Chapter 2      Liquids at rest

 

Introduction

This title covers several things that are of interest to the mechanical engineer. Our interest is in liquids that are at rest in a container under the effect of the gravitational field of the Earth. These liquids can exert very large forces even when they are stationary. Installations like flood barriers, river control devices and ships are subjected to large forces exerted by liquid in contact with them and we need some way of estimating these forces. Even when the liquid exerting the force is moving it might still be good enough for the purposes of engineering to regard the liquid as being at rest for calculations involving pressure. For example, even a cursory look at a ship under way shows that there are surface waves along the hull. I think that it is inescapable that the pressure under the hull must vary in some way along the length and that that pressure variation is closely connected with the profile of the surface as seen on the side of the ship. It is better to quantify this variation by calculating it from the wave profile than to do nothing.

 

So we need to make a start. We need to form a mental picture of a liquid at rest. The trouble is that any substantial body of liquid is usually on the move as a result of the wind or tide or perhaps an outflow and inflow. Whilst the flow within the liquid in this state can be both interesting and complicated my interest here comes from regarding it as a liquid at rest. However I think that I have only once seen a substantial body of liquid that was at rest in the sense meant in this chapter. It was a pool of water in a limestone cave. It was lit artistically for visitors to the cave and the surface was like a mirror and the reflection of the roof of the cave appeared to be identical to the cave. Of course someone could not just look at such perfection and dabbled fingers in it. I wondered just how long it would take to get back to its former state. Engineers do not deal with liquids that are at rest but with liquids that they think are near enough at rest for the purpose of analysis. It is a major step forward to understand a liquid at rest.

 

I think that it is helpful to have some idea of what goes on in a liquid at molecular level. Substances that are in a liquid state are well known to all of us. However the internal structure of a liquid is only recently being understood. The most common liquid is obviously water and water is also the liquid with the most extraordinary properties. The most important matter for us is its molecular behaviour. The molecule of water has a shape that stems from its structure. The two hydrogen atoms are bonded together and the oxygen atom is bonded to both of them to produce a T shaped arrangement. These T shaped molecules pack together closely and, under the action of intermolecular forces, the molecules form clusters. These clusters are very short lived and a single molecule may change the molecules with which it forms clusters as many as 10¹² times every second[1]. The size of the clusters varies but the dominant size is five molecules.

 

These facts do not concern us directly but they do tell us that the size of molecules is unimaginably small and the distance between them, when they form part of a liquid, is also very small. Even if we start to talk about large molecules like the long chain molecules of some plastics they are still very small. What does concern us is that if water, and any other liquid with a similar structure, is continually in a such a state of internal agitation it cannot retain a shape like a solid does. It follows that it can change shape in response to any force that might be applied to it. In fact I can see no way that it can offer any resistance except during a change in shape.

 

Liquids at rest

So a liquid is “runny” as we have seen above and can only be at rest in a gravitational field when it is wholly supported by solid boundaries having the shape of a bowl or a bottle. If we observe a liquid in, say a tea cup, (figure 2-1) we know from experience that the cup could be moved to any one of many positions and the liquid would always move until eventually it is at rest occupying whatever constitutes the bottom of the cup. Of course it is not really at rest internally as the ordinary molecular motion continues. However, for most purposes of engineering, we would regard the liquid as not moving and being in equilibrium with the forces exerted on it by the cup. Here we are with Newton’s “forces” all of which are the consequence of the gravitational attraction of the Earth. We shall have to think about how these forces are exerted by the cup and how they are distributed.

How can we proceed from this point? One thing that we can recognise is the fact that the free surface appears to be flat and also horizontal. Of course we know from our experience that the cup is just a small version of an ocean and that, when an ocean is viewed from space its surface is seen to be spherical. The tiny surface of the liquid in the cup must really be aligned with a more or less spherical surface of immense size compared with the cup and merely appears to be flat and horizontal. The gravitational field of the Earth is usually modelled as many concentric spherical surfaces and for each surface the strength of the gravitational field is the same all over it.[2] Our free surface is aligned with one of these spherical surfaces. In the way scientists have of describing fields the surface is said to be all at the same potential. We shall have to give more thought to the meaning of that word in the next chapter.

 

What else can we find out by reasoning from experience? That nice flat horizontal surface is actually a surface of separation between the water and the air above it. What then can we deduce? First, the water  must be in equilibrium under the gravitational force exerted on it, the forces exerted on it by the cup, and the force exerted on it by the air. It is the force on the surface of separation that is interesting. It can only be vertical and there is no influence that can make the force anything other than uniformly distributed all over the surface of separation. Now we have not just a force but a distributed force! We say that it produces a uniform pressure and find the intensity of this pressure by dividing the magnitude of the force by the area.

 

A further observation is that liquids can resist forces that tend to compress them. This is not a surprise in view of the molecular structure of the liquid with its closely packed molecules. In fact liquids mostly behave as if they are incompressible and only liquids under very high pressure contract measurably. However, even though the liquid is incompressible just like a solid, in one way it quite different to a solid. Most solids can resist the distributed gravitational force that acts on all its distributed mass. It follows that a solid can withstand a system of forces impressed on it without a significant change in shape. This is because it is held together by forces acting between its atoms or molecules that are “rigidly” set in some form of lattice. The internal forces in the solid may be of tension, compression or shear. By contrast the liquid must move and go on moving until the only forces on it are compressive. At rest it is free from tensile forces and from shearing forces as the molecular structure cannot sustain a resistance to such forces.

 

A cup is a very simple container for a liquid because it has a single free surface. What happens to a liquid in a vessel shaped as shown in figure 2-2? It is clear that there are now two free surfaces and that the liquid is continuous below these surfaces. The two separate surfaces an only be aligned with the same surface of uniform potential so that the surfaces have the same level.

