The Science Notebook
Gilbert Hydraulics - Part 5

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NOTE:  This book was published in 1920, and while many of the experiments and activitied here may be safely done as written, a few of them may not be considered particularly safe today.  If you try anything here, please understand that you do so at your own risk.  See our Terms of Use.

 Pages 101-125



Any liquid can be used in a barometer but liquids lighter than mercury require longer tubes. This is true of the water barometer. Mercury is 13.6 times as heavy as water and since the atmosphere supports a column of mercury 30 in. high it will support a column of water 13.6 x 30 = 408 in. high, that is, a column 408/12 = 34 feet high.

Otto von Guericke, the inventor of the Magdeburg hemispheres, made a water barometer in 1650, and had it so arranged that the top of the tube stuck up through the roof of his house. He had a small wooden figure floating on the water in the tube and in fine weather, when the water column rose, the figure rose above the roof, but in bad weather the figure retired from sight. This frightened and mystified his neighbors very much and they accused him of being in league with the evil one.


To show that the vertical height to which the atmosphere will lift water in a tube
is independent of the length or slant of the tube.

Make the experiments (1), (2) and (3), Fig. 134. Suck air out through the upper coupling on the tee and close the clip.

Is the vertical height of the water in one tube above the water in the tumbler always the same as that in the other?

Make experiments of your own.

To show that the height to which the atmosphere will lift water in a tube is independent
of the size or shape of the tube and of the water surface outside the tube.

Make the experiments (1), (2), (3) and (4) Fig. 135. Is the height of the water always the same in the two tubes?

Make experiments of your own.


To show that the atmosphere lifts heavy salt water to a less height,
and light gasoline to a greater height, than it lifts fresh water.

Make the experiments illustrated in Fig. 136.

To show that the atmosphere will lift weights.

Make the experiments illustrated in Fig. 137.


To show that the atmosphere will lift 15 lbs. per sq. in. but no more.


The plungers have an area of 3/10 sq. in. If then, the atmosphere will lift 15 lbs. on 1 sq. in., it will lift 3/10 x 15 = 4 1/ 2 lbs. on 3/10 sq. in.

Soap the plungers, have water between them but no air, pour an inch of water above the upper plunger to make it air-tight, attach a pail weighing less than 4 1/2 lbs. to the lower plunger, Fig. 138 and raise the upper plunger. Does the atmosphere lift the lower plunger and weight?

Add water to the pail until the total weight is 4 1/2 lbs. and raise the upper plunger. Do you find that the atmosphere does not lift the lower plunger? It does not do so because the atmospheric pressure on 3/10 sq. in. cannot lift 4 1/2 lbs. and also overcome the friction.

Hold the upper plunger and lift the tube. Does the atmosphere now lift 4 1/2 lbs. weight? It does so because the friction helps it in this case.


Repeat with the water and pail weighing 6 lbs. Do you find that the atmospheric pressure on 3/10 sq. in. will not lift 6 lbs. even with the help of the friction.

You have shown here roughly that the atmospheric pressure on 3/10 sq. in. will lift 4 1/2 lbs. but no more. This shows that the atmospheric pressure on 1 sq. in. will lift 4.5 x 10/3 = 15 lbs. but no more.

Make your own experiments.



The air-lift pump, Fig. 139, is operated by compressed air. It consists of two pipes one inside the other, both open at the bottom and without valves.  The pump is at least half-submerged, that is, the bottom is at least as far below the surface of the water in the well as the top is above it.

The air which is compressed in the storage tank passes into the outer pipe of the pump, forces the water down to the bottom of the inner pipe, and forces the water in the inner pipe up into the tank. After the first lot of water has been forced out of the inner pipe the pump settles down to its regular operation which is as follows. Compressed air from the outer pipe enters the inner pipe, the pressure in the outer pipe is thereby lowered and the water rises in the outer pipe above the bottom of the inner pipe, more compressed air comes from the tank and forces the water down in the outer pipe but up in the inner pipe. This operation takes place over and over again rapidly, and alternate layers of air and water are forced up the inner pipe as shown in Fig. 139. The water thus flows from the inner pipe into the tank in spurts as you will show in your next experiment.

