Mechanical Engineering Systems 2008 Part 6 pdf

h P X AA 1 H Measuring liquid Manometer tube Pipe Working liquid Fluid mechanics 117 Piezometer tube (or simple manometer tube) (Figure 3.1.10) The pressurized liquid in the horizontal pipe rises up the vertical glass tube until the pressure from the pipe is balanced by the pressure due to the column of liquid, and the liquid comes to rest. At that point the pressure head is simply the height of the liquid in the tube, measured from the centreline of the pipe. The pressure in the pipe can be found from the formula p = ␳gh where ␳ is the density of the liquid filling the pipe and the manometer tube. This is a beautifully simple device that is inherently accurate; as mentioned earlier it is only the vertical height that matters so any inclination of the tube or any variation in diameter does not affect the reading so long as the measuring scale itself is vertical. In practice the piezometer tube is limited to measuring heads of about 1 metre because otherwise 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 does not mix with the liquid in the pipe. Mercury is the most commonly used liquid 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 118 Fluid mechanics density of 13.6, i.e. mercury is 13.6 times denser than water) and it does not mix with common liquids since it is a metal. To prevent it escaping from 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. To solve any conversion problem with manometers it is usually best to work from the lowest level where the two liquids meet, in this case along the level AAЈ. Pressure at A is due to the left-hand column so p A = ␳ m gx Pressure at AЈ is partly due to the right-hand column and partly due to the pressure in the pipe, so p A Ј = ␳ wl gH + p wl We can interpret the pressure in the pipe as a pressure head using p wl = ␳ wl gh wl . Now we know that the pressure in a liquid is constant at a constant depth, so the pressure at A must equal the pressure at AЈ just inside the mercury. Therefore ␳ m gx = ␳ wl gH + ␳ wl gh wl To simplify this and find the pressure head: h wl =(␳ 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/m 3 is used to measure the pressure head in a pipe containing a liquid of density 850 kg/m 3 . Calculate the head in the pipe if the reading on the manometer is 245 mm and the lower meniscus is 150 mm vertically below the centreline of the pipe. The set-up is identical to the manometer shown in Figure 3.1.11 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 h–h 12 Air A H A 1 X 1 2 Mercury Flexible leads Flow of workin g liquid Fluid mechanics 119 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 two points and recording the differential head, as shown in Figure 3.1.12. Since it is the working liquid itself that fills the manometer tubes and there is no separate measuring liquid, the head difference is given directly by the difference in the heights in the tubes (h 1 – h 2 ). In this case the difference is limited again to about 1 metre because of the need for 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 can use mercury for the measuring liquid, giving a device which is by far the most common form of manometer because of its ability to be used on various flow measuring devices. Again we tackle the problem of converting from the reading x into the head difference of the working liquid (h 1 – h 2 ) by working from the lowest level where the two liquids meet. So we start by equating the pressures at A and AЈ (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 the pressure due to the vertical height of the column of working liquid in the left-hand tube: p 1 + ␳ wl gH. Figure 3.1.12 Differential inverted U-tube manometer Figure 3.1.13 Differential mercury U-tube manometer 120 Fluid mechanics Pressure at AЈ is equal to the pressure at point 2 in the pipe plus the pressure due to the vertical height of the short column of working liquid sitting on the column of mercury in the right-hand tube: p 2 + ␳ wl g(H – x) + ␳ m gx Rewriting the pressure terms to get heads h 1 and h 2 , and equating the pressures at A and AЈ we get: ␳ wl gh 1 + ␳ wl gH = ␳ wl gh 2 + ␳ wl g(H – x) + ␳ m gx Cancel g because it appears in all terms and cancel the term ␳ wl gH because it appears on both sides. ␳ wl h 1 = ␳ wl h 2 + ␳ m x – ␳ wl x Finally we have h 1 – h 2 =(␳ m /␳ wl – 1) x (3.1.3) 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 the difference in pressure between two points on a horizontal pipe containing liquid with a relative density of 0.79. If the reading on the manometer scale for the difference in height of the two levels is 238 mm, calculate the head difference and the pressure difference between the two points. We can use Equation (3.1.3) directly to solve this problem and give the head difference, but remember to convert millimetres to metres (divide by 1000) and convert relative density to real density (multiply by 1000 kg/m 3 ). 