Variation of pressure with depth in a liquid at rest

 

Before we can tackle this we have to look at a liquid at rest more carefully. Suppose that you could make a bag from thin polythene, fill it with, say, water, and then seal it. It would have a shape that was arbitrary in the sense that it is not a regular shape like a sphere. All that has been done is that a fixed quantity of water has been prevented from mixing when it is released in an expanse of water that can come to rest and, if the polythene has a density equal to that of water, the bag could end up anywhere because wherever it goes it is in equilibrium with the water surrounding it.

 

Figure 2-3 shows the plastic bag filled with water The horizontal line with two shorter lines indicates the  free surface. The water in the bag is in equilibrium under the attraction exerted on it by the Earth and the distributed compressive force exerted on it by the liquid surrounding it. I have tried to represent this distributed force as being made up of lots of small forces acting on the surface each shown by an arrow. These forces must, because there can be no shearing forces, be normal to the parts of the surface on which they act. They must also be, in every respect, equal and opposite to the internal forces in the element and they must combine to produce a force which is equal and opposite to the weight of the element. Furthermore this force must act vertically through the centre of mass of the element. But small forces acting on small areas produce pressures and so this enclosed liquid is subject to pressure over its whole surface but not to a uniform pressure. This enclosure containing the liquid could be of any shape and any size so it is possible to imagine it becoming smaller and smaller until the elemental forces are acting effectively at a point. Then the forces causing the pressure still act in all directions but in the end the pressures tend to become the same in all directions. It is what everyone expects anyway.

 

With these observations and deductions we can find an expression relating pressure and depth in a liquid at rest. Figure 2-4 shows a part of an extensive liquid at rest, with a vertical prism of liquid identified within it. The prism is of square cross-section, (although it could have any cross-section), with one end surface, of area A, in the free surface and the other at a depth h. The other surfaces are vertical. There is a downward vertical force on the top face equal to  where  is the atmospheric pressure.

 

We must now examine the lower face to make certain that it is not subject to any net force tending to make it move. If there were to be a force on it, the force could have vertical and horizontal components. If a horizontal component were to be in existence it could only be the result of a shearing action in the liquid and this we know to be impossible. Most liquids are quite mobile and come to rest quickly with no residual shearing stress. It follows that the horizontal component of any net force on the lower face must be zero. If there were to be a vertical component to this force it would have to be resisted either by shearing forces or by an acceleration impressed on the liquid. As the liquid is at rest we can discount both of these and must conclude that the lower face is in equilibrium under vertical forces.

 

It is easy to see that, in the absence of shearing forces, the downward force is due to gravity acting on the mass of liquid contained in the prism. If we use the concept of density (mass/unit volume), and take the density of a liquid to be independent of pressure, this force is equal to :-

                                               where  is the density of the liquid.

 

Recalling the cup with its free surface, it would be inconsistent with our concept of gravity to suppose that the pressure on this face was other than uniform. If this is accepted the net upward force on the prism is equal to :-

                                      where   is the absolute pressure at depth h in the liquid.

 

Then                              

                                           or

                                                                               

In practice pressures are most frequently measured above and below atmospheric pressure and called gauge pressures, and denoted . Then the gauge pressure  at depth  in a liquid of density  is given by:-

                                                                        

This simple expression opens the way to a whole range of applications that, at one time, depended on trigonometry and integration to find forces on immersed surfaces. The trigonometry is still with us but the integration is no longer a problem with mathematics packages. We need some applications.

 

Measurement of low pressures using manometers.

I wondered about including any mention of manometers even though they were in regular use when I was lecturing. I thought that they might have been displaced by electronic devices using diaphragms to detect pressure difference. I looked on the internet and it is clear that, whilst pressure measurement by electronic instruments is widespread, manometers have not disappeared. Certainly they are in use in laboratories where, I suppose, people still have a nagging doubt about using secondary instruments that depend on calibration for accuracy. So I decided to include manometers. They do give a useful source for examination questions.

 

The physical quantities  and  can all be measured accurately. It follows that devices designed to measure pressure by applying the expression  have the potential for equal accuracy. The devices in use are collectively called manometers. There are a few precautions to be observed when designing and using manometers. Liquids have surface tension and have different behaviours when in contact with solids like glass and plastic. We have all seen the meniscus formed between mercury and glass. Any unwanted effects like this can be minimised by using tubing of sufficiently large bore but not so large a bore that the response to a change in pressure is too slow for the intended purpose. The inner surfaces must be clean. There are published standards for the construction and use of manometers and where a manometer has been constructed and operated to these standards they can be accepted as accurate in law for financial and other transactions. Presumably there are now standards for electronic pressure measuring devices.

 

The most simple is the U-tube manometer that is shown in figure 2-5 as it might be used to measure the difference between the pressure of a gas in a pipe and atmospheric pressure. It is no more than two transparent tubes joined together at the bottom, set up vertically and close together. A liquid of known density partially fills the tubes. The top of one tube is connected to the tapping ring[3] and the top of the other is open to the atmosphere. The diagram shows the manometer measuring a pressure above atmospheric producing a difference of level of  between the levels in the two tubes. It could as easily measure a pressure below atmospheric.

 

If  can be measured accurately and, if we are confident that we have no unwanted forces affecting the value of , we can calculate the pressure in the pipe. In order to do so we must make use of the observation in figure 2-2. From this we see that the pressure on the free surface in the left hand limb of the tube is equal to the pressure at the same level in the right hand limb. So the pressures at A and B are equal because the liquid between them is continuous and at rest. If the density of the gas is  and pressure at A is  and is given by :-

                      , where p is the gauge pressure of the gas in the pipe

 

In the same way the gauge pressure at B,  is equal to  where  is the density of the fluid in the manometer.

