JPS5832973A - Method of taking out motive power by utilizing depth pressure of liquid during its forced circulation - Google Patents
Method of taking out motive power by utilizing depth pressure of liquid during its forced circulationInfo
- Publication number
- JPS5832973A JPS5832973A JP55115658A JP11565880A JPS5832973A JP S5832973 A JPS5832973 A JP S5832973A JP 55115658 A JP55115658 A JP 55115658A JP 11565880 A JP11565880 A JP 11565880A JP S5832973 A JPS5832973 A JP S5832973A
- Authority
- JP
- Japan
- Prior art keywords
- liquid
- power
- cylinder
- tank
- pressure receiving
- Prior art date
- Legal status (The legal status is an assumption and is not a legal conclusion. Google has not performed a legal analysis and makes no representation as to the accuracy of the status listed.)
- Pending
Links
Classifications
-
- F—MECHANICAL ENGINEERING; LIGHTING; HEATING; WEAPONS; BLASTING
- F03—MACHINES OR ENGINES FOR LIQUIDS; WIND, SPRING, OR WEIGHT MOTORS; PRODUCING MECHANICAL POWER OR A REACTIVE PROPULSIVE THRUST, NOT OTHERWISE PROVIDED FOR
- F03B—MACHINES OR ENGINES FOR LIQUIDS
- F03B17/00—Other machines or engines
- F03B17/02—Other machines or engines using hydrostatic thrust
- F03B17/04—Alleged perpetua mobilia
Abstract
Description
【発明の詳細な説明】
本発明は一定水位を維持する液槽内又は河川、湖％Ｎｉ
％貯水池等に開口受圧筒を設け、該液槽内の液体を強制
循環せしめて液深の深い位置での圧力を利用した動力取
出し方法に関するものである。DETAILED DESCRIPTION OF THE INVENTION The present invention is designed to maintain a constant water level in liquid tanks, rivers, and lakes.
This invention relates to a power extraction method that utilizes the pressure at a deep position by providing an open pressure receiving cylinder in a water reservoir or the like, forcing the liquid in the liquid tank to circulate, and utilizing the pressure at a deep position.
以下に図示の実施例に基きその内容について説明する０
１け液槽でその深さ■が約１２ｍのものを採用してあり
、液体としては例へば２０℃の水、塩水、プロピレング
リコール水溶液（寒冷地方で有効である）等を採用して
いる。The contents will be explained below based on the illustrated embodiment.A single liquid tank with a depth of approximately 12 m is used, and examples of liquids include water at 20°C, salt water, and propylene glycol aqueous solution (cold water). effective in rural areas).
２け液槽ｌ内に設置した開口受圧筒で高さＨｌが１ｍの
支柱３上に設けてあり、下方から液体が流入し易いよう
に考慮しであるＯ
この開口受圧筒２の筒底開口部分４の断面積ムは１♂に
採っである。It is an open pressure receiving cylinder installed in a 2-layer liquid tank 1, and is installed on a support 3 with a height Hl of 1 m, and is designed so that liquid can easily flow in from below. The cross-sectional area of portion 4 is 1♂.
５は動力取出しのための機器としてのタービンであって
前記筒底開口部分４に配置せしめである０
６Ｆｉ回転軸でタービｙ５と直結してあり、該タービン
５の回転力を動力室７の動力機器８へ伝動する機能を有
している０
９は開口受圧＠２の上部に設けた溜液槽で断面積もが１
１１１’で高さＩｔｓが３ｍである。Reference numeral 5 denotes a turbine as a device for extracting power, and it is directly connected to the turbine y5 by a rotating shaft arranged in the cylinder bottom opening 4, and the rotational force of the turbine 5 is transferred to the power of the power chamber 7. 09, which has the function of transmitting power to the device 8, is a reservoir tank installed at the top of the open pressure receiving @2, with a cross-sectional area of 1
111' and its height Its is 3 m.
１０は開口受圧筒２の上部開口部、１１は揚液管で開口
断面積Ａｐが１／、高さＨ８がＩｍである。Reference numeral 10 indicates an upper opening of the open pressure receiving cylinder 2, and 11 indicates a liquid pumping pipe, the opening cross-sectional area Ap of which is 1/, and the height H8 of which is Im.
１２は揚液ポンプで横軸々流上−ター１３１Ｃよって作
動し、開口受圧筒２及び溜液槽９内の液体を揚液し、放
出口！４より液槽１円に放出し液体を強制循環せしめて
いる０
冑、液面りと開口受圧筒２の筒底開口部分４との間の深
さ即ち液深Ｈａ＃Ｐｉ１０ ｍである。Reference numeral 12 denotes a liquid lift pump which is operated by a horizontal shaft upstream tank 131C, lifts the liquid in the open pressure receiving cylinder 2 and the stored liquid tank 9, and discharges the liquid from the discharge port! 4, the liquid is forced to circulate in the liquid tank 1, and the depth between the liquid level and the bottom opening 4 of the open pressure receiving cylinder 2, that is, the liquid depth Ha#Pi is 10 m.
次に叙上の構成より成る本発明の作用について説明する
。Next, the operation of the present invention constructed as described above will be explained.
先ず、液槽ｌ内の２０℃の水の水深Ｈａ＝ １０ｍ Ｋ
於ける静水圧は、
一般式 ｆ ａ ｆｇ＠ １！−・・（１）より求
めることが出来る。First, the depth of water at 20°C in the liquid tank L is Ha = 10m K
The hydrostatic pressure at is the general formula f a fg @ 1! - It can be obtained from (1).
舷ｆ／Ｃｆ＝水の密ｔ−９＊ｓ神／１ ｇ 冨重力の加速度ｘ９．８ｍ／ｓｅｅ”１：： １！讃水深冨ｌＩ４−１０ｍ とすると、 冨’３７８Ｇ４ （Ｎψｅｅり となる。Gender f/Cf=water density t-9*s god/1 g Gravity acceleration x9.8m/see”1:: 1! Sansui Fukatomi I4-10m Then, Tomi’378G4 (Nψeeri becomes.
一方、水蒙！！４ｘｉｏｍ即ち、開口受圧１２の筒底開
口部分４に於いて発生する力又は仕事Ｗは、一般式ＰＧ
・ム・ｖ−・・・・・（２）より求めることが出来る。On the other hand, Suimong! ! 4xiom, that is, the force or work W generated at the cylinder bottom opening portion 4 of the opening pressure receiving pressure 12 is expressed by the general formula PG
・mu・v−・・・・・・It can be obtained from (2).
鼓Ｋ ７３ｍ静水のゲージ圧−ｅｔｓｏａ（’５ｙ
、）ムＭ−曽底開口部分く４）の断面積−ＩＩＶ＝
流体の速度”１％。。（揚液ポンプ１２の揚液量をｌ／
、とし念場合、−・Ｃ
流速はＩＱ／ となる）
Ｓ・Ｃ
とすると、
仕事Ｗ−９７８０４（ｋｌ／Ｉ１．、。。り刈（♂）×
１悄４゜。）−９７８０４（凱嗜、。―）
となる。Tsuzumi K 73m still water gauge pressure - etsoa ('5y
, ) Mu M - cross-sectional area of the bottom opening part 4) - IIV =
Fluid velocity is 1%. (Liquid volume pumped by pump 12 is 1%.
