自由能源·德国KPP自由能浮力发电机原理（下）
博客：http://blog.sina.com.cn/u/2940475465
《Amethod of extraction (restitution) of the energy supply stored inliquid or gaseous mediums and transformation of the energy obtainedinto mechanical work》
《一种提取或还原存储在液态或气态介质中的能量，并将其转变成机械能量的方法》
刘中凯 译
Let us calculate a more powerful energy installation able to powera small urban village, military unit, a vessel etc. A 2VM10 - 63/9piston-type compressor with the following technicalspecifica-
tions will be used as the source of compressed air:
- Compressor output ——1.04 m3/sec
- Output pressure, MPa ——0.9 (9 Atm)
Compressor shaft power ——332 kW
Water cooling
现在让我们来分析一个更加强大的发电装置，该装置可以给一个村庄，一支部队，一辆汽车等等提供电力。一台型号为2VM10- 63/9的具有下述技术性能的活塞式压缩机，将用于提供压缩空气：
压缩机输出——1.04m3/sec
输出压力MPa——0.9(9Atm)
压缩机轴输出——332kW
水冷
A calculation will be carried out for an installation with a headof water column equaling 5m and with 10 working wheels installedinside at a distance of 500mm from each other. The capacity of thecompressor motor needed to feed air under a 5-meter water column,taking into consideration the atmospheric pressure, is:
5 m×(332kW/ 100 m) =16.6 kW
The installation capacity will total:
N=9.81 1.04 m3/sec×15m×10×1.2×0.9=1652KW
We obtained output energy exceeding the input energy by a factor of99.
被分析的装置的水柱高度是5米，里面安装了10组叶轮，每个叶轮的间距是500mm。考虑到大气压，将空气由5米高水柱下面泵入的压缩机电机所需要的功率是：
5 m×(332kW/ 100 m) =16.6 kW
该装置的总输出功率是：
N=9.81 1.04 m3/sec×15m×10×1.2×0.9=1652KW
这样，我们获得的输出能量是输入能量的99倍。
Thus, the obtaining of any amount of energy is possible,accompanied by an improvement in the gaseous water composition bymeans of an environmentally friendly method. This method impliesthe use of an inexhaustible energy source, when a naturalnon-equilibrium of water and air is used in any climatic zone.There is no need anymore to build expensive dams and sluices, whichleads to flooding of valuable agricultural lands.
这样，通过环保的手段，再加上对气-水构成的改进，获得任意数量的能量都是有可能的。当在任意气候条件下使用水和空气的一种自然状态的非平衡时，这种方法蕴含着一种永不耗竭的能量的利用。再不需要建造昂贵的大坝和水闸了，这种东西导致珍贵的耕地的被淹。
Calculation of an energy extracting pneumohydraulicengine
Source of compressed air ——VP2 10/9 piston-typecompressor.
Compressor output ——0.167 m3/sec
Output pressure, MPa——0.9 (9 Atm).
Compressor shaft capacity ——56.5 kW
Water cooling
下面来分析（另）一种提取能量的气动涡轮
压缩空气源——VP2-10/9活塞式压缩机
压缩机输出——0.167m3/sec
输出压力,MPa——0.9 (9 Atm).
压缩机轴输出能力——56.5kW
水冷
The efficiency of a pneumohydraulic engine will be evaluated bycomparing the power supplied and the power obtained, i.e. theamount of work per second. The compressor output is the volume ofair on the compressor's input, ie. the volume of air at atmosphericpressure. Then the value of 0. 167 m3/sec is the volumeof air on the compressor input and on the exit from the upper floatof the pneumohydraulic engine (Fig. 3).
气动涡轮的效率，是通过输入的功率和获得的功率的对比来计算的。
压缩机的输出，是指它输入的一个大气压下的空气量。因此，0.167 m3/sec这个数值，指的是压缩机的输入空气，和气动涡轮上部浮筒排出的空气（见图3）。
图3
Floats are released from the air and then filled with water at alevel that is situated below the level of water in the engine case.At an air pressure of 9 Atm it maybe fed under a water column witha head of 90m . If the air bubbles rise at a speed of 0.4 m/s itwill take 225 sec for a bubble to reach the surface. Moving airwill be present at all levels of the water column. This figure of0.4 m/ s is obtained during experimental evaluations.