 

Therefore                     

                                                          

There are only three liquids that are suitable for use in simple manometers. They are distilled water, mercury[4] and distilled paraffin and these have densities of 1000, 13,600 and about 780 kg/m. They are all mobile liquids, more or less non-toxic and can be used safely with simple precautions. The densities of the common gases are in the range of 1 to 2 kg/m³ at atmospheric pressure with air at 1.25 kg/m³ at atmospheric pressure. Given these figures it becomes evident that, even for paraffin in the manometer, putting  involves an error of only 2/780 or 0.25%. So, providing that z is not large, as it might be if the pressure at the top of a chimney was measured at ground level, we can, for most applications, write  :-

                                                                       

The practical details of the simple U-tube manometer depend on the application, but there are some difficulties. I have said that for the accuracy of the reading there should be no unwanted forces acting on the liquid. These forces may occur at the free surfaces of the liquid. Here we have the effects of surface tension and wetting to consider. The British Standards Institution recommend the use of tubes of a minimum bore of 13 mm to make these effects negligible. However we must recognise that the B.S.I. have to make recommendations which, if followed, are acceptable in legal arbitration and these may not be essential in some applications. Then the user must assess the situation and make a choice of diameter.

 

In order for a manometer to show the reading in figure 2-6 a quantity of the gas must flow from the pipe into the manometer. The volume of this gas will depend on  and the bore of the tubes. The rate at which the gas flows into the manometer will depend on the unbalanced pressure at any instant, which as equilibrium is approached tends to zero, and on the size of any restriction in the connection between the manometer and the pipe. A manometer may take an unacceptably long time to come to equilibrium if there is a serious restriction and the bore is large. The use of a B.S. manometer in conjunction with a small pitôt-static tube would give this combination. This gives an incentive to use much smaller bores than 13 mm. Provided that the tubes are clean, so that the wetting is not affected, mercury can be used with tubes of about 3 mm bore. Water is very troublesome, because it does not easily wet tubes that are dry, but paraffin, which wets dry tubes very easily, can be used down to 2 mm.

 

These simple manometers are robust and are used extensively by, for example, the gas supply industry. A further use is to test for leaks by noting the fall in pressure following the isolation of a pipe with a leak. Its response to a leak is quick enough for the manometer to be used to check that a pipe does not leak.

 

The user of a simple manometer must make two readings and subtract one from the other. This can lead to errors and mistakes. There is an incentive to design a direct reading manometer. Figure 2-6 shows such a device. One tube is enlarged to have a cross-sectional area many times that of the other and constructed as a closed pot that is connected to the other limb by a flexible tube. The pot is mounted so that it can be moved to bring the level in the long tube to the zero of the scale. (The levels in the tube and the pot may then differ because of surface tension effects but this does not matter if the tube is clean.) The pressure to be measured is connected to the pot and this leads to a rise in level in the tube. For a tube of small bore the volume displaced is small and as the area ratio can easily be 1000 to 1 the change in level in the pot can be negligible. Then either,  can be used to calculate the pressure or, the manometer may have a direct reading scale.

 

There is also a range of designs of inclined-tube manometers in which the vertical difference in level is magnified by inclining the tube. The arrangement is shown in figure 2-7 and the insert shows the way that the meniscus produces a quite sharp edge to act as a clear pointer to operate against the scale. It is enhanced by the colouring of the paraffin. A further adaptation is to have many parallel, inclined tubes connected to the pot (which must be further enlarged) and apply pressure to individual tubes rather than the pot. This gives a multi-tube manometer, for say wind tunnel work, and this can have the inclination adjustable.

 

 

 

 

 

 

 

 

There are two other manometers that are used to measure the pressure difference between two points in a pipe carrying gas or liquid. The application arises for use with certain designs of meter for measuring flow, for example, the Venturi-meter and the orifice meter.

 

The inverted U-tube manometer is used with liquids that may be under pressure. It is shown diagrammatically in figure 2-8. The U-tube is now upside down and the two pressure connections are made to the lower ends of the tubes. A non-return valve is fitted to the top of the inverted tube. In order to start the manometer the liquid in the pipe is first allowed to flow through the manometer and then air is pumped in with a bicycle pump through the non-return valve to separate the liquid in the two limbs.

 

Following the reasoning used in the simple U-tube the pressures at A and B are equal

because the air at rest above them is continuous.

 

Then     and  where   and  are the densities of the liquid and of the air.

 

Subtracting  from  gives :-

                                       , and,

                           

                                                            

This can now be reduced to  if  is small compared with .

 

The differential mercury manometer shown in figure 2-9 is used for gases or liquids in the pipe. The U-tube is now fitted with a cross connection at the top that incorporates a valve. The pressure connections are made to the tops of the limbs of the U-tube that is partly filled with mercury. In order to start the manometer the valve in the cross connection is opened and fluid allowed to flow through the connecting pipes to purge them of air. Then the valve is closed and the manometer will show a reading.

 

Once more  equals  because the mercury below them is continuous and at rest. Then:-

                       and

 where   and  are the densities of the fluid and of mercury.

Subtracting p_A from p_B gives :-

               .

As this manometer is most likely to be used in conjunction with a liquid in the pipe r_f will be significant compared with r_m and this leads to the inclusion of the descriptive word differential in the name.

 

There are many other designs of manometer including a whole range for measuring very small pressure differences. They use enlarged ends, two liquids of similar densities and other devices to improve the sensitivity. Whilst they are all primary devices, that is they do not need calibration because the accuracy depends on primary measurements, the problems of surface tension and wetting, of cleanliness, and in some cases of interaction between the surface of separation of two liquids and the walls of the containing tube, become more troublesome as the sensitivity increases. It becomes a field for the specialist supplier.

