WO2018190443A1 - Water pumping device using speed difference - Google Patents

Abstract

The present invention relates to a water pumping device for colliding two fluids (water) different in both speed and flow direction from each other to cause all the fluids to flow in one direction. The water pumping device is characterized in that, when water in a tank continuously flows out of the tank and simultaneously water continuously flows into the tank from the outside thereof, the speed of water flowing out of the tank is configured to be higher than the speed of water flowing into the tank to collide the two fluids outside the tank so as to allow all the water to flow in only one direction, i.e., toward only the tank, and, as a result, water flowing into the tank continuously pushes up water existing in the tank, thereby continuously raising the level of water in the tank.

Description

 Specification

 Name of invention: Pumping device using speed difference

 (1) Technical Field

 Continuous equation of fluid and correlation between potential energy and kinetic energy

 (2) Background Art

Conventional hydroelectric power generation was limited in the time and place of power generation. For many reasons, there was a lot of downtime, and no power plant could be built anywhere. I want to overcome them. Background literature

 [Document 1] Supervision of Power Generation Hydropower Practice: KEPCO Publisher: Gumi Technology, First Edition 94-07 一 10

[Ref. 2] Fluid Mechanics Fundamentals & Applications

 Kim Nae-Hyun Kim 6-Artist / Publisher: MCGRAW-HILL KOREA, MAGRO HILL KOREA Published Date: 2005—11-25 1 Printing

 Author: Yunus A. Cengel / John M. Cimbala

 (3) Disclosure of Invention

 (3) -l challenge: to allow both development and pumping at the same time, 24 hours a day

 To continue to develop 365 days a year

(3) -2 solution

 ************** Start of the solution to the problem ***************************

 Prerequisite: Figure 1 power generation unit (1) is not part of the present invention,

 In order to explain the present invention well, it is inevitably like part of the present invention.

 Treat it.

 1 ascending tank (16) top portion * received the fire passing upwards to FIG. 1 power generation section (1)

 1 upper tube 17 to lower and

1, the power generation unit (1) for generating water received from the upper pipe (17)

FIG. 1 stores the water falling after generating power in the power generation unit 1, and further stores the water.

FIG. 1 accelerated tank 6 sent to large pipe 7 and FIG.

1 reservoir tank (2) serving as a water source of the acceleration tank (6) and

 The water dropped into the water barrier 5 overflowing the top of FIG. 1 supply pump 4 and FIG. 1 speed tank 6 which supplies water from the reservoir tank 2 to the acceleration tank 6.

 1 is a recovery pump 3 to be recovered to the reservoir tank (2) and

 1 water tank (5) to prevent water falling over the top of the acceleration tank (6) and

 In the pipe for sending water from the accelerating tank (6) to FIG.

Figure 2 large pipe (7) directly connected to the acceleration tank (6) and 2 flow control valve (8) for adjusting the flow rate through the large pipe (7) and

 To connect the large pipe 7 of FIG. 2 and the small pipe 10 of FIG.

 2 inclined tube 9 and

 Fig. 2 small tube 10 for receiving and launching accelerated water through the inclined tube 9;

 9 FIG. 2 opening and closing valve 1 (11) for blocking and passing the water flowing through the small pipe (10) and

 From the small pipe (10) to the £ 2 pavement (14) in FIG.

 Figure 2 shows that the water launched into the small tube 10 is different from each other in flow velocity (as described later

 2, the firing speed in the small tube 10 is greater than the firing speed in the collision tube 14, Figure 2 (b) the wind blows, if all the dimensions and centerline of the two tubes equal

10 Water must enter the collision tube 14 of FIG. 2, but in reality the actual dimension of the collision tube 14 and FIG. The funnel can also be slightly different so that a small amount of water can escape the two tubes, while the funnel 1 (12) and the funnel 2 (13) and FIG.

 11 FIG. 2 impingement tube 14 connecting funnel 2 (13) to FIG. 2 ascending tank 16; and

 2 is a shut-off valve 2 (15) for blocking and passing the water flowing through the filling ball tube (14) and

 2 and 1 rising tank 16, which serves as a way for the water flowing from the layered pipe 14 to the rising tank 16 in FIG.

 1 is located at the top of the rising tank 16 for supplying fire to the rising tank 16

 12 water inlet (18) and

 Fig. 1 Pumping device consisting of a flow gauge 19 for checking the flow rate of the upper pipe 17. The principle of operation is as follows. First, the role of the claims in solving the problems in (3) -1 will be explained.

 13 Claim 1 Explanation of roles:

 In the case of two fluids flowing through the same diameter pipe with different speeds and different flow directions, if the two fluids collide in front of each other, the physical fact is adopted that after the collision all the fluids will flow in the flow direction of the high velocity fluid before the collision. It is characterized by.

 14 Fig. 2 Small pipe (10) If the outlet flow rate is greater than the layered pipe 14, the outlet flow rate, Fig. 2 Small pipe

 (10) Outlet and staircase tube (14) The water launched from the outlet stratifies in the lower part M of FIG. If the above process occurs continuously, the water entering the ascending tank (16) will continue to push up the existing water and the ascending tank (16) will continue to rise.

 15 Of course there is a limit to water level rise.

 As mentioned above, claim 1 plays a decisive role in solving (3) -1 problem.

By carefully reading this series, you will better understand the role of claim 1 in solving the problem. 16 Claim 2 Description of Roles:

When a constant flow rate is required at the outlet connected to the tank, if the water level is constantly maintained at the top of the tank instead of using a pump, the outlet flow rate at that connected pipe will always be constant. The outlet velocity is mass * _g * drop = 0.5 * mass * velocity ² and velocity is (¾ * drop)

 17 On the free fall side, the height from the pipe centerline to the top of the tank is always constant, so that the exit speed is constant if the level is maintained at the top of the tank. In other words, if the water level is kept at the top of the acceleration tank (6) in Fig. 2, the flow velocity down to the large pipe (7) is constant.

