WO2006021975A1

WO2006021975A1 - Ready hydel power system

Info

Publication number: WO2006021975A1
Application number: PCT/IN2005/000281
Authority: WO
Inventors: Ibrahim P. V. Mohamed
Original assignee: Mohamed Ibrahim P V
Priority date: 2004-08-26
Filing date: 2005-08-23
Publication date: 2006-03-02
Also published as: CA2577896A1; GB2431438A; EP1797243A1; GB0702088D0

Abstract

The Hali-mark of the invention Rinue Power System is the less consumption of water with minimum cost. The principle behind the Rinue Power System is very simple by which Electricity can be generated by installing pipe having particular measurements from a place having higher altitude from Sea Leavel to a place having lower altitude from sea leavel and water has to be pumped though this pipe and the generators will have to be installed at every 10 mtr distance. By this system the water needed to rotate one generator, we can operate fifty generators at 1500 rpm and can generate 100 units electricity in one hour. When the height of the place where the pipe has been installed increases, the distance of slope,, will also be increased. When the distance of slope increases, die number of generators to be connected will also be increased. This in turn, accelerates the production unit of electricity. When electricity production increases the partition of water molecules increase, thus the water needed to produce one unit of electricity will be very much reduced.

Description

4. DESCRIPTION:

READY HYDEL POWER SYSTEM

FieId of Technology

Theory of Present Hydel Power System

Power is one of the prime needs and its growth plays a vital role in the development of the country. Today almost all water power is generated as electricity. Hydel Power is generated by storing water in a reservoir which is built at the river origin. Water from the reservoir is carried through the pipe known as Penstock. Water from the Penstock rushes down in the leaf of the wafer turbines at a speed of 375, 500, 600,750 rpm. The force generated from the movement of turbine is responsible for generation of electrical power. This power is generated by using a generator, which connected to the shaft of the water turbine. Usually generators have capacity of 50, 100, 130 MW. (130 MW x 1000 = 13000 kilowatt, (Unit) 50 MW x 1000 = 50000 KW, 25 MW x 1000 = 25000 KW). The generated process ( 167 Ampire, 11000 Volts, 50 Hertz) is carried to a step down transformer by using 167 A, Mink 73 Sq.mm 11 KV. ALUMINIUM CABLE. Step down transformer reduces the generated power to 5A, 250 Volt and is distributed to domestic purposes, using 1.5 sqm .m, 5 A 250 V Aluminium Cable. For industrial purposes the power requirement varies.

Water from a height of 475 Meter failing down at a rate of 2.5 cubic meter per second (2500 Litre) produces 10000 Units of Electricity per hour. As the height of reservoir increase the requirement of water is less whereas lower the height higher the water requirement. In Kerala there are dams of different height of which Moozhiyaar darn is highest one having 750 Meter height and Malankara dam is the smallest one (14 Meter).

If 2500 Litres of water falls from a height of 475 Metre/second, in minute it will become 150000 litre (2500 x 60 = 150000) and in hour it will become 90 lakhs litre (150000 x 60 a 90,00,000), Thus the electricity generated from this amount of water is 10000 units per hour. Therefore for producing one unit of electricity, present Hydel Power System uses 900 Litre of water (90,00,000 / 10000 = 900 litre).

Since there are dams of different height (750 , 400m, 300 m & 14 m) the present system uses an average of about 600 litre of water for producing one unit of electricity. Construction cost of 100 MW (1,00,000 Unit) hydel (power plant is about 500 - 800 Crore rupees (100 MW x 1000 = 100000). Average cost for 100 MW plant is 650 Crores (6,50,00,00,000 / 100000 = 65000). The cost of one unit of electricity in Ready Hydel Power Plant system is Rs, 65000.

The Objectives of the invention "RINUE POWER SYSTEM"

> To generate electricity with minimum usage of water

> To generate electricity with minimum cost

> To generate electricity without any harm to environment

The quantity of water falling through a pipe particular measurement (length and width) near a reservoir is same as that of the quantity of water carried out through this pipe and falling at a distant place.

