e-ISSN: 2582-5208

International Research Journal of Modernization in Engineering Technology and Science

Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com

DESIGN AND DEVELOPMENT OF BUOYANCY GRAVITY POWER PLANT

Akshay Kumar*1
*1Master’s Thesis Researcher, Sapienza University Of Rome , Rome, Lazio , Italy.
ABSTRACT
The main aim of this thesis is to design a prototype of buoyancy gravity power plant and find out the
production capacity of the system. This new technology has been demonstrated with all necessary calculations
and a complete layout of a power plant that works entirely on buoyancy gravity power production technology
has been developed. CATIA has been used for designing the weight cell, magnetic valve, airbag, one-way bearing
and few other components and Auto-CAD for sketching the plant layout. With the help of some important
articles and journals, the author calculated and selected suitable components according to my needs like
braking system, transmission devices, selection of generator, air compressor, air compressor tank, rope size,
etc. The ideology behind designing this power plant is to produce power using buoyancy and gravity forces by
keeping the losses at a minimum level. The results that I need to find out are the velocity of the falling object,
generated forces, power produced at the driving shaft, total time taken by the system for the complete
operation, output power and efficiency of the system.
I. INTRODUCTION
Renewable energy is the most important topic in the world at present. It was identified that the fossil fuel
reserves in the world are diminishing rapidly and no reserves were identified. In addition to that, the usage of
fossil fuel costs more environmental problems like the emission of greenhouse gasses, global warming, and acid
rains. To prevent the pollution caused by fossil fuels, renewable energy sources play a major role. Electricity
can be produced by the accumulation of kinematic energy produced by the falling objects; we can call it as
gravitational energy. This is one of the new kinds of technology and a lot of new research are going in this field.
Most of them are on electricity storage process by gravitational power (examples hydro-energy storage system)
but not on electricity production because it requires lots of energy to restore the kinetic energy which is
produced by the falling objects. This project aims to design a gravitational energy production plant and to
enhance the restoring capacity of the system, I am implementing buoyancy power to restore the energy.
1.1. Buoyancy gravity power production system.
This is the new technology which can produce electricity with the help of kinetic energy produced by the object
which is falling underwater and in order to lift the object from the bottom surface of the water I am using
buoyancy force. The whole operation is taking place underwater, and the operation of the system takes place in
two cycles, one is the production cycle and another one is the recharge cycle. A heavyweight object is attached
to the rope and pulley arrangements which is in turn connected to the generators with the help of drives and
shaft arrangement. In the production cycles, the object is released from the surface of the water and due to
gravitational force, it will fall towards the bottom of the well and that produces some amount of kinetic energy.
We can use this to generate power with the help of a power generator. When the weighted cell reaches the
bottom, the system will be stopped with the help of the hydraulic breaks.
The operation of the recharge cycle starts from here. At the bottom, there is a magnetic valve, that can
automatically connect and supply’s the compressed air to the airbag through the hollow shaft (another special
application of this magnetic valve is that with the help of the magnetic pull force, it can hold the entire system
to stay connected until the airbag will reach its maximum level. When the buoyancy force increases in the
airbag, it will automatically separate). The air will start blowing the rubber airbag and when the balloon
reaches its maximum level, the entire system will separate from the magnetic valve and starts moving towards
the surface. This is because it will get a greater buoyancy force than the pulling force of the magnet.
1.2 System design and operation
The operation of the system takes place in two cycles, the production cycle and the recharge cycle. The
production cycle helps to produce electricity using the gravitational force generated by the falling object, while
the recharge cycle helps to restore the energy by lifting the weight from the bottom of the well. the production

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3234]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
efficiency of the plant has been estimated designing the plant layout of a possible prototype and calculating all
the main parameters such as velocity, acceleration of the falling object, the total time required to complete both
the cycles, the generated forces, and few other parameters. The sizes of all the components like weight cell,
magnetic valves, generator, drives, brakes, one-way bearings, have also been estimated.
The complete operation of the plant is underwater so forces acting on the weight cells like drag force ,
viscosity and buoyancy forces have been calculated.
1.3 Production cycle (cell discharge)
In the production cycle, the weight cell arrangement is attached to the rope and pulley system (to get effective
velocity and to reduce the drag loss, we are using a sphere-shaped weight cell). The weight cell is released from
the surface of the water and due to gravitational force, it will fall towards the bottom of the well and produces
an amount of kinetic energy. The rope is attached to the pulley and shaft arrangements so we can easily capture
that energy and we can use that to produce electricity. In order to design the plant layout, we need to assume
and fix some input parameters such as well depth, diameter, the mass of the weight cell, etc. and for this project,
we are giving the input parameters shown in table 1.
 Input parameters
Table 1-Input parameters
Parameter Input values Units
Depth (d) 250 M
The diameter of the well (D) 0.9144 M
Mass of the suspended weight cell (M) 300 Kg
The density of the water 1000 kg/m^3
The density of mild steel ( 7850 kg/m^3
Gravitational force (g) 9.81 m/s^2
The drag coefficient for a sphere 0.47 -
Dynamic viscosity of water at 20 °C 1.002 *10-6 m2/s
1.1.1. Designing of weight cell and frame arrangements

Figure 1 – Model of the weight cell

Based on the given parameters available in table 1, we have to design the weight cell and frame arrangements
based on its mass which is 300kg. In that, 260 kg is for the weight cell and the remaining 40 kg will be used for
the frame airbag design. For this project, we are choosing a spherical-shaped mild steel ball as a weight cell or
suspended weight because it has a lesser friction and drag coefficient compared to other shapes. On the top of
the weight cell, there is a properly designed frame arrangement that can help to connect the airbag and the
weight cell through a screw connector and the rope is attached to the frame with the help of a hook. On the
weight cell, there are 3 small wheel arrangements that are attached at an angle of 120 degrees to each other.
These wheels help to reduce the friction between the walls of the well and give a proper linear motion in the
falling and lifting cycle.