, then, -・C The flow velocity becomes IQ/) S・C Then, the work W-97804 (kl/I1.,... Rearing (♂)×
1° 4°. )-97804 (Kai Shu,.-) becomes.
而して、揚水ポンプ１２を作動せしめて溜液槽９円の液
体を毎秒１ｄｓｔｆＬすると仮定し、この液ｔｔ揚液ポ
ンプ１２の放出口１４より液面ＬＫ放流することＫよっ
て、開口受圧筒２内の液体は常に一定の流速で開口受圧
筒２、揚液ポンプ１２．液槽ｌと循環し、液体の自然損
失を無視すれば一定の深さＨ４−１０１１を維持するこ
ととなる。Assuming that the water pump 12 is activated and the liquid in the reservoir tank 9 is increased to 1 dstfL per second, the liquid level LK is discharged from the discharge port 14 of the liquid water pump 12. Therefore, the opening pressure receiving cylinder 2 The liquid inside the open pressure receiving cylinder 2, the liquid pump 12. If the liquid circulates with the liquid tank 1 and the natural loss of liquid is ignored, a constant depth H4-1011 will be maintained.
淘、現実には蒸発郷により液位が変動するので補充する
必要がある。In reality, the liquid level fluctuates due to evaporation, so it needs to be replenished.
一方、前記した開口受圧筒２の筒底開口部分４に設けた
タービン５は一定の速度で回転し、その回転力を回転軸
６ｔ−介して動力室７内の動力機器８に伝達される。On the other hand, the turbine 5 provided in the bottom opening portion 4 of the open pressure receiving cylinder 2 rotates at a constant speed, and its rotational force is transmitted to the power equipment 8 in the power chamber 7 via the rotating shaft 6t.
他方、前記した揚液ポンプ１２は毎秒１ｍＦの液体を高
さ７１ＨＷ １ ｍ％開ロ断面積ム９−１♂の揚液管に
より揚液しているので、横軸々流モー鼾１３が揚液に費
すカーは
％−Ｆ”ｇ−シ・・・・・・・・・（３）によって求め
ることが出来、従って、
ＷＭ−９９８（鵬令）×東８ （ｍ／ｓｅａ鵞）×１−
；９７８α４（大口、＃・Ｃり
となる。On the other hand, since the liquid pump 12 described above lifts liquid at a rate of 1 mF per second through the liquid pump with a height of 71 HW and a 1 m% open cross-sectional area M9-1♂, the transverse axis flow mode 13 is The car spent on liquid can be found by %-F''g-shi... (3), therefore, WM-998 (Peng Ling) x East 8 (m/sea) x 1-
;978α4 (large mouth, #/C).
故に、タービン５により砲り出し得る力、即ち仕事Ｗ？
は
ＷＴ冨Ｗ″′″町
−９７８０４−９７８０，４
＝ ５ｓｏｚａｓ（！λ１．。。り
となる・
要するに、開口受圧筒２内の液体全揚程して溜液槽９内
に貯った水１１ｍｍポンプ１２により連絡して揚液し次
いで液槽１の液面ＬＫ放出し、液面の高さを一定に維持
することｋより、筒底開口部分４に発生する力を一定に
することが出来、この力をタービンＳにより仕事Ｗとし
て引き出し、且つモーターの費し九仕事ＷＭｔ差し引く
ことＫよりＷ？なる大亀な動力を得ることが出来る。間
、上記の計算例では各部分の損失抵抗を無視したもので
あるが、この損失抵抗を考慮し九としても、その差し引
きは正（プラス）となり大、なる動力を得ることが可能
テアリ１童業上きわめて有益である０淘本実施例では液
槽での例奢示しであるが河川、湖、海、貯水池等で行え
ば更に好結果を生ずるものである。又本発明を複数個並
設して利用、すればより一層大きな動力を得ることがで
きることは勿論である。Therefore, the force that can be exerted by the turbine 5, that is, the work W?
is WT TomiW'''' Town-97804-9780,4 = 5sozas(!λ1...) In short, the total height of the liquid in the open pressure receiving cylinder 2 is 11 mm of water stored in the reservoir tank 9. By communicating with the pump 12 to pump the liquid and then discharging the liquid level LK in the liquid tank 1 to maintain the liquid level constant, the force generated in the cylinder bottom opening 4 can be made constant. , by extracting this force as work W by the turbine S and subtracting the work done by the motor WMt, we can obtain a large power of W? from K. In the above calculation example, the loss resistance of each part is ignored. However, even if this loss resistance is taken into account, the deduction is positive (plus), and it is possible to obtain a large amount of power. This is a modest example, but even better results will be obtained if it is carried out in rivers, lakes, seas, reservoirs, etc. Also, if multiple units of the present invention are installed in parallel and used, even greater power can be obtained. Of course it can be done.
