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CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers Pressure and Manometers b) 1.1 p gh What will be the (a) the gauge pressure and (b) the absolute pressure of water at depth 12m below the 10 3 9.81 0.4 3924 N / m 2 surface? water = 1000 kg/m3, and p atmosphere = 101kN/m2. [117.72 kN/m2, 218.72 kN/m2] c) g a) p gh p gauge gh 7.9 10 3 0.4 3160 N / m 2 1000 9.81 12 d) 117 720 N / m 2 , ( Pa ) p gh 117.7 kN / m2 , ( kPa ) 520 9.81 0.4 2040 N / m2 b) 1.4 pabsolute p gauge patmospheric A manometer connected to a pipe indicates a negative gauge pressure of 50mm of mercury. What is the (117 720 101) N / m2 , ( Pa ) absolute pressure in the pipe in Newtons per square metre is the atmospheric pressure is 1 bar? [93.3 kN/m2] 218.7 kN / m2 , ( kPa ) patmosphere 1bar 1 105 N / m2 1.2 At what depth below the surface of oil, relative density 0.8, will produce a pressure of 120 kN/m2? What pabsolute p gauge patmospheric depth of water is this equivalent to? gh patmospheric [15.3m, 12.2m] 13.6 10 3 9.81 0.05 105 N / m2 , ( Pa ) a) 93.33 kN / m2 , ( kPa ) water 1.5 0.8 1000 kg / m 3 What height would a water barometer need to be to measure atmospheric pressure? p gh [>10m] p 120 10 3 patmosphere 1bar 1 105 N / m2 h 15.29m of oil g 800 9.81 5 10 gh b) 105 h 1019 m of water . 1000 kg / m3 1000 9.81 5 120 10 3 10 h 12.23 m of water h 0.75 m of mercury 1000 9.81 (13.6 103 ) 9.81 1.3 What would the pressure in kN/m2 be if the equivalent head is measured as 400mm of (a) mercury =13.6 (b) water ( c) oil specific weight 7.9 kN/m3 (d) a liquid of density 520 kg/m3? [53.4 kN/m2, 3.92 kN/m2, 3.16 kN/m2, 2.04 kN/m2] a) water 13.6 1000 kg / m3 p gh 13.6 103 9.81 0.4 53366 N / m2 Examples: Answers CIVE1400: Fluid Mechanics 1 Examples: Answers CIVE1400: Fluid Mechanics 2 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 1.6 1.7 An inclined manometer is required to measure an air pressure of 3mm of water to an accuracy of +/- 3%. Determine the resultant force due to the water acting on the 1m by 2m rectangular area AB shown in the The inclined arm is 8mm in diameter and the larger arm has a diameter of 24mm. The manometric fluid diagram below. has density 740 kg/m3 and the scale may be read to +/- 0.5mm. [43 560 N, 2.37m from O] What is the angle required to ensure the desired accuracy may be achieved? O [12 39’] P p2 p1 diameter d 1.22m 1.0m 45° A C diameter D x d er Rea 2.0 m 2.0 m a le z2 Sc B D Datum line z1 The magnitude of the resultant force on a submerged plane is: R = pressure at centroid area of surface θ R gz A 1000 9.81 . 122 1 1 2 43 556 N / m2 p1 p2 man gh man g z1 z2 This acts at right angle to the surface through the centre of pressure. z2 Volume moved from left to right = z1 A1 A xA2 I OO 2nd moment of area about a line through O sin 2 Sc 2 2 2 Ax 1st moment of area about a line through O D z2 d d z1 x 4 sin 4 4 By the parallel axis theorem (which will be given in an exam), I oo I GG Ax 2 , where IGG is the 2nd z2 d 2 d2 moment of area about a line through the centroid and can be found in tables. z1 2 x 2 sin D D I GG Sc x d 2 Ax p1 p2 man gx sin D2 b 2 d water gh man gx sin D2 G G 0.008 2 water gh 0.74 water gx sin d 0.024 2 bd 3 For a rectangle I GG h 0.74 x (sin 01111) . 12 The head being measured is 3% of 3mm = 0.003x0.03 = 0.00009m As the wall is vertical, Sc D and x z, This 3% represents the smallest measurement possible on the manometer, 0.5mm = 0.0005m, giving 1 23 Sc . 122 1 0.00009 0.74 0.0005 (sin 01111) . . 12 1 2 122 1 2.37 m from O sin . 0132 7.6 [This is not the same as the answer given on the question sheet] Examples: Answers CIVE1400: Fluid Mechanics 3 Examples: Answers CIVE1400: Fluid Mechanics 4 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 1.8 Forces on submerged surfaces Determine the resultant force due to the water acting on the 1.25m by 2.0m triangular area CD shown in 2.1 the figure above. The apex of the triangle is at C. Obtain an expression for the depth of the centre of pressure of a plane surface wholly submerged in a [43.5 103N, 2.821m from P] fluid and inclined at an angle to the free surface of the liquid. b A horizontal circular pipe, 1.25m diameter, is closed by a butterfly disk which rotates about a horizontal axis through its centre. Determine the torque which would have to be applied to the disk spindle to keep the disk closed in a vertical position when there is a 3m head of fresh water above the axis. [1176 Nm] G G The question asks what is the moment you have to apply to the spindle to keep the disc vertical i.e. to d keep the valve shut? d/3 bd 3 For a triangle I GG 36 So you need to know the resultant force exerted on the disc by the water and the distance x of this force from the spindle. 2 Depth to centre of gravity is z 10 2 cos 45 1943m . . . We know that the water in the pipe is under a pressure of 3m head of water (to the spindle) 3 R gz A 2.0 125. . 1000 9.81 1943 2.375 h 3 2.0 2 h’ 23826 N / m Distance from P is x z / cos 45 2.748m F Distance from P to centre of pressure is x I oo Sc Diagram of the forces on the disc valve, based on an imaginary water surface. Ax I oo I GG Ax 2 h 3m , the depth to the centroid of the disc 3 I GG . 125 2 h’ = depth to the centre of pressure (or line of action of the force) Sc x 2.748 Ax . 36 125 2.748 Calculate the force: 2.829m F gh A 2 1.25 1000 9.81 3 2 36.116 kN Calculate the line of action of the force, h’. 2nd moment of area about water surface h' 1st moment of area about water surface I oo Ah By the parallel axis theorem 2nd moment of area about O (in the surface) I oo I GG Ah 2 where IGG is the 2nd moment of area about a line through the centroid of the disc and IGG = r4/4. Examples: Answers CIVE1400: Fluid Mechanics 5 Examples: Answers CIVE1400: Fluid Mechanics 6 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers I GG If we take moments from the surface, h' h Ah DR fd1 fd 2 fd 3 r4 3 D 3f f d1 d 2 d3 4( r 2 ) 3 12 d1 d 2 d3 r2 3 3.0326m Taking the first beam, we can draw a pressure diagram for this, (ignoring what is below), 12 So the distance from the spindle to the line of action of the force is 2H/3 x h' h 3.0326 3 0.0326m And the moment required to keep the gate shut is H F=58860 moment Fx 36.116 0.0326 1176 kN m . 