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3235]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
The frame weight will be 40 kg (including the weight of the airbag) and there is a 3-inch outer diameter, 1.5-
inch inner diameter and 6 feet long hallo shaft made up of mild steel. This shaft helps to hold the weight of the
complete weight cell and helps to supply compressed air to the airbag. Both sphere and the hollow shafts are
connected to each other through a screw connector. The entire rubber airbag is attached around this pipe and
there are several small holes on the shaft which helps to blow the higher-pressure air into the airbag. the upper
body, there is also a wheel arrangement which can help to reduce the vibration during sinking and lifting cycles.
On the top of the wheel arrangement, there is a 5-inch long solid rod which connect the entire set to the rope
arrangements with the help of a hook.
 Calculation
The weight cell is spherical in shape. In order to design it, we need to calculate the radius of the sphere. To find
the radius we need to calculate the volume of the sphere with help of the equation 1.

(1)

= 260/7850 m3
= 0.3312 m3

(1.1)

.03312=1.33*3.14*
= 0.1994 m
The radius of the sphere is 0.1994 m and the diameter will be 0.3988 m
1.1.2. The terminal velocity of the falling object
In order to find the kinetic energy of the falling object, firstly we need to calculate the velocity of the falling
weight and with the help of that, we can also find the time and acceleration of the falling weight cells. we need
to assume the initial velocity of the weight cell to be zero and we need to find the terminal velocity of the falling
object using the equation 2
To find the terminal velocity, we need to assume that all the forces will be equal to zero.
=0
(2)
We need to rewrite equation 2 by substituting all the known formulae and equation 3 is the evaluated form of
equation 2.
=0 (3)

Equation 3 is a quadratic equation, and the only unknown is terminal velocity. So, to find the terminal velocity,
we need to write it in the form of a quadratic equation.
=0
This is a quadratic equation in the form of (a ) where = x. So, we can use the quadratic
equation formula to find the unknown. Equation 4 is the formula to find out the unknown in the above
quadratic equation.
√
(3.2)

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3236]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
The solution of equation 3.2 gives us the terminal velocity of the falling object but before solving that, we need
to calculate and substitute the values of a, b, and c. With the help of equation 3 we can extract and calculate the
values of a, b and c. The formulas are listed below and numbered as equations 3.1.1, 3.1.2, and 3.1.3.
a=
(3.1.1)

We need to calculate the area of the circle using equation below (In this case, we need to use the formula for
the area of a circle because only one face is facing opposition).
A= π

A= 3.14*0.1994^2 m2
A = 0.1248 m2
The area of the circle will be 0.1248 m2 . By substituting this to equation 3.1.1 ,we will get the value of a.
a=
a = ½ *0.47*1000*0.1248
a = 29.328 kg/m
We know all the values of all the parameters in the equation 3.1.2 and 3.1.3 and by substituting that, we can get
the values of b and c.
b=

(3.1.2)
b= 6*3.14*1.002 * 10-6*0.1994
b = 3.7642 * 10^6 m3/s
and
c=

(3.1.3)
c= 300*9.81-(1000*.03312*9.81)
c= 2618.0928 N
We calculated the values of a, b, and c. To find the terminal velocity ,we need to substitute those values in
equation 3.2 and we need to assume that for the falling object, the downward direction will be always positive.
√

The terminal velocity of the falling object will be

1.1.3. Sink time calculation
To design the production cycle, we need to calculate the falling time of the weight cells. To calculate the falling
time, we first need to calculate the acceleration and with the help of equation 4 and 4.1, we can find the
acceleration of the falling object and falling time.
(3)

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3237]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
(4.1)

a = 0.17845 m/s^2
By substituting the known values to equation 4, we can calculate the falling acceleration which is 0.17845
m/s^2.

9.4452 =0 + 0.17845
= 52.96 s
We can’t calculate the exact velocity of the object, which is sinking underwater because while sinking, the
velocity of a falling object is not constant. It may vary because of water pressure but there is not much variation
in velocity and hence, our answer is considerable. We calculated the terminal velocity of the object but initially,
it takes a few seconds to reach the terminal velocity and there will be some losses due to friction. By
considering this factor, we can say it will take nearly 60s to reach the bottom.
1.1.4. Generated forces
When the weight cell is falling into the water, there will be some forces acting on both the directions. Force in
the downward direction is called force F and the opposing force is called buoyancy force. Based on the
properties of the fluid and the object shape, there will be a large amount of drag force and based on the fluid
viscosity, a small number of viscous forces also act in the upward direction. By substituting all the known
values to equation number 5.1, 5.2, 5.3, and 5.4 we can find all the generated forces.
downward force F= mg (4.1)
(5.2)

drag force (6.3)

viscous force = (7.4)

 Net downward force

To find the force generated in the downward direction, we need to substitute all the known values to equation
5.1
Downward force F= mg
F =300*9.81 N
 Buoyancy force
To find the buoyancy force, we need to substitute all the known parameters to equation 5.2.

1000*0.03312*9.81 N
=324.907 N
The generated bouncy force is 324.9072 N
If the buoyancy force is equal or greater than the downward force, the object will float. In our calculation, we
get a lesser buoyancy force and hence, the object will sink.
 Drag force
The drag force of a falling object depends on the shape of the object. To reduce this effect, we are using
spherical shaped weight cell because it has lesser drag coefficient (the generated drag force will be more but
this force is neglected here). To find the drag force, we need to substitute all the known parameters to equation
5.3.

= 29.328*9.44^2 N

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3238]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
Generated Drag force will be = 2618.09 N
The generated drag force will be 2618.09 N
 Viscous force
Based on the viscosity of the fluid, a force is generated. This force is called as viscous force and by substituting
the known parameters to equation 5.4, we can find its value.
Viscous force =
= 6*3.14*1.002*10-6 * 0.1962*9.44 N
Viscus force will be = 3.55*10-5 N
Generated viscous force will be 3.55*10-5 N
We calculated all the generated forces but according to equation 3, we stated that the sum of all the forces of the
object which is falling underwater will be equal to zero.
So to verify that statement, we need to substitute all the forces in equation 3
=0
= 0.0029 N
The net force will be closer to zero so we can say that our statement is correct.
 Net sinking force
To find the net sinking forces, we need to subtract buoyancy force, viscous force and the downward force. To
find the net downward or sinking force, we need to neglect the drag force because the terminal velocity which
we found previously is constant from the top till bottom of the well. So, the net sinking force will be equal to
drag force.

