ENGINEERING COUNCIL CERTIFICATE LEVEL ENGINEERING SCIENCE C103 TUTORIAL 11 HYROSTATICS Elements of this tutorial ma be skipped if ou are alread familiar with the subject matter. On completion of this tutorial ou should be able to do the following. efine the main fundamental properties of liquids. Explain Archimedes s Principle Calculate the pressure due to the depth of a liquid. Calculate the total force on a vertical surface and an inclined surface. efine and calculate the position of the centre of pressure for various shapes. Calculate the turning moments produced on vertical and inclined surfaces. Before ou start ou should make sure that ou full understand first and second moments of area. If ou are not familiar with this, ou should do that tutorial before proceeding. Let s start this tutorial b studing the fundamental properties of liquids..j.unn freestud.co.uk 1
1. SOME FUNAMENTAL STUIES 1.1 IEAL LIQUIS An ideal liquid is defined as follows. It is INVISCI. This means that molecules require no force to separate them. The topic is covered in detail in chapter 3. It is INCOMPRESSIBLE. This means that it would require an infinite force to reduce the volume of the liquid. 1. REAL LIQUIS VISCOSITY Real liquids have VISCOSITY. This means that the molecules tend to stick to each other and to an surface with which the come into contact. This produces fluid friction and energ loss when the liquid flows over a surface. Viscosit defines how easil a liquid flows. The lower the viscosit, the easier it flows. BULK MOULUS Real liquids are compressible and this is governed b the BULK MOULUS K. This is defined as follows. K = V p/ V p is the increase in pressure, V is the reduction in volume and V is the original volume. ENSITY ensit ρ relates the mass and volume such that ρ = m/v kg/ m 3 PRESSURE Pressure is the result of compacting the molecules of a fluid into a smaller space than it would otherwise occup. Pressure is the force per unit area acting on a surface. The unit of pressure is the N/m and this is called a PASCAL. The Pascal is a small unit of pressure so higher multiples are common. 1 kpa = 10 3 N/m 1 MPa = 10 6 N/m Another common unit of pressure is the bar but this is not an SI unit. 1 bar = 10 5 N/m 1 mb = 100 N/m GAUGE AN ABSOLUTE PRESSURE Most pressure gauges are designed onl to measure and indicate the pressure of a fluid above that of the surrounding atmosphere and indicate zero when connected to the atmosphere. These are called gauge pressures and are normall used. Sometimes it is necessar to add the atmospheric pressure onto the gauge reading in order to find the true or absolute pressure. Absolute pressure = gauge pressure + atmospheric pressure. Standard atmospheric pressure is 1.013 bar..j.unn freestud.co.uk
. HYROSTATIC FORCES.1 HYROSTATIC PRESSURE.1.1 PRESSURE INSIE PIPES AN VESSELS Pressure results when a liquid is compacted into a volume. The pressure inside vessels and pipes produce stresses and strains as it tries to stretch the material. An example of this is a pipe with flanged joints. The pressure in the pipe tries to separate the flanges. The force is the product of the pressure and the bore area. Fig.1 WORKE EXAMPLE No. 1 Calculate the force tring to separate the flanges of a valve (Fig.1) when the pressure is MPa and the pipe bore is 50 mm. SOLUTION Force = pressure x bore area Bore area = π /4 = π x 0.05 /4 = 1.963 x 10-3 m Pressure = x 10 6 Pa Force = x 10 6 x 1.963 x 10-3 = 3.97 x 10 3 N or 3.97 kn.1. PRESSURE UE TO THE WEIGHT OF A LIQUI Consider a tank full of liquid as shown. The liquid has a total weight W and this bears down on the bottom and produces a pressure p. Pascal showed that the pressure in a liquid alwas acts normal (at 90 o ) to the surface of contact so the pressure pushes down onto the bottom of the tank. He also showed that the pressure at a given point acts equall in all directions so the pressure also pushes up on the liquid above it and sidewas against the walls. The volume of the liquid is V = A h m 3 The mass of liquid is hence m = ρv = ρah kg The weight is obtained b multipling b the gravitational constant g. Fig. W = mg = ρahg Newton The pressure on the bottom is the weight per unit area p = W/A N/m It follows that the pressure at a depth h in a liquid is given b the following equation. p = ρgh The unit of pressure is the N/m and this is called a PASCAL. The Pascal is a small unit of pressure so higher multiples are common..j.unn freestud.co.uk 3
