Pressure and Manometers. Wat will be te (a) te gauge pressure and (b) te absolute pressure of water at dept m below te surface? ρ water 000 kg/m, and p atmospere 0kN/m. [7.7 kn/m, 8.7 kn/m ] a) b) p gauge 000 9. 8 p p + p 7 70 N / m, ( Pa) 7. 7 kn / m, ( kpa) absolute gauge atmosperic ( 7 70 + 0) N / m, ( Pa) 8. 7 kn / m, ( kpa). At wat dept below te surface of oil, relative density 0.8, will produce a pressure of 0 kn/m? Wat dept of water is tis equivalent to? [5.m,.m] a) b) ρ γρ water 08. 000 kg / m p p 0 0 5. 9m of oil 800 9. 8 ρ 000 kg / m 0 0. m of water 000 9. 8. Wat would te pressure in kn/m be if te equivalent ead is measured as 400mm of (a) mercury γ.6 (b) water ( c) oil specific weigt 7.9 kn/m (d) a liquid of density 50 kg/m? [5.4 kn/m,.9 kn/m,.6 kn/m,.04 kn/m ] a) ρ γρ water. 6 000 kg / m p ( ). 6 0 9. 8 0. 4 566 N / m CIVE400: Fluid Mecanics
b) c) d) p ( ) 0 9. 8 0. 4 94 N / m ω p ( ) 7. 9 0 0. 4 60 N / m p 50 9. 8 0. 4 040 N / m.4 A manometer connected to a pipe indicates a negative gauge pressure of 50mm of mercury. Wat is te absolute pressure in te pipe in Newtons per square metre is te atmosperic pressure is bar? [9. kn/m ] p bar 0 N / m atmospere p p + p absolute gauge atmosperic + p atmosperic 5 5. 6 0 9. 8 0. 05+ 0 N / m, ( Pa) 9. kn / m, ( kpa).5 Wat eigt would a water barometer need to be to measure atmosperic pressure? [>0m] p atmospere bar 0 N / m 0 5 5 5 0 m of water 000 9 8 09.. 5 0 0. 75m of mercury ( 6. 0 ) 9. 8 CIVE400: Fluid Mecanics
.6 An inclined manometer is required to measure an air pressure of mm of water to an accuracy of +/- %. Te inclined arm is 8mm in diameter and te larger arm as a diameter of 4mm. Te manometric fluid as density 740 kg/m and te scale may be read to +/- 0.5mm. Wat is te angle required to ensure te desired accuracy may be acieved? [ 9 ] p p diameter d diameter D Scale Reader x z Datum line z θ Volume moved from left to rigt ( ) p p ρ g ρ g z + z man man z A z A xa sinθ πd z d z x d π π 4 sinθ 4 4 z d z x d sinθ D D d p p ρman gx sinθ + D d ρwater g ρman gx sinθ + D 0. 008 ρwater g ( 0. 74 ρwater ) gx sinθ + 0. 04 0. 74 x (sin θ + 0. ) Te ead being measured is % of mm 0.00x0.0 0.00009m Tis % represents te smallest measurement possible on te manometer, 0.5mm 0.0005m, giving 0. 00009 0. 74 0. 0005 (sin θ + 0. ) sin θ 0. θ 7. 6 [Tis is not te same as te answer given on te question seet] CIVE400: Fluid Mecanics
.7 Determine te resultant force due to te water acting on te m by m rectangular area AB sown in te diagram below. [4 560 N,.7m from O] A B O.m.0m C.0 m.0 m D Te magnitude of te resultant force on a submerged plane is: R pressure at centroid area of surface R z A 000 9. 8 (. + ) ( ) 4 556 N / m Tis acts at rigt angle to te surface troug te centre of pressure. I OO Sc Ax nd moment of area about a line troug O st moment of area about a line troug O By te parallel axis teorem (wic will be given in an exam), Ioo I GG + Ax, were I GG is te nd moment of area about a line troug te centroid and can be found in tables. b I GG Sc + x Ax 45 P G G d bd For a rectangle I GG As te wall is vertical, Sc D and x z, Sc + (. + ) ( )(. + ). 7 m from O CIVE400: Fluid Mecanics 4
.8 Determine te resultant force due to te water acting on te.5m by.0m triangular area CD sown in te figure above. Te apex of te triangle is at C. [4.5 0 N,.8m from P] b G d/ G d bd For a triangle I GG 6 Dept to centre of gravity is z 0. + cos 45 94. m. R z A. 0 5. 000 9. 8 94.. 0 86 N / m Distance from P is x z / cos 45. 748m Distance from P to centre of pressure is I oo Sc Ax I I + Ax oo GG I GG 5. Sc + x Ax 6 5 748. 89m (. )(. ) + (. 748) CIVE400: Fluid Mecanics 5
Forces on submerged surfaces. Obtain an expression for te dept of te centre of pressure of a plane surface wolly submerged in a fluid and inclined at an angle to te free surface of te liquid. A orizontal circular pipe,.5m diameter, is closed by a butterfly disk wic rotates about a orizontal axis troug its centre. Determine te torque wic would ave to be applied to te disk spindle to keep te disk closed in a vertical position wen tere is a m ead of fres water above te axis. [76 Nm] Te question asks wat is te moment you ave to apply to te spindle to keep te disc vertical i.e. to keep te valve sut? So you need to know te resultant force exerted on te disc by te water and te distance x of tis force from te spindle. We know tat te water in te pipe is under a pressure of m ead of water (to te spindle).75 F x Diagram of te forces on te disc valve, based on an imaginary water surface. m, te dept to te centroid of te disc dept to te centre of pressure (or line of action of te force) Calculate te force: Calculate te line of action of te force,. F A 5. 000 9. 8 π 66. kn nd moment of area about water surface st moment of area about water surface I oo A By te parallel axis teorem nd moment of area about O (in te surface) Ioo IGG + A were I GG is te nd moment of area about a line troug te centroid of te disc and I GG πr 4 /4. CIVE400: Fluid Mecanics 6
I GG + A 4 πr + 4( πr ) r +. 06m So te distance from te spindle to te line of action of te force is And te moment required to keep te gate sut is x. 06 0. 06m moment Fx 6. 6 0. 06 76. kn m. A dock gate is to be reinforced wit tree orizontal beams. If te water acts on one side only, to a dept of 6m, find te positions of te beams measured from te water surface so tat eac will carry an equal load. Give te load per meter. [58 860 N/m,.m, 4.m, 5.47m] First of all draw te pressure diagram, as below: / f d d R f d f Te resultant force per unit lengt of gate is te area of te pressure diagram. So te total resultant force is R ρ g 0.5 000 9.8 6 76580 N ( per m lengt) Alternatively te resultant force is, R Pressure at centroid Area, (take widt of gate as m to give force per m) R ( ) 76580 N ( per m lengt) Tis is te resultant force exerted by te gate on te water. Te tree beams sould carry an equal load, so eac beam carries te load f, were f R 58860 N CIVE400: Fluid Mecanics 7