The net sinking force will be 2619 N and the drag force will be 2618.09 N. Here, both forces are almost equal
and hence it proved the statement explained earlier to be correct.
1.1.5. The power generated in the pulley shaft
When the object is falling, some amount of potential energy will be generated. To collect that, we use a rope and
pulley arrangement. The rope is attached to the falling weight and that rope is connected to the recoiler.
Between that, the rope is wound on a pulley and when the object is falling, the pulley will start to rotate. The
pulley is attached to the shaft through a one-way bearing so that the shaft rotates along with the pulley. This
continuous rotation produces some amount of power in the shaft. Our aim in the production cycle is to find out
how much power can be produced by the falling weight and to find out that power, we use the power equation.
(6)

 Tension in the rope ( )

= (300*9.81) – (1000*0.03312*9.81)
= 2943 – 324.9
2618.09 Nm
 RPM calculation
The rope is attached to the pulley and shaft arrangement. By using that, we can collect the kinetic energy in the
form of mechanical energy like rational motion. So, the RPM is the most important parameter in our project in
order to find out the power.
We can calculate RPM with the help of equation 6.2

RPM = (6.2)

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3239]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
C = 2πr (6.2.1)
C=2*3.14*0.0762
C = 0.478 m
The circumference of the pulley is 0.487m and to get the RPM, we need to substitute this to equation 6.2. By
doing so, we get
RPM =

RPM =
RPM= 523.012
The weight cell will take 60 seconds to reach the bottom and in one minute it gives 523 rotation in the pulley.
We calculated all the unknown values from equation 6.
To find the power, we need to substitute all those values.

*3.14*523.0125)/60)
= 10.901 kW
The power developed in the shaft during the production cycle is 10.9011 kW. To simulate it with the generator
for producing a continuous output, we must give the same amount of torque and RPM up to 60 minutes. To
produce the continuous output, we need to design the complete plant layout with a greater number of well
arrangements. Here the assumption that we need to consider is that all the wells that we are using in this plant
have the same production capacity. To find the required number of wells we need to design and calculate the
recharge cycle also.
1.2. Design of Recharge cycle (lifting cycle)
Energy restoration or lift of the weight cell process is done in this cycle. To lift the weight cell from the bottom
of the well, we are attaching an airbag to the frame arrangements. This airbag is specially designed to lift the
300 kg weight cell. At the beginning of this cycle, we will supply the compressed air to the airbag through a
magnetic valve, so the airbag starts to bloat. When it reaches its maximum level, its buoyancy force will increase
more than the downward force and hence it can effectively lift the weight cell. When the weight cell starts
moving upwards, the rope will lose its tension. We need to recoil the rope for which we are using auto rope
recoilers.
 Assumptions
1. Friction produced in the pulley bearing assembly is negligible.
2. Friction between cable and pulley is negligible.
3. Drag losses from float motion are negligible
1.2.1. Designing of the airbag
 Input parameters
Table 2- Input parameters to design the air bag
Parameter Value Unit
The diameter of the airbag ( ) 0.6096 m
Height of balloon 1.5 M
Frame height with weight 2.2 M
Balloon material Rubber
Density of air 1.204 at kg/m^3
 Model of the airbag

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3240]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com

Figure 2 – Model of the airbag

The above figure is the model of the airbag which we designed with the help of the given input parameters
mentioned in the table. In this airbag, there are two kinds of valves. One at the bottom is the air inlet valve and
two at the top are the air outlet valves. With the help of the input parameter, we designed the model of the air
bag, but we need to check if the airbag can lift the cell are not.
 Volume of the airbag
This equation represents the formula to find the volume of the balloon.
=π (7)
3.14*.3048^2*1.5
= 0.4375 m3
 Net lifting force of the airbag
In order to find the effective lifting force of the airbag, we need to substitute all the known parameters to the
buoyancy equation 7.2.
(7.2)

We already calculated the required upward force in the production cycle which is 2943 N.
Now we found the lifting capacity of the airbag to be 4285.43N which is clearly more than the required force.
Hence it can easily lift the weight cell at the bottom.
 Total weight that the airbag can lift
To get the effective lifting capacity of the airbag in the terms of weight, we need to divide the net upward force
with gravity g.
(7.3)

The lifting capacity of the airbag in terms of weight is 436.84kg.

1.2.2. Upward moment acceleration and time calculation
The distance between the magnetic valve from the top of the well is 260 m. We already designed the airbag to
lift the weight cell as fast as possible. To design the plant layout, we need to calculate the lifting speed and time.
We already calculated the effective buoyancy force which is acting on the weight cell and frame arrangement
that is 324.90 N.
=P

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3241]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com

 The net buoyancy force produced by airbag

The net buoyancy force produced by the airbag is calculated using the equation 8.1

= 394.49 N
We calculate the volume of the airbag, but the shape of the airbag is a hollow cylinder. To find the weight of the
airbag, we need to calculate the volume of the hollow cylinder with the help of equation 9.3 and the thickness of
the airbag will be 0.00381 m
The volume of the airbag

= 0.01087 m^3
By substituting the known values to the below equation, we get the mass of the airbag.
Mass of airbag is =
920*0.01087
= 10.00 kg
The net weight of the airbag is 10.kg. The mass of the frame arrangement will be 30 kg and the total weight of
the system will be 300 kg.
 Lifting speed
The lifting speed of the airbag and the lifting time is calculated by equation 9 and 9.1

By equation 9.2 we can find the acceleration due to force

By substituting the values of acceleration and all the known values to equation 9 we will get the lifting velocity.
Velocity

= - 26.16 m/s
The upward velocity will be 26.1645 m/s. By substituting the value of acceleration and velocity to equation 9.1,
we can find the lifting time.
Time
26.16= 0+ 1.31
= 19.88 second
The total time taken by the airbag to lift the system from the bottom of the well is 19.88 seconds; there are
some losses, so I approximated to 30 seconds.
1.3. Designing of plant layout
To develop the suitable working structure of the production plant, we need to calculate the required time to
complete both the cycles because the time is an important factor in the plant design process. Only after finding
the required time, we can fix the required number of wells and some other parameters. To calculate the time

