JP2015086851A - Differential pressure water power generation device - Google Patents

Differential pressure water power generation device Download PDF

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JP2015086851A
JP2015086851A JP2013235417A JP2013235417A JP2015086851A JP 2015086851 A JP2015086851 A JP 2015086851A JP 2013235417 A JP2013235417 A JP 2013235417A JP 2013235417 A JP2013235417 A JP 2013235417A JP 2015086851 A JP2015086851 A JP 2015086851A
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JP5612751B1 (en
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末夫 井手
Sueo Ide
末夫 井手
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    • Y—GENERAL TAGGING OF NEW TECHNOLOGICAL DEVELOPMENTS; GENERAL TAGGING OF CROSS-SECTIONAL TECHNOLOGIES SPANNING OVER SEVERAL SECTIONS OF THE IPC; TECHNICAL SUBJECTS COVERED BY FORMER USPC CROSS-REFERENCE ART COLLECTIONS [XRACs] AND DIGESTS
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Abstract

PROBLEM TO BE SOLVED: To utilize underwater pressure to be static pressure to take out new energy.SOLUTION: In a differential pressure water power generation device, the wide-angle diameter in the upper part of a release pipe 2 of a gradually extended shape with a pipe length of J m is the same level as a water level and sunk vertically; a pumping-up pump 3 is provided at the upper part of the release pipe; and a water turbine 4 and a dynamo 5 driven by the water turbine are provided at a pipe bottom part. By forcibly generating upward flow with pumping-up of the pump, water pressure in a vertical direction is forcibly made imbalance, the water turbine is driven by upward flow transformed by the water pressure, and the dynamo generates power. In order to minimum pipe resistance, the diameter of the pipe bottom part provided with the water turbine is made minimum, and the diameter of the upper part formed into a gradually extended shape is made maximum, and thereby the speed of upward flow is gradually accelerated.

Description

水力発電Hydropower

現状の発電装置は、環境問題を抱えて種々の方式があるが、実用された方式の風力発電装置、ソーラーパネルは、設置場所、設備費用の割りに発電力が少ないと言う重大な欠点があり、火力発電に頼る現状では、燃料費の増大で問題になつている。
この欠点を解決する手段として、燃料費無用で年間通して昼夜連続動作する発電装置であつて、水中の圧力差を利用した差圧水力装置を提供するものであるThe current power generation equipment has various methods with environmental problems, but the wind power generation equipment and solar panels that have been put to practical use have a serious disadvantage that they generate less power for the installation location and equipment costs. The current situation of relying on thermal power generation has become a problem due to an increase in fuel costs.
As a means for solving this drawback, a power generator that operates continuously day and night throughout the year without fuel costs, and provides a differential pressure hydraulic device that utilizes a pressure difference in water.

水中では、水面より深くなる方向（下方）に比例して水圧が上がるが、その事は位置エレルギーが上がる事を意味し、逆に、ある深い位置から水面方向への上昇流は、圧力が下がる方向、即ち位置エレルギーが下がる方向故、下がつた位置エレルギー分新たな動力が生まれる事になる。
処で、静圧である水中圧は、、自然の状態では、上昇流が生じないから、強制的に上昇流にさせれば、位置エレルギーの減少方向故、新たなエネルギーが生じる事になる。
そこで、開放管の一端を塞いだ状態で、塞いだ一端を下にして、他端を水面に合わせて沈めた状態で、塞いだ一端を開放させると、管底圧力が転じて上昇流になる訳だが、水圧は、水面で０圧で深さに応じて増大する事は周知の事実である。
この現象をベルヌーイ定理で説明すると、図１で示すと、沈めた開放管１の水面を基準にして、水面で外圧力；Ｐ１、水面上の水圧による速度；Ｖ１、位置水頭；Ｚ１
該１の管底では 外圧力；Ｐ２、水圧による速度；Ｖ２、位置水頭；Ｚ２ とすると