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 no further part. Pressure difference = 790 × 9.81 × 3.86 = 29 909 Pa A F F AA P Fluid mechanics 121 Hydrostatic force on plane surfaces In the previous section we looked at the subject of hydrostatic pressure variation with depth in a liquid, and used the findings to explore the possibilities for measuring pressure with columns of liquids (manometry). In this section we are going to extend this study of hydrostatic pressure to the point where we can calculate the total force due to liquid pressure acting over a specified area. One of the main reasons that we study fluid mechanics in mechanical engineering is so that we can calculate the size of forces acting in a situation where liquids are employed. Knowledge of these forces is essential so that we can safely design a range of devices such as valves, pumps, fuel tanks and submersible housings. Although we shall not look at all the many different situations that can arise, it is vital that you understand the principles involved by studying a few of the most common applications. Hydraulics The first thing to understand is the way that pressure is transmitted in fluids. 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 cross- sectional area of the rod is A and so a localized pressure of 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 lesser extent in the small surrounding region. It will not be experienced in the rest of the solid block towards the sides. Now look at Figure 3.1.15 where the same rod is used as a piston to push down with the same force on a sealed container of liquid. The result is very different because the pressure of P = F/A will now be experienced by all the liquid equally. If the container were to spring a leak we know that the liquid would spurt out normal to the surface that had the hole in it. If the leak were in the top surface of the container then the liquid would spurt upwards, in completely the opposite direction to the force that is being applied to the piston. This means that the liquid was being pushed in that upwards direction and that can only happen if the pressure acts normally to the inside surface. This simple imaginary experiment therefore has two very important conclusions: ᭹ Pressure is transmitted uniformly in all directions in a fluid. ᭹ Fluid pressure acts normal (i.e. at right angles) to any surface that it touches. These two properties of fluid pressure are the basis of all hydraulic systems 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 s f F Area a Area A d 122 Fluid mechanics force applied by the driver’s foot to the small cross-section piston and cylinder produces a large pressure in the hydraulic fluid that completely fills the system. This large pressure is transmitted along the hydraulic pipes under the car to the larger area pistons and cylinders at the wheels where it produces a large force that operates the brakes themselves. The force on the piston faces is at right angles to the face and therefore exactly in the direction where it can have most effect. The underlying principle behind hydraulics is that by applying a small force to a small area piston we can generate a large pressure in the hydraulic fluid that can then be fed to a large area piston to produce an enormous force. The reason that this principle works is that the hydraulic fluid transmits the applied pressure uniformly in all directions, as we saw above. It sometimes seems as though we are getting something for nothing in examples of hydraulics because small, applied forces can be used to produce very large output forces and shift massive objects such as cars on jacks. Of course we cannot really get something for nothing and the penalty that we have to pay is that the small piston has to be replaced by a pump which is pushed backwards and forwards many times in order to displace the large volume of fluid required to make the large piston move any appreciable distance. This means that the small force on the driving piston is applied over a very large distance (i.e. many times the stroke length of the pump) in order to make the large force on the big piston move through one stroke length, as illustrated in Figure 3.1.16. The volume of hydraulic fluid expelled by the small piston in one stroke 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: Work in = 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 Fluid mechanics 123 The work done on the load on the large piston is given by: Work out = Fd Now the volume expelled from the small cylinder is the same as the volume received by the large cylinder, because liquids are incompress- ible for all practical purposes, therefore: V = dA = sa and hence d = sa/A So Work out = Fsa/A = Psa = fs = Work in . (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 big advantage 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, is used to pump oil into a large vertical cylinder and piston, of diameter 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 the upper 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 that you were required to design a watertight housing for an underwater video camera that is used to traverse up and down the legs of a giant oil rig looking for structural damage, as shown in Figure 3.1.17. 