p atmospheric Statics : Pressure Hydrostatic Pressure: linear change in pressure with depth Measure depth, h, from free surface Pressure Head p gh


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1 IVE1400: n Introduction to Fluid Mechanics Statics : Pressure : Statics r P Sleigh: r J January 008 Module web site: Unit 1: Fluid Mechanics asics Flow Pressure Properties of Fluids Fluids vs. Solids Viscosity : Statics Hydrostatic pressure Manometry / Pressure measurement Hydrostatic forces on submerged surfaces Unit 3: ynamics The continuity equation. The ernoulli Equation. pplication of ernoulli equation. The momentum equation. pplication of momentum equation. Unit 4: Effect of the boundary on flow Laminar and turbulent flow oundary layer theory n Intro to imensional analysis Similarity 3 lectures 3 lectures 7 lectures 4 lectures Hydrostatic Pressure: linear change in pressure with depth p Measure depth, h, from free surface bsolute pressure auge pressure p1 g z z1 p gh p absolute g h + p atmospheric p gauge g h p atmospheric IVE1400: Fluid Mechanics Lecture 4 45 IVE1400: Fluid Mechanics Lecture 4 46 Pressure Head Examples of pressure head calculations: The lower limit of any pressure is the pressure in a What is a pressure of 500 knm  in head of water of density, 1000 kgm 3 Use p gh, Pressure measured above a perfect vacuum (zero) is known as pressure h In head of Mercury density kgm 3. Pressure measured relative to atmospheric pressure is known as pressure gauge pressure can be given using height of any fluid. h In head of a fluid with relative density 8.7. remember water ) h p gh This height, h, is referred to as the If pressure is quoted in head, the of the fluid also be given. IVE1400: Fluid Mechanics Lecture 4 47 IVE1400: Fluid Mechanics Lecture 4 48
2 Manometers for Measuring Pressure n Examples of a Piezometer. Manometers use the relationship between and to measure pressure 1. What is the maximum gauge pressure of water that can be measured by a Piezometer of height 1.5m? Piezometer Tube simple open tube attached to the top of a container with liquid at pressure. Liquid rises to a height, h, equal to the pressure in the container. p gh. If the liquid had a relative density of 8.5 what would the maximum measurable gauge pressure? The pressure measured is relative to so it measures pressure Problems with the Piezometer: IVE1400: Fluid Mechanics Lecture 4 49 IVE1400: Fluid Mechanics Lecture 4 50 Equality of Pressure t The Same Level In Static Fluid onsider these two tanks, one much larger than the other, and linked together by a thin tube: rea Fluid density ρ p l, p r, P Q Face L weight, mg Face Horizontal cylindrical element cross sectional area mass density left end pressure p l right end pressure p r For equilibrium the sum of the forces in the x direction is zero. Pressure in the horizontal direction is z L We have shown p l p r For a vertical pressure change we have p l and p r so p p p q z This true for any fluid. Pressure at the two equal levels are. IVE1400: Fluid Mechanics Lecture 4 51 IVE1400: Fluid Mechanics Lecture 4 5
3 n improved manometer: The U Tube U is connected as shown and filled with manometric fluid. Fluid density ρ U Tube enables the pressure of both liquids and gases to be measured Important points: 1. U Tube enables the pressure of both liquids and gases to be measured h 1 h. The manometric fluid density should be measured. man > Manometric fluid density ρ man Using the fact that pressure at two levels are equal: pressure at p For the left hand arm pressure at p For the right hand arm pressure at p p p p Subtract p atmospheric to give gauge pressure p IVE1400: Fluid Mechanics Lecture The two fluids should they must be immiscible. What if the fluid is a gas? changes. The manometric fluid is liquid (usually mercury, oil or water) nd Liquid density is much greater than gas, man >> gh is negligible and pressure is given by p IVE1400: Fluid Mechanics Lecture 4 54 n example of the UTube manometer. Using a utube manometer to measure gauge pressure of fluid density 700 kg/m 3, and the manometric fluid is mercury, with a relative density of What is the gauge pressure if: a) h 1 0.4m and h 0.9m? b) h and h 0.1m? Pressure ifference Measurement Using a U Tube Manometer. The U tube manometer can be connected at both ends to measure pressure difference between these two points Fluid density ρ E h b h h a Manometric fluid density ρ man IVE1400: Fluid Mechanics Lecture 4 55 IVE1400: Fluid Mechanics Lecture 4 56