Figure 2015086851
Ｐ１＝Ｐ２，水面上の水圧による速度；Ｖ１＝０、水面故に Ｚ１＝０
Figure 2015086851
この事は、重力加速度に逆らつた方向、即ち上昇速度水頭を示すから
水深；Ｘ＝Ｚ２ とすると、Ｘ＝Ｖ２／２Ｇ（ｍ）
Figure 2015086851
される為、効率が悪い。
効率良い方法は、管底部だけを水車を廻すための所定速度にして、管の他の部分の速度を出来る丈低くなる様な形状にすれば、管路抵抗を減少させる事になる。
その方法として、図２で示せば、管２の管底部を最小径にして漸拡形状にして、他端を最大径の形状管にする。
その管形状は、図２の如くで、管２の管底部では、径：Ｄφ、速度：Ｖａ、
該２の上部では、上部径：Ｅφ、長さ：Ａ、速度：ｕ
上部径：Ｅφは、管底部径：Ｄφのｎ倍にすると
上部径：Ｅ＝ｎ＊Ｄ ・・・・・・・・・・・・・・・・・・・・（１）
依つて 上部径の速度：ｕ＝Ｖａ／ｎ２ ・・・・・・・・・・・・・・・（２）
管路抵抗が最小時の漸拡部の角度：θ＝６°にする。
漸拡部の任意の位置をａ線よりｘ点の径：ｄｘとすると次の式が成りたつ。
（ｄｘ−Ｄ）／（Ｂ−ｘ）＝（Ｅ−Ｄ）／Ｂ
依つて ｄｘ＝（Ｅ−Ｄ）（Ｂ−ｘ）／Ｂ＋Ｄ ・・・・・・・・・・・・（３）
茲で、該２の管底部の速度：Ｖａ故に、ｘ点の速度：Ｖｘとして
Ｖｘ／Ｖａ＝Ｄ２／ｄｘ２ 依つて Ｖｘ＝Ｖａ＊Ｄ２／ｄｘ２
（３）式から Ｖｘ＝Ｖａ＊Ｄ２／（（Ｂ−ｘ）（Ｅ−Ｄ）／Ｂ＋Ｄ）２ ・・・・・・・・・・・・・・（４）
茲で ａ点 即ち ｘ＝０ の点で Ｖｘ＝Ｖａ＊Ｄ２／Ｅ２
（１）を代入して Ｖｘ＝Ｖａ＊Ｄ２／（ｎ＊Ｄ）２＝Ｖａ／ｎ２
この事は、ａ点の速度は、Ｄ径の速度：Ｖａの１／ｎ２である事を示し、ａ点より水面までは管径：Ｅは、長さ；Ａ間で一定だから、水面までは速度をｕとすると：ｕ＝Ｖａ／ｎ２
これを図３で示すと、線図２になる。
又、動力に付いて説明すると、
動力：Ｐ＝Ｆ＊Ｖｋｇｍ／ｓｅｃ Ｆ：作用する力（ｋｇ）、Ｖ：作用する方向の速度（ｍ／ｓｅｃ）
茲で ρ：密度（ｋｇ／ｍ３）
Ａ：管断面積（ｍ２）
Ｊ：管長＝水深（ｍ）とすると
Ｆ＝ρ＊Ａ＊Ｊ ｋｇ
依つて １式は Ｐ＝（ρ＊Ａ＊Ｊ）＊Ｖ ｋｇｍ／ｓｅｃ
依つて 動力は Ｐ＝ρ＊Ａ＊（Ｊ＊Ｖ）ｋｇｍ／ｓｅｃ ・・・・・・・・・（５）
この式から、（ρ＊Ａ）は定数，（Ｊ＊Ｖ）は面積を示す故、動力はＪ＊Ｖの面積に 比例する事を示す。
従つて、動力の大小は、面積で解るから、図３で、点ｅ，ｆ，ｇ，ｈ，ｉ，ｊとして、
点ｅ，ｆ，ｇで囲まれた面積：Ｓ１ｍ２／ｓｅｃは、汲上げポンプの動力を示すから
汲上げポンプの動力Ｌは、；Ｌ＝ρ＊Ａ＊Ｓ１ ｋｇｍ／ｓｅｃ・・・・・・・（６）
点ｅ，ｇ，ｈ，ｉ，ｊで囲まれた斜線部の面積：Ｓ２ｍ２／ｓｅｃは、水圧に依る動力を示すから
水圧に依る動力は、Ｐ１＝ρ＊Ａ＊Ｓ２ｋｇｍ／ｓｅｃ ・・・・・・・・（７）
茲で Ｓ２＞Ｓ１故，Ｐ１＞Ｌ
依つて 取得電力Ｐは Ｐ＝Ｌ−Ｐ１ ・・・・・・・・・・・・・・・・・・（８）
即ち、新しい電力が生じる事を証明する。Underwater, the water pressure increases in proportion to the direction deeper than the water surface (downward), which means that the position energy increases, and conversely the upward flow from a certain deep position toward the water surface decreases the pressure. Because the direction, that is, the direction in which the position energy is lowered, new power is generated for the position energy that has been lowered.
By the way, since the underwater pressure, which is a static pressure, does not generate an upward flow in a natural state, if it is forced to an upward flow, new energy is generated due to the decreasing direction of position energy.
Therefore, if one end of the open pipe is closed, the closed one end is down, the other end is submerged in accordance with the water surface, and the closed one end is opened, the pipe bottom pressure turns and becomes an upward flow. However, it is a well-known fact that the water pressure increases according to the depth at 0 pressure on the water surface.
This phenomenon is explained by Bernoulli's theorem. In FIG. 1, with reference to the water surface of the submerged open pipe 1, the external pressure on the water surface; P 1 , the velocity due to the water pressure on the water surface; V 1 , position head; Z 1
At the bottom of the tube 1, the external pressure; P 2 , velocity by water pressure; V 2 , position head; Z 2
Figure 2015086851
P 1 = P 2 , velocity due to water pressure on the water surface; V 1 = 0, because of the water surface, Z 1 = 0
Figure 2015086851
This indicates the direction against gravity acceleration, that is, ascending speed head, so water depth; if X = Z 2 , X = V 2 / 2G (m)
Figure 2015086851
Therefore, efficiency is bad.
An efficient method is to reduce the pipe resistance if the shape is such that only the bottom of the pipe is set to a predetermined speed for turning the water wheel and the speed of the other part of the pipe is made as low as possible.
As the method, as shown in FIG. 2, the bottom of the tube 2 is made to have a minimum diameter and gradually expanded, and the other end is made to have a maximum diameter.
The tube shape is as shown in FIG. 2. At the tube bottom of the tube 2, the diameter is D φ , the speed is Va,
In the upper part of 2, the upper diameter: Eφ , length: A, speed: u
Upper diameter: E φ is tube bottom diameter: D If φ is n times, upper diameter: E = n * D (1)
Therefore, the speed of the upper diameter: u = Va / n 2 (2)
The angle of the gradually expanding portion when the pipe resistance is minimum: θ = 6 °.
If the arbitrary position of the gradually expanding portion is the diameter x of the point x from the line a: dx, the following equation is established.
(Dx−D) / (B−x) = (ED) / B
Therefore, dx = (E−D) (B−x) / B + D (3)
Finally, the velocity of the bottom of the tube 2 is Va, and therefore the velocity at the point x: Vx is Vx / Va = D 2 / dx 2 Vx = Va * D 2 / dx 2
From the equation (3): Vx = Va * D 2 / ((B−x) (ED) / B + D) 2 (4)
At point a, ie at point x = 0 Vx = Va * D 2 / E 2
Substituting (1) Vx = Va * D 2 / (n * D) 2 = Va / n 2
This indicates that the speed at point a is 1 / n 2 of the speed of diameter D: Va, and the pipe diameter: E is the length from point a to the water surface; Where u is the speed: u = Va / n 2
This is shown in FIG.
Also, if you explain about power,
Power: P = F * Vkgm / sec F: acting force (kg), V: speed in acting direction (m / sec)
Boiled ρ: density (kg / m 3 )
A: Pipe cross-sectional area (m 2 )
J: Pipe length = water depth (m)
F = ρ * A * J kg
Therefore, Formula 1 is P = (ρ * A * J) * V kgm / sec
Therefore, the power is P = ρ * A * (J * V) kgm / sec (5)
From this equation, (ρ * A) is a constant, and (J * V) is an area, indicating that the power is proportional to the area of J * V.
Therefore, since the magnitude of the power is understood by the area, in FIG. 3, as points e, f, g, h, i, j,
Area surrounded by points e, f, and g: S 1 m 2 / sec indicates the power of the pump
The power L of the pump is: L = ρ * A * S 1 kgm / sec (6)
The area of the shaded portion surrounded by the points e, g, h, i, j: S 2 m 2 / sec indicates power depending on the water pressure.
Power depending on water pressure is P 1 = ρ * A * S 2 kgm / sec (7)
Boil S 2 > S 1 hence P 1 > L
Therefore, the acquired power P is P = LP 1 (8)
That is, it proves that new power is generated.