0.2m 0.4m 300m Sea water density 1030 kg/m 3 y dy F h dF W 124 Fluid mechanics In order to specify the thickness, and hence the strength, of the housing material, you would need to know the maximum force that each face of the housing would be likely to experience. This means that the pressure on the camera at maximum depth would need to be known, using Equation (3.1.1) p = ␳gh The maximum pressure on the camera is therefore: p max = 1030 × 9.81 × 300 = 3031 kPa The total force on one of the large faces of the camera housing would then be equal to this pressure times the area of the face. F = 3031 kPa × (0.2 × 0.4) m 2 = 242.5 kN This is clearly a very simple calculation and the reason why we do not have to employ anything more complicated is that the dimensions of the camera housing are tiny compared with the depth. This means that we do not need to consider the variation of pressure from top to bottom of the housing, or the fact that the camera might not be in an upright position. This is therefore a very special solution which would not apply to something like a lock gate in a shipping canal, where the hydrostatic pressure is enormous at the bottom but close to zero at the top of the structure 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; we need 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 of integration. We know that pressure varies with depth, but it is constant at a constant 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 y dy F D W h Fluid mechanics 125 The pressure at a depth y is: p = ␳gy Hence the small force dF is given by (pressure × area) dF =(␳gy) (w dy) The total force is now found by integrating from top to bottom. F = ␳gw ͵ h 0 y dy F = ␳gwh 2 /2 This is a final answer for this particular problem but we can interpret it 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 of pressure which can be thought to act over the total area. In fact it is the pressure which would be experienced half way down the rectangle, i.e. at the centroid or geometrical centre. Therefore: F = ¯ p × A (3.1.4) where ¯ p is the pressure at the centroid. It turns out that this expression is completely general for a flat surface and 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 would then be considered to act at a single point, rather like the distributed weight 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 edge along the surface of the liquid, as used above, this can be found quite easily. Again we need to use integration, but this time we are interested in taking moments because that is the way in which we generally locate the position 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 4m F 1 F 2 Gate 126 Fluid mechanics Therefore integrating T = ␳gw ͵ h 0 y 2 dy = ␳gwh 3 /3 This is the total torque about the top edge due to the hydrostatic force. 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 centre of pressure. Therefore T =(␳gwh 2 /2)D By comparison with the expression for T found above, it can be seen that D =2h/3 Again it must be stressed that this result applies only to the situation found in such devices as lock gates; it would not apply to any other shape or any other arrangement of a rectangle. Nevertheless it is always true that the centre of pressure is lower than the centroid of the immersed surface. To understand this, consider the predicament of the unfortunate waiter shown in Figure 3.1.20; someone has stacked the pots very unevenly on the left tray so that there is a linear distribution from one end to the other. Where should the waiter support the two trays? 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 heavily stacked end, and in fact the exact position is two-thirds of the way from the lightly stacked end. This is just the same sort of situation as the lock gate, but turned through 90° so that the load due to the water increases linearly 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 of the way down, showing that this must be the depth at which all the force distribution due to the water appears to act as a single force. Example 3.1.4 (a) The maximum resultant force that a lock gate (Figure 3.1.21) can safely withstand from hydrostatic pressure is 2 MN. The lock gate is rectangular and 5.4 m wide. If the height of water on the lower side is 4 m, how high can the water be allowed to rise on the other side? (b) When this depth is achieved, at what height above the bottom of the gate does the resultant force act? (a) Resultant force R is F 2 – F 1 = 2 × 10 6 So F 2 = F 1 + 2 × 10 6 Figure 3.1.20 Also finding the centre of pressure Figure 3.1.21 A side view of a lock gate [...]