If we take moments from te surface, DR fd + fd + fd ( ) ( + + ) D f f d d d d + d + d Taking te first beam, we can draw a pressure diagram for tis, (ignoring wat is below), H/ F58860 H We know tat te resultant force, F F ρ gh, so H F 58860 H ρ g 000 9. 8 46. m And te force acts at H/, so tis is te position of te st beam, position of st beam H. m Taking te second beam into consideration, we can draw te following pressure diagram, H H/ f d. d F 58860 f Te reaction force is equal to te sum of te forces on eac beam, so as before F ( 58860) H 000 9. 8 4. 9 m Te reaction force acts at H/, so H.7m. Taking moments from te surface, ( 58860). 7 58860. + 58860 d dept to second beam For te tird beam, from before we ave, d 4. m d + d + d dept to tird beam d. 4. 547. m CIVE400: Fluid Mecanics 8
CIVE400: Fluid Mecanics. Te profile of a masonry dam is an arc of a circle, te arc aving a radius of 0m and subtending an angle of 60 at te centre of curvature wic lies in te water surface. Determine (a) te load on te dam in N/m lengt, (b) te position of te line of action to tis pressure. [4.8 0 6 N/m lengt at dept 9.0m] Draw te dam to elp picture te geometry, F R a 60 F R R y F v 0sin 60 598. m a 0cos 60 50. m Calculate F v total weigt of fluid above te curved surface (per m lengt) F v ( area of sector - area of triangle) 60 598. 5 000 9.8 π0 60 775. kn / m Calculate F force on projection of curved surface onto a vertical plane F 05. 000 9. 8 598. 0. 68 kn / m Te resultant, F F + F 0. 68 + 775. R v 479. 7 kn / m acting at te angle CIVE400: Fluid Mecanics 9
Fv tan θ 0. 89 F θ 9. As tis force act normal to te surface, it must act troug te centre of radius of te dam wall. So te dept to te point were te force acts is, y 0sin 9. 9m.4 Te arc of a bridge over a stream is in te form of a semi-circle of radius m. te bridge widt is 4m. Due to a flood te water level is now.5m above te crest of te arc. Calculate (a) te upward force on te underside of te arc, (b) te orizontal trust on one alf of te arc. [6.6 kn, 76.6 kn] Te bridge and water level can be drawn as:.5m m a) Te upward force on te arc weigt of (imaginary) water above te arc. b) R volume of water π volume ( 5. + ) 4 4 6. 867 m R v v 000 9. 8 6. 867 6568. kn Te orizontal force on alf of te arc, is equal to te force on te projection of te curved surface onto a vertical plane..5.0 F pressure at centroid area 5+ 4 (. ) ( ) 7658. kn CIVE400: Fluid Mecanics 0
.5 Te face of a dam is vertical to a dept of 7.5m below te water surface ten slopes at 0 to te vertical. If te dept of water is 7m wat is te resultant force per metre acting on te wole face? [56.9 kn] 60 7.0 m, so 7.0-7.5 9.5. x 9.5/tan 60 5.485 m. Vertical force weigt of water above te surface, ( 05. ) F v x + x 659. kn / m (..... ) 980 7 5 5485+ 05 9 5 5485 Te orizontal force force on te projection of te surface on to a vertical plane. Te resultant force is And acts at te angle F 05. 000 9. 8 7 47. 545 kn / m F F + F 659. + 47. 545 R v 56. 9 kn / m Fv tan θ 0. 465 F θ 4. 94.6 A tank wit vertical sides is square in plan wit m long sides. Te tank contains oil of relative density 0.9 to a dept of.0m wic is floating on water a dept of.5m. Calculate te force on te walls and te eigt of te centre of pressure from te bottom of te tank. [65.54 kn,.5m] Consider one wall of te tank. Draw te pressure diagram: x CIVE400: Fluid Mecanics
d d F f d f f density of oil ρ oil 0.9ρ water 900 kg/m. Force per unit lengt, F area under te grap sum of te tree areas f + f + f f f ( 900 9. 8 ) 5974 N ( 900 9. 8 ) 5. 7946N ( 000 9. 8 5. ) 5. f 09 F f + f + f 65544 N To find te position of te resultant force F, we take moments from any point. We will take moments about te surface. DF f d + f d + f d 5. 65544 D 5974 + 7946 ( + ) + 09 ( + 5. ) D. 47m ( from surface) 5. m ( from base of wall) N CIVE400: Fluid Mecanics
Application of te Bernoulli Equation. In a vertical pipe carrying water, pressure gauges are inserted at points A and B were te pipe diameters are 0.5m and 0.075m respectively. Te point B is.5m below A and wen te flow rate down te pipe is 0.0 cumecs, te pressure at B is 475 N/m greater tan tat at A. Assuming te losses in te pipe between A and B can be expressed as k v were v is te velocity at A, g find te value of k. If te gauges at A and B are replaced by tubes filled wit water and connected to a U-tube containing mercury of relative density.6, give a sketc sowing ow te levels in te two limbs of te U-tube differ and calculate te value of tis difference in metres. [k 0.9, 0.0794m] d A 0.m A d B 0.m B Rp Part i) d 05. m d 0. 075m Q 0. 0 m / s p p 475 N / m A B f A kv g Taking te datum at B, te Bernoulli equation becomes: p u p u z z k u A A B B A + + A + + B + g g g B z A. 5 z 0 B giving u u A B ) ( π By continuity: Q u A A A u B A B ( π ) 0. 0 / 0. 075. m / s 0. 0 / 0. 075 4. 57 m / s CIVE400: Fluid Mecanics
p p u u z k u B A B A A A + 000g g g 5.. 5+ 045. 0. 065 0. 065k Part ii) p ρ gz + p xxl w B B k 0. 9 p ρ gr + ρ gz ρ gr + p xxr m p w A w p A. p xxl p ρ gz + p ρ gr + ρ gz ρ gr + p ( ) ( ) p B p A ρw g z A z B + grp ρm ρw 475 000 9. 8. 5+ 9. 8R ( 600 000) R xxr w B B m p w A w p A p 0. 079 m A Venturimeter wit an entrance diameter of 0.m and a troat diameter of 0.m is used to measure te volume of gas flowing troug a pipe. Te discarge coefficient of te meter is 0.96. Assuming te specific weigt of te gas to be constant at 9.6 N/m, calculate te volume flowing wen te pressure difference between te entrance and te troat is measured as 0.06m on a water U-tube manometer. [0.86 m /s] p d 0.m d 0.m Z Rp Z CIVE400: Fluid Mecanics 4
Wat we know from te question: Calculate Q. For te manometer: For te Venturimeter Combining () and () ρ g C g 9. 6 N / m d d d 0. 96 0. m 0. m u Q / 0. 0707 u Q / 0. 04 ( ) p + ρ gz p + ρ g z R + ρ gr g g p w p ( ) p p 9. 6 z z + 587. 4 < ( ) p u p u + + z + + z ρ g g ρ g g g g ( ) p p 9. 6 z z + 0. 80u < ( ) 0. 80u 587. 4 u ideal 7. 047 m / s 0. Qideal 7. 047 0. 85m / s π Q C Q 0. 96 0. 85 0. 86m / s d idea CIVE400: Fluid Mecanics 5
. A Venturimeter is used for measuring flow of water along a pipe. Te diameter of te Venturi troat is two fifts te diameter of te pipe. Te inlet and troat are connected by water filled tubes to a mercury U-tube manometer. Te velocity of flow along te pipe is found to be. 5 H m/s, were H is te manometer reading in metres of mercury. Determine te loss of ead between inlet and troat of te Venturi wen H is 0.49m. (Relative density of mercury is.6). [0.m of water] Z H Z For te manometer: For te Venturimeter ( ) p + ρ gz p + ρ g z H + ρ gh w w m p p ρ gz ρ gh + ρ gh ρ gz < ( ) w w m w p u p u + + z + + z + Losses ρ g g ρ g g w p Combining () and () w ρwu ρwu p + ρw gz ρw gz + Lρw g < ( ) p u p u + + z + + z + Losses ρ g g ρ g g w w ρ w Lρw g Hg( ρm ρw ) ( u u ) < ( ) but at. From te question CIVE400: Fluid Mecanics 6