図２で示す如く、管長Ｊｍの漸拡形状の開放管２の上部の広拡径を水面と同レベルにして垂直に沈め、該２の上部に汲上げポンプ３、該２の管底部に、水車４、該４で駆動する発電機５を設けた構造にして、該３で汲み上げる事に依つて強制的に上昇流を生じさる事で水圧の上下方向に強制的に不均衡にして、水圧を転換させた上昇流に依つて該４を駆動させ、該５に依つて発電させる方法であつて、発電効率が左右する管路抵抗を最小にするには、該４水車を設けた該２の管底部を最小径にした漸拡形状にして、該２の上部を最大径の形状管にして、上昇流の速度を、漸速させる構造で、式（８）に示す如く電力が取得出来るから、本発明は、発電装置の機能を果たせる。As shown in FIG. 2, the widened diameter of the upper part of the gradually expanding open pipe 2 having a pipe length of Jm is sunk vertically with the same level as the water surface, the pumping pump 3 is placed on the upper part of the pipe 2, and A structure in which a water turbine 4 and a generator 5 driven by the water turbine 4 are provided, and an upward flow is forcibly generated by pumping up by the water 3, so that the water pressure is forcibly imbalanced in the vertical direction. In order to minimize the pipe resistance that is affected by the power generation efficiency, the 4 is driven by the upward flow obtained by converting the current and the power is generated by the 5. With the structure where the bottom of the tube is gradually expanded with the minimum diameter, the upper part of the tube 2 is formed with the maximum diameter, and the speed of the upward flow is gradually increased, so that electric power can be obtained as shown in equation (8). Therefore, the present invention can fulfill the function of the power generation device.

所定の深さの海洋、湖であれば、日夜問わず年間連続発電が出来るから、海に囲まれた立地条件からして、潤沢な電力が得られる
他と比較して設備費が軽少で済み、運転経費は、燃料費不要だから、管理費だけで済むから消費者の負担が軽微で済み、社会的効果は絶大である。If the ocean and lake have a predetermined depth, continuous power generation is possible regardless of day and night, so the facility costs are low compared to others where ample power can be obtained due to the location conditions surrounded by the sea. In addition, since there is no need for fuel costs for operating costs, only management costs are required, so the burden on consumers is minimal and the social effects are enormous.

・・・・・・・・管内の上昇流エネルギーの説明図・ ・ ・ ・ ・ ・ ・ ・ Explanatory diagram of upward flow energy in the pipe ・・・・・・・・構成図········Diagram ・・・・・・・・水深と上昇流の流速の相関図・ ・ ・ ・ ・ ・ ・ ・ Correlation diagram of water depth and upward flow velocity

浮かべた浮胴体に本機を幾多併設し、浮胴体上に電気設備を設置して１セツトし、海底ケーブルで陸地に送電する。A lot of this machine is installed in the floating body, and electric equipment is installed on the floating body, and it is set for one set.