The force and speed of the water falling steeply down from a reservoir through a pipe of particular measurement will be same as that of the water flowing down through a pipe of 1° slant and length 50 times more than that of 1^st pipe. But it can be found that the linear force will be 50 times more than the initial linear/force. Like this way applicant found that the Hydel Power needed to rotate a generator of particular capacity of volt-ampere can rotate several thousands of same generators. By finding out the distance above the sea-level and the distance between seashore and place where the reservoir is situated, we can effectively use the above mentioned principle. A pipe generator should be drawn from Ooty Boat House Reservoir to Coimbatore. Like this way a pipe generator line should be drawn from Idukki Reservoir to the seashore town Vypin, where drinking water scarcity is more. If the water flowing down from a reservoir through a pipe of 110mm (4 inc diameter) width and length of 1m is 72,000 litre / hour, the water flowing out through th same pipehaving 1 lakhs metre will also be 72000 litre. That means the disposal of wate from the reservoir per hour will be the same.

The water in the reservoir and the water in the 1 lakh metre pipe will be of sam status. Never consider the water in the pipe which is being used. Amount of water effluxe from the reservoir does not depend upon the length of the pipe, (increase or decrease) This principle can be explained with the help of figures as shown in FIg. Nos. 1 & 2.

(FIg. No.1) Height 10 mtr.

Pipe 110mm (4 inch diameter)

Power 8 HP

Speed 10 M/s.

Weight 90 kgs.

Force 90 L/s .

RPM 1500

Connection 1 (2 Kwh. generator)

(FIg. No.2) Height 500 Mtr.

Pipe 110 mm (4inch diameter)

Power 400 HP

Speed 10 M/S.

Weight Kgs.

Force 90 L/s

RPM 1500

Connection 50 (2 Kwh. generator)

ONE IN EVERY 10 Mtr.

When water flows down steeply from a height of 10m through a pipe of 110mm ( inch diameter) (Pipe will be 90° and has 10m length). The speed of the water flowin down will be 10 m / sec. and the weight of this water is 90kgm. The force of this water wil be 90 litre / sec. and its rpm is 1500.

The force, speed and rpm of the water flowing through the pipe having 50 time length with 1° slant than the above mentioned pipe and falling at a distance of 500m. will be the same. But the power will be 50 times more (8 x 50 = 400) i .e. the power will be 400 HP (Horse Power)

The height of the straight pipe is 10 m whereas the length of the slope pipe is 500 m and this strength in length is the main cause for increase in the force. The force here is due to the compression of water molecules as a result of which a force will be developed due to flow. This force is similar to power. The formation of electricity is due to the magnetic power.

When the still water starts flowing, there arise the force and when the intensity of flow increases the power of water also increases i.e. its HP increase.

One degree slope is the 50 times the length of the perpendicular pipe. If the length of the pipe becomes 50,49,48,46 etc. up o times, thers won't be any change in force, speed and rpm, here only power will increase in every increase of length. That means when the length increases the power also increase. For example when the length is doubled, the power is also doubled and the increase in length is proportional to the increase in power.

The power of 10 m pipe is 8 HP. Therefore; the power of 20 m pipe is 16 HP. Like this the power of 500 m (i.e. 50 times) is 50 x 8 = 400 HP. In this way from 1° slant to 90° slant, except power all the other parameters will be same, but power decease at every increase in slope. When the slope increases the length of the pipe will decrease and vice versa. When length of pipe decrease, the linear force also decrease. When linear force decreases power will also decrease.

The slope will become less if the height is more than 50 times. When slope decreases the force, speed and RPM will also decrease. To make use of the increase in length after 50 times more length we should use weight. From here onwards weight of water is the main parameter taken in mind. Therefore, one should increase the volume of water. When the length increases more than the 50 times increase of the initial 10 m i.e. When it increases about 75 times the rpm will decrease to 1000 from 1500, if it increases 100 times rpm will become 750, when 150 times the rpm will again decrease to 500. When it increases up to 200 times, the rpm again decrease to 375 rpm. We can say 200 times increase but this is not practical.

Speed and force of water will also decrease when the rpm decrease. In order to compensate this, we must increase the diameter of the pipe.

1 - 50 times increase 110 mm 4 inch diameter 1500 rpm 50 - 75 times increase 140 mm 5 inch diameter 1000 rpm 75 - 100 times increase 160 mm 6 inch diameter 750 rpm 100 - 150 times increase 180 mm 7 inch diameter 500 rpm 150 - 200 times increase 220 mm 9 inch diameter 375 rpm

When rpm decreases the speed and force of water will also decrease, but there won't be change in usage of water. The amount of water disposed at 1500 rpm is 90 litre and it will be same in 375 rpm too. When the width of the pipe increases the force and speed will decrease but there won't be change in the amount of water.

