Mechanical Engineering Systems 2008 Part 6 pdf

This can often be measured directly by connecting a single manometer to the twopoints and recording the differential head, as shown in Figure 3.1.12.Since it is the working liquid itself

h P

p = gh

where  is the density of the liquid filling the pipe and the manometertube This is a beautifully simple device that is inherently accurate; asmentioned earlier it is only the vertical height that matters so anyinclination of the tube or any variation in diameter does not affect thereading so long as the measuring scale itself is vertical In practice thepiezometer tube is limited to measuring heads of about 1 metre becauseotherwise the glass tube would be too long and fragile

U-tube manometer (Figure 3.1.11)

For higher pressures we can use a higher density liquid in the tube.Clearly the choice of liquids must be such that the liquid in the tube doesnot mix with the liquid in the pipe Mercury is the most commonly usedliquid for the manometer tube because it has a high density (relative

Figure 3.1.10 A piezometer

tube

Figure 3.1.11 A U-tube

manometer

density of 13.6, i.e mercury is 13.6 times denser than water) and it doesnot mix with common liquids since it is a metal To prevent it escapingfrom the manometer tube a U-bend is used.

Note that in the diagram the height of the mercury column is labelled

as x and not as h This is because the head is always quoted as the height

of a column of the working liquid (the one in the pipe), rather than the measuring liquid (the one in the manometer tube) We therefore need to convert from x to obtain the head of working liquid that would be

obtained if we could build a simple manometer tube tall enough Tosolve any conversion problem with manometers it is usually best towork from the lowest level where the two liquids meet, in this casealong the level AA

Pressure at A is due to the left-hand column so

mgx = wlgH + wlghwl

To simplify this and find the pressure head:

hwl = (m/wl)x – H metres of the working liquid (3.1.2)

Example 3.1.1

A U-tube manometer containing mercury of density

13 600 kg/m3is used to measure the pressure head in a pipecontaining a liquid of density 850 kg/m3 Calculate the head inthe pipe if the reading on the manometer is 245 mm and thelower meniscus is 150 mm vertically below the centreline ofthe pipe

The set-up is identical to the manometer shown in Figure3.1.11 and so we can use Equation (3.1.2) directly Remem-ber to convert any measurements in millimetres to metres

Head h = (13 600/850) × (245/1000) – 150/1000

= (16 × 0.245) – 0.15

= 3.92 – 0.15

= 3.77 m

Differential inverted U-tube manometer

In many cases it is not just the pressure that needs to be measured, it is

the pressure difference between two points that is required This can

often be measured directly by connecting a single manometer to the twopoints and recording the differential head, as shown in Figure 3.1.12.Since it is the working liquid itself that fills the manometer tubes andthere is no separate measuring liquid, the head difference is given

directly by the difference in the heights in the tubes (h1– h2) In thiscase the difference is limited again to about 1 metre because of the needfor a glass tube in order to see the liquid levels

Differential mercury U-tube manometer (Figure 3.1.13)

To make it possible to record higher differential pressures, again we canuse mercury for the measuring liquid, giving a device which is by far themost common form of manometer because of its ability to be used onvarious flow measuring devices

Again we tackle the problem of converting from the reading x into the head difference of the working liquid (h1 – h2) by working from thelowest level where the two liquids meet So we start by equating thepressures at A and A (i.e at the same depth in the same liquid).Pressure at A is equal to the pressure at point 1 in the pipe plus thepressure due to the vertical height of the column of working liquid in the