2.2 A dock gate is to be reinforced with three horizontal beams. If the water acts on one side only, to a depth 1 2F of 6m, find the positions of the beams measured from the water surface so that each will carry an equal We know that the resultant force, F gH 2 , so H 2 g load. Give the load per meter. [58 860 N/m, 2.31m, 4.22m, 5.47m] 2F 2 58860 H 3.46 m First of all draw the pressure diagram, as below: g 1000 9.81 And the force acts at 2H/3, so this is the position of the 1st beam, d1 2 h 2h/3 position of 1st beam H 2.31m 3 f d2 Taking the second beam into consideration, we can draw the following pressure diagram, R d3 f d1=2.31 H 2H/3 f f d2 The resultant force per unit length of gate is the area of the pressure diagram. So the total resultant force is F=2 58860 f 1 R gh 2 = 0.5 1000 9.81 62 176580 N ( per m length) 2 The reaction force is equal to the sum of the forces on each beam, so as before Alternatively the resultant force is, R = Pressure at centroid Area , (take width of gate as 1m to give 2F 2 (2 58860) force per m) H 4.9 m g 1000 9.81 h R g h 1 176580 N ( per m length) The reaction force acts at 2H/3, so H=3.27m. Taking moments from the surface, 2 (2 58860) 3.27 58860 2.31 58860 d 2 This is the resultant force exerted by the gate on the water. depth to second beam d 2 4.22 m The three beams should carry an equal load, so each beam carries the load f, where For the third beam, from before we have, R f 58860 N 12 d1 d 2 d 3 3 depth to third beam d 3 12 2.31 4.22 5.47m Examples: Answers CIVE1400: Fluid Mechanics 7 Examples: Answers CIVE1400: Fluid Mechanics 8 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 2.3 As this force act normal to the surface, it must act through the centre of radius of the dam wall. So the The profile of a masonry dam is an arc of a circle, the arc having a radius of 30m and subtending an angle depth to the point where the force acts is, of 60 at the centre of curvature which lies in the water surface. Determine (a) the load on the dam in N/m y = 30sin 39.31 =19m length, (b) the position of the line of action to this pressure. [4.28 106 N/m length at depth 19.0m] 2.4 The arch of a bridge over a stream is in the form of a semi-circle of radius 2m. the bridge width is 4m. Draw the dam to help picture the geometry, Due to a flood the water level is now 1.25m above the crest of the arch. Calculate (a) the upward force on R the underside of the arch, (b) the horizontal thrust on one half of the arch. [263.6 kN, 176.6 kN] a The bridge and water level can be drawn as: 60° 1.25m FR y h R 2m Fh a) The upward force on the arch = weight of (imaginary) water above the arch. Rv g volume of water Fv 22 volume (125 2) 4 . 4 26.867 m 3 2 h 30 sin 60 2598 m . Rv 1000 9.81 26.867 263568 kN . a 30 cos 60 150 m . b) Calculate Fv = total weight of fluid above the curved surface (per m length) The horizontal force on half of the arch, is equal to the force on the projection of the curved surface onto Fv g (area of sector - area of triangle) a vertical plane. 60 . 2598 15 = 1000 9.81 302 1.25 360 2 2711375 kN / m . Calculate Fh = force on projection of curved surface onto a vertical plane 1 Fh gh 2 2.0 2 05 1000 9.81 25982 . . 3310.681 kN / m The resultant, Fh pressure at centroid area FR F 2 F 2 3310.681 2 2711375 . 2 g 125 1 . 2 4 v h 176.58 kN 4279.27 kN / m acting at the angle 2.5 Fv tan 0.819 The face of a dam is vertical to a depth of 7.5m below the water surface then slopes at 30 to the vertical. Fh If the depth of water is 17m what is the resultant force per metre acting on the whole face? 39.32 [1563.29 kN] Examples: Answers CIVE1400: Fluid Mechanics 9 Examples: Answers CIVE1400: Fluid Mechanics 10 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers d1 d2 h2 d3 f1 F f2 f3 h1 60 density of oil oil = 0.9 water = 900 kg/m3. Force per unit length, F = area under the graph = sum of the three areas = f1 + f2 + f3 x (900 9.81 2) 2 h2 = 17.0 m, so h1 = 17.0 - 7.5 = 9.5 . x = 9.5/tan 60 = 5.485 m. f1 3 52974 N 2 Vertical force = weight of water above the surface, f2 (900 9.81 2) 15 3 79461 N . Fv g h2 x 0.5h1 x . (1000 9.81 15) 15 . f3 3 33109 N 9810 7.5 5.485 0.5 9.5 5.485 2 F f 1 f 2 f 3 165544 N 659.123 kN / m To find the position of the resultant force F, we take moments from any point. We will take moments The horizontal force = force on the projection of the surface on to a vertical plane. about the surface. 1 Fh gh 2 DF f 2 d 2 f 3d 3 f 1d1 2 2 15 . 2 0.5 1000 9.81 17 2 165544 D 52974 2 79461 (2 ) 33109 (2 . 15) 3 2 3 1417.545 kN / m D 2.347m ( from surface) The resultant force is 1153m ( from base of wall) . FR Fv2 Fh2 659.1232 1417.5452 1563.29 kN / m And acts at the angle Fv tan 0.465 Fh 24.94 2.6 A tank with vertical sides is square in plan with 3m long sides. The tank contains oil of relative density 0.9 to a depth of 2.0m which is floating on water a depth of 1.5m. Calculate the force on the walls and the height of the centre of pressure from the bottom of the tank. [165.54 kN, 1.15m] Consider one wall of the tank. Draw the pressure diagram: Examples: Answers CIVE1400: Fluid Mechanics 11 Examples: Answers CIVE1400: Fluid Mechanics 12 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 2 2 2 Application of the Bernoulli Equation pB p A u B u A uA zA k 3.1 1000 g 2g 2g In a vertical pipe carrying water, pressure gauges are inserted at points A and B where the pipe diameters are 0.15m and 0.075m respectively. The point B is 2.5m below A and when the flow rate down the pipe is 15 2.5 1045 0.065 . . 0.065k 0.02 cumecs, the pressure at B is 14715 N/m2 greater than that at A. v2 Assuming the losses in the pipe between A and B can be expressed as k where v is the velocity at A, k 0.319 2g find the value of k. Part ii) If the gauges at A and B are replaced by tubes filled with water and connected to a U-tube containing p xxL gz B pB w mercury of relative density 13.6, give a sketch showing how the levels in the two limbs of the U-tube differ and calculate the value of this difference in metres. p xxR m gR p w gz A w gR p pA [k = 0.319, 0.0794m] p xxL p xxR w gz B pB m gR p w gz A w gR p pA dA = 0.2m A pB pA w g zA zB gR P m w 14715 1000 9.81 2.5 9.81R p 13600 1000 Rp 0.079 m dB = 0.2m B 3.2 A Venturimeter with an entrance diameter of 0.3m and a throat diameter of 0.2m is used to measure the volume of gas flowing through a pipe. The discharge coefficient of the meter is 0.96. Assuming the specific weight of the gas to be constant at 19.62 N/m3, calculate the volume flowing when Rp the pressure difference between the entrance and the throat is measured as 0.06m on a water U-tube manometer. [0.816 m3/s] Part i) d2 = 0.2m dA 0.15m dB 0.075m Q 0.02 m 3 / s d1 = 0.3m pB pA 14715 N / m 2 kv 2 hf 2g Taking the datum at B, the Bernoulli equation becomes: Z2 pA u2 A pB 2 uB u2 A zA zB k g 2g g 2g 2g Z1 hRp zA 2.5 zB 0 By continuity: Q = uAAA = uBAB uA 0.02 / 0.075 2 1132 m / s . uB 0.02 / 0.0375 2 4.527 m / s giving Examples: Answers CIVE1400: Fluid Mechanics 13 Examples: Answers CIVE1400: Fluid Mechanics 14 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers What we know from the question: 3.3 A Venturimeter is used for measuring flow of water along a pipe. The diameter of the Venturi throat is gg 19.62 N / m 2 two fifths the diameter of the pipe. The inlet and throat are connected by water filled tubes to a mercury Cd 0.96 U-tube manometer. The velocity of flow along the pipe is found to be 2.5 H m/s, where H is the d1 0.3m manometer reading in metres of mercury. Determine the loss of head between inlet and throat of the d2 0.2 m Venturi when H is 0.49m. (Relative density of mercury is 13.6). [0.23m of water] Calculate Q. u1 Q / 0.0707 u2 Q / 0.0314 For the manometer: p1 g gz p2 g g z2 Rp w gR p p1 p2 19.62 z 2 z1 587.423 (1) For the Venturimeter 2 2 p1 u1 p2 u2 z1 z2 Z2 gg 2g gg 2g 2 p1 p2 19.62 z 2 z1 0.803u 2 (2 ) Z1 H h Combining (1) and (2) 2 0.803u 2 587.423 u 2 ideal 27.047 m / s 2 0.2 For the manometer: Qideal 27.047 0.85m 3 / s 2 p1 gz1 p2 g z2 H gH w w m Q Cd Qidea 0.96 0.85 0.816m 3 / s p1 p2 gz2 gH gH gz1 (1) w w m w For the Venturimeter p1 u12 p2 2 u2 z1 z2 Losses wg 2g wg 2g 2 2 u w 2 u w 1 p1 p2 w gz2 w gz1 L w g ( 2) 2 2 Combining (1) and (2) p1 u12 p2 2 u2 z1 z2 Losses wg 2g wg 2g w 2 L w g Hg m w u2 u12 (3) 2 but at 1. From the question u1 2.5 H 175m / s . u1 A1 u 2 A2 2 d2 2d 175 . u2 4 10 u2 10.937 m / s Examples: Answers CIVE1400: Fluid Mechanics 15 Examples: Answers CIVE1400: Fluid Mechanics 16 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 2 Substitute in (3) u3 . 