As described above, claim 2 is a prerequisite for solving the problem (3) -1.

 Fig. 2 The constant top flow rate of the large pipe (7) serves to solve basic but important points in solving the problem.

 By carefully reading this series, you will better understand the role of claim 2 in solving the problem.

 19

 Start this description. This is the description below-

(The principle of circulation is explained later.)

 Water continues to circulate up and down according to the CCR below (see Tip below).

 20 And all the flow of CCR occurs continuously at the same time. The water continues to circulate.

 Fig. 1 Acceleration tank (6) ᅳ-> rising tank (16) 상> upper pipe (17) —> generator (1)

 —―> Acceleration tank (6) ..... ― .—. Since the flow rate is the same for CCR and the whole process of CCR,

 21 That is the flow rate of power generation of the power generation unit (1).

Tip: up to just before the CCR, even ≤ standing between the two lower first funnel 12 and a funnel 2 _(13), half

The two fluid flows in a layer ^■, and finally with a one-way flow of

 There is a process of becoming a fluid.

 Introducing species on the water cycle **************

22

 2, only the flow regulating valve 8 is fully opened, and the opening and closing valve 1 (11) and the opening and closing valve 2 (15) are completely closed, and supply water to the top of the acceleration tank 6 in FIG. 6) Fill the water completely to the top level and supply water to the top of the ascending tank 16 through the FIG. 1 water inlet 18.

 23

And even after the above-mentioned CCR starts, the top of FIG. To keep the water level high

 In the case where the water level of the acceleration tank 6 is somewhat lower than the top, a sufficient amount of water is stored in advance in the reservoir tank 2 so that the water level can be quickly raised to the top of the acceleration tank 6.

24 Must be stored and quickly supplied to FIG. 1 Acceleration Tank (6) 6

 (The water storage of the reservoir tank 2 is described in the best mode for carrying out the invention or the embodiment for carrying out the invention.)

 Then, when water is rapidly supplied to the acceleration tank 6 of FIG. 1 and the top portion of the acceleration tank 6 overflows, the overflowed water immediately falls into the water barrier 5 of FIG. 1,

 The administrator can immediately collect the dropped water into the reservoir tank (2) of FIG. 1. From now on, we will start explaining how it works.

 To explain the details of the water flow,

 Set the reference point as Figure 1 Acceleration Tank (6). Benchmark

26 As stated above

 Fig. 2 With the flow regulating valve 8 fully open

 Figure 1 Acceleration tank (6) Fill the water completely with the top level

 1 Ascending tank 16 is completely filled with water to the top

 1 After sufficient water is stored in the reservoir tank 2

 27

 When opening and closing the valve 1 (11) and the valve 2 (15) at the same time,

 The water starts from the above-described reference point 2 acceleration tank 6 and continuously passes through the large pipe 7, the inclined pipe 9, and the small pipe 10, and the water flows toward the collision pipe 14 of FIG. 2. Fired, according to the continuity equation, the water is accelerated much faster than the passage speed of FIG.

Fired at 28 and descends into FIG. 2 collision tube (14).

 On the other hand, water is launched from the collision tube 14 in FIG. 2 and rises to the small tube 10 in FIG.

 Eventually the funnel with two fluids moving in opposite directions as shown in Figure 2

 There is a frontal collision between 1 (12) and funnel 2 (13) (Fig. 2, the large dashed arrow)

 And small dotted arrows indicate the direction and magnitude of the two fluid velocities)

 With 29 the two fluids become one fluid and move in only one direction, which is of course the direction of the stratified velocity.

 ***** Tip: As you will see later, the direction from the small pipe (10) to the masonry pipe (14), that is, the small pipe (10), the collision pipe (14) and the rising tank at the top of the acceleration tank (6) of FIG. The direction toward the upper tube (17) via (16) becomes the final flow direction after the collision.

30 **** ^{^} ip end * * * * *

Then, after the water projected from the small pipe 10 to the masonry pipe 14 in FIG. 2 and the water fired from the collision pipe 14 to the small pipe 10 are collided, all the water is released in FIG. Even if the inner diameter and the thickening line of the two-layer stone pipe 14 and the small pipe 10 are equal to each other in order to flow naturally inside, the actual dimension and the actual thickening line are slightly different from each other, 31 A small amount of water can escape the tube (FIGS. 2 10 and 14), in which case funnel 2 (13) and funnel 2 (13) are used to keep the water from leaving the two tubes. Fig. 2 Fluid velocity at the outlet of the lower coffin 10 (direction: downward) 210 FIG. 2 Fluid velocity at the outlet of the lower impingement tube 14 (direction: upward) -214 2 Fig. 2 With the funnel 1 (12) As indicated by the dashed arrows between the funnels 2 (13), 210 and 214 are in opposite directions.

 If we find 210 and 214 for Naze, 210 is much larger, and as shown in ^ 2, the flow direction after the stratification of the two fluids is in the direction of 210, that is, the direction from the small tube 10 to the collision tube 14, Water will only flow from the acceleration tank (6) in FIG. 2 towards the ascension tank (16).

 3 But already at the beginning of FIG. 2 Fully filled with water to the top of the rising tank 16

 Therefore, the amount of water moved from the accelerating tank 6 to the ascending tank 16 moves over the top of the ascending tank 16 to the upper tube 17 and eventually in the upper tube 17. 1 Water falls to the power generation unit (1), and water drops to the acceleration tank (6) after power generation.

 Eventually, water returned to Acceleration Tank (6). 1. The wheel is gentle.