Pressure at A is equal to the pressure at point 2 in the pipe plus thepressure due to the vertical height of the short column of working liquidsitting on the column of mercury in the right-hand tube:

p2+ wlg(H – x) + mgx Rewriting the pressure terms to get heads h1and h2, and equating the

pressures at A and A we get:

wlgh1+ wlgH = wlgh2+ wlg(H – x) + mgx Cancel g because it appears in all terms and cancel the term wlgH

because it appears on both sides

wlh1 = wlh2+ mx – wlx

Finally we have

This equation is well worth learning because it is such a common type

of manometer, but remember that it only applies to this one type

Example 3.1.2

A mercury U-tube manometer is to be used to measure thedifference in pressure between two points on a horizontalpipe containing liquid with a relative density of 0.79 If thereading on the manometer scale for the difference in height ofthe two levels is 238 mm, calculate the head difference andthe pressure difference between the two points

We can use Equation (3.1.3) directly to solve this problemand give the head difference, but remember to convertmillimetres to metres (divide by 1000) and convert relativedensity to real density (multiply by 1000 kg/m3)

Head difference = {(13 600/790) – 1} × 238/1000

= (17.215 – 1) × 0.238

= 16.215 × 0.238

= 3.86 m of the pipeline liquid

The pressure difference is found by using p = gh and

remembering that the density to use is the density of the

liquid in the pipeline; once we have converted from x to h in

the first part of the calculation then the mercury plays nofurther part

Pressure difference = 790 × 9.81 × 3.86

= 29 909 Pa

A F

F

AA

P

Hydrostatic force on plane surfaces

In the previous section we looked at the subject of hydrostaticpressure variation with depth in a liquid, and used the findings toexplore the possibilities for measuring pressure with columns ofliquids (manometry) In this section we are going to extend this study

of hydrostatic pressure to the point where we can calculate the totalforce due to liquid pressure acting over a specified area One of themain reasons that we study fluid mechanics in mechanical engineering

is so that we can calculate the size of forces acting in a situationwhere liquids are employed Knowledge of these forces is essential sothat we can safely design a range of devices such as valves, pumps,fuel tanks and submersible housings Although we shall not look at allthe many different situations that can arise, it is vital that youunderstand the principles involved by studying a few of the mostcommon applications

Hydraulics

The first thing to understand is the way that pressure is transmitted influids Most of this is common sense but it is worthwhile spelling it out

so that the most important features are made clear

Look at Figures 3.1.14 and 3.1.15 In Figure 3.1.14 a short solid rod

is being pushed down onto a solid block with a force F The sectional area of the rod is A and so a localized pressure of

cross-P = F/A

is felt underneath the base of the rod This pressure will be experienced

in the block mainly directly under the rod, but also to a much lesserextent in the small surrounding region It will not be experienced in therest of the solid block towards the sides

Now look at Figure 3.1.15 where the same rod is used as a piston topush down with the same force on a sealed container of liquid Theresult is very different because the pressure of

P = F/A

will now be experienced by all the liquid equally If the container were tospring a leak we know that the liquid would spurt out normal to thesurface that had the hole in it If the leak were in the top surface of thecontainer then the liquid would spurt upwards, in completely the oppositedirection to the force that is being applied to the piston This means thatthe liquid was being pushed in that upwards direction and that can onlyhappen if the pressure acts normally to the inside surface

This simple imaginary experiment therefore has two very importantconclusions:

 Pressure is transmitted uniformly in all directions in a fluid

 Fluid pressure acts normal (i.e at right angles) to any surface that ittouches

These two properties of fluid pressure are the basis of all hydraulicsystems and are the reason why car brakes work so successfully The

Figure 3.1.14 Force applied to

a solid block

Figure 3.1.15 Force applied to

a liquid

Equal volumes swept out

The underlying principle behind hydraulics is that by applying asmall force to a small area piston we can generate a large pressure in thehydraulic fluid that can then be fed to a large area piston to produce anenormous force The reason that this principle works is that thehydraulic fluid transmits the applied pressure uniformly in alldirections, as we saw above

