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CHAPTER I
The Problem and Its Background
1.1 Background of the Study
Since the dawn of time, the use of windmills has been one of the most
reliable methods of converting wind energy to mechanical energy. Due to
the enormous amount of wind energy around the globe, people have
developed different methods on how to utilize this kind of energy that is seen
on wind turbines and wind pumps.
In the Southeast Asian region specifically the Philippines, the first
windmill park was built in the windy seashores of Bangui Bay, Ilocos Norte.
Thus, it officially started the advent of renewable energy sources. With this
technology, Filipinos have taken full advantage of the abundant wind energy
provided by the Asia-Pacific monsoon and have built different kinds of
windmills around the archipelago. One of the most widely used types of
windmill in the Philippines is the wind pump. Windmill driven pumps are
usually used to sustain the need of water when there is shortage of supply in
certain areas. It has mainly benefited famers in providing water for their
animals and crops.
The Caleruega Retreat House, an escape from the bustling metro district
of Manila, is a spiritual retreat house of the Dominicans located at Nasugbu,
Batangas just after Tagaytay City. It has almost the same elevation with
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Tagaytay City which is 2,100 ft. (640m) above sea level. Thus, installing a
wind pump in the area will be useful in transporting rain water from a
reservoir to a water tank which they use to water their plants. The rain water
will be collected from a man-made pond and will be pumped to a water tank
located at a higher elevation.
1.2 Conceptual Framework
Gathering related topics and other information needed is the initial
procedure in conducting the study of this project. During the data gathering,
the most important step is to survey the location and collect all necessary
information in windmill construction. Also, actual background studies with
the windmill fabricators are essential in this study. Computation and analysis
of gathered data is required to come up with a design and forecast possible
remedies in different situation that the windmill may experience. Lastly,
designing the structure and looking for errors so that researchers could come
up with improvements and recommendations.
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1.3 Theoretical Framework
The theoretical properties of this research study are based on the following
principles;
1. The available materials for the construction of the wind pump and
the multi bladed vertical windmill are found and bought locally.
2. The research study done applied wind driven pumps method and
can be associated to the standard mechanical water pump method.
3. The researchers have adequate information about wind energy,
fluid mechanics, machine design and strength of materials.
1.4 Research Paradigm
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1.5 Paradigm
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1.6 Statement of the Problem
The main problem of this study is to design a wind pump that will utilize
the wind at Caleruega Retreat House, Nasugbu, Batangas. The wind pump
design should be able to produce a certain shaft speed that will meet the
required power of the centrifugal pump to transport rainwater from a
reservoir to an existing tank at the area. This study should be able to answer
the following questions:
1. What would be the optimum pump capacity and power requirement
for the given application?
2. Is the nature of the wind (power and speed) enough to run a
centrifugal pump based on the design?
1.7 Scope and Delimitation of the study
The objective of the study is to design a wind pump that will utilize the
abundant wind energy at Caleruega Retreat House, Nasugbu, Batangas. The
researchers will focus on determining the amount of power that can be
transmitted by the rotor to the centrifugal pump. With the calculated amount
of power transmitted by the rotor, the researchers will then determine if the
power supplied is sufficient enough to operate a centrifugal pump that will
supply water to the existing water tank at Caleruega Retreat House,
Nasugbu, Batangas. The researchers will limit their study on designing that
wind pump alone. By designing a wind pump, the researchers will consider
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the materials needed to construct the wind pump to come up with the
desired power that will operate the system. Thus, fabrication and installation
of the designed wind pump will not be the concern of the researchers.
1.8 Significance of the study
The design of the wind pump for Calaruega, Nasugbu, Batangas is very
significant for the people who reside there and for the students who come
there for their retreat. This is to show that the Dominican Community is
supporting the use of renewable energy. Once fabricated, it can be a good
example for the other communities so that they will also follow suit in using
renewable energy sources.
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Definition of Terms
Blade. The part of a wind generator rotor that catches the wind.
Braking System. A device to slow a wind turbines shaft speed down to safe
levels electrically or mechanically.
Check Valve. A valve allowing flow in one direction only.
Power.The rate of energy output.
Pump. A machine, usually with rotary action or reciprocal action of a piston,
for raising or moving liquids, compressing gases, inflating types etc.
Pump Efficiency. Ratio of power out and power in of a pump.
Reciprocating Pump. Reciprocating type pumps use a piston and cylinder
arrangement with suction and discharge valves
integrated in the pump.
Rotor.The blade and hub assembly of a wind generator.
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Shaft. The rotating part in the centre of a wind generator or motor that
transfers power.
Static Head. It is the maximum height that can be delivered in a pump.
Total Head. It is the measure of the energy increase per pound imparted to
the liquid by the pump and is therefore the algebraic difference
between the total discharge head and the total suction lift exists.
Torque.The moment of a system of forces tending to cause rotation.
Tower. A structure that supports a wind generator, usually high in the air.
Wind. Air moving from a area of high pressure to an area of low pressure.
Windmill. A device converting wind power to mechanical rotation with a low
velocity turbine designed for compressible fluids.
Wind Velocity. It is referred to as the rate of motion of wind.
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CHAPTER II
Review of Related Literature
2.1 History of Windmill
According to Wailes (1954), a windmill is a machine that is powered by
the energy of the wind. It is designed to convert the energy of the wind into
more useful forms using rotating blades or sails. Windmill also refers to the
structure it is commonly built on. Windmills served originally to grind grain
(hence the "mill" derivation), though later applications included pumping
water and, more recently, generation of electricity. Recent electricity-
generating versions are referred to as wind turbines.
Most windmills have a wheel of blades or sails that is turned by the wind.