電力は、産業、社会の基盤故、効果は絶大である。Electricity is extremely effective because it is the foundation of industry and society.

１・・・・・・開放管
２・・・・・・斬拡管
３・・・・・・汲上げポンプ
４・・・・・・水車
５・・・・・・発電機1 ·································· 2

水中では、水面より深くなる方向（下方）に比例して水圧が上がるが、その事は位置エレルギーが上がる事を意味し、逆に、ある深い位置から水面方向への上昇流は、
圧力が下がる方向、即ち位置エレルギーが下がる方向故、下がつた位置エレルギー分新たな動力が生まれる事になる。
静圧である水中圧は、、自然の状態では、上昇流が生じないから、強制的に上昇流にさせれば、位置エレルギーの減少方向になるから、新たなエネルギーが生じる事になる。
そこで、開放管の一端を塞いだ状態で、塞いだ一端を下にして、他端を水面に合わせて沈めた状態で、塞いだ一端を開放させると、管底圧力が転じて上昇流になる訳だが、
その速度は、トリチエリー定理で説明出きる。
トリチエリー定理とは「速度は、水面から出口までの高さと重力加速度に依つてのみ決

Figure 2015086851
図１で示すと、開放管１のａ点を塞いだ状態で沈めて、ａ点を開放するとａ点は、大気への出口となつて上昇流に転ずる事になるから、トリチエリー定理そのものである。
Figure 2015086851
この事は、上昇流は、深さに依つて決まる事を意味する。
Figure 2015086851
管の形状を考察すると、直管の場合、管路抵抗：Ｒ＝ｆ＊（ｌ／ｄ）＊ｖ２／２Ｇで示す如く、管路抵抗は、管長ｌと速度ｖ２に比例して増大するから、直管では、動力の殆どが、管路抵抗に費やされる為、効率が悪い。
従つて、効率良い方法は、管内流速を低くする事にあるから、管の速度を出来る丈低くなる様な形状にし、管底部だけを水車を廻すに必要な高速度になる様な管径にすれば良い。
その方法として、図２で示せば、水車４設置部の管２の管底部を最小径にして最大流速にして、漸拡形状にして、最上部を最大径の形状管にして速度を減速する。
その管形状は、図２の如くで、管２の管底部は、最小部にして径：Ｄφ、速度：Ｖａ、該２の上部では、上部の最大径：Ｅφ、長さ：Ａ、速度：ｕ Ｅφを、管底部径：Ｄφのｎ倍にすると 上部径：Ｅ＝ｎ＊ Ｄとすると ‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥（１）
速度は、径の二乗に反比例するから 上部径Ｅφの速度：ｕ＝Ｖａ／ｎ２‥‥‥‥（２）
漸拡部の任意の位置をａ線とすると、ａ線よりｘ点の径：ｄｘとして次の式が成りたつ。
（ｄｘ−Ｄ）／（Ｂ−ｘ）＝（Ｅ−Ｄ）／Ｂ
依つて ｄｘ＝（Ｅ−Ｄ）（Ｂ−ｘ）／Ｂ＋Ｄ ‥‥‥‥‥‥‥‥‥‥‥‥‥‥（３）
茲で、該２の管底部の速度：Ｖａ故に、ｘ点の速度：Ｖｘとして
Ｖｘ／Ｖａ＝Ｄ２／ｄｘ２ 依つて ｘ点の速度： Ｖｘ＝Ｖａ＊Ｄ２／ｄｘ２
（３）式を代入して Ｖｘ＝Ｖａ＊Ｄ２／（（Ｂ−ｘ）（Ｅ−Ｄ）／Ｂ＋Ｄ）２ ‥‥‥‥‥‥‥‥（４）
茲で ａ点 即ち ｘ＝０ の点で Ｖｘ＝Ｖａ＊Ｄ２／Ｅ２
（１）を代入して Ｖｘ＝Ｖａ＊Ｄ２／（ｎ＊Ｄ）２＝Ｖａ／ｎ２
この事は、ａ点（ｘ＝０）即ち、Ｄφ径の速度は、管底部の速度Ｖａの１／ｎ２である事を示し、ａ点より水面までの長さ；Ａ間は、管径：Ｅで一定だから、ａ点より水面までは速度ｕで一定であるから、管径Ｅφの速度：ｕ＝Ｖａ／ｎ２
漸拡部Ｂ間の速度は、管底部の速度：Ｖａ、で上昇して上部径Ｅの速度：ｕに減速する事になる。
これを図３で示すと、線図２になる。
動力：Ｐ＝Ｆ＊Ｖｋｇｍ／ｓｅｃ Ｆ：作用する力（ｋｇ）、Ｖ：作用する方向の速度（ｍ／ｓｅｃ）
茲で ρ：密度（ｋｇ／ｍ３）Ａ：管断面積（ｍ２）Ｊ：管長＝水深（ｍ）とすると
Ｆ＝ρ＊Ａ＊Ｊ ｋｇ
依つて 動力：Ｐ＝（ρ＊Ａ＊Ｊ）＊Ｖ ｋｇｍ／ｓｅｃ
依つて 動力は Ｐ＝ρ＊Ａ＊（Ｊ＊Ｖ）ｋｇｍ／ｓｅｃ ‥‥‥‥‥‥‥‥‥‥（５）
この式から、（ρ＊Ａ）は定数，（Ｊ＊Ｖ）は面積を示す故、動力はＪ＊Ｖの面積に 比例する事を示す。
従つて、動力の大小は、面積で解るから、図３で、点ｅ，ｆ，ｇ，ｈ，ｉ，ｊとして、
点ｅ，ｆ，ｇで囲まれた面積は、汲上げポンプの動力を示し、ｇ，ｈ，ｉ間では、汲上げ速度より水圧によつて出し得る上昇速度が大であるから汲上げ動力は、水圧による上昇動力に打ち消される事になるから汲上げ動力は、ｇ，ｈ，ｉ間では無い事を意味する。
従つて、点ｅ，ｆ，ｇで囲まれた面積：Ｓ１ ｍ２／ｓｅｃは、汲上げポンプの動力を示すから
汲上げポンプの動力Ｌは、；Ｌ＝ρ＊Ａ＊Ｓ１ ｋｇｍ／ｓｅｃ‥‥‥‥（６）
点ｅ，ｇ，ｈ，ｉ，ｊで囲まれた斜線部の面積：Ｓ２ ｍ２ｓｅｃは、水圧に依る動力を示すから
水圧に依る動力は、Ｐ＝ρ＊Ａ＊Ｓ２ ｋｇｍ／ｓｅｃ ‥‥‥‥‥‥‥（７）
茲で Ｓ２＞Ｓ１ 故，Ｐ＞Ｌ
依つて 取得電力Ｐ０は Ｐ０＝Ｐ−Ｌ ‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥‥（８）
即ち、新しい電力が生じる事を証明する。Underwater, the water pressure increases in proportion to the direction deeper than the water surface (downward), but this means that the position energy increases, and conversely, the upward flow from a certain deep position toward the water surface is
Because the pressure decreases, that is, the position energy decreases, new power is generated for the position energy that has been decreased.
The underwater pressure, which is a static pressure, does not generate an upward flow in a natural state. Therefore, if the upward flow is forcibly increased, the position energy decreases and new energy is generated.
Therefore, if one end of the open pipe is closed, the closed one end is down, the other end is submerged in accordance with the water surface, and the closed one end is opened, the pipe bottom pressure turns and becomes an upward flow. In translation,
The speed can be explained by the Tritiery theorem.
What is the Tritiery theorem? “Velocity is determined only by the height from the surface to the exit and the acceleration of gravity.
Figure 2015086851
As shown in FIG. 1, the point a of the open pipe 1 is sunk in a closed state, and when the point a is opened, the point a becomes an outlet to the atmosphere and turns into an upward flow. .
Figure 2015086851
This means that the updraft is determined by the depth.
Figure 2015086851