 4 However, the reference point is set as an acceleration tank (6) for convenience. However, the following points can be used as reference points.

 Fig. 1 Acceleration tank (6) ᅳ> Sangsong tank (16) —> Upper pipe (17) ᅳ> Generator (1) ——> Acceleration tank (6) in the water circulation process, ie in CCR,

 The same thing happens simultaneously in the ascending tank 5 (16), the upper pipe (17), and the power generation unit (1) in the east and west as the water leaves the acceleration tank (6) and returns to the acceleration tank (6). In the end, the water returned to the reference 1 acceleration tank (6).

 And since all valves are already open, the above-mentioned fluid circulation process (CCR) continues.

 6 It should be noted here that water of low height, ie, the water in the acceleration tank (6), has risen to the upper pipe (17) beyond the top of the ascending hank (16) in FIG.

 That is, water continuously flows from the acceleration tank (6) below the rising tank (16) to the bottom of the rising tank (16) by the amount of water coming in continuously.

 7

 From now on, let's look at 210 and 214 and see that 210 is bigger.

 First find 210.

 The outlet velocity 210 of the small tube 10 of FIG. 2 is indicated by a large dotted arrow between the bottom 12 and 13 of FIG.

 38 210Calculation Order and Solution:

 First, the exit speed of Fig. 2 small pipe (10) ignoring head loss is obtained.

2, the head loss from the top of the large pipe (7) to the outlet of the small pipe (10) Find the speed that corresponds to it (as described in the section "Meaning the Pox Loss" in #).

 Third 210 = Speed according to the first—Speed according to the second—-一 — 210 So, first we find the exit speed of Fig. 2 and (10) ignoring the head loss occurring in the fluidized bed.

 1. FIG. 2 Large Pipe (7) Bottom Flow Rate

According to the law of conservation of energy, the mass * g * drop) is 5 * mass * velocity ^2, so the velocity is

(2g * fall): the fire is always full at the acceleration tank (6) at the top of the S.2 large pipe (7) the top flow rate is V (2 * 9.8 * KSH) because the fall is 2 KSH, (2) The lowest flow velocity of the large pipe (7) is equal to the velocity corresponding to the drop P of FIG. 2, that is, the velocity is equal to (2 * 9.8 * 1) (2 * 9.8 * 1 11) + (2 * 9.8 * ^' :?) Fig. 2 Large Pipe (7) Bottom Flow Rate—— PS

2. Fig. 2 (9) Bottom flow velocity

 The uppermost flow velocity of the inclined tube 9 of FIG. 2 is PS, which is the lowest flow velocity of the large tube 7 of FIG.

 2, the inclined tube 9 has a water tank area decrease, so that the inclined tube 9

 The speed increases. As shown in Fig. 3, the top area / bottom area = 5, so that the bottom flow velocity of the inclined tube (9) is the highest flow velocity of the inclined tube (9) based on the continuous equation.

5PS becomes 5PS 5PS And, the bottom flow velocity of the inclined tube 9 is further increased due to the drop Q of FIG. 2 (see FIG. 3): mass * g * drop = 0.5 * mass * velocity ² Additional velocity increase of inclined tube (9) due to velocity = (2g * fall) = 2Q Q:

 V (2g * 2Q) —— 一 —IQ Thus, the bottom flow rate of the final FIG. 2 slope tube (9) is the sum of the two above. That is, IQ + 5PS is the bottom flow rate of the final Fig. 2 inclined tube (9). IQ + 5PS 3. FIG. 2 Small Tube (10) Bottom Flow Rate

 2, the top flow rate of the small pipe 10 is IQ + 5PS, which is the bottom flow rate of the inclined pipe 9.

The lowest flow rate of the small pipe (10) can be added to the above flow rate by the speed increase due to the free fall R. Mass * g * fall = 0.5 * mass * speed ² at speed = V (2g * fall), so increase in velocity due to freefall R = V (2g * R)

5 Fig. 2 Small Tube (10) Bottom Flow = IQ + 5PS + (2g * R) ——— IGB This is the outlet flow of the Small Pipe (10) in Fig. 2, ignoring the head loss occurring during the flow of the IGB. From now on, we will find the head loss that causes the flow rate and the corresponding speed. 2, the head loss from the top of the large pipe (7) to the outlet of the small pipe (10). Calculate. The speed at which head loss is calculated is the speed at which head loss is applied.

1. Find the head loss at the boundary point in FIG.

 Background literature [Distribution 2] Fluid Mechanics Fundamentals & Applications Kim, Hyun-Ho and 6th Airspace Publisher: MAGRO HILL CO., LTD.

 Head loss = loss factor * speed * speed /(2*9.8)

 By Note 1 below

 Fig. 2 Speed just before the boundary point = V (2g * Fig. 2 KSH) — based on SS loss head equation

 Head loss = 0.03 * SS * SS / (2 * 9.8) Figure 2 Boundary head head loss Note) 1. Energy conversion (interconversion between potential energy and velocity energy)

Mass * g * fall = 0.5 * mass * speed ² , velocity = V (2g * fall)

2. Friction Head Loss of Large Pipe 7

 According to Darcy-Weisbach formula and Manning formula

 Background Documents [Document 1] Hydropower Practice (KEPCO), pp. 77, 78

hf ^ (f * L / D) * V ² / 2g

 Welded steel pipe n value: 0.013 ， f = 124.5 * n * n / (^ D)

 Head loss: (f * L / D) * (V * V) / (2 * 9.8), V: Average flow velocity

 f is the coefficient of friction, L is the length of the pipe (Fig. 2 P: length of Fig. 2), D is the diameter of the pipe: the diameter V of Fig. 2 is the velocity and end point of the starting point (inlet) due to the acceleration of gravity The speeds are different so you need to apply the average of those two speeds.