In most cases, the wheel is set on a horizontal shaft. The shaft is mounted on
a tower, mast, or other tall structure. The shaft is turned by the movement of
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the wheel, and it transmits power, through a series of gears, to a vertical
shaft. The vertical shaft then carries power to a water pump, electric
generator or other device (World Book, 1995)
The majority of windmills had four sails. An increase in the number of
sails meant that an increase in power could be obtained, at the expense of
an increase in the weight of the sail assembly. The earliest record of a multi-
sailed mill in the United Kingdom was the five sail Flint Mill, Leeds,
mentioned in a report byJohn Smeaton in 1774. Multi-sailed windmills were
said to run smoother than four sail windmills. In Lincolnshire, more multi-
sailed windmills were found than anywhere else in the United Kingdom.
There were five, six and eight sail windmills.
Fig 1. An example of eight sailed windmill, Wailes 1954
If a four sail windmill suffers a damaged sail, the one opposite can be
removed and the mill will work with two sails, generating about 60% of the
power that it would with all four sails. A six sail mill can run with two, three,
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four or six sails. An eight sail mill can run with two, four, six or eight sails,
thus allowing a number of options if an accident occurs. A five sail mill can
only run with all five sails. If one is damaged then the mill is stopped until it
is replaced (Wailes, 1954).
One type of a windmill is the wind pumpa type of windmill used for
pumping water from a well or draining land. Wind pumps are used
extensively in Southern Africa and Australia and on farms and ranches in the
central plains of the United States. In South Africa and Namibia thousands of
wind pumps are still operating. These are mostly used to provide water for
human use as well as drinking water for large sheep stocks.
Fig. 2(a) Fig.2(b)Figure 2(a) and (b). Examples of wind pumps
Kenya has also benefited from the African development of wind pump
technologies. At the end of Eventually steel blades and steel towers replaced
wooden construction, and at their peak in 1930, an estimated 600,000 units
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were in use, with capacity equivalent to 150 megawatts (Wiley and Sons,
1995)
Eldridge (1980) explains that early wind pumps directly operated the
pump shaft from a crank attached to the rotor of the windmill; the
installation of back gearing between wind rotor and pump crank allowed the
pump to function at lower wind speeds. Today water is primarily raised by
electric pumps, and only a few wind pumps survive as unused relics of an
environmentally sustainable technology. This type of wind pump can be
found worldwide and is still manufactured in the United States, Argentina,
New Zealand, and South Africa. A six-foot diameter wind pump rotor can lift
up to 180 gallons per hour of water with a 15 to 20 mile per hour wind,
according to a modern manufacturer (about 700 litres/hour by a 1.8 metre
rotor in 2432 km/hour wind). Wind pumps require little maintenance, only
requiring gear oil changes about once per year. An estimated 60,000 wind
pumps are still in use in the United States. They are particularly economical
in remote sites distant from electric power distribution.
2.2 Wind Machines and Rotors
Also, Eldridge (1980) enumerates the classification of machines using
rotor as wind energy collectors in terms of the orientation of their axis of
rotation, relative to the wind stream:
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1. Horizontal axis rotors the axis of rotation is parallel to the
direction of the wind stream; typical of conventional windmills.
2. Crosswind Horizontal axis rotors the axis of rotation is both
horizontal to the surface of the earth and perpendicular to the direction
of the wind stream; somewhat like a mater wheel.
3. Vertical axis rotors the axis of rotation is perpendicular to the
surface of the earth and the wind stream.
The blade of the windmill is to be computed by following the Blade
Element Theory. The Blade Element Theory is a mathematical process which
is originally founded by William Froude, David W. Taylor, and Stefan
Drzewiecki to be able to determine the concept of propellers. The process is
done by breaking a single blade down to several small partitions and then
finding the forces in each of these small blade elements. The forces are
within integration along the entire blade over one rotor revolution to be able
to obtain the forces and the moments that is produced by the entire
propeller blades. The hard part lies within modelling the induced velocity on
the rotor disk. As a solution to this, the Blade Element Theory is often
combined with the Momentum Theory to further provide additional
relationships necessary to describe the induced velocity on the rotor disk.
2.3 Wind Pump and the Blade of a Windmill
During the 1970s, the UKNGOIntermediate Technology Development
Group provided engineering support to the Kenyan company Bobs Harries
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Engineering Ltd for the development of the Kijito wind pumps. Bobs Harries
Engineering Ltd is still manufacturing the Kijito wind pumps, and more than
300 Kijito wind pumps are operating in the whole ofEast Africa.
Many wind pumps were built inThe Broads, ofEast Anglia in the United
Kingdom for the draining of land. They have since been mostly replaced by
electric power, many of these wind pumps still remain, mainly in a derelict
state (pictured), however some have been restored.
On US farms, particularly in the Midwest, wind pumps were used to
pump water from farm wells for cattle. The self-regulating farm wind pump
was invented by Daniel Holladay in 1854.
Figure 3. Velocities and Forces Acting on the Blade Component Diagram
Where:
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dT = Thrust dL = Lift
dF = Force dN = Torque / radius
dD = Drag U = Induced speed
UA = Axial induced speed UT = Tangential induced speed
VA = Translation speed nP = Pitch speed
r = Blade tip speed = Angle of attack
= Inflow angle
The blade of a windmill can also be computed by using the Momentum
Theory. This theory describes a mathematical model of an ideal propeller.
The originators of the momentum theory are W.J.M. Rankine, Alfred George
Greenhill, and R.E. Froude. Under certain principles in fluid dynamics, there
can be a mathematical connection between power, radius of the rotor,
torque, and induced velocity. This theory is best described by the formula:
Where:
T = the thrust
= the density of air
A = the area of the rotor disc
2.4 Betz Law
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According to Betz Law, wind turbine obtains its power input by
converting the force of the wind into torque (turning force) acting on the
rotor blades. The amount of energy that the wind transfers to the rotor
depends on the density of the air, the rotor area, and the wind speed. The
kinetic energy of a moving body is proportional to its mass (or weight). The
kinetic energy in the wind thus depends on the density of the air, or its mass
per unit of volume. So the "heavier" the air, the more energy is received by
the turbine. Also, the air is denser when it is cold than when it is warm. At
high altitudes, (in mountains) the air pressure is lower, and the air is less
dense. Also, the rotor area is an important factor because it determines how
much energy the wind turbine harvests from the wind (www.windpower.org).