第１図は本発明の詳細な説明図、第２図は第１図の要部
の横断平面図である。
１・・・液槽 ２・・・開口受圧筒
４・・・筒底開口部分 ５・・・タービン８・・・動
力機器 ９・・・溜液井１１・・・揚液管
１２・・・揚液ポンプ第２図
明 細 書 ＩＬ 発
明の名称
液体の常時強制循環時に於いて発生する液体の流速と液
深圧力とを利用した動力取出し方法２、特許請求の範囲
液体中に、筒底開口部分４及び溜液槽９を有する開口受
圧筒２ｔ−配置し、前記筒底開口部分４から液体を流入
せしめ、上部開口部１０を経程すると筒内の液体は、溜
液槽９の液体が機器により強制揚柳さｎて液Ｉｉ！に放
出される液量と比例する液１１％流速を得て受圧筒２内
を上昇して溜液槽９に至り、更に同層液槽９より機器に
より液体表面へ放出されて、液面を常に一定の高さに維
持すべく該液体を常に強制循環せしめて液圧を不変にす
る一方、前記筒底開ロ部分４附近に動力取出しのための
機器−を配設した゛液体の常時強制循環時に於いて発生
する液体の流速と液深圧力とを利用した動力蛾出し方法
発明の詳細な説明
本発明は一定水位を維持する液槽内又は河川、湖、海、
貯水池等に開口受圧筒を設け、該液槽内の液体を強制循
環せしめて液深の深い位置での圧力を利用し九動力暖出
し方法に関するものである。
以下に図示の実施例に基きその内容について説明する。
１は液槽でその深さ■が約１２ｍのものを採用してあり
、液体としては例へば２０℃の水、塩水、プロピレング
リコール及びプロピレングリコール水溶液（共に寒冷地
方で有効である。）等を採用している。
２は液槽１円に設置した高さがＩｏｎの開口受圧筒でそ
の上部は液面りより１ｍ下に位置するようにし又高さＨ
Ｉが１ｍの支柱３上に設けてあって、該開口受圧筒２ｔ
−固定し安定を図り且つ下方から液体が流入し易いよう
に考慮しである。
この開口受圧筒２の筒底開口部分４の断面積Ａ＃−１ｔ
ｌ♂に採っである。
５は動力取出しの九めの機器としてのタービンであって
、前記筒底開口部分４に配置せしめである。
６Ｆｉ回転軸でタービン５と直結してあり、該タービン
５の回転力を動力憲７の動力機器８へ伝動する機能を有
している。
９は開口受圧筒２の上部に設けた溜液槽で断面積ムｐが
１ｍ”で高さＨｌが３ｍである。１０は開口受圧筒２の
上部開口部、１１は揚液管で開口断面積ム９が１ｍ／、
高さＨａが１ｍである。
１２は揚液するための機器としての揚液ポンプで横軸々
流上−ター１３によって作動し、溜液槽９内の液体を揚
液し、放出口１４より液槽１内に放出し開口受圧筒２並
びに溜液槽９円の液体を強制循環せしめている・
尚、液面りと開口受圧筒２の筒底開口部分４との間の深
さ即ち液深Ｈ４はｌ１ｍである・次に叙上の構成より成
る本発明の作用について説明する。
先ず、液槽１円の２０Ｃの水の水＠Ｈ４−ｔｔｍに於け
る静水圧は、
一般式 ｒ−１・Ｈ・・・・・・（１）より求めること
が出来る。
鼓にＰ−水の密度−９９８１，４ｔｈ
６−動の加速度−９−８ｍ／Ｂ”
Ｈ−水深！ Ｈ，ｗ ｌ ｌ Ｈ
とすると
静水の圧力Ｐは
Ｐ ロ
Ｐ＝−９９８（Ｋｆ／ｊ）Ｘ９．８（ｚ／／ｓすＸｌｌ
（ｍ）中ｔｏ７ｓｓ＋（’４／ｎ、ｓｌ ）となる。
今、鼓に社団法人日本機械学会発行（昭和５４年８月２
０日発行）の機械工学Ｂエマニュアル「（５）流体工学
・流体機械におけるＳ工の使い方」と称する文献のＰ４
に於ける表５・６を単位の換算に採用すると
（！！ ＳＳ Ｓ一部抜すい）
Ｐ＝１０７５８４（Ｐａ）
＝１０７．５８４（ＫＰａ）
中１０７．６（ＫＰａ）
となる。
一方、水深１１４”−１１Ｂ即ち、開口受圧筒２の筒底
開口部分４に於いて発生する動力又は仕事率Ｆｉ
一般式 ＰａＡ−Ｖ−・・・・・（２）より求めること
が出来る。
鼓Ｋｐ−静水の圧力−１０７５８４（ｈ／ｍａｄ” ）
ム＝筒底開ロ部分４の断面積＝＝１ｍ１Ｖ；流体の速度
−１％（揚液ポンプ１２の揚液量ｔ−ｅｄ１９ｍとした
場合、流速は１ ｒｒＸｓとなる。）
とすると、
動力又は仕事率Ｗ筐Ｐ＠Ａ”Ｖ
＝１０７５８４（−・♂企）
となる。
この動力又は仕事率ｗｔ表５．６により単位の換算を行
うと
Ｗ−１０７５８４（Ｗ）
＝１０７．５８４（Ｋｗ）
＋１０７．５８（Ｋｗ）
となる。
而して、揚水ポンプ１２を作動せしめて溜液檜９円の液
体を毎秒１−揚液すると仮定し、この液量を揚液ポンプ
１２の放出口１４より液面りに放流することによって、
開口受圧筒２同の液体は常に一定の流速で開口受圧筒２
、揚液ポンプ１２．液槽１と循環し、液体の自然損失を
無視すれば、一定の深さＨ４＝１１ｍを維持することと
なる。
陶、現実には蒸発等圧より液位が変動するので補充する
必要があり、又液槽ｌに蓋をして蒸発を防ぐようにして
もよい・
一方前記した開口受圧筒２の筒底開口部分４に設けたタ
ービン５は一定の速度で回転し、その回転力を回転軸６
１介して動力室７内の動力機器８に伝達さする。
他方、前記した揚液ボンダ１２Ｆｉ、毎秒１１１／の液
体を高さ■・−１ｍ、開口断面積ムｐｍ ｌ ＠″の揚
液管により揚液しているので、横軸々流上−ター１３が
揚液に費す動力ＷＭは
ＷＭ” Ｉ’ｓ ｌ−Ｈａ ｅ Ａｐｌｌ ｖ；Ｐ
・ムｐ＠ｖ
＝ ９９８ （’ｌ／１）Ｘ９．８（ｍ／８１ ）Ｘ
１（＋１１１）ＸＩ（ＩＩう刈（ちし≦］＝ ９７＆
０．４ （ｆｆ＠ｌ／／ｉｓ ）となる。
これを前記した表器６により単位の換算を行うと
ＷＭ−９７８０，４（７）
中９４８ （ＫＷ）
となる。
故にタービン５により増り出し得る力、即ち動力ＷＴは
ＷＴ口ｗ −ＷＭ
コ１０７．５８（聰）−９，７８（ＫＷ）諺ｅ ７．８
（ｍ）
となる。
要するに、開口受圧筒２円に流入し溜液槽９内に貯った
水ｔ−揚液ポンプ１２により揚液し次いで液槽１の液面
ＬＫ放出し、液面の高さを一定に維持することにより、
筒底開口部分４に発生する力を一定にすることが出来、
この力をタービン５により仕事Ｗとして引き出し、且つ
モーターの費した仕事ＷＭを差し引くことによりＷＴな
る大きな動力を得ることが出来る。
尚、上記の計算例では各部分の損失抵抗を無視したもの
であるが、この損失抵抗ｔ−考慮したとしても、その差
し引きは正（プラス）となり大きな動力を得ることが出
来、産業上きわめて有益である。
本実施例では液槽での例を示しであるが河川、湖、海、
貯水池等で行えば更に好結果を生ずるものである。
又本発明装置即ち開ロ受圧筒２＠：複数個並設して利用
すれば一層大きな動力を得ることができ頁にｇａ図に示
す如く開口受圧筒２の筒底開口部分４を広くすればより
高効率となる。
未 図面の簡単な説明
第１図は本発明の詳細な説明図、第２図は第１図の要部
の横断平面図、第３図は他の実施例としつ開口受圧筒の
筒底開口部分の拡大図である。
１・・・液槽 ２・・・開口受圧筒 ４・・・筒底開口
部分１１・・・揚液管 １２・・・揚液ポンプ第・う
図
７１．６よ、１□１□８１
特許庁長官島田春樹殿
１、事件の表示
２゜
昭和５５年特 許 願第１１５６５８号３、 補正を
する者
事件との関係 特 許 出 願 人４、代理
人
発明の名称
液体の常時強制循環時に於いて発生する液体の流速と液
深圧力とを利用した動力権出し方法特許請求の範囲
（１）
有
一層
流
表
２条
時強制循環時に於いて発生する液体の流源液深圧力とを
利用した動力窄めし方法
■ 特 求の範 第１項の配 に於いて、動設した液