Substitute in () u A u. 5 H 75. m / s u A d d 75. π uπ 4 0 u 0. 97 m / s ( ) ( )( ) 0. 49 9. 8 600 000 000 / 0. 97. 75 Losses L 9. 8 000 0. m.4 Water is discarging from a tank troug a convergent-divergent moutpiece. Te exit from te tank is rounded so tat losses tere may be neglected and te minimum diameter is 0.05m. If te ead in te tank above te centre-line of te moutpiece is.8m. a) Wat is te discarge? b) Wat must be te diameter at te exit if te absolute pressure at te minimum area is to be.44m of water? c) Wat would te discarge be if te divergent part of te mout piece were removed. (Assume atmosperic pressure is 0m of water). [0.075m, 0.066m /s, 0.08m /s] From te question: Apply Bernoulli: If we take te datum troug te orifice: Between and d 0. 05m p minimum pressure. 44m p m p 0 p u p u p u + + z + + z + + z g g g z 8. m z z 0 u negligible CIVE400: Fluid Mecanics 7
Between and p p u 0 + 8.. 44 + g If te mout piece as been removed, p u 57. m / s 0 05 57 0 0665. Q u A. π. m / s u 8. g u 599. m / s Q u A d 0. 0665 599. π 4 d 0. 075m p p p u + z + g u gz 599. m / s Q 5 99 0.. π 05 0. 08m / s 4.5 A closed tank as an orifice 0.05m diameter in one of its vertical sides. Te tank contains oil to a dept of 0.6m above te centre of te orifice and te pressure in te air space above te oil is maintained at 780 N/m above atmosperic. Determine te discarge from te orifice. (Coefficient of discarge of te orifice is 0.6, relative density of oil is 0.9). [0.0095 m /s] P 780 kn/m 0.66m oil d o 0.05m From te question CIVE400: Fluid Mecanics 8
Apply Bernoulli, Take atmosperic pressure as 0, ρo σ 0. 9 ρ ρ o C d 900 0. 6 p u p u z z + g + + g + w 780 0 6 ρ g +. o ug u 6. 5m / s 0 05 Q 0 6 6 5 π 0 0095m s.... /.6 Te discarge coefficient of a Venturimeter was found to be constant for rates of flow exceeding a certain value. Sow tat for tis condition te loss of ead due to friction in te convergent parts of te meter can be expressed as KQ m were K is a constant and Q is te rate of flow in cumecs. Obtain te value of K if te inlet and troat diameter of te Venturimeter are 0.0m and 0.05m respectively and te discarge coefficient is 0.96. [K060].7 A Venturimeter is to fitted in a orizontal pipe of 0.5m diameter to measure a flow of water wic may be anyting up to 40m /our. Te pressure ead at te inlet for tis flow is 8m above atmosperic and te pressure ead at te troat must not be lower tan 7m below atmosperic. Between te inlet and te troat tere is an estimated frictional loss of 0% of te difference in pressure ead between tese points. Calculate te minimum allowable diameter for te troat. [0.06m] d 0.5m d From te question: d 05. m Q 40m / r 0. 667m / s u Q / A. 77m / s p p 8m 7m Friction loss, from te question: CIVE400: Fluid Mecanics 9
Apply Bernoulli: f 0. ( p p ) p u p u + + + + g g p p u u + f g g u 5. 77. 5 g g u. 46 m / s Q u A d 0. 0667. 46 π 4 d 0. 06m f.8 A Venturimeter of troat diameter 0.076m is fitted in a 0.5m diameter vertical pipe in wic liquid of relative density 0.8 flows downwards. Pressure gauges are fitted to te inlet and to te troat sections. Te troat being 0.94m below te inlet. Taking te coefficient of te meter as 0.97 find te discarge a) wen te pressure gauges read te same b)wen te inlet gauge reads 570 N/m iger tan te troat gauge. [0.09m /s, 0.04m /s] d 0.5m d 0.076m From te question: CIVE400: Fluid Mecanics 0
Apply Bernoulli: a) p p By continuity: d 05. m A 0. 084m d 0. 076m A 0. 00454m ρ 800 kg / m C d 0. 97 p u p u + + z + + z g g u u + z + z g g Q u A u A u u A u A 4 u 6u + 0. 94 g g 0. 94 9. 8 u 094. m / s 5 Q C A u d Q 0. 96 0. 084 094. 0. 09 m / s b) p p u p p 570 u g (.. ) 570 Q 0 4 55 g ( ) 55. 8577 Q 0. 4 55. Q 0. 05m / s 0. 94 0. 94 CIVE400: Fluid Mecanics
Tank emptying 4. A reservoir is circular in plan and te sides slope at an angle of tan - (/5) to te orizontal. Wen te reservoir is full te diameter of te water surface is 50m. Discarge from te reservoir takes place troug a pipe of diameter 0.65m, te outlet being 4m below top water level. Determine te time for te water level to fall m assuming te discarge to be 0. 75a gh cumecs were a is te cross sectional area of te pipe in m and H is te ead of water above te outlet in m. [5 seconds] 50m r H x 5 From te question: H 4m a π(0.65/) 0.m Q 0. 75a g 096. In time δt te level in te reservoir falls δ, so Qδt Aδ A δt δ Q Integrating give te total time for levels to fall from to. T A Q d As te surface area canges wit eigt, we must express A in terms of. A πr But r varies wit. It varies linearly from te surface at H 4m, r 5m, at a gradient of tan - /5. r x + 5 5 x + 5(4) x 5 so A π( 5 + 5 ) ( 5π + 5π + 50π ) Substituting in te integral equation gives CIVE400: Fluid Mecanics
T 5π 096. From te question, 4m 5π + 5π + 50π d 096. + + d 764. + + d / / / 764. + + d 4 764. + + 5 / 5/ / m, so [( ) ( )] [ ] T 764 4 + + + + 5 4 4 4 5 4. 764. 4 +. 8 + 0. 667. 88 +. 6+. 77 764. 7. 467 8. 86 sec / 5/ / / 5/ / 4. A rectangular swimming pool is m deep at one end and increases uniformly in dept to.6m at te oter end. Te pool is 8m wide and m long and is emptied troug an orifice of area 0.4m, at te lowest point in te side of te deep end. Taking C d for te orifice as 0.6, find, from first principles, a) te time for te dept to fall by m b) te time to empty te pool completely. [99 second, 66 seconds].0m.0m L.6m Te question tell us a o 0.4m, C d 0.6 Apply Bernoulli from te tank surface to te vena contracta at te orifice: p p and u 0. u p u p u z z + g + + g + g We need Q in terms of te eigt measured above te orifice. CIVE400: Fluid Mecanics