 That is, V = (start point speed + end point speed) / 2.

 The starting point velocity is SS, which is the velocity applied at 1.

 That is, V (2g * is also 2 KSH)

 The endpoint velocity is determined by the starting point by the length (P) of FIG.

 Increases the speed by the speed corresponding to

 2, that is, the speed increases by (2 * 9.8 * degrees 2 Ρ) according to 1)

 Thus, the endpoint velocity is V (2 * g * degrees 2 KSH) + V (2 * g * degrees 2 P)

 ED = end point (outlet) speed ED blast furnace ， V = SS + ED

 2

3 Finally, head loss is calculated using the following equation.

Head loss: (f * L / D) * (V * V) / (2 * 9.8) Figure 2 Friction head loss of large pipe (7) . Fig. 2 Inclined tube 9 perspective axle friction head loss

According to Darcy Weisbach formula and Manning formula

 Background Documents [Document 1] Generating Hydropower Practice (P84, P85, P77, P78)

 Tube shrinkage friction head loss ᅳ>

13/3>— (1/D1 /V13/3)) n: 0.013 Dl: 도 2 7의 안지름 즉，큰안지

13/3> — (1 / D1 / V13 / 3)) n: 0.013 Dl: inner diameter of FIG.

 D2: Fig. 2 Small inner diameter = 10, inner diameter = 14

 Relationship between D1 and D2 (D1 ^ 5 * D2): The large diameter = ^ 5 * small diameter relation is based on the case where the area ratio is 0.2 and the angle is 30 degree in Figure 3.11 of the above-mentioned P84, P85.

 The distance is the length of FIG. 3 9 (see the description of FIG. 3) —— Since the Q flow rate is constant, apply the flow rate according to the flow rate of the starting point of 2 (large inlet) flow rate.

 Flow start area of large pipe (7) in Fig. 2 * 2 Starting point (large pipe inlet)

 '= {3.14 * (FIG. 3 a / 2) * (FIG. 3 a / 2)} * SS Based on the above, the inclined tube (9) perspective axle friction head loss can be calculated. . Figure 2 Tube loss head loss of the inclined pipe (loss due to vortex, noise, etc.)

3.19 equation (hgc = ¾c * V ² / 2g, where V is the flow rate after shrinking gradually) and the book P84 Garden loss factor table under Darcy-Weisbach official certificate: Loss factor * Flow immediately after shrinkage Flow rate /(2*9.8)

 Reduction Factor: Experimental Values of Change

 0.04 (^ 3 Narrowing angle: 30 degrees / Area ratio: 0.2) Background literature [Reference 1] Hydropower Practice (KEPCO) P83, P84, P77, P78

 The end point of 2 (large outlet) is ED, which has a large radius and V5 times as mentioned in FIG.

 Large diameter through area / Small diameter through area

 Passing area of = 7 / Passing area of 10 = 5

 Therefore, the flow rate immediately after reduction is determined according to the continuous equation.

^~ > 5 * Direct velocity before contraction = 5 * ED 一 AP

AP is the increased velocity with fluid passage width change

 Fig. 2 shows the speed increase due to the drop (Q mentioned in 3) of the inclined tube 9

It must be added to become the flow rate immediately after the final reduction. 61 That is, since ¾ is equal to zero as the starting point (inlet) of FIG. 3, the inclination pipe needs to be added to the AP.

 So according to note 1 of 1, V (2 * g * is also added by 2 Q).

 Blast furnace, flow rate immediately after final shrinkage: AP + V (2 * g * Q) -— BP blast furnace, Ο.04 * βΡ * ΒΡ / (2 * 9.8) ——- Fig. 2 Head shrinkage loss of the inclined tube (9)

62

 5. Fig. 2 small head (10) friction head loss

 According to Darcy-Weisbach formula and Manning formula

 Background literature [Reference 1] Power generation hydro exercise (KEPCO) See pages 77,78,

hf = (f * L / D) * V ² / 2g

 63 Welded steel pipe n value: 0.013 ， f = 124.5 * n * n / (^ D)

 Head loss: (f * L / D) * (V * V) / (2 * 9.8), V; average speed

 f is the coefficient of friction, L is the tube length (length of Fig. 2 10: Fig. 2 R),

 D is the tube diameter: Fig. 3 A / 5

 V is the velocity of the starting point (entrance) and the speed of the end point (entrance) due to the acceleration of the layer force.

Since 64 is different, we need to apply the average of those two speeds.

 That is, V = (start point speed + end point speed) / 2.

 The starting point speed is BP of 4, which is the velocity immediately after the final reduction, and the end speed is falling as much as the corresponding speed since the drop has occurred as much as the length (R) of FIG.

5 So according to note 1 of 1, V (2 * g * also increases by 2 R).

 Blast furnace, end point (outlet) velocity = 4 BP + V (2 * g * also 2 R), blast furnace, N = 4 BP + V (2 * g * also 2 R)

 2

 Above 66, the friction head loss of the small pipe 10 can also be seen.

 Abnormal Head Loss Calculation Through 1 ᅳ 2, 3, 4, 5, the head loss occurred during the flow from the top of the large pipe (7) to the outlet of the small pipe (10).

 If the sum of head loss (total head loss) of 1 ~ 5 is L5 (unit: m),

 ： Total head loss of Γ5 (total head loss) — ᅳ--ᅳ L5

****** ***** The difference between chickenpox loss Λ] ^ |-** ************ ******* Background literature [2] Fluid Mechanics Fundamentals & Applications Kim Nae-Hyeon and Six Persons Airspace / Publisher: MCGRAW— HILL KOREA, MAGRO HILL KOREA Published Date: 2005-11—25 1 Printing

 Author: Yunus A. Cengel / John M. Cimbala

As mentioned at the bottom of page 311 above, "Head loss refers to the lift the pump must raise to overcome frictional losses inside the pipe." When pumping water, to pump water to the target point, in addition to the flow rate that overcomes the height difference, more flow rate that overcomes the head loss is needed.