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3242]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
requirement of the recharge cycle, initially, we need to select the suitable air compressor and air tank. The
complete details on the air compressor and air tank is listed below.
1.3.1. Air Compressor and air tank selection
An air compressor is a pneumatic device that converts power into potential energy stored in pressurized air. By
one of several methods, an air compressor forces more and more air into a storage tank thereby increasing the
pressure. A storage tank will help to receive and store the pressurized air from the compressor.
For our project, we are selecting a standard air compressor with tank arrangement and the specification is
listed in below table.
Specification of the compressor and air tank
Table 3- specification of the air compressor.
Parameter Values Unit
Horsepower of compressor 8 Hp
The output pressure of the compressor 500 Psi
Output capacity of the compressor 40 Cfm
The capacity of an air tank 500 Gallon
The output pressure of the air tank 400 Psi
Output capacity of an air tank 20 Cfm
Maximum capacity of an air tank 40 Bar
 The time required to fill the air tank
The airbag which we are using is 500 gallons and the capacity is 40 bars. This air tank is assisting to give a
continuous supply of compressed air to the airbag during operation.
The detailed calculation is listed below. We know that one gallon is about 231 cubic inches one cubic foot is
exactly equal to 1728 cubic inches and one cubic foot occupies 7.48 gallons.
To find the volume of air that is present in the tank before compression, we need the following equation.
Volume in air tank = air tank capacity / 7.48
Volume in air tank= 500/7.48= 66.84 cubic feet of air in it
Therefore, there are 66.8449 cubic feet of atmospheric air in the tank at the beginning. Now we need to
calculate the maximum air storage capacity of the air tank by substituting the known values to equation 10.1.
We can calculate the capacity of the air tank as follows.
The capacity of air tank = (air tank pressure capacity in psi/cfm) *air in the tank
Air filling capacity of air tank = (580/40) *66.8449= 969.25 cubic feet
The total air storage capacity of the air tank is 969.25 cubic feet. Initially, there was 66.8449 cubic feet of air in
it. By subtracting this with the obtained value, we get 902.40615 cubic feet of air can that is filled in the tank.
We know 40 Cfm of air compressor produces 40 cubic feet of air every minute. To find the time, we need to
substitute all the calculated values to equation 10.2
Time to fill 500-gallon air tank = air filling capacity of the air tank / compensate output capacity
Time to fill 500-gallon air tank = 902.4061/40=22.56 minutes
For easier calculations, we can approximate this value to be 23 minutes, which is the time required to fill the
500-gallon air tank.
 The time required to fill the airbag
We already designed the structure of the airbag but now, we need to calculate the time requirement and the
capacity of the airbag. With the help of the pressure control valve, we can regulate the outlet pressure and flow
at the air tank level. Based on the elasticity and material of the airbag, the pressure capacity ranges from 20-25
psi. For our calculations, we fixed it to the maximum value that is 25 psi. But the water pressure acting on the
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[3243]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
airbag is 370 psi. By adding both the values, we can fix the inlet pressure which is 400 psi. Hence, we need to
regulate the output pressure of the air tank to 400 psi and air at 25 cfm. We already know that the volume of
the airbag is 0.49 m^3. So, the capacity of the airbag will be129.444 gallons.
We know that 1 cubic foot receives 7.5 gallons of air and the airbag capacity is 129.44 gallons. For calculation
purposes, we need to convert it to cubic feet which is 17.2572 cubic feet or in other words, the airbag receives
17.25 cubic feet of atmospheric air in it. At the tank level, we need to regulate the output pressure value to 400
psi. In that, 370 psi of pressure is required to push the water away from the airbag and the remaining is the
airbag capacity which we already know it to be 25-30 psi. Initially, there is no air present in the airbag. To find
the amount of air that will be filled in the airbag we need equation 8.3.
Amount of air that will be filled in the airbag = (airbag pressure capacity in psi/cfm) *air present
in airbag
Amount of air that will be filled in the airbag = 30/25 * 17.2572=20.708 cubic feet. The amount of air that will
be filled in the airbag is 20.70864 cubic feet and the Air tank delivers 25 cfm at 400 psi in one minute. To find
the required time to fill the airbag, we can use equation 8.4.
Time to fill = Amount of air that will be filled in the airbag/cfm
Time to fill is = 20.70864/25 = 0.8283 minutes or 50 seconds.
The time required to fill the air into the airbag is 50 seconds.
 Total time Requirement to complete both the cycles
In the production cycle, we calculated the time taken by the air bag to lift the weight cell which is 60 seconds.
We also found that the air filling time is 50 seconds. The lifting time of the airbag is 20 seconds and the output
pressure of the airbag is 20 psi. So, it hardly takes 30 seconds to remove all the air from the airbag.
By summing all calculated time, we can get the total time required to finish one cycle
One cycle takes = 60+50+20+30+3= 153 seconds
One cycle takes 2.7 minutes. By considering the losses and errors we can extend the time to 3 minutes .To
maintain the continuous production output, we are planning to design the plant layout with 5 wells
arrangements because weights are dropped in one by one in separated by a time gap of 1-minute.
The complete cycle for one well takes 3 minutes and the resting time for each well is 1 minute. Hence, every
well will repeat its working cycle again in 4 minutes of time interval. Therefore, we feel that 5 wells are enough
to produce continuous output.
1.3.2. Plant layout
With the help of all the calculations which we made in the above pages, we designed the complete plant layout
and we represented the layout in the figure below. In this figure there, are five production wells, each well is
having a separately mounted suspended weight cell. The weight cells arrangement is attached to the rope and
pulley arrangement. All the pulleys are attached to a single shaft through one-way bearings. The shaft is
mounted between each well with the help of supportive bearings. Another end of the rope is attached to the
rope coiler. In the recharge cycle, when the airbag starts moving towards the surface, this rope coiler will help
to automatically recoil the rope. When the weight cell reaches the top of the well, it will lock the system till next
operation. A generator is attached to the shaft through the belt drive and a fly wheel is attached in between the
generator. The fly wheel helps to prevent the vibrations coming from the driving shaft and gives a smooth
movement to the generator. On the other side of the layout, there is an air compressor and air tank
arrangement which supplies air to each well. A magnetic valve is present in each well connected to the air tank
with the help of pipe arrangement. During the production cycle, with the help of these pipes, the compressed air
will be supplied to the magnetic valve. The effective floor area of the plant will be less than 400 f^2. It occupies
less surface area. We can use this arrangement on the ground and, we can use this construction on the water
reservoir or in the ocean. The sketch is done by AutoCAD 2D software.