Normally, the wind turbine deflects all the wind that is coming towards
it, so this means that it will never catch all the energy in the wind. This only
means that it could only catch part of the wind energy as stated in Betz Law.
Its because the more kinetic energy we absorb, more wind will be slowed
down as it goes away from the left side of the turbine (see fig. 4). As usual
one would think that extracting all the wind will make more energy but that
is untrue because if this is done, no air would leave the turbine as to air
would be prevented from entering the rotor. So, there must be some
breaking of the wind that must be done and it is explained in Betz Law.
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Figure 4. The Stream Tube, www. windpower.dk
Betz Law is a theory stating that no turbine can capture more than 60%
of the kinetic energy in the wind. This theory was developed by German
physicist Albert Betz in 1919. To further understand Betz Law, we assume
that the average wind velocity through the rotor area is the average of the
undistributed wind speed before the wind turbine which is V1, and the after
passing of the wind velocity leaving the turbine which is V2.
The mass of air flowing through thee rotor per second is equal to the
formula:
M = F (V1 + V2)/2
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Where:
M = the mass per second
= the density of air
F = the swept rotor area
(V1 + V2)/2 = the average wind velocity in the rotor zone
Thus the power that is coming from the wind by the rotor is equal to the
mass times the wind velocity drop squared (based on Newtons Second Law):
P = (1/2) m (V12 * V22)
Then we substitute the mass into this formula from the first expression
we will arrive at the following expression for the power that is extracted from
the wind
P = /4 (V12 * V22) (V1 + V2) F
Comparing the result with the total power in the undisturbed wind
streaming exactly the same area F, with no rotor blocking the way of the
wind. Lets call this Power Po:
Po = /2 V13F
So the ratio between the extracted power and the undisturbed power is:
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P/Po = (1/2)(1- {V2/V1}2)(1+{V2/V1})
Plotting their relationship in a graph, we can see that the function
reaches its maximum for V2/V1 which is equal to 1/3 and the maximum
value extracted from the wind is 0.59 or 16/27 of the total power of the wind
(Betz, 1966).
Figure 5. Graph showing ratio of Power and Velocity of the wind
2.5 History of Pumps
Many pumps were invented around the world during the old times.
Among the ones that was the first to be used is the shadoof. A shadoof was
originally developed in ancient Mesopotamia, which also appeared on a
Sargonid Seal of c2000BC (Needham, 1965). It was used as an irrigation tool
during 2000BC. The shadoof consists of an upright frame on which is
suspended a long pole or branch, at a distance of about one-fifth of its length
from one end. At the end of the pole lies a bucket or skin bag while on the
other side carries a clay which acts as a balance. Typically this works like
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scooping up the water from a well. With an almost effortless swinging and
lifting motion, the waterproof vessel is used to scoop up and carry water
from one body of water (typically, a river or pond) to another. At the end of
each movement, the water is emptied out into runnels that convey the water
along irrigation ditches in the required direction. It is estimated that a
shadoof can raise over 2,500 litres per day. Typical discharge of the well is
around 2 litres per second. Maximum water depth may be up to 3 meters.
They are still used up to now in some rural parts of the world.
Fig ure 6.
Shadoof in Egypt Figure 7. Shadoof in Romania
Then during the year 200BC, a Greek inventor and mathematician in
Alexandria, Egypt whose name was Ctesibius invented the first reciprocating
pump. Ctesibius wrote the first treatises on the science of compressed air
and its uses in pumps. Also, the principle of the siphon was attributed to him.
He also described one of the first force pumps for producing a jet of water, or
lifting water from wells, examples of these have been found in various
Roman sites such as the Silchester in Britain (Encyclopaedia Britannica).
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Also, during the 250 BC, pistons pumps were already present. They
were invented by the Romans. Authors like Vitruv and Heron of Alexandria
gave descriptions of pumps and other mechanisms whose function is based
on cylinders and pistons.
Figure 8. An example of a Piston Pump during 250BC
In the 3rd Century BC, the invention of the water screw pump was
attributed to the Greek Archimedes. So in honor of Archimedes it was named
after him therefore calling it the Archimedes Screw. This machine was
historically used for transferring water from a low lying body of water into
irrigation ditches (Oleson, 1984).
Figure 9. Archimedes screw
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This screw consists of a screw inside a hollow pipe that is usually
turned by hand or ran upon during the old times. At the bottom end, the
screw scoops up water which is then brought upward by the rotating motion
of the screw, after that the water when it reaches the top of the screw it
pours out thus feeding the irrigation systems.
In 1650, a German scientist, namely Otto von Guericke invented the
vacuum pump. This vacuum pump consisted of a piston and an air gun
cylinder with two way flaps designed to pull air out of whatever vessel it was
connected to, and used it to investigate the properties of the vacuum in
many experiments (Wikipedia).
For folks with hand dug, rock or wood lined wells, a chain pump
appeared on the market along about 1870. It had a series of rubber washers
attached to an endless chain attached to the crank. The rubber washers fit
snugly into a metal tube on the upward travel, carrying water to the top of
the well, where it spilled into the spout or outlet. This was a refinement of an
ancient Roman system for carrying water to the surface with rags tied to a
chain and the water squeezed out at the spout by drawing the rag through a
short tube (A History of Mans Progress, 1978).
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Figure 10. An example of an old chain pump
The common well or cistern pump came out about 1880 and was a
refinement of a huge old English reciprocating pump of the sixteenth
century. It has a cylinder and a plunger. Both the cylinder and the leather
sealed plunger carry check valves that allow the water to go only one way,
usually straight up. Early day reciprocating pumps had wooden pipe or
casing and a glass lined cylinder down in the well. The cylinder must not be
placed more than twenty-two feet above the water level, for this is the
maximum height to which water can be sucked by vacuum. It can be pushed
hundreds of feet, however, the weight of the water itself being the only
limitation (A History of Mans Progress, 1978).