Considering the shape of the pipe, in the case of a straight pipe, the pipe resistance increases in proportion to the pipe length l and the velocity v 2 as indicated by pipe resistance: R = f * (l / d) * v 2 / 2G. Therefore, in the straight pipe, since most of the power is spent on the pipe resistance, the efficiency is poor.
Therefore, an efficient method is to reduce the flow velocity in the pipe, so that the pipe speed is made as low as possible and the pipe diameter is adjusted to the high speed necessary for turning the water turbine. Just do it.
As the method, as shown in FIG. 2, the bottom of the pipe 2 of the water turbine 4 installation part has a minimum diameter to obtain a maximum flow velocity, a gradually expanding shape, and a top part having a maximum diameter shaped pipe to reduce the speed. .
The tube shape is as shown in FIG. 2, and the tube bottom portion of the tube 2 has a minimum portion with a diameter: Dφ , a speed: Va, and an upper portion of the upper portion with a maximum diameter: Eφ , length: A, Velocity: u E When φ is made n times the tube bottom diameter: D φ Upper diameter: E = n * When D is set …………………………………………………………………………………… (1)
Since the speed is inversely proportional to the square of the diameter, the speed of the upper diameter E φ : u = Va / n 2 (2)
Assuming that the arbitrary position of the gradually expanding portion is a line, the following expression is established as the diameter dx of point x from line a.
(Dx−D) / (B−x) = (ED) / B
Therefore, dx = (E−D) (B−x) / B + D ………………………………………………………………………… (3)
The speed at the bottom of the tube: Va, therefore, the speed at the x point: Vx, Vx / Va = D 2 / dx 2 and therefore the speed at the x point: Vx = Va * D 2 / dx 2
Substituting the equation (3), Vx = Va * D 2 / ((B−x) (E−D) / B + D) 2 (4)
At point a, ie at point x = 0 Vx = Va * D 2 / E 2
Substituting (1) Vx = Va * D 2 / (n * D) 2 = Va / n 2
This indicates that the point a (x = 0), that is, the speed of the diameter φ is 1 / n 2 of the pipe bottom speed Va, and the length from the point a to the water surface; diameter: because constant E, because until the water surface than a point which is fixed at a rate u, the speed of the pipe diameter E φ: u = Va / n 2
The speed between the gradually expanding parts B increases at the pipe bottom speed: Va, and decreases to the speed of the upper diameter E: u.
This is shown in FIG.
Power: P = F * Vkgm / sec F: acting force (kg), V: speed in acting direction (m / sec)
茲: Density (kg / m 3 ) A: Pipe cross-sectional area (m 2 ) J: Pipe length = depth of water (m) F = ρ * A * J kg
Power: P = (ρ * A * J) * V kgm / sec
Therefore, the power is P = ρ * A * (J * V) kgm / sec ………………………………………………………………………………………… (5)
From this equation, (ρ * A) is a constant, and (J * V) is an area, indicating that the power is proportional to the area of J * V.
Therefore, since the magnitude of the power is understood by the area, in FIG. 3, as points e, f, g, h, i, j,
The area surrounded by the points e, f, and g indicates the power of the pump, and between g, h, and i, the pumping power is higher than the pumping speed because the rising speed that can be generated by the water pressure is larger. This means that the pumping power is not between g, h, and i because it is canceled out by the rising power due to the water pressure.
Therefore, since the area surrounded by the points e, f, and g: S 1 m 2 / sec indicates the power of the pump, the power L of the pump is: L = ρ * A * S 1 kgm / sec (6)
The area of the shaded area surrounded by the points e, g, h, i, j: S 2 m 2 sec indicates the power depending on the water pressure, so the power depending on the water pressure is P = ρ * A * S 2 kgm / sec (7)
B> S 2 > S 1 and P> L
Therefore, the acquired power P 0 is P 0 = P−L ………………………………………………………………………………………………………………………………………………………………………………………………………… (8)
That is, it proves that new power is generated.