体の常時強制循環時に於いて発生する液体の流速と液深
圧力とを利用した動力申出し方法
本発明は一定水位を維持する液槽内又は河川、湖、海、
貯水池等に開口受圧筒を設け、該液槽内の液体を強制循
環せしめて液深の深い位置での圧力を利用した動力申出
し方法に関するものである。
以下に図示の実施例に基きその内容について説明する。
第一の実施例（第１図乃至第２図）について。
１＃′ｉ液槽でその深さＨが約１２ｍのものを採用して
あり、液体としては例へば２０℃の水、塩水、プロピレ
ングリコール及ヒプロビレングリコール水溶液（共に寒
冷地方で有効である。）等を採用している。
２＃′ｉ液槽１内に設置し九高さが１０ｍの開口受圧筒
でその上部は液面りより１ｍ下に位置するようにし又高
さｌｉｔが１ｍの支柱３上に設けてあって、該開口受圧
筒２゛を固定し安定を図り且つ下方から液体が流入し易
いように考直しである。
この開口受圧筒２の筒底開口部分４の断面積Ａは１ｍ”
に採っである。
５Ｆｉ動力な出しのための機器（動力申出し装置ｆ）と
しての水中タービンであって、前記筒底開口部分４に配
着せしめである。又この水中タービン５を筒底開口部分
４と上部開口Ｗ６１０との間の任意の位置に配設しても
よい。
６は回転軸で水中タービン５と直結してあり、該水中タ
ービン５の回転力を動力室７の動力機器８へ伝動する機
能をも有している。
９は開口受圧筒２の上＠に設けた溜液槽で断面積Ａｏが
１１１”以上で高さＨ！が３ｍ以上である。
この溜液槽９の容積は状況により増量することは勿論で
ある・
１０ＩＩ′ｉ開口受圧筒２の上部開口部、１１Ｆｉ揚液
管で開口断面積Ａｐが１−１″、高さＨ３が１ｍである
。
崗、この揚液管１１ｔ−第４図に示す如く揚液ポンプの
一部分とみなしてもよい・
１２Ｆｉ揚液するための機器としての揚液ポンプでモー
ター１３によりて作動し、舗液槽９内の液体を揚液し、
放出口１４より液槽１内に放出し溜液槽９内の液体を強
制循環せしめている・尚、液面りと開口受圧筒２の筒底
開口部分４との間の深さ即ち液深ＨａＦ１１１ｍである
。
次に斜上の構成より成る本発明の作用について説明する
。
先ず、液槽１内の２０℃の水の水深Ｈ４＝１１ ｔｎに
於ける静水圧は、
一般式 ρ・ｆ−Ｈ・−（１）
より求めることが出来る。
誠に ρ＝水の密度＝９９８Ｅｔ／、／ｌ＝重力の加速
度＝９．８ｍ１５２
Ｈ＝水深＝Ｈａ＝１１ｍ
とすると
静水の圧力Ｐは
Ｐ＝ρ・ｆ＠Ｈ
Ｐ＝９９８（−Ｋｇｌｄ）Ｘ９．８（ｍ／ＳりＸ１１−
中１（）７５８４（Ｋｅ／ｍ＠３り
となる・
今、舷に社団法人日本機械学会発行（昭和５４年８月２
０日発行）の機械工学ＳＩマニュアルｒ（５１！Ｉ体工
学・流体機械におけるＳＩの使い方Ｊと称する文献のＰ
４に於ける表５・６を単位の換算に採用すると
（表５．６一部数すい）
Ｐ＝１０７５８４（Ｐａ）
＝１０７．５８４（ＫＰａ）
中１０７．６（ＫＰａ）
となる。
一方、水深Ｈ４＝ １１ ｍ即ち、開口受圧筒２の筒底
開口部分４に於いて発生する動力又は仕事率Ｗは
一般式 Ｐ−Ａ＠Ｖ・・・・・・■ □より求め
ることが出来る。
１ＫＰ＝静水の圧力＝１０７５８４（ｈ／ｍ、Ｓ”）Ａ
−筒底開口部分４の断面積＝１＝１″■−流体の速度＝
１呵（揚液ポンプ１２の揚液量を１−Ｈｌとした場合、
流速は
１ｍ／Ｓとなる・）
とすると、
ａ力又ａ仕事率ｗ＝ ｐ ＠Ａ −Ｖ
＝ １０７５８４（Ｋｇ／ｍ１１ｓ冨）ｘｌＧ４ｘ１（
ｍ／５）
＝１［１７５８４（Ｋ４−膳”／Ｓ３）となる０
この動力又は仕事率ｗｔ表５．６により単位の換算を行
うと
Ｗ＝ １０７５８４ （Ｗ）
＝１０７．５８４（Ｋｗ）
中１０７．５８（Ｋｗ）
となる。
而して、揚水ポンプ１２’を作動せしめて溜液槽９内の
液体を毎秒ｉＩＩ／揚液すると仮定し、この液量を揚液
ポンプ１２の放出口１４より液面りに放流することによ
りて、開口受圧筒２内の液体は常に一定の流速で開口受
圧ｗＩ２、揚液ポンプ１２．液槽１と循環し、液体の１
然損失を無視すれば、一定の深さ）ｉ４＝１１ｍを維持
することとなる。
尚、現実に社蒸宛等により液位が変動するので補充する
必要があり、又液槽１ｒＩＣ！をして蒸発を防ぐように
してもよい。
一方、前記した開口受圧筒２の筒底開口部分４に設は几
タービン５ｒｔ一定の速度で回転し、その回転力ｆｔ回
転軸６を介して動力室Ｚ内の動力機６８に伝達される。
陶、水中タービン５の回転軸６Ｖｃ発電＊ｔ−附設して
おいて、８ｍ電力を曜り出してもよい。
他方、前記した揚液ポンプ１２１ｊ、毎秒１１Ｉ／の液
体の高さＨ＠＝ １ｍ、開口断面積Ａｐ＝ Ｉ ＩＩＩ
’の揚液管によ０揚液しているので、構簗麩嶽モーター
１３が揚液に費す動力ＷＭは
Ｗｙｌ ＝ρｍ ｆ ｅ ｌｉ、 ｊ Ａ、＋１ Ｖ＝
Ｐ＃Ａ、６Ｖ
＝９９８（Ｗｅ／ｄ＞Ｘ９．８（ｍ／Ｓ”）ｘ１ｈ＃Ｘ
１←）Ｘｌ（弓４）
＝９７８０．４（Ｉｔ−＋／／Ｓｊ）
となる。
これを前記し念表５．．５ Ｋ ｘり単位の換算を行う
と
ＷＭ＝９７８０．４ｖ＠
６９．７８ （ＩＣり
となる・
故にタービン５により申り出し得る力、即ち動力ＷＴＦ
ｉ
ＷＴ＝Ｗ−ＷＭ
＝ １０７５８ （ｆｆ） −９，７８（ＫＷ）＝９７
．８（ＫＷ）
となる。
要するに、開口受圧筒２内に流入し溜液槽９内７；に貯
った水ｔ−＊＊ポンプ１２により揚液し次いで液槽１の
液面ＬＫ放出し、液面の高さを一定に維持することによ
り、筒底開口部分４１ｆＣ発生する力を一定にすること
が出来、この力をタービン５により仕事Ｗとして引曇出
し、且つモーターの費した仕事ＷＭを差し引くことによ
りＷＴなる大きな動力を得ることが出来る。
尚、上記の計算例では各部分の損失抵抗を無視したもの
であるが、この損失抵抗を考慮したとしても、その差し
引きは正（プラス）となり大きな動力を得ることが出来
、産業上きわめて有益である。
本実施例では液槽での例を示しであるが河川、湖、海、
貯水池等で行えば更に好結果を生ずるものである。
又、本発明装置即ち開ロ受圧筒２ｔ−複数個並設して利
用すれば一層大きな動力を得ることができ更に第３図に
示す如く開口受圧筒２の筒底開口部分４ｔ−広くすｎば
より高効率となる。
第二の実施例（＠４図）について。
本実施例に於いて、第一の実施、、例と同じ部分には同
じ番号を附しである。
次に本実施例の具体例について説明する。
液槽１の深さＨｔｌｌ ５ｍに採っである。
開口受圧筒２の断面積５Ｆｉ１１／、高さｈは１１ｍに
して、核部２の頂部は液面りと同じにしである。即ち液
深１１ｍということである。
開口受圧筒２の筒底開口部分４の断面積は１♂である。
動力窄めし装置としての水中タービン５には発電機を内
蔵するが、あるいは回転軸６によってその回転力を動力
室７の動力機器８へ伝えてもよい。
開口受圧筒２の上部開口部１０と連通しであるＭ液槽９
は前記上部開口部１０より流入する液量よりも大容量で
ありて、その容積は状況によって決定される・
開口受圧筒２からの上昇液は上部開口部１０より溜液槽
９内へ流入し該溜液槽９内の液面Ｌ（水槽１の液面りと
同じ）Ｋまで達する。この上部開口部１０の断面積は１
１１１″に成しである。
揚液管１１は揚液ポンプ１２と直結してあって溜液槽９
中に１ｍの深さに沈下して、溜液槽９内の液体１に咳揚
液ポンプ１２により揚程するためのものであって、その
断面積Ｆｉ１．”である。
揚液ポンプ１２＃ｉモーター１３４Ｃより作動し、溜液
槽９内の液体を揚液して水槽１の液面しに放出するもの
である・
１４ＡＦｉ揚液ポンプ１２からの液体を液槽１内の液面