Figure 11. An example of the first common reciprocating pump, 1880
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The first centrifugal pump prototype which was characterized as a
water or mud lifting machine appeared as early as 1475 in a treatise by the
Italian Renaissance engineer Francesco di Giorgio Martini. True centrifugal
pumps were developed during the late 1600s. This is the time when Denis
Papin, who was a French physicist known for his invention of the steam
digester, made one with straight vanes. Then the curved vane was
introduced by British inventor John Appold at the Great Exhibition at the
Crystal Palace in 1851.
2.6 Dynamic Pumps
Dynamic pumps are one category of pumps under which there are
several classes, two of which are: centrifugal and axial. These pumps
operate by developing a high liquid velocity and converting the velocity to
pressure in a diffusing flow passage. Dynamic pumps usually have lower
efficiencies than positive displacement pumps, but also have lower
maintenance requirements. Dynamic pumps are also able to operate at fairly
high speeds and high fluid flow rates.
2.7 Centrifugal Pumps
This type of pump is a rotodynamic pump that uses a rotating impeller
to increase the pressure of the fluid. They are commonly used to move
liquids through a piping system. The fluid enters the pump impeller along or
near to the rotating axis and is accelerated by the impeller, flowing radially
outward into a diffuser or volute chamber, from where it exits into the
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downstream piping system. Centrifugal pumps are used for large discharge
through smaller heads. Some of the advantages of cenrtifugal pumps are,
smooth flow through the pump and uniform pressure in the discharge pipe,
low cost, and an operating speed that allows for direct connection to steam
turbines and electric motors.The centrifugal pump accounts for not less then
80% of the worlds pump production because it is more suitable for handling
large capacities of liquids than the positive-displacement pump.
Figure12. Warman Centrifugal Pump in a Coal Handling and Preparation Plant
(CHHP) Application
A centrifugal pump works by the conversion of the rotational kinetic
energy, typically from an electric motor or turbine, to an increased static
fluid pressure. This action is described by Bernoullis principle. The rotation
of the pump impeller imparts kinetic energy to the fluid as it is drawn in from
the impeller eye and is forced outward through the impeller vanes and to the
periphery. As the fluid exits the impeller, the fluid kinetic energy is then
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converted to static pressure due to the change in area the fluid experiences
in the volute section. Typically the volute shape of the pump casing
(increasing in volume), or the diffuser vanes (which serve to slow the fluid,
converting to kinetic energy in to flow work) are responsible for the energy
conversion. The energy conversion results in an increased pressure on the
downstream side of the pump, causing flow.
Figure 13. Single Stage Radial Flow
Centrifugal Pump
Meanwhile a centrifugal pump containing two or more impellers is
called a multistage centrifugal pump. The impellers may be mounted on the
same shaft or on different shafts. The advantages to using a multistage
centrifugal pump or a single stage pump include:
o Producing a higher pressure head
o Discharging a larger amount of liquid
If a higher pressure head is to be developed then the impellers are
mounted on the same shaft (series) while for large discharge of liquid the
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impellers are mounted on different shafts (parallel). In getting a high head,
the number of impellers are connected in series. And in getting high
discharge, the impellers should be connected in parallel.
Figure 14. A Multistage
Centrifugal Pump
2.8 Energy Usage in Pumps
The energy usage in a pumping installation is determined by the flow
required, the height lifted, the length, and the characteristics of the pipeline.
The power required to drive a pump (Pi) is simply defined simply using SI
units by:
Where:
Pi - is the input power requirement in watts
is the fluid density in kg/m3
g is the gravitational constant (9.81m/s2)
H is the energy Head added to the flow in meters
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Q is the flow rate in m3/s
pump plant efficiency
The head added by the pump is a sum of the static lift, the head loss
due to friction and any losses due to valves or pipe bends all expressed in
meters of water. Power is more commonly expressed as KW (103 W) or
horsepower (multiply KW by 0.746). The value for the pump efficiency may
be stated for the pump itself as a combined efficiency of the pump and
motor system. The energy usage is determined by multiplying the power
requirement by the length of time the pump is operating.
2.9 Head Developed by Pump
The head developed HT, by a pump consists of the difference between
the head at the discharge flange and that at the pump suction flange. The
head can be solved by the formula:
P (hp) = Q(ft3/s) y (lbf/ft3) H (ft)
550 ft lbf/sec hp
Since the power requirement of any pump is a function of the capacity
and total head against which pump works, the total head HT, should be
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determined accurately. In determination of the total head, the following
statements should be borne in mind:
1. The datum line is the center line of the pump. All
vertical distances above the datum line have a plus sign
and all vertical distances below the datum line have a minus
sign.
2. The reference pressure is the atmospheric pressure.
At sea level, it is zero psi gage or 14.7 psia. All pressures above
atmospheric have a plus sign and all pressures below
atmospheric have a minus sign.
3. All loses of head due to friction turbulence in the
suction line have a minus sign and those losses of head due
to friction turbulence on the discharge line have a plus sign.
Based on the foregoing statements, the general equation for
determining the total head, HT is:
HT = Hd Hs
where: Hd = total discharge head in ft., = Zd + Pd / y + Vd2 / 2g
and Hs = total suction head in ft., = Zs + Ps / y + Vs2 / 2g
The total discharge head or sometimes called the dynamic discharge
head consists of the static discharge head plus velocity head on discharge
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plus head lost due to friction and turbulence (which is generally included in
friction head).
Hd = Zd + Pd / y + Vd2
/ 2g
The total suction head meanwhile, or sometimes called dynamic
suction head consists of static suction head or lift and the velocity head and
the friction head from source of supply to pump suction nozzle corrected to
datum line condition.
Hs = Zs + Ps / y + Vs2 / 2g
2.10 Centrifugal Pump Performance Curves with Different Impeller Sizes
Pump manufacturers use this graph in order to show that impellers of
several different diameters can be used in the same housing.
Pump Size & Type Ingersoll-Rand Pumps Curve no.
4x13LPO-F
Ring clearance 0.015in RPM 1170 Nss = 7000
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Figure 15. Actual Performance Curves for a Centrifugal Pump
Three different impeller diameter pump head curves are presented in
Figure 9 along with corresponding curves of constant values of efficiency and
input shaft power. BHP stands for Brake Horsepower, which is the same as
input shaft power.