 69 will. You need to create that flow rate using a pump.

If the head of the head loss side pump is to be raised h (m) and the flow rate to overcome friction loss is v (m / sec), the correlation between potential energy and kinetic energy mgh = 0.5mv ² so v = 2gh (m / sec). At least v must be left speed so that the water

You can climb up to 70 targets.

 Again,

 If you have head loss, you need additional flow rate as much as V (2 * g * head loss). If you do not have head loss, you do not need additional flow rate.

 In other words, the head loss means to reduce the flow rate by V (2 * g * head loss).

71. therefore,

 2 The total head loss ^ (L5) in the flow from the top of the large pipe (7) to the small pipe (10) exit reduces the exit speed of the small pipe (10) ignoring head loss, i.e., IGB,

 The speed reduction amount is V (2 * g * L5). L5 (Total Head Loss)

72 ******************* This End of Head Loss: Head Loss According to the meaning of the present invention, the speed reduction according to the head loss of the present invention is V (2 * 9.8 * L5) Also, since the outlet flow rate of the small pipe 10 of FIG. 2 ignoring the previously obtained head loss is IGB, the outlet flow rate of the small pipe 10 of FIG. 2 to which the head loss is applied, that is, 210 is determined according to the head loss.

73 210 = IGB— (2 * 9.8 * 1)

 = IGB- (2 * 9.8 * L5) 210 obtained by 210 TOZ or more, 210 is indicated by a large dotted arrow between the lower part 12 and 13 of FIG.

74, then the flow direction is the opposite of 210

 The exit velocity of the exit tube 14, ie 214, is indicated by a small dot line arrow between the bottom 12 and 13 of FIG. I will save 214 from now on.

 Fig. 2 Flow velocity at the exit point of the impingement 14 in Fig. 2 when the speed drop due to gravity and the speed decrease due to friction when the water rises in the impingement tube 14 are ignored.

Find 75 214.

 The reason for applying velocity without neglecting velocity reduction is to emphasize that even if 214 is stronger than it actually is, it will be larger than 210 to overcome 214 and the final flow direction will be 210 after two fluid collisions.

And when we get 214, we use SSH for Figure 2, which also applies the maximum value of 214 to 76. 214 Find the flow rate indicated by the small dashed arrows between the bottom 12 and 13 of FIG. Energy conversion (conversion to potential energy <-ᅳ> speed energy)

That is, mass * g * drop = 0.5 * mass * velocity ² at speed = V (2g * drop)

 In the end, we get 214

 77 (2g * FIG. 2 SSH) = 2 Outlet Point Flow Rate (214) As shown in Figure 14 and above, the two exit points (FIG. 2 bottom 10 and 14 outlets) velocity 210 and 214 have been written (flow). The directions are opposite to each other.) ， You cannot judge at a glance which side is big. So, let's calculate the actual values so that we can feel the difference between 210 and 214. In other words, we'll prove by using real numbers that 210 is bigger.

^ | ^^ ： ¾ ： ¾ <> 11! ¾ ^■ X] ************** ******** The following figures are just examples and you can create a lot of data by substituting many different values. There is.

 3 Inside diameter of large pipe (7) ： a, a = 5 m

Inner diameter of small tube 10: aA / 5 = inner diameter of collision tube 14

 Fig. 3 relationship between two diameters: the inner diameter of 7 = V5 * 10, the background literature [1], power generation training (KEPCO)

 Figure 3.1 of P84 and P85, P84 Gardel's fgc loss factor table), which is a narrowing angle of 30 degrees and an area ratio of 0.2.

80

 Fig. 2 KSH: 20m Fig. 2 SSH: 50m

 Fig. 2 large pipe (7) length P = 2, Fig. 2 small pipe (10) length R = 4,

 Length of the inclined tube 9, ie Q —> see FIG.

 If you calculate with the above conditions

81 Degree IGB flow rate at exit 10: 149.21 m / s (ignore headloss)

 Figure 2 is the flow rate applying the head loss of the exit 10 according to the TOZ

 110.9 m / sec 210 The flow velocity at the exit point in Fig. 2 14 depends on the TOF: 31.30 m / sec 214

82 The calculations under these conditions result in 210 being greater than 214. Thus, the final flow direction after the two lower fluids in FIG. 2 collide in M becomes the direction from the small pipe 10 in FIG.

 Then, the final fluid movement speed of the two fluids after the stratification is 79.54 m / sec, which is 210 minus 214, and the flow direction is one direction, which is the direction from 83 to 14 in FIG.

 That is, the final flow after the two fluids collide in FIG. 2M: FIG. 2 through the top tube 17 through the small tube 10, the masonry tube 14 and the ascending tank 16 in the acceleration tank 6; 1 shot

Fluid moves to all (1). 84 Post-collision flow rate = laminar post-travel speed (210-214) * Fig. 2 Passing area of the crash tube 14 = 79.54 * 3.14 * (5 / (2 * V5)) * (5 / (2 * 75)) = 312 m ³ / sec

Eventually, 312 m ³ (312 ton / sec) of water will continue to move and ascend ᅳ Movement direction: Moving from 10 to 14 in Fig. 2 to 16 and to 16 at the top 16 down 2 Fig. 6> Fig. 2 16, the lower -> 16 2 top - ^o> 2 17

 85 Falling Water Falling Water

 After the two fluid collisions above, the fluid moving speed and the flow rate are only one of many cases, and if the actual value is changed variously, there are many different values.