Q C a u C a g d o d o 0. 6 0. 4 9. 8 0. 595 And we can write an equation for te discarge in terms of te surface eigt cange: Qδt Aδ A δt δ Q Integrating give te total time for levels to fall from to. A T Q d A 68. d < ( ) a) For te first m dept, A 8 x 56, watever te. So, for te first period of time: T 68. 56 d [ ] [ ] 40. 08 40. 08. 6 6. 99 sec b) now we need to find out ow long it will take to empty te rest. We need te area A, in terms of. So A 8L L A. 6 60 T 68. 60 d / / [( ) ( ) ] 68. 9 68 9 [( ) ( ) ] 6 / 0 /.. 6. 67 sec Total time for emptying is, T 6 + 99 66 sec CIVE400: Fluid Mecanics 4
4. A vertical cylindrical tank m diameter as, at te bottom, a 0.05m diameter sarp edged orifice for wic te discarge coefficient is 0.6. a) If water enters te tank at a constant rate of 0.0095 cumecs find te dept of water above te orifice wen te level in te tank becomes stable. b) Find te time for te level to fall from m to m above te orifice wen te inflow is turned off. c) If water now runs into te tank at 0.0 cumecs, te orifice remaining open, find te rate of rise in water level wen te level as reaced a dept of.7m above te orifice. [a).4m, b) 88 seconds, c) 0.5m/min] Q 0.0095 m /s d o 0.005m From te question: Q in 0.0095 m /s, d o 0.05m, C d 0.6 Apply Bernoulli from te water surface () to te orifice (), p p and u 0. u p u p u + + z + + z g g g. Wit te datum te bottom of te cylinder, z, z 0 We need Q in terms of te eigt measured above te orifice. Q C a u C a g out d o d o 0. 05 0. 6π 9. 8 0. 005 < ( ) For te level in te tank to remain constant: inflow out flow Q in Q out 0. 0095 0. 005 4. m (b) Write te equation for te discarge in terms of te surface eigt cange: CIVE400: Fluid Mecanics 5
Qδt Aδ A δt δ Q Integrating between and, to give te time to cange surface level T and so A 608. 06. Q d / d / [ ] / / [ ] 06. T 88 sec c) Q in canged to Q in 0.0 m /s From () we ave Qout 0. 005. Te question asks for te rate of surface rise wen.7m. i.e. Qout 0. 005 7. 0. 0068m / s Te rate of increase in volume is: Q Q Q 0. 0 0. 0068 0. 0m / s in As Q Area x Velocity, te rate of rise in surface is Q Au out Q 0. 0 u 0. 004m / s 0. 5m / min A π 4 4.4 A orizontal boiler sell (i.e. a orizontal cylinder) m diameter and 0m long is alf full of water. Find te time of emptying te sell troug a sort vertical pipe, diameter 0.08m, attaced to te bottom of te sell. Take te coefficient of discarge to be 0.8. [70 seconds] m d m From te question W 0m, D 0m d o 0.08m C d 0.8 d o 0.08 m CIVE400: Fluid Mecanics 6
Apply Bernoulli from te water surface () to te orifice (), p u p u + + z + + z g g p p and u 0. u g. Wit te datum te bottom of te cylinder, z, z 0 We need Q in terms of te eigt measured above te orifice. Q C a u C a g out d o d o 0. 8π 0. 078 0. 08 9. 8 Write te equation for te discarge in terms of te surface eigt cange: Qδt Aδ A δt δ Q Integrating between and, to give te time to cange surface level T But we need A in terms of A Q d.0m a L Surface area A 0L, so need L in terms of + L a a ( ) + L ( ) ( ) L A 0 Substitute tis into te integral term,.0m. CIVE400: Fluid Mecanics 7
T 6. 6. 0 0078. [ ] d. 6 d 6. d d / [( ) ] 749. 07. 88 69. 6sec 4.5 Two cylinders standing uprigt contain liquid and are connected by a submerged orifice. Te diameters of te cylinders are.75m and.0m and of te orifice, 0.08m. Te difference in levels of te liquid is initially.5m. Find ow long it will take for tis difference to be reduced to 0.66m if te coefficient of discarge for te orifice is 0.605. (Work from first principles.) [0.7 seconds] d.75m d.0m.5m d o 0.08m by continuity, 75 A 4m A 0785m. π. π. 0 08 do 008m ao 00050m Cd 0605.., π.. A δ A δ Qδ t < ( ) defining, - δ δ + δ Substituting tis in () to eliminate δ CIVE400: Fluid Mecanics 8
A δ A ( δ δ) A δ A δ δ A A δ A + A Aδ A + A Qδt < ( ) From te Bernoulli equation we can derive tis expression for discarge troug te submerged orifice: So Integrating Q C a g d o A A δ A + A δt T ( + ) C a g δt d o A A δ A A C a g d o A A A A C a g d ( + ) A A ( + ) A A C a g ( ) d d o o ( ). 4 0. 785. 4 + 0. 785 0. 605 0. 0050 9. 8 0. 7sec ( 084. 69. ) 4.6 A rectangular reservoir wit vertical walls as a plan area of 60000m. Discarge from te reservoir take place over a rectangular weir. Te flow caracteristics of te weir is Q 0.678 H / cumecs were H is te dept of water above te weir crest. Te sill of te weir is.4m above te bottom of te reservoir. Starting wit a dept of water of 4m in te reservoir and no inflow, wat will be te dept of water after one our? [.98m] From te question A 60 000 m, Q 0.678 / Write te equation for te discarge in terms of te surface eigt cange: CIVE400: Fluid Mecanics 9
Qδt Aδ A δt δ Q Integrating between and, to give te time to cange surface level T A Q d 60000 0. 678 8849558. / d / [ ] From te question T 600 sec and 0.6m / / [ ] 600 76995. 0. 6 0. 585m Total dept.4 + 0.58.98m CIVE400: Fluid Mecanics 0
Notces and weirs 5. Deduce an expression for te discarge of water over a rigt-angled sarp edged V-notc, given tat te coefficient of discarge is 0.6. A rectangular tank 6m by 6m as te same notc in one of its sort vertical sides. Determine te time taken for te ead, measured from te bottom of te notc, to fall from 5cm to 7.5cm. [99 seconds] From your notes you can derive: Q 8 θ Cd gh 5 5/ tan For tis weir te equation simplifies to Q 44H 5 /. Write te equation for te discarge in terms of te surface eigt cange: Qδt Aδ A δt δ Q Integrating between and, to give te time to cange surface level T A Q d 6 6 44. 66. 67 0.5m, 0.075m / 5 / [ ] d / / [ ] T 44. 44 0. 075 05. 99sec CIVE400: Fluid Mecanics
5. Derive an expression for te discarge over a sarp crested rectangular weir. A sarp edged weir is to be constructed across a stream in wic te normal flow is 00 litres/sec. If te maximum flow likely to occur in te stream is 5 times te normal flow ten determine te lengt of weir necessary to limit te rise in water level to 8.4cm above tat for normal flow. C d 0.6. [.4m] From your notes you can derive: From te question: Q Cdb g Q 0. m /s, / x Q.0 m /s, x + 0.84 were x is te eigt above te weir at normal flow. So we ave two situations: / / 0. Cdb gx 80. bx < ( ) / / 0. C ( + 0. 84) 80. ( + 0. 84) db g x b x < ( ) From () we get an expression for b in terms of x b 0x /. Substituting tis in () gives, So te weir breadt is x + 0. 84 0. 80. 0. x / x + 0. 84 5 x x 0996. m ( ) b 0. 0996. 4. m / / CIVE400: Fluid Mecanics