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3244]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com

Figure 3 – Layout of the buoyancy gravity power plant

1.4. Power Output
1.4.1. Generator selection based on produced power
The main aim of our project is to find out how much power it produces and the efficiency of the plant. We
already designed the plant layout that can produce a continuous supply of power to the generator and we
already calculated the power produced in the driving shaft to be 10.9011 kW. By considering our output power,
we need to select a suitable generator for our power plant.
To convert the kW into kVA we need to use equation 11
Produced power in kVA = produced power at the shaft in kW/ power factor
Produced power in kVA = produced power at the shaft in kW/ power factor
Produced power in kVA = 10.9011/0.8
Produced power in kVA = 13.62
The power capacity of the generator that is selected is 10- 20% more than the one used in this project. We are
selecting a generator whose capacity is 15% higher and 15% of 13.62 is 2.04 kVA.
Now the required generator capacity is obtained by adding the produced power and its 15%. The calculation is
mentioned below
Required generator capacity is = 13.6263+2.0439 =15.67 kVA
We calculated the required power capacity of the generator to be 15.60 kVA rounded off to 16 kVA. Now we
need to convert it to the kW, so we need to rearrange the equation 11 again
Required generator power in kW = produced power at the shaft in kVA*power factor
Required generator power in kW = 15.67023 * 0.8 = 12.53 kW
The required generator power is 12.5361 kW, so we can choose a 13-kW generator for our project.
 generator specification

Figure 4 – 13 kW generator
The generator is an electrical device that can help to convert mechanical energy into electrical energy. For our
project, we are using a standard 13 kW generator and the specification of the generator is listed in the below
table.
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[3245]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
Table 4- specification of the generator
Parameter Values Unit
Rated power 13 kW
Capacity 16 Kva
Rating 1500-1800 Rpm
Rated torque 80 Nm
Efficiency 93 %
Rated voltage 190-480 V
Frequency 50 Hz
Based on our need and calculation, we selected the generator but to transmit the power from the driving shaft
to the generator shaft.
We need to design and select the suitable transmission drive.
1.4.2. Designing of Transition drives
The transmission drives are the devices that can help to transmit mechanical power from one shaft to another
shaft. There are so many kinds of transmission drives such as gear drives, chain drives, belt drives etc but for
our project, we are using v-shaped belt drives and pulley arrangements. The required RPM in the generator
shaft will be 1500 rpm, but we are getting 524 rpm in the driving shaft. To increase the RPM, we need to reduce
the size of the generator pulley by 3 times and to get a smooth and vibration-free motion, we are using a
flywheel in between the generator and driving shaft. Based on the above assumption, we fix the size of the
pulley. A 9-inch pulley is attached to the driving shaft and connected that to another 9-inch pulley with the help
of belt drives. This new 9-inch pulley is attached to the standard flywheel. The flywheel helps to reduce the
vibrations coming from the driving shaft and gives a smooth and clean rational force to the generator. The
flywheel pulley is attached to the generator pulley through belt drives, but flywheels receive and deliver the
same amount of energy.
 Calculations of transmission drives
Driving pulley transmits the same amount of power to flywheel pulley. Hence the size of both the pulleys will be
same. We need to design and calculate the output power of flywheel pulley and generator pulley. We can find
that using equation 12

= 9*524/1500
=3.14 inch
is the diameter of generator pulley which is 3.14 inches
 Torque delivered at generator pulley
We know the torque developed in the flywheel pulley. To find the torque developed in the generator pulley, we
need to use the following equation 12.1.

By substituting the values to equation number 12.1, we will get the torque at the generator shaft.
= 10901.1/ (2*3.14*1500/60)
= 69.4337 Nm
The generated torque at the generator shaft will be 69.43 Nm

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3246]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
1.4.3. The output power of the generator
To find the output power of the generator, we need to use the following equation 13.
= *
By substituting the values to equation number 13, we will get the power output of the generator
= 0.93 * 10901.1
=10138.023 W
If the generator is continuously running for 1 hour, we get 10.1380 kWh of energy.
 Power loss during transmission
In mechanical devices, there will always some losses happening during transmission. To find that transmission
loss, we are using the following equation.

= 753.077 W
Power loss during transmission is 753.077 W
Simply assume the working time of the plant to be 24 hours continuously. In the below calculations, we find the
daily, monthly, and yearly energy output of the power plant.
In one day = 24*10.1380 = 243.312 kWh of energy per day
In one month, it can produce 30*243.312= 7299.36 kWh of energy.
In one year, it can produce 87.5923 MWh of energy
 Output current
The power is defined as the product of voltage and current. We already found the power and the voltage rating
as given in the generator specification and therefore, we can regulate it to any level. Now we need to calculate
the current. To find it, we can use Equation 13.2, and this is the formula for active power.
= VI cos )
When we substitute all the known parameters to equation 13.2, we get the out-put current.
= VI cos )
10138.023= 400 I cos )
I = 25.34 A
1.4.4. The efficiency of the system
To find the efficiency and effective output of the power plant, we need to calculate the output power
consumption happening in the power plant. In this power plant, the consumption will be happening in two
cases: one by an air compressor and another one by automatic rope coiler. The power consumption details of
both the instruments are listed below.
 Power consumed by the compressor
We already calculated the effective working capacity of the standard air compressor. Now we are calculating
the power consumption by the compressor based on its working operation time.
We are using a standard 8 HP oil cooled compressor. It can consume around 5.968 kW of energy per hour, but
in our system, to reduce the power consumption, we are using an air tank whose capacity is 500 gallons. In the
starting stage, it takes 22 minutes to fill the tank. Then it IS turned off till the pressure in the tank will decrease
to some level and then it will be restarted. Air-filled in the tank is 902 cubic feet and the capacity of the airbag is
20.708 cubic feet. So, the air tank fills the airbag up by 43.55 times. We can say it 43 times. We already
calculated the time required to fill the airbag that is 1 minute. So, to maintain the pressure in the tank, we need
to restart the compressor every 20 minutes and the working time of the air compressor will be 10 minutes.
With the help of the above calculation that we made previously, we can find the power consumption by the air
compressor. In the starting stage, air compressor takes 22 minutes to fill the air tank and consumes 2.1853

