4) | (@ — The 1/7 power law for turbulent boundary layer is not applicable very neaglto fM solid | voundary. Do you agree? If yes, why ? Justify your answer. “~*~Example S_ A cylinder contains 0.35 m’ of air at 0°C and 276 kN/m? absolute. The air is compressed to 0.071 m?, (a) Assuming isothermal conditions, what is the Pressure at the new volume and what is the isothermal bulk modulus of elasticity at the new state. (b) Assuming isentropic conditions, what is the pressure and what is the isentropic bulk modulus of elasticity? (Take the ratio of specific heats of air y= 14) Solution (a) For isothermal conditions, PV = pay Then (2.76 x 10°)0.35 = (p,).071 which gives, Py = 13.6 x 10° Nim? = 1.36 MN/m? The isothermal bulk modulus of elasticity at any state of an ideal gas equals to its pressure at that state. Hence £ = p, = 1.36 MNim?. (b) For isentropic conditions Pvis= Pans Then — (2.76 x 10°) (0.35)'4= (p,) (.071)!# From which, P= 258X105 N/m? 2.58 MN/m? The isentropic bulk modulus of elasticity E= yp=140x 25.8 x 10° N/m? = 3.61 MN/m?@.30 A solid block of 2.0 kg mass slides steadily at a velocity V along a vertical wall as Ans. shown in the figure below. A thin ol film of thickness fh = 0.15 mm provides lubrication between the block and the wall. The surface area of the face of the block in contact with the oil film is 0.04 m?. The velocity distribution within the oil fim gap is linear as shown in the figure. Take dynamic viscosity of oil as 7 x 10-° Pas and acceleration due to gravity as 10 m/s?, Neglect weight ofthe oil. The terminal velocity V (in m/s) of thes block is. (correct to one decimal place). Af neos5 ne A= 0.04 mi? Ingenio (10.714) HA v Terminal velocity is a constant velocity ie. the net acceleration is zero, So, BF yg = mg - tA A=mg v BAA =m v 7x109x—Y_ x004 = 2x 018x105 Beng V= 10.714 misBr 2yh @ Olea # yi ) e @ 2 36, A20em cubical box slides on oil (mass density = 00 kg/m), over a large plane surface with a steady state velocity of 0.4 m/s. The plane surface ig jnclined at an angle of 30° with the horizontal plane, The oil film between the block and the plane surface is 0.4 mm thick. The weight of the cubical pox is 64 N. The kinematic viscosity of the oil is @) 0.001 m%s @) 0.002 m/s | Common Data Questions eaadienika a )2.56 Fluids of viscosities 1 = 0.1 N - sim? and jp = 0.15 N - s/m? are contained between two plates (each plate is 1 m? in area). EV ———b Iy hy P2.56The thicknesses are h; = 0.5mm and hy = 0.3 mm, respectively. Find the force F to make the upper plate move at a speed of 1 m/s. What is the fluid velocity at the interface between the two fluids? 2.57 Fluids of viscosities yp, =0.15 N-s/m?, 2 =0.5 N-sim?, and p3=0.2 N-s/m? are contained between two plates (each plate is 1 m? in area). The thicknesses are h; = 0.5 mm, hy = 0.25 mm, and h3 = 0.2 mm, respectively. Find the steady speed V of the upper plate and the velocities at the two inter- faces due to a force F = 100 N. Plot the velocity distribution.245 Two mnie, camprotbe, vicous Pa baring same denis at Aan cals cote ate fet pl lee 2m paso elo. The boom nei ied enh pe te mows tothe i consane ‘woacy of? mi, With he wanton of Newton, sey ntl developed Tate No wit eo pessoal hci, tment sy ee So pam cto eeu in te ea ee ‘esc tb eace (um ef two cial ls} (248 Abe ose 100 mms lad htm nny onaner on ane a ses Gey oth nora wea 80 kg Liquor dent m0 kg pow p18 tt one Fs nits hgh rw yd fe oe onan Fert che ou aps an tl Oo <2.73 A viscometer is built from a conical pointed shaft that turns in a conical bearing, as shown. The gap between shaft |. and bearing is filled with a sample of the test oil. Obtain an f algebraic expression for the viscosity j« of the oil as a func- ld tion of viscometer geometry (H, a, and 8), turning speed w, | and applied torque 7. For the data given, find by referring to a Figure A.2 in Appendix A, the type of oil for which the applied torque is 0.325 N-m. The oil is at 20°C. Hint: First obtain an expression for the shear stress on the surface of the conical shaft as a function of z.to measure the viscosity of a liquid sample. The upper disk rotates at height above the lower disk. The viscosity of the liquid in the gap is to be calculated from measurements of the torque needed to turn the upper disk steadily. Obtain an algebraic expression for the torque needed to turn the disk. Could we use this device to measure the viscosity of a non- ‘Newtonian fluid? Explain. l t tt! 2.66 2.67 The cone and plate viscometer shown is an instrument used frequently to characterize non-Newtonian fluids. It consists of a flat plate and a rotating cone with a very obtuse angle (typically @ is less than 0.5 degrees). The apex of the cone just touches the plate surface and the liquid to be tested fills the narrow gap formed by the cone and plate. Derive an expression for the shear rate in the liquid that fills the gap in terms of the geometry of the system. Evaluate the torque on the driven cone in terms of the shear stress and geometry of the system. i[MADEEASY | Mochine Design [411 Whichone olin folowing erteronin he fhysrdynanic uma bearngs? {@) Sommerelérumber (©) Rating ie (@) Speci cyeamie capacty (@ Potaton ctor [2008 : 1 Marg 4.12 Aballbeating operating atalod Fes Gite rtiot tte nfs see loed|s doubled te 2s (9 800 (©) e000 © «00 {6 1000 (2007 4 Mak 4.19 Anatura fed journal bearing of clameter50rm andlength 60mm operating st 20reveuienec ‘cartes aload of20KN. The Woreantusedhas a viacosty of 20 mPas. The racial clearanco is 04mm. The Soret number tribe bearings (@) 0062 (b) 0125 (0250 (o) 0785 (2007 : 2 Marks} 414 journal beating has a shalt diameter of 40. ‘and a length of 40mm. The shalt is rotating at 20 tad and the viscosty ofthe lubricant is ‘2omPas, The clearance 0.020 nen. To oss ot torque duo tothe viscosity ofthe lubricant ‘approximately (@) 0040Nvm () —0252Nm (6) 040m (@) —0652Nm [2008 : 2 Marks] 44.48 Aight loaded ful Journal beering as journal daetar of 50 mm, bush bore of 50.05 mm and beshlengthot 20mm. fotational speed of our is 1200 pm and average viscosity of quia |Ubcant is 0.03 Pa, the power lass (1 ) wil be @x ore wm (e257 [2010 : 2 Marks) 4.38 Twoldentcal bel bearings Pand Qare operating atloads 20 kN and 45 kN respectively, Th aio Ate tot bearing Pothole of bearing ie @ove oz oom a2 (eott : 2 Maka] | 235 4.17 A sald cteuar shat needs tobe designed tran torque of 0m healloabe shear sires ofthe matral is 140WPa, assuming 2 laclr of salty o! 2, minimum allowable design Careterinmmis os oa (we (oo 2ore: 2 Marks} 4.18 Bal bearings ate rated by a manufacturer fr a Heol 10% revolutions, The eatalogue rating ofa patticulr bearing ie 16KN. the design load is 2M, the He ofthe bearing will be P x 10 ‘evoons, where Pi equal 0 [20142 7 Mark, Sota] 4.19 A hycredyramie jounal bearing is subjected 0 2000 N oad at a rotational speed of 2000 rpm Bath nearing bore ciameter and length a 20mm, ras earance s 20 xm and Dearing ienicstes wih an oravng vscosiyO03PA.s, the Sommereld numberof the bearings 2014 : 2 Merks, Set-1] 4,20 Forbalibesrings, tn fetiguelite L measured in ruber ol ovoutions and ne racial load Fare ralated by FL'9 = K, whore Kis 2 constant tt wstands radia foad of 2 fora if of 540 milion voto. The load (i kN) fora He of ‘one ion revoutons is _ [2015 : 2 Marks, Sot 4.21 Whichofhe bearings gen below SHOULD NOT be subjected toa trust load? {@) Deep grcovebal bearing (@) Angular conact ball boaring (6) Oyindveslstrsigh role Dearing (6) Sino ow apered taller bearing [2016 : 1 Mark, Sets) 4,22 Aseltalignng ball bearing has basic dynamic load rating (Ci, for 10 revolutions) of 35 KN. ‘he equivalent acialloadon the bearing is 45 KN, the expocte ie (in 108 rovckions) I (a peiw0S (0) 051008 (908010 (above t.0 (2018 : 2 Marks, Sot]Example 1.2 Acylinder of 0,12 m radius rotates concentrically inside a fixed hollow cylinder of 0.13 m radius. Both the cylinders are 0.3 m long. Determine the viscosity of the liquid which fills the space between the cylinders if a torque of 0.88 Nm is required to ‘maintain an angular velocity of 2rad/s. Solution The torque applic The resisting torque by the fluid = (Shear stress) x (Surface area) x (Torque arm) Hence, at any radial location rfrom the axis of rotation. 088= 12 2rx0.3)r 0.467 r Now, according to Eq. (1.1), a t_ 0467 dy wr Rearranging the above expression and substituting - dr for dy (the minus si cates that r, the radial distance, decreases as V increases), we obtain Vaya Jav= The velocity of the inner cylinder, Vignse= 2.4%0.12=0.754 ms 0467 (11) 754-0)= | —_- Hence, 0754-0)= la - os} From which H= 0.397 PasScanned with CamSci "oO‘Scanned with CamScanner ® Eu Hu? iE rt Eo aEforces acting on the surface. Considering the eq surface, a force balance in the direction perpendicular to the surface results Fig. 1.11 State of stress and force balance on a curved liquld Interface In equllbrium with surrounding due to surface tension 2 ora a (2%) 42.07 88028) ype Focanal ais 2 2)" 2 Hence, from the above equation of force balance we can write (2) a(t) ay and gis the surface tension of the liquid in contact with the specified fluid at its convex side. Ifthe liquid surface coexists with another immiscible fluid, usually '2s, on both the sides, the surface tension force appears on both the concave and ‘convex interfaces and the net surface tension force on the surface willbe twice as ‘that described by Eq. (1.13). Hence the equation for pressure difference in this case becomes p= (242) aay Special Cases For a spherical liquid drop, the Eq. (1.13) is applicable with r, =r (the radius of the drop) to determine the difference between the pressure inside and outside the drop 38, p= 2oir (14s) ‘The excess pressure in a cylindrical liquid jet over the pressure of the surrounding atmosphere can be found from Eq, (1.13) with 7 => « and r, =r (the radius of the jet) as* agnring Fd Macha ner son ‘om tah Pl te amp ten fe yh Ti Ral reas eng om ta, ne tng od Toye ase an B40 mom sma se i pri ‘cn Pe wreak Bars ‘amr XVINOLASH NON ‘BAIT 30 NOUVOLIREVIO WODOTOGTE 4 ere te tuo ae fo xml, use condi ew ant i hone fess oh Sarasin Sa ‘ample LI Dering the flow of 1 fi flowing he er venta? 1 eae hat th cy ata wi th id fei gen by = 2S eh he fl tas ann he ma lacy Th vont 5 Wo (a) Clete he shear et esd ce we Ung = 02m ond 1 om (2) Ema sera af Newent nce heme shar ef amt ste pfs som mata cy[ae gan f © 2 sree aeak SHEAR3:33 < Sagar fmt convection 46 a7 49 ereuinnn rome evr Rye VARIO UF HL {ETON coaticiant can be obtained by mulipiying the focal heat transter coatticiont atthe aling edge ‘by the tactor (@ 075 () 10 18 (@ 20 IESE : 1998) ‘When thera is tow ol fluid over a flat plate of langth ‘L.the average heat tanstar coallicient is given by a fina o 2m © Pfinar Efinuas {ESE : 1997) ‘Whan all the conditions are identical, in the case ‘ol flow through pipes with heat transfer, the velocity profile will be identical for {a) liquid heating and tiquid ‘cooling {b) gas heating and gas cooling (€) liquid heating and gas cooling {d) heating and cooling of any fluid {ESE : 1997] |i the cage of turbulent flow through a horizontal isothermal cylinder of diemeter'D tree convection heat transfer costficiont forthe cylinder will (6) be independent of lameter (0) vary as OM {c) vary as DM (@) vary as Ov (ESE : 1997) Given that Pr Prandiinumiber Nu= Nusselt number ‘Sh= Sherwood number Re = Reynolds number So Schmidt number and Gr= Gresholbascibsaies we nyta mechani end tysraute Machines Ser sess onthe hat += SH =0.2x452.4= 90.48 Pa Shear force F, = 1x2mrxL =s048( =12791N Ter Torque 12min 22 -oenenin Pome P= 28M _ nx 240 0.6756 = 145 W "1.12. Two vertical plates are placed with a gap ‘of 15 mm between them and the gap is ‘led with (Gy = 1260 kg/m? and y= 1.5 kg. s/m), An 80.cm square, 3 mm thick steel plate weighing 110N ts placed exactly midway in the alycerin-flled gapof the vertical plates. itis ‘required to pull the steet plate vertically upwards at a constant velocity of 15cm.s, estimate the force required. (Neglect the resistance at the ‘edges ofthe pate). Ve 15m ‘Steel plate We We sw —s0m— Fig. 1.7 Example 1.12 Gap= 6mm. Solution: The thickness of the glycerin layer is ‘same on either side of the plate, 11 = thickness of glycerin layer =(5-3/2=60mm Shear stress on one side ofthe plate = r= AZ F, = Total shear force (considering both sides of the plate) Here, A = reaofplate= 08x08 = 064m? ‘Weight of steel plate = W,= 110 ‘Volume of the plate = 0.64 x 0.003 = 0.00192 m? 2x0.64x1.5%015 ‘Shear force F, = 0.006 =48 Up thrust on submerged plate ~ 7, x (volume of the plate) 7, = (1260 x 9.81) x0,00192 = 23.73 N Effective weight of plate = We= W,—W,= 0-23.73 =8627N ‘Total force required to pull the plate = F=F,+W,=48 +8627 = 13427 "1.13 The gap between a horizontal shaft ‘and a concentric sleeve is filled with ‘Viscous oll. The sleeve moves with a constant ‘velocity of 1.5 m/s when a force of 1250 N is applied parallel to the axis of the shaft. If It was required to move the sleeve at a velocity (of 1.8 m/s, what should have been the force? “The temperature can be assumed to be constant throughout bi je aa i to Fig. 1.8 Example 1.13is immersed in oil of mass density 940 kg/m} to a depth of 9 mm. If a pressure of 148 N/m? is needed to form a bubble which is just released, what is the surface tension of the oil? 50. A glass tube of 2:5 mm intemal diameter | a ame 7 (ay. 0-041 N/m ue won - en) 0.043 Nh eon” ave fy 0-04 im - sno? ~% 21, Se (0-046 Nim SB. 9:050 Nin -12:40 al TED < Sagar jas obj fluids (©) halt (0) V2 times more [CSE-Pre : 2001) When a flat plate of 0.1 m? area is pulled at a Constant velocity of 30 cm/s parallel to another stationary plate located at a distance 0.01 om from it and the space in between is filed with a fluid of dynamic viscosity = 0.001 Ns/m?, the force required to be applied is @ 03N (©) 3N (©) 10N (a) 16N ICSE-Pre : 2004) Consider the following statements : The state of stress in a fluid consists of normal pressure only if the fluid 1. isatrest 2. isin uniform motion 3. has non-uniform velocity profile 4, has zero viscosity Z Which ofthe statements given above are correct? () 1.2and3—(b) 1, 2and4 ©) 1,3and4 (0) 2,3and4 ICSE-Pre : 2005] The pressure inside a soap bubble of 50 mm diameter is 25 N/m? above the atmospheric pressure. Whatis the surface tension in soap film? (@) O.186Nim — (b) 0.312NN/m (6) 0624Nm —(d):0.948N/m [CSE-Pre : 2005] Which one of the following is the correct staterient? For the case of laminar flow between two fixed parallel plates, the shear stress is (@) constant across the passage (b) maximum at the centre and zero at the boundary (©) zeroall through the passage (¢) maximen=etthesissereies nnd zero at the(0) viscous Shear stress in liquid varies linearly across the gap {ESE : 2010) 1.41 Athin plane lamina, of area A and weight W, slides down a fixed plane inclined to the vertical at an angle and maintains a uniform gap e trom the surface of the plane, the gap being filed with oil of constant viscosity 4. The terminal velocity of the plane is ecosa aul © UWA () Asie eWsina uWsina (e) An (d) eA [ESE : 2010] 2 In a quiescent sea, density of water at free surface os Pp and at a point much below the surface @ensity is p. Neglecting variation in gravitational acceleration gand assuming a constant value of bulk modulus &, the depth of the point from the free surface is Kf1i1 ®) ans) Form} s(2-2) K fe © Glo o\p+Po [ESE : 2010] 1,43 Pseudo plastic is a fluid for which (a) dynamic viscosity decreases as the rate of shear increases (b) Newton's law of viscosity holds good (c) dynamic viscosity increases as the rate of shear increases (4) dynamic viscosity increases withthe time for which shearing forces are applied [ESE : 2010) 1.44 If angle of contact of a drop of liquid is acute then (@) adhesion is more than cohesion (b) cohesion is more than adhesion 1.468 14 14 \S CHXAMPLE 2.1 (GS GIVEN Because of a leak in a buried gasoline storage tank, water hes seeped into the depth shown in Fig. E2.1. The specific gravity ofthe gasoline is SG = 0.8. FIND Determine the pressure atthe gasoline-water interface and at the bottom of the tank. Express the pressure in units of Ib/R?,Ib/in?, and as a pressure head in feet of water. SowuTion Scene nnd ope a we w— = @— a FIGURE E21 ‘Since we are dealing wit liquids at rest, the pressure distribution will be hydrostatic, and therefore the pressure variation can be found from the equation: p= rh py With p, comesponding tothe pressure a the free surface of the sgzsotne, then the pressure a the interface is P= SCyyoh + my (0.68)(62.4 Ib/T°\(17 A) + py 21 + py Ib/FP) If we measure the pressure relative to atmospheric pressure (gage pressure) follows tat py = and therefore p= 721 te (ans) aa byte i Ta ~ 501Ib/nt (Ans) TAWA ea (Ans) 2.4 1b Ho Its noted that a rectangular column of water 11.6 fall and 1 {in cross section weighs 721 Ib. A similar column with a 1-in? cross sectlon welghs 5.01 Ib, ‘We can now apply the same relationship to determine the pres- sure a the tank bottom: thats, Pe= Moho +B (62.4 b/f0\(3 ft) + 721 bM@ (Ans) = 908 lb/ft? 908 I/F? 2 Tag inzyne ~ 631 Ibn (Ans) Pe, 208 IME sen (Ans) Yo 624 1b/ COMMENT Observe that if we wish to express these pres- sures in terms of absolute pressure, we would have to add the lo- cal atmospheric pressure (in appropriate units) to the previous results. A further discussion of gage and absolute pressure is given in Section 2.5.2.14 THE HYDROSTATIC PARADOX From equation (2.17) it can be seen that the pressure exerted by a fluid is dependent only on the vertical head of fluid and its mass density p; it is not affected by the weight of the fluid present. Thus, in Fig. 2.12 the four vessels all have the same base area A and are filled to the same height A with the same liquid of density p. Pressure on bottom in each case, p= pgh, Force on bottom = Pressure x Area = pA = pgh. Thus, although the weight of fluid is obviously different in the four cases, the force on the bases of the vessels is the same, depending on the depth / and the base area A. ania ania ani ink2.5 Amerculy manometer is used to meee the static pressure at a pointin a water pipe as shown in figure. The level difference of mercury in the two limbs is 10 mm. The gauge pressure at that point is H,0 Manometer Hg (@) 1236 Pa (b) 1333 Pa i) Ze (d) 98 Pa [1996 : 1 Mark] 2.6 Tha tarnn Comaf ia are cee 0 gems meh alo estes Kl eA inte ie Pye Ag pete gn =D Sn ene 1 mse. Accratien det grav = 0m, (20 0 ins vinin w= 28% 30 i ane hf 5 wa wih 19 10a in wig ah ee ney. he vency of ae (081 Any ann fond Ines famed ovr thom i te pe, Tee ae seit ote fab Te wy oie te ct eb ones nea 3-} davaaeua eine de pny QA Abe sm ner 0 pine dl 0 yim in = fB anave ens (@) Saat S1Me- Sy hg + Spy (0) Sata~ Si Ma Sp (ha~ hg) + 8, Ma~ Sp hy (©) Sahar Sy Ma~ S2(ha~ he) Shp 4 Sag (0) Salat 5: Me + Sy hg~ Sy hg [CSE-Pre ; 2001) The pressure difference between point B and A (as shown in the above figure) in centimetres of water is (a) - 44 (b) 44 (c) -76 (d) 76 [CSE-Pre : 2002)6 A three-fluid system (immiscible) is connected to a vacuum pump. The specific gravity values of the fluids (Si, S2) are given in the figure. To Vacuum Pump Unit weight of water, y, = 9.81 kN/m? Atmospheric Pressure, Pam = 95.43 kPa The gauge pressure value (in kN/m?, up to two decimal places) of p, isP245 Determine the gage pressure at point A in Fig. P2.45, in pascals. Is it higher or lower than Paamosphere? Solution: Take y = 9790 N/m’ for water and 133100 Nim’ for mercury. Write the hydrostatic formula between the atmosphere and point A: Pam * (0.85}(9790)(0.4 m) —(133100)(0.15 m)—(120.30 m) +(9790(0.45 m)=p,,Question Number : 45 Correct : 2 Wrong: 0 The figure shows a U-tube having a 5 mm = 5 mm square cross-section filled with mercury (specific gravity = 13.6) up to a height of 20 cm in each limb (open to the atmosphere). Meme | 20cm If 5 cm’ of water is added to the right limb, the new height (in cm, up to two decimal places) of mercury in the LEFT limb will be3.18 A partitioned tank as shown contains water and mercury. Whatis the gage pressure in the air trappedin the left chamber? What pressure would the air on the left need to be pumped to in order to bring the water and mercury free surfaces level? 0.75 m = Water, 3m lm 29m on P3.18, P3.19 3-19 In the tank of Problem 3.18, if the opening to atmo- sphere on the right chamber is first sealed, what pressure would the air on the left now need to be pumped to in order to bring the water and mercury free surfaces level? (Assume the air trapped in the right chamber behaves isothermally.)eT td Machanie and Machings IID ERS Use i j Frictoniess andincompressie Viscosity s invariant with shear stress Viscosity increasos at higher shear stess Viscosity decreases athigher ear atese Codes: @ © ey @ sesce (ese: 2002), E29 nic cveyra ecetn rene SY > 4, is held between two parle plates a distance‘ apart. the ste8s applied tothe top plates 3, then the velocity wth which the top plate moves relate 10 the bottom plate would be w Axle ow ofeye o(gle ooffye 0.15 The veloiycistibution for flow over a patois given by u=05 y- y? where U's the velocity in ms at adistance 'y meter above the plate tne dynamic viscosity ofthe uid 0. Nin? then what isthe shear sires at 0.20 m fromthe (@) ro stress should be acting on 10) tone svoss acing on must bo 259 {c) shear stoxs acing on must Be Zor {)noppoln on sexi be under any sve8s (GATE : 2006) 10.17 Consider 0 uid of viscosity W Between two crear paral patent radh'F” Separated ty Sihetanee The upper plate rotated a an fanaa velocity whore ase Dost lat, Fut etatonary, The velit Polls Detwoan the two plates istinear, Thatorque experienced by the bottom pata 0.18 Velocity profefr aud between two flat patos Yor as shown in igure given by “2 indicate which ona of the following is FALSE. Fans Pase (2) Shear svess is constant in the gap every ‘whore andis independent o'y. (©) Shear svessactsntheregatve directon onthe lower plate, (6) Shear svess acts nthe negatve'y rection ‘onthe upper pate (@) Shear stress inversely proportonal to M 0.19 A spherical water drop of radius A’ spits up in ianaye ain’ smal ops of equal size. Tre we (@ 9Nm® —@) 18Nn? required in spting up he drop (he sue () 225M? (@) GOBNim? tension of water = 6) Lia) (0) donh’n —(e) doxn*(o"-1) €.16 The necessary and sulicient condton for a 8 ° surface o be caled 2s a‘toe suo! is fe) dannin'® (a) donto(o!