0333779061

1Hydrostatics

This chapter introduces some of the fundamental quantities involved in hydraulics, such as pres-

sure, weight, force, mass density and relative density. It then considers the variation of pressure

intensity with depth below the surface of a static liquid, and shows how the force on a sub-

merged surface or body can be calculated. The principles outlined are used to calculate the

hydrostatic forces on dams and lock gates, for example. These same principles are applied in

Chapter 2 in connection with pressure measurement using piezometers and manometers, and

in Chapter 3 to the analysis of floating bodies. Thus the sort of questions that are answered in

this chapter are:

What is meant by pressure?

What is the difference between force and weight?

What is the difference between mass and weight?

How and why does pressure intensity vary with depth in a liquid?

How can we calculate the pressure intensity at any depth?

How can we calculate the force on a flat immersed surface, such as the face of a dam?

How can the hydrostatic force be calculated when the immersed surface is curved?

Does hydrostatic pressure act equally in all directions, and if it does – why?

How can the buoyancy force on a body be calculated?

What do we do if the liquid is stratified with layers of different density?

CH

APT

ER

1

2 Understanding Hydraulics

Figure 1.1 Illustration of the pres-sure exerted on a floor by two typesof shoe. The stiletto is the more damaging because the weight is dis-tributed over a small area, so giving arelatively large pressure

1.1 Fundamentals

1.1.1 Understanding pressure and force

❝ Have you ever asked yourself why a trainer will not damage a soft wooden floor,but a stiletto heel will? ❞

The answer is because the average pressure, PAV, exerted on the floor is determined by theweight of the person, W, and the area of contact, A, between the sole of the shoe and thefloor. Thus:

(1.1)

So, because a trainer has a flat sole with a large area of contact, it exerts a relatively smallpressure on the floor (Fig. 1.1). On the other hand, the sharp point of a stiletto means thatmuch of the weight is transmitted to the floor over a small area, giving a large pressure.Similarly a drawing pin (or a ‘thumb tack’ in American) creates a large, penetrative pressureby concentrating a small applied force at a sharp point.

❝ I understand that, but can you now tell me what is the difference between weight and force? ❞The answer is basically ‘none’. Weight is simply one particular type of force, namely that

resulting from gravitational attraction. So equation (1.1) can also be written as PAV = F/A,where F is the force. This can be rearranged to give:

F = PAVA (1.2)

The unit of force is the Newton (N), named after Sir Isaac Newton, so pressure has the unitsN/m2. A Newton is defined as the force required to give a mass of 1kg an acceleration of 1m/s2. Hence:

P W AAV =

Force = mass ¥ accelerationF = Ma (1.3)

where M represents mass and a is the acceleration. For weight, W, which is the force causedby the acceleration due to gravity, g, this becomes:

Weight = mass ¥ gravity

W = Mg (1.4)

On Earth, gravity, g, is usually taken as 9.81m/s2.

1.1.2 Understanding the difference between mass and weight

❝ OK, so what is the essential difference between mass and weight, and why is it important? ❞It is important to have a clear understanding of the difference between mass and weight,

because without it you will make mistakes in your calculations. The essential difference isthat mass represents the amount of matter in a body, which is constant, so mass stays thesame everywhere in the universe, while weight varies according to the local value of gravitysince W = Mg (equation (1.4) and Fig. 1.2).

❝ So what is mass density and weight density What is meant by relative density? And how heavy is water? ❞

Density, r, is the relationship between the mass, M, of a substance and its volume, V.Thus:

(1.5)r = M V

Hydrostatics 3

Figure 1.2 The concept of weight, which varies according to the local valueof gravity

The density of fresh water (r) is 1000kg/m3. This can be thought of as the mass densityof the water, since it gives the mass per unit volume. Alternatively, the weight (W) per unit volume may be quoted, which is the weight density, w (also called the specific weight). Using equations (1.4) and (1.5), weight density can be expressed in severalways:

(1.6)

Thus the weight density of fresh water is 1000 ¥ 9.81N/m3.Another term you may come across is the relative density (or specific gravity) of a liquid,

s. This is the ratio of the density of a substance, rS, to the density of fresh water, r. Of course,the same value can be obtained by using the ratio of the weight densities (equation (1.6)),since g is the same for both substances. Thus:

(1.7)

where wS is the weight density of the substance. Since s represents a ratio of the mass orweight of equal volumes of the two substances, it has no dimensions. For example, waterhas a relative density of 1.0 while mercury has a relative density of 13.6.

s s w w= =r rS Sor

w W V w Mg V w g= = =or or r


RememberIt is important to realise that water is heavy! Each cubic metre of water weighs 9.81¥ 103 N, that is one tonne. Thus every cubic metre weighs about the same as a largecar.

Box 1.1

Figure 1.3 Illustration of the weight of water

Using relative densityIt is important to remember that s usually has to be multiplied by the density of waterbefore it can be used in your calculations, otherwise the answer you obtain will bewrong, both numerically and dimensionally. For example, the density ofmercury (rM) is 13.6 ¥ 1000kg/m3. Quoting the relative density as 13.6is just a shorter and more convenient way of writing this.

Box 1.2

1.1.3 An application of what you have learned so far – the hydraulic jack

❝ You may not realise it, but you now have a sufficient understanding of hydrostatics tounderstand how a hydraulic jack works. ❞

The hydraulic jack uses two cylinders (Fig. 1.4), one with a large cross-sectionalarea (CSA), A, and one with a small area, a. By using a handle, or somethingsimilar, a small force, f, is applied to the piston in the small cylinder. From equation (1.2),it can be seen that this generates a pressure in the liquid of PAV = f /a. Now one of the prop-erties of a liquid is that it transmits pressure equally in all directions (more of this later), sothis means that the same pressure PAV acts over the whole cross-sectional area (A) of thelarge piston. As a result, the force exerted on the large piston is F = PAVA (equation (1.2)).Because A > a, the output force F > f, even though the pressure of the liquid is the same.Thus the jack acts as a kind of hydraulic amplifier. This simple but extremely useful effectcan be used to lift weights of many tonnes while applying only a relatively small force tothe input end of the jack.

1.2 Hydrostatic pressure and force

❝ Now let us try to determine how we can work out the hydrostatic force, F, on a dam, or on a lock gate, or on the flap gate at the end of a sewer. ❞

The term ‘hydrostatic’ means, of course, that the liquid is not moving. Con-sequently there are no viscous or frictional resistance forces to worry about (seesection 4.1). Also, in a stationary liquid there can be no shear forces, since this would implymovement. The water pressure must act at right angles to all surfaces with which the liquidcomes into contact. If the pressure acted at any other angle to the surface, then there would

Hydrostatics 5

Figure 1.4 A hydraulic jack. The hydraulic pressure that results from applying a smallforce to the small piston is transmitted to the large piston, so enabling a relatively heavyload to be raised

be a component of force along it which would cause the liquid to move. However, this com-ponent is zero when the pressure is normal to the surface since cos90° = 0.Hence in a static liquid the pressure acts at right angles to any surface. Thisfact comes in useful later.

❝ OK, so the pressure acts at 90° to the surface.Please can you now explain why a submarine can only dive to a certain depth, as in all those old war movies? ❞The answer is quite simple. The pressure intensity increases with depth. Beyond a certain

depth the water pressure would crush the hull of the submarine.

❝ But what causes the pressure, and how can you calculate what it is? After all, if you were inthe submarine you would want to know, right? ❞The weight of the water above the submarine causes the pressure. Remember, every cubicmetre of fresh water equals 1 tonne, which is 9810N (that is rg N with r = 1000kg/m3

and g = 9.81m/s2). This makes it quite easy to calculate the pressure. Try thinking of it likethis.

Imagine a large body of fresh water. Then consider a column of the liquid with a planarea of 1m2 extending from the surface all the way to the bottom, as in Fig. 1.6. Now,suppose we draw horizontal lines at one metre intervals from the surface, so that the columnis effectively separated into cubes with a volume of 1m3. Every cube weighs 9.81 ¥ 103 N.Since the pressure on the base of each of the cubes is equal to the weight of all the cubesabove it divided by 1m2 (PAV = W/A), it can be seen that the pressure increases uniformlywith depth. Similarly, if the column of liquid has a total depth, d, then the total weight ofall the cubes is 9.81 ¥ 103 ¥ dN. Dividing this by 1m2 to obtain the pressure on the base ofthe column gives 9.81 ¥ 103 ¥ dN/m2. Therefore, at any depth, h, below the water surfacethe pressure is:

(1.8)

Equation (1.8) shows that there is a linear relationship between pressure, or pressureintensity, and depth. This pressure–depth relationship can be drawn graphically to obtain

P gh= r N m2


Figure 1.5 Typical examples of situations where the hydrostatic force may have to be calculated

a pressure intensity diagram like that in Fig.1.7. This diagram shows the pressure intensityon a vertical surface that is immersed in a staticliquid and which has the same height, h, as thedepth of water. The arrows can be thought of asvectors: they are drawn at 90° to the surface indi-cating the direction in which the pressure acts,while the length of the arrow indicates the relative magnitude of the pressure intensity.When analysing a problem, a pressure intensitydiagram is used to help visualise what is hap-pening, while equation (1.8) provides the meansto calculate the pressure intensity.

The relationship described by equation (1.8)is very useful; it can be used to calculate the pres-sure at any known depth, or alternatively, to calculate the depth from a known pressure. The fact there is a precise relationship betweenpressure and depth forms the basis of manyinstruments that can be used to measure pres-sure, such as manometers, which are describedin Chapter 2.

Now one important point. Figure 1.7 onlyshows the pressure caused by the weight of thewater. This is called the gauge pressure, and is

Hydrostatics 7

Figure 1.6 Variation of pressure withdepth

Figure 1.7 A pressure intensitydiagram corresponding to Fig.1.6

the pressure most often used by engineers. For convenience, gauge pressure measures thepressure of the water relative to atmospheric pressure, that is it takes the pressure of the air around us as zero. Now in reality, the atmosphere exerts a pressure of about 101 ¥ 103 N/m2 on everything at sea level (this is equivalent to the pressure at the bottomof a column of water about 10.3m high, that is a ‘head’ of 10.3m of water). So if we wantto obtain the absolute pressure measured relative to an absolute vacuum, that is the totalpressure exerted by both the water and the atmosphere, we have to add atmospheric pres-sure, PATM, to the gauge pressure (Fig. 1.8). Thus the absolute pressure, PABS, is:

(1.9)

A good way to think of this is that you can measure the height of a table top either fromthe floor, which is the most convenient way, or above sea level (ordnance datum). Similarly,it is more convenient to measure temperature above the freezing point of water than aboveabsolute zero. Consequently in this book we will always use gauge pressures (unless statedotherwise). For future reference, note that under some circumstances, such as in pipelines,a pressure less than atmospheric may occur (Fig. 1.8). This is a negative gauge pressure, –rgh, but equation (1.9) is still valid. Note also that if absolute pressure is used then thegauge pressure intensity diagram shown in Fig. 1.7 will have to have PATM added to it, asshown in Fig. 1.9.

Now try Self Test Question 1.1. A short guide solution is given in Appendix 2, if you need it.

SELF TEST QUESTION 1.1

Oil with a weight density, wO, of 7850N/m3 is contained in a vertically sided, rectangular tankwhich is 2.0m long and 1.0m wide. The depth of oil in the tank is 0.6m.

(a) What is the gauge pressure on the bottom of the tank in N/m2?

(b) What is the weight of the oil in the tank?

P gh PABS ATM2N m= +r


Visualising the size of unitsYou can easily visualise a metre, because it is just over three feet in length, and, ofcourse, you know how long a second is. You may also be aware that a kilogramme isabout 2.2 lb, that is about the equivalent of a bag of sugar.

But do you know how large or small a Newton is?

If you use equation (1.8) to work out the pressure at a depth of 0.3mof fresh water you get P = 1000 ¥ 9.81 ¥ 0.3 = 2943N/m2. So everytime you have a bath at home, parts of your body are being subjectedto almost 3000N/m2. It does not cause any discomfort, in fact you do not even notice.So you may deduce that a Newton is a relatively small unit of force. For this reason itis frequently not worthwhile quoting a value to less than a Newton (the exceptionbeing if you are dealing with very, very small values where accuracy may be affectedby rounding off).

Box 1.3

Hydrostatics 9

Figure 1.8 Relationship between gauge pres-sure and absolute pressure

Figure 1.9 Pressure intensity diagramincluding atmospheric pressure

(c) If the bottom of the tank is resting not flat on the ground but on two pieces of timberrunning the width of the tank, so that each piece of timber has an area of contact withthe tank of 1.0m ¥ 0.1m, what is the pressure on the timber?

1.3 Force on a plane (flat), vertical immersed surface

❝ How do you work out the force on something as a result of the hydrostatic pressure? Say, something like a rectangular gate at the end of a sewer or culvert? ❞OK, there are two thing to remember. First of all, equation (1.2) tells us that F = PAVA, so

a force is a pressure multiplied by an area. However, the second thing we have to remem-ber is that the pressure varies with depth. So, on a vertical surface such as the gate in Fig.1.10, the pressure at the top of the gate is rgh1. At the bottom of the gate the pressure isrgh2. Hence the average pressure on the gate is PAV = (rgh1 + rgh2)/2. Now if we multiplythis by the area of the gate in contact with the water, A, we get the force, F:

(1.10)

For a rectangle, (h1 + h2)/2 is the depth to the centre of the area, that is the vertical depthto the centroid, G, of the immersed surface. This depth is represented by hG, so the expres-sion for the resultant hydrostatic force, F, becomes:

F = rghGA (1.11)

This equation can be applied to surfaces of any shape. For geometrical shapes other than arectangle, the depth to the centroid can be found from Table 1.1. For the full derivation ofequation (1.11), see Proof 1.1 in Appendix 1.

F g h h A= +( )[ ]r 1 2 2

The next paragraph can be helpful in some circumstances, since it reconciles whatcan appear to be different ways to solve a particular problem. However, you may omitit the first time you read the chapter, or if it confuses you.

From equation (1.10), the resultant force, F = average pressure intensity ¥ area of the immersed surface (A). For simple, flat surfaces like that in Fig. 1.10, the average pressure intensity is (rgh1 + rgh2)/2. If A = DL, then equation (1.10) can be written as F = rg[(h1 + h2)/2]DL. The same expression can be obtained by calculating the area of thetrapezoidal pressure intensity diagram in contact with the gate, rg[(h1 + h2)/2]D and multi-plying by the length of the gate, L. This can sometimes provide a useful check that whatyou are doing is correct, or a means of remembering the equation. However, your bestapproach initially is usually to go straight to equation (1.11).


Table 1.1 Geometrical properties of some simple figures

Shape Dimensions Location of the Second momentcentroid, G of area, IG

Rectangle breadth L D/2 from base LD3/12height D

Triangle base length L D/3 from base LD3/36height D

Circle radius R centre of the circle pR4/4

Semicircle radius R 4R/3p from base 0.1102R4

G G

G

Figure 1.10 A vertical gate at the end of a sewer which discharges to a river. The gatehangs from a hinge at the top: (a) side view, (b) front view, (c) pressure intensitydiagram. Note that only the part of the pressure intensity diagram at the same depthas the gate contributes to the hydrostatic force acting on it

1.4 Location of the resultant force on a vertical surface

❝ How do you know where the resultant force, F, acts?I assume that there must be some way of working it out? ❞

Yes, there is a way of calculating wherethe resultant force acts, and normallyyou would work this out at the same timeas the magnitude of the force itself.However, the proof is a bit complicated,so I have put it in Appendix 1 (the secondhalf of Proof 1.1). You can go through itlater if you want to. For the time being,though, let us try to deduce somethingabout where the force must act.

Consider the dam in Fig. 1.11. In thiscase the pressure intensity diagram is triangular, since the gauge pressure variesfrom zero (atmospheric pressure) at thesurface to rgh at the bottom. The averagepressure intensity on the dam is therefore(0 + rgh)/2 or rgh/2. This pressure occursat G, half way between the water surfaceand the bottom of the dam.

But where would the resultant force act?

At G, half way down? Above? Below?

Can you deduce where it would be?

Think of it this way. The resultant force on the dam is the result of the average pressureintensity acting over the area of the dam face in contact with the water. Thelonger the arrows of the pressure intensity diagram, the greater the pressure.The larger the area of the pressure intensity diagram, the greater the force.

Hydrostatics 11

RememberWhenever you are faced with calculating the horizontal hydrostatic force on aplane, vertical immersed surface, the equation F = rghGA is the one to use. This simpleequation can solve a lot of problems. We will also use it later on when we progress tothe force on inclined and curved immersed surfaces. Remember that A is the area ofthe immersed surface in contact with the liquid.

Box 1.4

G

Figure 1.11 Pressure intensity on adam. G is the centroid of the wettedarea, P is the centre of pressure wherethe resultant force acts

Look at the triangular area that forms the top half of the pressure intensity diagram, andcompare it with the area of the trapezoidal bottom half. The area of the bottom part of thediagram is much larger, indicating that the resultant force would act below half depth. Infact, the resultant force acts horizontally through the centroid of the pressure intensitydiagram. For the triangular pressure intensity diagram in Fig. 1.11, this is located at h/3from the base (but note that this is only the case when the pressure intensity diagram istriangular). The point, P, at which the resultant force acts is called the centre of pressure(Fig. 1.11).

With more complex problems, like that in Fig. 1.10, there is no simple rule to give thelocation of P, but if hP is the vertical depth to the centre of pressure then this can be calculated from:

(1.12)

where the value in the brackets gives the vertical distance of P below the vertical depth tothe centroid of the surface, hG. The appropriate expression for the second moment of areacalculated about an axis through the centroid, IG, can be found from Table 1.1. For a rec-tangle IG = LD3/12, where L is the length of the body and D its height. A is the surface areaof the body. The derivation of equation (1.12) can be found in Appendix 1.

Examples 1.1 and 1.2 show how equations (1.11) and (1.12) are used to solve a coupleof typical problems, one involving the flap gate at the end of a sewer and the other a lockgate. Study these carefully and then try Self Test Question 1.2 (a short solution is given inAppendix 2).


A rectangular culvert (a large pipe) 1.8m wide by 1.0m high discharges to a river. At the endof the culvert is a rectangular gate which seals off the culvert when the river is in flood (as inFig. 1.10). The gate hangs vertically from hinges at the top. If the flood level in the river rises to1.9m above the top of the gate, calculate the magnitude and location of the resultant hydro-static force on the gate caused by the water in the river.

EXAMPLE 1.1

A rectangular gate is 2m wide and 3m high. It hangs vertically with its top edge 1m below thewater surface. (a) Calculate the pressure at the bottom of the gate. (b) Calculate the

h I Ah hP = ( ) +G G G


Note that the centre of pressure, P, is always below the centroid, G, ofthe surface in contact with the water. In many problems it is not obviouswhere P is located, so this has to be calculated using equation (1.12). However,as the depth of immersion of the surface increases, P moves closer to G. This isapparent from equation (1.12): the distance between P and G is (hP - hG). If A andIG have constant values, then the equation can be rearranged as (hP - hG) = C/hG whereC represents the value of the constants. Thus (hP - hG) decreases as hG increases.

Box 1.5

resultant hydrostatic force on the gate. (c) Determine the depth at which the resultant force acts.

(a) From equation (1.8), P = rgh

(b) From equation (1.11), F = rghGA

(c) From equation (1.12)

EXAMPLE 1.2

A lock on a canal is sealed by a gate that is 3.0m wide. The gate is perpendicular to the sidesof the lock. When the lock is used there is water on one side of the gate to a depth of 3.5m,and 2.0m on the other side. (a) What is the hydrostatic force of the two sides of the gate? (b)At what height from the bed do the two forces act? (c) What is the magnitude of the overallresultant hydrostatic force on the gate and at what height does it act?

(a) Using F = rghGA

(b) Since both pressure intensity diagramsare triangular, both forces act at one-third depth from the bed:

(c) Overall resultant force FR = F1 - F2

FR = 121.40 ¥ 103 N

Taking moments about O to find theheight, YR, of the resultant:

121.40 ¥ 103 ¥ YR = 180.26 ¥ 103 ¥ 1.17 - 58.86 ¥ 103 ¥ 0.67

YR = 1.41m above the bed.

Y

Y1

2

3 5 3 1 17

2 0 3 0 67

= == =

. .

. .

m

m

F

F

1

3

2

3

1000 9 81 3 5 2 3 5 3 0

180 26 10

1000 9 81 2 0 2 2 0 3 0

58 86 10

= ¥ ¥ ( ) ¥ ¥( )= ¥= ¥ ¥ ( ) ¥ ¥( )= ¥

. . . .

.

. . . .

.

N

N

h Ah h

LD

A h

h

P G G G

G

G

P

where m

and are as above

so

m

= ( ) += = ¥ =

= ¥( ) +=

I

I 3 3 412 2 3 12 4 50

4 50 6 2 50 2 50

2 80

.

. . .

.

Now m

m

Thus

N

Gh

A

F

= + ( ) == ¥ == ¥ ¥ ¥= ¥

1 3 2 2 50

2 3 6

1000 9 81 2 50 6

147 15 10

2

3

.

. .

.

Therefore

N m

P = ¥ ¥ +( )= ¥

1000 9 81 3 1

39 24 103 2

.

.

Hydrostatics 13

G

G

Figure 1.12

Figure 1.13

The value of YR obtained in part (c) of theabove example may have surprised you. Possibly you expected YR to be somewherebetween 0.67m and 1.17m, whereas it isactually 1.41m. This is a situation where thepressure intensity diagrams (which are notreally needed to conduct the calculations)can be used to visualise what is happening.In Fig. 1.13 the slope of the two pressureintensity triangles is the same, since thewater has the same density on both sides ofthe gate. Thus if the triangle on the right issubtracted from the triangle on the left, theresult is as in Fig. 1.14. This is the net pres-sure intensity on the gate. The diagram ismore rectangular than either of the trianglesso, employing a similar argument to thatused with Fig. 1.11, this indicates that YR would be higher above the base than either Y1 or Y2.


Figure 1.14 Net pressure intensitydiagram for Example 1.2

Figure 1.15 The dam on the bottom left of the photograph isholding back a considerable quantity of water. The force exerted bythe water on the structure must be calculated before the dam can bedesigned. Many lay people believe, incorrectly, that the greater thevolume of water stored behind the dam, the larger the force on thestructure. This is not the case. Equation (1.8) indicates that the pres-sure on the dam is related to the depth of water, while the force isthe product of the average pressure and the area of the dam incontact with the water (equation (1.2))

1.5 Force on a plane, inclined immersed surface

❝ I understand how to work out the force on a flat vertical surface, buthow about one that is inclined at an angle to the water surface? Surely this is much more difficult? ❞The answer is ‘no’. The calculations are still very simple and almost identical to those

above. There are three things that you should remember when analysing these situations:

(1) The resultant force acts at right angles to the immersed surface.

(2) The hydrostatic pressure on the inclined surface is still caused only by the weight ofwater above it, so P = rgh.

(3) When calculating the location of the resultant force on an inclined surface, alwaysuse equation (1.13) (never equation (1.12), see below).

To illustrate simply that the resultant force can be calculated in the same way as for avertical surface, consider this. The pressure at the top of the rectangular, inclined surface inFig. 1.16a is rgh1 while that at the bottom is rgh2. Thus the average pressure intensity onthe surface is rg(h1 + h2)/2, or rghG since hG = (h1 + h2)/2. The resultant force is the averagepressure intensity multiplied by the area of the surface, and since the pressure acts at rightangles to the inclined surface the actual area, A, should be used. Thus F = rghGA, as in equa-tion (1.11). Note that the inclination of the surface is automatically taken into account bythe value of hG. For example, if h1 in Fig. 1.16a is fixed, and the surface rotated upwardsabout its top edge, then hG will decrease so that hG = h1 when it is horizontal. Similarly, themaximum possible value of hG would be obtained when the surface is vertical.

One other important point, the resultant force on the inclined surface, F, has compo-nents in both the vertical and horizontal directions. These can be calculated separately, asin section 1.6 and Example 1.4, but the procedure outlined above is quicker for flat (plane)surfaces.

To calculate the location of the resultant force, the following equation should be used:

Hydrostatics 15

G

G G

G

Figure 1.16 (a) Force on an inclined surface. (b) When the surface is inclinedalways use the dimensions LG and LP with equation (1.13) (never the verticaldimensions hG and hP with equation (1.12))


Using equations (1.12) and (1.13)Remember that when you have an inclined surface, always use equation (1.13) to findthe location of the resultant force. You can then calculate the vertical depth of thecentre of pressure, P, below the surface from LP if you want to (see Example 1.3). Nevertry to do this by using equation (1.12) instead of equation (1.13). The reason for thisis that IG is calculated in the plane of the surface. For example, with a rectangularinclined surface, IG is still taken as LD3/12 where D is the actual inclined dimension ofthe surface, so the remainder of the terms in equation (1.13) must have the same ori-entation for consistency (see the derivation of the equation in Appendix 1). The sameargument applies to vertical surfaces and equation (1.12).

Box 1.6

(1.13)

This is similar to equation (1.12), but the inclined lengths, LP and LG, are used to denotethe location of the centre of pressure and centroid of surface (Fig. 1.16b), not the verticaldepths.

EXAMPLE 1.3

A sewer discharges to a river. At the end of the sewer is a circular gate with a diameter (D) of0.6m. The gate is inclined at an angle of 45° to the water surface. The top edge of the gate is1.0m below the surface. Calculate (a) the resultant force on the gate caused by the water in theriver, (b) the vertical depth from the water surface to the centre of pressure.

(a) Vertical height of gate = 0.6sin45° = 0.424m

Vertical depth to G = hG = 1.000 + 0.424/2 = 1.212m

Area of gate, A = pD2/4 = p0.62/4 = 0.283m2

F = rghGA = 1000 ¥ 9.81 ¥ 1.212 ¥ 0.283 = 3365N

L I AL LP = ( ) +G G G

G G

G

Figure 1.17 An inclined, circular gate at the end of a sewer

(b) Slope length to G, LG = 1.212/sin45° = 1.714m

For a circle (Table 1.1) IG = pR4/4 = p(0.3)4/4 = 0.0064m4

Vertical depth to P, hP = LP sin45° = 1.727sin45° = 1.221m

L AL LP G G G m= ( ) + = ¥( ) + =I 0 0064 0 283 1 714 1 714 1 727. . . . .

Hydrostatics 17

Figure 1.18 This vertical lift gate on the Old Bedford River pro-vides another example of where the engineer may be requiredto calculate the resultant hydrostatic force. If the horizontal forceis large it may be difficult for a vertical lift gate to slide up anddown, the gate being pushed hard against the guide channels.In the Fens of East Anglia much of the drainage is controlled byman, using pumps and sluice gates like the one above

1.6 Force on a curved immersed surface

❝ I suppose that you are now going to tell me that working out the force on a curvedsurface is just as easy as calculating the force on a flat or inclined surface? ❞Well, the calculations are perhaps a little longer, but no more difficult. Let me clarify this

by breaking the analysis of the force on an immersed curved surface down into steps.

(1) The resultant force (F) acts at right angles to the curved surface. This force can bethought of as having both a horizontal (FH) and a vertical (FV) component (Fig. 1.19).

(2) To calculate the horizontal component of the resultant force (FH), project the curvedsurface onto a vertical plane, as in Fig. 1.20. This effectively is what you would see ifyou looked at the curved surface from the front. Calculate the force on this projectedvertical surface as you would any other vertical surface using FH = rghGA, where A is thearea of the projected vertical surface (not the area of the actual curved surface).

(3) Calculate the vertical component of the resultant force (Fig. 1.21) by evaluating theweight of the volume (V) of water above the curved surface, that is:

FV = rgV (1.14)

(4) The resultant force, F, is given by:

(1.15)

(5) The direction of the resultant force (Fig. 1.22) can be found from:

(1.16)

This gives the angle, f, of the resultant to the horizontal.

Remember, the resultant also acts at 90° to the curved surface, so it passes through thecentre of curvature (for example, the centre of the circle of which the surface is a part).

tanf = F FV H

F F F= +( )H V2 2 1 2


Figure 1.19 Pressure intensityon a curved surface. F passesthrough the centre of curvature,C

Figure 1.20 Projection of the curved surface ontoa vertical plane

(6) The above steps enable the resultant force on the upper side of the surfaceto be calculated. Always remember that there is an equal and opposite forceacting on the other side of the surface. This fact comes in useful later,because it is always easier to calculate the force on the upper surface, even if this is notthe surface in contact with the water.

EXAMPLE 1.4

A surface consists of a quarter of a circle of radius 2.0m (Fig. 1.22). It is located with its top edge1.5m below the water surface. Calculate the magnitude and direction of the resultant force onthe upper surface.

Step 1 Project the curved surface onto a vertical plane and calculate FH

FH = rghGA where A is the area of the projected vertical surface.Since the length of the gate is not given, calculate the force per metre length with L = 1.0m.Thus A = 2 ¥ 1.0 = 2.0m2 per metre lengthThe value of hG is that for the projected vertical surface: hG = 1.5 + (2.0/2) = 2.5m.

Step 2 Calculate FV from the weight of water above the surfaceFV = rgV where V is the volume of water above the curved surface. Again using a 1m length:

Step 3 Calculate the magnitude and direction of the resultant force

The resultant passes through the centre of curvature, C, at an angle of 50.8°.

F F F

F F

= +( ) = +( ) = ¥

= ( ) = ( ) = ∞- -

H V

V H

N m2 2 1 23 2 2 1 2 3

1 1

10 49 05 60 23 77 68 10

60 23 49 05 50 8

. . . .

tan tan . . . .f

V

F

= ¥ ¥( ) + ¥ ¥( ) == ¥ ¥ = ¥

1 4 2 0 1 0 2 0 1 5 1 0 6 14

1000 9 81 6 14 60 23 10

2 3

3

p . . . . . .

. . .

m per metre length

N m.V

FH N m= ¥ ¥ ¥ = ¥1000 9 81 2 5 2 0 49 05 103. . . .

Hydrostatics 19

Figure 1.21 The verticalcomponent of force, FV,caused by the weight ofwater above the surface

Figure 1.22 The direction ofthe resultant force, F, whichmust also pass through C


An open tank which is 4.0m wide at the top containsoil to a depth of 3.4m as shown in Fig. 1.24. Thebottom part of the tank has curved sides which haveto be bolted on. To enable the force on the bolts to bedetermined, calculate the magnitude of the resultanthydrostatic force (per metre length) on the curved sur-faces and its angle to the horizontal. The curved sec-tions are a quarter of a circle of 1.5m radius, and theoil has a relative density of 0.8.

❝ I understand Example 1.4, but when you described the steps used to analyse the force on a curved surface, in point 6 you said something about always analysing the upper side of the surface. You said that we should do this even if the upper side of thesurface was not in contact with the water. How can this be right? No water, no hydrostatic force I would have thought. ❞I suppose this is one of the tricks you have to learn to make hydraulics easy. Think of it

like this. The curved surface in Fig. 1.25 is an imaginary one, drawn in a large body of staticliquid. Now it is possible to calculate the force on the upper side of this imaginary surface


G

G

Figure 1.23

Figure 1.24 Tank for SelfTest Question 1.3.

using the same procedure as in Example 1.4.However, the surface is only imaginary, so whatresists this force? Something must because theliquid is static, that is not moving. The answer isthat there is an equal and opposite force acting onthe underside of the imaginary surface, so thatthis balances the force on the top. It does notmatter which force you calculate, because they arenumerically equal, but it is easier to calculate thaton the upper surface. The same is true with realsurfaces. Remember this when you encounterproblems like Example 1.5 with air on the uppersurface and water underneath.

Something to note from Example 1.5 is that thevertical component of the resultant force acts upwards, which means that itis a buoyancy force. Sometimes there is a tendency to think of a buoyancyforce as being different from the hydrostatic force, but in fact they are thesame thing. The buoyancy force on a body, such as a ship, is the result of the hydrostaticpressure acting on the body. This will be explored in more detail in section 1.7.

EXAMPLE 1.5

A radial gate whose face is part of a circle of radius 5.0m holds back water as shown in Fig. 1.26.The sector of the circle represented by the gate has an angle of 30° at its centre. Water standsto a depth of 2.0m above the top of the upstream face of the gate. The other side of the gateis open to the atmosphere. Determine the magnitude and direction of the resultant hydrostaticforce. The gate is 3.5m long.

Step 1 Project the curved surface onto a vertical plane and calculate FH

Vertical height of projection = BC = 5.0cos60° = 2.5m.

Hydrostatics 21

Figure 1.25 Equal and oppo-site forces on a surface

Figure 1.26

FH = rghGA = 1000 ¥ 9.81 ¥ 3.25 ¥ 8.75 = 278.97 ¥ 103 N.

Step 2 Calculate the vertical component, FV, from the weight of water above the surfaceIn this case calculate the weight of water that would be above the gate if it was not there, thatis the weight of the water displaced by the gate. This is shown in the diagram as AEFH. The widthof this area, DE, can be calculated as follows:

AB = 5.0sin60° = 4.33m, so DE = 5.00 - 4.33 = 0.67m.

The area of ADE (and subsequently AEFH) can be found using geometry, as follows.Area sector ACE = (30/360)ths of a 5.0m radius circle = (30/360)p5.02 = 6.54m2.Area triangle ACD = (1/2) ¥ 4.33 ¥ 2.5 = 5.41m2.Area ADE = 6.54 - 5.41 = 1.13m2.Therefore, the total area AEFH = 1.13 + (0.67 ¥ 2.00) = 2.47m2.Volume of water displaced, V = 2.47 ¥ 3.5 = 8.65m3.

FV = rgV = 1000 ¥ 9.81 ¥ 8.65 = 84.86 ¥ 103 N.

Step 3 Calculate the magnitude and direction of the resultant force

Resultant acts at 16.9° to the horizontal passing upwards through the centre of curvature, C.

1.7 Variation of pressure with direction and buoyancy

❝ We have already discussed the fact that hydrostatic pressure acts at right angles to anysurface immersed in it, so it follows that on the underside of a horizontal surface the resultant

force is acting vertically upwards. This is a buoyancy force, and it is caused simply by thehydrostatic pressure on the surface. Try thinking it through like this. ❞

Imagine a sphere some distancebelow the water surface as in Fig.1.27. The hydrostatic pressure acts at90° to the surface of the sphere.Looking at this two-dimensionally, asin the diagram, then the smallestpressure intensity is rgh1 at the top,and the largest is rgh2 at the bottom.

Now, consider what would hap-pen if the diameter of the spheregradually decreased so that the dif-ference between h1 and h2 decreased.This would cause the two pressuresrgh1 and rgh2 to become closernumerically. If the diameter of thesphere continued to decrease until it

f = tan V H- -( ) = ( ) = ∞1 1 84 86 278 97 16 9F F tan . . .

F F F= +( ) = +( ) = ¥H V N.2 2 1 23 2 2 1 2 310 278 97 84 86 291 59 10. . .

h AG m and m= + ( ) = = ¥ =2 0 2 5 2 3 25 2 5 3 5 8 75 2. . . , . . . .


h3

Figure 1.27 Pressure on a sphere

Hydrostatics 23

became infinitesimally small then the difference between h1 and h2 would be negligible sothat rgh1 = rgh2. By the same argument, the pressure intensity in any other direction, suchas rgh3 acting horizontally, would also have the same value (see Proof 1.2, Appendix 1).

Thus the pressure at a point in a static liquid acts equally in all directions,up, down, sideways or whatever.

❝ That’s all very interesting, but does it have any practical purpose, and how can I work out the value of the buoyancy force? ❞Yes it has a practical purpose, and working out the value of the buoyancy force is quite

easy. In fact you can do so using what you have already learnt. Let me illustrate by using asimilar situation to the sphere in Fig. 1.27, but this time we will make the body a cubebecause it simplifies the calculations. The cube is shown in Fig. 1.28. The pressure intensi-ties on the vertical sides cancel each other out, so only the pressure acting on the top andbottom faces need be considered. Let the area of each face of the cube be A. Then:

Assuming the top and bottom faces arein a horizontal plane, then the pressureis constant over the face so:

Pressure on the top face = rgh1

Pressure on the bottom face = rgh2

The force on the face is equal to the pres-sure multiplied by the area of the face,A. So:

Force on the top face = rgh1AForce on the bottom face = rgh2A

Since h2 > h1 there will be a net forceacting vertically upwards, F. This is:

Now (h2 - h1)A is the volume of thecube, V, so:

F = rgV (1.14)

F gh A gh A g h h A= - = -( )r r r2 1 2 1Figure 1.28 Buoyancy force, F

RememberThe buoyancy force, F, acts vertically upwards through the centre of gravity of thedisplaced liquid (such as the centre of the cube). The point at which F acts is calledthe centre of buoyancy, B. The force, F, is equal to the weight of the volume of liquiddisplaced by the body, that is rgV. This is known as Archimedes’ Principle. Now goback and look at Step 2 of Example 1.5. You should be able to see that a buoyancyforce is just the vertical force caused by hydrostatic pressure. See also Chapter 3 andBox 3.1.

Box 1.7

❝ When we analysed the buoyancy force on the cube in Fig. 1.28 we only considered thehydrostatic forces acting vertically on it. The weight of the cube was irrelevant. However, if wewanted to know whether or not the completely immersed cube would float or sink, we wouldhave to compare the weight of the cube (W ) with the buoyancy force (F), remembering that

weight is a force. ❞

W = weight density of cube material ¥ volume = rSgV N acting vertically downwards.

F = weight of liquid displaced by the cube = rgV N acting vertically upwards.

Since g and V are the same, it follows that if the density of the substance, rS, that forms thecube is greater than the density of the liquid, r, then the cube would sink (W > F). Conversely, if rS < r, then the cube would float (F > W). If rS = r then the cube has neutralbuoyancy and would neither float nor sink, but would stay at whatever depth it was located(F = W).

The analysis above explains why a concrete or steel cube would sink, and a cork or poly-styrene cube would float. However, this assumes that the cube is solid. If the cube washollow, its average density would have to be used in the calculations, not the density of thematerial from which it was made. Submarines provide aninteresting example, because they must be able to sinkand, more importantly, rise to the surface again. This canbe achieved by adjusting the average density of the sub-marine, by changing its weight by admitting or expellingwater from tanks on the outside of the hull.

Floating bodies, such as ships and the pontoon inExample 1.7, are quite easy to analyse. If the depth ofimmersion is constant, then obviously W and F in Fig.1.29 are exactly equal (otherwise the body would moveup or down). Hence the starting point for many calcula-tions involving floating bodies is:

(1.17)

Thus a floating body of weight W displaces a volume of water (V) that has a weight (rgV)equal to its own. Since W = Mg this can also be written as:

Mg = rgV (1.18)

or M = rV (1.19)

Therefore it is also true to say that a floating body of mass M displaces a volume of water(V) that has a mass (rV) equal to its own. Of course, equation (1.19) is a rearrangement ofequation (1.5). Remember to use W with the weight density (rg) and M with the massdensity (r). Typically the body’s weight or mass is known, so the relationships above allowthe volume of water (V) displaced by a floating body to be calculated. Then for pontoonswhich are rectangular in plan and cross-section like those in Fig. 3.1:

depth of imersion = V/plan area (1.20)

By now it should be apparent that a solid steel cube sinks, but a ship made from steel platesfloats because it is hollow and can displace a much larger volume of water (V) that has a

W F

W gV

==or r


Figure 1.29 Floatingbody

Hydrostatics 25

mass (or weight) equal to that of the ship. This is why we say that a ship has a displace-ment of 10000 tonnes, for instance. When W = F the depth of immersion is constant, butif W is increased by adding cargo the ship settles deeper in the water, increasing its dis-placement and consequently F, until W = F again.

From 1876 onwards, British ships have had a Plimsoll line painted on their hull to indi-cate the maximum safe loading limit. Since the density of water changes according to tem-perature and salt content, the Plimsoll line includes marks for sea or fresh water, winter orsummer, in tropical or northern waters.

Figure 1.30 Lock gates provide another exampleof where it may be necessary to calculate hydrosta-tic forces. The buoyancy, depth of immersion andfreeboard (the distance from the deck to the water-line) of the barge may also be the subject of an engi-neer’s calculations


Try this – amaze your friendsGet an empty fizzy drink bottle, fill it completely with water and put a sachet ofketchup in it (Fig. 1.31a). You need one that just floats, so you may have to try a fewdifferent types until you find one that works. Now challenge your friends to concen-trate their minds and use the power of thought to make the sachet sink. Unless theyreally do have telekinetic powers, they won’t be able to do it of course. Now here’sthe trick. When it is your turn, make sure you have your hands around the bottle, andgently squeeze it. Try to disguise the fact you are doing this. If you squeeze hardenough the sachet will sink, and you can claim to have a better brain than all of yourfriends combined.

The reason the sachet sinks is as follows. A body in water has two forces acting onit: its weight (W) acting vertically down and the buoyancy force (F = rgV ) acting ver-tically up. The weight of the sachet cannot change, so W is constant. However, Fdepends upon the volume (V ) of water displaced by the sachet. When you squeezethe bottle you are exerting pressure on the water inside. The water is incompressible,but the air in the sachet can be compressed. So by compressing the air, V is reducedand so is F. When W > F the sachet sinks. When you stop squeezing F > W so the sachetrises.

Human divers can control their buoyancy and move up and down like this, eitherby inflating or deflating their dry suits or by controlling the amount of air in theirlungs. Usually a Cartesian diver consists of a small length of open ended glass tubingwith a bubble at one end (Fig. 1.31b). It can be used instead of the sachet and worksin the same way.

Box 1.8

W

Air

F

Figure 1.31 (a) Alternative Cartesian diver using asachet of sauce. Squeezing and releasing the bottlemakes the diver sink and then rise. (b) Conventionalglass diver

(b)(a)

EXAMPLE 1.6

A pipe which will carry natural gas is to be laid across an estuary which is open to the sea. Theweight of the pipe is 2360N per metre length and its outside diameter is 1.0m. The weight ofthe gas can be ignored. The density of sea water is 1025kg/m3. Determine whether the pipe willremain on the sea bed or float. If it does float,what force would be required to hold the pipe onthe sea bed?

The maximum buoyancy force occurs when thepipe is fully submerged. Thus:

Buoyancy force, F = rgV.

The net force acts upwards since F > W.The pipe would float and a force of at least5533N/m would be required to hold it down.

EXAMPLE 1.7

A pontoon which is being used to conduct some construction work on a pier built into the seahas a mass of 50 tonnes (1 tonne = 1000kg). The pontoon is rectangular in plan and cross-section. Its length is 10m, its width 5m, and its sides are 2m high. The density of sea water(rSW) is 1025kg/m3. (a) Determine the volume of water displaced by the pontoon. (b) Deter-mine the depth of immersion and the freeboard of the pontoon. (c) Determine the buoyancyforce on the pontoon.

(a) A floating body displaces its ownmass (or weight) of water, so fromequation (1.19), volume displaced =mass of pontoon/rSW = 50000/1025= 48.78m3.

(b) Depth of immersion (the depth inthe water) = volume displaced/planarea = 48.78/(10 ¥ 5) = 0.98m.

Freeboard = (2.00 - 0.98) = 1.02m.

(c) Buoyancy force = weight of waterdisplaced = weight of pontoon = 50000 ¥ 9.81 = 490.5 ¥ 103 N.

(Alternatively F = rgV = 1025 ¥ 9.81¥ 48.78 = 490.5 ¥ 103 N)

Therefore, net force on pipe

N m

= -( )=

7893 2360

5533

Weight of pipe, N m.W = 2360

F = ¥ ¥ =1025 9 81 0 785 7893. . .N m

V = ¥ ( ) =1 1 0 4 0 7852 3p . . m m length.

Hydrostatics 27

Figure 1.32

Figure 1.33

1.8 The hydrostatic equation

The hydrostatic equation is really a statement of what, by now, should be obvious to you.Nevertheless, it can be useful, so the meaning of the equation is considered below whilethe derivation of the equation can be found in Appendix 1.

Basically, the hydrostatic equation states that the change in pressure intensitybetween two levels of a homogeneous (uniform) liquid is proportional to the verticaldistance between them.

Consider points 1 and 2 at some dis-tance below the surface as in Fig. 1.34. Thistime let us measure the depth of the pointsfrom the bottom (not from the surface) andlet these distances be denoted by z1 and z2.

The difference in pressure between the twopoints is (P2 - P1) where:

(1.21)

Equation (1.21) shows that the dif-ference in pressure between two points isequal to the vertical distance (z2 - z1)between them. However, the equation ismore useful when rearranged so:

(1.22)

This equation contains four of the six terms of the Bernoulli (or energy) equation that willbe discussed in Chapter 4. The two terms that are missing from the Bernoulli equation arethe velocity heads (V2/2g), which is logical since the velocity (V) is zero in a static liquid.

When we start considering pressure measurement using manometers in the next chapterwe will be using equations (1.21) and (1.22), or at least the meaning of the equations if notthe actual equations themselves. Perhaps you can see from the equations that if you knowthe pressure (say P1) at some point in a static liquid, then you can calculate the pressure (P2)at any other point so long as you can measure the vertical distance between them. Mano-meters are designed to enable the pressure difference (P2 - P1) to be determined from thedifference in the height of two columns of liquid (z2 - z1), knowing the weight density ofthe liquid rg. Do not worry if you do not fully understand this at the moment, sincemanometers are explained in the next chapter.

1.9 Stratified fluids

❝ How do you calculate the pressure if you have two liquids of different density? Does this make things more difficult? ❞

P g z P g z2 2 1 1r r( ) + = ( ) +

P P g z z2 1 2 1-( ) = - -( )r

P P g d z pg d z

g d z d z

g z z

2 1 2 1

2 1

2 1

-( ) = -( ) - -( )= - - +( )= - +( )

rrr

Pressure at point 1,

Pressure at point 2,

P gh g d z

P gh g d z1 1 1

2 2 2

= = -( )= = -( )

r rr r


Figure 1.34 Pressure intensity at twopoints

Well, again this is nothing new. If you look back to section 1.2 you will see that we dis-cussed the fact that hydrostatic pressure is caused simply by the weight of the liquid abovethe point (or surface) that we are considering. This is still true when you have two or moreliquids of different densities. All youhave to do is work out the weights (orpressures) of the liquids one at a timethen add them together. Let us analysethe situation in Fig. 1.36.

Say that the column of liquid has aplan area of A m2. The weight of theupper block of liquid is given by:

W1 = weight density ¥ volume

= r1gh1A

Similarly, the weight of the lowerblock is:

W2 = r2gh2A

Total weight WT = W1 + W2

= r1gh1A + r2gh2A

Now equation (1.1) told us that:

pressure = weight area

Hydrostatics 29

The equal level, equal pressure principleOne final thing about the hydrostatic equation, which again is a statement of theobvious, is that at a constant depth (or height z in the case of Fig. 1.35) the pressureis constant. It has to be since P = rgh. However, this gives rise to the ‘equal level, equalpressure’ principle. This simply states that if you draw a horizontal line in a continu-ous body of static, uniform fluid then the pressure is the same anywhere on that line.The meaning and significance of this will be clearer if you look at Fig. 1.35. Again, thisprinciple is used with manometers and will be used in the next chapter. However,remember the liquid must have a uniform density (otherwise see section 1.9).

Box 1.9

Figure 1.35 Equal level, equal pressure prin-ciple. The broken line is horizontal and theliquid has a constant density, so the hydro-static pressure is constant along the line andP1 = P2 = P3

Figure 1.36 A stratified liquid with layersof density r1 and r2


RememberBecause the liquid is stratified and has two different densities, the pressure inten-sity diagram does not have the same gradient over the whole depth (as it did in Fig.1.7, for instance). Instead, there is a change in gradient at the interface between thetwo liquids. However, within a particular liquid the gradient is uniform. Figure 1.37bin Example 1.8 provides an illustration of this.

Box 1.10

so the total pressure, PT, at the base of the column of liquid is:

(1.23)

❝ We have analysed many different situations in this chapter. To help you to remember how to approach the different types of problem, I have provided a summary for you at

the end of the chapter. This may prove useful when revising or when tackling the revision questions. ❞

EXAMPLE 1.8

A tank with vertical sides contains both oil and water. The oil has a depth of 1.5m and a rela-tive density of 0.8. It floats on top of the water, with which it does not mix. The water has adepth of 2.0m and a relative density of 1.0. The tank is 3.0m by 1.8m in plan and open to theatmosphere. Calculate (a) the total weight of the contents of the tank; (b) the pressure on thebase of the tank; (c) the variation of pressure intensity with depth; (d) the force on the side ofthe tank.

(a) WT = (r1gh1 + r2gh2)A

Plan area A = 3.0 ¥ 1.8 = 5.4m2

WT = (0.8 ¥ 1000 ¥ 9.81 ¥ 1.5 + 1.0 ¥ 1000 ¥ 9.81 ¥ 2.0)5.4= (11772 + 19620)5.4 = 169517N

(b) Total pressure at base of tank = WT/A = 169517/5.4 = 31392N/m2

(c) Pressure at the surface = atmospheric = 0

Pressure at the bottom of the oil = r1gh1 = 11772N/m2

Total pressure at the bottom of the tank = 31392N/m2

The pressure intensity diagram is shown in Fig. 1.37b.

(d) The side of the tank is 3.0m long. The force on the side of the tank can be obtained fromequation (1.2) by multiplying the area of the tank in contact with each of the liquids by theaverage pressure intensity of the particular liquid.

P W A

P gh ghT T

T

== +r r1 1 2 2

Average pressure of the oil = (0 + 11772)/2 = 5886N/m2

Force due to the oil = 3.0 ¥ 1.5 ¥ 5886 = 26487N

Average pressure of the water = (11772 + 31392)/2 = 21582N/m2

Force due to the water = 3.0 ¥ 2.0 ¥ 21582 = 129492N

Total force on the side = 26487 + 129492 = 155979N

Hydrostatics 31

Figure 1.37 (a) Tank containing a stratified liquid, and (b) the corre-sponding pressure intensity diagram


G

GG

GG

G

G

G G G

G

G

G

G

G

G

Summary

G

G

G

G

G

G

G

G

G G G

G G

Hydrostatics 33

Revision questions

the river. The gate has a radius of 0.5m, andduring a flood the hinge is 3.5m below the watersurface. What is the force exerted by the flood-water on the gate, and at what depth from thesurface does it act?

[(a) 163.16 ¥ 103 N at 4.461m; (b) 30.82 ¥ 103 N at 4.015m]

1.7 A gate at the end of a sewer measures 0.8mby 1.2m wide. It is hinged along its top edge andhangs at an angle of 30° to the vertical, this beingthe angle of the banks of a trapezoidal riverchannel. (a) Calculate the hydrostatic force on the gate and the vertical distance between the centroid of the gate, G, and the centre of pressure,P, when the river level is 0.1m above the top of thehinge. (b) If the river level increases to 2.0m abovethe hinge, what is the force and the distance GPnow? (c) Has the value of GP increased ordecreased, and why has it changed in this manner?

[(a) 4.21 ¥ 103 N, 0.090m; (b) 22.10 ¥ 103 N, 0.017m]

1.8 A circular gate of 0.5m radius is hinged so thatit rotates about its horizontal diameter, that is itrotates about a horizontal line passing through thecentroid of the gate. The gate is at the end of apipe discharging to a river. Measured above thecentroid of the gate, the head in the pipe is 6.0mwhile the head in the river is 2.0m. Assuming thatthe gate is initially vertical: (a) calculate the forceexerted by the water in the pipe on the gate, andthe distance GP between the centre of the gate, G,and the centre of pressure, P; (b) calculate theforce exerted by the river water on the gate, andthe distance GP; (c) by taking moments about thehinge, using the results from above, determine thenet turning moment on the gate caused by thetwo forces acting at their respective centres of pressure on opposite sides of the gate. Explain youranswer.

[(a) 46228.5N at 0.0104m; (b) 15409.5N at 0.0312m;

(c) 0 exactly, allowing for rounding errors]

1.1 Define clearly what is meant by the following,and give the appropriate units in each case: (a)pressure; (b) force; (c) weight; (d) gravity; (e) mass;(f) mass density; (g) weight density; (h) relativedensity; (i) hydrostatic pressure; (j) buoyancy force.

1.2 (a) Explain what is meant by gauge pressureand absolute pressure. (b) What is the approximatenumerical value of atmospheric pressure expressedin N/m2 and as a head of water? (c) Calculateatmospheric pressure expressed as a head ofmercury (the relative density of mercury is 13.6)

[(c) 0.76m]

1.3 A rectangular tank is 1.0m long and 0.7mwide and contains fresh water to a depth of 0.5m.(a) What is the gauge pressure at the bottom ofthe tank in N/m2? (b) What is the absolute pressureat the bottom of the tank?

[4905N/m2; 105905N/m2]

1.4 For the tank in question 1.3, using gauge pres-sure, calculate (a) the mean pressure intensity onthe 0.7m wide end of the tank; (b) the mean pres-sure intensity on the 1.0m long side of the tank;(c) the force on the end of the tank; and (d) theforce on the side.

[2453N/m2; 2453N/m2; 858N; 1226N]

1.5 A dam that retains fresh water has a verticalface. Over a one metre length of the face at thecentre of the valley the water has a depth of 38m.(a) Calculate the resultant force on this unit lengthof the face. (b) At what depth from the surfacedoes the resultant force act?

[7083 ¥ 103 N; 25.33m]

1.6 (a) A rectangular culvert 2.1m wide by 1.8mhigh discharges to a river channel as in Fig. 1.10.At the end of the culvert is a vertical flap gatewhich is hinged along its top edge, the gatehaving the same dimensions as the culvert. Duringa flood the river rises to 3.5m above the hinge.What is the force exerted by the floodwater on thegate, and at what depth from the surface does itact? (b) A circular gate, also hinged at the top,hangs vertically at the end of a pipe discharging to


1.9 A gate which is a quarter of a circle of radius4.0m holds back 2.0m of fresh water as shown inthe diagram.

board? (b) What additional weight will be requiredto sink the structure onto the sea bed if the depthof water is 5.3m, assuming the structure is water-tight? (c) If the additional weight is to be providedby a blanket of sand (density 2600kg/m3), how thick must the layer of sand be? (1 tonne =1000kg).

[(a) yes, 4.07m, 3.93m; (b) >22352 ¥ 103 N;

(c) >0.5m]

1.12 (a) Explain what is meant by a stratified fluid.(b) A pressure transducer is used to measure thehydrostatic pressure on the sea bed in a tidalestuary. The water in the estuary is stratified at the point where the measurement is taken, withfresh water (1000kg/m3) overlying saline water(1025kg/m3). Water sampling shows that the freshwater extends from the water surface to a depth of2.7m. If the transducer indicates a gauge pressureof 69.73 ¥ 103 N/m2, how thick is the layer of salinewater?

[4.3m]

Fig. Q1.10

Fig. Q1.9

Calculate the magnitude and direction of the resul-tant hydrostatic force on a unit length of the gate.

[52.05 ¥ 103 N/m at 67.9° to the horizontal, actingupwards through the centre of curvature, C]

1.10 The dam in Fig. Q1.10 has a curved face,being part of a 40m radius circle. The dam holdsback water to a depth of 35m. Calculate the mag-nitude and direction of the resultant hydrostaticforce per metre length.

[7840.6 ¥ 103 N/m at 40° to the horizontal, actingdownwards through the centre of curvature, C]

1.11 A 7500 tonne reinforced concrete lock struc-ture has been constructed in a dry dock. The lock is60m long by 30m wide in plan and is shaped likean open shoe box. The side walls are 8m high. (a)Will the lock structure float in sea water of density1025kg/m3, and if so, what is its draught and free-

0333779061

Documents

Transcript of 0333779061