りへ放出するための放出管であって断面積は１１／以上
にしである。１５Ｆｉ液槽上端部、１６は溜液槽上端部
、１７は開口受圧筒２の筒底開口部分４管閉じるための
蓋で蝶番等により開閉自在になしてあり、万−何等かの
事故が生じて修繕するような場合に使用するものである
。
１８Ｆｉ液槽底部である。
４、図面の簡単な説明
第１図は本発明の詳細な説明図、第２図は第１図の要部
の横断平面図、第３図は開口受圧筒の筒底開口部分の他
の実施例の拡大図である。
第４図は本発明の第二の実施例である。
第５図は第４図の要部の横断平面図である。
１・・・液槽 ２・・・開口受圧筒４・・・筒底開
口部分 ５・・・動力窄めし装置８・・・動力機器
９・・・溜液槽１１・・・揚液管 １２・・
・揚液ボンプ手続補正書
昭和５７年９月２Ｂ日
特許庁長官 若 杉 和 夫 殿
Ｌ 事件の表示
昭和５５年特許願第１１５６５８号
２ 発明の名称
し方法
ａ 補正をする者
事件との関係 特許出願人
未代理人
東京都文京区本郷四丁目１２番１２号
２．４１Ｉ許請求の範囲
（１） 液体中に、筒底開口部分及び上部開口部を有
する開口受圧部を配設し、該上部開口部と溜液槽とを連
通すると共に開口受圧部の筒底開口部分より上昇する液
体を前記溜液槽内へ流入すべく成す一方、溜液槽の上端
を液体表面と同高若しくはその上方に位置せしめ、然る
後、液体を流入して溜液槽内の液位を液体表面と同一に
する他方、溜液槽内の液体を機器を用いて連続して揚液
し、前記液体表面へ放出して液体表ｍｔ常に一定の高さ
Ｋｍ持する一方、前記筒底開口部分附近の上昇液体中に
動力窄めしのための機器を配設した液体の常時強制循環
時に於いて発生する液体の流速と液深圧力とを利用した
動力申出し方法（零 体中に、筒底開口部分及び上部
開口部をいて発生する液体の流速と１［＃圧力とを利用
した動力申出し方法FIG. 1 is a detailed explanatory diagram of the present invention, and FIG. 2 is a cross-sectional plan view of the main part of FIG. 1. 1... Liquid tank 2... Open pressure receiving cylinder 4... Cylinder bottom opening part 5... Turbine 8... Power equipment 9... Reservoir well 11... Lifting pipe
12...Liquid Pump Figure 2 Specification IL Name of the Invention Method 2 for extracting power using the flow rate and deep pressure of liquid generated during constant forced circulation of liquid, Claims: , an open pressure receiving cylinder 2t having a cylinder bottom opening part 4 and a reservoir liquid tank 9 is disposed, and when the liquid is allowed to flow in from the cylinder bottom opening part 4 and passes through the upper opening part 10, the liquid in the cylinder flows into the liquid storage tank. The liquid No. 9 is forcibly pumped up by the device and becomes liquid II! The liquid has a flow rate of 11%, which is proportional to the amount of liquid released into the tank, and rises inside the pressure receiving cylinder 2 to reach the reservoir liquid tank 9, and is further discharged from the same layer liquid tank 9 to the liquid surface by a device to raise the liquid level. The liquid is constantly forced to circulate in order to maintain it at a constant height so that the liquid pressure does not change, while a device for extracting power is installed near the cylinder bottom opening part 4. Detailed Description of the Invention: A Powered Moth Removal Method Using the Liquid Flow Velocity and Liquid Deep Pressure Generated During Circulation The present invention is applicable to a liquid tank that maintains a constant water level, a river, a lake, the sea, etc.
This invention relates to a nine-power warming method in which an open pressure-receiving cylinder is provided in a reservoir or the like, and the liquid in the liquid tank is forcedly circulated to utilize the pressure at a deep position of the liquid. The contents will be explained below based on the illustrated embodiment. 1 is a liquid tank with a depth of approximately 12 m, and the liquid used includes, for example, water at 20°C, salt water, propylene glycol, and a propylene glycol aqueous solution (both of which are effective in cold regions). are doing. 2 is an open pressure receiving cylinder with a height of Ion installed in a liquid tank of 1 yen, the upper part of which is located 1 m below the liquid level, and a height of H.