浮筒中的空气释放掉之后，又在涡轮机外壳内的水面下充满水。当空气压力是9Atm时，它可以从90米深的水柱下被泵入。如果空气泡上升速度是0.4m/s，气泡要达到水面就需要225秒的时间。运动的空气会在水柱的所有层面出现，0.4m/ s这个数值是在实验评估中获得的。
If the water column and compressor output remain stable, anincrease or decrease in speed at which the bubbles rise, resultsonly in a change in the horizontal dimensions of the floats (theirlength and width), since it is the air volume that increases ordecreases. This, in turn ,only increases or decreases the force,not influencing the capacity of the pneumohydraulicengine.
如果水柱和压缩机的输出能保持稳定，气泡上升速度的改变——快或慢，只能由于浮筒水平尺度（长或宽）的改变而产生，原因是浮筒内的气量增加或减少了。这种改变，反过来只能改变力的大小，不会影响气动涡轮的功率。
The possibility to change the horizontal dimensions of the floatsallows making floats of a needed volume preserving the watercolumn. The volume of air on the output of the compressor'spressure tube at a depth of 90m will total (taking intoconsideration the atmospheric pressure) :
0.167 (m3/sec) / 10 Atm = 0.0167m3/sec
改变浮筒水平尺寸的可能性，可以使我们将浮筒制成合适的尺寸来存留水。90米深处压缩机压力管出口处的空气量（要考虑到大气压）是：0.167(m3/sec) / 10 Atm = 0.0167 m3/sec
since the pressure of a 10-meter water column will equal 1Atm anddue to the fact that an increase in the volume of air by the valueof the initial volume takes place every 10 meters the air rises. Ifthe air volume remained permanent, at the moment of reaching thesurface its volume would be:
0.0167 (m3/sec)×225sec = 3.757 m3
因为10米水柱的压力等于1Atm，另一个原因是空气的体积每上升10米，它的体积和最初的体积相比都要变大一些。如果空气体积始终保持不变，那么在空气到达水面时，它的体积是：0.0167(m3/sec)×225sec = 3.757 m3
Taking into consideration the volume of air at the moment itreaches the surface, its total volume will amount to:
3.757 m3×10Atm = 37.57 m3
Taking into consideration the coefficient of thermal expansion, itsvolume will total:
37.57 m3×1.2= 45.084 m3
将这个到达水面的空气体积（假设它的体积没变）代入，空气的总的体积则是：
3.757 m3×10Atm = 37.57 m3
将热膨胀系数代入，空气的总体积将是：
37.57 m3×1.2= 45.084 m3
The buoyancy force of a 1m3 of air equals 1000 kgf. The amountof work performed by this volume of air as it rises will amountto:
45084 kgf×0.4m/sec =18 033 kgf×m/sec
or 18033 kg×fm/sec
Since 1kgf = 9.81 W, the result of recalculation is thefollowing:
18033 kg×fm/sec×9.81=176903.73 W or 176.9kW
1m3空气的浮力是1000kgf，那么这些空气上浮时所做的功是：
45084 kgf×0.4m/sec =18033 kgf×m/sec
或是：18033kgf m/sec
因为1kgf= 9.81 W，（1kfg*m/S= 9.81w），重新计算的结果是：
18033 kgf m/sec×9.81=176903.73W 或 176.9kW
By adding no less than 30% of the energy that is returned, obtaineddue to the reactive force that is created during the filling of afloat with air, to the energy that is received, we get:
176.9 kW +18 kW = 194 kW
We obtained the output energy exceeding the input energy by afactor of 3.4.