Another form of air-lift pump is illustrated in Fig. 140. Here the air enters through the inner pipe and the mixture of water and air is


forced out through the outer pipe. The water comes out as a continuous heavy spray because the air is mixed with the water in bubbles rather than in layers.

These are called air-lift pumps but the water is not raised by the air pressure. It is raised by the weight of the water in the well outside the pump, because the water rising in the pump is really a mixture of air and water and is lighter than a water column of the same height.

You can illustrate this by means of experiments shown in Fig. 141. In (1) both sides of the U tube are filled with water and you know from your experiments that the water will be at the same level in both sides. In (II) one side is filled with kerosene oil which is only 8/10 as heavy as water, and you know that a column of water 8 in. high will support a column of oil 10 in high. Similarly in (III) a depth of water of 8 inches will support a column of oil 10 inches high. If now the oil in (III) were replaced by a mixture of air and water which was only 1/2 as heavy as water, you can see that the 8 inch depth of water would support a column of the mixture 16 inches high, and so on.

The bottom of the air-lift pump is always placed at least as far below the sur-


face of the water as the top is above, and the water outside the pump lifts the lighter mixture of air and water to the top. You will illustrate this in the next experiment.

To make and operate two air-lift pumps.

Make an air-lift pump. (1) Fig. 142. Use a quart sealer to represent the well, fill it to the top with water, and insert the air-lift pump until it is half submerged, that is, until the water in the sealer is at a point half way between the bottom of the wide tube and the top of the elbow of the discharge pipe.

Force air in through the hose and observe what takes place near the bottom of the pump.

Do you observe that the water level in the pump moves alternately down below the end of the discharge pipe and then up above it, and that alternately water and air are forced up the discharge pipe?

Do you observe further that when you force air in at just the right rate the pump works steadily and the water comes up the discharge pipe in


spurts at regular intervals.

In the other type of air-lift pump the compressed air passes down the inside pipe and the mixed air and water move up the other pipe.

Make a pump of this kind, (2) Fig. 142 and blow air in through the inside pipe.

Do you find that air and water are forced up over the top of the outside pipe?

Repeat the experiment with the pump deeper in the water.

Do you find that it works better the deeper it is in the water?


In the remaining pages of this book you will study three laws which apply to gases and you will illustrate many practical applications of these laws. They are Pascal's law, Archimedes' law, and Boyle's law. You will begin with Pascal's law.

You learned on pages 49, 50 and 51, Pascal's law which states one property of liquids; namely, pressure exerted on a liquid is transmitted by the liquid equally and undiminished in all directions. This law also states a property of gases as follows: pressure applied to a gas is transmitted by the gas equally and undiminished in all directions.

You are very familiar with one application of this law, namely in the pneumatic tire. The air in a bicycle or automobile tire exerts pressure outward equally at every part of the tire.

To illustrate Pascal's law as it applies to gases.

Shove the plunger in (1) Fig. 143, down, and feel the air at the nozzles. Are the pressures equal?

Blow a soap bubble (2). Is it a perfect sphere? This shows that the air exerts pressure equally in all directions against the inside of the bubble.

Make a three legged siphon filled with air (3), place two legs in tumblers of water, place the third leg in the wide tube partly filled with water, and raise and lower the wide tube.


The water in the wide tube exerts pressure on the air in the third leg. Is this pressure exerted equally and undiminUhed by the air, that is, is the water level in the three legs always at the same distance below the water outside?

Repeat this with the apparatus (4). Is the result the same?

You have here proved Pascal's law, namely that a gas transmits pressure equally and undiminished in all directions.



The Law of Archimedes applied to Gases

Balloons float in air and this fact is due to a property of air which is expressed by the law of Archimedes.

You have already made experiments on this law with liquids and you have shown that the buoyant force of a liquid on a body is equal to the weight of the liquid displaced by the body. This is the law of Archimedes as it applies to liquids.