Figure 15 provides pump head curves for impeller diameters of 5in.,
5.5in., 6in., 6.5in., and 7in. Also superimposed are constant efficiency curves
of 45%, 55%, 60%, 65%, and 68% and constant input power curves of 1/2
hp, hp, 1 hp, 1 hp, 2 hp, and 3hp. Pump head values range up to 190
gallons per minute (gpm).
2.11 Head versus Flow Rate Curves for Centrifugal Pumps
Figure 16. Head versus flow rate for a centrifugal pump for water and for a more
viscous liquid.
Figure 16 shows pump head versus flow rates curves for a centrifugal
pump. The maximum head produced by a centrifugal pump is called pump
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shutoff head because an external system is closed and there is no flow. We
can see in Figure 16 that as the external system resistance decreases (which
occurs when a system valve is opened more fully), the flow rate increases at
the expense of reduced pump head. Because of this the outflow rates
change significantly with external system resistance, centrifugal pumps are
rarely used in fluid power systems. Zero pump head exists if the pump
discharge port were reopened to the atmosphere such as when filling a
nearby open tank with water.
2.12 Performance Characteristic Curve for Centrifugal Pumps
When centrifugal pump manufacturers test their pumps, they typically
produce (for given geometry and speed) performance curves of head.
Overall efficiency, and input shaft power versus flow rate of the specified
fluid. Figure 17 shows these three curves plotted on the same graph. Not
that as the flow rates increases from zero, the efficiency increases from zero
until it reaches a maximum, and then it decreases as the maximum flow rate
is approached. The point where the maximum efficiency occurs is the best
efficiency point (BEP), and the corresponding flow rate is the design flow
rate. When selecting a pump for a given application, it is usually desirable to
use a pump that will operate near its best efficiency point. Maximum
efficiency values for centrifugal pumps typically range from 60% to 80%.
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Figure 17. Typical Performance Characteristic for Centrifugal Pumps
CHAPTER III
Research Methodology
3.1 Research Design
This study focuses on a design of a wind pump for the community of
UST Dominican Fathers at Caleruega, Nasugbu Batangas. Such study will
determine the conditions in the area that will be very important in designing
the components of the wind pump.
There were a number of factors that were considered in the design
such as the atmospheric conditions of the area and the location of the wind
pump. Some existing wind pumps were also analyzed by the researchers in
order to gain more knowledge in designing.
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3.2 Research Instrument
Design of existing windmill both local and foreign was evaluated based
on available research materials. This was done to determine the general
design of a windmill and applying to it the local conditions in the area and
the functional requirement. The group used research materials such as
books, encyclopedias, existing studies, and the internet to be able to get
useful information.
The group also conducted inspection trips to Caleruega in order to
make ocular assessments of the area. The group used their knowledge in the
subjects of Machine Design and Fluid Machinery to be able to solve the
power done by the windmill and the total dynamic head.
CHAPTER IV
Presentation, Analysis and Interpretation of Data
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Figure 18. Windmill and Pump Nomenclature
Figure 18 shows the researchers design of the wind pumps
mechanism. The rotor, which is a standard design of an American multi-
bladed windmill, is the prime mover of the whole system. It is composed of
18 blades (A) and has a diameter of 16 feet. The power that it can extract
from the wind will be transmitted through shaft 1 (N) which connects the
rotor and bevel gear 1 (B). Bevel gear 1 which is meshed with bevel gear 2
(C) will then transmit the shaft power from shaft 1 to shaft 2. Shaft 2, which
links bevel gears 2 and 3, supplies power to shaft 3 through bevel gears 3
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and 4. Sheave 1, which is connected to bevel gear 4, is belted to sheave 2 to
create a high speed rotation needed by the centrifugal pump to attain the
required power and head that will transport the rain water to the elevated
storage tank.
To prove that this mechanism design is efficient in operating a
centrifugal pump, the researchers made certain calculations and
assumptions based on mathematical theories and existing studies about
windmills and wind energy.
The pump power requirement to deliver the rain water from a man-
made pond to the elevated storage tank was first considered. The formula
used for calculating the pump power is shown below:
Parameters:
Volume of existing storage tank is 1335ft3
Assuming that the minimum wind pump operation per day with
maximum wind speed of 8m/s would be 3 hours only.
Since only 50% of the tank volume is used for irrigation per day;
therefore, water flow rate is:
Q = (Virrigation)(hours of operation) = 0.5 (1335ft3)(1 day/3 hrs of
operation)(1 hr/60 min) Q = 3.71 ft3/min
Output power of the pump
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(1) Pout = water HT Q
where:
Pout is the output power needed by the centrifugal pump
wateris the specific weight of water, 62.4 lb/ft3
HTis the total dynamic head
Q is the water flow rate of the existing storage tank, 3.71 ft3/min
To compute for the total dynamic head:
(2) HT = Hdischarge Hsuction
where:
Hdischarge is the total discharge head
o Hdischarge = Zd + Pd/ + Vd2/2g + Hfd
Zd is the static discharge head, 32ft
Pressure head (Pd/ ) and velocity (Vd2/2g ) head are
negligible
Hfd is the head loss due to friction turbulence at the
discharge line
Hsuctionis the total suction head
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o Hsuction = Zs + Ps/ + Vs2/2g + Hfs
Zs is the static suction lift at low water level, -4ft
Pressure head (Ps/ ) and velocity (Vs2/2g ) head are
negligible
Hfs is the head loss due to friction turbulence at the suction
line
To get the total discharge head and total suction head, the head losses
on discharge and suction lines are estimated using the Darcy-Weisbach
Equation:
(3) Hfs = fLs V2/Ds 2g
where:
fis called the friction factor (a dimensionless parameter)
Ls is the estimated length of suction straight pipe, 50ft
Ds is the actual inside diameter of pipe
o Since the nominal pipe size is designed to be 1.5 in GI pipe
schedule 40, the actual inside diameter is 1.610in.