For e.g. - If the water flowing down in 5 minutes at 1500 rpm is 450 litre, this will be same for 375 rpm.

If the distance traveled by 1500 rpm is 10 meter in one second it will be 2½ M in 375 rpm. Then the water inside the 1500- rpm pipe of 10 mtr will be 90 litre. Then the water that can fill 10 metre pipe used for 375 rpm will be 360 litre. The water passing through 1500-rpm pipe in 4 second is 360 litre. Then the water flowing through 375 rpm pipe in 4 second will also be the same 360 litre. The water inside the 40 metre long pipe of 1500 rpm 110 mm is 360 litre. By 4 second this water will move 90 metres and flow out:. The water enclosed inside the 10 metre long pipe of 375 rpm 220 mm pipe is 350 litre. In 4 second this water will travel 10 metre and will flow out. Therefore the water flew out in every 4-second will be the same in all pipes. Here the water used by both the lines for making electricity will be same and the electricity generated is also the same.

When rpm increases the flowing speed of water also increases and when rpm decreases the flowing speed decreases but the amount of water increases. That is why the amount of water is same when rpm is increased or decreased.

The force and speed of water will be same as that of increase in slope length of the 50 times more length of the one-degree slant at a particular height. (See Fig .3). Here 2000 m heights, with one-degree slope, increase in 50 times and ultimately became 1 lakh metre.

When we draw a straight line perpendicular to 2000 M height, the distance between the slope and the straight line will increase by every increases of length. I.e. For E.g., the distance between 1 M slope in the slope line and the straight line perpendicular to height in 2 M. When it is 10 M it is 20 cm. When it is 100 M it is 2 M. When it is 1000 M it is 20 M and when it is 10,000 M it is 200 M. When the total slope length is 1 lakh meter, the distance will be initial height 2000 M.

Water falling at that point will have the force, speed and power of water falling from a height of 2000 M.

The power of water flowing out from a height of 10 M through a pipe of 110 mm (4 inch diameter) will be 8 HP. Assuming that the water will be full in the pipe and the pipe will be steeply placed, using this force we can make 2 unit of electricity by rotating a generator of 2 kilowatt in 1500 rpm.

We can get this same power in the one-degree slope for about 50 times more length ie. Up to 500 mtr we get the same power (10 x 50 = 500 metre, 8 x 500 = 4000). Therefore 4000 HP will be the power of the 50 times more slope line.

Therefore, we can give 50 generators of 2 k.w.h. at this slope line. The generator at height line will get 8 HP power , out of this 4 HP at 1500 rpm is used to rotate this generator, ie. It is changed to electricity, 2 units of electricity is generated and the balance 4 HP power is flown out through the pump.

In 4000 HP line, for eg. In the 500 Mtr. Long iron rod has an 90 kg. Cast Iron weight attached at every 10 Mtr. Distance. Total number of cast iron weights attached will be 50 Nos. when the 1^st cast iron weight is moved to a distance of 10 mtr. In 10 seconds all the other cast iron weights will also move same distance that means all the 50 cast iron weights will move the same distance which the 1^st cast iron weight has moved.

Therefore by using the power to rotate one generator of 2 kw. we can rotate the 50 generators attached with the slope line.

In other terms the total power in 500 m slope line is 4000 HP, If 50 generators are connected to the slope line the resistance put forward by each generator is equal to

4 HP.

Therefore it is equal to the total power 2000 HP produced by all the 50 generators. There also 2000 HP power remains as balance. That means, by the water needed to rotate one generator we can operate 50 generators at 1500 rpm and can generate 100 units of electricity in one hour.

Here is the 500 metre slope line, the water used to produce 100 units of electricity is 3,24,000 litre. Therefore water needed to generate one unit electricity is 3240 litre. (324000 / 100 = 3240) (According to rinue power system water flowed out through the pipe of 110 mm (4 inch diameter) pipe is 90 litre in one second. Therefore, in one minute it will be (90 x 60 = 5400) 5400 litre and in one hour it will be (5400 x 60 = 324000) 324000 litres. Now, when the length of this line is one lakh metre (100 km), in every 10 metre distance, there will be total 10000 generators having the capacity of 2 kW/Hr. Electricity generated by 10000 generators will be 20000 unit. Therefore, here the water used to generate one unit of electricity (324000 / 20000 = 16,2 ) will be 16 litres and 200 milli litre.