The law of Archimedes in regard to gases is: the buoyant force of a gas on a body is equal to the weight of the gas displaced by the body.

How the Total Lift of a Balloon is Calculated

The weight of air is about 1 1/4 ounces per cubic foot at ordinary temperatures and at the surface of the earth. If then a balloon displaces 1,000,000 cubic feet of air, its total lift or buoyancy is 5/4 x 1,000,000 = 1,250,000 ounces = 1,250,000/16 lbs. = 78,125 lbs. and so on. The useful load a balloon can lift is its total lift minus the weight of the envelope, of the gas in the envelope, of the cars, of the engines, and of the fuel.

In Fig. 145 we show the relative strengths in dirigible balloons of Germany, France and Great Britain at the beginning of the war. Britain


and France built many dirigibles during the war and one of the latest built by Britain displaces 1,600,000 cubic feet of air. Its total lift therefore is 1,600,000/16 x 5/4 = 125,000 lbs.

The balloon is filled with hydrogen which weighs about 1/14 as much as air, and therefore 1/14 of the total lift is used up in lifting the hydrogen gas. The weight of the hydrogen is 125,000/14 = 8928 lbs.

Hydrogen gas has been used in balloons because it is the lightest gas known. It has one great drawback, however, in that it burns very readily. There is another gas called helium which is twice as heavy as hydrogen but which has the great advantage that it does not burn.

Before the war helium was very expensive but during the war it was found that the helium which occurs in some of the natural gases of the United States could be separated at a reasonable cost. It is expected that the dirigibles of the future will be filled with helium, and since it does not burn, it will be possible to put the engines in a room inside the balloon as shown in Fig. 146.


Although helium is twice as heavy as hydrogen its lifting power is only 1/13 less because the lift of a balloon depends primarily on the weight of air displaced. You can show this as follows :

If a balloon displaces 140,000 lbs. of air and it is filled with hydrogen, it holds 140,000/14 = 10,000 lbs. of hydrogen, since hydrogen weighs 1/14 as much as air.

If the balloon is filled with helium it holds 140,000/7 = 20,000 lbs. of helium, since helium weighs 1/7 as much as air.

The lift minus the weight of hydrogen = 140,000 - 10,000 = 130,000 lbs.

The lift minus the weight of helium = 140,000 - 20,000 = 120,000 lbs.

That is, the lift with helium is only 1/13 less.

To illustrate the buoyant force of air.

Blow a soap bubble with illuminating gas (1) Fig. 147. Does the bubble rise?



Blow up a balloon with illuminating gas by means of the force pump (2) Fig. 147. Does the balloon rise?

The bubble and balloon rise because they displace a greater weight of air than their own weight plus the weight of the gas in them.

To illustrate the buoyant force of air by means of a balloon filled with hydrogen.


If the metal zinc is placed in an acid, the metal is dissolved by the acid and hydrogen gas is produced. You can make hydrogen gas and fill the large balloon with it as follows.

Purchase at a drug store 2 ounces of strong hydrochloric acid (also called muriatic acid) which should cost about 5 or 10 cents; also purchase at an electrical shop a zinc rod for a Leclanché battery, which will also cost about 5 or 10 cents, or purchase two zinc strips.

Pour the acid into the bottle and add an equal volume of water. This dilutes the acid and slows up the production of the hydrogen. If the hydrogen is produced too fast it will bubble acid up into the balloon.

Now to make hydrogen and to fill the balloon, proceed as follows: Arrange the bottle as shown in Fig. 148 and attach the large balloon to the elbow by means of a short piece (about 1 1/2 in.) of a stretched rubber band. When you have done this place the zinc rod or zinc strips gently in the bottle, insert the stopper at once, and allow the hydrogen to fill the balloon. It will take about 5 minutes to fill the large balloon completely.

When the balloon floats well in the air, slip it off the elbow with its stretched rubber band. The band will contract and close the balloon.

Now release the balloon. Do you find that it floats up to the ceiling? Precautions. Be very careful not to get any of the acid on your hands or clothes. It will burn very bad holes if it does.