因为在对浮筒充气的过程中出现一种反作用力，这个力至少使浮筒做的功增加30%，这样得到的总功率就是：
176.9 kW +18 kW = 194 kW
这样，我们获得的输出能量，就是输入能量的3.4倍。
The mechanical energy efficiency of a pneumohydraulic engine willbe rather high since during operation the engine is well lubricatedby water, while the floats are mutually balanced. The energyefficiency of the compressor is taken into account duringconsideration of the compressors engine capacity.
气动引擎的效率是较高的，原因是机器在运转过程中有水做充分的润滑，而浮筒相互间都是平衡的。在考虑压缩机的功率时，要考虑到它的效率。
The pneumohydraulic engine is equipped with a brake that makes itstop during operation. When the engine stops, air is still presentin the floats, which means that no energy will be consumed on thenext start-up since the engine will be put in operation by the airleft in the floats.
这种气动引擎安装了刹车，这样可以使它在运行中停下来。当引擎停住后，空气仍然存留在浮筒中，这意味着，再次启动引擎不需耗费任何能量，因为浮筒中存留的空气可以带动引擎运转。
In our calculations, we proceeded from parameters of a seriallyproduced compressor, able to feed air under a water column with ahead of 90m. This is a way to increase the effectiveness ofhydroelectric stations by means of installing pnerunohydraulicengines in pontoons at water——storageponds. Increasing the effectiveness of hydroelectric stations byusing tail ponds is considered in the description of the patentsclaim in Germany.
通过对系列生产的，可以在90米深的水下泵气的压缩机的各种参数进行的仔细分析可知，通过在存储池水面的浮船上安装气动引擎，是一种提高水电站效能的方法。用尾塘提高水电站的效能这种方法，已被写入在德国申请的专利中。
The design of the pneumohydraulic engine is remarkable for its lowsteel intensity, thus makmg it very light. Any river, pond, spring,thermal source or cooling tower may become a source of energy. Aleveling of the water temperature at hydroelectric stations willbecome possible by means of blending lower, more warm water sheetsand cold upper water sheets. The process will be accompanied by asimultaneous extraction of heat from the water.
用钢量少，是气动引擎的一大特色，这使它变得非常轻巧。任何河流、池塘、泉水、热源或冷却水，都可以成为能量的来源。通过混合水面下层温暖的水和上层较冷的水这种方法，可以使水电站的整体水温持平。这个过程，同时还会伴随着从水中吸热的过程。
The most important point is that there will be no need to economizeenergy, since we do not amplify the natural energy imbalance byusing a natural non-equilibrium. On the contrary, we restore it bygetting rid of the consequences of thermal pollution. As for thesolar energy, we do not spend more of it than we obtain.
最重要的一点，是你再也不需要节约什么能源了，因为我们并没有用破坏自然界平衡的方法，放大自然界能量的失衡。相反，我们通过去除热污染的种种恶果，重新恢复了自然界能量的平衡。至于说到太阳能，只不过是所得即所用而已，它不会增值能量。
We considered a method of obtaining energy in industrialconditions, but there is a great need for energy installations witha wattage of 3-4 kW. Let us try to estimate their sizes. Take aninstallation with a head of water column equaling 2m. Using thesame type of compressor (only for calculation) we may find out thecapacity of the compressor engine needed to feed air under a 2meter water column:
N=(2m×56.5kW)/(90m+10m)=1.131 Kw
The compressor output ——0.167m3/sec
考虑到工业条件下获得能量的方法，那就需要大量的3-4kW的发电装置了。让我们估算一下它们的尺寸。如果一个装置的水柱高度是2米，用同样牌子的压缩机（只是为了计算），可以算出向2米深水柱下泵气的压缩机电机所需要的功率是：
N=(2m×56.5kW)/(90m+10m)=1.131 Kw
压缩机的输出是：0.167m3/sec
A 2-meter water column creates pressure equaling 0.2 Atm. Then thewater volume at a depth of 2m will amount to (taking intoconsideration the atmospheric pressure) :
0.167 (m3/sec) / 1.2 Atm= 0.139m3/sec
