three forms: hard drawn, medium-hard drawn, and soft drawn (annealed). Hard-
drawn copper wire has the greatest strength of the three and is, therefore, mainly used
for distribution circuits of long spans (60m or more). Soft-drawn wire is the weakest
of the three. Its use is limited to short spans and for tying conductors to pin-type
insulators. It used widely for services to buildings and some distribution circuits. [11]
Aluminum Conductors
Aluminum has lesser conductivity than copper; it is used because of its light weight
and low cost for transmission line systems, except for short distribution spans. It
hasn‘t good mechanical properties, so aluminum wires are stranded on a core of steel
wire. Such steel reinforced aluminum wire has great strength for the weight of the
conductor and is especially suitable for long spans.
Steel Conductors
Steel wire is rarely used alone. However, where very cheap construction is needed
steel offers an economic advantages. Because steel wire is three to five times as
strong as copper, it permits longer spans and requires fewer supports.
Our range of conductors includes:
All Aluminum Conductors (AAC)
Aluminum Conductor Steel Reinforced (ACSR)
All Aluminum Alloy Conductors (AAAC)
Insulated Copper Conductors
Insulated Aluminum Conductors
In manufacturing this measure take in considerations:
Capacity to have higher load resistance
Durable & tension
Corrosion resistant properties
Resistivity and Temperature Coefficients of Some Conductors are shown in the table
below:
Resistivity at 20 Temperature Coefficients at
Material
C0(Ω.m) (20C0)
Silver 1.59 x10-8 0.00380
Annealed copper 1.72 x10-8 0.00393
Hard-drawn copper 1.77 x10-8 0.00381
Aluminum 2.83 x10-8 0.00377
Table (3.2) [12]
Conductor Stranding
As conductors become larger, they become too rigid for easy handling. Bending can
damage a large solid conductor. For these practical reasons, the stranded conductor
was developed. Stranded conductor consists of a group of wires twisted into a single
conductor. The more wires in the conductor‘s cross section, the greater will be its
54
Chapter 3 supply systems
flexibility. Usually, all the strands are of the same size and same material (copper,
aluminum, or steel).
Aerial Bunched Cables (ABC)
This type of cable is used for LV distribution in
residential areas. These cables are not practical more
than 500m distance [6]. Most utilities commonly
install triplex secondary for overhead service to single- Figure (3.26)
phase customers, where three insulated phase conductors are wrapped around the
neutral. The neutral supports the weight of the conductors. These cables have high
capacitance and low inductance leading to low impedance of lines, lower voltage
drop, higher current carrying capacity, better voltage regulation, easy maintenance
and many more.
Applications:
As replacement of bare lines where reliability of supply is important.
As replacement of bare lines where high degree of stability of supply voltage
is important.
In hilly terrains where cost of erection of underground cable becomes very
high.
As reinforcement of existing system without increasing voltage.
3.4.5.2 Insulators
Overhead conductors are electrically insulated from each other as well as from the
pole or tower by nonconductors which are called insulators. We use different type of
insulator according to their mechanical strength and electrical properties. There are
three practical insulator materials; porcelain, glass and polymer insulators. Polymer
insulators are not so restricted and have the advantage also of being lighter in weight
than porcelain or glass. Porcelain can stand compression force and high temperature
and cannot stand tension force. Polymer may be used both in tension and
compression.
There are many types of insulators we will discuss only six types:
Pin-Type Insulator
The pin-type insulator is designed to be mounted on a pin which in
turn is installed on the cross arm of the supporting structure. This
type of insulator is applicable for rural and urban distribution
circuits, and it is usually constructed as one solid piece of porcelain Figure (3.27)
or glass. Larger and stronger pin-type insulators are used for high-voltage
transmission lines. These differ in construction in that they consist of two or three
pieces of porcelain cemented together. These pieces form what are called petticoats.
They are designed to shed rain and sleet easily.