4 pressure at p p p n example using the utube for pressure difference measuring In the figure below two pipes containing the same fluid of density 990 kg/m 3 are connected using a utube manometer. a) What is the pressure between the two pipes if the manometer contains fluid of relative density 13.6? Fluid density ρ Fluid density ρ iving the pressure difference h a 1.5m h 0.5m E h b 0.75m p  p gain if the fluid is a gas man >>, then the terms involving can be neglected, Manometric fluid density ρ man 13.6 ρ p  p IVE1400: Fluid Mechanics Lecture 4 57 IVE1400: Fluid Mechanics Lecture 4 58 dvances to the U tube manometer Problem: Two reading are required. Solution: esult: p 1 p diameter z 1 atum line diameter d z If the manometer is measuring the pressure difference of a gas of (p 1  p ) as shown, we know p 1  p volume of liquid moved from the left side to the right The fall in level of the left side is z 1 z d / 4 / 4 d z Putting this in the equation, d p1 p g z z d gz 1 If >> d then (d/) is very small so p p 1 IVE1400: Fluid Mechanics Lecture 4 59 IVE1400: Fluid Mechanics Lecture 4 60
5 Inclined manometer Problem: Small changes difficult to see Incline the arm: same height change but bigger movement. p p 1 diameter d z 1 diameter θ Scale eader x z atum line Example of an inclined manometer. n inclined tube manometer consists of a vertical cylinder 35mm diameter. t the bottom of this is connected a tube 5mm in diameter inclined upward at an angle of 15 to the horizontal, the top of this tube is connected to an air duct. The vertical cylinder is open to the air and the manometric fluid has relative density a) etermine the pressure in the air duct if the manometric fluid moved 50mm along the inclined tube. a) What is the error if the movement of the fluid in the vertical cylinder is ignored? The pressure difference is still given by the height change of the manometric fluid. but, p p p gz 1 z p 1 The sensitivity to pressure change can be further by a inclination. IVE1400: Fluid Mechanics Lecture 4 61 IVE1400: Fluid Mechanics Lecture 4 6 hoice Of Manometer Take care when fixing the manometer to vessel urrs cause local pressure variations. Lecture 5: Forces in Static Fluids : Statics isadvantages: response  only really useful for very slowly varying pressures  no use at all for fluctuating pressures; For the U tube manometer measurements must be taken to get the h value. It is often difficult to measure variations in pressure. It cannot be used for pressures unless several manometers are connected in series; For very accurate work the and relationship between temperature and must be known; dvantages of manometers: They are very. No is required  the pressure can be calculated from first principles. From earlier we know that: 1. static fluid can have acting on it.. ny force between the fluid and the boundary must. F n Pressure force normal to the boundary. True also for curved surfaces any imaginary plane in the fluid n element of fluid at rest is in equilibrium: 3. The sum of forces in any direction is 4. The sum of the moments of forces about any point is F n F F 1 1 IVE1400: Fluid Mechanics Lecture 4 63 IVE1400: Fluid Mechanics Lecture 5 64
6 eneral submerged plane Horizontal submerged plane F 1 p 1 δ 1 F n p n δ n F p δ The total or resultant force,, on the plane is the sum of the forces on the small elements i.e. and This force will act through the centre of. For a plane surface all forces acting can be represented by one single force, acting at rightangles to the plane through the centre of. The pressure, p, will be at all points of the surface. The resultant force will be given by urved submerged surface Each elemental force is a different magnitude and in a different direction (but still normal to the surface.). It is, in general, not easy to calculate the resultant force for a curved surface by combining all elemental forces. The sum of all the forces on each element will always be than the sum of the individual forces, p. IVE1400: Fluid Mechanics Lecture 5 65 IVE1400: Fluid Mechanics Lecture 5 66 esultant Force and entre of Pressure on a general plane surface in a liquid. Fluid density ρ esultant Force P z z area δ x S c θ Q area Take pressure as zero at the surface. s O d O elemental area δ x z is known as the 1 st Moment of rea of the plane PQ about the free surface. nd it is known that z is the area of the plane z is the distance to the centre of ( ) Measuring down from the surface, the pressure on an element, depth z, p So force on element F esultant force on plane (assuming and g as constant). In terms of distance from point O z (as z x sin ) The resultant force on a plane Pressure at centre of gravity X rea IVE1400: Fluid Mechanics Lecture 5 67 IVE1400: Fluid Mechanics Lecture 5 68