86 Reference) Fig. 2 Conduit (7) As the inner diameter is increased, the circulation flow rate of CCR, that is, the generation flow rate, is greatly increased as follows.

 1 Power generation flow rate of the power generation unit 1 = circulation flow rate of the CCR

Fig. 2 Large pipe (7) Inside diameter: 5m ^~ > Power flow rate: 312 (m ³ / s-ton / s)

Fig. 2 Large pipe (7) Inner diameter: 10m〉 Power generation flow rate: 1,383 (m ³ / s = ton / s)

87 Fig. 2 Large pipe (7) Inner diameter: 12m ^~ > Generation flow rate: 2 ᅳ 035 (m ³ / s = ton / s) Even if the generation flow rate is 1,000 (ton / sec), a large amount of electric power is produced.

 **************** Calculation based on actual numerical conditions End: 1

88 Calculations based on actual numerical conditions confirmed that 210 is greater than 214.

 After the two fluids collide in M at the lower part of FIG. 2, the final flow is from the small tube 10 to the collision tube 14, that is, the small tube 10 and the masonry tube at the top of the acceleration tank 6. 14) through the bottom and top of the ascension tank (16), towards the upper tube (17).

 89

 The conclusion is as follows.

 Fig. 1 After filling the tank with enough water, the water will continue to circulate up and down, so that the generator 1 will continue to operate even without water supply from the outside.

 90 That is, hydroelectric power continues without external water class.

 Therefore, when the pumping device of the present invention is used, hydro power generation can continue without water supply, and the location of the power plant is infinitely wide, thereby creating infinite hydro power plants. In other words, there are countless hydroelectric power plants in the mountains, in the desert, in factories, on the docks, in schools, in hospitals, and other areas.

 In addition, the amount of power generated by each power plant can also be ultra-ultra large.

Moreover, it is possible to continue to develop 365 days a year 24 hours a day. Illlll or more completed the task solution. That is, complete (3) -2 IIIII

92 (3) -3 Effects and benefits of the solution

 1. Hydro power can be increased more than 100 times.

 Using the present invention as mentioned in the solution to the problem, since the water is continuously circulating up and down at the same time, it is possible to continue to generate power only by the water of the power plant itself without supply from outside the power plant. That is, any bot

93 It is possible to build hydro power plants.

 In other words, the location restrictions of hydropower plants are eliminated, which means that very large / large / layered / small hydro power plants can be created innumerable / without number.

 Power plant ramp time is also unlimited (24 hours a day, 365 days a year). In conclusion, we increase hydroelectric power more than 100 times.

 94

 2. Provides low power.

 By using a small amount of energy and pumping energy, it is possible to drastically reduce costs.

 Power plant (factory, home, school, pier, hospital farm, etc.) You can build a power plant right next to it, which greatly reduces the transmission cost and lowers the electricity bill.

 95

 3. The pollution of the earth is improved and the environment of the earth is improved.

 As a result of 1, the pollution-free hydroelectric power is produced almost indefinitely, which causes the earth to breathe. Significantly reduces thermal power generation, significantly reducing C02 and fine dust. Eventually, extreme weather will disappear and the earth will be cool and clear. So the earth becomes healthy.

4. A safe planet is created.

 As a result of 1 and 2, nuclear power disappears, creating a safer planet.

97 5. Extended Earth Life

 As a result of 1 ~ 4, a clear and clean earth for humankind will last forever. (4) Brief description of the drawings and code description

 Fig. 1: Fluid flow chart and power plant

 98 FIG. 2: FIG. 1 Detailed view of the connection between the acceleration tank 6 and the rising tank 16

 3: Reference diagram around the inclined tube 9

FIG. 4: Figure 1 illustrating the amount of water that is always present between the top of the rising tank 16 and the top of the accelerating tank ₆ after the CCR of the task solution occurs. FIG. 5: FIG. 1 of the accelerating tank 6 With two or more joints of the tank and lift tank (16) 99

 Description

 1. Power Generation Unit 2. Reservoir Tank 3. Recovery Pump 4 Supply Pump

 5.Fireproof 6.Acceleration Tank

 7. Large pipe 8. Flow control valve 9. Inclined pipe 10. Small pipe

 100 11. Opening and closing valve 1 12. Funnel 1 13. Funnel 2

 14. Layer pipe 15. Valve 2 16. Lift tank

 17. Upper pipe 18. Water inlet 19 Flow gauge

(5) Best mode for carrying out the invention or form for carrying out the invention

 101

 1. Implementation and Application of the Invention:

 Start of basic dimension ——Start of basic dimension Fig. 1 KSH = 20m SSH = 80m Known Fig. 2 Large pipe (7) If inside diameter = 5

 3 Inside diameter of large pipe (7) ： a, a = 5 m

 102 FIG. 3 Inside Diameter of Small Tube 10: a / 5 = Inside Diameter of Impinging Tube 14

 Fig. 3 Relationship between two diameters: 7 diameter = 5 * 10 The relationship between the diameter of the narrower angle in the background literature [Reference 1], the development of hydropower practice (Korea Electric Power Plant) P84, P85 Figure 3.1 P84 Gardel fgc loss coefficient table) 30 degree and area ratio is 0.2 FIG. 2 Large pipe (7) length P = 3, FIG. 2 small pipe (10) length R = 4,

 103 Fig. 2 Length of the inclined tube 9, ie Q> see Fig. 3

 The diameter of the upper tube 17 in Fig. 1 is equal to the diameter of the large tube 7 in Fig. 2.

 The diameter of the top of the rising tank 16 is larger than the diameter of the upper pipe 17 of FIG. The height of the two-layered pipe 14 is set to 2 m.

