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1 Pressure Variation with Depth Pressure in a static fluid does not change in the horizontal direction as the horizontal forces balance each other out. However, pressure in a static fluid does change with depth, due to the extra weight of fluid on top of a layer as we move downwards. Consider a column of fluid of arbitrary cross section of area, A: Column of Fluid Pressure Diagram Considering the weight of the column of water, we have: Σ F y = 0: p 1 A+ γ A(h 2 - h 1 ) - p 2 A= 0 Obviously the area of the column cancels out: we can just consider pressures. If we say the height of the column is h=h 2 h 1 and substitute in for the specific weight, we see the difference in pressure from the bottom to the top of the column is: 31

2 p 2 - p 1 = ρgh This difference in pressure varies linearly in h, as shown by the Area 3 of the pressure diagram. If we let h 1 = 0 and consider a gauge pressure, then p 1 = 0 and we have: p 2 = ρgh Where h remains the height of the column. For the fluid on top of the column, this is the source of p 1 and is shown as Area 1 of the pressure diagram. Area 2 of the pressure diagram is this same pressure carried downwards, to which is added more pressure due to the extra fluid. To summarize: The gauge pressure at any depth from the surface of a fluid is: p = ρgh Summary 1. Pressure acts normal to any surface in a static fluid; 2. Pressure is the same at a point in a fluid and acts in all directions; 3. Pressure varies linearly with depth in a fluid. By applying these rules to a simple swimming pool, the pressure distribution around the edges is as shown: 32

3 Note: 1. Along the bottom the pressure is constant due to a constant depth; 2. Along the vertical wall the pressure varies linearly with depth and acts in the horizontal direction; 3. Along the sloped wall the pressure again varies linearly with depth but also acts normal to the surface; 4. At the junctions of the walls and the bottom the pressure is the same. Problems - Pressure 1. Sketch the pressure distribution applied to the container by the fluid: 2. For the dam shown, sketch the pressure distribution on line AB and on the surface of the dam, BC. Sketch the resultant force on the dam. 33

4 3. For the canal gate shown, sketch the pressure distributions applied to it. Sketch the resultant force on the gate? If h 1 = 6.0 m and h 2 = 4.0 m, sketch the pressure distribution to the gate. Also, what is the value of the resultant force on the gate and at what height above the bottom of the gate is it applied? 3.3 Pressure Measurement Pressure Head Pressure in fluids may arise from many sources, for example pumps, gravity, momentum etc. Since p =ρgh, a height of liquid column can be associated with the pressure p arising from such sources. This height, h, is known as the pressure head. Example: The gauge pressure in water mains is 50 kn/m2, what is the pressure head? The pressure head equivalent to the pressure in the pipe is just: p =ρgh h = 34

5 = 50 x x m So the pressure at the bottom of a 5.1 m deep swimming pool is the same as the pressure in this pipe. Manometers A manometer (or liquid gauge) is a pressure measurement device which uses the relationship between pressure and head to give readings. In the following, we wish to measure the pressure of a fluid in a pipe. Piezometer This is the simplest gauge. A small vertical tube is connected to the pipe and its top is left open to the atmosphere, as shown. 35

6 The pressure at A is equal to the pressure due to the column of liquid of height h 1 : p A = ρ gh 1 Similarly, p B = ρ gh 2 The problem with this type of gauge is that for usual civil engineering applications the pressure is large (e.g. 100 kn/m2) and so the height of the column is impractical (e.g.10 m). Also, obviously, such a gauge is useless for measuring gas pressures. U-tube Manometer To overcome the problems with the piezometer, the U-tube manometer seals the fluid by using a measuring (manometric) liquid: Choosing the line BC as the interface between the measuring liquid and the fluid, we know: Pressure at B, p B = Pressure at C, p c For the left-hand side of the U-tube: 36

7 p B =p A + ρgh 1 For the right hand side: p c = ρ max gh 2 Where we have ignored atmospheric pressure and are thus dealing with gauge pressures. Thus: p B = p c p A +ρgh 1 = ρ max gh 2 And so: p A = ρ man gh 2 - ρgh 1 Notice that we have used the fact that in any continuous fluid, the pressure is the same at any horizontal level. Differential Manometer To measure the pressure difference between two points we use a u-tube as shown: 37

8 Using the same approach as before: Pressure at C, p c = Pressure at D, p D p A +ρga = p B +ρg (b h) +ρ man gh Hence the pressure difference is: p A p B = ρg (b a) +hg (ρ man ρ) A differential manometer can be used to measure the difference in pressure between two containers or two points in the same system. Again, on equating the pressures at points labeled (2) and (3), we may get an expression for the pressure difference between A and B: p A + p B = γ 2 h 2 + γ 3 h 3 γ 1 h 1 In the common case when A and B are at the same elevation (h 1 = h 2 + h 3 ) and the fluids in the two containers are the same (γ 1 = γ 3 ), one may show that the pressure difference registered by a differential manometer is given by p =( -1)ρgh Where is the density of the manometer fluid, ρ is the density of the fluid in the system, and h is the manometer differential reading. 38

9 Inclined-tube manometer As shown above, the differential reading is proportional to the pressure difference. If the pressure difference is very small, the reading may be too small to be measured with good accuracy. To increase the sensitivity of the differential reading, one leg of the manometer can be inclined at an angle θ, and the differential reading is measured along the inclined tube. As shown above, h 2 = L 2 sinθ, and hence p A - p B = γ 2 L 2 sinθ + γ 3 h 3 γ 1 h 1 Obviously, the smaller the angle θ, the more the reading L 2 is magnified. Multifluid manometer The pressure in a pressurized tank is measured by a multifluid manometer, as is shown in the figure. Show that the air pressure in the tank is given by p air = p atm + g( ρ mercury h 3 ρ oil h 2 ρ water h 1 ) 39

10 Problems Pressure Measurement 1. What is the pressure head, in metres of water, exerted by the atmosphere? (Ans m) 2. What is the maximum gauge pressure of water that can be measured using a piezometer 2.5 m high? (Ans kn/m2) 3. A U-tube manometer is used to measure the pressure of a fluid of density 800 kg/m3. If the density of the manometric liquid is kg/m3, what is the gauge pressure in the pipe if (a) h 1 = 0.5 m and D is 0.9 m above BC; (b) h 1 = 0.1 m and D is 0.2 m below BC? (Ans kn/m2, kn/m2) 4. A differential manometer is used to measure the pressure difference between two points in a pipe carrying water. The manometric liquid is mercury and the points have a 0.3 m height difference. Calculate the pressure difference when h = 0.7 m. (Ans kn/m2) 5. For the configuration shown, calculate the weight of the piston if the gauge pressure reading is 70 kpa. 40

11 (Ans kn) 6. A hydraulic jack having a ram 150 mm in diameter lifts a weight W = 20 kn under the action of a 30 mm plunger. What force is required on the plunger to lift the weight? 41 Hydrostatic pressure Consider a tank of fluid which contains a very thin plate of (neutrally buoyant) material with area A. This situation is shown in Figure below. If the plate is in equilibrium (it does

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28 multiple choice, 4 wrong answers will be dropped Covers everything learned in Phys 105 and 106 Final exam 2:30-5:00 pm, Tuesday 5/10/2011 FMH 310 28 multiple choice, 4 wrong answers will be dropped Covers everything learned in Phys 105 and 106 About 7 problems from Phys 105 About 8-9 problems are

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