Disc-type insulator
These disc type porcelain insulators have high force tension, they
are used in the power transmission and distribution systems in
which there are a change in the level of the support structure,
Sometimes a line must withstand great strain, for instance at a Figure (3.28)
55
Chapter 3 supply systems
56
Chapter 3 supply systems
The length of poles depends on the required clearance above the surface of the
ground, the number of crossarms to be attached, and other equipment which may be
installed. Required pole strength is determined by the weight of crossarms, insulators,
wires, transformers, and other equipment it must carry, as well as by ice and wind
loadings. There are factors must be considered in deciding how deep a pole must be
planted in the ground: Soil conditions, the height of the pole, weight and pull. [12]
Guying pole
Earlier great detail was given about how carefully poles are chosen to carry the load
placed on them by the conductors. Careful specifications have been drawn regarding
the length, strength, measurements, and setting depth of a pole for every individual
situation. In spite of all this care and planning, situations arise where the conductor
tries to force the pole from its normal position. This happens because of abnormal
loads of ice, sleet, snow, and wind as well as because of uneven spans, corners, dead-
ends, and hills.
In this work anchor type guy sets are to be used. These guys shall be provided at
angle locations
dead end locations
Steep gradient locations.
Double Pole, & four poles
Figure (3.35)
57
Chapter 3 supply systems
excessive current causes the fuse-element to melt and the current path is interrupted.
Technological developments have served to make fuses more predictable, faster and
safer. Drop out fuse is implied to power system with voltage being 35kv or lower,
used as overload and short protector of wire and power transformer as well as
separating or closing rated load current. [12]
3.4.5.8 Capacitors
Capacitors are applied as an economic tool to reduce system
losses by supplying kilovars locally. Shunt capacitor banks
including fixed and switched banks are used on primary feeders
to reduce voltage drop, reduce power loss, and improve power
factor. The closer to the load they can be installed the greater the Figure (3.37)
economic benefit. Capacitors are not only an economic tool for the distribution
system, but they can eliminate the need for adding reactive sources in the bulk power
system. Kilovars supplied directly to load areas reduce the current in all portions of
the system. [12]
3.4.5.9 Switches
Switches are used to interrupt the continuity of a circuit. They classified to: air
switches, oil, vacuum and gas (SF6) switches. In distribution air switches are used. In
58
Chapter 3 supply systems
an air switch the interruption of the circuit occurs in air. In this process air is the
insulation medium between contacts. [4]
3.4.5.10 Reclosers
A recloser consists essentially of an oil switch or breaker
actuated by relays which cause it to open when
predetermined current-values flow through it. The switch or
breaker is arranged to reclose after a short interval of time
and re-open again if the fault or overload which caused the
excess current-flow persist. The recloser can be set for three
or four operations before it locks itself open for manual Figure (3.38)
operation. [12]
59
Chapter 3 supply systems
60
Chapter 3 supply systems
The table below shows the average span in the same level for different conductor‘s
cross section area (11kV) according to (sector of rural electricity).
61
Chapter 3 supply systems
length. If an overhead line is classified as short shunt capacitance is small that it can
be omitted entirely with little loss of accuracy and we need to consider only the series
resistance R and the series L for the total length of the line.
A medium length line can be represented sufficiently by R and L as lumped
parameters as shown in the figure (3.41) with half the capacitance to neutral of the
line lumped at each end of the equivalent circuit shunt conductance G as a mentioned
previously, is usually neglected in overhead power transmission lines when
calculating voltage and current. The same circuit represents the short line if capacitors
are omitted. [7]
According to the standards length of the transmission lines the short line up to 80 km,
medium length lines are roughly between 80Km and 240Km long. Lines longer than
240Km require in terms of distributed constants if a high degree of accuracy is
required.
L R
+ +
Z=R+JwL
Vs C/2 C/2 VR
- -
IS IR (3-35)
VS V R I R Z (3-36)
Where Z is the total series impedance of the line
IS L R
+
+ Z=R+JwL
G Vs VR load
- -
62
Chapter 3 supply systems
3.4.7 Calculations
3.4.7.1 Voltage selection
The design of electric supply and distribution systems can proceed only after a
distribution voltage level has been determined. The electrical impact of the
installation or facility as well as its location will influence the selection. A new
service may be necessary or extension of an existing service may be acceptable.
Before discussing selection of the system voltage, system voltage terminology and
preferred voltage ratings need to be defined, refer to the glossary for definitions of
standard voltage terms.