When you are through empty out the liquid left in the bottle, as it is of no further use, and rinse the bottle and rod very thoroughly with water.

You must not use the zinc in small pieces because it produces the hydrogen too fast and makes the acid bubble up into the balloon. Use the zinc in the shape of a rod or strips.


To shoot down a balloon.


We show in Fig. 149 three views of a balloon shot down by means of incendiary bullets. These bullets set the hydrogen on fire, the envelope burns, and the car and machinery fall to the ground.

A toy balloon filled with hydrogen as in the last experiment floats up to the ceiling. It will come down by itself in a few hours because the hydrogen gradually passes out through the rubber.

If you are in a hurry to get the balloon down, and if you have an air rifle, you can shoot a hole through the balloon: the hydrogen will then escape and the balloon will fall at once. This method, however, spoils the balloon.


You can shoot the balloon down with a syringe without destroying it as shown in Fig. 150. The water on the balloon will make its weight greater than the buoyancy of the air displaced by the balloon and this will bring it down.

If you let the water evaporate, the balloon will rise again because it again becomes lighter than the air it displaces. You can then shoot it down again with water.

To illustrate the buoyant force of a gas heavier than air by means of a soap bubble filled with air.

Purchase at a drug store one ounce of ether and pour it into an empty 12 qt. pail, cover the pail with a newspaper and allow it to stand for about 10 minutes.

The ether will evaporate and produce ether gas. This being heavier than air will remain in the bottom of the pail and force the lighter air out at the top.

Now dip the end of the wide tube in the soap suds and shake off the excess soapy water. Blow a large bubble and detach it about 6 in. above the bottom of the pail.

Do you find that the soap bubble filled with air floats on the heavy ether gas?

The buoyant force of the ether gas is the weight of this gas displaced by the bubble. This buoyant force is sufficient to support the soap bubble film and the air inside of it.



You will now illustrate Boyle's law and then make a number of appliances which make use of this law, namely, the air brake, flame thrower, fire extinguisher, air pump, bicycle pump, sand blast, pneumatic paint brush, diving bell, pneumatic caisson, and submarine air supply.

Boyle's law is: The volume of a gas varies inversely as the pressure on it.

This is illustrated in Fig. 152. In (1) the tube is full of air and the pressure on the air is one atmosphere because the tube is open to the atmosphere. In (2) the pressure on the air is 2 atmospheres and the volume of the air is 1/2 what it was in (1). In (3) the pressure on the air is 3 atmospheres and 1/3 what is was in (1) and so on.

In (4) the air in the tube below the plunger is under 1 atmosphere pressure because the tube is open to the atmosphere. In (5) the tube is closed, the plunger is raised until the pressure on the air is 1/2 atmosphere and its volume is two times what it was in (4). In (6) the plunger is raised until the pressure on the air . in only 1/3 and its volume is three times what it was in (4).

These illustrate Boyle's law.



Boyle's law is usually illustrated by means of the apparatus shown in Fig. 153. The glass tube A is closed at the top and is partly filled with air, the second glass tube B is open at the top, and the two tubes are connected by a rubber tube filled with mercury.


The mercury surfaces at the beginning are at the same level, (1) Fig. 154, and since the pressure on the mercury surface in B is 1 atmosphere, the pressure on the air in A is also 1 atmosphere.

If now B is raised until its mercury surface is 30 in. above that in A, the air in A is under 2 atmospheres pressure and it is compressed to 1/2 its first volume, (2).

If B is raised until its mercury surface is 60 in. above that in A, the air in A is under 3 atmospheres pressure and it is compressed to 1/3 its first volume (3), and so on. If on the other hand, B is lowered, (5), until its mercury surface is 15 in.


below that in A, the air in A is under a pressure of only 1/2 atmosphere and it expands until its volume is 2 times its volume in (4).

If B is lowered (6) until its mercury surface is 20 in. below that in A, the air in A is under a pressure of only 1/3 atmosphere and it expands until its volume is 3 times its volume in (4), and so on.