5. Sow tat te rate of flow across a triangular notc is given by QC d KH 5/ cumecs, were C d is an experimental coefficient, K depends on te angle of te notc, and H is te eigt of te undisturbed water level above te bottom of te notc in metres. State te reasons for te introduction of te coefficient. Water from a tank aving a surface area of 0m flows over a 90 notc. It is found tat te time taken to lower te level from 8cm to 7cm above te bottom of te notc is 4.5seconds. Determine te coefficient C d assuming tat it remains constant during is period. [0.65] 8 θ Te proof for Q Cd g H Cd KH 5 5/ 5/ tan is in te notes. From te question: So A 0m θ 90 0.08m 0.07m T 4.5sec Q.6 C d 5/ Write te equation for te discarge in terms of te surface eigt cange: Qδt Aδ A δt δ Q Integrating between and, to give te time to cange surface level T 0. 6C A Q d 4. C [ / ] d. 8 45. 0 07 0 08 C C d d 0. 65 d d 5/ 0. 08 0. 07 / / [.. ] 5.4 A reservoir wit vertical sides as a plan area of 56000m. Discarge from te reservoir takes place over a rectangular weir, te flow caracteristic of wic is Q.77BH / m /s. At times of maximum rainfall, water flows into te reservoir at te rate of 9m /s. Find a) te lengt of weir required to discarge tis quantity if ead must not exceed 0.6m; b) te time necessary for te ead to drop from 60cm to 0cm if te inflow suddenly stops. [0.94m, 09seconds] From te question: A 56000 m Q.77 B H / Q max 9 m /s a) Find B for H 0.6 9.77 B 0.6 / B 0.94m CIVE400: Fluid Mecanics
b) Write te equation for te discarge in terms of te surface eigt cange: Qδt Aδ A δt δ Q Integrating between and, to give te time to cange surface level T A Q d 56000 77. B 56000 77. T 09sec / d / 0. [ ] 0. 6 B / / [.. ] 5784 0 0 6 5.5 Develop a formula for te discarge over a 90 V-notc weir in terms of ead above te bottom of te V. A cannel conveys 00 litres/sec of water. At te outlet end tere is a 90 V-notc weir for wic te coefficient of discarge is 0.58. At wat distance above te bottom of te cannel sould te weir be placed in order to make te dept in te cannel.0m? Wit te weir in tis position wat is te dept of water in te cannel wen te flow is 00 litres/sec? [0.755m,.8m] Derive tis formula from te notes: Q 8 θ Cd gh 5 5/ tan From te question: θ 90 C d 0.58 Q 0. m /s, dept of water, Z 0.m giving te weir equation: Q 7H 5 /. a) As H is te eigt above te bottom of te V, te dept of water Z D + H, were D is te eigt of te bottom of te V from te base of te cannel. So Q 7. ( Z D) 0. 7. (. D) D 0. 755m b) Find Z wen Q 0. m /s 5/ 5/ 0. 7. ( Z 0. 755) Z 8. m 5/ CIVE400: Fluid Mecanics 4
5.6 Sow tat te quantity of water flowing across a triangular V-notc of angle θ is 8 5/ Q Cd tanθ gh. Find te flow if te measured ead above te bottom of te V is 8cm, wen 5 θ45 and C d 0.6. If te flow is wanted witin an accuracy of %, wat are te limiting values of te ead. [0.6m /s, 0.77m, 0.8m] Proof of te v-notc weir equation is in te notes. From te question: H 0.8m θ 45 C d 0.6 Te weir equation becomes: Q 47. H 5/ 47. ( 0. 8) 06. m / s 5/ Q+% 0.9 m /s 09. 47. H H 0. 8m 5/ Q-% 0.4 m /s 04. 47. H H 0. 77m 5/ CIVE400: Fluid Mecanics 5
Application of te Momentum Equation 6. Te figure below sows a smoot curved vane attaced to a rigid foundation. Te jet of water, rectangular in section, 75mm wide and 5mm tick, strike te vane wit a velocity of 5m/s. Calculate te vertical and orizontal components of te force exerted on te vane and indicate in wic direction tese components act. [Horizontal.4 N acting from rigt to left. Vertical 4.6 N acting downwards] 45 5 From te question: a u 0. 075 0. 05 875. 0 5m / s Q 875. 0 5m / s a a, so u u m Calculate te total force using te momentum equation: ( cos 5 cos45) F Q u u T ρ x 000 0. 0469( 5cos 5 5cos 45) 44. N ( sin 5 sin 45) F Q u u T ρ y 000 0. 0469( 5sin 5 5sin 45) 4. 6 N Body force and pressure force are 0. So force on vane: R F. 44N x t x R F 4. 6N y t y CIVE400: Fluid Mecanics 6
6. A 600mm diameter pipeline carries water under a ead of 0m wit a velocity of m/s. Tis water main is fitted wit a orizontal bend wic turns te axis of te pipeline troug 75 (i.e. te internal angle at te bend is 05 ). Calculate te resultant force on te bend and its angle to te orizontal. [04.044 kn, 5 9 ] y u x u θ From te question: a m d m m π 0. 6 0. 8 0. 6 0 u u m / s Q 0848. m / s Calculate total force. ( ) F Q u u F + F + F Tx ρ x x Rx Px Bx F ( ) Tx 000 0. 848 cos 75 886. kn ( ) F Q u u F + F + F Ty ρ y y Ry Py By F ( ) Ty 000 0. 848 sin 75 0. 457kN Calculate te pressure force p p p 0 000 9.8 94. kn/m FTx pa cosθ pa cosθ 9400 0. 8( cos 75) 67. kn F p a sinθ p a Ty sinθ 9400 0. 8( 0 sin 75) 80. 76 kn Tere is no body force in te x or y directions. CIVE400: Fluid Mecanics 7
F F F F Rx Tx Px Bx 886. 6. 7 0 6. 66 kn F F F F Ry Ty Py By. 457 + 80. 76 0 8. 8kN Tese forces act on te fluid Te resultant force on te fluid is F F + F 04. 44 kn R Rx Ry FRy θ tan 5 9 F Rx 6. A orizontal jet of water 0 mm cross-section and flowing at a velocity of 5 m/s its a flat plate at 60 to te axis (of te jet) and to te orizontal. Te jet is suc tat tere is no side spread. If te plate is stationary, calculate a) te force exerted on te plate in te direction of te jet and b) te ratio between te quantity of fluid tat is deflected upwards and tat downwards. (Assume tat tere is no friction and terefore no sear force.) [8N, :] y u x u u θ From te question a a x0 - m u 5 m/s Apply Bernoulli, p u p u p u + + z + + z + + z g g g Cange in eigt is negligible so z z z and pressure is always atmosperic p p p 0. So u u u 5 m/s By continuity Q Q + Q u a u a + u a CIVE400: Fluid Mecanics 8
so a a + a Put te axes normal to te plate, as we know tat te resultant force is normal to te plate. Q a u 0-5 0.0 Q (a + a ) u Q a u Q (a - a )u Calculate total force. ( ) F Q u u F + F + F Tx ρ x x Rx Px Bx F ( ) Tx 000 0. 0 0 5sin 60 90 N Component in direction of jet 90 sin 60 8 N As tere is no force parallel to te plate F ty 0 F u a Ty u a ρ ρ ρu a θ cos 0 a a a cosθ 0 a a + a a + a cosθ a a 4 4a a a a a Tus /4 of te jet goes up, /4 down 6.4 A 75mm diameter jet of water aving a velocity of 5m/s strikes a flat plate, te normal of wic is inclined at 0 to te jet. Find te force normal to te surface of te plate. [.9kN] y u x u u θ CIVE400: Fluid Mecanics 9