水中では、水面より深くなる方向（下方）に比例して水圧が上がるが、その事は位置エネルギーが上がる事を意味し、逆に、ある深い位置から水面方向への上昇流は、圧力が下がる方向、即ち位置エネルギーが下がる方向故、下がつた位置エネルギー分新たな動力が生まれる事になる。
静圧である水中圧は、自然の状態では、上昇流が生じないから、強制的に上昇流にさせれば、位置エネルギーの減少方向になるから、新たなエネルギーが生じる事になる。
そこで、開放管の一端を塞いだ状態で、塞いだ一端を下にして、他端を水面に合わせて沈めた状態で、塞いだ一端を開放させると、管底圧力が転じて上昇流になる訳だが、その速度は、トリチエリー定理で説明出きる。
トリチエリー定理とは「速度は、水面から出口までの高さと重力加速度に依つてのみ決

Figure 2015086851
図１で示すと、開放管１のａ点を塞いだ状態で沈めて、ａ点を開放するとａ点は、大気への出口となつて上昇流に転ずる事になるから、トリチエリー定理そのものである。
Figure 2015086851
この事は、上昇流は、深さに依つて決まる事を意味する。
Figure 2015086851
管の形状を考察すると、直管の場合、管路抵抗：Ｒ＝ｆ＊（ｌ／ｄ）＊ｖ２／２Ｇで示す如く、管路抵抗は、管長ｌと速度ｖ２に比例して増大するから、直管では、動力の殆どが、管路抵抗に費やされる為、効率が悪い。
従つて、効率良い方法は、管内流速を低くする事にあるから、管の速度を出来る丈低くなる様な形状にし、管底部だけを水車を廻すに必要な高速度になる様な管径にすれば良い。
その方法として、図２で示せば、水車４設置部の管２の管底部を最小径にして最大流速にして、漸拡形状にして、最上部を最大径の形状管にして速度を減速する。
その管形状は、図２の如くで、管２の管底部は、最小部にして径：Ｄφ、速度：Ｖａ、該２の上部では、上部の最大径：Ｅφ、長さ：Ａ、速度：ｕ Ｅφを、管底部径：Ｄのｎ倍にすると 上部径：Ｅ＝ｎ＊Ｄとすると …………………（１）
速度は、径の二乗に反比例するから 上部径Ｅの速度：ｕ＝Ｖａ／ｎ２ ………（２）
漸拡部の任意の位置をａ線とすると、ａ線よりｘ点の径：ｄｘとして次の式が成りたつ。
（ｄｘ−Ｄ）／（Ｂ−ｘ）＝（Ｅ−Ｄ）／Ｂ
依つて ｄｘ＝（Ｅ−Ｄ）（Ｂ−Ｘ）／Ｂ＋Ｄ ………………（３）
茲で、該２の管底部の速度：Ｖａ故に、ｘ点の速度：Ｖｘとして
Ｖｘ／Ｖａ＝Ｄ２／ｄｘ２ 依つて ｘ点の速度： Ｖｘ＝Ｖａ＊Ｄ２／ｄｘ２
（３）式を代入して Ｖｘ＝Ｖａ＊Ｄ２／（（Ｂ−ｘ）（Ｅ−Ｄ）／Ｂ＋Ｄ）２ ……………（４）
茲で ａ点 即ち ｘ＝０ の点で Ｖｘ＝Ｖａ＊Ｄ２／Ｅ２
（１）を代入して Ｖｘ＝Ｖａ＊Ｄ２／（ｎ＊Ｄ）２＝Ｖａ／ｎ２
この事は、ａ点（ｘ＝０）即ち、Ｄφ径の速度は、管底部の速度Ｖａの１／ｎである事を示し、ａ点より水面までの長さ；Ａ間は、管径：Ｅφで一定だから、ａ点より水面までは速度ｕで一定であるから、管径Ｅの速度：ｕ＝Ｖａ／ｎ２
漸拡部Ｂ間の速度は、管底部の速度：Ｖａ、で上昇して上部径Ｅの速度：ｕに減速する事になる。
これを図３で示すと、線図２になる。
動力：Ｐ＝Ｆ＊Ｖｋｇｍ／ｓｅｃ Ｆ：作用する力（ｋｇ）、Ｖ：作用する方向の速度（ｍ／ｓｅｃ）
茲で ρ：密度（ｋｇ／ｍ３）Ａ：管断面積（ｍ２）Ｊ：管長＝水深（ｍ）とすると
Ｆ＝ρ＊Ａ＊Ｊ ｋｇ
依つて 動力：Ｐ＝（ρ＊Ａ＊Ｊ）＊Ｖ ｋｇｍ／ｓｅｃ
依つて 動力は Ｐ＝ρ＊Ａ＊（Ｊ＊Ｖ）ｋｇｍ／ｓｅｃ ……………（５）
この式から、（Ｊ＊Ｖ）は面積を示す故、動力はＪ＊Ｖの面積に 比例する事を示す。
従つて、動力の大小は、面積で解るから、図３で、点ｅ，ｆ，ｇ，ｈ，ｉ，ｊとして、
点ｅ，ｆ，ｇで囲まれた面積は、汲上げポンプの動力を示し、ｇ，ｈ，ｉ間では、汲上げ速度より水圧によつて出し得る上昇速度が大であるから汲上げ動力は、水圧による上昇動力に打ち消される事になるから汲上げ動力は、ｇ，ｈ，ｉ間では無い事を意味する。
従つて、点ｅ，ｆ，ｇで囲まれた面積：Ｓ１ ｍ２／ｓｅｃは、汲上げポンプの動力を示すから
汲上げポンプの動力Ｌは、；Ｌ＝ρ＊Ａ＊Ｓ１ ｋｇｍ／ｓｅｃ ………（６）
点ｅ，ｇ，ｈ，ｉ，ｊで囲まれた斜線部の面積：Ｓ２ ｍ２／ｓｅｃは、水圧に依る動力を示すから
水圧に依る動力は、Ｐ＝ρ＊Ａ＊Ｓ２ ｋｇｍ／ｓｅｃ ………（７）
茲で Ｓ２＞Ｓ１ 故，Ｐ＞Ｌ
依つて 取得電力Ｐ０は Ｐ０＝Ｐ−Ｌ ………………（８）
即ち、新しい電力が生じる事を証明する。Underwater, the water pressure increases in proportion to the direction deeper than the water surface (downward), which means that the potential energy increases, and conversely the upward flow from a certain deep position toward the water surface decreases the pressure. Because the direction, that is, the direction in which the potential energy decreases, new power is generated by the amount of potential energy that has been lowered.
Underwater pressure, which is a static pressure, does not generate an upward flow in a natural state, and if it is forced to an upward flow, the potential energy decreases and thus new energy is generated.
Therefore, if one end of the open pipe is closed, the closed one end is down, the other end is submerged in accordance with the water surface, and the closed one end is opened, the pipe bottom pressure turns and becomes an upward flow. However, the speed can be explained by the Tritiery theorem.
What is the Tritiery theorem? “Velocity is determined only by the height from the surface to the exit and the acceleration of gravity.
Figure 2015086851
As shown in FIG. 1, the point a of the open pipe 1 is sunk in a closed state, and when the point a is opened, the point a becomes an outlet to the atmosphere and turns into an upward flow. .
Figure 2015086851
This means that the updraft is determined by the depth.
Figure 2015086851
Considering the shape of the pipe, in the case of a straight pipe, the pipe resistance increases in proportion to the pipe length l and the velocity v 2 as indicated by pipe resistance: R = f * (l / d) * v 2 / 2G. Therefore, in the straight pipe, since most of the power is spent on the pipe resistance, the efficiency is poor.