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3247]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
kWh of energy and during operation, for every hour, it will work 20 minutes and consumes 1.98933 kWh of
energy and the daily energy consumption of a compressor will be 45.956 kWh.
 Power consumption by the rope recoiler
The rope coiler helps to recoil the rope during the lifting cycle. To achieve the recoiling process, we are
attaching it to a 1 HP motor, and it will work continuously and helps to recoils the entire rope one by one. The
lifting time of the airbag is 30 seconds, so it works continuously and consumes 0.746 kWh of energy per hour.
The operation time of the plant is 24 hours so the 1HP motor consumes 17.907 kWh per day.
The total energy consumption by the air compressor and the rope re coiler will be 63.86 kWh
 The efficiency of the power plant
To find the effective efficiency of the system, we need to use equation 14

By substituting all the known parameters to equation 14, we will get the efficiency of the plant.
*100

The efficiency of the power plant will be 73.75%

1.5. Designing and modelling of the components
1.5.1. Modelling of complete weight cell and well arrangements

Figure 5 – complete model of the system

The above figure shows the complete model of the weight cell arrangement and it will be attached to a properly
designed frame. This model was developed using CATIA software. Here, we can see there are three major parts
in the model. The explanation for components is given below.
 Weight cell.

Figure 6 – model of the weight cell

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[3248]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
It weighs around 300 kg including the weight of the airbag. On the top of the weight cell, there is a properly
designed frame arrangement that can help to connect the airbag and the weight cell through a screw connector.
The rope is attached to the frame with the help of a hook. On the weight cell, there are 3 small wheel
arrangements that are attached at an angle of 120 degrees to each other. Those wheels help to reduce the
friction between the walls of the well and gives proper linear motion in the falling and lifting cycle. At the centre
of the weight cell, there is a cylindrical shaped rubber airbag. Above the airbag, there is also a wheel
arrangement that can help to reduce the vibration during sinking and lifting cycles. On the top of the wheel
arrangement, there is a 5-inch long solid rod that can help to connect the entire set to the rope arrangements
through the hook. We already showed the designing and calculation of both the components in the above pages.
 Model of the airbag
The figure below is the model of the airbag that we designed with the help of the given input parameters. We
already did the calculations for the airbag capacity and its dimensions. With the help of those calculations, I
designed it. In this airbag, there are two kinds of valves that are placed in this airbag. One at the bottom is the
air inlet valve, and two at the top are the air outlet valves. The airbag is attached to the weight cell in its centre
with the help of a nut and bolts.
 Well arrangements

Figure 7 – model of the frame arrangement

The above figure is the model of the frame arrangements which we are using inside the well. If we are
constructing it in the abandoned bore wells, we can use this configuration on the wall of the wells. If we are
constructing it on a water reservoir or onshore places, we can use it directly to construct the plant
arrangements. There are 3 small grooves made on the wall of well at an angle of 120 from each other. These
grooves will help to give a proper linear motion to the system and reduces friction between the walls. We
attached a pipe in the frame arrangement which helps to carry air pipe through it and connect it to the
magnetic valve.
1.5.2. Magnetic valve design

Figure 8 – model of the magnetic valve

The above figure represents the sketch of the magnetic valve. Here we can see that there are two major parts
and the explanation for them are given below.

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3249]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
Upper part: It has a semi sphere-shaped magnet with 0.5-inch rubber coating around it (this can help to
prevent friction when it attached to the cavity cup). In the lower part of the sphere, there is a pneumatic
plunger with a one-way pneumatic valve (plunger helps to open the pneumatic valve which is inside the cavity
cup). In the upper part of the sphere, there is a suspension pipe which gives an up and down movement and it
connects air to the upper body.
Lower part: There is a metallic cavity cup that can help to hold the magnetic sphere in it. At the centre of the
cup, there is a plunger that is attached to a 400-psi pressure valve which helps to open and close the pneumatic
valve.
Operation
The magnetic sphere arrangement is attached to the frame arrangements. When the frame reaches the bottom,
it will be stopped by the mechanical brakes. With the help of rope coilers, we can fix the metallic cup
arrangements at the distance of 5-6 inches from the magnetic sphere. When the magnetic flux in the sphere
attracts the metallic cup, it makes the magnetic sphere move downwards. When it fixes inside the metallic cup,
the plunger will push the pneumatic plunger which is in the metallic cup. That opens the pneumatic valve and
air starts to flow inside the sphere through the one-way pneumatic valve. When the airbag reaches the
maximum level due to the upward force, the magnetic sphere will automatically be separated from the metallic
cup that closes the pneumatic valve.
 Calculations
Magnetic force is at the distance of 0.5 inches from the metallic cup because there is a rubber coating.

F=

To find the magnetic flux density, we need to substitute all the known parameters to equation 15.2.
=
= 0.3686 tesla
The magnetic flux density is 0.3686 Tesla. We calculated the output parameters in equation 15. By substituting
it, we will find the net magnetic force produced by the magnet.
= =
= 876.49 N
1.5.3. Designing of brakes
When the 300 kg-weight cell reaches 250 m of depth, we need to stop it within 10m with the help of brake
arrangements. We need to design and select the suitable braking system. The shaft of the rope coiler is
connected to the braking disk and a sensor are attached to the rope coiler. When the weight cell reaches the
250 m of depth, that sensor sends the information to the brake system to apply the brakes. This will reduce the
speed of the falling weight and stop it at 10m. We use the following calculations to design the brakes.
 Calculation
Kinetic energy

Here,
www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[3250]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
By substituting the known parameters to equation 16, we will get the required kinetic energy in Joules (J).