The time needed for a bubble to rise equals:
2 m/ 0.4 (m/sec) = 5 sec
2米深的水柱产生的压力等于0.2Atm，那么，考虑到大气压，2米深的水柱是：0.167(m3/sec) / 1.2 Atm= 0.139 m3/sec
气泡上升所需时间是：2m/ 0.4 (m/sec) = 5 sec
The volume of moving air that will be present in the floats of apneurnohydraulic engine in 5 seconds (taking into account theincrease in volume as the air rises and the thermal expansioncoeffi-
cient) will total:
0.139 (m3/sec) -5 sec×1.2Atm×l.2= 1 m3
5秒内进入气动引擎浮筒的移动空气总量（考虑到空气上升后的膨胀和热膨胀系数）是：
0.139 (m3/sec) -5 sec×1.2Atm×l.2= 1m3
The amount of work performed will amount to:
1000 kgf×0.4m/sec = 400 kgf m/sec
The amount of work per second equals the power. Since 1kgf = 9.81 W, thecapacity will be:
N = 9.81W×400= 3924 W = 3.924 kW
By adding 30% of the power returned, we get:
3.924 kW + 0.34 kW = 4.263 kW
空气所做的功是：
1000 kgf×0.4m/sec = 400 kgf m/sec
每秒做的功等于功率，因为1kgf= 9.81 W，所以功率是：
N = 9.81W×400= 3924 W = 3.924 kW
加上返回的30%的能量，我们得到：
3.924 kW + 0.34 kW = 4.263 kW
If the mechanical energy efficiency equals 0.9, we get thefollowing capacity:
N = 4.263 kW×0.9= 3.84 kW
We obtained output energy exceeding the input energy by a factor of3.4: 3.84 kW / 1.13 kW = 3.4
如果机械效率等于0.9，我们得到下面的功率：
N = 4.263 kW×0.9= 3.84 kW
于是，我们得到的输出能量是输入能量的3.4倍：
3.84 kW / 1.13 kW = 3.4
In order to once again make sure of the effectiveness of theproposed method of obtaining energy, let us compare itseffectiveness with that of a storage plant, in which water ispumped to a high-level storage pond by means of a pump or areversible hydroset and then used at a lower level in a turbine. Inthis case, if the energy efficiency factor amounts to100%，itmeans that we obtained an amount of energy that equals the amountof the energy consumed. Let us calculate the capacity of a pumpengine needed to lift water to the level of 90m, the output ofwhich is 0.167 m3/sec:
N=(9.81 x 0.167m3/sec×90m)/0.75 = 196.5 kw
为了进一步确认这种推荐的获取能量的方法的有效性，我们可以将之与一座蓄能式水电站的效能做一对比，在蓄能水电站中，水通过一个水泵或一台可逆的水电（或流体）装置，被泵到一个高处的蓄水池中，然后用于下面的一台水轮机。在这种情况下，如果能量效率达到100%，意味着我们得到的能量等于消耗的能量。让我们计算一台泵的功率，泵的输出是0.167m3/sec，泵水高度是90米，它的功率是：
N=(9.81×0.167m3/sec×90m)/0.75 = 196.5 kw
Let us compare the power obtained by a pump engine to that obtainedby a compressor engine with a capacity of 56.5 kW and air output of0.167m3/sec. The latter can displace the same amount ofwater, lifting it to a level of 90m and feeding it to a turbine. Anamount of power equaling 196.5 kW is obtained, which means that 3.5less energy is spent.
我们将这台泵获得的能量，与一台额定功率56.5kW、气体输出也是0.167m3/sec的压缩机获得的能量做一个对比，就可以看到，后者可以将同样多的水提升到90米的高度，并将之用于驱动一台涡轮。这样可以得到相当于196.5KW的功率，而付出的能量却很少，只是这台泵的3.5分之一。
Besides, the moving air that remains throughout the head of thewater column will also perform work, which is confirmedby the aforecited calculation. The possibilities of implementationof the proposed method are reflected in the diagram shown below(Fig2a、2b)
It can be seen from this diagram that the buoyancy force manifestsitself starting from the volume V0. The cross-hatched part of thepicture is a water column H, to overcome which the energy generatedby the compressor is spent. V0 is the volume of water at a depth ofH; Vk is the volume of air, expanded due to the fall in pressure asair rises ,Vq is the active air volume.