183 0.49 9.81 13600 1000 1000 / 2 10.937 2 1.75 2 2g Losses L u3 5.99 m / s 9.81 1000 0.233m Q u 3 A3 2 3.4 d3 0.02665 5.99 Water is discharging from a tank through a convergent-divergent mouthpiece. The exit from the tank is 4 rounded so that losses there may be neglected and the minimum diameter is 0.05m. d3 0.0752m If the head in the tank above the centre-line of the mouthpiece is 1.83m. a) What is the discharge? b) What must be the diameter at the exit if the absolute pressure at the minimum area is to be 2.44m of If the mouth piece has been removed, p1 p2 water? c) What would the discharge be if the divergent part of the mouth piece were removed. (Assume 2 p1 p2 u2 atmospheric pressure is 10m of water). z1 [0.0752m, 0.0266m3/s, 0.0118m3/s] g g 2g u2 2 gz1 5.99 m / s 0.052 Q 5.99 0.0118 m 3 / s 4 3.5 A closed tank has an orifice 0.025m diameter in one of its vertical sides. The tank contains oil to a depth h of 0.61m above the centre of the orifice and the pressure in the air space above the oil is maintained at 13780 N/m2 above atmospheric. Determine the discharge from the orifice. (Coefficient of discharge of the orifice is 0.61, relative density of oil is 0.9). 2 [0.00195 m3/s] 3 P = 13780 kN/m2 From the question: d2 0.05m p2 0.66m minimum pressure 2.44 m oil g p1 p3 10m g g Apply Bernoulli: do = 0.025m p1 u12 p2 2 u2 p3 2 u3 z1 z2 z3 From the question g 2g g 2g g 2g o If we take the datum through the orifice: 0.9 w z1 183m . z2 z3 0 u1 negligible 900 o Between 1 and 2 Cd 0.61 2 u2 Apply Bernoulli, . 10 183 2.44 2g p1 u12 p2 2 u2 u2 1357 m / s . z1 z2 g 2g g 2g 2 0.05 Q u2 A2 . 1357 0.02665 m 3 / s Take atmospheric pressure as 0, 2 Between 1 and 3 p1 p3 Examples: Answers CIVE1400: Fluid Mechanics 17 Examples: Answers CIVE1400: Fluid Mechanics 18 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 2 2 2 13780 u2 p1 u 1 p2 u2 0.61 hf og 2g g 2g g 2g 2 2 u2 6.53 m / s p1 p2 u 1 u 2 hf 0.025 2 g g 2g 2g Q 0.61 6.53 0.00195 m 3 / s 2 2 3.77 2 u2 25 2.5 3.6 2g 2g The discharge coefficient of a Venturimeter was found to be constant for rates of flow exceeding a certain u2 21.346 m / s value. Show that for this condition the loss of head due to friction in the convergent parts of the meter can Q u 2 A2 be expressed as KQ2 m where K is a constant and Q is the rate of flow in cumecs. d 22 Obtain the value of K if the inlet and throat diameter of the Venturimeter are 0.102m and 0.05m 0.0667 21.346 respectively and the discharge coefficient is 0.96. 4 [K=1060] d2 0.063m 3.7 3.8 A Venturimeter is to fitted in a horizontal pipe of 0.15m diameter to measure a flow of water which may A Venturimeter of throat diameter 0.076m is fitted in a 0.152m diameter vertical pipe in which liquid of be anything up to 240m3/hour. The pressure head at the inlet for this flow is 18m above atmospheric and relative density 0.8 flows downwards. Pressure gauges are fitted to the inlet and to the throat sections. the pressure head at the throat must not be lower than 7m below atmospheric. Between the inlet and the The throat being 0.914m below the inlet. Taking the coefficient of the meter as 0.97 find the discharge throat there is an estimated frictional loss of 10% of the difference in pressure head between these points. a) when the pressure gauges read the same b)when the inlet gauge reads 15170 N/m2 higher than the Calculate the minimum allowable diameter for the throat. throat gauge. [0.063m] [0.0192m3/s, 0.034m3/s] d1 = 0.15m d1 = 0.152m d2 From the question: d1 = 0.076m d1 015m . Q 240 m 3 / hr 0.667 m3 / s u1 Q / A 3.77 m / s p1 p2 18m 7m g g Friction loss, from the question: p1 p2 hf 0.1 g From the question: Apply Bernoulli: d1 0.152 m A1 0.01814 m d2 0.076m A2 0.00454 m 800 kg / m 3 Cd 0.97 Apply Bernoulli: Examples: Answers CIVE1400: Fluid Mechanics 19 Examples: Answers CIVE1400: Fluid Mechanics 20 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 2 2 p1 u1 p2 u2 Tank emptying z1 z2 g 2g g 2g 4.1 A reservoir is circular in plan and the sides slope at an angle of tan-1(1/5) to the horizontal. When the a) p1 p2 reservoir is full the diameter of the water surface is 50m. Discharge from the reservoir takes place u12 2 u2 through a pipe of diameter 0.65m, the outlet being 4m below top water level. Determine the time for the z1 z2 water level to fall 2m assuming the discharge to be 0.75a 2 gH cumecs where a is the cross sectional 2g 2g area of the pipe in m2 and H is the head of water above the outlet in m. By continuity: [1325 seconds] Q u1 A1 u 2 A2 50m A1 u2 u1 u1 4 A2 u12 16u12 r 0.914 2g 2g H 0.914 2 9.81 u1 10934 m / s . 15 x Q Cd A1 u1 Q 0.96 0.01814 10934 . 0.019 m 3 / s 1 b) 5 p1 p2 15170 2 From the question: H = 4m a = (0.65/2)2 = 0.33m2 p1 p2 u2 u12 0.914 g 2g Q 0.75a 2 gh 15170 Q 2 220.432 55112 . 10963 h . 0.914 g 2g In time t the level in the reservoir falls h, so 55.8577 Q 2 220.432 55112 . Q t A h Q 0.035 m 3 / s A t h Q Integrating give the total time for levels to fall from h1 to h2. h2 A T dh h1 Q As the surface area changes with height, we must express A in terms of h. A = r2 But r varies with h. It varies linearly from the surface at H = 4m, r = 25m, at a gradient of tan-1 = 1/5. r = x + 5h 25 = x + 5(4) x=5 so A = ( 5 + 5h )2 = ( 25 + 25 h2 + 50 h ) Substituting in the integral equation gives Examples: Answers CIVE1400: Fluid Mechanics 21 Examples: Answers CIVE1400: Fluid Mechanics 22 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 2 h2 25 25 h 50 h And we can write an equation for the discharge in terms of the surface height change: T dh h1 10963 h . Q t A h 2 25 h2 1 h 2h A dh t h 10963 . h1 h Q h2 1 h2 2h Integrating give the total time for levels to fall from h1 to h2. 71641 . dh h1 h h h h2 A h2 1/2 T dh 71641 . h h 3/ 2 2h 1/2 dh h1 Q h1 h2 h2 A 2 53/ 2 4 3/ 2 168 . dh (1) 71641 2h 1/2 . h h h1 h 5 3 h1 a) For the first 1m depth, A = 8 x 32 = 256, whatever the h. From the question, h1 = 4m h2 = 2m, so So, for the first period of time: 2 4 2 4 T 71641 2 4 1/2 . 4 53/ 2 4 3/ 2 2 2 1/2 2 53/ 2 2 3/ 2 h2 256 5 3 5 3 T 168 . dh h1 h 71641 4 12.8 10.667 . 2.828 2.263 3.77 430.08 h1 h2 71641 27.467 8.862 . 1333 sec 430.08 2.6 . 16 299 sec 4.2 A rectangular swimming pool is 1m deep at one end and increases uniformly in depth to 2.6m at the other end. The pool is 8m wide and 32m long and is emptied through an orifice of area 0.224m2, at the lowest b) now we need to find out how long it will take to empty the rest. point in the side of the deep end. Taking Cd for the orifice as 0.6, find, from first principles, a) the time for the depth to fall by 1m b) the time to empty the pool completely. We need the area A, in terms of h. [299 second, 662 seconds] A 8L 32.0m L 32 h 1.6 A 160h 1.0m So 2.6m h2 160h T 168 . dh h1 h L 2 3/ 2 3/ 2 268.9 h h2 3 1 2 3/ 2 3/ 2 The question tell us ao = 0.224m2, Cd = 0.6 268.9 16 . 