 Etc: height of other items to reduce effective drop: 2m

 104 Fig. 2 Small Tube (10) Outlet and Collision Tube (14) Outlet Distance: Fig. 2 Small Tube (10) 2.5 times the inner diameter

—— _ End of basic dimension End of basic dimension

1-1. Strategy before building the invention

Based on the above basic dimensions and problem solving means, the flow rate after CCR generation is 300ton (fn ³ ) / sec

It becomes 105. This flow rate may be somewhat different from the power generation flow rate (hereinafter referred to as "actual power flow rate" or "actual flow rate") when the present invention is actually operated after construction. The utility company is responsible for the above expected generation flow rate, basic dimensions,

106 Calculate the expected maximum power generation by carefully considering all matters related to power generation. If it is 1.5 times larger than the required power generation (see Tip under 3), adjust the following items. Repeated adjustment calculation over and over again Find several dimensions close to 1.5 times the volume.

 Tip: The reason why the estimated maximum generation amount is 1.5 times that of the urea generation amount is to increase the marginal margin.

107 to watch.

 ***** Adjustments: Figure 2 Large pipe (7)

 Of course, KSH and SSH, other basic dimensions, Fig. 2 flow control valve (8), all other matters related to the invention, and of course, the expected generation flow will change.

 If possible, Figure 2 KSH and SSH should remain unchanged.

 108 ****** ADJUSTMENT "END OF SECTION **************************************

 And I'll tell you about the various tanks in Figs. 2 The area of the acceleration tank 6 is set to be much larger than the water passage area of the large pipe 7.

 :！ Later, you will see later "2. Amount of water to store in the reservoir tank 2"

 Set up the reservoir tank 2 in FIG.

 1, the rise tank 16 is set structurally stable in consideration of height and flow rate. The top of the rising tank 16 in FIG. 1 is made larger than the diameter of the top pipe 17 in FIG.

 상부 1 The diameter of the upper tube (17) is equal to the diameter of the large tube (7) of FIG.

 The above is the strategy before constructing the present invention.

 110

 1-2. Construction and Post-Construction Strategies of the Invention

 The utility company constructs the present invention according to the contents determined in 1-1.

 (See Figure 5 above for at least two basic connections.)

 The production and installation of the present invention is the pie that withstands the corresponding pressure and water stratification for each field

Preparing and installing tanks, tanks and other components makes it easy to manufacture and install. 112 This is very easy for plant 112 construction professionals and utilities.

 After constructing all of FIG. 1 except for the power generation unit 1 of FIG. 1, the present invention is actually operated to measure the actual flow rate of the upper pipe 17 through the flow gauge 19 of FIG. 1. When the actual flow rate is not enough for the utility company to generate the required generation amount, the basic flow rate 1 and the basic connection 2 are simultaneously operated to increase the actual flow rate.

 113 Fig. 5 When there is no problem with only the basic connection 1, only the basic connection 1 is used for a period of time, and then only the basic connection 2 is used. That is, two basic connections are used alternately. On the other hand,

 5 If only the basic connection 1 is used, if the actual flow rate of the upper pipe 17 of FIG. 1 is large enough to generate the required power generation amount, the power generation is performed separately with the excess flow rate. You can adjust it to get the actual flow rate.

Finally, when the actual flow rate that can generate the required power generation is confirmed by the power flow gauge 19 in FIG. 2, the flow control valve 8 is fixed and the opening / closing valve is locked, so construction is suspended. Whatever you do, build the Power Plant (1) to complete the plant. Over 115, the pumping device side to produce the required power generation, the construction of the present invention was completed, ^ Han, the power generation unit (1) also completed the construction of the hydroelectric power plant applying the present invention was completed.

2. The amount of water to be stored in the reservoir tank 2 in FIG.

 116

 After the water cycle begins, as with CCR in the task solution, water continues to flow in all pipes and tanks in the water cycle. Even in space, water always flows. This means that water is always present in the space.

 1 The rising tank 16 and the accelerating tank 6 are basically watered up to the top of each other in advance.

You don't have to worry about filling 117 with H,

 1 The amount of water to be stored in the reservoir tank 2 is calculated as follows.

<A> Fig. 1 Defect as large as the pipe volume of the entire upper tube 17

 <B> Figure 1 rising tank (16) top extension line and acceleration tank (6) top part shown in FIG.

118 Water falling in this space

 Fig. 2 The space between the small pipe (10) exit and the collision pipe (14) exit The amount of water falling down (2 times the amount of water required because of the basic connection of Fig. 5 according to 1-2)

<a> and <da> are insignificant compared to <b>, so ignore them and instead store more water than <b>.

 119 First, find out how much water you have.

 The amount of water falling in the space between the top of the rising tank (16) top extension line and the acceleration tank (6) top, shown in Fig. 4, i.e., the total amount of water always present in the space.

The estimated power flow rate of 1-1, that is, the circulation flow rate is FM (m ³ / sec;), and the air gap between the uppermost extension line of the rising tank 16 and the acceleration tank 6, as shown in FIG.

If the time for free fall of the liver is t seconds, the water continues to drop continuously as much as FM (m ³ ) per second, so when it reaches t seconds after the start of drop, the whole space is filled with water and the whole space The amount of water filled is FM * t (m ³ ). That is, because this amount of FM * t (m ³ ) of fire is always present in the space, it can be stored in reservoir tank 2 as much as this amount.

 Since the water does not fall completely vertically in the above drop space, it is sufficient to store three times the FM * t calculated above in Figure 1 reservoir tank (2).

 The dimensions of each tank and tube shown in Figs. 1 and 2 make it easy to find Fig. 4t (total 122 hours of free fall), so the amount of water to be obtained is also easy.