 

Unless the user of a manometer is prepared to use a stepladder and a kneeling mat the maximum value of  is limited to about 700 mm. This means that the maximum pressures that can be measured are about 0.07, 0.93 and 0.05 bar for water, mercury and paraffin as the manometric fluids. When compared with pressures of 300 bar, that are commonly met in oil hydraulic systems, these are low pressures.

 

High pressures are usually measured using pressure gauges. These have a pressure-sensing element, which is most commonly a Bourdon tube or a diaphragm, and some magnifying and display system. These systems are either mechanical or electrical and pressure gauges have evolved to be both reliable and accurate. However these gauges are secondary measuring instruments and they depend on some other device for the accuracy of their calibration.

 

The device used is the dead-weight tester and this uses the ideas of pressures in liquids at rest. In this tester the required known pressure is created by applying a known force to a known area. The area is that of an accurately machined piston working in a vertical cylinder. The known force is that exerted by the gravitational field of the Earth acting on a mass. This means that the test masses must be made to suit the strength of the gravitational field where the tester is to be used. The arrangement of the tester is shown in figure 2-10. In practice this simple device must be designed and operated so that the effects of mechanical friction between the piston and cylinder is eliminated. In the dead weight tester the gauge to be tested is connected to a closed system containing oil that can be pressurised by screwing in a plunger. The piston is a very good fit in the cylinder allowing a very small leakage of oil. The piston and the weights are carefully machined so that, however they are assembled, they interlock to form a single mass with its centre of mass on the axis of the cylinder. Then, if the cylinder is set upright using the levelling screws, there are no horizontal forces between the piston and cylinder and the piston and its masses can then be made to "float" on the oil. When the piston and masses are rotated slowly, the system is as free from friction as we know how to make it and the pressure at the face of the piston equals the applied force divided by the area. The gauge connection is regarded as the level at which the gauge measures the pressure and during calibration the level of the piston face is adjusted to that of the gauge connection.

 

Calculation of forces on immersed surfaces

There are many surfaces that have to be designed and built by engineers to achieve purposes that require the surface to be submerged. These surfaces are frequently submerged in liquid that is either at rest or effectively at rest and, almost without exception, they are made to simple geometrical shapes. As the forces exerted on such surfaces may have to be calculated for design purposes this makes calculation relatively easy.

 

Surfaces that work when submerged, such as dams, sluices, weirs, flow control gates, lock gates and entraining walls can be seen in rivers and canals. Examples of submerged surfaces that are not so obvious are tanks holding water or chemicals, ships, (which are inside out tanks), shuttering for shaping concrete and moulds for casting metal.

 

These surfaces do not share a mode of construction. The plates of a ship or a water tank have to be supported at intervals both horizontally and vertically and deeply submerged plates need stronger support than similar plates at a smaller depth. Ships and water tanks need internal support structures. In order to design such structures we need to be able to calculate the force exerted on a part of the surface and we may need to know both the magnitude and direction of the resultant force on this part of the surface. By contrast, in civil engineering, water is ducted and stored using surfaces that have to be anchored into the ground. Indeed a dam that may be made by dumping earth and rock across a river and then sealing the water face has no formal structure as such. Clearly we need methods to calculate forces that are exerted on submerged surfaces but which forces we need to calculate depends on the application.

 

We need some basic method for calculating these forces and then find ways to apply it sensibly. The method is simple but the application is part of the skill that an engineer brings to design and is best learnt by example. For the purpose of examination the question setter will effectively define the application and expect the examinee to use whatever methods appear to be appropriate. All sorts of methods have evolved, mainly it must be said, to avoid integration but now software like Mathcad make this unnecessary in practice. Examiners still have to devise examination questions for students who may not in examination have access to computers. In this text I will use Mathcad and concentrate on the basic methods.

 

The method can be illustrated in stages as follows. Usually this involves dividing the surface into elemental strips, finding elemental forces and then integrating. Some thought is needed to find a suitable way of subdividing but the process is usually straightforward.

 

Figure 2-11 shows a vertical surface that is rectangular with one edge parallel to the free surface. At a depth h below the free surface the pressure p is given by , where r is the density of the liquid and, if we take an elemental strip of the surface of area , all at this depth h it will have a force on it . This elemental force acts at the mid-point of the length  and at right angles to the surface. The surface can be regarded as being made up of many such strips and we can sum the elemental forces to give the total force exerted on the surface or sum the moments of such forces about some point to locate the resultant force or make any other summation that suits our purpose

 

For the surface in figure 2-11 the force on the element  so the total force exerted on the surface is given by:-               

 

For this surface this can be integrated[5] to give .

 

Putting some values to the dimensions and to  for water Mathcad can be used to integrate and give the force in Newtons.[6]

 

 

 

Any force that acts over an area in this way might usefully be called a distributed force. We have found its value and there could be some advantage in knowing where a single concentrated force equal to the distributed force[7] would act. The depth at which this concentrated force acts can be found by taking moments of the elemental forces about the free surface and dividing by . The moment of the elemental force =  and the total moment M about the free surface is given by:-  which equals  Dividing the moment by the force gives the depth of the point of action of the total force =2/3.H.