0 3 Apply Bernoulli from the tank surface to the vena contracta at the orifice: 362.67 sec p1 u12 p2 2 u2 Total time for emptying is, z1 z2 g 2g g 2g T = 363 + 299 = 662 sec p1 = p2 and u1 = 0. u 2 2 gh We need Q in terms of the height h measured above the orifice. Q Cd a o u 2 Cd a o 2 gh 0.6 0.224 2 9.81 h 0.595 h Examples: Answers CIVE1400: Fluid Mechanics 23 Examples: Answers CIVE1400: Fluid Mechanics 24 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 4.3 Q t A h A vertical cylindrical tank 2m diameter has, at the bottom, a 0.05m diameter sharp edged orifice for A which the discharge coefficient is 0.6. t h Q a) If water enters the tank at a constant rate of 0.0095 cumecs find the depth of water above the orifice when the level in the tank becomes stable. Integrating between h1 and h2, to give the time to change surface level b) Find the time for the level to fall from 3m to 1m above the orifice when the inflow is turned off. c) If water now runs into the tank at 0.02 cumecs, the orifice remaining open, find the rate of rise in water h2 A T dh level when the level has reached a depth of 1.7m above the orifice. h1 Q [a) 3.314m, b) 881 seconds, c) 0.252m/min] h2 1/ 2 . 6018 h dh h1 1/ 2 h2 1203.6 h Q = 0.0095 m3/s h1 1 1203.6 h2 / 2 h11/ 2 h1 = 3 and h2 = 1 so T = 881 sec h c) Qin changed to Qin = 0.02 m3/s From (1) we have Qout 0.00522 h . The question asks for the rate of surface rise when h = 1.7m. i.e. Qout 0.00522 17 . 0.0068 m3 / s do = 0.005m The rate of increase in volume is: Q Qin Qout 0.02 0.0068 0.0132 m 3 / s From the question: Qin = 0.0095 m3/s, do=0.05m, Cd =0.6 As Q = Area x Velocity, the rate of rise in surface is Apply Bernoulli from the water surface (1) to the orifice (2), Q Au p1 u12 p2 2 u2 Q 0.0132 z1 z2 u 0.0042 m / s 0.252 m / min g 2g g 2g A 22 p1 = p2 and u1 = 0. u 2 2 gh . 4 With the datum the bottom of the cylinder, z1 = h, z2 = 0 4.4 A horizontal boiler shell (i.e. a horizontal cylinder) 2m diameter and 10m long is half full of water. Find We need Q in terms of the height h measured above the orifice. the time of emptying the shell through a short vertical pipe, diameter 0.08m, attached to the bottom of the Qout Cd a o u2 Cd a o 2 gh shell. Take the coefficient of discharge to be 0.8. 2 [1370 seconds] 0.05 0.6 2 9.81 h d = 2m 2 0.00522 h (1) 32m For the level in the tank to remain constant: inflow = out flow Qin = Qout 0.0095 0.00522 h h 3.314 m (b) Write the equation for the discharge in terms of the surface height change: do = 0.08 m From the question W = 10m, D = 10m do = 0.08m Cd = 0.8 Examples: Answers CIVE1400: Fluid Mechanics 25 Examples: Answers CIVE1400: Fluid Mechanics 26 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers Apply Bernoulli from the water surface (1) to the orifice (2), h2 20 2h h 2 T dh p1 u12 p2 2 u2 h1 01078 h . z1 z2 g 2g g 2g h2 2h h 2 1123.6 dh p1 = p2 and u1 = 0. u 2 2 gh . h1 h h2 2h h 2 With the datum the bottom of the cylinder, z1 = h, z2 = 0 1123.6 dh h1 h We need Q in terms of the height h measured above the orifice. h2 1123.6 2 h dh h1 Qout Cd a o u2 Cd a o 2 gh 2 2 3/ 2 h2 0.08 1123.6 2 h 0.8 2 9.81 h 3 h1 2 749.07 2.828 1 1369.6 sec 0.0178 h 4.5 Write the equation for the discharge in terms of the surface height change: Two cylinders standing upright contain liquid and are connected by a submerged orifice. The diameters Q t A h of the cylinders are 1.75m and 1.0m and of the orifice, 0.08m. The difference in levels of the liquid is A initially 1.35m. Find how long it will take for this difference to be reduced to 0.66m if the coefficient of t h discharge for the orifice is 0.605. (Work from first principles.) Q [30.7 seconds] Integrating between h1 and h2, to give the time to change surface level d1 = 1.75m d2 = 1.0m h2 A T dh h1 Q But we need A in terms of h h = 1.35m 2.0m 1.0m a L h do = 0.108m . 2 2 Surface area A = 10L, so need L in terms of h . 175 1 A1 2.4m2 A2 0.785m2 2 2 2 L 12 a2 0.08 2 2 do 0.08m, ao 0.00503m2 Cd 0.605 a (1 h) 2 L 2 by continuity, 12 (1 h) 2 2 A1 h1 A2 h2 Q t (1) 2 L 2 2h h defining, h = h1 - h2 A 20 2h h 2 h h1 h2 Substitute this into the integral term, Substituting this in (1) to eliminate h2 Examples: Answers CIVE1400: Fluid Mechanics 27 Examples: Answers CIVE1400: Fluid Mechanics 28 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers A1 h1 A2 ( h1 h) A2 h1 A2 h Q t A h A2 h A h1 t h A1 A2 Q A2 h Integrating between h1 and h2, to give the time to change surface level A1 Q t ( 2) A1 A2 h2 A T dh From the Bernoulli equation we can derive this expression for discharge through the submerged orifice: h1 Q Q Cd ao 2 gh 60000 h2 1 dh 0.678 h1 h 3/ 2 So 1/ 2 h2 2 8849558 h . h1 A h A1 2 Cd a o 2 gh t A1 A2 From the question T = 3600 sec and h1 = 0.6m A1 A2 1 3600 17699115 h2 1/ 2 . 0.6 1/ 2 t h A1 A2 Cd a o 2 g h h2 0.5815m Integrating Total depth = 3.4 + 0.58 = 3.98m A1 A2 h2 1 T dh A1 A2 Cd a o 2 g h1 h 2 A1 A2 h2 h1 A1 A2 Cd a o 2 g 2 2.4 0.785 . 0.8124 11619 2.4 0.785 0.605 0.00503 2 9.81 30.7 sec 4.6 A rectangular reservoir with vertical walls has a plan area of 60000m2. Discharge from the reservoir take place over a rectangular weir. The flow characteristics of the weir is Q = 0.678 H3/2 cumecs where H is the depth of water above the weir crest. The sill of the weir is 3.4m above the bottom of the reservoir. Starting with a depth of water of 4m in the reservoir and no inflow, what will be the depth of water after one hour? [3.98m] From the question A = 60 000 m2, Q = 0.678 h 3/2 Write the equation for the discharge in terms of the surface height change: Examples: Answers CIVE1400: Fluid Mechanics 29 Examples: Answers CIVE1400: Fluid Mechanics 30 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers Notches and weirs 5.2 Derive an expression for the discharge over a sharp crested rectangular weir. A sharp edged weir is to be 5.1 constructed across a stream in which the normal flow is 200 litres/sec. If the maximum flow likely to Deduce an expression for the discharge of water over a right-angled sharp edged V-notch, given that the occur in the stream is 5 times the normal flow then determine the length of weir necessary to limit the rise coefficient of discharge is 0.61. in water level to 38.4cm above that for normal flow. Cd=0.61. A rectangular tank 16m by 6m has the same notch in one of its short vertical sides. Determine the time [1.24m] taken for the head, measured from the bottom of the notch, to fall from 15cm to 7.5cm. [1399 seconds] From your notes you can derive: From your notes you can derive: 2 Q C b 2 gh 3/ 2 8 3 d Q C tan 2 g H 5/ 2 15 d 2 From the question: For this weir the equation simplifies to Q1 = 0.2 m3/s, h1 = x 3 Q 144 H . 5/ 2 Q2 = 1.0 m /s, h2 = x + 0.384 Write the equation for the discharge in terms of the surface height change: where x is the height above the weir at normal flow. Q t A h So we have two situations: A 2 t h 0.2 C b 2 g x 3/ 2 1801bx 3/ 2 . (1) Q 3 d 2 3/ 2 3/ 2 Integrating between h1 and h2, to give the time to change surface level 10 . C b 2 g x 0.384 1801b x 0.384 . ( 2) 3 d h2A From (1) we get an expression for b in terms of x T dh h1 Q 3/ 2 b 0111x . 