When the height increases the distance of slope will also increase. When the distance of slope increase the connections will also he increased. When connections are increased the number of generators used will also be increased. This in turn accelerating the production unit of electricity. When electricity production increases the partition of water increases, that means the water needed to produce one unit of electricity will be much reduced.

From here onwards we are dealing with the practical aspects of the basic principles which were mentioned above, practical aspects are fulfilled along the pathways. Following the principles mentioned above, two projects are explained hereunder.

OOTY - COIMBΑTORE LINE:-

Ooty boat house is located at 2286 metre above ths seal level. From this Ooty boat house water body a 110 mm (4 inch diameter) G.I. Pipe is installed and in that pipe line at every 10 rn distance 10400 generating capacity of 2 kw. Generators are to be installed. This pipeline ends at a distance of 104 km, At Coimbatore, which is situated about 16 m. above saa level.

Total distance is 104 km, (104000 m) and the distance between each generator is 10 m. Therefore total no. of generators used will be 10400 (104000 / 10 = 10400). Water used by one Generator in one hour will be 162000 litre. The water used to rotate this generator can be used to rotate, all the other 10400 2 kw. Generator (162000 / 10400 = 15.57) Therefore, now the water needed to rotate one generator will be 15 litre and 600 millilitre. The electricity produced by one generator will be 2 unit, therefore, electricity generated by all the 10400 generator will be 20800 units (10400 x 2 = 20800). The water used to produce this much unit (i.e. 20800 unit) will be 16200 litre. Therefore, the water needed to produce 1 unit will be 7 litre and 800 milli litre. (162000 / 20800 = 7.78)

Information Table at a glance

Ooty boat house reservoir is located at about 2286 Metre above seal level. This is recorded as height: 2286 M (A stick having 2286 M if put in a straightly in vertical position, can be termed as height.) The double of this height is referred to as multiple. Coimbatore City is located at about 16 m above sea level. Therefore Ootγ is located at about 2270 metre from Coimbatore. (2286 - 16 = 2270) . The distance between Ootγ and Coimbatore is 104420 Metre. This will be 46 times the height of Ooty Coimbatore Height. 2270 x 46 = 104420) ie. 104 ½ km. This is referred to as multiple x 46.

The slope of this line is one degree. This is referred as slope: 1 degrea 2070 m. The total distance of this slope line is 104420 M and this is termed sε Length; 104 km. (104000 Metre.)

Ths speed of water flowing down through the pipe of 110 mm (4 inch Diameter) Straightly without any obstacle, will be 10 metre per second, (i.e. in one second it will travel 10 metre.) That means in 227 second (3 minute 47 sec,) it will cover 2270 metre. This is known as the speed of this water line and is referred to as the speed: 10 m/s. unconnected.

If all the 10400 generators are installed in this like 270 m line, the resistance produced by these will reduce the speed, to half. This is termed as Speed : 5 M/s. connected.

The amount of the water flowing down steeply from the height of 2270 m through the 104000 m. long pipe of 110 mm (4 inch dia meter) pipe will be 20430 Litre. The force of water flowing down from through this pipe at a height of 1 metre will be 9 litre/second. As height increases multiples of 9 will be increased. Eg. (10 m x 9 = 90) 90 Litre. Therefore, from a height of 2270 Mtr. (Ooty) force of water flowing down per second will be 20430 litre (2270 x 9 = 20430) 20430 Litre. This is referred to as 20430 L/H 2270 M. unconnected.

If 1400 generators are fitted in this lines the force will be reduced to 90 litre at every 10 metre. Height. But every Generator feels this 90 litre/h 10 M force. Then the force of the line will be Force II 90 L/H 10 M. connected. The above is noted as Force II in the table. RPM : 1500 mentioned in the table is the rotation per minute of 4 inch pipe fitted to this line. In this 104000 metre line, the force at every 10 M is 8 HP. Like this way if we divide 104000 metre by 10 we get 10400 m (104000 / 10 = 10400). If the force is 8 HP at one point, the total force of this line will be 83200 (10400 x 8 = 83200). (This force will be there if generator is fitted or not). This is mentioned as HP : 83200 in the table.

Water disposed through this line in one hour is 324000. This is when the generators are fitted in this line. This is mentioned as water 324000 L/H unconnected in the table.

If 10400 alternators & 10400 pumps are fitted at every 10 M length of the line, the water flown out will be reduced to half ie. 162000 litre. This is mentioned as Water: 162000 litre per hour connected in the table.

