Pneumatics

Author: Hero of Alexandria

The Siphon

Hero of Alexandria

1. Let ABC be a bent siphon, or tube, of which the leg AB is plunged into a vessel DE containing water. If the surface of the water is in FG, the leg of the siphon, AB, will be filled with water as high as the surface, that is, up to H, the portion HBC remaining full of air. If, then, we draw off the air by suction through the aperture C, the liquid also will follow from the impossibility, explained above, of a continuous vacuum.2 And, if the aperture C be level with the surface of the water, the siphon, though full, will not discharge the water, but will remain full: so that, although it is contrary to nature for water to rise, it has risen so as to fill the tube ABC; and the water will remain in equilibrium, like the beams of a balance, the portion HB being raised on high, and the portion BC suspended. But if the outer mouth of the siphon be lower than the surface FG, as at K, the water flows out; for the liquid in KB, being heavier,3 overpowers and draws toward it the liquid in BH. The discharge, however, continues only until the surface of the water is on a level with the mouth K, when, for the same reason as before, the efflux ceases. But if the outer mouth of the tube be lower than K, as at L, the discharge continues until the surface of the water reaches the mouth A. If then we wish all the water in the vessel to be drawn out, we must depress the siphon so far that the mouth d may reach the bottom of the vessel, leaving only a passage for the water.

2. Now some writers have given the above explanation of the action of the siphon, saying that the longer leg, holding more, attracts the shorter. But that such an explanation is incorrect, and that he who believes so would be greatly mistaken if he were to attempt to raise water from a lower level, we may prove as follows. Let there be a siphon with its inner leg longer and narrow, and the outer much less in length but broader so as to contain more water than the longer leg. Then, having first filled the siphon with water, plunge the longer leg into a vessel of water or a well. Now, if we allow the water to flow, the outer leg, containing more than the inner, should draw the water out of the longer leg, which will at the same time draw up the water in the well; and the discharge having begun will exhaust all the water or continue forever, since the liquid without is more than that within. But this is not found to be the case; and therefore the alleged cause is not the true one. Let us then examine into the natural cause. The surface of every liquid body, when at rest, is spherical and concentric with that of the earth; and, if the liquid be not at rest, it moves until it attains such a surface.1 If then we take two vessels and pour water into each, and, after filling the siphon and closing its extremities with the fingers, insert one leg into one vessel plunging it beneath the water, and the other into the other, all the water will be continuous, for each of the liquids in the vessels communicates with that in the siphon. If, then, the surfaces of the liquids in the vessels were at the same level before, they will both remain at rest when the siphon is plunged in. But if they were not, as soon as the water is continuous it must inevitably flow into the lower vessel through the channel of communication, until either all the water in both vessels stands at the same height, or one of the vessels is emptied. Suppose that the liquids stand at the same height; they will of course be at rest, so that the liquid in the siphon will also be at rest. If, then, the siphon be conceived to be intersected by a plane in the surface of the liquids in the vessels, even now the liquid in the siphon will be at rest, and, if raised without being inclined to either side, it will again be at rest, and that, whether the siphon is of equal breadth throughout or one leg is much larger than the other. For the reason why the liquid remained at rest did not lie in this, but in the fact that the apertures of the siphon were at the same level. The question now arises why, when the siphon is raised, the water is not borne down by its own weight, having beneath it air which is lighter than itself. The answer is that a continuous void cannot exist;2 so that, if the water is to descend, we must first fill the upper part of the siphon, into which no air can possibly force its way. But if we pierce a hole in the upper part of the siphon, the water will immediately be rent in sunder, the air having found a passage. Before the hole is bored, the liquid in the siphon, resting on the air beneath, tends to drive it away, but the air having no means of escape does not allow the water to pass out: when however the air has obtained a passage through the hole, being unable to sustain the pressure of the water, it escapes. It is from the same cause that, by means of a siphon, we can suck wine upwards, though this is contrary to the nature of a liquid; for, when we have received into the body the air which was in the siphon, we become fuller than before, and a pressure is exerted on the air contiguous to us, and this in turn presses on the atmosphere at large, until a void has been produced at the surface of the wine, and then the wine undergoing pressure itself will pass into the exhausted space of the siphon;1 for there is no other place into which it can escape from the pressure. It is from this cause that its unnatural upward movement arises. . . .

[A Siphon of Uniform Flow]

4. It is evident from what has been proved above that as long as the siphon is stationary the stream through it will be of irregular velocity, for the result is the same as in a discharge through a hole pierced in the bottom of a vessel, where the stream is irregular from the pressure of a greater weight on the discharge at its commencement, and, of a less, as the contents of the vessel are reduced. In like manner, in proportion as the excess of the outer leg of the siphon is greater, the velocity of the stream is greater; for a greater pressure is exerted on the discharge than when the projection of the outer leg below the surface of the water in the vessel is less. Therefore we have said that the discharge through the siphon is always of variable velocity. But we must contrive a siphon in which the velocity of the discharge shall be uniform.

Let there be a vessel, AB, containing water, on which a small basin, CD, floats, having its mouth covered with the lid CD. Through this lid and the bottom of the basin insert one leg of the siphon soldering it into the holes with tin. Let the other leg be outside the vessel AB, having its mouth lower than the surface of the water in AB. If we draw the air in the siphon through the outer extremity, the water will at once follow because of the impossibility of a continuous vacuum in the siphon; and the siphon, having begun to flow, flows on until it has exhausted all the water in the vessel: but the discharge will be uniform, since the projection of the outer leg below the surface of the water does not vary; for, as the vessel becomes empty, the basin sinks with the siphon. The greater the excess of the outer leg the greater will be the velocity of the discharge, yet still uniform.1 In the figure, EFG is the siphon described, and the surface of the water is in the line HK.

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Chicago: Hero of Alexandria, Pneumatics in A Source Book in Greek Science, ed. Morris R. Cohen and I. E. Drabkin (Cambridge, MA: Harvard University Press, 1966), 242–245. Original Sources, accessed May 18, 2024, http://www.originalsources.com/Document.aspx?DocID=1DQKHPJNUVMFTIP.

MLA: Hero of Alexandria. Pneumatics, in A Source Book in Greek Science, edited by Morris R. Cohen and I. E. Drabkin, Cambridge, MA, Harvard University Press, 1966, pp. 242–245. Original Sources. 18 May. 2024. http://www.originalsources.com/Document.aspx?DocID=1DQKHPJNUVMFTIP.

Harvard: Hero of Alexandria, Pneumatics. cited in 1966, A Source Book in Greek Science, ed. , Harvard University Press, Cambridge, MA, pp.242–245. Original Sources, retrieved 18 May 2024, from http://www.originalsources.com/Document.aspx?DocID=1DQKHPJNUVMFTIP.