Therefore, an efficient method is to reduce the flow velocity in the pipe, so that the pipe speed is made as low as possible and the pipe diameter is adjusted to the high speed necessary for turning the water turbine. Just do it.
As the method, as shown in FIG. 2, the bottom of the pipe 2 of the water turbine 4 installation part has a minimum diameter to obtain a maximum flow velocity, a gradually expanding shape, and a top part having a maximum diameter shaped pipe to reduce the speed. .
The tube shape is as shown in FIG. 2, and the tube bottom portion of the tube 2 has a minimum portion with a diameter: Dφ , a speed: Va, and an upper portion of the upper portion with a maximum diameter: Eφ , length: A, Speed: u E When φ is made n times the tube bottom diameter: D Upper diameter: E = n * D ………………… (1)
Since the speed is inversely proportional to the square of the diameter, the speed of the upper diameter E: u = Va / n 2 (2)
Assuming that the arbitrary position of the gradually expanding portion is a line, the following expression is established as the diameter dx of point x from line a.
(Dx−D) / (B−x) = (ED) / B
Therefore, dx = (ED) (BX) / B + D (3)
The speed at the bottom of the tube: Va, therefore, the speed at the x point: Vx, Vx / Va = D 2 / dx 2 and therefore the speed at the x point: Vx = Va * D 2 / dx 2
(3) Vx = Va * D 2 / by substituting equation ((B-x) (E -D) / B + D) 2 ............... (4)
At point a, ie at point x = 0 Vx = Va * D 2 / E 2
Substituting (1) Vx = Va * D 2 / (n * D) 2 = Va / n 2
This indicates that the point a (x = 0), that is, the speed of the Dφ diameter is 1 / n of the pipe bottom speed Va, the length from the point a to the water surface; : E Since it is constant at φ , since the speed from the point a to the water surface is constant at the speed u, the speed of the pipe diameter E: u = Va / n 2
The speed between the gradually expanding parts B increases at the pipe bottom speed: Va, and decreases to the speed of the upper diameter E: u.
This is shown in FIG.
Power: P = F * Vkgm / sec F: acting force (kg), V: speed in acting direction (m / sec)
茲: Density (kg / m 3 ) A: Pipe cross-sectional area (m 2 ) J: Pipe length = depth of water (m) F = ρ * A * J kg
Power: P = (ρ * A * J) * V kgm / sec
Therefore, the power is P = ρ * A * (J * V) kgm / sec ……………… (5)
From this equation, (J * V) indicates the area, so the power is proportional to the area of J * V.
Therefore, since the magnitude of the power is understood by the area, in FIG. 3, as points e, f, g, h, i, j,
The area surrounded by the points e, f, and g indicates the power of the pump, and between g, h, and i, the pumping power is higher than the pumping speed because the rising speed that can be generated by the water pressure is larger. This means that the pumping power is not between g, h, and i because it is canceled out by the rising power due to the water pressure.
Therefore, since the area surrounded by the points e, f, and g: S 1 m 2 / sec indicates the power of the pump, the power L of the pump is: L = ρ * A * S 1 kgm / sec ……… (6)
The area of the shaded part surrounded by the points e, g, h, i, j: S 2 m 2 / sec indicates the power depending on the water pressure, so the power depending on the water pressure is P = ρ * A * S 2 kgm / sec ......... (7)
B> S 2 > S 1 and P> L
Therefore, the acquired power P 0 is P 0 = P−L (8)
That is, it proves that new power is generated.