Note: A column of mercury 30 inches high exerts a pressure equal to that of one atmosphere. Similarly 15 in. = 1/2 atmosphere and 10 in. = 1/3 atmosphere.

To illustrate Boyle's law.

If you have a spring balance you can prove Boyle's law as follows: Use the apparatus (1) Fig. 155 and compress the air to one half its volume as in (2). Is the average pull on the balance 4 1/2 lbs.?

Note: Friction opposes the plunger when it is moving in, but it helps the plunger to remain in. You will find that it takes more than 4 1/2 lbs. to compress the gas, but less than 4 1/2 lbs. to hold it after it is compressed, the average is 4  l /2 lbs.


The area of the plunger is 3/10 sq. in., therefore the pressure per square inch is 4.5 x 10/3 = 15 lbs. or 1 atmosphere, but the air on the outside exerts a pressure of 1 atmosphere on the plunger, therefore the total pressure the plunger exerts on the air in the tube is 1 + 1 = 2 atmospheres.

You have shown here that when you double the pressure on a gas you compress the gas to one half its first volume.

To show that when you halve the pressure on a gas its volume doubles, use the apparatus (3) Fig. 155.

Start with a distance of 2 inches between the plungers, (3) then pull up the spring balance until the distance is 4 inches, (4). Is the average pull on the balance 2 1/4 lbs.?

A pull of 2 1/4 lbs. on 3/10 sq. in. is 2.25 x 10/3 = 7.5 lbs. per sq. in. or 1/2 atmosphere. Since the pull of the balance is only 1/2 atmosphere, the air in the tube must be exerting the other 1/2 atmosphere.



One of the commonest applications of compressed air is in the air brakes on trains. The air compressor A, on the side of the engine boiler, is operated by steam from the boiler. It compresses air in the large tank B, on the locomotive, and this compressed air is carried through the train pipe under the cars to the air brake under each car. The air brake on each car consists of a triple valve F, an air tank E and a cylinder C containing a piston P. The brake beam is attached to D.

The operation of the air brakes is as follows: Air is pumped into the locomotive tank B until its pressure is about 75 tbs. per sq. in. This compressed air moves through the train pipe, through the triple valve F, and into the car tanks E.

When the train is running, the pressure in each car tank E is equal to that in the locomotive tank B ; but there is no air in the cylinder C and the brakes are "off", because the spring holds the piston P in the position shown.

When the engineer puts "on" the brakes, he turns a lever which closes the valve between B and the train pipe, and which at the same time, lets the air out of the train pipe. When the air pressure in the train pipe


decreases, the triple valve shifts in such a way that compressed air passes from the tank E into the cylinder C; this compressed air drives the piston out with a pressure of 75 lbs. per sq. in. and puts the brakes "on."

When the engineer wishes to take the brakes "off" again, he turns the lever back. This closes the train pipe and at the same time allows air to flow from tank B through the train pipe to the triple valve F. When the pressure in the trian pipe increases, the triple valve shifts back in such a way that it lets air pass from B into E, also it closes the passage from E to C, and lets the air out of C. The spring then forces the plunger in and takes the brakes "off".

It will be noticed that if the train should break in two by the breaking of a coupling, the rubber air hose connection on the train pipe is broken and the air is let out of the train pipe. This automatically sets the brakes on each car and both parts of the train are brought to a standstill.

You will now make and operate an air brake and illustrate the working of the cylinder, triple valve, and air tank.

To make and operate an air brake and to illustrate the working
of the triple valve, cylinder, air tank, and train pipe.


Use the apparatus as shown in Fig. 157, open clip 1, and blow air into the rubber tube.

Your mouth here represents the compressor and air tank on the locomotive, and while you are blowing air into the tank E you are representing the conditions when the train is running and the brakes are "off". You will notice here that when clip 1 is open and 2 is closed the triple valve is admitting air to the tank E, the cylinder C is open, and the brakes are "off". Clips 1 and 2 represent the triple valve. Now close clip 1 and open clip 2. Do you observe that the compressed air in E forces the piston out? This is exactly what happens when the


engineer puts the brakes "on". You will notice here that when clip 1 is closed and 2 is opened, the triple valve has closed the passage between the cylinder and train pipe, and has opened the pipe between E and the cylinder. This is the condition when the brakes are "on".