Connections: 10400 mentioned in the table referred to as the No. of 2 kw. alternators fitted at every 10 M (104000/10 = 10400). A group consisting of 100 alternators is hereby referred to as centi group. In this line total electricity produced in one hour is 20.8 mega watt per hour (one 2 kv. Generator = 2 units. Therefore 10400 x 2 = 20800 units). This is referred to as Megawatt per hour: 20.8 mw/h (20800 units psr hour.)

The water disposed in one hour is 152000 litre. Total unit of electricity produced in one hour is 20800 units (Therefore 162000/21800 = 7.78). Now water needed for producing one unit of electricity is 7 litre and 800 milli litre. This is referred to as Unit of water used: 7.800 litre per hour.

The Unit cost for the construction of plant is 15000 rupees. This is referred to as amount Unit : Rs.15000

The Ooty Coimbatore line, is power system constructed using Rinue Power System. It needs about 31 Crore and 50 lakhs rupees. This is referred to as Plant Amount : R Rs. 31,50,00,000/- only. The quantitγ of electricity produced in the plant daily is 504000 unit. This is referred to as Megawatt par day. : 504/ D KW. (50400 Units/ D)

The distance between 2 generators is referred to as Velocity: 10 Metre,

Ooty water body is located at a height of 2270 Mtr, from Coimbatore. The distance between Ooty and Coimbatore is 46 times. The Height between these two places (2270 x 46 * 104200) ie. The distance is 104420 Metre (104.420 km.)

The 50 times of 2270 metre is 113500 metre. Since this distance is lying in a slope, there will be 2270 metre deepness from the height at 104420 metre distance, (ie. At Coimbatore), the deepness' will be 2088 Metre. In one lakh metre there will be 2000 metre deepness' and 5000 metre there will be 100 Metre deepness' at 100 Metre distance there will be 2 metre and at one metre height distance there will be 2 cm. deepsness.

IDUKKI - VYPIN LINE

Project Summary

Idukki, Cheruthoni dam is located at about 600 metre above sea-level. From Cheruthoni dam a line has to be drawn and concluded in Vypin, a small town located at a distance of 100 km and 200 metre near sea shore.

Idukki dam is located at 600 metre above sea level. The distance between Idukki and Vypin is 100200 Metre. Here the slope is 167 times the height (167 x600 = 100200). Here the slope is more than 60 times. The height ie. 167 times the height distance, the speed, force and rpm of water will be reduced. Therefore we should increase the diameter of the pipe. Therefore this line must be installed with 200 mm x 8 inch dismeter pipe having 500 rpm, 2 KW per hour generator. The total distance is 100 km. and 200 m is. 100200m. Distance between the Generator is 10 M. Therefore, the total No. of generator to be installed are: 10020 (100200/10=10020). The Electricity production by this 10020 generators will be 20040 unit. The water used to rotate one generator and 10000 generators will be the same 162000 litres. Therefore, water needed to produce one unit of electricity is 8 litre and 80 M.L (162000 - 20040 = 8.03)

IDUKKI - VYPEEN LINE Information Table at a Glance

Idukki - Vypeen line height is about 600 M above sea level, 50 times more than this height is 30,000 M (600 x 50 = 30,000, 30 km.) When 30,000 M is traveled down from the slope, the deepness will be equal to the height of 600 m, . From this point onward, then the deepeness will be same at any stretch of distance. This is because 50 times slopes will be ending at that point. For every 50 times height slope the deepness can be calculated by multiplying 2 cm. for every 1 metre, for example, in a 50 times more slopes at 1500 metre. The deepness will be 30,000 cm. ie. 300 m (1500 x. 2 = 30000)

Special features of the Invention

Rinue power system plant is to be installed on the banks of river side. So therefore, to find out a suitable place we have to inspect both the river side at the river origin.

We have to make a total estimate of how much length of pipe is needed how much alternators must be installed at every 10 m. distance, fittings (pumps, boxes, valves, L-bows, Connectors). There should be a tank constructed at the end point of the line to store fresh water which is disposed by this line.

Tank Construction

A rectangular pit of 35 m length and 30 m, breadth having 2 feet depth has to be constructed and granite stone is to be laid in that pit for strengthening the foundation. Six inch thickness of concrete is to be laid above the foundation. Above the concrete a rectangular wall is to be built using granites stone. The dimension of the tank is 35 m length, 30 m breadth and 2 feet width and a height of about 1.5 m. Inside portion and bottom portion of the tank has to be well cemented and light green colour ceramic tile must be pasted.