A preferred nominal system voltage such as 11kV, 12.5 kV, 13.2 kV, 13.8 kV, or 15
kV, will be selected for the primary distribution system. On sizable installations
where distances to loads are considerable or loads are large, the use of 33 kV or
24.9kV primary distribution systems may be more economical. Primary distribution
voltages of the nominal 11kV class and under will not be used, unless an off-site
supply of a higher voltage is not available. For such cases, the size of the installation
and the distances involved must make the use of voltages below 11 kV more
economical in order to justify the selection. [6]
63
Chapter 3 supply systems
64
Chapter 3 supply systems
Regulation Techniques
Distribution utilities have several ways to control steady-state voltage. The most
popular regulation methods include:
Substation load tap-changing transformers
Bus voltage regulators
Line voltage regulators
Fixed and switched capacitors
65
Chapter 3 supply systems
3.4.7.5 Losses
Systems have two types of energy losses: losses in the conductors and feeders due to
the magnitude of the current and transformer core losses that are independent of
current. Current related losses are equal to the current squared time the resistance of
the feeder or transformer. Accompanying these losses are reactive losses,
For single phase:
PLoss I 2 R (3-46)
QLoss I 2 X (3-47)
For three phase:
PLoss 3I 2 R (3-48)
QLoss 3I 2 X (3-49)
66
Chapter 3 supply systems
The core losses result from the energy used in transformer cores as a result of
hysteresis and eddy currents. These losses depend on the magnetic material used in
the core. Core losses in a power system can exceed 3% of the power generated
constituting as much as 40% of the total loss on the system. The capacity of
generation and reactive sources must be sufficient to supply these losses. [12]
67
Chapter 3 supply systems
transformer. This value expresses the impedance value as a percentage of the base
value. This is illustrated in the following calculation:
Assume
Vp: Primary voltage, Vs: Secondary voltage, S: Capacity, Z: Impedance,
Vb: base voltage
Calculation of base value, Zb:
2 2
V V
S b Zb b
Z S
(Referred to secondary side of transformer)
Z actual Z % Z b (3-50)
VS
I fault (3-51)
3 Z actual
Manipulation of the above formulas will give a quick reference formula to determine
the fault current at the secondary terminals of the transformer:
S
I fault (3-52)
3 Z % VS
For lines we use this formula:
VL
I fault (3-53)
3 ZL
Where:
VL: line voltage, ZL: The impedance of the line
3.4.8 Protection
The design optimization of power systems tolerates the disturbance of mentioned
process in term of evaluated risk based on probability theory. Therefore disturbances
may occur in any time and anywhere. The disturbances can be caused by: Shunt
Faults, lightning over voltages leading to insulation breakdown, generating unit trips,
etc. From previous section should have learned fault analysis and fault calculation
techniques, types of faults, fault levels and fault current calculations (transient and
steady state), system parameters affecting fault levels, system voltages during the
faults, etc. Usually, the shunt faults cause large currents, which can lead to great
damages to the power system due to localized release of a considerable amount of
energy. The rapid isolation of these faults will minimize the damages and disruption
to the system. The protection can be achieved by using relays or fuses. This is the
ultimate objective of the protection system. There are three types of protection:
68
Chapter 3 supply systems
3.4.9 Reliability
Reliability can be measured by the frequency of interruption of the power supply as
well as the duration of this interruption. Making the system more reliable may involve
more investment due to the duplication of the cables, switchgears, transformers,...etc.
The increase of cost may be justified if the estimated cost of the power outage is high.
Complete reliability is unattainable but loads which require higher reliability can be
provided with one of the following:
- Use of duplicated feeders.
- Feed the distribution system from multiple bulk supply points.
- Install enough on-line local generation.
- Use of uninterruptible power supply to feed very critical loads.
69
Chapter 3 supply systems
3.5 Reference
PV supply system
1. TECHNICAL TRAINING OF PV SUPPLIERS, Boniface Gissima Nyamo-Hanga -
(Energy Consultant)
2. RENEWABLE ENERGY DEVELOPMENT STRATEGY; Lahmeyer International
GmbH ; Dr. Romeo Pacudan, Dr. Andreas Wiese; February 2007, Bad Vilbel
3. Solar Electricity, practical guide to designing and installation small
photovoltaic systems; Simon Roberts; 1991
4. Optimal Wire Size for Photovoltaic Systems Operating at Maximum Power
Point: A Closed Form Approach Version 2.1 march 2006
5. Lead-Acid Battery Guide for Stand-Alone Photovoltaic Systems; IEA Task III
Report IEA-PVPS 3-06:1999 December 1999.
6. Photovoltaic Power Systems and The 2008 National Electrical Code AN
UPDATE; John C. Wiles
7. A GUIDE TO PHOTOVOLTAIC PV system design installation. Endecon
Engineering 347 Norris Court
8. Use of photovoltaic systems for rural electrification in Thailand N.
Rapapate and O. Goln Ubon Ratchathani Raja hat University Ratchathani
Road, Muang, Ubon Rachathani 34000. Thailand.