It sometimes seems as though we are getting something for nothing inexamples of hydraulics because small, applied forces can be used toproduce very large output forces and shift massive objects such as cars onjacks Of course we cannot really get something for nothing and thepenalty that we have to pay is that the small piston has to be replaced by apump which is pushed backwards and forwards many times in order todisplace the large volume of fluid required to make the large piston moveany appreciable distance This means that the small force on the drivingpiston is applied over a very large distance (i.e many times the strokelength of the pump) in order to make the large force on the big pistonmove through one stroke length, as illustrated in Figure 3.1.16

The volume of hydraulic fluid expelled by the small piston in onestroke is given by:

V = sa where s and a are the stroke length and area of the small piston

respectively

The work done by the piston during this stroke is given by:Workin = force × distance

= fs

This volume of fluid expelled travels to the large cylinder, which

moves up by some distance d such that the extra volume in the cylinder

is given by:

V = dA

Figure 3.1.16 A simple

hydraulic system

The work done on the load on the large piston is given by:

Workout = Fd

Now the volume expelled from the small cylinder is the same as thevolume received by the large cylinder, because liquids are incompress-ible for all practical purposes, therefore:

V = dA = sa and hence d = sa/A

So Workout= Fsa/A = Psa = fs = Workin (Note: pressure is uniform so

P = f/a = F/A.)

In practice there would be a slight loss of energy due to friction and

so it is clear that hydraulic systems actually increase the amount of work

to be done rather than give us something for nothing, but their bigadvantage is that they allow us to carry out the work in a manner which

is more convenient to us

Example 3.1.3

A small hydraulic cylinder and piston, of diameter 50 mm, isused to pump oil into a large vertical cylinder and piston, ofdiameter 250 mm, on which sits a lathe of mass 350 kg.Calculate the force which must be applied to the small piston

in order to raise the lathe

The pressures in the two cylinders must be equal and so

P = f/a = F/A

Where the lower case refers to the small components and theupper case refers to the large components We are trying to

find the small force f and we do not need to calculate P, so

this equation can be rearranged:

f = Fa/A

The large force F is equal to the mass of the lathe multiplied

by the acceleration due to gravity, i.e it is the lathe’s weight.The two areas do not need to be calculated because the ratio

of areas is equal to the square of the ratio of the diameters.Note that we do not need to convert the millimetres to metres

in the case of a ratio Therefore

The force on the small piston = f = (350 × 9.81) × (50/250)2

= 137.3 N

Pressure on an immersed surface

We now need to look at fluid forces in a more general way Suppose thatyou were required to design a watertight housing for an underwatervideo camera that is used to traverse up and down the legs of a giant oilrig looking for structural damage, as shown in Figure 3.1.17

p = gh

The maximum pressure on the camera is therefore:

pmax = 1030 × 9.81 × 300 = 3031 kPaThe total force on one of the large faces of the camera housing wouldthen be equal to this pressure times the area of the face

F = 3031 kPa × (0.2 × 0.4) m2

= 242.5 kNThis is clearly a very simple calculation and the reason why we do nothave to employ anything more complicated is that the dimensions of thecamera housing are tiny compared with the depth This means that we donot need to consider the variation of pressure from top to bottom of thehousing, or the fact that the camera might not be in an upright position.This is therefore a very special solution which would not apply tosomething like a lock gate in a shipping canal, where the hydrostaticpressure is enormous at the bottom but close to zero at the top of thestructure close to the surface of the liquid We must look a little further

to find a general solution that will accurately predict the total force in

any situation, taking into account the variation of all the factors.

Consider the vertical rectangular surface shown in Figure 3.1.18; weneed to add up all the pressure forces acting over the entire area in order

to find the total force F This means that we will have to carry out an

integration, and so the first step is to identify a suitable area element ofintegration

We know that pressure varies with depth, but it is constant at aconstant depth Therefore if we use a horizontal element with an

infinitesimal height dy, we can treat it just like the underwater camera

above and say that the pressure on it is a constant

Figure 3.1.17 An inspection

system on an oil rig

Figure 3.1.18 Finding the total

force on a submerged plane

dy F

in a more meaningful way as

F = ( gh/2)(wh)

Any force can be thought of as a pressure times an area, so by taking out

the total area of the flat rectangle wh we must be left with a ‘mean’

pressure as the other term Therefore gh/2 must be an average value ofpressure which can be thought to act over the total area In fact it is thepressure which would be experienced half way down the rectangle, i.e

at the centroid or geometrical centre

Therefore:

where ¯p is the pressure at the centroid.