=13382.9039 J
The required kinetic energy to operate the brake is 13382.9039 J
Required braking force
The equation below is the formula to find out the required braking force.
= /D
By substituting the known parameters to equation number 16, we will get the required braking force.
= 13382.9039/10
= 1338.29 N
Braking time
To find the braking time we need to calculate the acceleration. We know the terminal velocity of the falling
object. To find out the acceleration, we can use the velocity formula given by equation 16.2. By substituting all
the known parameters, we can find out the acceleration.

a = -4.4634 m/s^2
We find out the acceleration to be -4.4634 m/s. To find out the breaking time, we need equation 16.3.
,
0=9.4482 - 4.4634*
= 2.116 s
The required braking time is 2.116 seconds
1.5.4. Other components
 Rope design
It’s a cable line made of metallic wires and it is in the shape of a helix. To select the suitable cable for our
project, we need to calculate its thickness and lifting capacity. We need to select the rope to lift the 300kgs
weight but for safety purposes, we are designing it for 400 kgs. The equation below is the formula to find out
the diameter of the rope.
Safest working load =
Safest working load = 400 kg
D = diameter of the rope
By substituting the known parameters to equation number 16.4 we will get the required diameter of the rope
400 =

D= 7.077 mm
The required diameter for the cable is 0 .007m. We can choose a suitable rope in the market that has more
capacity than this.
 Driving shaft

Figure 9 – shaft arrangement

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science
[3251]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
In the above figure, we can observe how the shaft is fixed. The material of the shaft is mild steel and whose
length is 22 feet. Each weight cells are attached to the same shaft with the help of a pulley and a one-way
bearing. The supports are fixed between each well at the distance of 4 feet to each other. At one end of the shaft,
the output pulley system is attached. The transmission of the power takes place through this pulley.
 One way bearing

Figure 10 – Model of the one way bearing

The above figure is the model of the one-way baring. The pulley which carries the rope will be attached to this
bearing. The advantages of this bearing are that on one direction, it rotates freely and when it rotates in the
reverse direction, it will rotate along with the shaft. When the weight cell is falling, this one-way bearing will
rotate along with the shaft. When the weight cell is moving towards the surface, this bearing will rotate freely
and helps to recoil the rope without affecting the production process because in our arrangement, the weight
cells are attached to the same shaft and they are falling one by one and so when the weight cell is moving
toward the surface, the bearing will rotate in one direction and the shaft will rotate in another direction.
 Rope recoiler
The rope recoiler is the specially designed component. It places a major role in both cycles. The braking system
is mainly attached to the roller of the rope recoiler with the help of a disk and a motor attached to it which in
turn will help to recoil the rope. In the production cycle, the rope recoiler will help to release the weight cell
from the top of the well. When the weight cell reaches the bottom of the well, it helps to stop the weight cell
with the help of the braking system attached to it. In the recharge cycle, when the weight cell is coming towards
the surface, it will automatically recoil the rope with the help of the motor and when the weight cell reaches the
top of the well, it will lock the entire system till the next operation with the help of pneumatic locking system.
 Sensors
Each well is fitted with few sensors such as pneumatic and few other electric sensors. These help to receive and
send the information to the working system to guide actions such as when to realize the weight, when to apply
the brakes, when to switch on the motors, when to start the recoilers and when to lock the system till the next
operation.
II. RESULTS
2.1 Production cycle
The main aim of our project is to design and model the configuration of the buoyancy gravity power plant. To
achieve that, initially I assumed few parameters such as the dimension of the well, mass of the weight cell, the
density of water and object and some other parameters. With the help of these parameters, I designed and
modelled the weight cell and calculated the velocity of the falling object using of quadratic equations.
During the velocity calculation, initially I assumed that the net forces to be equal to zero and I proved this
assumption to be correct by finding the difference among the generated forces. With the help of terminal
velocity, I estimated the acceleration and time of the falling object. Later I found the total amount of the power
and torque generated in the driving shaft. All the calculated values in the production cycle are listed in the table
below.

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3252]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
Table 5- Results of the production cycle calculations
Parameter Values Unit
Terminal velocity m/s
Acceleration a 0.178 m/s^2
Net sinking time 52.96 s s
Downward force F 2943 N
Buoyancy force 324.907 N
Generated Drag force will be 2618.09 N
Viscus force will be 3.55*10-5 N
The torque generated in shaft T 199.13 Nm
The power generated in the shaft 10.9011 kW
1.6. Recharge cycle
Table 6- Results of the recharge cycle calculations

Parameter Values Unit

Lifting force of the airbag N
Net upward velocity 26.16 m/s
Acceleration m/s^2
Force acting on the system due to N
pressure
Mass of the airbag 10.00635 Kg

In the recharge cycle, to design an effective model of the airbag, initially I fixed some input parameters. With the
help of that, I calculated and modelled the airbag using CATIA. I then found out the total lifting capacity of the
airbag and with the help of this calculation, I deduced the velocity, acceleration and time taken by the system to
reach the surface of the water. Later I designed and calculated the mass of the airbag and calculated the of loss
in force due to water pressure. All the calculated values are listed in the table above.
1.7. Plant layout
To select the suitable working structure for the plant, I calculated the total time consumption of the complete
cycle of a single well by calculating and summing the falling time, lifting time, braking time, time taken in filling
the airbag and air removal time of airbag. With these calculations, I also fixed the required number of wells.
Here I designed the air compressor and air tank capacity and with the help of this, I calculated the power
consumption of the air compressor. All the calculated values are listed in the table below.
Table 7- Details of time requirement to complete the cycle
Parameter Values Unit
Sinking time after reaching the terminal velocity 53 s
Braking time of the system 3 s
Air filling time of the airbag 50 s
Lifting time of the airbag 30 s
Air outlet time of the airbag 20 s
The total time is taken by the system 143 s

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3253]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com

sinking time after reaching

0, 0%
20, 13% the terminal velocity t_sink
Braking time of the system
53, 34%
30, 19% Air filling time of the airbag