CIVE400: Fluid Mecanics From te question, d jet 0.075m u 5m/s Q 5π(0.075/) 0. m /s Force normal to plate is F Tx ρq( 0 - u x ) F Tx 000 0. ( 0-5 cos 0 ).9 kn 6.5 Te outlet pipe from a pump is a bend of 45 rising in te vertical plane (i.e. and internal angle of 5 ). Te bend is 50mm diameter at its inlet and 00mm diameter at its outlet. Te pipe axis at te inlet is orizontal and at te outlet it is m iger. By neglecting friction, calculate te force and its direction if te inlet pressure is 00kN/m and te flow of water troug te pipe is 0.m /s. Te volume of te pipe is 0.075m. [.94kN at 67 40 to te orizontal] y p u A x p m u 45 A & Draw te control volume and te axis system p 00 kn/m, Q 0. m /s θ 45 d 0.5 m d 0. m A 0.77 m A 0.0707 m Calculate te total force in te x direction ( ) F ρq u u T x x x ( cosθ u ) ρq u by continuity A u A u Q, so CIVE400: Fluid Mecanics 40
u u 0. 6. 98m / s π ( 05. / 4) 0. 4. 4m / s 0. 0707 F ( ) T 000 0. 4. 4cos 45 6. 98 x 49. 68 N and in te y-direction ( ) F ρq u u T y y y ( sinθ ) ρq u 0 ( ) 000 0. 4. 4sin45 899. 44 N 4 Calculate te pressure force. F P pressure force at - pressure force at F p A cos0 p A cosθ p A p A cosθ P x F p A sin 0 p A sinθ p A sinθ P y We know pressure at te inlet but not at te outlet. we can use Bernoulli to calculate tis unknown pressure. p u p u + + z + + z + g g were f is te friction loss In te question it says tis can be ignored, f 0 Te eigt of te pipe at te outlet is m above te inlet. Taking te inlet level as te datum: z 0 z m So te Bernoulli equation becomes: 00000 6. 98 000 9 8 9 8 0 p 4. 4 + + + + 000 9 8 9 8 0..... p 564. N / m f CIVE400: Fluid Mecanics 4
F P x 00000 0. 077 564. cos 45 0. 0707 770 66. 4 9496. 7 kn F P y 564. sin 45 0. 0707 66. 7 5 Calculate te body force Te only body force is te force due to gravity. Tat is te weigt acting in te y direction. F B volume y 000 9. 8 0. 075 9056. N Tere are no body forces in te x direction, F 0 B x 6 Calculate te resultant force F F + F + F T x R x P x B x F F + F + F T y R y P y B y F F F F R x T x P x B x 49. 6 + 9496. 7 50. 7 N F F F F R y T y P y B y 899. 44 + 66. 7 + 75. 75 9056. N And te resultant force on te fluid is given by F Ry F Resultant φ F Rx CIVE400: Fluid Mecanics 4
CIVE400: Fluid Mecanics F F F R R x R y 50. 7 + 9056.. 95kN And te direction of application is φ tan F R y tan 9056. 67. 66 F 50. 7 R x Te force on te bend is te same magnitude but in te opposite direction R F R 6.6 Te force exerted by a 5mm diameter jet against a flat plate normal to te axis of te jet is 650N. Wat is te flow in m /s? [0.08 m /s] u y u x u From te question, d jet 0.05m F Tx 650 N Force normal to plate is F Tx ρq( 0 - u x ) 650 000 Q ( 0 - u ) Q au (πd /4)u 650-000au -000Q /a 650-000Q /(π0.05 /4) Q 0.08m /s CIVE400: Fluid Mecanics 4
6.7 A curved plate deflects a 75mm diameter jet troug an angle of 45. For a velocity in te jet of 40m/s to te rigt, compute te components of te force developed against te curved plate. (Assume no friction). [R x 070N, R y 5000N down] u u y x θ From te question: a u π 0. 075 / 4 4. 4 0 40m / s m Q 4. 4 0 40 0767. m / s a a, so u u Calculate te total force using te momentum equation: ( cos45 ) F Q u u F T ρ x T y 0707. N ( ) 000 0767. 40cos45 40 ( sin ) ρq u 45 0 4998 N Body force and pressure force are 0. So force on vane: ( ) 000 0767. 40sin45 R F 070N x t x R F 4998N y t y CIVE400: Fluid Mecanics 44
6.8 A 45 reducing bend, 0.6m diameter upstream, 0.m diameter downstream, as water flowing troug it at te rate of 0.45m /s under a pressure of.45 bar. Neglecting any loss is ead for friction, calculate te force exerted by te water on te bend, and its direction of application. [R4400N to te rigt and down, θ 4 ] y ρ u A x ρ u A θ & Draw te control volume and te axis system p.45 0 5 N/m, Q 0.45 m /s θ 45 d 0.6 m d 0. m A 0.8 m A 0.0707 m Calculate te total force in te x direction ( ) F ρq u u T x x x ( cosθ u ) ρq u by continuity A u A u Q, so u u 0. 45 59. m / s π ( 0. 6 / 4) 0. 45 6. 65m / s 0. 0707 CIVE400: Fluid Mecanics 45
F ( ) T 000 0. 45 6. 65cos 45 59. x 0 N and in te y-direction ( ) F ρq u u T y y y ( sinθ ) ρq u 0 800 N 4 Calculate te pressure force. ( ) 000 0. 45 6. 65sin45 F P pressure force at - pressure force at F p A cos0 p A cosθ p A p A cosθ P x F p A sin 0 p A sinθ p A sinθ P y We know pressure at te inlet but not at te outlet. we can use Bernoulli to calculate tis unknown pressure. p u p u + + z + + z + g g were f is te friction loss In te question it says tis can be ignored, f 0 Assume te pipe to be orizontal z z So te Bernoulli equation becomes: 45000 59. p 6. 65 + + 000 9. 8 9. 8 000 9. 8 9. 8 p 6007 N / m f F P x 45000 0. 8 6000cos 45 0. 0707 405 600 475 N F P y 6000sin 45 0. 0707 600 N CIVE400: Fluid Mecanics 46
CIVE400: Fluid Mecanics 5 Calculate te body force Te only body force is te force due to gravity. Tere are no body forces in te x or y directions, F B x F 0 B y 6 Calculate te resultant force F F + F + F T x R x P x B x F F + F + F T y R y P y B y F F F F R x T x P x B x 0 475 45 N F F F F R y T y P y B y 800 + 600 800 N And te resultant force on te fluid is given by F Ry F Resultant φ F Rx F F F R R x R y 45 + 800 49 kn And te direction of application is F R y 800 φ tan tan. 6 F 45 R x Te force on te bend is te same magnitude but in te opposite direction R F R CIVE400: Fluid Mecanics 47
Laminar pipe flow. 7. Te distribution of velocity, u, in metres/sec wit radius r in metres in a smoot bore tube of 0.05 m bore follows te law, u.5 - kr. Were k is a constant. Te flow is laminar and te velocity at te pipe surface is zero. Te fluid as a coefficient of viscosity of 0.0007 kg/m s. Determine (a) te rate of flow in m /s (b) te searing force between te fluid and te pipe wall per metre lengt of pipe. [6.4x0-4 m /s, 8.49x0 - N] Te velocity at distance r from te centre is given in te question: u.5 - kr Also we know: µ 0.0007 kg/ms r 0.05m We can find k from te boundary conditions: wen r 0.05, u 0.0 (boundary of te pipe) 0.0.5 - k0.05 k 6000 u.5-600 r a) Following along similar lines to te derivation seen in te lecture notes, we can calculate te flow δq troug a small annulus δr: δq u A b) Te sear force is given by From Newtons law of viscosity r annulus A π( r + δr) πr πrδr annulus F τ (πr) (. ) δq 5 6000r πrδr 0. 05 (. ) Q π 5r 6000r dr 0. 5r π 64. m / s 6000 4 du τ µ dr du 6000r 000r dr F 0. 0007 000 0. 05 ( π 0. 05) 848. 0 N r 4 0. 05 0 CIVE400: Fluid Mecanics 48
