# How does a garden hose swell up?

In my garden I have a long, black hose attached to an electric pump at one end and provided with a nozzle at the other. The hose is used mainly for watering, but sometimes also for heating water on a warm day. That is why I know the exact volume of the water in the hose: it's 12 litres (that's quite a lot - enough to take a shower). I can also use the hose as a small water-supplier. After the pump is turned off with the nozzle closed, the hose remains swollen and it can throw out up to 3 l of water (that's a lot, too - one can easily wash hands several times without using the pump). In other words, under the pressure of the pump (about 2 atm) the volume of the hose increases by 1/4. That is a fact coming from observation. The volume of the cylinder which is the hose is equal to the product of its length and the area of its cross-section. One day I the following question intrigued me: to what degree is the 25% increase in volume due to an increase of the section area and to what degree is it due to an increase in length. If a cube swelled up, then the approximate equality

would imply that each linear dimension would have increased by 8% approximately. Taking into account the length of my hose, which is 60 m, this 8 percent makes almost 5 m. This is quite a change and it cannot pass unnoticed. However, when I measured the extension of the hose after filling it with water, the result was...zero! If a dimensional analysis leads to an estimation of the extension of the order of the hose length, i.e. in meters, and in practice there are only centimetres, then some dimensionless parameter must be involved. With a false feeling of understanding I at first related it to the disproportion between the diameter of the hose and its length. Nevertheless I still felt uneasy about it and decided to give it some more thought.

To explain the observation we must refer to Hooke's law. This law has been formulated almost as long ago as Newton's laws and its practical significance for technological problems cannot be overestimated, but it is generally known only for the specific case of rod expansion and contraction. The formula to recall at this moment is F=k D(l). In this formula F is obviously the expansion force and D(l) represents the extension of the rod. The proportionality constant k is different for different kinds of rod. It can be easily shown that this constant, considered for rods made of the same material, should be directly proportional to the cross-section area s and inversely proportional to the initial rod length l. Taking this observation into account, we may rewrite Hooke's law in the following way:

where E now only depends on the kind of material and not on the dimensions of the rod. The constant E has the name of Young's module.

For further considerations it is more convenient to consider the force corresponding to a unit of the cross-section area rather than the force acting on the whole cross-section. The former is called stress and it can be represented by N=F/s. When N is negative, it is also called pressure. Similarly, we will consider relative extension x=D(l)/l rather than the absolute value of the extension and this relative extension will be called deformation. Using these two magnitudes we write

Thus deformation is directly proportional to stress or, as Hooke had originally put it - to each stress its deformation.

If we take a closer look at a rod being expanded, we may notice that together with a lengthwise deformation caused by a longitudinal stress there is also a transversal deformation - without any transversal stress occurring! This phenomenon has particular significance in the case of a rather short rod, particularly when it is a cuboid or just a cube. A complete Hooke's law should thus also describe what happens with transversal dimensions. The matter is simple. It is almost evident that the value of each transversal deformation (along the y axis and along the z axis) should be a certain constant (for a given material) fraction of the longitudinal deformation:

In the above equations, the Greek letter sigma denotes the second constant besides E which characterizes elasticity properties of the material. It is called Poisson's constant.

Strictly speaking, all that has been said above is true of so-called isotropic substances, the structure of which does not distinguish any direction. This could be glass or a plain piece of metal, which is in fact a conglomerate of an enormous number of chaotically oriented microcrystals. An example of a nonisotropic substance is supplied by wood, which under constant stress undergoes different deformations depending on whether the force is applied along its fibres or across them. The same happens with monocrystals; we shall not conmsider generalizations of Hooke's law to anisotropic substances here.