... of radius 0 .6 m and length 2.3 m Volume of pillar submerged = ␲0 .62 × 2.3 = 2 .60 m3 Weight of water displaced = 2 .60 × 1000 × 9.81 = 25.5 06 kN The tension in the cable with the pillar in air = 5 × 1000 × 9.81 = 49.050 kN Hence the resultant tension in the cable with the pillar partly submerged is: 49.050 – 25.5 06 = 23.54 kN Buoyancy This phenomenon of solid objects being able to float is part of a larger... when a particle in the liquid can be traced and shown to move along a smooth line in the direction of flow Suppose we could take a series of very high-speed photographs of a single particle in the fluid and build up a multiple exposure record of what the particle does In streamlined flow we would see that the particle followed exactly the line of flow of the liquid, as in this illustration of a particle... gate over is 2 × 1 06 × d, but this could also be calculated from the individual moments due to the forces F1 and F2 Therefore 2 × 1 06 × d = (F2 × 9.57/3) – (F1 × 4/3) = (2 423 800 × 9.57/3) – (423 800 × 4/3) which finally gives d = 3.59 m Properties of some common fluids Fluid Density ␳ kg/m3 Relative density RD Air Water Mercury Mineral oil 1.3 1000 13 60 0 900 1.3 × 10–3 1 13 .6 0.9 Dynamic viscosity... a tube In this case the line of flow of the fluid is known as a streamline because the individual particles stream along it Streamline Figure 3.2.1 Streamlined flow Turbulent flow is when the individual particles do not follow any regular path but their overall motion gives rise to the liquid flow The particles are clearly being stirred around in the fluid and undergo many collisions with their colleagues... continuity law represented in Equation (3.2.3) We must remember to convert the volume flow rate to cubic metres per second from cubic metres per minute Q = A1 v1 = A2 v2 15 /60 = 0.05 v1 = 0.02 v2 v1 = 15/ (60 × 0.05) = 5 m/s v2 = 15/ (60 × 0.02) = 12.5 m/s Laminar flow Viscosity Before we can consider flow of fluids in much more detail, we must take a closer look at the property of fluids known as viscosity... of a cylinder of radius 1.5 m, pivoted so that it can be rotated upwards to open Its centre of gravity is at G, its mass is 60 00 kg and its length (into the paper) is 3 m Calculate: Pivot (a) (b) 0.6m 0.6m 1.5m G the magnitude of the resultant force on the gate due to the water; the turning moment required to start to lift the gate (18) Figure 3.1.31 Sluice gate 10m 0.9m 2.4m Figure 3.1.32 A dolphin... width 0.9 m If the pool is filled with seawater of relative density 1. 06, calculate the magnitude and direction of the hydrostatic thrust on one panel (19) A cruise liner is to be fitted with a hemispherical glass observation window, which will be let into the side of the hull below the waterline The radius of the hemisphere will be 2 .6 m and its top edge will be 3 m below the water surface If the sea... weight of water displaced Example 3.1 .6 A crane is to lower a vertical cylindrical pillar, of diameter 1.2 m, into position on a plinth submerged in water The pillar has a mass of 5 tonnes What will be the tension experienced in the supporting cable when the lower end of the pillar is submerged to a depth of 2.3 m? We need to calculate the weight of water displaced by the partly submerged pillar in order... Clearly we only need to consider any water which is directly above any part of the wall, i.e in the volume BCD, since gravity always acts vertically Any water beyond BC only presses down on the bed of the reservoir, not on the wall Fvertical = weight of water W = (volume BCD) × ␳ × g Calculation of the volume BCD may seem complicated but in engineering the shapes usually have clearly defined profiles such... same direction as that in the example above and so we know that the particles must ultimately travel along the tube from left to right The route taken by the fluid at any point is called a pathline In the example in Figure 3.2.2 the pathline looks identical to the streamline in Figure 3.2.1, but the difference is that the individual particles do not follow it The difference between the two extreme types . volume of a cylinder of radius 0 .6 m and length 2.3 m. Volume of pillar submerged = ␲0 .6 2 × 2.3 = 2 .60 m 3 Weight of water displaced = 2 .60 × 1000 × 9.81 = 25.5 06 kN The tension in the cable with. once we have converted from x to h in the first part of the calculation then the mercury plays no further part. Pressure difference = 790 × 9.81 × 3. 86 = 29 909 Pa A F F AA P Fluid mechanics 121 Hydrostatic. real density (multiply by 1000 kg/m 3 ). Head difference = {(13 60 0/790) – 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