Vs is the average fluid velocity at suction pipe
g is the acceleration due to gravity, 32.2ft/s2
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Computing the average fluid velocity in the suction pipe:
(4) Vs = Q/As
where:
As is the cross-section area of the suction pipe
o As = (/4) (Ds2)
Solving eq. (4)
Vs = (3.71ft3/min)(1min/60s)/( /4)(1.610in)2(1ft/12in)2
Vs = 4.37 ft/s
To compute for the friction factor, the type of fluid flow is determined
through the Reynolds Number, NR:
(5) NR = VsDs/vwater
where:
vwater is the kinematic viscosity of water at based on surrounding
temperature,
0.926x10-5 ft2/s
Solving eq. (5)
NR = (4.37ft/s)(1.610in)(1ft/12in)/( 0.926x10-5 ft2/s)
NR = 63,316
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Since 63,316 > 4000, then the flow is classified as turbulent.
Relative Roughness = GI pipe/Ds
= 0.0005ft/(1.610in)(1ft/12in)
= 0.00373
Therefore, plotting the intersection of the Relative Roughness and Reynolds
Numberon the Moody Diagram: f= 0.0295
Solving eq. (3)
Hfs = (0.0295)(50ft)(4.37ft/s)2/(1.610in)(1ft/12in)(2)(32.2ft/s2)
= 3.26ft
(6) Hfd = fLd V2/Dd 2g
where:
fis called the friction factor (a dimensionless parameter)
Ld is the estimated length of discharge straight pipe, 180ft
Dd is the actual inside diameter of pipe
o Since the nominal pipe size is designed to be 1 in GI pipe
schedule 40, the actual inside diameter is 1.049in.
Vd is the average fluid velocity at discharge pipe
g is the acceleration due to gravity, 32.2ft/s2
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Computing the average fluid velocity in the discharge pipe:
(7) Vd = Q/Ad
where:
Ad is the cross-section area of the discharge pipe
o Ad = (/4) (Ds2)
Solving eq. (7)
Vd = (3.71ft3/min)(1min/60s)/( /4)(1.049in)2(1ft/12in)2
Vd = 10.30 ft/s
To compute for the friction factor, the type of fluid flow is determined
through the Reynolds Number, NR:
(8) NR = VdDd/vwater
where:
vwater is the kinematic viscosity of water at based on surrounding
temperature,
0.926x10-5 ft2/s
Solving eq. (8)
NR = (10.30ft/s)(1.049in)(1ft/12in)/( 0.926x10-5 ft2/s)
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NR = 97,234
Since 97,234 > 4000, then the flow is classified as turbulent.
Relative Roughness = GI pipe/Dd
= 0.0005ft/(1.049in)(1ft/12in)
= 0.00572
Therefore, plotting the intersection of the Relative Roughness and Reynolds
Numberon the Moody Diagram: f= 0.0325
Solving eq. (6)
Hfd = (0.0325)(180ft)(10.30ft/s)2/(1.049in)(1ft/12in)(2)(32.2ft/s2)
= 110.24ft
Solving eq. (2)
Since:
Hsuction = Zs + (-Hfs)
= -4ft 3.26ft
= -7.26ft
Hdischarge = Zd + Hfd
= 32ft + 110.24ft
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= 142.24ft
Therefore:
HT = 142.24ft (-7.26ft)
HT = 149.5ft
Solving eq. (1)
Pout = [(62.4lb/ft3)(149.5ft)(3.71ft3/min)]/[33,000 ftlb/min HP]
Pout = 1.05 HP
Input power needed by the pump
Considering a conservative pump efficiency of 70%:
pump = Pout/Pin
Pin = 1.05 HP/.70
Pin = 1.5 HP
Since the power input needed by the pump is 1.62 HP, the power to be
produced by the rotor less all the losses in the mechanism should be greater
than or equal to the needed power input by the pump. Also, a certain driver
speed will be considered to reach the desired head. Therefore, to determine
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the power and shaft speed that the rotor can produce such calculations
should be done:
To harness the power from the wind
(9) Pwind = Arotor Vwind3
where:
Pwind is the wind power
is the air density based on 26C temperature and 2100ft elevation at
Caleruega, 0.0679lb/ft3
Arotor is the swept rotor area
o Arotor = (/4) Drotor2
o Drotor = diameter of rotor =16ft
Vwind is the average wind speed at Caleruega based on wind map, 8m/s
Solving eq. (9)
Pwind = (0.0679lb/ft3)(1kg/2.2lb)(3.28ft/1m) 3(/4)
(16ft)2(1m/3.28ft)2(8m/s)3
= 5210.64 kgm2/s3 since: kg = Ns2/m
= 5210.64 (Ns2
/m)(m2
/s3
) since: Nm/s = J/s = W
= (5210.64 W)(1HP/746W)
Pwind = 6.98 hp
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Although the computation above showed the power in the wind, the
actual power that can be extracted from the wind is significantly less than
what that figure suggests. The actual power will depend on several factors,
such as the type of machine and rotor used, the sophistication of blade
design, friction losses, the losses in the pump or other equipment connected
to the wind machine, and there are also physical limits to the amount of
power which can be extracted realistically from the wind. Thus, according to
the Betz limit theory, it is shown that any windmill can only possibly extract a
maximum of 59.3% of the power from the wind. But, in reality, for a wind
pump, that figure usually ranges from 25% to 40% only.
The said percentage is the proportion of the power in the wind that the
rotor can extract depending on the several technical parameters that are
used to characterize windmill rotors. This proportionality constant is termed
as the coefficient of performance (or power coefficient or efficiency; symbol
Cp) and its variation as a function of tip-speed ratio is commonly used to
compute the actual power that a certain type of windmill rotor can get from
the wind. Hence, the following calculations thereafter have been done as
proof for the presented theory on wind power extraction.