We shall soon introduce stresses along the remaining two axes y and z, so let's rename our stress N, which acts along the x axis; we shall denote it by N(x). Reassembling formulas for all the three seformations corresponding to a stress N(x) we get

If a stress N(y) is added to N(x), then new deformations will apear: (1/E)(N(y)) in the direction y and -sigma(1/E)(N(y)) in each of the directions x and z (which are now transversal with respect to the stress N(y). Finally, if there is still another stress N(z), the complete deformation is

Before we return to our hose let's spend one more minute on general considerations. Let's add the above three equations sidewise. We get:

The reader will easily verify that for small defomrations the sum x+y+z represents precisely the change of volume:

A body compressed on each side cannot increase its volume (a spontaneous explosion would occur), so

For typical metals Poisson's constant is close to 1/3, and it approaches the upper limit, i.e. 1/2, for substances the elasticity of which with respect to a change in volume is much greater than their elasticity with respect to changes of shape. Rubber is such a substance and it has sigma= 0.49. For simplicity we may even assume the value sigma=1/2.

Let's return to the hose now. We shall try to determine stresses on rubber in a hose full of water. Imagine a small square lying on the hose surface so that two of its sides are parallel to the axes. What is the force acting on the sides of this square? Or, putting it differently, what would the strength needed to keep together each centimetre of the "wound" which would appear on the hose surface if it were cut in full depth with a lancet be, if we were to prevent loss of water? To find the longitudinal strength the easiest way is to cut the hose on the whole perimeter. In this case this strength is equal to the pressure of water on the bottom of the cylinder. Denoting the hose radius by r and the water pressure (exceeding the atmospheric pressure; we still neglect the latter as immaterial for our problem) by p and ignoring for the moment the thickness h of rubber we get that the strength needed for a perimeter of length (2 Pi r) is equal to (Pi r^2 p). Thus pr/2 corresponds to each unit of length and pr/2h corresponds to each unit of the area of the material cut.

Now, to determine transversal stress imagine that we cut the hose along its entire length d (like a French baguette, which we want to use for a very long sandwich). The force pushing the two halves of the hose apart is again equal to the product of pressure and the cross-section area (and not of the area of the cylinder's side surface!), i.e. 2rdp, and it has to be evenly distributed along the length of two "wounds", equal to 2d. Thus for each unit of length we have a force of pr and a stress of pr/h. Observe that the longitudinal stress is exactly twice as small as the transversal one. No wonder that sausages always crack along their axis when cooked, if ever.

It may be somewhat less evident how to cope with the stress (negative now) acting perpendicularly to the hose surface. This stress varies from zero on the outer surface to -p on the inner surface. Fortunately this stress does not depend on the value h, whereas the other two had h in the denominator. For a hose with thin walls (as compared to the radius) we can assume that the perpendicular stress is equal to zero.

Suppose now that the axis of the hose coincides with the x-axis, the direction perpendicular to the hose surface is z and the direction on the surface tangent to the perimeter is y. Then we have

Substituting these values into Hooke's law we can immediately compute

Thus when the stress ratio is 1:2, the deformation ratio is, in our case, equal to

Without the "sigma" it would also be 1:2 and upon a 25% increase in volume this would mean a 5% increase in length (about 3 m). But for rubber sigma is almost 1/2! With this approximation the ratio becomes 0:2. Taking a more realistic value 0.49 for Poisson's constant, we get a deformation ratio equal to 0.02:1.51, which yields an extension of the hose of about 8 cm, not very noticeable in a 60 m hose. The unexpectedly small parameter which modifies the swelling process completely contrary to our intuition is the difference between the true value of Poisson's constant for rubber and the theoretically maximal value 1/2.

Finally, let's try to figure out how should a cylinder with thick walls behave in a similar situation. Take its inner radius to be one-half the outer one and assume for simplicity that the extreme value of sigma for the material is taken, i.e. sigma = 1/2. If we mentally (or even physically) divide the cylinder into many concentric cylinders one within the other, each of them becomes very thin and therefore satisfies the assumptions of our previous considerations. The only diference lies in the fact that, except for the innermost and outermost cylinders, the remaining ones are compressed by the adjacent rubber surfaces rather then by water or air. Even if rubber is not subject to Pascal's law, the symmetry of the problem guarantees that in this case, too, the action is performed along the radii! Hence each of the coats is in the mechanical situation we have already considered. We know that none of them will change its length, so there is no reason for the hose as a whole to do so.

Andrzej SZYMACHA
Department of Physics, Warsaw University