9. Power conditioning in photovoltaic system; Brian Johnson-
10. Characterization of solar batteries through extended time; tests in
accordance with IEC 614271Prof. Dr.-Ing. Wolfgang Wiesner FH-Köln
11. ELECTRONIC CONTROL CIRCUIT for solar battery charging; A.A.
AZOOZ, J.M. SULAYMAN March 4, 2005
12. Dimensioning of a solar/battery backup system; May 2008
13. Solar Battery Charging Stations an analysis of viability and best
practices Submitted BySGA energy limited Ottawa ,Ontario Canada;
February 1999
14. 1990 -- تكنىلىجيا الطاقة البذيلة –د يىسف عياش
70
Chapter 3 supply systems
71
Chapter 3 supply systems
72
Chapter 4
Project
Information
Chapter (4) Project information
4.1 Introduction
In every design for any project we must know all information about the site of the
project, so in this chapter we will summarize this information for project position,
environmental description, general information about the supply systems in the
country and special information for all supply systems in this site. Understanding the
project information gives knowledge for the function, importance, form of the site,
position.
71
Chapter (4) Project information
Healthy
Villages Population houses mosques schools
center
Kauman 588 74 3 1 1
Al_sabalah 154 22 1 ----- -----
Bani juma’ah 63 10 1 ----- -----
Bait al_esal 140 20 2 ----- -----
Al_mianeed 138 19 1 ----- -----
72
Chapter (4) Project information
73
Chapter (4) Project information
Wind energy
Results of the resource assessment study show that the Republic of Yemen is
endowed with significant wind resources. These resources were found to sustain
large-scale commercial power development and small-scale isolated system to meet
the energy needs of rural and remote communities.
The spatial distribution of the annual mean wind speed at the height of 50 meters
within the whole area of Yemen is shown in the map, figure (4.3)
Figure (4.3) average wind speed in Yemen (50m above ground level) [3]
Power densities of the wind energy resource (50 meters above ground level) are
shown in the map, figure (4.3). Regions with very good to excellent wind conditions
have power densities higher than 200 W/m2. Power densities above this level are
considered to be suitable for power development. Figure (4.3) clearly indicates the
regions where wind resources could be potentially exploited for power development.
Figure (4.4) Wind power density (50 meters above ground level) [3]
74
Chapter (4) Project information
75
Chapter (4) Project information
Note:
According to our available information we have seen that it can’t be developed
electrical power by using wind sources due to the following reasons:
The direction of wind speed is not stable during the seasons of the year.
Transportation process of wind farms components is very difficult such as wind
turbines due to unavailability of the roads required for this purpose.
Lack of technical or commercial skills and information: in Yemen markets,
skilled personnel who can install, operate, and maintain wind energy.
76
Chapter (4) Project information
For the villages, the total power can be found by using table (2.2) as follow:
Schools
The existing school consists of 11 classes, four room for teachers and four bathrooms.
Assume each class and room need average power =180 W (three lamp 60W),
Four bathrooms, they need 60W in each for lighting,
The total power of the school for lighting = (15×180+4×60) = 2940W.
Pump = 450W
Water heater = 1200W
77
Chapter (4) Project information
Table (4.6)
There are three circuits, from table (2.3), KS = 0.9
Stotal = 4957×0.9 = 4.462 kVA
The school is in kauman village, so the contribution of school loads with houses is
multiple by KS = 0.4 according to table (2.2)
S =4.462×0.4 = 1.78 kVA
Healthy centre
There are four rooms for patients and two for doctors and each room needs average
power = 180 W for lighting.