I is provided on the support column 3 with a length of 1 m, and the open pressure receiving cylinder 2t
- It is designed to be fixed and stable, and to allow liquid to easily flow in from below. Cross-sectional area A#-1t of the cylinder bottom opening portion 4 of this open pressure receiving cylinder 2
It was taken by l♂. Reference numeral 5 denotes a turbine as the ninth device for extracting power, which is disposed in the bottom opening portion 4 of the cylinder. It is directly connected to the turbine 5 through a 6Fi rotating shaft, and has the function of transmitting the rotational force of the turbine 5 to the power equipment 8 of the power chain 7. Reference numeral 9 denotes a storage liquid tank provided at the upper part of the open pressure receiving cylinder 2, and has a cross-sectional area Mp of 1 m" and a height Hl of 3 m. 10 is the upper opening of the open pressure receiving cylinder 2, and 11 is a liquid lifting pipe with an open cutoff. Area m9 is 1m/,
The height Ha is 1 m. Reference numeral 12 denotes a liquid pump as a device for lifting liquid, which is operated by a transverse shaft upstreamer 13, lifts the liquid in the accumulated liquid tank 9, and discharges it into the liquid tank 1 from the discharge port 14. The liquid in the pressure receiving cylinder 2 and the reservoir tank 9 yen is forced to circulate.The depth between the liquid level and the bottom opening 4 of the open pressure receiving cylinder 2, that is, the liquid depth H4, is 11 m.Next The operation of the present invention having the above configuration will be explained below. First, the hydrostatic pressure in 20C water @H4-ttm of 1 yen liquid tank can be calculated from the general formula r-1·H (1). P - Density of water - 9981, 4th 6 - Acceleration of motion - 9 - 8 m/B" H - Depth of water! If H, w l l H then the pressure of static water P is P loP = -998 (Kf/j )X9.8(z//s
(m) Medium to7ss+('4/n, sl). Published by Japan Society of Mechanical Engineers (August 2, 1978)
P4 of the document entitled "(5) How to use S-engine in fluid engineering/fluid machinery" of Mechanical Engineering B Emmanual (published on 0)
If Tables 5 and 6 are used for unit conversion (!! SS S is partially omitted), P = 107584 (Pa) = 107.584 (KPa) Medium 107.6 (KPa). On the other hand, when the water depth is 114"-11B, the power or power generated at the bottom opening 4 of the open pressure receiving cylinder 2 can be determined from the general formula PaA-V- (2). Kp - Hydrostatic pressure - 107584 (h/mad”)
Mu = Cross-sectional area of cylinder bottom open part 4 = = 1 m 1 V; Velocity of fluid - 1% (If the liquid volume t-ed of the liquid pump 12 is 19 m, the flow rate will be 1 rrXs.) Then, the power or Power rate W casing P@A”V = 107584 (-・♂). When converting the unit using this power or power wt Table 5.6, W-107584 (W) = 107.584 (Kw) +107.58 (Kw). Therefore, assuming that the water pump 12 is operated to pump up 1 - 1 yen of liquid per second by operating the water pump 12, this amount of liquid will be pumped from the discharge port 14 of the liquid pump 12. By discharging onto the liquid level,
Open pressure receiving cylinder 2 The same liquid always flows through the open pressure receiving cylinder 2 at a constant flow rate.
, liquid pump 12. If the liquid circulates with the liquid tank 1 and the natural loss of liquid is ignored, a constant depth H4=11 m will be maintained. However, in reality, the liquid level fluctuates due to the equal pressure of evaporation, so it needs to be replenished, and the liquid tank 1 may be covered with a lid to prevent evaporation. A turbine 5 provided in the section 4 rotates at a constant speed and transmits its rotational force to a rotating shaft 6.
1 to the power equipment 8 in the power chamber 7. On the other hand, since the liquid pumping bonder 12Fi mentioned above lifts liquid at a rate of 111/sec through a pumping pipe with a height of 1 m and an opening cross-sectional area of pm l @'', the horizontal axis of the upstream tank 13 The power WM spent on pumping the liquid is WM” I's l-Ha e Apll v;P
・MP@v = 998 ('l/1)X9.8(m/81)X
1 (+111)
0.4 (ff@l//is). When this is converted into units using the table 6 described above, it becomes 948 (KW) in WM-9780, 4 (7). Therefore, the power that can be increased by the turbine 5, that is, the power WT, is WT (w - WM) 107.58 (聰) - 9,78 (KW) 7.8
(m) becomes. In short, the water flowing into the open pressure receiving cylinder 2 and stored in the reservoir tank 9 is pumped up by the liquid pump 12, and then the liquid level LK in the liquid tank 1 is released, keeping the liquid level constant. By doing so,
The force generated in the cylinder bottom opening part 4 can be made constant,
By extracting this force as work W by the turbine 5 and subtracting the work WM spent by the motor, a large power WT can be obtained. In addition, in the above calculation example, the loss resistance of each part is ignored, but even if this loss resistance t- is taken into account, the deduction will be positive and a large amount of power can be obtained, which is extremely useful for industry. It is. In this example, a liquid tank is shown as an example, but rivers, lakes, oceans, etc.
Even better results will be obtained if this is done in a reservoir, etc. In addition, the device of the present invention, that is, the open pressure receiving cylinder 2 @: If a plurality of open pressure receiving cylinders 2 are used in parallel, even greater power can be obtained. Higher efficiency. Brief description of the drawings Fig. 1 is a detailed explanatory diagram of the present invention, Fig. 2 is a cross-sectional plan view of the main part of Fig. 1, and Fig. 3 shows another embodiment and the bottom opening of the open pressure receiving cylinder. It is an enlarged view of a part. 1...Liquid tank 2...Open pressure receiving cylinder 4...Cylinder bottom opening part 11...Liquid lift pipe 12...Liquid pump No. 71.6, 1□1□81 Patent Haruki Shimada, Director-General of the Agency 1. Display of the case 2. Patent Application No. 115658 of 1980 3. Relationship with the case of the person making the amendment Patent applicant 4. Name of the agent's invention During constant forced circulation of liquid Claims (1) Claims (1) A power output method that utilizes the liquid flow velocity and liquid deep pressure generated during forced circulation. Power Constriction Method ■ Scope of Specific Requirements In the arrangement set forth in item 1, the present invention provides a power supply method that utilizes the liquid flow velocity and liquid deep pressure generated during constant forced circulation of a moving liquid. In a liquid tank that maintains the water level, or in a river, lake, sea,
This invention relates to a method of applying power by providing an open pressure receiving cylinder in a reservoir or the like, forcing the liquid in the tank to circulate, and utilizing the pressure at a deep position of the liquid. The contents will be explained below based on the illustrated embodiment. Regarding the first embodiment (FIGS. 1 and 2). A 1#'i liquid tank with a depth H of about 12 m is used, and examples of liquids include water at 20°C, salt water, propylene glycol and hypropylene glycol aqueous solutions (both of which are effective in cold regions). etc. are adopted. 2#'i It is an open pressure receiving cylinder with a height of 10 m installed in the liquid tank 1, and its upper part is located 1 m below the liquid level, and it is installed on a support 3 with a height of 1 m. , the open pressure receiving cylinder 2' was fixed to ensure stability, and it was reconsidered so that liquid could easily flow in from below. The cross-sectional area A of the bottom opening portion 4 of this open pressure receiving cylinder 2 is 1 m''