If you have a bicycle pump, use it instead of your mouth and pump more air into the tank E. You will then find that the piston is driven out with greater force.

At the next opportunity examine the air brakes under a box car or flat car on a railway siding. Identify the air tank, cylinder, piston rod end, the triple valve, and the train pipe. Notice that the outward movement of the piston rod, moves a lever, and that this lever in turn sets the brakes.



You have read of the flame throwers, which were used during the war. You will illustrate their action in the next experiment.

A flame thrower is a strong metal tank with a pipe and nozzle leading from the bottom. It contains a mixture of inflammable oils in the lower part and above this, hydrogen gas under great pressure.

The tank is carried on the back of the soldier, as shown in Fig. 158, and when the nozzle is opened the compressed hydrogen drives the oil out with great force. The oil is set on fire by a pilot light just beneath the nozzle and the moving stream becomes a stream of flame or liquid fire.


To illustrate the action of the flame thrower.

It is dangerous to illustrate the action of a flame thrower with oil and you will use water instead.

Arrange the apparatus as shown in Fig. 159. To load the flame thrower, place a clip on the rubber tube, put a stopper and elbow on the end, insert the stopper into a water faucet, open the faucet gently, open the clip, and allow water to enter the bottle until it is one half full, then close the clip.

The flame thrower is now loaded; the water represents the oil and the compressed air represents the compressed hydrogen.

Now to use the flame thrower; replace the elbow and stopper at the end of the rubber tube by a nozzle, turn the bottle upside down, point the nozzle at the thing you wish to hit, and open the clip.


The common household fire extinguisher, Fig. 160, is a strong brass cylinder with a short piece of hose attached at the top; this hose and its nozzle are open at all times. The extinguisher is charged as follows: In the bottom there is a solution of 1 1/2 lbs. of sodium carbonate (Na2CO3) and 2  l/2 gal. of water, and above this there is an 8 oz. bottle containing 4 ozs. of strong sulphuric acid (H2SO4). This bottle is fitted with a loose lead stopper which falls out when the extinguisher is turned upside down.

To use the extinguisher, you carry it right side up to the fire, then turn it upside down and direct the stream of water and gas upon the fire by means of the short hose. Use all of the water, because once you have turned the extinguisher upside down, the liquids are mixed, and the extinguisher is of no further use until you have re-


charged it. You should do this at once in order to be prepared again for a fire. In recharging you should follow the directions printed on the case.

The action which takes place in the extinguisher is as follows: when you turn it upside down, the sulphuric acid and sodium carbonate react chemically and produce a large quantity of carbon dioxide gas. The volume of carbon dioxide gas produced is much greater than the volume of the cylinder and therefore the gas exerts pressure on the water and drives it out with great force.

The fire is extinguished, partly by the water, and partly by the gas. It seems strange to speak of putting out a fire by means of gas, but carbon dioxide gas has three properties which make it very valuable for this purpose: first, it does not burn; second, it does not support combustion, that is, it does not help other things to burn; third, it is heavier than air. The carbon dioxide gas surrounds the fire and smothers it, because it does not support combustion and it takes the place of the air which does support combustion.

To make and operate a fire extinguisher.

You will not use strong sulphuric acid because it burns practically everything it touches, but instead you will use a dilute acid, vinegar; also you will use baking soda which is sodium carbonate. Arrange the apparatus as shown in Fig. 161. Pour six tablespoonsful of vinegar into the bottle, fill the bottle four fifths full of water and shake, measure out one level tablespoonful of baking soda and place it on a piece of paper ready for use.

Now to use the fire extinguisher, go outside and let one experimenter hold the bottle and stopper while the other holds the baking soda and the nozzle. Dump the soda into the bottle, put in the stopper quickly and hold it very

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