Nroom = 6 rooms » Plighting= 6×180= 1080 W,
water heater =1200 W, and pump = 450W
Table (4.7)
78
Chapter (4) Project information
79
Chapter (4) Project information
Healthy
Villages Population houses mosques schools Ks
center
Kauman 588 74 3 1 1 0.4
Al_sabalah 154 22 1 ----- ----- 0.49
Bani juma’ah 63 10 1 ----- ----- 0.63
Bait al_esal 140 20 2 ----- ----- 0.49
Al_mianeed 138 19 1 ----- ----- 0.53
Table (4.13)
Houses
Electrical Instruments Demand
No Qty AC(V) hours/day
Appliances PRating (W) Factor(Kd)
1 lighting 10 220 600 1 6
2 TV+ receiver 1 220 120 1 6
3 Radio 1 220 20 1 3
4 Water heater 1 220 1200 1 0.5
5 Pump 1 220 450 0.6 0.5
6 Refrigerator 1 220 200 1 12
7 Washer 1 220 450 0.8 0.6
8 Blender 1 220 250 1 0.04
9 Clothes iron 1 220 1200 1 0.15
10 Vacuum cleaner 1 220 1000 1 0.5
11 Computer 1 220 400 0.7 6
12 Mobile charger 4 220 20 1 2
13 Electric Heater 1 220 1100 0.8 8
14 Others appliances ---- 220 500 1 0.15
Table (4.14)
80
Chapter (4) Project information
Schools
Kind of load P (watt) Demand Factor( Kd) hours/day
Lighting 4020 1 6
Water heater 1200 1 0.5
Pump 450 0.6 1
Other appliances 500 1 0.15
Table (4.15)
Healthy center
Kind of load P (watt) Demand Factor( Kd) hours/day
Lighting 1200 1 12
Water heater 1200 1 0.5
Pump 450 0.6 0.5
Other appliances 500 1 0.15
Table (4.16)
Mosques
Kind of load P (watt) Demand Factor( Kd) hours/day
Lighting 660 1 4
speaker 500 1 3
Pump 450 0.6 2
Other appliances 500 1 0.15
Table (4.17)
The total energy demand required for the whole village:
ETotal Ehouses Eschools EHcenters Emosques
The loads that have supplied directly from the PV cells are all the predetermined loads
Eload1 ET otal
The loads that have supplied from lead acid batteries are equal to the last
predetermined:
Eload2 ET otal
The total energy demand for the whole system per day:
E PV Eload2 Eload2
By using the above equations we found the following table:
Villages Kauman Al_sabalah Bani Bait al_esal Al_mianeed
al_hamra juma’ah
Ehouses(Wh/day) 510777.6 186019.68 108712.8 169108.8 173767.92
Emosques(Wh/day) 5706 2329.95 2995.65 2329.95 2520.15
Eschools (Wh/day) 10026 0 0 0 0
EH.centers(Wh/day) 6084 0 0 0 0
ETotal (Wh/day) 532593.6 188349.63 111708.45 171438.75 176288.07
Table (4.18)
81
Chapter (4) Project information
4.8 References
82
Chapter 5
Supply
Systems Design
Chapter 5 supply systems design
5.1.2 Sizing
5.1.2.1 Estimation the required PV arrays
For accuracy in calculation we have to determine the factors that affect on panels as a
mentioned at solar generation part (temperature, manufacture, dirt, losses) and they are
the following:
The climate in general is quite in summer and cold in winter the average temperature in
summer is between (20, 31) C0 and in winter is between (18,-2) C0 in night and morning.
And according to the manufactures data sheets
0.05%/c , Man 91%
And the temperature factor determined by the following formulas:
According to the international standard the allowable losses from the PV modules to the
inverter is 3% and the distribution losses is about 6.9%, so;
loss 90.4%
83
Chapter 5 supply systems design
VBb
NMS (5-4)
VMP
The number of strings in parallel is determined by using the following formula:
NM
NMP (5-5)
NMS
EB (Wh/days)
CAh (5-10)
V Bb cc B DOD
The number of required batteries in parallel is determined by using the following
formula:
CAh (5-11)
NBP
BAh
The number of required batteries in series is determined by using the following formula:
84
Chapter 5 supply systems design
VBb
NBS (5-12)
VB
The total number of batteries is determined by using the following formula:
85
Chapter 5 supply systems design
5.2.2 Sizing
The suitable available diesel generators ratings for all villages are shown below:
Villages S (kVA) Available ratings (kVA) No. diesel generator
kauman 234.3 261 DG1
Al_sabalah al_hamra 83.8 85 DG2
Bani juma’ah 50.4 55 DG3
Bait al_esal 77.8 85 DG2
Al_mianeed 76.3 85 DG2
Table (5.7) [1]
The generators above have the following special features:
Compact size and light weight.
Sound proof type.
Performance elevation at heavy loading injection.
Convenient use from outside.
Automatic generation with load variation.
Voltage regulator.
Over current protection (air circuit breaker).
Frequency and unbalance load protection.