另外，存留在整个水柱中移动的空气，仍然在做功，这一点已为前面的计算所确认。这种推荐的方法的可行性反映在图2a、2b中。
在这张图中可以看到，浮力从水柱 V0处便出现了，图中交叉影线部分是水柱H，为克服该水柱的压力，压缩机以它产生的能量为代价。V0是水柱在深度为H的位置。Vk是空气体积，当它上升时，由于压力下降而膨胀，Vq是源的或实际的空气体积。
图2a、2b
Thediagram shows that the volume of active air in a pneumohydraulicengine equals Vq, while the volume Vk is essential for apneumohydraulic turbine, since it operates on a displaced volume ofwater. This fact explains the difference in theireffectiveness.
这张简图显示，在一台气动引擎中的源的或实际的空气体积等于Vq，而空气体积Vk，对一台气动引擎来说是至关重要的，因为它运行在一个移动的水柱之中。这个现象，正解释了它们效能上的差异。
The inexhaustibility of the energy source, its absoluteenvironmental friendliness, the ease of production and quickpayback due to the ever-growing need for energy provide foreffective marketing of the proposed construction, while thediversity of designs provide for a wide area ofapplication.
由于永远增长的对能源的需求，这种永不会耗竭的能源，它的绝对的环境亲和性，和易于生产及快速的回报，给它的发展提供了实实在在的市场，而多样化的设计更给它提供了广阔的应用前景。
原文链接：
http://www.lenr.com.cn/index.php?m=content&c=index&a=show&catid=13&id=50
有关该项技术的更多的参考资料请见我的博文：
《后石油时代与明天的能源战国·你准备好了吗？》（二）
【译后寄语】
1. 首先，希望本文对开阔国内自由能源爱好者、研究者的视野、激发他们的想象力、创造力和实践的勇气，能有所裨益。
2. 至于文章内容，不要迷信，用能量守恒的理念、观点、思路解释永动机，永远不过是一种徒劳之举，如果能解释清楚，爱因斯坦早就解释清楚了，也就不用劳烦各位了。
3. 文中使用的计算公式在内在深层的逻辑上是错误的、荒谬的，甚至是荒唐的，纯粹是一种拼凑的产物，其目的只有一个：就是使计算结果“符合”或“凑合”实验观察所得出的数据（令人讶异的是，这一点他们竟然做到了！当然，肯定也和爱因斯坦的所有公式一样，只是做到了“近似”，或者更准确的说，是“形似神不似”）。完全是一种实用主义的做法，也许只是一种工程设计上的需要，（如果纯粹是为了工程设计及生产上的需要，而发明、拼凑这样一套计算公式，倒也无可厚非）。
4. 这篇文章的做法也使人不禁想到，在正桶派的那些海量的、繁琐的计算、公式，和无穷无尽的推导中，到底有多少是这样拼凑和发明出来的东西？
5. 有关这种“反能量守恒”的装置的深层机理问题，可以阅读我的有关著作和文章。
6. 这个KPP，确实是个好东西，是我们目前所知道的最简单有效的永动机之中的一种，值得大力推广，特别是在我国广大的农村。可以肯定地说，磁能自由能全面普及和发展的千载一遇的时机已迫在眉睫，好男儿应志在四方，希望他们能像老一辈的革命家那样，踊跃投身到宣传、普及、推广自由能源的伟大革命洪流当中，为国家的改革和人类的进步做出自己应有的贡献。
对本博客研究探讨的内容，你看后如果有兴趣，欢迎到磁能时代QQ群:468704593，加入我们，继续我们的梦想之旅。
祝各位愉快！
刘中凯
2016.10.9于北京