7 This resultant force acts at right angles through the centre of pressure,, at a depth. How do we find this position?. Sum of moments Moment of about O Equating s the plane is in equilibrium: The moment of will be equal to on all the elements about the same point. It is convenient to take moment about O The force on each elemental area: Force on the moment of this force is: Moment of Force on about O g s sin s gsins, g and are the same for each element, giving the total moment as The position of the centre of along the plane measure from the point O is: s Sc x How do we work out the summation term? This term is known as the Second Moment of rea, I o, of the plane (about an axis through O) nd moment of area about O I o It can be easily calculated for many common shapes. s IVE1400: Fluid Mechanics Lecture 5 69 IVE1400: Fluid Mechanics Lecture 5 70 The position of the centre of pressure along the plane measure from the point O is: nd S c nd Moment of area about a line through O Moment of area about a line through O How do you calculate the nd moment of area? nd moment of area is a geometric property. It can be found from tables  UT only for moments about an axis through its centroid I. and epth to the centre of pressure is We need it for an axis through O Use the parallel axis theorem to give us what we want. The parallel axis theorem can be written I o IVE1400: Fluid Mechanics Lecture 5 71 IVE1400: Fluid Mechanics Lecture 5 7
8 We then get the following equation for the position of the centre of pressure The nd moment of area about a line through the centroid of some common shapes. S c Shape rea nd moment of area, I, about an axis through the centroid ectangle b bd bd 3 h 1 (In the examination the parallel axis theorem and the I will be given) Triangle ircle b h h/3 bd bd Semicircle (4)/(3π) IVE1400: Fluid Mechanics Lecture 5 73 IVE1400: Fluid Mechanics Lecture 5 74 Example 1: etermine the resultant force due to the water acting on the 1m by m rectangular area shown in the diagram below. [ N,.37m from O O P Example : etermine the resultant force due to the water acting on the 1.5m by.0m triangular area shown in the example above. The apex of the triangle is at. [ N,.81m from P] 1.m 1.0m 45.0 m.0 m IVE1400: Fluid Mechanics Lecture 5 75 IVE1400: Fluid Mechanics Lecture 5 76
9 Example 3: Find the moment required to keep this triangular gate closed on a tank which holds water..0m 1.m Lecture 6: Pressure iagrams and Forces on urved Surfaces : Statics Pressure diagrams 1.5m For vertical walls of constant width it is possible to find the resultant force and centre of pressure graphically using a pressure diagram. We know the relationship between pressure and depth: p So we can draw the diagram below: IVE1400: Fluid Mechanics Lecture 5 77 IVE1400: Fluid Mechanics Lecture 6 78 This is know as a pressure diagram. Pressure increases from zero at the surface linearly by p, to a maximum at the base of p. The area of this triangle represents the resultant force on the vertical wall, Units of this are per metre. rea The force acts through the centroid of the pressure diagram. For a triangle the centroid is at its height i.e. the resultant force acts horizontally through the point z. For a vertical plane the depth to the centre of pressure is given by esultant force per unit width IVE1400: Fluid Mechanics Lecture 6 79 IVE1400: Fluid Mechanics Lecture 6 80
10 heck this against the moment method: The same technique can be used with combinations of liquids are held in tanks (e.g. oil floating on water). For example: The resultant force is given by: oil ρ o 0.8m water ρ 1.m and the depth to the centre of pressure by: and by the parallel axis theorem (with width of 1) I I x o ρg0.8 ρg1. Find the position and magnitude of the resultant force on this vertical wall of a tank which has oil floating on water as shown. epth to the centre of pressure IVE1400: Fluid Mechanics Lecture 6 81 IVE1400: Fluid Mechanics Lecture 6 8 Forces on Submerged urved Surfaces E If the surface is curved the resultant force must be found by combining the elemental forces using some vectorial method. F O H alculate the and components. ombine these to obtain the resultant force and direction. (lthough this can be done for all three dimensions we will only look at one vertical plane) v The fluid is at rest in equilibrium. So of fluid such as is also in. In the diagram below liquid is resting on top of a curved base. IVE1400: Fluid Mechanics Lecture 6 83 IVE1400: Fluid Mechanics Lecture 6 84