All diesel generators have all specifications shown in appendix (A)
86
Chapter 5 supply systems design
5.2.5 Accessories
The accessories include the following:
Water tank
Diesel tank (for one week)
Steel pipes
Valves
Accessories Water tank Diesel tank** Pipes (6m) Valves (3 pieces)
villages (barell*) (barell) (inch) (inch)
kauman 10 60 ¾ ¾
Al_sabalah al_hamra 10 20 ¾ ¾
Bani juma’ah 10 15 ¾ ¾
Bait al_esal 10 20 ¾ ¾
Al_mianeed 10 20 ¾ ¾
Table (5.10)
*One barell = 200 litre
** using equation (3.34)
87
Chapter 5 supply systems design
88
Chapter 5 supply systems design
1.32 km
1.74 km
0.52
km
G
75
km
0.
Bani juma’ah
10 house kauman0.42
63 person 74 house km B Al_mianeed
588 person
19 house
1.6
138 person
7k
m
2.97
km
5.43
km
1.
35
km
A
PEC exist line
89
Chapter 5 supply systems design
5.3.3 Sizing
5.3.3.1 Voltage selection
The grid voltage of the nearest point to the location is 11kV, so we will choose the
standard values of calculations according to this voltage.
Table (5.12)
*See section (5.3.3.4) table (5.15)
To calculate the nominal currents for the villages we use equations (3-40):
S
I
3 *V
The rating of the fuses must be double of the full load or maximum currents.
The amperes ratings and types of fuses shall be as in table (5.13)
90
Chapter 5 supply systems design
By using second method „Permissible voltage drop‟ of conductor selection in chapter (3)
section (3.4):
From voltage drop standard the allowable voltage drop percentage is 5%, this value is for
all transmission portions, so we will find the voltage drop for each portion by using this
formula:
Xf
Vd f Vd (5-14)
X max
X max : Maximum path of conductors in one direction
X f : Portion distance of feeder
Vd : Maximum allowable voltage drop
V d f : Voltage drop for feeder
From dimensions map X max is the distance between point „A‟ and „F‟ » X max =15.91 km
91
Chapter 5 supply systems design
We will find the loads at the points in the dimensions map by using transmission losses
factor in table (2.6) which is equal (1.001). For example to find the load in point „E‟ the
load in point „F‟ will multiple by transmission losses factor ‟77.52×1.001+79.04=156.64‟
and so on, from table (5.18) the loads at the points shown in dimension map are tabled
below:
Table (5.16)
By using these equations (3-40, 3-42):
S
To find the current » I
3 V
V
To find the resistance » R
I
L
To find the cross section area » A where aluminum = 2.83x10 -8
.m
R
We find the table (5.17) in the next page
92
Chapter 5 supply systems design
93
Chapter 5 supply systems design
Feeder Feeder
Post insulators Pin insulators Disc insulators
portions length (m)
A-B 11420 15 405 45
B-C 1320 3 42 6
B-G 420 --- 12 6
C-D 2490 3 84 15
C-E 2170 3 72 15
E-F 1000 3 30 9
Total insulators 27 645 96
Table (5.19) Quantities of the insulators
All insulators shall meet the specifications shown in appendix (D)
94
Chapter 5 supply systems design
5.3.4 Spacing
5.3.4.1 Sag and span
From table (3.6) the minimum span between poles for 11kV, (35 mm2) ACSR conductors
equal (80 m) and from table (3.7) the minimum sag equal (0.32m).
5.3.4.2 Clearance
From table (3.4 and 3.5) we find the following table which gives the minimum clearance
of the 11 kV O.H.T.L between the conductors and supports and various objects.
95
Chapter 5 supply systems design
5.3.5 Calculations
5.3.5.1 Voltage drop
According to equation (3-37) we find this formula:
I ( R jX L )
V % 100%
11 103
The resistance and inductance of the conductors from appendix (C) is equal 0.784 Ω/km
0.4 Ω/km respectively, so we found the following table:
96
Chapter 5 supply systems design
The total losses for the system include the losses of the feeders and transformers, so from
table (5.12) for 0.86 power factor we find the following table:
From table (5.23, 5.24) we found the total losses of the system equal (28220+j14900).