 

This could be evaluated for the dimensions above by using Mathcad again. It gives:-

 

 

 

Note that the numerical answer agrees with the outcome of the mathematics.[8]

 

These simple calculations are valuable in the design of tanks for holding liquid and the inverse for tanks used as floating vessels, both of which must be stiffened. It is also useful in dealing with other surfaces which are immersed in a liquid at rest. We can also note that the force F can be written as F = b.H.(r.g.H/2) which can be interpreted as the area multiplied by the pressure at the depth of the centre of area. Whilst this is of no special value here, it turns out to be a general statement for all plane surfaces. As areas and the position of centres of area are known for many geometrical plane shapes the expression can be useful.

 

We have looked at the most simple surface that we are likely to use in engineering. As we have seen it might form part of a tank. Now we must look at sloping or curved surfaces because the weight of the water that is supported by the surface is used as an aid to holding the surface in place. By way of exercise one might choose to calculate any one of the forces that might be regarded as acting on the face but the important ones to know are the horizontal and vertical forces that act on the surface. Let us look at the surface shown in figure 2-13 that could be a length b of one wall of a trapezoidal channel. The wall is inclined at q  to the horizontal.

 

An elemental area, all at depth h, is shown in figure 2-13. For convenience the vertical height of the element is dh although the slant height is greater than this. The horizontal force on the element is given by :-F = å r.g.h.b.dh as before  this can be integrated to give the same answer.

 

In figure 2-14 The triangle of forces for the element shows that the vertical force on the elemental area is equal to the horizontal force/tanq  which can also be evaluated using say Mathcad.

 

This force will be equal to the gravitational force on the water that is vertically above it. This can be found quite easily by calculating the volume and multiplying by .

 

 

 

 

 

 

 

Figure 2-15 Shows a curved surface replacing the plane wall.

 

This surface could be produced by moving a part of a curve such as y = a.xⁿ along a horizontal line. We have to describe the curve in some way. Let the upper point  have co-ordinates X and E relative to the vertex of the curve. Given this we can find a value for the constant a for any value of n. It will be . Now we have a surface where the distributed force is made up of lots of elemental forces all acting in different directions. This complicates the finding of the forces on the surface. The magnitude and position of the resultant force on the whole face could be found as in figure 2-15 provided that the volume and the position of the centre of mass of the water above the curved face is known. However this is unlikely and will not help if the force on a part of the surface is required. It is more likely that we would need to find the vertical and horizontal forces on the surface. We have to go back to the elemental areas.

 

An elemental area all at depth h and therefore subject to the same pressure is shown in figure 2-15 and in end elevation in figure 2-16 The plan area of this element is b.dx and the pressure on it is  Then the horizontal force on the elemental strip s given by  We have to decide how to calculate. If we choose to use a computer package we can easily make it general.

 

In frame 2-17,  which is a copy of a Mathcad calculation, I have put values to the various lengths and constants so that they can be changed easily. Then the first calculation gives the horizontal force in Newtons.

 

In order to find the vertical force on the curved face we must find the projected area of the elemental strip. It is. . Then the vertical force on the elemental strip is:-

                                                 

This is evaluated in the second calculation also in Newtons. It is a very simple extension of this computation to give a graph of the curved face and to show the water level. This has been added.

 

The system can then be explored for any depth, any value of n and for any values of the other dimensions. If this were to be a really significant application the ability to do this would be very powerful.

 

When looking at examples of submerged surfaces it is sometimes important to realise that they have been designed to withstand collision rather than water forces. When looking at control works in rivers it is useful to have observed that the resultant force on any part of a submerged cylinder passes through the axis of the cylinder. This simple fact permits a rotating drum gate to be mounted on a shaft so that the force exerted by the water has no moment about the shaft.

 

Using manometers with fluctuating pressures.

Don’t is the best advice. Sometimes the pressure in a pipe will fluctuate as a result of pumping. Then a manometer connected to a measuring device will produce a reading that might appear to indicate flow but, in fact, because of the poor time-response of the manometer, that reading could be wildly inaccurate. As the connecting pipes cannot be  a part of any standard there can be no panacea for this. However an electronically operated gauge could be attached in a standard way and read remotely and it is conceivable that it could have some averaging circuitry that could produce an accurate measurement of flow.

 

The stability of floating bodies.

 

We cannot discuss the design of floating bodies, such as ships, floating cranes and oil rigs, until we understand the physics of a simple body floating at rest in a liquid, typically water, that is also is at rest.

 

I want to explain the principles governing stability and not get tangled up with hulls of many shapes and for my purpose the cuboid will be most suitable. Let me start with a cuboid having a uniform density less than that of water. If it has the correct proportions, not long and thin, it will float with two opposite faces parallel to the free surface as shown in figure 2-18

 

We could do things to this cuboid. We could push it further into the water and release it, or we could tilt it to one side and release it, and, if we did either or both of these things it would ultimately recover its original attitude. It follows that the cuboid must to be in vertical equilibrium and that equilibrium must be stable both vertically and laterally. We can examine the requirements for these two states of equilibrium separately. Let us start with the vertical equilibrium.

If the body has mass m the gravitational force exerted on it is m.g and this acts at the centre of mass G. For the body to be in vertical equilibrium a force that is equal and opposite to the weight of the cuboid and passes through G must be exerted on the body. This force must be exerted on the body by the water and must be the consequence of the pressure exerted on the part of the surface of the body that is submerged (wetted). This is shown in figure 2-19. This force can usefully be called the upthrust U and figure 2-20 shows the equal forces  and U. We know the position of G because the body is of uniform density but the point of action of U is not immediately obvious. However in connection with the element shown in figure 2-4 we said that the external forces exerted on the element must equal the weight of the element and be equal and opposite and act at the centre of mass of the element. The fact that we now have similar external forces acting on a solid surface which cuts the free surface makes no difference, the water behaves as if the space occupied by the ship is full of water and the pressure distribution results in a force equal to the weight of the water which could occupy the space acting vertically where the centre of mass of this water would be.