Claims (1)

図２で示す如く、管長Ｊｍの漸拡形状の開放管２の上部の広拡径を水面と同レベルにして垂直に沈め、該２の上部に汲上げポンプ３、該２の管底部に、水車４、該４で駆動する発電機５を設けた構造にして、該３で汲み上げる事に依つて強制的に上昇流を生じさる事で水圧の上下方向に強制的に不均衡にして、水圧が転換した上昇流に依つて該４を駆動させ、該５に依つて発電させる方法であつて、発電効率が左右する管路抵抗を最小にして、上昇流の速度を漸速させる為、該４水車を設けた該２の管底部を最小径にした漸拡形状にした上部を最大径の形状管にした構造。As shown in FIG. 2, the widened diameter of the upper part of the gradually expanding open pipe 2 having a pipe length of Jm is sunk vertically with the same level as the water surface, the pumping pump 3 is placed on the upper part of the pipe 2, and A structure in which a water turbine 4 and a generator 5 driven by the water turbine 4 are provided, and an upward flow is forcibly generated by pumping up by the water 3, so that the water pressure is forcibly imbalanced in the vertical direction. Is a method of driving the 4 according to the converted upward flow and generating power according to the 5 in order to minimize the pipe resistance that is affected by the power generation efficiency and gradually increase the speed of the upward flow. A structure in which the top of the two pipe bottoms provided with four water turbines is formed in a gradually expanding shape with the smallest diameter, and the shape with the largest diameter is formed on the upper part.
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US1527090A (en) * 1923-03-21 1925-02-17 Simard Wilfrid Thrashing machine
GB1527090A (en) * 1978-03-20 1978-10-04 Tinawy C Turbofloat
JPH06101621A (en) * 1992-09-24 1994-04-12 Shunichi Matsutani High-vacuum differential-pressure pumping type power generation system
CN1161411A (en) * 1997-01-06 1997-10-08 邸文学 Mechanism for power-producing using deep water hydrostatic pressure energy
JP2001165029A (en) * 1999-12-09 2001-06-19 Port & Harbour Research Inst Ministry Of Land Infrastructure & Transport Negative pressure utilizing type pumping structure and pumping method utilizing negative pressure
JP2003227451A (en) * 2002-02-04 2003-08-15 Yasuhiko Shinjo Fluid engine driven by buoyancy
WO2007009192A1 (en) * 2005-07-22 2007-01-25 Stephen John Hastings Power generation system
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* Cited by examiner, † Cited by third party
Publication number Priority date Publication date Assignee Title
US1527090A (en) * 1923-03-21 1925-02-17 Simard Wilfrid Thrashing machine
GB1527090A (en) * 1978-03-20 1978-10-04 Tinawy C Turbofloat
JPH06101621A (en) * 1992-09-24 1994-04-12 Shunichi Matsutani High-vacuum differential-pressure pumping type power generation system
CN1161411A (en) * 1997-01-06 1997-10-08 邸文学 Mechanism for power-producing using deep water hydrostatic pressure energy
JP2001165029A (en) * 1999-12-09 2001-06-19 Port & Harbour Research Inst Ministry Of Land Infrastructure & Transport Negative pressure utilizing type pumping structure and pumping method utilizing negative pressure
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