WORKE EXAMPLE Calculate the pressure and force on an inspection hatch 0.75 m diameter located on the bottom of a tank when it is filled with oil of densit 875 kg/m 3 to a depth of 7 m. SOLUTION The pressure on the bottom of the tank is found as follows. p = ρ g h ρ = 875 kg/m 3 g = 9.81 m/s h = 7 m p = 875 x 9.81 x 7 = 60086 N/m or 60.086 kpa The force is the product of pressure and area. A = π /4 = π x 0.75 /4 = 0.44 m F = p A = 60.086 x 10 3 x 0.44 = 6.55 x 10 3 N or 6.55 Kn.1.3 PRESSURE HEA When h is made the subject of the formula, it is called the pressure head. h = p/ρg Pressure is often measured b using a column of liquid. Consider a pipe carring liquid at pressure p. If a small vertical pipe is attached to it, the liquid will rise to a height h and at this height, the pressure at the foot of the column is equal to the pressure in the pipe. Fig.3 This principle is used in barometers to measure atmospheric pressure and manometers to measure gas pressure. In the manometer, the weight of the gas is negligible so the height h represents the difference in the pressures p 1 and p. p 1 - p = ρ g h Barometer Fig.4 Manometer In the case of the barometer, the column is closed at the top so that p = 0 and p 1 = p a. The height h represents the atmospheric pressure. Mercur is used as the liquid because it does not evaporate easil at the near total vacuum on the top of the column. P a = ρ g h.j.unn freestud.co.uk 4
WORKE EXAMPLE No.3 A manometer (fig.4) is used to measure the pressure of gas in a container. One side is connected to the container and the other side is open to the atmosphere. The manometer contains oil of densit 750 kg/m 3 and the head is 50 mm. Calculate the gauge pressure of the gas in the container. SOLUTION p 1 - p = ρ g h = 750 x 9.81 x 0.05 = 367.9 Pa Since p is atmospheric pressure, this is the gauge pressure. p = 367.9 Pa (gauge) 3. ARCHIMEES PRINCIPLE Consider a clinder floating in a liquid as shown. The pressure on the bottom is p = ρ g h The force pushing upwards is F = pa = ρgha and this must be equal to the weight of the clinder. ha is the volume of the liquid that is displaced b the clinder. ρgha is the weight of the liquid displaced b the clinder. It follows that a floating bod displaces its own weight of liquid. This is Archimedes principle. Since g is a constant it also follows that it follows that a floating bod displaces its own mass of liquid..j.unn freestud.co.uk 5
WORKE EXAMPLE No. 4 A ship is made from steel. It rests in a dr dock as shown. The dr dock is 80 m long and 40 m wide. When seawater is allowed into the dr dock, it is found that the ship just starts to float when the level reaches 10 m from the bottom. The volume of water that was allowed in was estimated to be 0000 m 3. Calculate: The mass of the ship. The volume of steel used to make the ship. The pressure on the bottom of the ship. The densit of sea water is 1036 kg/m 3 The densit of steel is 7830 kg/m 3 SOLUTION Volume of water without the ship = 80 x 40 x 10 = 3000 m 3 Fig.5 Volume of water with ship = 0000 m 3 Volume displaced = 1 000 m 3 Mass of water displaced = 1 000 x 1036 = 14300 kg Mass of the ship is hence 14300 kg This is all steel so the volume of steel is V = Mass/densit = 14300 /7830 = 158.77 m 3 The pressure on the bottom = ρ g h = 1036 x 9.81 x 10 = 101.63 kpa SELF ASSESSMENT EXERCISE No.1 1. A mercur barometer gives a pressure head of 758 mm. The densit is 13 600 kg/m3. Calculate the atmospheric pressure in bar. (1.0113 bar). A manometer (fig.4) is used to measure the pressure of gas in a container. One side is connected to the container and the other side is open to the atmosphere. The manometer contains water of densit 1000 kg/m 3 and the head is 50 mm. Calculate the gauge pressure of the gas in the container. (.45.5 kpa) 3. Calculate the pressure and force on a horizontal submarine hatch 1. m diameter when it is at a depth of 800 m in seawater of densit 1030 kg/m 3. (8.083 MPa and 9.14 MN).J.UNN freestud.co.uk 6