This tank must have the capacity to store the 39 lakhs litre water which is disposed by the line in 24 hours. The water filled in this tank par day must be disposed on that day itself.

Material Required for Construction

Materials needed for construction of plant are fittings, instruments needed for fitting, G.I, Pipe of 10 Mtr. Length, 2 kwh/h alternators (must be ordered and made), 4 inch suction 4 inch distribution pumb with air valve. Model of Tank and Specially made part instruments required for this invention are shown in drawing ( Fig. Nos. 7 to 22).

Fixing Procedure of Materials

Bulb shaped foot valve and one connector has to be connected together and after that joining 10 Mtr. length pipe has to be connected to this connector. Then one., more connector is to be connected with pipe. To this fixting again one more 10 Mtr. length pipe is to be connected. Then connect an L bow to this along with a connector. To this L bow connect a one Mtr. pipe above. This is to be taken and to be installed perpendicular to the water body (Foot Valve portion should be immersed inside the water).

Fixing procedure of materials are shown as Figure Nos. 23 to 30.

The deepness of the water body should be noted and according to the deepness the length pipes are to be connected. For Eg. Here two length pipe is to be fitted. When the length pipe is fitted to the water body, care must te taken to place the foot valve 3 mtr. from the clay level. This is in order to prevent the suction of the clay. Foot Valve is placed in the bottom and in order to prevent the variations in the water leve! through the pipe, it can be seen from Figure No.31.

Then fit 4 inch valve to the one Mtr. pipe which is protruding out way (See Fig. No.32).

To this valve (above one) connect a pipe of 4 inch length (See Fig. No.33)

Connect a connector to the above fitted 6 " pipe (See Fig. No. 34) To this connector, connect 4 x 2 dimension T in upward direction, the length of 'T' must be two inch, (See Fig. No.35)

Connect a 2 inch valve to the 2 inch pipe on the 'T' (See Fig. No. 36)

To the above 'T' connector, connect a connector at the other end (See Fig. No. 37}

Take an L bow place it in a 4 inch planch. The holes in the pump's planch (pump is fitted in the bottom) and holes of the 1^st Planch must be equal in position and the L bow must be placed correctly. The pump is fitted at a particular level. The L- bow Planch is connected to the Connector, connected after the 2 inch value. The Planch position must be placed downwards. (See Fig. No. 39)

Now take the first pump and connects its upwards delivery portion planch to the L- bow planch which is facing downwards. Using washer, shellak without the leakage of vvater, should carefully joint it. It should be fixed air tightly with nut and bolts, (See Fig. No. 40)

Take an 10 m long pipe and connect a connector to one and and fit a L-bow and Planch. (See Fig. No. 41).

This is connected to the Receiver portion of first pump. (See Fig. No. 42)

To this fix a L-Bow fitted with connectors at both end in such a way that it should be in the opposite direction of the L-Bow which is connected to the receiver portion. (See Fig. No. 43)

To this connector 4 x 2 - 'T' inch portion of T connect 2 inch value. To the 4 inch portion of "T" and to this connector, connect a L Bow with planch. Planch must be position, (See Fig, No. 44)

To this Planch connect the second The delivery portion is connected to the Planch. (See Fig. No. 45) Like the pattern given in the figure line coming out from the second pump is connected to the receiver end of the Third Pump and the line corning from the receiver portion of Third Pump is connected to the delivery portion of fourth pump. That means line commencing from top is fitted to Delivery portion and line, going out is fitted to the receiver portion. Or the water coming down is entered to the pump through the delivery portion and flown out through the receiver portion.

Like this manner, continuously place the pump till the fresh water lake. The end portion of the last length pipe should be placed inside the lake.

Schematic diagram of pumps fixed from Reservoir to Tank can be seen in Fig. No.

46.

Since the topography of land is different in different places. (There may be places with more curvature and less Curvature, also steep and low lying lands). There will be difference in the degrees of cone of L.Bows.

Since the Line pass through different types of path (zig zag, slope, steep paths, plains) the fittings there and pipe lengths will also be changed.

A 4 inch valve is connected to the L bow coming out from the Receiver portion of 5^th Pump. In this manner at the receiver portion of 10^th pump, 15^th pump, 20^th pump and up to reservoir we should place the 4 Inch valve by keeping the specific distance.