16 6 h2 1 dh Substituting this in (2) gives, 144 h1 h 5/ 2 . 2 3/ 2 h2 x 0.384 3/ 2 66.67 h h1 . . . 10 1801 0111 3 x h1 = 0.15m, h2 = 0.075m x 0.384 52 / 3 T 44.44 0.075 3/ 2 015 . 3/ 2 x x 01996 m . 1399 sec So the weir breadth is 3/ 2 b . . 0111 01996 124m . Examples: Answers CIVE1400: Fluid Mechanics 31 Examples: Answers CIVE1400: Fluid Mechanics 32 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 5.3 b) Write the equation for the discharge in terms of the surface height change: Show that the rate of flow across a triangular notch is given by Q=CdKH5/2 cumecs, where Cd is an Q t A h experimental coefficient, K depends on the angle of the notch, and H is the height of the undisturbed water level above the bottom of the notch in metres. State the reasons for the introduction of the A t h coefficient. Q Water from a tank having a surface area of 10m2 flows over a 90 notch. It is found that the time taken to Integrating between h1 and h2, to give the time to change surface level lower the level from 8cm to 7cm above the bottom of the notch is 43.5seconds. Determine the coefficient Cd assuming that it remains constant during his period. h2A T dh [0.635] h1Q 8 56000 h2 1 The proof for Q C tan 2 g H 5/ 2 Cd KH 5/ 2 is in the notes. dh 15 d 2 177 B h1 h 3/ 2 . From the question: 2 56000 1/ 2 0.3 h 177 B . 0. 6 A = 10m2 = 90 h1 = 0.08m h2 = 0.07m T = 43.5sec 1/ 2 5784 0.3 0.6 1/ 2 So T 3093 sec Q = 2.36 Cd h5/2 5.5 Write the equation for the discharge in terms of the surface height change: Develop a formula for the discharge over a 90 V-notch weir in terms of head above the bottom of the V. Q t A h A channel conveys 300 litres/sec of water. At the outlet end there is a 90 V-notch weir for which the A coefficient of discharge is 0.58. At what distance above the bottom of the channel should the weir be t h placed in order to make the depth in the channel 1.30m? With the weir in this position what is the depth Q of water in the channel when the flow is 200 litres/sec? Integrating between h1 and h2, to give the time to change surface level [0.755m, 1.218m] A h2 8 T dh Derive this formula from the notes: Q C tan 2 g H 5/ 2 Q h1 15 d 2 10 h2 1 From the question: dh 2.36Cd h1 h 5/2 = 90 Cd 0.58 Q = 0.3 m3/s, depth of water, Z = 0.3m 2 4.23 3/2 0.08 h giving the weir equation: 3 Cd 0.07 Q 137 H 5/ 2 . 2.82 435 . 0.07 3/2 0.08 3/2 Cd a) As H is the height above the bottom of the V, the depth of water = Z = D + H, where D is the height Cd 0.635 of the bottom of the V from the base of the channel. So 5/2 5.4 Q 137 Z . D A reservoir with vertical sides has a plan area of 56000m2. Discharge from the reservoir takes place over 0.3 137 13 D . . 5/2 a rectangular weir, the flow characteristic of which is Q=1.77BH3/2 m3/s. At times of maximum rainfall, D 0.755m water flows into the reservoir at the rate of 9m3/s. Find a) the length of weir required to discharge this quantity if head must not exceed 0.6m; b) the time necessary for the head to drop from 60cm to 30cm if b) Find Z when Q = 0.2 m3/s the inflow suddenly stops. 0.2 137 Z 0.755 . 5/ 2 [10.94m, 3093seconds] Z 1218m . From the question: A = 56000 m2 Q = 1.77 B H 3/2 Qmax = 9 m3/s a) Find B for H = 0.6 9 = 1.77 B 0.63/2 B = 10.94m Examples: Answers CIVE1400: Fluid Mechanics 33 Examples: Answers CIVE1400: Fluid Mechanics 34 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 5.6 Application of the Momentum Equation Show that the quantity of water flowing across a triangular V-notch of angle 2 is 6.1 8 The figure below shows a smooth curved vane attached to a rigid foundation. The jet of water, Q Cd tan 2 g H 5/ 2 . Find the flow if the measured head above the bottom of the V is 38cm, when 15 rectangular in section, 75mm wide and 25mm thick, strike the vane with a velocity of 25m/s. Calculate =45 and Cd=0.6. If the flow is wanted within an accuracy of 2%, what are the limiting values of the the vertical and horizontal components of the force exerted on the vane and indicate in which direction head. these components act. [0.126m3/s, 0.377m, 0.383m] [Horizontal 233.4 N acting from right to left. Vertical 1324.6 N acting downwards] Proof of the v-notch weir equation is in the notes. From the question: H = 0.38m = 45 Cd = 0.6 45 The weir equation becomes: 25 Q 1417 H 5/ 2 . 1417 0.38 . 5/ 2 From the question: 0126 m3 / s . a1 0.075 0.025 1875 10 3 m 2 . u1 25 m / s Q 1875 10 3 25 m 3 / s . Q+2% = 0.129 m3/s a1 a 2 , so u1 u 2 0129 1417 H 5/ 2 . . H 0.383m Calculate the total force using the momentum equation: FT x Q u 2 cos 25 u1 cos 45 3 1000 0.0469 25 cos 25 25 cos 45 Q-2% = 0.124 m /s 0124 1417 H . . 5/ 2 233.44 N H 0.377m FT y Q u 2 sin 25 u1 sin 45 1000 0.0469 25 sin 25 25 sin 45 1324.6 N Body force and pressure force are 0. So force on vane: Rx Ft x 233.44 N Ry Ft y 1324.6 N Examples: Answers CIVE1400: Fluid Mechanics 35 Examples: Answers CIVE1400: Fluid Mechanics 36 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 6.2 FRy FTy FPy FBy A 600mm diameter pipeline carries water under a head of 30m with a velocity of 3m/s. This water main is 2.457 80.376 0 82.833 kN fitted with a horizontal bend which turns the axis of the pipeline through 75 (i.e. the internal angle at the bend is 105 ). Calculate the resultant force on the bend and its angle to the horizontal. These forces act on the fluid [104.044 kN, 52 29’] The resultant force on the fluid is u2 y FR FRx FRy 104.44 kN 1 FRy tan 52 29 ' FRx x 6.3 A horizontal jet of water 2 103 mm2 cross-section and flowing at a velocity of 15 m/s hits a flat plate at 60 to the axis (of the jet) and to the horizontal. The jet is such that there is no side spread. If the plate is stationary, calculate a) the force exerted on the plate in the direction of the jet and b) the ratio between the u1 quantity of fluid that is deflected upwards and that downwards. (Assume that there is no friction and therefore no shear force.) [338N, 3:1] θ y x u2 From the question: 2 0.6 a 0.283 m 2 d 0.6 m h 30 m 2 u1 u2 3m / s Q 0.848 m 3 / s u1 Calculate total force. FTx Q u2 x u1x FRx FPx FBx θ FTx 1000 0.848 3 cos 75 3 1.886 kN FTy Q u2 y u1 y FRy FPy FBy u3 FTy 1000 0.848 3 sin 75 0 2.457 kN From the question a2 = a3 =2x10-3 m2 u = 15 m/s Calculate the pressure force Apply Bernoulli, p1 = p2 = p = h g = 30 1000 9.81 = 294.3 kN/m2 FTx p1 a 1 cos p 2 a 2 cos p1 u12 p2 2 u2 p3 2 u3 1 2 z1 z2 z3 g 2g g 2g g 2g 294300 0.283 1 cos 75 6173 kN . FTy p1 a1 sin p 2 a 2 sin Change in height is negligible so z1 = z2 = z3 and pressure is always atmospheric p1= p2 = p3 =0. So 1 2 294300 0.283 0 sin 75 u1= u2 = u3 =15 m/s 80.376 kN By continuity Q1= Q2 + Q3 There is no body force in the x or y directions. u1a1 = u2a2 + u3a3 so a1 = a2 + a3 FRx FTx FPx FBx 1886 61.73 0 . 