11 onsider the Horizontal forces The sum of the horizontal forces is zero. The resultant horizontal force of a fluid above a curved surface is: H F No horizontal force on as there are no shear forces in a static fluid Horizontal forces act only on the faces and as shown. H We know 1. The force on a vertical plane must act horizontally (as it acts normal to the plane).. That H must act through the same point. So: H acts horizontally through the of the of the curved surface onto an vertical plane. F, must be equal and opposite to H. is the projection of the curved surface onto a vertical plane. We have seen earlier how to calculate resultant forces and point of action. Hence we can calculate the resultant horizontal force on a curved surface. IVE1400: Fluid Mechanics Lecture 6 85 IVE1400: Fluid Mechanics Lecture 6 86 onsider the Vertical forces The sum of the vertical forces is zero. E esultant force The overall resultant force is found by combining the vertical and horizontal components vectorialy, esultant force v There are no shear force on the vertical edges, so the vertical component can only be due to the of the fluid. So we can say The resultant vertical force of a fluid above a curved surface is: V. It will act vertically down through the centre of gravity of the mass of fluid. nd acts through O at an angle of. The angle the resultant force makes to the horizontal is The position of O is the point of intersection of the horizontal line of action of H and the vertical line of action of V. IVE1400: Fluid Mechanics Lecture 6 87 IVE1400: Fluid Mechanics Lecture 6 88
12 typical example application of this is the determination of the forces on dam walls or curved sluice gates. Find the magnitude and direction of the resultant force of water on a quadrant gate as shown below. 1.0m ate width 3.0m What are the forces if the fluid is below the curved surface? This situation may occur or a curved sluice gate. F O H Water ρ 1000 kg/m 3 v The force calculation is very similar to when the fluid is above. IVE1400: Fluid Mechanics Lecture 6 89 IVE1400: Fluid Mechanics Lecture 6 90 Horizontal force Vertical force F O H The two horizontal on the element are: The horizontal reaction force H The force on the vertical plane. The resultant horizontal force, H acts as shown in the diagram. Thus we can say: The resultant horizontal force of a fluid below a curved surface is: H v What vertical force would keep this in equilibrium? If the region above the curve were all water there would be equilibrium. Hence: the force exerted by this amount of fluid must equal he resultant force. The resultant vertical force of a fluid below a curved surface is: v Weight of the volume of fluid the curved surface. IVE1400: Fluid Mechanics Lecture 6 91 IVE1400: Fluid Mechanics Lecture 6 9
13 The resultant force and direction of application are calculated in the same way as for fluids above the surface: Example 1: m wide wall as shown in the figure below is fitted with a sluice gate at the base. Find the force on the gate. esultant force.5m nd acts through O at an angle of. The angle the resultant force makes to the horizontal is 1m 1m Force on a submerged surface, pressure at centroid rea of surface IVE1400: Fluid Mechanics Lecture 6 93 IVE1400: Fluid Mechanics Lecture 6 94 Exampel : What would be the force if the gate was changed to a circular one arranged as below?.5m 1m Vertical force weight of (real or imaginary) fluid above the surface. v gvolume above surface v v h 1m ( h + v ) 1/ Horizontal force on curved surface force on projection on to a vertical surface v i.e. equivalent to this figure:.5m h 1m h gh h IVE1400: Fluid Mechanics Lecture 6 95 IVE1400: Fluid Mechanics Lecture 6 96
14 Example 3: curved sluice gate which experiences force from fluid below. 1.5m long cylinder lies as shown in the figure, holding back oil of relative density 0.8. If the cylinder has a mass of 50 kg find a) the reaction at b) the reaction at E IVE1400: Fluid Mechanics Lecture 6 97
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