4. FORCES ON SUBMERGE SURFACES 4.1 TOTAL FORCE Consider a vertical area submerged below the surface of liquid as shown. The area of the elementar strip is da = B d You should alread know that the pressure at depth h in a liquid is given b the equation p = ρgh where ρ is the densit and h the depth. In this case, we are using to denote depth so p = ρg The force on the strip due to this pressure is Fig.6 df = p da =ρb g d The total force on the surface due to pressure is denoted R and it is obtained b integrating this expression between the limits of 1 and. It follows that 1 R = ρgb ( 1)( + 1) This ma be factorised. R = ρgb ( - 1 ) = so B( - 1 ) = B =Area of the surface A ( + 1 )/ is the distance from the free surface to the centroid. It follows that the total force is given b the expression R = ρga The term A is the first moment of area and in general, the total force on a submerged surface is R = ρg x 1st moment of area about the free surface..j.unn freestud.co.uk 7
4. CENTRE OF PRESSURE The centre of pressure is the point at which the total force ma be assumed to act on a submerged surface. Consider the diagram again. The force on the strip is df as before. This force produces a turning moment with respect to the free surface s s. The turning moment due to df is as follows. dm = df = ρgbd The total turning moment about the surface due to pressure is obtained b integrating this expression between the limits of 1 and. M = ρgb d = ρgb B definition I Hence ss = B d d M = ρgiss This moment must also be given b the total force R multiplied b some distance h. The position at depth h is called the CENTRE OF PRESSURE. h is found b equating the moments. M = h R = h ρ g A = ρ g I ρ g Iss Iss h = = ρ g A A nd ss moment of area h = about s - s st 1 moment of area In order to be competent in this work, ou should be familiar with the parallel axis theorem (covered in part 1) because ou will need it to solve nd moments of area about the free surface. The rule is as follows. Iss = Igg + A Iss is the nd moment about the free surface and Igg is the nd moment about the centroid. You should be familiar with the following standard formulae for nd moments about the centroid. Rectangle I gg = B3/1 Rectangle about its edge I = B 3 /3 Circle I gg = π4/64.j.unn freestud.co.uk 8
WORKE EXAMPLE No.5 Show that the centre of pressure on a vertical retaining wall is at /3 of the depth. Go on to show that the turning moment produced about the bottom of the wall is given b the expression ρgh 3 /6 for a unit width. SOLUTION Fig. 7 For a given width B, the area is a rectangle with the free surface at the top edge. h = A = bh 1 st nd h moment of area about the top edge is A = B h moment of area about the top edge is B 3 3 h nd B moment 3 h h = = = st 1 moment h 3 B It follows that the centre of pressure is h/3 from the bottom. The total force is R = ρga = ρgbh / and for a unit width this is ρgh / The moment bout the bottom is R x h/3 = (ρgh /) x h/3 = ρgh 3 /6 3 SELF ASSESSMENT EXERCISE No. 1. A vertical retaining wall contains water to a depth of 0 metres. Calculate the turning moment about the bottom for a unit width. Take the densit as 1000 kg/m 3. (13.08 MNm). A vertical wall separates seawater on one side from fresh water on the other side. The seawater is 3.5 m deep and has a densit of 1030 kg/m 3. The fresh water is m deep and has a densit of 1000 kg/m 3. Calculate the turning moment produced about the bottom for a unit width. (59.1 knm).j.unn freestud.co.uk 9