Construction of Control Room and filling of water,

Firstly construct a control room near the reservoir. From there a single line (1 positive and 1 negative line) is drawn till the end of Plant line near the fresh water tank. At every 10 Mtr. distance pump, fit a single light near to it.

A controller should be appointed to give signals at each stage. The ' On - off ' switch of this lights will be in the control room. This signal light line will be connected to a eiactronic key board to check the disorders in the alternators fitted or other parts by blowing the bulbs or by-making alarm. The signal line can be operated by using electricity from the existing line or from a battery operated inverter.

Model diagram of Pumps fitted with Signal light can be seen in Fig. No. 47.

We can also fit the modem signal light system.

Filling of Water:

Firstly open all the air valves of pumps fitted to this line and also the 4 inch valve and 2 inch valves. After that tightly close the 4 inch valve near the tank at the end of the Plant line. Then fill the water in the 2 inch valve on the top of the pump. Check that no air is inside the line from main valve to this 2 inch valve and if there is no air close the air valve of pump and then carefully close the 2 inch valve.

After that fill water in the 2 inch valve above this point and do the same procedures as done before. (Check for water leakage in any part of the of the line and if there is any leakage it should be removed). This procedure is done for the next 2 inch valve above this point and like this manner every 2 inch valve is filled with water and the air valve of pump is closed followed by the closure of 2 inch valve. When water reaches every main valve, it should be carefully closed. Water should be poured and filled in the lines up to the pipe which is immersed into the reservoir (ie. Up to foot valve we have to fill the water) Schematic diagram of control room can be seen in Fig. No. 48.

Like the above mentioned way the 2 consecutive main valves (4 inch valves) are closed and water is filled in the 5 valves in between this valves and close the air valve of pumps and the valves are also closed by filling water in full amount.

Long plant line there will be about 10000 pumps, 2000 main valves and 10000 2 inch valves (valves are fitted only for filling water). After filling water completely in all the valves, an individual should be appointed at the main valve (1 person for 5 pumps). They are placed in order to check the working of the pumps. Only during the initial time we need this much sequence of Personnel. When plant begins to operate the plant line will be connected to the computer placed in the computer room. Therefore in a minute disorder can be detected at the control room so, we need only 3 Engineers, 2 Controllers. Therefore, total manpower required is 5 persons.

After that signal light switch is ON from the Control Room, when the light is ON all the 1000 valves are opened. When the Water reaches the tank from the reservoir by the working of pumps the Signal lights are OFF and all valves are closed.

Alternator fittings,

There is a chance for the Alternator (cumulative compound 2 kw/hr. 8 Ampere 250 volt (2000 volt ampere single phase 1500 rpm) to burn out due to the intense heat caused by continuous working. Therefore an alternator with cooling system is placed in the fibre box

The shaft of the alternator protruding outward the box is tightly connected to the bearing hole of pump. After that the cover of the box is token and, the 2 electric lines (+ve, -ve) are drawn outwards through the 2 holes in the cover of the box.

The box is covered tightly. (See Fig No. 49 , 50 & 51).

The box is covered in order to prevent moisture, like this way connect the alternator to the each pump in the line. The capacity and volt ampere of all the alternators fitted must be the same. The electricity produced from a single phase alternator is grouped to 1000 alternators, after trial parallely connect all the 1000 alternators in a group. A special type of cable is manufactured for 8000 ampere of current. To one cable the positive wire of alternator and to another cable the negative wire of alternator is connected using alkaparsol (black lead, white lead, cadmium, there are the mixture of alkoparsol) These two lines are connected to the 250 V secondary and therefrom 11,000 volt primary of the step up transformer. For every 1000 alternator group a set up transformer to b e installed.

8000 A current is converted to 180 A and 11000 Volt (UG x LPE 180 A 11000 V). This is passed through a underground cable and distributed to industries and to the houses which are far away from generator. This is reduced by using step down transformer. This system can be connected to the present current supplying system.

By fitting a 3 phase alternator the electricity produced from each alternator can be supplied to different houses and shops. By using this technique (3 phase alternator) there will not be any use of step up and step down transformer.

Current Production and Control System

Details of Plant accessories fitted from reservoir to fresh water tank (Group the alternators to a 100 Nos. group and give code No. to each group, group of valves, speed of water, pressure of water, working condition of pumps and alternators) should be fed into the most modem computer placed in the Control Room. All the works, should be connected to the computer using wirelessly or by using wire. The computer should have the facility to know the minute defect of the line.