Lifting time of the airbag

4, 2%

50, 32% Air outlet time of the airbag

Figure 11 – Total time requirement chart

1.8. Output power and efficiency
To achieve the continuous power output, I designed the suitable plant layout and estimated the power and
torque developed in the driving shaft. With the help of this estimations I did suitable calculations to fix a
suitable driving shaft and with the help of this, I selected a suitable generator for my project.
Later, I estimated the output power of the plant and its effective efficiency. To achieve this, I found out the total
power consumption happening in the system due to air compressors and motors. The estimated values are in
the below table.
The required generator power is 12.5361 kW. So, I chose a 13-kW generator for this project.
Table 8- details of the output power and efficiency
Parameter Values Unit
Torque developed at generator shaft 69.43 Nm
Generated power 10138.023 W W
Power loss during transmission is 753.077 W
Produced current 25.34 A
The energy produced per day 243.312 kWh
The energy consumption by the compressor per day 45.956 kWh
Energy consumption by the rope recoiler 17.907 kWh
The efficiency of the power plant

500 243.312 179.449

45.95617.907
0
Energy analysis chart Energy remaining

The energy produced per day

The energy consumption by the compressor per day
Energy consumption by the rope recoiler

Figure 12 – Chart for output power and efficiency

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3254]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
1.9. Modelling of the component
I calculated all the necessary data I require and finally I designed the plant layout using Auto CAD. With the help
of these estimated data, I calculated and designed the necessary components using which I constructed the
gravitational energy production system. The figures of the modelled components such as weight cell, magnetic
valves, generator, drives, brakes, one-way bearings, and few other components and the explanations for them
are also attached in the modelling chapter. The designing and modelling of the component are done using
CATIA software.
III. CONCLUSIONS
The buoyancy gravity power production technology is one of the new topics in the renewable energy field.
Here, I am implemented both buoyancy and gravitational forces to develop this technology. With the help of
this, I designed a complete plant layout and calculated the production output of the plant. To achieve this, I
calculated certain unknown parameters with help of some other assumed input parameters and they are the
velocity of the falling object, generated forces, power produced at the driving shaft, total time taken by the
system to complete operation, output power, and efficiency of the system.
After finding out all the required unknown parameters, I designed the suitable working structure of the power
plant and to develop the system, I modelled some important components which I used in the plant. They are the
weight cell, magnetic valve, airbag, one-way bearing etc. I also designed components such as braking system,
transmission devices, selection of generator, air compressor, air compressor tank, rope size, etc.
In conclusion, the results and research on this new technology are promising as the operation principle of
buoyancy gravity production technology has been confirmed and based on that I developed a simple power
production plant.
Further research is required to investigate on how this technology can be applied on a utilization scale.
Future scope
 In this thesis, I designed the complete plant layout structure with the multiple-well arrangement for the
study purposes. But in future, instead of digging several wells, we can use buoyancy gravity technology on
the water reservoirs or on onshore platforms. This will cut costs and we can also expect a good power
production output.
 Here I designed a complicated structure of the system. To make it simpler, in the future we can design using
the rope pipes which means that the air pipe is fixed inside the rope and that will easily supply the
compressed air to airbag and reduce the construction cost.
 We can see that their ample number of abandoned oil and gas hollow shafts available in the world. If we
redesign and install this technology properly into them, we can provide good output, and this will help to
reduce the cost of drilling holes.
IV. REFERENCE
[1] Integration of buoyancy-based energy storage with utility-scale wind energy generation K.P. Bassett*,
R. Carriveau, D.S.-K. Ting Turbulence and Energy Laboratory, University of Windsor, Canada,
https://isiarticles.com/bundles/Article/pre /pdf/145433.pdf
[2] Underwater energy storage through the application of Archimedes principle Kyle Bassett*, Rupp
Carriveau, David S.-K. Ting Turbulence and Energy Laboratory, University of Windsor, Canada
[3] Underwater Energy Storage - Emphasis on Buoyancy Technique. See discussions, stats, and author
profiles for this publication at: https://www.researchgate. Net/publication/323225265.
[4] Vertical holding capacity of torpedo anchors in underwater cohesive soils wenkai wanga,b, xuefeng
wanga, Guoliang yua,∗a skloe, Cisse, school of naval architecture, ocean & civil engineering, shanghai
jiao tong university, shanghai, 200240, china b china waterborne transport research institute, Beijing,
100088, China
[5] A review of marine renewable energy storage zhiwen wang1 | Rupp carriveau2 | David s.‐k. Ting2 | wei
xiong1 zuwen wang1 IEREK Interdisciplinary Series for Sustainable Development
https://www.youtube.com/watch?v=MNkVhdJVm3w&list=WL&index=163&t=0s

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3255]
 e-ISSN: 2582-5208
International Research Journal of Modernization in Engineering Technology and Science
Volume:03/Issue:05/May-2021 Impact Factor- 5.354 www.irjmets.com
[6] Analytical and experimental evaluation of energy storage using work of buoyancy force Abdul Hai
Alami Citation: Journal of Renewable and Sustainable Energy 6, 013137 (2014); DOI:
10.1063/1.4866036 View online: http://dx.doi.org/10.1063/1.4866036
[7] Physics - Fluid Dynamics (24 of 32) Force Due to Drag Coefficient and Terminal
Velocity,https://www.youtube.com/watch?v=GNqWtWV2mRs&list=WL&index=158
[8] Physics - Fluid Dynamics (12 of 25) Viscosity & Fluid Flow: Terminal Velocity of an Air Bubble
https://www.youtube.com/watch?v=MNkVhdJVm3w&list= WL&index=163&t=0s
[9] Stoke's law. Terminal velocity. Law of flotation. https://www.youtube.com/watch?v=
hI_Ze6KbUZw&list=WL&index=167&t=162s
[10] nasa aeronautical limited, terminal velocity and drag force, https://www.grc.nasa.gov/ www/k-
12/airplane/termv.html
[11] Physics - Fluid Dynamics (23 of 32) Buoyancy, Viscosity, and Drag Forces Compared: Trial ,
https://www.youtube.com/watch?v=A91JaSOVP5s&list=WL& index 167&t=25s
[12] Underwater energy storage through the application of Archimedes principle Author links open overlay
panelKyleBassettRuppCarriveauDavid S.-K.Ting https://doi.org/ 10.1016/j.est.2016.07.005
[13] Australian magnetic solution, force calculation of magnate
http://www.magneticsolutions.com.au/magnet-force-calc.html
[14] Physical Magnet Data, https://www.supermagnete.de/eng/physical-magnet-data.

www.irjmets.com @International Research Journal of Modernization in Engineering, Technology and Science

[3256]