WORKE EXAMPLE No. 6 A concrete wall retains water and has a hatch blocking off an outflow tunnel as shown. Find the total force on the hatch and the position of the centre of pressure. Calculate the total moment about the bottom edge of the hatch. The water densit is1000 kg/m3. SOLUTION Fig. 8 Total force = R = ρ g A For the rectangle shown = (1.5 + 3/) = 3 m. A = x 3 = 6 m. R = 1000 x 9.81 x 6 x 3 = 176580 N or 176.58 kn h = nd mom. of Area/ 1st mom. of Area 1 st moment of Area = A = 6 x 3 = 18 m3. nd mom of area = Iss = (B3/1) + A = ( x 33/1) + (6 x 3) Iss = 4.5 + 54 = 58.5 m4. h = 58.5/18 = 3.5 m The distance from the bottom edge is x = 4.5 3.5 = 1.5 m Moment about the bottom edge is = Rx = 176.58 x 1.5 = 0.75 knm..j.unn freestud.co.uk 10
WORKE EXAMPLE No. 7 Find the force required at the top of the circular hatch shown in order to keep it closed against the water pressure outside. The densit of the water is 1030 kg/m3. = m from surface to middle of hatch. Fig. 9 Total Force = R = ρ g A = 1030 x 9.81 x (π x /4) x = 63487 N or 63.487 kn Centre of Pressure h = nd moment/1 st moment nd moment of area. Iss = Igg + A =(π x 4/64) + (π x /4) x Iss =13.3518 m4. 1 st moment of area A = (π x /4) x = 6.83 m3. Centre of pressure. h = 13.3518/6.83 =.15 m This is the depth at which, the total force ma be assumed to act. Take moments about the hinge. F = force at top. R = force at centre of pressure which is 0.15 m below the hinge. For equilibrium F x 1 = 63.487 x 0.15 Fig. 10 F = 7.936 kn.j.unn freestud.co.uk 11
WORKE EXAMPLE No. 8 The diagram shows a hinged circular vertical hatch diameter that flips open when the water level outside reaches a critical depth h. Show that for this to happen the hinge must be located 8h - 5 at a position x from the bottom given b the formula x = 8h - 4 Given that the hatch is 0.6 m diameter, calculate the position of the hinge such that the hatch flips open when the depth reaches 4 metres. SOLUTION Fig.11 The hatch will flip open as soon as the centre of pressure rises above the hinge creating a clockwise turning moment. When the centre of pressure is below the hinge, the turning moment is anticlockwise and the hatch is prevented from turning in that direction. We must make the centre of pressure at position x. = h - h = h - x second moment of area h = about the surface first moment of area I h = + = h - x 16 x = h - = h - h - 16 16 h - x = - + A A Equate for h gg ( 16h - 8) ( 16h - 8) x = 1- π π + = 64 4 π 4 + 8h - 4 4 = - 8h - 4 - = 8h - 4 = + 16 = Putting = 0.6 and h = 4 we get x = 0.5 m 8h - 5 8h - 4.J.UNN freestud.co.uk 1
SELF ASSESSMENT EXERCISE No.3 1. A circular hatch is vertical and hinged at the bottom. It is m diameter and the top edge is m below the free surface. Find the total force, the position of the centre of pressure and the force required at the top to keep it closed. The densit of the water is 1000 kg/m3. (9.469 kn, 3.08 m,4.5 kn). A large tank of sea water has a door in the side 1 m square. The top of the door is 5 m below the free surface. The door is hinged on the bottom edge. Calculate the force required at the top to keep it closed. The densit of the sea water is 1036 kg/m3. (7.11 N) 3. A culvert in the side of a reservoir is closed b a vertical rectangular gate m wide and 1 m deep as shown in fig. 11. The gate is hinged about a horizontal axis which passes through the centre of the gate. The free surface of water in the reservoir is.5 m above the axis of the hinge. The densit of water is 1000 kg/m3. Assuming that the hinges are frictionless and that the culvert is open to atmosphere, determine (i) the force acting on the gate when closed due to the pressure of water. (55.897 kn) (ii) the moment to be applied about the hinge axis to open the gate. (1635 Nm) Fig.1.J.UNN freestud.co.uk 13