7. A liquid wose coefficient of viscosity is m flows below te critical velocity for laminar flow in a circular pipe of diameter d and wit mean velocity u. Sow tat te pressure loss in a lengt of pipe is um/d. Oil of viscosity 0.05 kg/ms flows troug a pipe of diameter 0.m wit a velocity of 0.6m/s. Calculate te loss of pressure in a lengt of 0m. [ 50 N/m ] See te proof in te lecture notes for Consider a cylinder of fluid, lengt L, radius r, flowing steadily in te centre of a pipe δr r r R Te fluid is in equilibrium, searing forces equal te pressure forces. τ πr L p A pπr p r τ L Newtons law of viscosity τ µ du dy, We are measuring from te pipe centre, so τ µ du dr Giving: p r du µ L dr du p r dr L µ In an integral form tis gives an expression for velocity, p u L µ r dr Te value of velocity at a point distance r from te centre p r ur + C L 4µ At r 0, (te centre of te pipe), u u max, at r R (te pipe wall) u 0; p R C L 4µ At a point r from te pipe centre wen te flow is laminar: CIVE400: Fluid Mecanics 49
u r p L 4µ ( R r ) Te flow in an annulus of tickness δr δq u A r annulus A π( r + δr) πr πrδr annulus So te discarge can be written p δq ( R r ) πrδr L 4µ R p π Q ( R r r ) dr L µ p Q π 4 d L 8 µ 0 4 4 p πr pπd L 8µ L8µ To get pressure loss in terms of te velocity of te flow, use te mean velocity: u Q / A pd u µ L µ Lu p d µ u p d per unit lengt b) From te question µ 0.05 kg/ms d 0.m u 0.6 m/s L 0.0m 0. 05 0 0. 6 p 50 N / m 0. CIVE400: Fluid Mecanics 50
CIVE400: Fluid Mecanics 7. A plunger of 0.08m diameter and lengt 0.m as four small oles of diameter 5/600 m drilled troug in te direction of its lengt. Te plunger is a close fit inside a cylinder, containing oil, suc tat no oil is assumed to pass between te plunger and te cylinder. If te plunger is subjected to a vertical downward force of 45N (including its own weigt) and it is assumed tat te upward flow troug te four small oles is laminar, determine te speed of te fall of te plunger. Te coefficient of velocity of te oil is 0. kg/ms. [0.00064 m/s] plunger cylinder Q F 45N d 5/600 m 0. m 0.8m Flow troug eac tube given by Hagen-Poiseuille equation p Q π 4 d L 8 µ Tere are 4 of tese so total flow is 4 4 p πd 4π ( 5/ 600) Q 4 p p60. 0 L 8µ 0. 8 0. Force pressure area 0. 08 5 / 600 F 45 p π 4π p 9007. 06 N / m Q 4. 0 6 m / s Flow up troug piston flow displaced by moving piston Q Av piston 0 CIVE400: Fluid Mecanics 5
.4 0-6 π 0.04 v piston v piston 0.00064 m/s 7.4 A vertical cylinder of 0.075 metres diameter is mounted concentrically in a drum of 0.076metres internal diameter. Oil fills te space between tem to a dept of 0.m. Te rotque required to rotate te cylinder in te drum is 4Nm wen te speed of rotation is 7.5 revs/sec. Assuming tat te end effects are negligible, calculate te coefficient of viscosity of te oil. [0.68 kg/ms] From te question r - 0.076/ r 0.075/ Torque 4Nm, L 0.m Te velocity of te edge of te cylinder is: u cyl 7.5 πr 7.5 π 0.075.767 m/s u drum 0.0 Torque needed to rotate cylinder T τ surface area ( πr L) 4 τ τ 654. N / m Distance between cylinder and drum r - r 0.08-0.075 0.005m Using Newtons law of viscosity: du τ µ dr du 767. 0 dr 0. 0005 τ 65. µ 54 µ 0. 64 kg / ms ( Ns / m ) CIVE400: Fluid Mecanics 5
Dimensional analysis 8. A stationary spere in water moving at a velocity of.6m/s experiences a drag of 4N. Anoter spere of twice te diameter is placed in a wind tunnel. Find te velocity of te air and te drag wic will give dynamically similar conditions. Te ratio of kinematic viscosities of air and water is, and te density of air.8 kg/m. [0.4m/s 0.865N] Draw up te table of values you ave for eac variable: variable water air u.6m/s u air Drag 4N D air ν ν ν ρ 000 kg/m.8 kg/m d d d Kinematic viscosity is dynamic viscosity over density ν µ/ρ. ρud ud Te Reynolds number Re µ ν Coose te tree recurring (governing) variables; u, d, ρ. From Buckingams π teorem we ave m-n 5 - non-dimensional groups. ( u, d,, D, ) φ( π, π ) φ ρ ν 0 0 π u d π u d a b c ρ D ρ ν a b c As eac π group is dimensionless ten considering te dimensions, for te first group, π : (note D is a force wit dimensions MLT - ) ( ) ( ) ( ) 0 0 0 a b c M L T LT L ML MLT M] 0 c + c - L] 0 a + b - c + -4 a + b T] 0 -a - a - b - CIVE400: Fluid Mecanics 5
π u d ρ D D ρu d And te second group π : ( ) ( ) ( ) 0 0 0 a b c M L T LT L ML L T M] 0 c L] 0 a + b - c + - a + b T] 0 -a - π u d ν ud a - b - ρ ν 0 So te pysical situation is described by tis function of nondimensional numbers, ν φ( π, π ) φ, D ρ u d ud 0 For dynamic similarity tese non-dimensional numbers are te same for te bot te spere in water and in te wind tunnel i.e. For π π π π air water π air water D ρu d D ρu d air water Dair 4 8. 0. 4 ( d) 000 6. d D 0. 865 N For π ν ν ud ud ν ν u d 6. d air u air air air water 0. 4m / s CIVE400: Fluid Mecanics 54
8. Explain briefly te use of te Reynolds number in te interpretation of tests on te flow of liquid in pipes. Water flows troug a cm diameter pipe at.6m/s. Calculate te Reynolds number and find also te velocity required to give te same Reynolds number wen te pipe is transporting air. Obtain te ratio of pressure drops in te same lengt of pipe for bot cases. For te water te kinematic viscosity was. 0-6 m /s and te density was 000 kg/m. For air tose quantities were 5. 0-6 m /s and.9kg/m. [447, 8.4m/s, 0.57] Draw up te table of values you ave for eac variable: variable water air u.6m/s u air p p water p air ρ 000 kg/m.9kg/m ν. 0 6 m /s 5. 0 6 m /s ρ 000 kg/m.8 kg/m d 0.0m 0.0m Kinematic viscosity is dynamic viscosity over density ν µ/ρ. ρud ud Te Reynolds number Re µ ν Reynolds number wen carrying water: Re water To calculate Re air we know, Re water ud 6. 0. 0 6 447 ν. 0 Re air uair 0. 0 447 6 5 0 u 8. 44m / s air To obtain te ratio of pressure drops we must obtain an expression for te pressure drop in terms of governing variables. Coose te tree recurring (governing) variables; u, d, ρ. From Buckingams π teorem we ave m-n 5 - non-dimensional groups. φ ( u, d, ρ, ν, p) φ( π, π ) π u d π 0 0 u d ρ ν a b c ρ a b c p As eac π group is dimensionless ten considering te dimensions, for te first group, π : CIVE400: Fluid Mecanics 55