63.616 kN Put the axes normal to the plate, as we know that the resultant force is normal to the plate. Examples: Answers CIVE1400: Fluid Mechanics 37 Examples: Answers CIVE1400: Fluid Mechanics 38 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers -3 Q1 = a1u = 2 10 15 = 0.03 FTx = Q( 0 - u1x ) Q1 = (a2 + a3) u FTx = 1000 0.11 ( 0 - 25 cos 30 ) = 2.39 kN Q2 = a2u 6.5 Q3 = (a1 - a2)u The outlet pipe from a pump is a bend of 45 rising in the vertical plane (i.e. and internal angle of 135 ). The bend is 150mm diameter at its inlet and 300mm diameter at its outlet. The pipe axis at the inlet is Calculate total force. horizontal and at the outlet it is 1m higher. By neglecting friction, calculate the force and its direction if the inlet pressure is 100kN/m2 and the flow of water through the pipe is 0.3m3/s. The volume of the pipe FTx Q u2 x u1x FRx FPx FBx is 0.075m3. FTx 1000 0.03 0 15 sin 60 390 N [13.94kN at 67 40’ to the horizontal] Component in direction of jet = 390 sin 60 = 338 N y p2 u 2 A2 As there is no force parallel to the plate Fty = 0 x 2 2 FTy u2 a 2 u3 a 3 u12 a1 cos 0 a2 a 3 a1 cos 0 a1 a2 a3 p1 1m a 3 a1 cos a1 a 3 u1 45° 4 4a 3 a1 a 3 2 1 A1 a3 a 3 2 Thus 3/4 of the jet goes up, 1/4 down 6.4 1&2 Draw the control volume and the axis system A 75mm diameter jet of water having a velocity of 25m/s strikes a flat plate, the normal of which is inclined at 30 to the jet. Find the force normal to the surface of the plate. [2.39kN] p1 = 100 kN/m2, Q = 0.3 m3/s = 45 y d1 = 0.15 m d2 = 0.3 m A1 = 0.177 m2 A2 = 0.0707 m2 x u2 3 Calculate the total force in the x direction u1 FT x Q u2 x u1 x Q u2 cos u1 θ by continuity A1u1 A2u2 Q , so u3 From the question, djet = 0.075m u1=25m/s Q = 25 (0.075/2)2 = 0.11 m3/s Force normal to plate is Examples: Answers CIVE1400: Fluid Mechanics 39 Examples: Answers CIVE1400: Fluid Mechanics 40 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 0.3 FP x 100000 0.0177 2253614 cos 45 0.0707 . u1 16.98 m / s 0152 / 4 . 1770 11266.34 9496.37 kN 0.3 u2 4.24 m / s 0.0707 FP y 2253614 sin 45 0.0707 . 11266.37 FT x 1000 0.3 4.24 cos 45 16.98 1493.68 N 5 Calculate the body force The only body force is the force due to gravity. That is the weight acting in the y direction. and in the y-direction FB y g volume FT y Q u2 y u1 y 1000 9.81 0.075 1290156 N . Q u2 sin 0 There are no body forces in the x direction, 1000 0.3 4.24 sin 45 FB x 0 899.44 N 6 Calculate the resultant force 4 Calculate the pressure force. FP pressure force at 1 - pressure force at 2 FT x FR x FP x FB x FT y FR y FP y FB y FP x p1 A1 cos 0 p2 A2 cos p1 A1 p2 A2 cos FR x FT x FP x FB x FP y p1 A1 sin 0 p2 A2 sin p2 A2 sin 4193.6 9496.37 5302.7 N We know pressure at the inlet but not at the outlet. we can use Bernoulli to calculate this unknown pressure. FR y FT y FP y FB y 899.44 11266.37 735.75 2 2 1290156 N . p1 u1 p2 u2 z1 z2 hf g 2g g 2g And the resultant force on the fluid is given by where hf is the friction loss FRy In the question it says this can be ignored, hf=0 FResultant The height of the pipe at the outlet is 1m above the inlet. Taking the inlet level as the datum: z1 = 0 z2 = 1m φ So the Bernoulli equation becomes: 100000 16.982 p2 4.242 FRx 0 . 10 1000 9.81 2 9.81 1000 9.81 2 9.81 2 p2 2253614 N / m . Examples: Answers CIVE1400: Fluid Mechanics 41 Examples: Answers CIVE1400: Fluid Mechanics 42 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers FR F2 F 2 6.7 Rx R y A curved plate deflects a 75mm diameter jet through an angle of 45 . For a velocity in the jet of 40m/s to 5302.7 2 12901562 . the right, compute the components of the force developed against the curved plate. (Assume no friction). 13.95 kN [Rx=2070N, Ry=5000N down] u2 y And the direction of application is 1 FR y 1 . 1290156 tan tan 67.66 x FR x 5302.7 The force on the bend is the same magnitude but in the opposite direction R FR u1 6.6 The force exerted by a 25mm diameter jet against a flat plate normal to the axis of the jet is 650N. What θ is the flow in m3/s? [0.018 m3/s] y u2 From the question: a1 0.0752 / 4 4.42 10 3 m2 u1 40 m / s x 3 Q 4.42 10 40 01767 m3 / s . a1 a2 , so u1 u2 u1 Calculate the total force using the momentum equation: FT x Q u2 cos 45 u1 . 1000 01767 40 cos 45 40 207017 N . FT y Q u2 sin 45 0 u2 . 1000 01767 40 sin 45 4998 N From the question, djet = 0.025m FTx = 650 N Body force and pressure force are 0. Force normal to plate is So force on vane: FTx = Q( 0 - u1x ) Rx Ft x 2070 N 650 = 1000 Q ( 0 - u ) Ry Ft y 4998 N Q = au = ( d2/4)u 650 = -1000au2 = -1000Q2/a 650 = -1000Q2/( 0.0252/4) Q = 0.018m3/s Examples: Answers CIVE1400: Fluid Mechanics 43 Examples: Answers CIVE1400: Fluid Mechanics 44 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 6.8 FT x . 1000 0.45 6.365 cos 45 159 A 45 reducing bend, 0.6m diameter upstream, 0.3m diameter downstream, has water flowing through it 1310 N at the rate of 0.45m3/s under a pressure of 1.45 bar. Neglecting any loss is head for friction, calculate the force exerted by the water on the bend, and its direction of application. [R=34400N to the right and down, = 14 ] and in the y-direction ρ2 u2 FT y Q u2 y u1 y y A2 Q u2 sin 0 1000 0.45 6.365 sin 45 x 1800 N 4 Calculate the pressure force. ρ1 FP pressure force at 1 - pressure force at 2 u1 A1 θ FP x p1 A1 cos 0 p2 A2 cos p1 A1 p2 A2 cos FP y p1 A1 sin 0 p2 A2 sin p2 A2 sin 1&2 Draw the control volume and the axis system We know pressure at the inlet but not at the outlet. we can use Bernoulli to calculate this unknown pressure. 5 2 3 p1 = 1.45 10 N/m , Q = 0.45 m /s = 45 p1 u12 p2 2 u2 z1 z2 hf d1 = 0.6 m d2 = 0.3 m g 2g g 2g 2 2 A1 = 0.283 m A2 = 0.0707 m where hf is the friction loss In the question it says this can be ignored, hf=0 3 Calculate the total force Assume the pipe to be horizontal in the x direction z1 = z 2 So the Bernoulli equation becomes: FT x Q u2 x u1 x 145000 159 2 . p2 6.365 2 Q u2 cos u1 1000 9.81 2 9.81 1000 9.81 2 9.81 p2 126007 N / m 2 by continuity A1u1 A2u2 Q , so FP x 145000 0.283 126000 cos 45 0.0707 41035 6300 34735 N 0.45 u1 159 m / s . 0.6 2 / 4 FP y 126000 sin 45 0.0707 0.45 u2 6.365 m / s 6300 N 0.0707 5 Calculate the body force Examples: Answers CIVE1400: Fluid Mechanics 45 Examples: Answers CIVE1400: Fluid Mechanics 46 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers The only body force is the force due to gravity. Laminar pipe flow. There are no body forces in the x or y directions, 7.1 FB x FB y 0 The distribution of velocity, u, in metres/sec with radius r in metres in a smooth bore tube of 0.025 m bore follows the law, u = 2.5 - kr2. Where k is a constant. The flow is laminar and the velocity at the pipe 6 Calculate the resultant force surface is zero. The fluid has a coefficient of viscosity of 0.00027 kg/m s. Determine (a) the rate of flow in m3/s (b) the shearing force between the fluid and the pipe wall per metre length of pipe. [6.14x10-4 m3/s, 8.49x10-3 N] FT x FR x FP x FB x The velocity at distance r from the centre is given in the question: FT y FR y FP y FB y u = 2.5 - kr2 Also we know: = 0.00027 kg/ms 2r = 0.025m FR x FT x FP x FB x We can find k from the boundary conditions: 1310 34735 when r = 0.0125, u = 0.0 (boundary of the pipe) 33425 N 0.0 = 2.5 - k0.01252 k = 16000 FR y FT y FP y FB y u = 2.5 - 1600 r2 1800 6300 a) 8100 N Following along similar lines to the derivation seen in the lecture notes, we can calculate the flow Q through a small annulus r: Q ur Aannulus And the resultant