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Figure 19. Windmill rotor
Blade Specification
Thickness of GI sheet: gage 15
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Power transmitted by the rotor
(10) Protor = Cp Arotor Vwind3
where:
Protor is the actual power extracted by the rotor from the wind
Cp is the coefficient of performance of a windmill
o Since the design is a typical multi-bladed rotor, Cp = 0.4
Solving eq. (10)
Protor = (0.4)(0.0679lb/ft3)(1kg/2.2lb)(3.28ft/1m) 3(/4)
(16ft)2(1m/3.28ft)2(8m/s)3
= 2084.25 kgm
2
/s
3
since: kg = Ns
2
/m
= 2084.25 (Ns2/m)(m2/s3) since: Nm/s = J/s = W
= (2084.25 W)(1HP/746W)
Protor = 2.79 hp
Since the rotor is attached to shaft1, the shaft power is equal to the rotor
power.
Protor= 2.79 hp = Pshaft1
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The resulted power of the rotor from the computation is still not the
actual power that will drive the centrifugal pump. Losses in the whole
mechanism are still to be considered. Calculations of power losses are shown
below:
Figure 20. Power transmission from shaft 1 to shaft 2
According to Kent (1976), studies say that rigidly mounted and well
designed bevel gears are 98% efficient in transmitting power; therefore:
Power transmission from shaft 1 to shaft 2
Pshaft2 = 0.98Pshaft1 = (0.98) (2.79 hp)
Pshaft2 = 2.73 hp
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Speed of Bevel Gear 1
Vrotor = DrotorNrotor
* transforming the formula to get Nrotor
(11) Nrotor = Vrotor/Drotor
where:
Nrotor is the speed of rotor
Vrotoris the velocity of rotor
*to get the velocity of rotor, Vrotor
Vrotor = (TSR) (Vwind)
where:
TSR (Tip Speed Ratio) is 2 based on Cp of 0.4
Vrotor = (2) (8m/s)
Vrotor = 16 m/s
Solving eq. (11)
Nrotor =[16 m/s] [60s/1min] / [()(16ft)(1m/3.28ft)]
Nrotor = 62.64 rpm
Since the rotor and bevel gear 1 are on the same shaft, therefore:
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Nrotor = Nbg1
Nbg1 = 62.64 rpm
Speed of Bevel Gear 2
Dbg2Nbg2 = Dbg1Nbg1
* transforming the formula to get Nbg2
(12) Nbg2 = (Dbg1Nbg1)/( Dbg2)
where:
Nbg2 is the speed of bevel gear 2
Dbg1 is the diameter of bevel gear 1 based on the design, 12in
Nbg1is the speed of bevel gear 1, 62.64 rpm
Dbg2 is the diameter of bevel gear 2 based on the design, 4in
Solving eq. (12)
Nbg2 = [(12in)(62.64 rpm)]/[4in]
Nbg2 = 187.92 rpm
Since the bevel gear 2 and bevel gear 3 are on the same shaft, therefore:
Nbg2 = Nbg3
Nbg3 = 187.92 rpm
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To prove that shaft 1 exerts enough force to drive shaft 2, the torque on both
shafts should be compared.
Torque available on shaft 1
Tshaft1 = Pshaft1/2Nshaft1
Nshaft1 = Nbg1 = Nrotor = 62.64 rpm
Tshaft1 = [2.79 hp(33000 ft-lb/min-hp)]/[2(62.4 rpm)]
Tshaft1 = 234.83 ft-lb
Torque available on shaft 2
Tshaft2 = Pshaft2/2Nshaft2
Nshaft2 = Nbg2 = 187.92 rpm
Tshaft2 = [2.73 hp(33000 ft-lb/min-hp)]/[2(187.92 rpm)]
Tshaft2 = 76.3 ft-lb
Since the torque on shaft 1 is greater than the torque on shaft 2, therefore
shaft 1 is capable of driving shaft 2.
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Figure 21. Power transmission from shaft 2 to shaft 3
Considering 98% efficiency of Power transmission from shaft 2 to shaft
3 which is connected to spur gear 1:
Pshaft3 = 0.98Pshaft2
= (0.98)(2.73 hp)
Pshaft3 = 2.68 hp
Speed of Bevel Gear 4
Dbg4Nbg4 = Dbg3Nbg3
* transforming the formula to obtain Nbg4
(13) Nbg4 = (Dbg3Nbg3)/( Dbg4)
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where:
Nbg4 is the speed of bevel gear 4
Dbg4 is the diameter of bevel gear 4 based on the design, 4in
Nbg3is the speed of bevel gear 3, 187.92 rpm
Dbg3 is the diameter of bevel gear 3 based on the design, 12in
Solving eq. (13)
Nbg4 = [(12in)(187.92 rpm)]/[4in]
Nbg2 = 563.76 rpm
Since the bevel gear 4 and sheave 1 are on the same shaft, therefore:
Nbg4 = Nsheave1
Nsheave1 = 563.76 rpm
To prove that shaft 2 exerts enough force to drive shaft 3, the torque on both
shafts should be compared.
Torque available on shaft 3
Tshaft3 = Pshaft3/2Nshaft3
Nshaft3 = Nbg4 = Nsheave1 = 563.76 rpm
Tshaft3 = [2.68 hp(33000 ft-lb/min-hp)]/[2(563.76 rpm)]
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Tshaft3 = 24.97 ft-lb
Torque available on shaft 2
Tshaft2 = 76.3 ft-lb
Since the torque on shaft 2 is greater than the torque on shaft 3, therefore
shaft 2 is capable of driving shaft 3.
Torsional stress on shaft 1
St1 = 16Tshaft1/dshaft13
Where:
dshaft1 is the diameter of shaft 1 based on the design, 1.5in
Su is the ultimate strength of the material used for the shaft, 122,000
psi-AISI 4340
St1 = [16(234.83 ft-lb)]/[(1.5in)/(12in/ft)]3
= 612,341.437 lb/ft2 (1ft/12in)2
St1 = 4,252.37 psi
Since:
Su (122,000 psi) St1 (1,252.37 psi)
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Therefore, the diameter of shaft 1 and the material used can endure the
torsional stress induced by the shaft torque.
Torsional stress on shaft 2
St2 = 16Tshaft2/dshaft23
Where:
dshaft2 is the diameter of shaft 2 based on the design, 1.5in
Su is the ultimate strength of the material used for the shaft, 122,000
psi-AISI 4340
St2 = [16(76.3 ft-lb)]/[(1.5in)/(12in/ft)]3
= 198,959.467 lb/ft2 (1ft/12in)2
St2 = 1,381.66 psi
Since:
Su (122,000 psi) St2 (1,381.66 psi)
Therefore, the diameter of shaft 2 and the material used can endure the
torsional stress induced by the shaft torque.
Torsional stress on shaft 3
St3 = 16Tshaft3/dshaft33
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Where:
dshaft3 is the diameter of shaft 3 based on the design, 1.5in
Su is the ultimate strength of the material used for the shaft, 122,000
psi-AISI 4340
St3 = [16(24.97 ft-lb)]/[(1.5in)/(12in/ft)]3
= 65,111.637 lb/ft2 (1ft/12in)2
St3 = 452.16 psi
Since:
Su (122,000 psi) St3 (452.16 psi)
Therefore, the diameter of shaft 3 and the material used can endure the
torsional stress induced by the shaft torque.
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Figure 22. V-Belt drive
Speed of Sheave 2
Dsheave2Nsheave2 = Dsheave1Nsheave1
* transforming the formula to get Nsheave2
(14) Nsheave2 = (Dsheave1Nsheave1)/( Dsheave2)
where:
Nsheave2 is the speed of sheave 2
Dsheave2 is the diameter of sheave 2 based on the design, 3in
Nsheave1is the speed of sheave 1, 563.76 rpm
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Dsheave1 is the diameter of sheave 1 based on the design, 18in
Solving eq. (14)
Nsheave2 = [(18in)(563.76 rpm)]/[3in]
Nsheave2 = 3,382.56 rpm
Since sheave 2 is the driver of the pump, therefore:
Nsheave2 = 3,382.56 rpm = Rated Speed Output of the windmill
mechanism
Number of belts needed:
Since Design hp is 2.68 and rpm of smaller sheave is 3,382.56, section A V-
belt is to be used. (Figure 17.14 of Design of Machine Elements by Faires)
(15) Vbelt = Dsheave1Nsheave1
where:
Vbeltis the belt speed
Solving eq. (15)
Vbelt = (18in)(1ft/12in)(563.76 rpm)
Vbelt = 2,656.66 ft/min
(16) Lbelt = 2C + (/2)(Dsheave1 + Dsheave2) + (Dsheave1 Dsheave2)2/4C
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where:
Lbeltis the length of V-belt
C is the center line distance of sheave 1 and sheave 2 based on the
design, 18in
Solving eq. (16)
Lbelt = [2(18) + (/2)(18in + 3in) + (18in 3in)2]/ 4(18in)
Lbelt = 72.095 in
Therefore, standard length of belt is 76.3inA75 (Table 17.3 of Design of
Machine Elements by Faires)
Considering the design factors of section A V-belt on Design of Machine
Elements by Faires:
(17) Rated hp = [a (103/Vbelt)0.09 (c/KdDsheave2) b(Vbelt2/106)][Vbelt/103]
where:
a is 2.684, b is 0.0136, c is 5.326
Kdis 1.14
Solving eq. (17)
Rated hp = [2.684 (103/2,656.66)0.09 (5.326/(1.14)(3))
0.0136(2656.662/106)][2656.66/103]
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Rated hp = 2.14 hp/strand
Adjusted hp = Rated hp KKL = 2.14hp/strand(0.87)(1.02)
Adjusted hp = 1.90
No. of V-belts = Design hp/Adjusted hp = (2.68 hp)/(1.90 hp/strand)
No. of V-belts = 1.4 2 strands
Considering the calculations made with the required pump power input
to transport the rain water from the man-made pond to the elevated tank,
the wind pumps mechanism can operate a centrifugal pump with a standard
specification of:
Pump Power Input = 1.5 hp 2 hp
Pump Speed 3,382 rpm
Maximum head of at least 150 ft
]
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MATERIALS AND COSTING
(Gears, shafting, bars, and rods are all AISI 4340 and rods and bars are
available 20 ft max
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TOTAL COSTING: Php 108,630.00
Chapter V
Summary, Conclusions and Recommendations
5.1 Summary
The researchers have gathered data and information from different
references and existing studies to broaden their ideas and wind energy and
how to harness it efficiently using a wind pump. Since typical windmills
seldom operate a centrifugal pump, the group studied and considered
numerous theories and principles to come up with a patented and feasible
design of the mechanism.
Also, ocular inspections were done by the researchers at Caleruega so
that actual conditions of the site will be used to arrive at more accurate
results. Moreover, the researchers collated formulas on aerodynamics and
machine design to support their design through mathematical computations.
5.2 Conclusion
To justify the design of the windmills mechanism, the pump power
requirement was compared to the power that the mechanism can supply.
Hence, based on the calculations made, the power transmitted by the
mechanism met the power requirement of the centrifugal pump. Also, the
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mechanism produced a high shaft speed that is needed to require a
centrifugal pump and attain the required total dynamic head.
5.3 Recommendation
The researchers recommend a more precise and accurate data
gathering in the location. Testing wind speeds must be done over a period
within the location in order to get better results in computations. Also, it is
important to know the time and duration when the wind velocity is at its
peak.
The researchers also recommend conducting experiments and tests.
Since most of the formulas are based on theories, conducting experiments
will further support the wind pump design. Also, the researchers recommend
some additional set ups on the design when it is needed. Upgrading the
design when needed may be costly but it will save money in the future.
Examples of these upgrades are the overrunning clutch and the gear train
mechanism.
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INDICATE REFERENCE SOURCES OF THE PICS/IMAGES IN THE REVIEW
OF RELATED LIT
BIBLIOGRAPHY AFTER CHAPTER V FOLLOWED BY THE APPENDICES
AND CVs