 

There is a concept of a centre of volume and this concept is useful here, indeed, it is hard to see any other application for it. The centre of volume is the same point as the centre of mass of a solid of uniform density and, as the centre of mass is the point through which an accelerating force can be exerted on the body without causing rotation when the body is free to move, it can only be found by taking moments of volume. If we decide to use the idea of a centre of volume then we can locate the point of action of U as being on the axis of symmetry of the cuboid and at half the depth of immersion and this is shown in figure 2-20. This point of action is called the centre of buoyancy and denoted B. (A body which floats is said to be buoyant.) Clearly for this cuboid G is above B.

 

This gives the condition for vertical equilibrium but does not tell us whether it is stable. The normal test for stability is to disturb the equilibrium and see whether a system of restoring forces come into existence. So if we push the cuboid down the upthrust exceeds the weight of the cuboid and a restoring force does comes into existence. If the cuboid is raised,  is greater than U, and a downward force tends to restore the body to its original position. The body is in stable vertical equilibrium.

 

However we have G above B and this arrangement is not what one might associate immediately with stable lateral equilibrium. We can now apply the ordinary test for stability. Figure 2-21 shows the cuboid tilted to one side in some way that still leaves us with U equal to . (There is no way to do this but that does not affect the argument.) The position of G relative to the body is unchanged but, as U still equals , the volume displaced by the cuboid is not altered. But the shape of the displaced volume has altered and the position of the centre of volume, that is the centre of buoyancy, has moved to B'. Clearly B' and G are now no longer in the same vertical line and there is a couple exerted on the body tending to restore it to its original position. So we can see that the equilibrium will be stable if the relative positions of G and B after tilting combine to produce a restoring or righting couple.

 

Figure 2-22 shows the relative positions of G, B and B' and the centre line of the cuboid when the cuboid is in the tilted position. The forces  and U are equal. If P is the point of intersection of the line of action of U and the centre line and q is the angle of tilt, the restoring couple equals . If PG were to be large we would think of the cuboid as being very stable and if it were to be small not very stable. It is evident that we have a concept of the "degree" of stability even if we have no word for it. The degree of stability is in some way linked to the magnitude of PG. We need to find a useful expression for PG.

 

Now P and G are two independent points. The position of G for this cuboid is fixed because we let it be of uniform density but it could have been anywhere on the axis of symmetry.[9] P is a point that will depend on the value of q and the consequent position of B¢. P has a position even when q is zero and that position turns out to be very important. It is so important that it has special name, the metacentre[10], and it is given the symbol M. We can now draw a diagram showing the relative positions of B, M and G.

 

However we cannot expect to find MG directly because P is a function of the shape of the cuboid and G depends on its content, we must find the distance PM that is a function of shape, instead. For this we must find the horizontal shift of B to B' and, because the practical application requires a position for P when  is zero, we must find BP for small values of .

 

In figures 2-23 a and 2-23 b the cuboid is shown when it is floating freely and when it is tilted through a small angle q. As the volume of water displaced is unchanged the original and final positions of the waterline intersect on the centreline of the cuboid. The change in shape of the displaced volume is equivalent to cutting off a wedge of volume from one side and adding it to the other. In figure 2-23c the displaced volume is shown in the upright position. The shape can be considered to be made up of two parts, a part that is symmetrical about the centre line and a part that is equivalent to the sum of the volume of the cuboid that was raised above the water by the tilt and the volume that was submerged. For our purpose all we need is the horizontal shift of the centre of volume. For this we need only to find the moment of the two wedges of volume about the line represented in figures 2-23 b and 2-23 c as OO and divide by the displaced volume. There is an advantage in dealing with this by dividing one of the wedges into elemental volumes as shown in Figure 2-24 because the outcome will be general and not limited to this cuboid.

 

Provided that the angle q  is small, the volume  of the element is given by :-

                                                               

and its moment about OO is:-

                                                               

or we can say that the moment of the element is:-

                                                                    

Then the horizontal shift of the centre of volume, which equals the moment of two wedges of volume about OO divided by the displaced volume V is given by:-

                                                            

 where b/2 is half the width of the cuboid.

 

Now  is really the summation of the elemental quantities made up of an elemental area  multiplied by x twice. This occurs sufficiently often in mathematics and engineering to have a name. It is called the second moment of area and is given the symbol I. So the horizontal shift of the centre of volume of the displaced volume is given by :-

                                            , where I is the second moment of area of the water plane section about its longitudinal axis of symmetry (line OO)

 

However, as , , which, when q is small sinq =tanq can be written :-

 

                                                                      BP = I/V.

This is now a special case of P being the position of P when the cuboid is floating without heel. It is called the metacentre and denoted M.

 

From figure 2-21 we found that the righting couple was given by  and now, because P is the metacentre M, we can write this as  and go on to give the righting couple as:-

                             righting couple = , from which  

 

                            righting couple = m.g.(I/V—BG).sinq

 

This result is interesting. It looks to be of little value because it applies to this cuboid floating in still water and to small angles of heel. But its value does not lie in calculating righting couples. We have derived it for a cuboid for which it is clearly correct. But boats are not shaped like cuboids at all. They have sides that might well be vertical over the middle part of the length but they have shaping at the bow and stern and generally appear to be so far removed from the cuboid to require a whole new analysis. However boats, whatever their shape still have a centre of buoyancy and a displaced volume and a section at the water plane. It follows that the value of BM can be calculated for a boat or any other floating vessel at the design stage. It is a dimension of the hull just like its beam and its length. We must see how it is used.

 

We can start by considering the stability of ships or, more correctly, the ability of ships to survive at sea, that is, their sea keeping qualities.