force on the fluid is given by Aannulus (r r)2 r2 2 r r FRy Q 2.5 16000r 2 2 r r FResultant 0.0125 Q 2 2.5r 16000r 3 dr 0 0.0125 2.5r 2 16000 4 2 r φ 2 4 0 614 m3 / s . FRx b) FR FR2 x FR2 y The shear force is given by F= (2 r) 334252 8100 2 From Newtons law of viscosity 34392 kN du dr And the direction of application is du 2 16000r 32000r dr FR y 8100 tan 1 tan 1 13.62 F 0.00027 32000 0.0125 (2 0.0125) FR x 33425 8.48 10 3 N The force on the bend is the same magnitude but in the opposite direction R FR Examples: Answers CIVE1400: Fluid Mechanics 47 Examples: Answers CIVE1400: Fluid Mechanics 48 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 7.2 p 1 A liquid whose coefficient of viscosity is m flows below the critical velocity for laminar flow in a circular ur R2 r2 L 4 pipe of diameter d and with mean velocity u. Show that the pressure loss in a length of pipe is 32um/d2. Oil of viscosity 0.05 kg/ms flows through a pipe of diameter 0.1m with a velocity of 0.6m/s. Calculate the The flow in an annulus of thickness r loss of pressure in a length of 120m. Q ur Aannulus [11 520 N/m2] Aannulus (r r )2 r2 2 r r See the proof in the lecture notes for p 1 Consider a cylinder of fluid, length L, radius r, flowing steadily in the centre of a pipe Q R2 r2 2 r r L 4 R p δr Q R 2 r r 3 dr L 2 0 p R4 p d4 r r L 8 L128 R So the discharge can be written p d4 Q L 128 The fluid is in equilibrium, shearing forces equal the pressure forces. To get pressure loss in terms of the velocity of the flow, use the mean velocity: 2 r L pA p r2 u Q/ A p r pd 2 L 2 u 32 L du 32 Lu Newtons law of viscosity , p dy d2 du 32 u We are measuring from the pipe centre, so p per unit length dr d2 Giving: b) From the question = 0.05 kg/ms d = 0.1m p r du u = 0.6 m/s L = 120.0m L 2 dr 32 0.05 120 0.6 du p r p 11520 N / m2 012 . dr L 2 In an integral form this gives an expression for velocity, p 1 u r dr L 2 The value of velocity at a point distance r from the centre p r2 ur C L 4 At r = 0, (the centre of the pipe), u = umax, at r = R (the pipe wall) u = 0; p R2 C L 4 At a point r from the pipe centre when the flow is laminar: Examples: Answers CIVE1400: Fluid Mechanics 49 Examples: Answers CIVE1400: Fluid Mechanics 50 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 7.3 vpiston = 0.00064 m/s A plunger of 0.08m diameter and length 0.13m has four small holes of diameter 5/1600 m drilled through 7.4 in the direction of its length. The plunger is a close fit inside a cylinder, containing oil, such that no oil is A vertical cylinder of 0.075 metres diameter is mounted concentrically in a drum of 0.076metres internal assumed to pass between the plunger and the cylinder. If the plunger is subjected to a vertical downward diameter. Oil fills the space between them to a depth of 0.2m. The rotque required to rotate the cylinder in force of 45N (including its own weight) and it is assumed that the upward flow through the four small the drum is 4Nm when the speed of rotation is 7.5 revs/sec. Assuming that the end effects are negligible, holes is laminar, determine the speed of the fall of the plunger. The coefficient of velocity of the oil is 0.2 calculate the coefficient of viscosity of the oil. kg/ms. [0.638 kg/ms] [0.00064 m/s] From the question r-1 = 0.076/2 r2 = 0.075/2 Torque = 4Nm, L = 0.2m F = 45N d = 5/1600 m Q The velocity of the edge of the cylinder is: ucyl = 7.5 2 r = 7.5 2 0.0375 = 1.767 m/s udrum = 0.0 plunger Torque needed to rotate cylinder T surface area 0.13 m 4 2 r2 L 226354 N / m2 . Distance between cylinder and drum = r1 - r2 = 0.038 - 0.0375 = 0.005m cylinder Using Newtons law of viscosity: du dr du 1767 0 . 0.8m dr 0.0005 22635 . 3534 0.64 kg / ms ( Ns / m 2 ) Flow through each tube given by Hagen-Poiseuille equation p d4 Q L 128 There are 4 of these so total flow is p d4 4 (5 / 1600) 4 10 Q 4 p p3601 10 . L 128 013 128 0.2 . Force = pressure area 2 2 0.08 5 / 1600 F 45 p 4 2 2 p 9007.206 N / m2 Q 3.24 10 6 m 3 / s Flow up through piston = flow displaced by moving piston Q = Avpiston 3.24 10-6 = 0.042 vpiston Examples: Answers CIVE1400: Fluid Mechanics 51 Examples: Answers CIVE1400: Fluid Mechanics 52 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers Dimensional analysis And the second group 2 : 8.1 M LT0 0 0 LT 1 a2 L b2 ML 3 c2 L2 T 1 A stationary sphere in water moving at a velocity of 1.6m/s experiences a drag of 4N. Another sphere of twice the diameter is placed in a wind tunnel. Find the velocity of the air and the drag which will give M] 0 = c2 dynamically similar conditions. The ratio of kinematic viscosities of air and water is 13, and the density L] 0 = a2 + b2 - 3c2 + 2 of air 1.28 kg/m3. [10.4m/s 0.865N] -2 = a2 + b2 Draw up the table of values you have for each variable: T] 0 = -a2 - 1 variable water air a2 = -1 u 1.6m/s uair b2 = -1 Drag 4N Dair 2 u 1d 1 0 13 ud 1000 kg/m3 1.28 kg/m3 So the physical situation is described by this function of nondimensional numbers, d d 2d D , , 0 1 2 u 2 d 2 ud Kinematic viscosity is dynamic viscosity over density = For dynamic similarity these non-dimensional numbers are the same for the both the sphere in water and ud ud The Reynolds number = Re in the wind tunnel i.e. 1air 1 water Choose the three recurring (governing) variables; u, d, 2 air 2 water From Buckinghams theorem we have m-n = 5 - 3 = 2 non-dimensional groups. For 1 u, d , , D, 0 D D , 0 1 2 u2d 2 air u2d 2 water u a1 d b1 c1 D 1 Dair 4 a2 b2 c2 2 u d 128 10.4 2 (2d ) 2 . 1000 16 2 . d2 As each group is dimensionless then considering the dimensions, for the first group, 1: Dair 0.865 N -2 (note D is a force with dimensions MLT ) For 2 0 0 0 1 a1 b1 3 c1 2 M LT LT L ML MLT ud air ud water M] 0 = c1 + 1 13 c1 = -1 uair 2d 16 d . L] 0 = a1 + b1 - 3c1 + 1 uair 10.4 m / s -4 = a1 + b1 T] 0 = -a1 - 2 a1 = - 2 b1 = -2 1 u 2d 2 1 D D u2d 2 Examples: Answers CIVE1400: Fluid Mechanics 53 Examples: Answers CIVE1400: Fluid Mechanics 54 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 8.2 M] 0 = c1 Explain briefly the use of the Reynolds number in the interpretation of tests on the flow of liquid in pipes. L] 0 = a1 + b1 - 3c1 + 2 Water flows through a 2cm diameter pipe at 1.6m/s. Calculate the Reynolds number and find also the velocity required to give the same Reynolds number when the pipe is transporting air. Obtain the ratio of -2 = a1 + b1 pressure drops in the same length of pipe for both cases. For the water the kinematic viscosity was T] 0 = -a1 - 1 1.31 10-6 m2/s and the density was 1000 kg/m3. For air those quantities were 15.1 10-6 m2/s and 1.19kg/m3. a1 = -1 [24427, 18.4m/s, 0.157] b1 = -1 Draw up the table of values you have for each variable: 1 u d 1 1 0 variable water air u 1.6m/s uair ud p pwater pair 1000 kg/m3 1.19kg/m3 And the second group 2 : m s m s (note p is a pressure (force/area) with dimensions ML-1T-2) 1 a1 b1 3 c1 1000 kg/m3 1.28 kg/m3 M 0 L0 T 0 LT L ML MT 2 L 1 d 0.02m 0.02m M] 0 = c2 + 1 c2 = -1 Kinematic viscosity is dynamic viscosity over density = L] 0 = a2 + b2 - 3c2 - 1 ud ud -2 = a2 + b2 The Reynolds number = Re T] 0 = -a2 - 2 Reynolds number when carrying water: a2 = - 2 ud . 