It was taken in This is an underwater turbine as a device for generating 5Fi power (power supply device f), and is arranged in the cylindrical bottom opening portion 4. Further, this underwater turbine 5 may be arranged at any position between the cylinder bottom opening portion 4 and the upper opening W610. 6 is a rotating shaft that is directly connected to the underwater turbine 5 and also has the function of transmitting the rotational force of the underwater turbine 5 to the power equipment 8 in the power chamber 7. Reference numeral 9 denotes a reservoir tank provided above the open pressure receiving cylinder 2, and has a cross-sectional area Ao of 111" or more and a height H! of 3 m or more. The volume of the reservoir tank 9 can of course be increased depending on the situation. The upper opening of the 10II'i opening pressure receiving cylinder 2 is a 11Fi liquid pumping pipe with an opening cross-sectional area Ap of 1-1'' and a height H3 of 1 m. As shown in Fig. 4, this liquid lift pipe 11t may be regarded as a part of the liquid pump.12Fi is a liquid pump as a device for lifting liquid, and is operated by a motor 13, The liquid is fried,
The liquid in the reservoir tank 9 is discharged from the discharge port 14 into the liquid tank 1 for forced circulation.The depth between the liquid level and the bottom opening 4 of the open pressure receiving cylinder 2, that is, the liquid depth. HaF111m. Next, the operation of the present invention having the slanted configuration will be explained. First, the hydrostatic pressure of water at 20° C. in the liquid tank 1 at a depth H4=11 tn can be obtained from the general formula ρ・f−H・−(1). Sincerely, ρ = Density of water = 998Et/, /l = Acceleration of gravity = 9.8m152 H = Water depth = Ha = 11m Then, the pressure P of static water is P = ρ・f@H P = 998(-Kgld)X9. 8 (m/SriX11-
Junior High School 1 () 7584 (Ke/m@3) Published by Japan Society of Mechanical Engineers (August 2, 1978)
P of the document called Mechanical Engineering SI Manual R (51! I How to Use SI in Body Engineering/Fluid Machinery J) of
If Tables 5 and 6 in 4 are used for unit conversion (Table 5.6 is a portion), P = 107584 (Pa) = 107.584 (KPa) medium 107.6 (KPa). On the other hand, water depth H4 = 11 m, that is, the power or power W generated at the bottom opening portion 4 of the open pressure receiving cylinder 2 can be determined from the general formula P-A@V...■ □ . 1KP = static water pressure = 107584 (h/m, S”)A
- Cross-sectional area of cylinder bottom opening 4 = 1 = 1'' ■ - Velocity of fluid =
1 呵 (If the amount of liquid pumped by the liquid pump 12 is 1-Hl,
The flow velocity is 1m/S.) Then, a force or a power w= p @A −V = 107584 (Kg/m11s) xlG4x1(
m/5) = 1 [17584 (K4-zen"/S3) 0 This power or power wt When the unit is converted according to Table 5.6, W = 107584 (W) = 107.584 (Kw) Medium 107.58 (Kw). Therefore, assuming that the water pump 12' is operated to pump up the liquid in the reservoir tank 9 at a rate of iII/second, this amount of liquid is calculated from the discharge port 14 of the liquid pump 12. By discharging the liquid closer to the liquid level, the liquid in the opening pressure receiving cylinder 2 circulates with the opening receiving pressure wI2, the liquid pump 12, and the liquid tank 1 at a constant flow rate, and the liquid 1
If natural loss is ignored, a constant depth (i4 = 11 m) will be maintained. In addition, in reality, the liquid level fluctuates depending on the steam supply, so it needs to be replenished, and the liquid tank 1rIC! may be used to prevent evaporation. On the other hand, a turbine 5rt installed in the bottom opening portion 4 of the open pressure receiving cylinder 2 rotates at a constant speed, and its rotational force is transmitted to the power machine 68 in the power chamber Z via the rotating shaft 6. It is also possible to install a rotating shaft 6Vc power generation *t- of the underwater turbine 5 and output 8m electric power. On the other hand, the above-mentioned pump 121j has a liquid height of 11 I/sec, H@= 1 m, and an opening cross-sectional area Ap= I III
Since zero liquid is being pumped through the liquid pumping pipe of
P#A, 6V =998(We/d>X9.8(m/S")x1h#X
1←)Xl (bow 4) = 9780.4 (It-+//Sj). Please note this as a reminder 5. ．． 5 K x unit conversion becomes WM = 9780.4v @ 69.78 (IC unit) Therefore, the power that can be offered by the turbine 5, that is, the power WTF
i WT=W-WM=10758 (ff) -9,78(KW)=97
．． 8 (KW). In short, the water flowing into the open pressure receiving cylinder 2 and stored in the reservoir tank 9 is pumped up by the pump 12, and then the liquid level LK in the liquid tank 1 is discharged to keep the liquid level constant. By maintaining this, the force generated at the cylinder bottom opening 41fC can be made constant, and this force is extracted as work W by the turbine 5, and by subtracting the work WM spent by the motor, a large power WT is obtained. can be obtained. In addition, in the calculation example above, the loss resistance of each part is ignored, but even if this loss resistance is taken into account, the deduction will be positive and a large amount of power can be obtained, which is extremely useful industrially. be. In this example, a liquid tank is shown as an example, but rivers, lakes, oceans, etc.
Even better results will be obtained if this is done in a reservoir, etc. Further, if the device of the present invention, that is, a plurality of open pressure receiving cylinders 2t, is used in parallel, even greater power can be obtained, and as shown in FIG. This results in higher efficiency. Regarding the second embodiment (@Figure 4). In this embodiment, the same parts as in the first embodiment are given the same numbers. Next, a specific example of this embodiment will be explained. The depth of the liquid tank 1 is 5 m. The cross-sectional area of the open pressure receiving cylinder 2 is 5Fi11/, the height h is 11 m, and the top of the core part 2 is at the same level as the liquid level. In other words, the liquid depth is 11 meters. The cross-sectional area of the bottom opening portion 4 of the open pressure receiving cylinder 2 is 1♂. The underwater turbine 5 serving as a power reducing device has a built-in generator, or its rotational force may be transmitted to the power equipment 8 in the power room 7 through the rotating shaft 6. M liquid tank 9 communicating with the upper opening 10 of the open pressure receiving cylinder 2
is larger than the amount of liquid flowing in from the upper opening 10, and its volume is determined depending on the situation.The rising liquid from the open pressure receiving cylinder 2 flows into the reservoir tank 9 from the upper opening 10. The liquid level L in the reservoir tank 9 reaches K (same as the liquid level in the water tank 1). The cross-sectional area of this upper opening 10 is 1
111". The liquid pump 11 is directly connected to the liquid pump 12 and is connected to the liquid tank 9.
The device sinks to a depth of 1 m in the liquid storage tank 9 and lifts the liquid 1 in the liquid storage tank 9 by the cough liquid pump 12, and has a cross-sectional area Fi1. Lift pump 12 #i It is operated by motor 134C to lift the liquid in the reservoir tank 9 and discharge it to the liquid level in the water tank 1. This is a discharge pipe for discharging to the liquid level in the liquid tank 1, and the cross-sectional area is 11/2 or more.15Fi is the upper end of the liquid tank, 16 is the upper end of the reservoir tank, and 17 is the cylinder of the open pressure receiving cylinder 2. The bottom opening part is a lid to close the 4 pipes, and it can be opened and closed freely with a hinge, etc., and is used in the event that an accident occurs and needs to be repaired. This is the bottom of the 18Fi liquid tank. 4. BRIEF DESCRIPTION OF THE DRAWINGS FIG. 1 is a detailed explanatory diagram of the present invention, FIG. 2 is a cross-sectional plan view of the main part of FIG. It is an enlarged view. Fig. 4 is a second embodiment of the present invention. Fig. 5 is a cross-sectional plan view of the main part of Fig. 4. 1... Liquid tank 2... Open pressure receiving cylinder 4... Cylinder bottom opening part 5... Power constriction device 8... Power equipment
9...Storing liquid tank 11...Liquid lift pipe 12...