 

Ships are designed and built to make long voyages across any of the oceans. They must have an acceptably high probability of arriving undamaged at their destination although, as we hear of ships foundering almost daily, it is evident that the totally safe operation of ships is unlikely to be possible.

 

Osborne Reynolds put the problem in simple terms in a lecture on the safety of life boats in 1886. He said “.......the peculiar construction of boats of all sizes is the result of a long process of trial and failure, and that, although certain general principles, connecting the (sea-keeping) qualities of ships to their shapes, have been discovered and recognised in the last thirty years, still the recognition of these principles has not resulted in any considerable improvement to be effected in what were before high class vessels, such as yachts and fast sailing vessels, but rather have confirmed the form previously arrived as the best, and led to their being copied in larger vessels.”

 

He went on    “(Whilst) the discovery and recognition of certain general principles have undoubtedly been of immense service in improving large modern vessels the improvement is not as great as might have been expected. But this is mainly because with large ships there is not the same opportunity for trial and failure as with the small, the number being so much smaller, and experiments are so much slower and more costly; but the main reason is that the circumstances which call out the highest qualities of the large vessels become so extremely rare. There is no doubt that many large vessels pass through their whole lives without meeting weather which tests their sea-going qualities in the way in which those of fishing boats are tested many times every winter.”

 

He went on to argue that there was a case for building scale models of new designs for large ships to a size that permitted them to be operated as vessels in their own right so that they would meet the “circumstances which call out their highest qualities” more quickly. This argument has not been accepted even now as the proponents of the wide beam frigate as a replacement for the long thin frigates lost during the Falklands war discovered.

 

Reynolds is saying that ship design is difficult and that it proceeds by evolution. It is difficult because the water in which the ship must operate is not at rest. Surface waves may frequently be 20 metres in height with wavelengths up to 600m and very occasionally as much as 35 metres in a single rogue wave. These conditions cannot be quantified and so we cannot construct mathematical models of a ship and the sea conditions it might meet. A strategy is needed to facilitate the evolutionary process. The physics above has helped in this.

 

The starting point is that we must accept that a ship has an underwater shape which has been chosen as a compromise between various requirements and that it will only work properly when the ship is loaded to the intended depth. In this condition the shape of the displaced volume is known and this means that the displaced volume V and the position of the centre of buoyancy B can be calculated. In addition the value of I, the second moment of area of the water-plane section of the ship about the longitudinal axis, can be calculated and this means that the position of M can be found. The positions of B and M are as much a part of the geometry of the ship as the length and breadth. Now, in some way, the stability of a ship is linked to the distance between G and M and this distance is called the metacentric height (and even in legal proceedings called MG for short). The strategy involves the keeping of records, by interested bodies such as the insurers, of values of MG for types of ship and records of their service. This leads to a knowledge of the values of MG which have proved to be successful. Typical values are :-      for ocean going ships.........0.3 to 1.5 metres[11]      for river craft ..............4 metres.[12]

 

The value for ocean going ships seems to be ludicrously small but there are good reasons for this low value. It is well known that ships pitch and roll simultaneously in response the fluctuating forces caused by wave motion. Further the water forces may combine to subject the ship to both twisting and bending. All ships have concentrated masses in them such as the engines, the fuel bunkers, and the cargo. As the ship moves these masses have the same accelerations impressed on them as the motion of the water is impressing on the ship. Nowhere else will masses of this magnitude be tossed about in this way. The forces involved are very large indeed and they are exerted by the water acting through the structure of the ship. The structure of the ship has to be strong enough to withstand these forces and Reynolds was observing that eventually every ship will meet conditions which are so severe as to lead to failure.

 

However we can see that the magnitude of the forces on a given ship are in some way connected with the value of the metacentric height and that they increase as MG is increased and so an upper limit must be placed on MG.

 

There must also be a lower limit for MG because with a low value of MG the roll rate is so slow that the ship cannot right itself sufficiently quickly to shed one lot of seawater from the deck before the next wave breaks over it.

 

The best design of ship is the one that makes the best compromise between strength and therefore cost to build, cost of operation, and safety. Perhaps it is not just accidental that the value of MG that suits the structure of the ship also gives the best ride, in the circumstances, for passengers. Now we must consider the factors that determine the value of MG for a given ship when it leaves port. Ships have marks painted on the hull at bow and stern to show the depth to which it should be loaded. If they are loaded to these marks the position of M is known. But the matacentric height is the distance between the metacentre and the centre of mass and the position of the centre of mass can be altered. In a passenger liner the scope for moving the centre of mass is small but for a ship carrying mixed cargo the position of the centre of mass will depend on the disposition of the cargo. As the best value of MG is small, a small shift of G can materially affect the value of MG. It is not surprising to hear claims for the provision of computers to help load-masters on vehicle ferries.[13]

 

In order to operate these computer programmes for a container ship it is necessary to have a position for the centre of mass of every container. This is not possible but nor is opting out possible. Container ships have to be loaded and be safe to operate. I think that the way that this has been managed is interesting. Initially each container is assumed to have a centre of mass at one of two or three positions eg mid way up, 1/3 up and 1/4 up from the bottom. Then each container is assigned a position for its centre of mass according to the content of the container from these pre-selected positions. Just knowing that a container is loaded with motor-cycles might be enough to make the choice as might a knowledge that it holds overstuffed furniture. Then the ship is loaded using the computer programme to oversee the loading so that its calculated centre of mass is known. The fact that the ship carries many containers almost certainly means that errors tend to average out. From a record over a significant time of the behaviour of the ship and of other ships using the same loading procedure the assigning parameters are refined as part of an on-going process to give better loading. It is what I would call a strategy but I find others pooh-pooh this idea just because the word is not recognised in engineering.