( ) ( ) ( ) 0 0 0 a b c M L T LT L ML L T M] 0 c L] 0 a + b - c + - a + b T] 0 -a - a - b - π u d ν ud ρ ν 0 And te second group π : (note p is a pressure (force/area) wit dimensions ML - T - ) ( ) ( ) ( ) 0 0 0 a b c M L T LT L ML MT L M] 0 c + c - L] 0 a + b - c - - a + b T] 0 -a - a - b 0 π u p ρu ρ p So te pysical situation is described by tis function of nondimensional numbers, φ( π, π ) φ ν, p ud ρu 0 For dynamic similarity tese non-dimensional numbers are te same for te bot water and air in te pipe. π π π air water π air water We are interested in te relationsip involving te pressure i.e. π CIVE400: Fluid Mecanics 56
p ρu p p p ρu air water air ρwateru ρ u air water water air 000 6. 9 8 44... 6. 7 058 Sow tat Reynold number, ρud/µ, is non-dimensional. If te discarge Q troug an orifice is a function of te diameter d, te pressure difference p, te density ρ, and te viscosity µ, sow tat Q Cp / d /ρ / were C is some function of te non-dimensional group (dρ / d / /µ). Draw up te table of values you ave for eac variable: Te dimensions of tese following variables are ρ ML - u LT - d L µ ML - T - i.e. Re is dimensionless. Re ML - LT - L(ML - T - ) - ML - LT - L M - LT We are told from te question tat tere are 5 variables involved in te problem: d, p, ρ, µ and Q. Coose te tree recurring (governing) variables; Q, d, ρ. From Buckingams π teorem we ave m-n 5 - non-dimensional groups. ( Q, d,,, p) φ( π, π ) φ ρ µ a b c π Q d ρ µ π 0 0 Q d ρ a b c p As eac π group is dimensionless ten considering te dimensions, for te first group, π : ( ) ( ) ( ) 0 0 0 a b c M L T L T L ML ML T M] 0 c + c - L] 0 a + b - c - - a + b T] 0 -a - a - b CIVE400: Fluid Mecanics 57
π Q d ρ µ dµ ρq And te second group π : (note p is a pressure (force/area) wit dimensions ML - T - ) ( ) ( ) ( ) 0 0 0 a b c M L T L T L ML MT L M] 0 c + c - L] 0 a + b - c - - a + b T] 0 -a - a - b 4 4 π Q d ρ p 4 d p ρq So te pysical situation is described by tis function of non-dimensional numbers, 4 µ φ( π, π ) φ, 0 d d p Qρ ρq or 4 dµ φ ρ ρ d p Q Q d p d p Te question wants us to sow : Q f ρ µ ρ / / / Take te reciprocal of square root of π : π / ρ Q π d p, / a Convert π by multiplying by tis number π / dµ ρ Q µ π π a Qρ d p dρ p a ten we can say / / / CIVE400: Fluid Mecanics 58
/ / / p ρ d d p φ( / π a, πa ) φ, / 0 µ ρ or p ρ d d p Q φ / µ ρ / / / 8.4 A cylinder 0.6m in diameter is to be mounted in a stream of water in order to estimate te force on a tall cimney of m diameter wic is subject to wind of m/s. Calculate (A) te speed of te stream necessary to give dynamic similarity between te model and cimney, (b) te ratio of forces. Cimney: ρ.kg/m µ 6 0-6 kg/ms Model: ρ 000kg/m µ 8 0-4 kg/ms [.55m/s, 0.057] Draw up te table of values you ave for eac variable: variable water air u u water m/s F F water F air ρ 000 kg/m.kg/m µ 8 0 4 kg/ms 6 0 6 kg/ms d 0.6m m Kinematic viscosity is dynamic viscosity over density ν µ/ρ. ρud ud Te Reynolds number Re µ ν For dynamic similarity: Re water Re air 000uwater 06.. 4 6 8 0 6 0 u 55. m / s water To obtain te ratio of forces we must obtain an expression for te force in terms of governing variables. Coose te tree recurring (governing) variables; u, d, ρ, F, µ. From Buckingams π teorem we ave m-n 5 - non-dimensional groups. ( u, d,,, F) φ( π, π ) φ ρ µ 0 0 a b c π u d ρ µ π u d a b c ρ F As eac π group is dimensionless ten considering te dimensions, for te first group, π : CIVE400: Fluid Mecanics 59
( ) ( ) ( ) 0 0 0 a b c M L T LT L ML ML T M] 0 c + c - L] 0 a + b - c - - a + b T] 0 -a - a - b - π u d ρ µ µ ρud i.e. te (inverse of) Reynolds number And te second group π : ( ) ( ) ( ) 0 0 0 a b c M L T LT L ML ML T M] 0 c + c - L] 0 a + b - c - - a + b T] 0 -a - a - b - π u d ρ F F u dρ So te pysical situation is described by tis function of nondimensional numbers, µ φ( π, π ) φ, F ρud ρdu 0 For dynamic similarity tese non-dimensional numbers are te same for te bot water and air in te pipe. π π π air water π air water To find te ratio of forces for te different fluids use π CIVE400: Fluid Mecanics 60
π π air water F F ρu d ρu d air F F ρu d ρu d F F air air water. water water 000 55. 06. 0. 057 8.5 If te resistance to motion, R, of a spere troug a fluid is a function of te density ρ and viscosity µ of te fluid, and te radius r and velocity u of te spere, sow tat R is given by µ ρur R f ρ µ Hence sow tat if at very low velocities te resistance R is proportional to te velocity u, ten R kµru were k is a dimensionless constant. A fine granular material of specific gravity.5 is in uniform suspension in still water of dept.m. Regarding te particles as speres of diameter 0.00cm find ow long it will take for te water to clear. Take k6π and µ0.00 kg/ms. [8mins 9.sec] Coose te tree recurring (governing) variables; u, r, ρ, R, µ. From Buckingams π teorem we ave m-n 5 - non-dimensional groups. ( u, r,,, R) φ( π, π ) φ ρ µ a b c π u r ρ µ π 0 0 u r ρ a b c R As eac π group is dimensionless ten considering te dimensions, for te first group, π : ( ) ( ) ( ) 0 0 0 a b c M L T LT L ML ML T M] 0 c + c - L] 0 a + b - c - - a + b T] 0 -a - a - b - π u r ρ µ µ ρur i.e. te (inverse of) Reynolds number CIVE400: Fluid Mecanics 6
And te second group π : ( ) ( ) ( ) 0 0 0 a b c M L T LT L ML ML T M] 0 c + c - L] 0 a + b - c - - a + b T] 0 -a - a - b - π u r ρ R R u rρ So te pysical situation is described by tis function of nondimensional numbers, µ φ( π, π ) φ, R ρur ρru 0 or R µ φ ρru ρur µ ρur e question asks us to sow R f ρ µ or R ρ ρur f µ µ Multiply te LHS by te square of te RHS: (i.e. π (/π ) ) So R ρru ρ u r Rρ µ µ Rρ ρur f µ µ Te question tells us tat R is proportional to u so te function f must be a constant, k Rρ µ ρ k ur µ R µ kru Te water will clear wen te particle moving from te water surface reaces te bottom. At terminal velocity tere is no acceleration - te force R mg - uptrust. From te question: σ.5 so ρ 500kg/m µ 0.00 kg/ms k 6π CIVE400: Fluid Mecanics 6
r 0.0000m dept.m 4 mg π 0. 0000 9. 8 500 000 66. 0 ( ) µ kru 0. 00 6π 0. 0000u 66. 0 4 u. 5 0 m / s. t. 5 0 4 8min 9. sec CIVE400: Fluid Mecanics 6