・Liquid pump procedure amendment September 2B, 1980 Director of the Patent Office Kazuo Wakasugi L. Indication of the case 1982 Patent Application No. 115658 2 Method of naming the invention a Person making the amendment Relationship to the case Patent Applicant, not represented, 4-12-12 Hongo, Bunkyo-ku, Tokyo 2.41I Claims (1) An open pressure receiving part having a cylinder bottom opening and an upper opening is disposed in a liquid, and the upper The opening part communicates with the reservoir tank, and the liquid rising from the cylindrical bottom opening of the open pressure receiving part flows into the reservoir tank, while the upper end of the reservoir tank is at the same level as the liquid surface or above it. After that, liquid is flowed in to make the liquid level in the reservoir tank the same as the liquid surface, and the liquid in the reservoir tank is continuously pumped using a device to reach the liquid surface. While the liquid surface mt is always kept at a constant height Km, a device for power constriction is installed in the rising liquid near the opening at the bottom of the cylinder, and the liquid is generated during constant forced circulation of the liquid. Power application method using the flow velocity and liquid deep pressure (zero Power application method using the flow velocity of the liquid generated through the cylinder bottom opening and the top opening in the body and the 1 [# pressure)
Claims (1)
圧筒２を配置し排液ポンプ１２により前記筒底開口部分
４から液体を揚種し、上部開口部１０を経て溜液槽９な
一旦貯め次いで、該溜液槽９内の液体を再び揚液し液体
表面へ放出し液面を常に一定の高さに維持すぺ〈該液体
を強制循環せしめる一方、前記筒底開口部分４附近に動
力暖めしの九めの機器５を配設し液体の強制循環時に於
ける液深圧力を利用した動力戦出し方法An open pressure receiving cylinder 2 having a cylinder bottom opening 4 and a reservoir tank 9 is disposed in the liquid, and the liquid is pumped up from the cylinder bottom opening 4 by a drainage pump 12, and passes through the upper opening 10 to the reservoir tank. 9 is once stored, and then the liquid in the reservoir tank 9 is pumped up again and released onto the liquid surface to maintain the liquid level at a constant level. A method for generating power using the deep pressure of the liquid during forced circulation of the liquid by installing a ninth power heating device 5 near 4.
Priority Applications (1)
| Application Number | Priority Date | Filing Date | Title |
|---|---|---|---|
| JP55115658A JPS5832973A (en) | 1980-08-22 | 1980-08-22 | Method of taking out motive power by utilizing depth pressure of liquid during its forced circulation |
Applications Claiming Priority (1)
| Application Number | Priority Date | Filing Date | Title |
|---|---|---|---|
| JP55115658A JPS5832973A (en) | 1980-08-22 | 1980-08-22 | Method of taking out motive power by utilizing depth pressure of liquid during its forced circulation |
Publications (1)
| Publication Number | Publication Date |
|---|---|
| JPS5832973A true JPS5832973A (en) | 1983-02-26 |
Family
ID=14668090
Family Applications (1)
| Application Number | Title | Priority Date | Filing Date |
|---|---|---|---|
| JP55115658A Pending JPS5832973A (en) | 1980-08-22 | 1980-08-22 | Method of taking out motive power by utilizing depth pressure of liquid during its forced circulation |
Country Status (1)
| Country | Link |
|---|---|
| JP (1) | JPS5832973A (en) |
Cited By (3)
| Publication number | Priority date | Publication date | Assignee | Title |
|---|---|---|---|---|
| US5074746A (en) * | 1989-10-31 | 1991-12-24 | Kubota Corporation | Constant speed vertical pump with aeration |
| ES2060531A2 (en) * | 1992-11-20 | 1994-11-16 | Borda Juan Parellada | Hydroelectric generator with closed-circuit operation |
| JP2012237241A (en) * | 2011-05-11 | 2012-12-06 | Takeo Terauchi | Hydraulic power generating device |
-
1980
- 1980-08-22 JP JP55115658A patent/JPS5832973A/en active Pending
Cited By (3)
| Publication number | Priority date | Publication date | Assignee | Title |
|---|---|---|---|---|
| US5074746A (en) * | 1989-10-31 | 1991-12-24 | Kubota Corporation | Constant speed vertical pump with aeration |
| ES2060531A2 (en) * | 1992-11-20 | 1994-11-16 | Borda Juan Parellada | Hydroelectric generator with closed-circuit operation |
| JP2012237241A (en) * | 2011-05-11 | 2012-12-06 | Takeo Terauchi | Hydraulic power generating device |
Similar Documents
| Publication | Publication Date | Title |
|---|---|---|
| Anderson | Centrifugal pumps and allied machinery | |
| US5169595A (en) | Reactor core isolation cooling system | |
| JP2011501811A (en) | Reactor with improved cooling in the event of an accident | |
| JPH05501901A (en) | water flow energy converter | |
| KR20120050985A (en) | Low-drag hydro-pneumatic power cylinder and system | |
| FR2502417A1 (en) | ELECTRIC POWER PRODUCTION FACILITY | |
| US9330796B2 (en) | Stable startup system for a nuclear reactor | |
| JPS5832973A (en) | Method of taking out motive power by utilizing depth pressure of liquid during its forced circulation | |
| US20140348641A1 (en) | Low-drag hydro-pneumatic power cylinder and system | |
| US2105624A (en) | Water power | |
| US4257819A (en) | Method for flushing out a narrow gap | |
| CN111883270B (en) | Heat discharging system, pool type reactor and pool type reactor heat discharging method | |
| CN114593026A (en) | Cooling system of wind generating set | |
| JP6462872B2 (en) | Intake equipment for nuclear power plants | |
| JPS6244692A (en) | Decay heat removing device for nuclear reactor | |
| CN111447811A (en) | Circulating water cooling device | |
| CN217354787U (en) | Dry-wet dual-purpose sewage-discharging submersible electric pump | |
| JPH11351118A (en) | Generator using water located at low position without dropping from high position and using wind force, air pressure and water pressure in combination | |
| JPS5958167A (en) | Simple automatic power-generator applying specific gravity | |
| JPS5813172A (en) | Power-generation method by pumping-up of capillary phenomenon | |
| US2161158A (en) | Pressure loaded tank for hydraulic transmissions | |
| CN206709056U (en) | A kind of boiler condensation apparatus for recovering | |
| JP3214319U (en) | Water pressure power generation system | |
| JPS6363757B2 (en) | ||
| KR0180771B1 (en) | Vacuum degassification system and method for nuclear reactor system filling and venting |