5.4 References
1. Juma'an company
2. Al sharq company for PV system
3. Rural electricity sector
4. Al ashwal company for electricity
The other references of this chapter is the same of the references of chapter (3)
97
Chapter 6
Economic Evaluation
and Comparison
Chapter 6 Economic evaluation and comparison
6.1 Introduction
The main objective of the supply system design is to provide an adequate supply of
acceptable quality whilst keeping the cost as low as possible.
Cost is the common factor which affects the selection of supply system, For example
there are some loads which have special requirements, such as loads that must be kept in
operation under all conditions (loads like computers for traffic control or supply to
hospital operation theatres). These special requirements are addressed in a number of
ways depending on the economies of the situation.
Supply system equipment incurs two types of cost:
Capital cost (fixed cost), which includes the equipment, land, labor for site preparation
construction and installation … etc.
Operating costs, this includes labor and equipment for operation, maintenance and
services as well as power losses, although, equipment losses may be small compared to
the power it delivers, its life time losses may be much higher than its capital cost.
98
Chapter 6 Economic evaluation and comparison
Where:
i = interest rate (i % for renewable energy systems in Yemen) = 1%.
The cost of the required batteries CB1 is given by the following equation:
CB1 N B1 Cb1 N B 2 Cb2 (6-5)
C B1 114 3000 285 2650 1097250$
Because that warranty of batteries is 5 year and the life time of the project is 20 year and
according to the interest rate (i) so that the total capital cost of the batteries can be
calculated by using the following equation:
C B C B1 C B1 (1 i ) 5 C B1 (1 i ) 10 C B1 (1 i ) 15 (6-6)
CB =1878151.548 $
99
Chapter 6 Economic evaluation and comparison
Control cost:
According to the references and the last developed projects for some areas the cost of the
requirements of control equal to one percent (1%) of the whole panels and batteries cost
and it is calculated as the following:
C cont. 1% (CB1 C Array ) (6-7)
100
Chapter 6 Economic evaluation and comparison
From table (4.19) and table (5.6), EPV(F,P) = 1416.465 + 2360.757 = 3777.222 kWh/day
So EPV(F,P) 3777.222[ KWh / day] 365[day / Year] 1378686.03[ KWh / Year]
And the cost of KWh is calculated by the following formula:
A
CE n (6-16)
E PV ( F , P )
487315.89
CE 0.3535 $ / KWh
1378686.03
101
Chapter 6 Economic evaluation and comparison
102
Chapter 6 Economic evaluation and comparison
Maintenance cost
The maintenance of diesel generator includes changing diesel filter ‘every month’,
changing oil filter ‘every month’ and changing pistons and its accessories ‘five years'.
The cost of diesel filter (7$), oil filter (9$) and the pistons and its accessories (1500$), so
the maintenance cost will be represented as ‘$/kWh’ as follows:
103
Chapter 6 Economic evaluation and comparison
From table above the total cost of all equipments and its accessories equal ‘178812.1$’.
The cost of erection for the transmission line equal (3200 $/km, the length of the transmission
line 18.82 km), for transformers, (50 kVA = 900 $), (100 kVA = 1000 $), (200 kVA = 1100 $), so
the total cost of erection equal (63224 $). [4]
So the capital cost equal (242036.1 $)
104
Chapter 6 Economic evaluation and comparison
An
CE
E
164946.55
CE 0.0419 $ / kWh
3937059.36
105
Chapter 6 Economic evaluation and comparison
6.5 Comparison
For comparison between various studied supply systems we use the following factors:
Reliability
Environmental impacts
Cost
6.5.1 Reliability
The reliability for any system depend on various factors faults, equipment failures’,
outage of loads due to the partial or the whole maintenance for the main components of
the intended system.
For diesel power plant: Faults: the resulted problems from the faults will not effect on
the equipment or any other component of the system due to existing of protection system
but for the last design the protection system will disconnect the supply of the customers
then this is considered disadvantage for this design because there is no stand by supply.
Equipment failure (diesel generator or transformer): if any of mentioned parts of system
generation has failed the supply will be disconnected of the loads for repair the faulted
part so this problem is considered problem for the reliability.
Maintenance: it is most important for any design where includes diesel generator and any
part of the system but at different levels in our design the maintenance of the main
components of the system will force us for disconnecting the supply of loads so this is
considered disadvantage for the reliability of the system.
For PV power plant: Faults: the faults in PV system don’t effect on the system
component as compared AC systems and the protection system isolate any faulted part
and save the remain system to operate as before the fault so this considered the most
important advantage for this system as compared with the other system designs.