After that switch on the signal light and open the main valve, now the Rinue Power System has began to work. In case if the system is to be stopped, we can do it by the closure of valve near the reservoir, valve in the center(half the total No. of valves) and last valve fitted near the tank. We can completely stop the system by closing the above mentioned 3 valves and also we can again restart the system by opening the 3 valves. This system does not cause any harm to the environment and also it can be used for centuries. (Fig. No.52.) CONCLUSION

40% of the electricity in our country is produced by Hydel Power. If we increase this production to one and half times more we can save the money which we spend for fuels and also we will have surplus amount of electricity. But Rinue Power system will increase the current production up to 75 times. If this system is used in our country after deducting the daily usage of current, 72 and a half times daily usage of current is left.

If this system is placed along the Himalayan Originating Rivers like river Sindhu, Ganga, Bhramaputhra, which are situated at about 7000 feet, above sea level, the water required to produce one unit of electricity will be 2 litre and 320 millilitre. That means after the usage of current by our country 184 and a half times current will be balance.

This balance electricity can be used in place of cooking by in households and hotels,, by replacing LPG, Kerosene, Fire-woods, coal etc. Thus we can save the money spend for the fuels. So the money spend for fuels can be saved and can be used for the progress of our country.

Claims

5. CLAIMS:CLAIM

1. Rinυe Power System by which Electricity can be generated by installing pipe having particular measurement from a place having higher altitude from Sea Level to a place having lower altitude from Sea Level and water has to be pumped through this pipe and generators will have to be installed at every 10 Mtrs. Distance.

2. The Rinue Power System as claimed in Claim No.1 wherein the pipe to be installed has to be in slanted position having a particular slope so as to flow the water through the pipe and to rotate the generator. By this system the water needed for production of the Electricity will be less than the present Hydel Power System.

3. The Rinue Power System as claimed in Claim No.1 wherein the length of the pipe increases, along with the increase in width of the pipe, then the force of the water will decrease. The amount of water will remain same where the length and width of the pipe increase or decrease.

4. The Rinue Power System as claimed in Claim No.1 wherein the water needed to rotate one generator, we can operate 50 generators at 1500 rpm. and can generate 100 Units of Electricity in one hour.

For eg. If there is a 500 m slope line, the water used to produce the 100 units of electricity is 3,24,000 litres. Therefore the water needed to generate one unit of electricity is 3240 litres.

5. The Rinue Power System as claimed in Claim No.1, when the height of the place where the pipe has to be installed increases, the distance of slope will also be increased. When the distance of slope increase, the number of generators to be connected will also be increased. This in turn accelerate the production Unit of electricity, When Electricity production increases, the partition of water molecules increase; that means, the water needed to produce one unit of electricity will be much reduced. 6. The Rinue Power System as claimed in Claim No.1 by vvhich the principle can be made use of by installing a pipe having 4 inch dia is to be installed from Ooty Boat House to a place 204 k.m. away at Coimbatore, the altitude of Ooty Boat House is 2286 mtr. from the Sea level and that of Coimbatore is 16 mtr. from the Sea Level. The total number of generators to be connected in this 104 km, line will be 10,400 numbers. The water to be used by one Generator in one hour will be 162000 Litres. The water used to rotate this generator can be used to rotate all other 10400 generators having the capacity of 2 kw. each., water needed to rotate one generator will be 15 litres and 600 ml. The electricity produced by one generator will be 2 unit and therefore electricity generated by all the 10400 generators will be 20,800 Units. The water used to produce this much unit (20800 Units) will be 1,62,000 litres. Therefore the water needed to produce one unit will be 7.800 litres.

7. The Rinue Power System as claimed in Claim No. 1 by which the principle can be made use of by installing a pipe having 4 inch diameter is to be installed from Idukki to a place 100.2 k.m. away at Vypin. The total number of Generators to be connected in this distant line will be 10020 Nos. The water to be used by one Generator in one hour is: 1,62,000/- litrs. The Electricity produced by one Generator will be 2 units and therefore electricity generated by all the generators (10200 x 2) will be 20,040 Units. The water used to produce this much Unit will be 1,52,000 liters. The water needed to produce one Unit of electricity is 8,08 liters.

6. DATE AND SIGNATURE: