ES2A7 - Fluid Mechanics Example Classes Example Questions (Set IV)

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Edward Hodge
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1 ESA7 - Flui Mechanics Eample Classes Eample Quesions (Se IV) Quesion : Dimensional analysis a) I is observe ha he velociy V of a liqui leaving a nozzle epens upon he pressure rop P an is ensiy ρ. Show ha he relaionship beween hem is of he form V= C Pρ b) I is observe ha he frequency of oscillaion of a guiar sring f epens upon he mass m, he lengh l an ension F. Show ha he relaionship beween hen is f= C F ml c) Fin he imension of he bulk moulus K, knowing is relaionship wih he spee of he soun a in a liqui an he ensiy ρ : a= C Kρ a) Dimension: V = L T, P = M LT, ρ = M L [ ] [ ] [ ] The relaionship shoul verify: [ ][ ][ ρ] a b c V P = a b 3c b+ c b a L M T = 3 This leas o he following sysem: a b 3c b b 3b = + = a= b+ c= b= c b= b a b a = = c= The relaion is hus verifie b) Dimension: f = T, m= M, l= L, T = ML T [ ] [ ] [] [ ] The relaionship shoul verify: [ ][ ][][ ] a b c f m l T = c+ b+ a L M T = This leas o he following sysem: a= c+ = = c b b b = + = = c= a a = = = The relaion is hus verifie c) From he relaion, we have : K= ρa C M L M = = = L T LT so [ K] [ ρ][ a] 3

2 Quesion : Dimensional analysis Waer flows hrough a cm iameer pipe a.6m/s. Calculae he Reynols number an fin also he velociy require o give he same Reynols number when he pipe is ransporing air. For he waer he kinemaic viscosiy was.3-6 m/s an he ensiy was kg/m3. For air hose quaniies were m/s an.9kg/m3. Kinemaic viscosiy is ynamic viscosiy over ensiy = ν = µ/ρ. The Reynols number is : ρu U Re= = µ ν Reynols number when carrying waer: 6. Rewaer= = To calculae Re air we know: Re = Re air waer Uair = 5 U = 8.44m.s air Quesion 3: Momenum conservaion The figure below shows a smooh curve vane aache o a rigi founaion. The je of waer, recangular in secion, 75mm wie an 5mm hick, srike he vane wih a velociy of 5m/s. Calculae he verical an horizonal componens of he force eere on he vane an inicae in which irecion hese componens ac. 3 From he quesion: a.875 = m, u= 5m.s Since he je secion is consan a =a, he velociy is also consan u =u (Mass conservaion in he case of an incompressible flow) 3 The bulk flow is hus Q= a.u = a.u =.469m.s Calculaion of he oal force using he momenum equaion: F= ρq(v V )

3 Projecing his relaion along an y: F = ρqu(cos5 cos 45) = (cos5 cos 45) = 33N F = ρqu(sin5 sin45) y = (sin5 sin45) = 35N Boy force an pressure force are. F is hus equal o he reacion force applie on he flui. F= F + F+ F = F R B P R So force on vane: R = F = 33N y R = F = 35N y R is irece owar he lef an R y owar he boom. Quesion 4: Momenum conservaion A 6mm iameer pipeline carries waer uner a hea of 3m wih a velociy of 3m/s. This waer main is fie wih a horizonal ben which urns he ais of he pipeline hrough 75 (i.e. he inernal angle a he ben is 5). Calculae he resulan force on he ben an is angle o he horizonal.

4 From he quesion: P= P = ρgh= 9.8 3= 94.3 kn/m a = a = π.6 =.83m u= u= 3m.s The bulk flow is hus 3 Q= a.u = a.u =.848m.s Calculaion of he oal force using he momenum equaion: F= ρq(v V ) Projecing his relaion along an y: F = ρqu(cos75 cos) = (cos75 ) = 886N F = ρqu(sin75 sin) y = (sin75 ) = 457N Calculaion of he pressure force: F = F + F P P P Projecing his relaion along an y: F = ap(cos cos75) P = ( cos75) = 6N F = ap(sin sin75) Py = ( sin75) = 8376N There is no boy force in he or y irecions. F is hus equal o sum of he reacion force an he pressure force applie on he flui. F= FR+ FB+ FP= FR+ FP so FR = F FP So force on ben: R = F = F + F = = 63886N R P R = F = F + F = = 8833N y Ry y Py R is irece owar he righ an R y owar he boom. R= R + R = 4kN y R θ= = R y an 5

5 Quesion 5: Mass conservaion + Bernoulli A waer clock is an aisymmeric vessel wih a small ei pipe in he boom. Fin he shape for which he waer level falls equal heighs in equal inervals of ime. ɺ ɺ ou in There is no inflow an he flui is incompressible. Le a be he cross-secion of he ei pipe: ( υ ( ) ) ρ = mɺ ou = ρau ( ), Coninuiy equaion: ( ρυ ) + m m = ( ) υ = a U Bernoulli equaion beween he upper free surface an he ei secion: () ρu ρ gh() + = + () = gh() U So using he firs relaion: = υ a gh = υ a g h From he quesion, we know ha he variaion of h is linear in ime: h = h So () υ = a g h a 3 υ ( ) = g ( h ) 3 a 3 υ ( ) = g ( h( ) ) 3 The volume of flui can always be wrien as π R z h h υ = rrθz = π R z z h So h a a R z z = g ( h( ) ) R h( ) = g h 3π π R ( ) 3

6 Quesion 6: Momenum conservaion Because he flui is conrace a he nozzle forces are inuce in he nozzle. Anyhing holing he nozzle (e.g. a fireman) mus be srong enough o wihsan hese forces. Deermine hese forces. The analysis akes he following proceure: ) Draw a conrol volume ) Decie on co-orinae ais sysem 3) Calculae he oal force 4) Calculae he pressure force 5) Calculae he boy force 6) Calculae he resulan force & Conrol volume an Co-orinae ais have been one for you an are shown in he figure below. Noice how his is a one imensional sysem which grealy simplifies maers. 3) Calculaion of he oal force: F= F = ρq( u u) From he coninuiy equaion, he bulk flow is : Q= Au = Au so F= ρq A A 4) Calculaion of he pressure force F = F + F P P P F = F = F F P P P P We use he Bernoulli equaion o calculae he pressure P u P u + + z= + + z ρg g ρg g The nozzle is horizonal ( z =z ) an he pressure ousie is amospheric (P =). Wih coninuiy, i gives : ρq p= A A

7 ρq A FP= A 5) Calculaion of he boy force The only boy force is he weigh ue o graviy in he y-irecion - bu we nee no consier his as he only forces we are consiering are in he -irecion. 6) Calculaion of he resulan force F= F + F R P ρq A F = F F = Q R P ρ A A A ρq A F R= + A A A So he fireman mus be able o resis he force of ρq A R= FR = + A A A This force increases when A increase or A ecrease, which correspon o he epece behaviour. Quesion 7: Momenum conservaion Consier a rocke of mass m r raveling a a spee u r as measure from he groun. Ehaus gases leave he engine nozzle (area A e ) a a spee U e relaive o he nozzle of he rocke, an wih a pressure ha is higher han local amospheric pressure by an amoun p e. The aeroynamic rag force on he rocke is D. Derive an equaion for he acceleraion of he rocke.

8 We selece a conrol volume enclosing he rocke an moving wih he rocke. Because a reference frame aache o he rocke woul be noninerial, we chose a saionary reference frame fie o he launch pa. We selece he y-irecion momenum equaion. The force iagram shows hree forces: weigh, aeroynamic rag force, an he pressure force acing a he nozzle ei. From he force iagram F = P A D W y e e The momenum iagram shows an accumulaion erm an an ouflow erm. The momenum accumulaion erm is no zero because he momenum of he rocke is changing wih ime as he rocke acceleraes. The momenum of he rocke is he mass of he rocke imes he velociy of he rocke, an he accumulaion erm is given by (/)(m r u r ). This can be formally evelope by using inegraion: uyρv ur ρv = (mru r) = CV CV In his evelopmen, we assume ha all pars of he rocke were raveling a he same spee u r. The ouwar momenum flow is mu ɺ, where u o is relaive o he saionary reference frame: u = ( U u) So : mu ɺ = mɺ ( U u) o e r From he momenum iagram CV o e r u V m u mu = (m u )+ m( U u) ρ + ɺ ɺ ɺ y o oy i iy r r e r cs cs Subsiuing force an momenum erms ino he momenum equaion gives: PeA e D W= (mu r r)+ mɺ ( Ue ur) The accumulaion erm can be epane using he prouc rule for iffereniaion. This erm may be evaluae by applying coninuiy o he conrol volume: m r m = ɺ Combining he wo previous equaions gives: (mu r r)=m r (u r) ɺ mur Subsiuing his resul ino he momenum equaion yiels: m r (u r) = mu ɺ e+ PeA e D W The mass flow rae eiing he rocke nozzle is: mɺ =ρeaeue Combining he above wo equaions resuls in: m r (u r)= ρ eaeue+ PeA e D W The acceleraion of he rocke (u r /) epens on he insananeous mass of he rocke an he 4 erms on he righ sie of he righ sie of his equaion. The erm T= ρeaeue+ PeA e is known as he hrus of he rocke moor. The ei pressure of he ehaus je (P e ) is ofen epresse using absolue pressure, so he hrus is T= ρ A U + P A P P, where P is he local amospheric pressure, given by e e e e e e which will change wih aliue as a rocke ascens. o

1. VELOCITY AND ACCELERATION

1. VELOCITY AND ACCELERATION 1.1 Kinemaics Equaions s = u + 1 a and s = v 1 a s = 1 (u + v) v = u + as 1. Displacemen-Time Graph Gradien = speed 1.3 Velociy-Time Graph Gradien = acceleraion Area under