Equipment failure (diesel generator or transformer): for PV system due to varying of
system parts any faulted component such as module or battery failure may be change
without effects on continuity of supply or may be disconnect small part of the connected
load as mentioned in the reliability of PV system in chapter(3).
Maintenance: from the most important advantages is maintenance of any part of the
system without affecting on system continuity of supply or sometimes part of loads may
be outage at the inverter maintenance state.
For extension with the national grid: The more effect factor on the extension with the
grid is faults that may cause disconnecting of the loads due to the protection system such
as faults in distribution transformers or distribution lines. The maintenance and
environmental impacts are related with the power generation stations that supply the
national grid.
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Chapter 6 Economic evaluation and comparison
6.5.3 Cost
The cost includes the cost of the electricity as follows:
Supply systems Cost of energy ($/kWh)
Solar PV system 0.3535
Diesel generator system 0.3296
Extend with national grid 0.0419
Table (6.7)
Due to that the loads are not important (the continuity of supply is not necessary) so the
reliability will be excluded from comparison between alternatives.
And also the environmental impacts will be excluded from comparison due to the
quantity of impacts at environment are small.
The most important factor that will be used for comparison between different supply
systems is the cost which includes some last mentioned factors.
The results are shown at the below table:
From the above table we found the extension with national grid is the best supply systems
because it is the lowest cost of energy, so the cost of energy in the bill will be '0.0419
$/kWh'.
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Chapter 6 Economic evaluation and comparison
Tension set
Earthing set
New lines
Exist line
Figure (6.1)
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Chapter 6 Economic evaluation and comparison
6.7 Conclusion
Most of Yemeni governorates have many alternatives to supply their by electricity. More
reliability system is higher cost of electricity. The economical way to supply villages
located near to the national grid is the extension of the national grid because of the cost
of the electricity in extension of the national grid is lesser than the cost of electricity in
other supply systems and other reason is Yemen Government supporting for the
electricity generation than real costs of generation.
6.9 Recommendations
We recommend dealing with special sectors which support the projects by more
information.
We use individual design for each village, so we recommend studying central supply
system for all villages and comparing it with our designs.
In solar PV system we recommend individual designing for each house and compare
it with our design.
Taken into account the cost of the land to calculate the capital cost for all designs.
Taken into account the reliability and environmental impact as essential parts in
comparison between different alternatives.
We recommend asking support form economical experts.
6.10 Reference
1. Juma'an company
2. Hiziaz Power Station
3. Al sharq company for PV system
4. Rural electricity sector
5. Al ashwal company for electricity
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Appendix A _
Transformer specifications
Specification \ transformers T1 T2 T3
Applicable Standards IEC 60076
Manufacture EAGLERISE ELECTRIC &ELECTRONIC CO
Country of manufacture CHINA INDIA INDIA
vector group Dyn11 Dyn11 Dyn11
No - phases 3-phase 3-phase 3-phase
Rating (kVA) 200 kVA 75kVA 50 kVA
Over load at 10C0 (6hours) 20% 20% 20%
Primary Voltage 11 kV 11 kV 11 kV
Secondary Voltage 0.415kV 0.415kV 0.415kV
Frequency 50 Hz 50 Hz 50 Hz
Connection Delta / Star Delta / Star Delta / Star
Primary winding 38 kV 38 kV 38 kV
Secondary
3 kV 3kV 3 kV
Applied winding
test, No load current 1.6 % 1.1 % 0.9 %
1min.50Hz No load losses 0.29 kW 0.115kW 0.08 kW
Load losses (%) 0.95 1.6 1.9
Impedance 4% 4% 4%
tapping of primary winding (±3×2.5%) (±3×2.5%) (±3×2.5%)
Noise level 46 dB 44 dB 43 dB
0 0
Temperature rise 65C for winding, and 55C for oil
Cooling Type ONAN
Oil Capacity 172 Litre 105 Litre 95 Litre
Oil weight 155 kg 100 kg 100 kg
Dimensions 1110×730×1400 1000×650×1270 900×650×1270
Total weight 645 kg 350 kg 260 kg
Appendix C _
Poles specifications
Fuses specifications
Specification \ Type F1 F2 F3 F4
Applicable Standards IEC 62271-107
Manufacture TACHUAN CO, China
Rated voltage 11kV
Rated current 25A 10A 5A 63A
Breaking current 20kA
Impulse voltage 70kV
Power frequency withstand voltage 55kV