It turns out that this expression is completely general for a flat surfaceand we could apply it to triangles, circles, etc

Location of the hydrostatic force

Having calculated the size of the total hydrostatic force it would be an

advantage now if we could calculate where it acts so that we could treat

fluid mechanics as we would solid mechanics The total force wouldthen be considered to act at a single point, rather like the distributedweight of a solid body that is taken to act through the centre of gravity

This corresponding single point for a fluid force is the centre of pressure

(Figure 3.1.19) For the case of the vertical rectangle with one edgealong the surface of the liquid, as used above, this can be found quiteeasily

Again we need to use integration, but this time we are interested intaking moments because that is the way in which we generally locate theposition of a force We will use the same element of integration as in the

first case, and will consider the small turning moment dT caused by the small force dF about the top edge.

dT = y dF

= y(gy)(w dy)

Figure 3.1.19 Finding the

centre of pressure

h d R

We could also have calculated it by saying that it is equal to the total

force F (calculated earlier) multiplied by the unknown depth D to the

The answer for the evenly stacked tray is easy – he should support it

in the middle The other tray must be supported closer to the heavilystacked end, and in fact the exact position is two-thirds of the way fromthe lightly stacked end This is just the same sort of situation as the lockgate, but turned through 90° so that the load due to the water increaseslinearly from the top to the bottom If we had to support the gate with

a single horizontal force then we would need to locate it two-thirds ofthe way down, showing that this must be the depth at which all the forcedistribution due to the water appears to act as a single force

(b) When this depth is achieved, at what height above thebottom of the gate does the resultant force act?

(a) Resultant force R is F2 – F1= 2 × 106

Now F1 = mean pressure × area

Hydrostatic forces on curved surfaces

In the previous section we found out how to calculate the totalhydrostatic force acting on any immersed flat surface, using theequation

F = ¯p × A

We now have to consider the way to calculate the force acting on acurved surface, because many engineering applications such as fuel

This sounds as though we are going to need to employ some fairlycomplicated integration to add up the effect of the variation of pressure

with depth and direction, but fortunately there are some short cuts we

can take to make this into quite a simple problem

Resolving the force on a curved surface (Figure 3.1.22)

The key to this simple treatment is the resolution of the total force into

horizontal and vertical components which we can then treat completely

separately Suppose we have the problem of calculating the totalhydrostatic force acting on a large dam wall holding the water back in

a reservoir

We are looking for the single force F which represents the sum of all

the little forces acting over all the dam wall We can find this if we cancalculate separately the horizontal and vertical forces acting on the walldue to the water Let us begin with the vertical force

Vertical force component

The vertical force caused by a static liquid which has a free surface open

to the atmosphere results solely from gravity In other words, thevertical force on this dam wall is simply the weight of the water pressingdown on it Clearly we only need to consider any water which is directlyabove any part of the wall, i.e in the volume BCD, since gravity alwaysacts vertically Any water beyond BC only presses down on the bed ofthe reservoir, not on the wall

Fvertical = weight of water W = (volume BCD) ×  × g

Calculation of the volume BCD may seem complicated but inengineering the shapes usually have clearly defined profiles such asellipses, parabolas or circles, which greatly simplify the problem

Horizontal force component

The horizontal force component is what is left of the total force once wehave removed the vertical component It is therefore exactly the same asthe force acting on a flat vertical wall of the same height and width All

we need to do then is calculate the force which would act on animaginary vertical wall erected along BC, using the method developed

in the previous section

Figure 3.1.22 Forces on a

curved surface