4. The diagram shows a rectangular vertical hatch breadth B and depth. The hatch flips open when the water level outside reaches a critical depth h. Show that for this to happen the hinge must be located at a position x from the bottom given b the formula 6h - 4 x = 6h - 3 Given that the hatch is 1 m deep, calculate the position of the hinge such that the hatch flips open when the depth reaches 3 metres. (0.466 m) Fig.13 5. This question is much more advanced than required for the module and should be attempted onl if ou reall wish to test ourself. Fig.14 shows an L shaped spill gate that operates b pivoting about hinge O when the water level in the channel rises to a certain height H above O. A counterweight W attached to the gate provides closure of the gate at low water levels. With the channel empt the force at sill S is 1.635 kn. The distance l is 0.5m and the gate is m wide. etermine the magnitude of H. (i) when the gate begins to open due to the hdrostatic load. (1 m) (ii) when the force acting on the sill becomes a maximum. What is the magnitude of this force. (0.5 m) Assume the effects of friction are negligible. Fig.14.J.UNN freestud.co.uk 14
5. INCLINE SURFACES Consider a sloping surface as shown. For a vertical surface the total force is R = ρga This is exactl the same for the inclined surface so long as A is the actual area. If we wish to use distances measured from point O we have the following. Figure 15 G = /sin θ so R = ρg A G sin θ IO For the inclined surface cp = AG It is convenient to use the parallel axis theorem I O = I G + A G Figure 15 IG + AG IG cp = = AG AG + G For a rectangular surface B b with the top edge at O the centre of pressure is at /3 the vertical depth as before. WORKE EXAMPLE No.9 A circular hatch 1.5 m diameter is immersed in water and inclined at 46 o to the horizontal. The top edge is 0.6 m below the surface. Calculate the total force and the vertical depth to the centre of pressure. SOLUTION Figure 16 A = π /4 = π x 1.5 /4 = 1.8 m istance OA = 0.6/sin 46 = 0.834 m G = 0.834 + 1.5/ = 1.46 m = 1.46 sin 46 = 1.05 m R = ρga = 1000 x 9.81 x 1.8 x 1.05 = 1650 N IO cp = AG Using the parallel axis theorem I O = I G + A G I G = π 4 /64 = 0.1198 m 4 I O = 0.1198 + 1.8 x 1.46 =.737 m 4 A G = 1.8 x 1.46 = 1.793 m 3 IO.737 cp = = = 1.56 m A 1.793 G h = 1.56 sin 46 = 1.098 m.j.unn freestud.co.uk 15
WORKE EXAMPLE No. 10 A rectangular surface of width B is inclined at angle θ as shown. The top edge is distance h from the free surface and the depth to the bottom is d. etermine the expression for total force and the vertical depth to the centre of pressure. Figure 17 R = ρga A = B x length length = L = (h h 1 )/sin θ A = B (h h 1 )/sin θ = (h + h 1 )/ R = (ρgb) (h h 1) (h + h 1 )/sin θ ( h + h1 ) IG + AG IG R = ρgb w = = + G sinθ Az AG I G = BL 3 /1 G = /sin θ 3 BL w = 1 BL G w = 6 sinθ ( h - h ) ( h - h ) 1 w = 1 sin θ w = 1 sinθ h = wsinθ L + z = 1 ( h - h1) ( h + h1) ( h - h1) ( h + h ) ( h h1) 6( h + h ) G 1 G + + + ( h + h ) sinθ ( h + h ) G 1 + z = + 1 sinθ ( h h ) + 1 h = + 1 sinθ Note if h = 0 this gives h = h 3 1 1 sinθ 3 3 ( h h1 ) Also note that that deriving from first principles ields h 3( h + h ) the same result. = and this apparentl gives 1.J.UNN freestud.co.uk 16
SELF ASSESSMENT EXERCISE No.4 1. The inclined surface in worked example 10 is hinged about the bottom edge. Show that the moment of force about this point is given b the following expression. 3 3 ( d d h + dh ) ρgb M = h 6sin θ. A dam has a wall inclined at 60 o to the horizontal and is filled with water to a depth of 0 m. Calculate the total force and the overturning moment on the wall per unit width about the base. (.66 MN and 17.44 MNm) 3. A large tank of water has a wall sloping outwards at 80 o to the horizontal and a circular hatch seals a hole m diameter in this wall. Calculate the total force on the hatch when it is filled to a depth of 3 m above the top edge of the hatch. The hatch pivots about its horizontal centre line. Calculate the turning moment about the pivot. (1.9 kn and 7.59 knm).j.unn freestud.co.uk 17