Three times of quantity of water ^ FM * t of <me> For reference, the water does not completely fall vertically in the above space, but as shown in FIG. 4, after free fall of the water for a certain time, the space occupied by the dropped water is symbolically represented as a vertical rectangular shape.

3. Here is an example of practical application of the present invention.

All dimensions should be determined by the utility company. In order to help the utility's understanding of the present invention, the inventors have arbitrarily defined all dimensions (somewhat rough).

 According to the basic dimension of 1

 Fig. 1 KSH = 20m SSH = 80m and Fig. 2 Large pipe (7) Inside diameter = 5

 The inner diameter of 125 degree 3 large pipe 7 ： a, a = 5 m

 Fig. 3 Inside diameter of the small tube 10: a / V5 = Fig. 2 Inside diameter of the canal tube 14

 Fig. 2 tube 7 length 1 ^ 3, Fig. 2 tube 10 length R = 4,

 Length of the inclined tube 9, ie Q —> see FIG.

 The diameter of the upper tube 17 of FIG. 1 is equal to the diameter of the large tube 7 of FIG.

 126 FIG. 1 The diameter of the upper portion of the rising tank 16 is made larger than the diameter of the upper tube 17 of FIG. The height of the impingement tube 14 in FIG. 2 shall be 2 m.

 Etc: Height of other items to reduce effective drop: It is to be set to 2m.

 Fig. 2 Distance between the small tube 10 outlet and the collision tube 14 exit: Fig. 2 small tube 10 of inner diameter

 2.5x

 127

 Background Document [Document 1] Supervision of Power Generation Hydropower Practice: KEPCO Power Agency Issuer: Gumi Technology First Edition 94-07-10 P2, P4, P5

 Flow rate unit: mVsec I Effective drop unit: m

 According to (1.2) of Page 2, theoretical hydraulic pressure = 9.8 * flow rate * Effective drop (KW) — PA

!8

 According to (1.5) of Page 4, generator output =

 Theoretical Hydraulic Power * Aberration Efficiency * Generator Effectiveness (KW)-— PB According to Table 1.1 of PB, the aberration efficiency * Generator Effectiveness = 0.85 PC

129 The inventor determines the effective free fall based on the following calculation

 (The exact effective drop can be understood by electric power company.)

Inventor's effective free fall calculation according to the basic dimension of 1:

 Effective drop of inventor = SSH— KSH -P-Q-R

130 — Fig. 2 Distance between small (10) exit and masonry (14) exits – Fig. 2 Collision (14) height-Effective drop by Etc inventors = 80-20—3-5.15-4—5.59—2 2 = 38.26 m Ex) If the required power generation is 60 MW,

 According to the basic dimension of 1, PA,: PB, PC and inventor's effective drop (38.26 m) and the solution, the estimated power generation and the maximum power generation after CCR of 1-1

131 Estimated generation flow rate = 300 (m ³ / sec) and depends on PA, PB and PC

 Estimated maximum generation = 96 MW This corresponds to about 1.5 times the required generation, so the current dimensions are left to 1-2. If the expected maximum generation

 . If it is too different from 1.5 times the required generation, adjust the dimensions to bring it closer to 1.5 times the required generation. In this case, of course, adjust the dimensions to 1-2.

132

 Through all the steps 1-1, depending on the dimensions

 Except for the power generation unit 1 of Fig. 1 to 1-2, everything is installed. Then, the present invention is actually operated to check the flow rate of the flow gauge 19 in FIG. The amount of electricity generated is checked as the effective drop of the checked flow rate and the nominal (the exact effective drop can be known by the utility company).

133 If the calculated value is larger than the required power generation amount, gradually adjust the flow control valve (8) of FIG. This will eventually create the required generation. After that, the power generation unit 1 of FIG. 1 is finally installed and pumped, and at the same time, electric power is produced.

134 Conversely,

 1, the flow rate checked in the flow rate gauge 19 and the inventor's effective drop (the exact effective drop can be known by the low-power company) are smaller than the required generation so that the value is close to the required generation amount. Additional basic operation 2 of FIG. 5 may be properly operated to gradually increase the flow rate of the FIG. 1 flow gauge 19.

 135 This will eventually generate the required amount of power. After that, the power generation unit (1) of FIG. 1 is finally installed, pumped, and produced at the same time.

The dimensions and methods applied in 3. are intended to help the utility company understand.

 As a result of the

You can judge well at 136 companies.

(6) Industrial Applicability

 1. Very easy to construct

 2. Basically, the construction cost is small, and the construction cost is much smaller than the effect.

 137 3. The power plant to which the present invention is applied has no limit in the installation place, the site cost is small, and there can be numerous power plants (more than 1,000,000),

 There is no environmental pollution because it is hydro power.

 It is easily used and welcomed anywhere in the world for the above 1-3 reasons. (3) -2 solutions and (5) best practice for the implementation of the invention.

138 As described in the form or form for carrying out the invention, it is omitted here. (7) Sequence Listing

^8, SEQ ID _NO. Sequence Listing Free Text

Claims

A bill of war  139 Blue ^ t 1  Different speed, different flow direction  Colliding two fluids, resulting in all fluids flowing in only one direction 140 and  When water in the tank continues to flow out of the tank and at the same time continuously flows out of the tank toward the moving tank.  Make the inflow into the tank larger than the outflow from the tank: fill the two fluids outside the funnel so that the water flows in only one direction, 141 flows only into H, and consequently the water entering the S ^ tank continues to push up the existing water in the tank.  Claim 2, wherein the water level in the tank to continuously rise.  142 Pumping device characterized in that the water level is always kept at the highest level of the tank, that is, the full water level, so that the outlet flow rate from the pipe connected to the tank is always kept constant.