 

It will be noted that the metacentric height of river craft is much greater than that for ocean going ships. This merely reflects the fact that river craft do not meet excessively rough seas. However there is a group of vessels which are used at sea but only when conditions are calm. These are the floating cranes, pontoons, dry docks and floating drilling rigs. For these it is necessary to note that in addition to the requirement for stability there is the need to use the condition for vertical equilibrium. When, say, a floating crane lifts a load its centre of mass moves laterally and in order to restore vertical equilibrium the vessel must tilt so that the under-water shape changes to bring the centre of buoyancy below the centre of mass. As an extension of this problem the sequence of events required to tilt a drilling rig, which has been built on its side and floated to position, into the upright position is quite complicated.

 

Cruise liners and tankers

Cruise liners and tankers can be about the same length and some are very long at 300 m or more. Cruise liners are built like a floating block of flats yet tankers are almost wholly submerged. Tankers do not put to sea with empty tanks so there is some aspect of sea keeping that is not just obvious. We can explore the mechanics of the cuboid to learn more.

I have drawn a cuboid as it might float. I have given it a length , a beam of  and a draught of . I propose to give the cuboid an aspect ratio that I shall call  so that . Then the draught will be  where  is some number less than 1. With these ratios I can alter the proportions of the submerged volume at will.

 

All that I want to do is explore the expression  for different aspect ratios and for different values of . We know that the second moment of area for a rectangle about its long axis is   and that  so Mathcad can be used to draw some graphs of  against length for values of  and .

 

I have to choose some practical values for the aspect ratio. Generally speaking, aspect ratios of 4, that is having a length equal to four times the beam, is more usual for smaller boats. Cruise liners and super-tankers will be up in the region of eight. Queen Mary 2 has an aspect ratio of about 8.5. I have drawn three graphs of the variation of MB with length of my cuboid for aspect ratios of 4,6, and 8. They are graphs 2.1, 2 and 3.

 

 

I have said that the range of values for the metacentric height of ships is from about 0.3 to 1.5 metres. Now I have values of BM that for large ships are much greater. As B is about half the draught above the bottom of the ship the metacentre is a long way above the waterline, in some cases several times the desirable metacentric height.

 

This raises the question of the validity of my cuboid when thinking about cruise liners and super-tankers In truth the shape of the displaced volume is not so much different from my cuboid. For a cruise liner the value of k would be about 0.25 that is the draught is about ¼ of the beam Then the value of BM for the cruise liner of 300 metres in length is about 25 metres. That means that the centre of gravity is about 24 metres above the centre of buoyancy. No wonder the cruise liner has so many decks and two swimming pools on the top one.

 

By comparison the super-tanker will have a draught equal to about half the beam and an aspect ratio of about 8. Then the value of BM is much smaller. I suppose that having a double bottom by law helps in getting the centre of gravity up to the required level.

 

One could go on exploring the stability of this cuboid to take into account the position of the centre of buoyancy etc. but I think that this is sufficient for this text.

 

Effect of moving ballast

The whole question of ship stability is further complicated by the problem of moving ballast. The most common form of moving ballast is liquid that is free to move, either without constraint, for example bilge water, or liquid that must be carried but is free to move in a tank which is not completely full[14].  Figure 2-25a shows a vessel that is shaped like a cuboid with an inner tank which is partly filled with liquid of density r_b and is floating level in a liquid of density r. If the vessel is made to tilt through a small angle q the liquid moves to the new position shown in figure 2-25b. As a consequence of the tilt the centre of mass of the vessel moves from G to G' and the centre of buoyancy moves from B to B¢, as shown in figure 2-26.

 

The righting couple is now :-

                                                                     

where Q is the point of intersection of the line of action of  and the centre line. The vessel is less stable than it would have been if the liquid had been prevented from moving. Using the method we used to find the horizontal shift of B we can see that the horizontal shift of G is given by :-

                                                               

 where  is the second moment of area of the free surface of the moving liquid ballast about the axis of the vessel. Then the righting couple is equal to:-

                                     or

 

                       righting couple = m.g.((I/V) - (r_b.I_b/r.V).tanq)                                            

 

The importance of this equation is that it tells us that it is not the quantity of moving liquid contained in the ship that matters but how much can move from side to side. Liquids are carried in ships and, as liquid can move fore and aft just as easily as it can move transversely, the liquid is carried in many tanks each having a small cross section and for preference these are filled. This problem is also a headache for designers and operators of vehicle ferries because the vehicles are carried on an unobstructed floor in the ship. Should water get on to this floor, as it has in several well-known incidents[15], the ship becomes unstable with only a little water and may capsize.

 

The metacentric height as a measurable quantity is used very extensively to quantify the stability of vessels of all sorts. It can obviously be used to classify vessels as stable, unstable or having neutral equilibrium according to whether MG  is positive, negative or zero. Some vessels are clearly stable both when upright and capsized but lifeboats are designed to have a negative metacentric height when inverted so that they are self-righting.

 

The metacentric height can also be used to give a quite good estimate of the natural frequency of rolling of a ship. If the motion is limited to small amplitudes and if it is regarded as taking place about a longitudinal axis through the centre of mass and the ship has a second moment of mass about this axis of   we can say that :-

                               righting couple =  times the angular acceleration or:-

                                                           

 and, as q  is small and can be put equal to sinq this is the expression for S.H.M.for which the period of roll T will be given by :-

                                                   T = 2pÖ(I_G/(m.g.MG))                                                

It will be obvious that the whole process of rolling involves the movement of the water as well as the ship and this expression could be regarded as much too simple to be of any value. Nevertheless, in practice, it gives a very useful guide to the observed period of rolling.