16 0.02 b2 = 0 Re water 24427 131 10 6 . u 2 1 p 2 To calculate Reair we know, p Re water Re air u2 uair 0.02 So the physical situation is described by this function of nondimensional numbers, 24427 15 10 6 uair 18.44m / s p 1 , 2 , 0 ud u2 To obtain the ratio of pressure drops we must obtain an expression for the pressure drop in terms of governing variables. Choose the three recurring (governing) variables; u, d, For dynamic similarity these non-dimensional numbers are the same for the both water and air in the pipe. From Buckinghams theorem we have m-n = 5 - 3 = 2 non-dimensional groups. 1air 1 water u, d , , , p 0 2 air 2 water 1 , 2 0 We are interested in the relationship involving the pressure i.e. 2 a1 b1 c1 1 u d 2 u a2 d b2 c2 p As each group is dimensionless then considering the dimensions, for the first group, 1: 0 0 0 1 a1 b1 3 c1 2 1 M LT LT L ML LT Examples: Answers CIVE1400: Fluid Mechanics 55 Examples: Answers CIVE1400: Fluid Mechanics 56 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 1 1 1 p p 1 Q d u2 air u2 water d pwater u 2 Q water water 2 pair air u air 1000 16 2 . 1 And the second group 2 : 6.327 119 18.44 2 . . 0158 (note p is a pressure (force/area) with dimensions ML-1T-2) Show that Reynold number, ud/ , is non-dimensional. If the discharge Q through an orifice is a function 1 a1 b1 3 c1 of the diameter d, the pressure difference p, the density , and the viscosity , show that Q = Cp1/2d2/ 1/2 M 0 L0 T 0 L3 T L ML MT 2 L 1 where C is some function of the non-dimensional group (d 1/2d1/2/ ). M] 0 = c2 + 1 Draw up the table of values you have for each variable: c2 = -1 The dimensions of these following variables are L] 0 = 3a2 + b2 - 3c2 - 1 ML-3 -2 = 3a2 + b2 u LT-1 T] 0 = -a2 - 2 d L a2 = - 2 ML-1T-1 b2 = 4 Re = ML-3 LT-1L(ML-1T-1)-1 = ML-3 LT-1 L M-1LT = 1 2 Q 2d 4 1 p i.e. Re is dimensionless. 4 d p Q2 We are told from the question that there are 5 variables involved in the problem: d, p, , and Q. So the physical situation is described by this function of non-dimensional numbers, Choose the three recurring (governing) variables; Q, d, d d4p From Buckinghams theorem we have m-n = 5 - 3 = 2 non-dimensional groups. , , 0 1 2 Q Q2 Q, d , , , p 0 or 1 , 2 0 d d4p 1 Q a1 d b1 c1 Q 1 Q2 Q a2 d b2 c2 p 1/ 2 2 d p 1/2 d 2 p 1/2 The question wants us to show : Q f As each group is dimensionless then considering the dimensions, for the first group, 1: 0 0 0 3 1 a1 b1 3 c1 1 1 1/ 2 M LT LT L ML ML T 1 Q Take the reciprocal of square root of 2: 2a , d p1/2 2 M] 0 = c1 + 1 2 c1 = -1 Convert 1 by multiplying by this number 1/ 2 L] 0 = 3a1 + b1 - 3c1 - 1 d Q 1a 1 2a -2 = 3a1 + b1 Q d 2 p 1/ 2 d 1/ 2 p1/ 2 T] 0 = -a1 - 1 then we can say a1 = -1 b1 = 1 Examples: Answers CIVE1400: Fluid Mechanics 57 Examples: Answers CIVE1400: Fluid Mechanics 58 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 1/ 2 1/ 2 2 1/ 2 1 a1 b1 3 c1 p d d p M 0 L0 T 0 LT L ML ML 1T 1 1/ 1a , 2a , 1/2 0 M] 0 = c1 + 1 or c1 = -1 p 1/2 1/ 2 d d 2 p 1/ 2 Q 1/ 2 L] 0 = a1 + b1 - 3c1 - 1 -2 = a1 + b1 8.4 A cylinder 0.16m in diameter is to be mounted in a stream of water in order to estimate the force on a tall T] 0 = -a1 - 1 chimney of 1m diameter which is subject to wind of 33m/s. Calculate (A) the speed of the stream a1 = -1 necessary to give dynamic similarity between the model and chimney, (b) the ratio of forces. b1 = -1 Chimney: = 1.12kg/m3 = 16 10-6 kg/ms 1 u 1d 1 1 Model: = 1000kg/m3 = 8 10-4 kg/ms [11.55m/s, 0.057] ud Draw up the table of values you have for each variable: i.e. the (inverse of) Reynolds number variable water air And the second group 2 : u uwater 33m/s 0 0 0 1 a2 b2 3 c2 M LT LT L ML ML 1T 2 F Fwater Fair M] 0 = c2 + 1 1000 kg/m3 1.12kg/m3 c2 = -1 kg ms kg/ms L] 0 = a2 + b2 - 3c2 - 1 d 0.16m 1m -3 = a2 + b2 T] 0 = -a2 - 2 Kinematic viscosity is dynamic viscosity over density = a2 = - 2 ud ud The Reynolds number = Re b2 = -1 2 u 2d 1 1 F For dynamic similarity: F Re water Re air u2d 1000uwater 016 . 112 33 1 . So the physical situation is described by this function of nondimensional numbers, 8 10 4 16 10 6 uwater 1155m / s . F , , 0 To obtain the ratio of forces we must obtain an expression for the force in terms of governing variables. 1 2 ud du 2 Choose the three recurring (governing) variables; u, d, F, From Buckinghams theorem we have m-n = 5 - 3 = 2 non-dimensional groups. For dynamic similarity these non-dimensional numbers are the same for the both water and air in the pipe. u, d , , , F 0 1air 1 water 2 air 2 water 1, 2 0 1 u d a1 b1 c1 To find the ratio of forces for the different fluids use 2 2 u a2 d b2 c2 F As each group is dimensionless then considering the dimensions, for the first group, 1: Examples: Answers CIVE1400: Fluid Mechanics 59 Examples: Answers CIVE1400: Fluid Mechanics 60 CIVE1400: Fluid Mechanics Examples: Answers CIVE1400: Fluid Mechanics Examples: Answers 2 air 2 water And the second group 2 : F F 0 0 0 1 a2 b2 3 c2 M LT LT L ML ML 1T 2 u2 d air u2 d water M] 0 = c2 + 1 F F c2 = -1 u2 d air u2 d water L] 0 = a2 + b2 - 3c2 - 1 Fair 112 332 1 . 0.057 -3 = a2 + b2 Fwater 1000 11552 016 . . T] 0 = -a2 - 2 8.5 If the resistance to motion, R, of a sphere through a fluid is a function of the density and viscosity of a2 = - 2 the fluid, and the radius r and velocity u of the sphere, show that R is given by b2 = -1 2 ur 2 u 2r 1 1 R R f R Hence show that if at very low velocities the resistance R is proportional to the velocity u, then R = k ru u2r where k is a dimensionless constant. So the physical situation is described by this function of nondimensional numbers, A fine granular material of specific gravity 2.5 is in uniform suspension in still water of depth 3.3m. Regarding the particles as spheres of diameter 0.002cm find how long it will take for the water to clear. R , , 0 Take k=6 and =0.0013 kg/ms. 1 2 ur ru 2 [218mins 39.3sec] or Choose the three recurring (governing) variables; u, r, R, R From Buckinghams theorem we have m-n = 5 - 3 = 2 non-dimensional groups. ru 2 1 ur u, r , , , R 0 2 ur R ur 1 , 2 0 he question asks us to show R f or 2 f 1 u a1 r b1 c1 2 a2 b2 c2 Multiply the LHS by the square of the RHS: (i.e. 2 (1/ 1 )) 2 u r R 2 R u2r 2 R As each group is dimensionless then considering the dimensions, for the first group, 1: ru 2 2 2 1 a1 b1 3 c1 M 0 L0 T 0 LT L ML ML 1T 1 So M] 0 = c1 + 1 R ur 2 f c1 = -1 L] 0 = a1 + b1 - 3c1 - 1 The question tells us that R is proportional to u so the function f must be a constant, k -2 = a1 + b1 R ur 2 k T] 0 = -a1 - 1 a1 = -1 R kru b1 = -1 1 u 1r 1 1 The water will clear when the particle moving from the water surface reaches the bottom. At terminal velocity there is no acceleration - the force R = mg - upthrust. ur From the question: i.e. the (inverse of) Reynolds number = 2.5 so = 2500kg/m3 = 0.0013 kg/ms k=6 Examples: Answers CIVE1400: Fluid Mechanics 61 Examples: Answers CIVE1400: Fluid Mechanics 62 CIVE1400: Fluid Mechanics Examples: Answers r = 0.00001m depth = 3.3m 4 mg 0.000013 9.81 2500 1000 3 616 10 11 . 11 kru 0.0013 6 0.00001u . 616 10 u 2.52 10 4 m / s 3.3 t 4 218 min 39.3 sec 2.52 10 Examples: Answers CIVE1400: Fluid Mechanics 63

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