* 1) MADE EASY i»‘= Moment of inertia sbout indicated axis I, = Moment of inertia shout indicated axis passing throuth the centre of gravity of the area.5m aise (e) 76.5 ms @) 925 we 11, Consider an L-shaped gate with water level above Uehingeas shows Atoproimaly what height Dot the water evel wl he gate opest Neglect ‘the mass of the gate. Assume g = 10 mist. D2 TH =346™M‘sy Co co a =) ares fu th tne al coo oe on @20 35 (8¢:20001 2p an pare hes n Th hg oo ss (ese 200) 211 A dam ig ing a curved ution a own 6 uso girte henoraenaoes acing (6) 1962 AOE () 8024 10 (SE: 2002) S12.consider he flloing statements about Pramanethe sume even renmensace " » 2 et vocal even whan fe uc fo ys 1) On (ESE 20 1.18 Te vera argon fre on eure (6) weg iqudcctem tou ns CSc 18) ight ffi tow tha. (@ poset fpessuea C5, rupted by he [ESE 200) cing utes suorarged nla w(oymeas (ese 203) wt mt (ese 05) 2.36 Aasaion (A Feravenenty nsec, Faso (Carr frase Ie bo. fo) Alston Ris abe {a Altace Ree ese 2000)EXAMPLE 3.10 FORCE TO OPEN AN ELLIPTICAL GATE ‘An clpical gate cover the end of pipe 4m in diameter If the gt is hinged tthe top, what normal force Ff required toopen the gate when waters 8m decp above the 1p ofthe Pie and th pipe open to the atmosphere onthe other side? ‘epic the weight ofthe gate Problem Definition Sleution: Water pretsre i sting onan elliptical gt, Find: Normal fore (in newtons) required to open gat Properties: Water (10°C), Table A. 9 = 9810 Nin ‘Assumptions: 1. Neglect the weight ofthe gate 2. Neplestfioton between the Boom on the gate and the pipe wal. Plan 1. Cakculate sunt hydrostatic fore using F = pi 2. Find the location of he center of pressure using Fa. (3.28) 3. Draw an FBD ofthe gate 4. Apply moment equilibrium about te hinge. Solution 1, Hydrostate (estan) free Pressure at depth ofthe centroid (8810 N/sm)(10 m) = 98.1 kPa P= Crea Eom + A= area ofelpical panel (sing Fig. A.1tofid formula) A=nab = n(25 m2 m) = 15.71? + Calouate restart force = pA = (98.1 kPa)(18.71 m2) [ESM] revere 4 dana 2, Cente of pressure + = 123 m, where fis the slot distance fom the ‘rater surface To he setoid. + Arca moment of neta T of nlite! panel using formula fom Fig A.) ab. 205 mYOm) «94548 ? + i + Finding emer of pressure TZ 25s4mt (125 myis.7im) = 0125 m 3. FBD ofthe gate: 4, Moment equilibria TMinae = 0 1581 10° Nx 2625 m= Fx Sm = on](UERRABIBIIC| Figure 2.29 shows a Mash board, Find the depth of water f at the instant when the water is just ready to tip the flash board. I rarreraererne ner ME NITION TTT TTOTT TOTTI B Fig. 2.29 A flash board in water Solution Tae fash board will tip ifthe hydrostatic force on the board ae at a point away from the hinge towards the free surface. Therefore, the depth of water h for which the hydrostatic force F, passes through the hinge point O isthe required depth when water is just ready to tip the board, Let be the centre of gravity ofthe submerged partof the board (Fig. 2.29). An k Th BG = 2 = win 60° V5 If yp and y, are the distances of the pressure centre (point of application of the hydrostatic fore and the centre of gravity respectively from the fee surface along the board, then from Eq, (24a) nny sone Ba ow em I 2) a) (considering unit length of the board) ‘Again from the geometry, BG - Be (o8) =1 e280) Bguating thet pte fats Ea (278) and (2.80), have wu (8) = nv =1 3v3_ 2 from which 6mExample 2.21 ‘AVetabe manometer with diameter D (<< L) acolertestowards th right ax abown in Fig. 243 Find out the val off, when 0, = Sav Solution ‘resco om pom te Fi (24S)Q39 The large vessel shown in the figure contains oil and water. A body is submerged at the interface of oil and water such that 45 percent of its volume is in oil while the rest is in water. The density of the body is kg/m’, The specific gravity of oil is 0.7 and density of water is 1000 kg/m’. Acceleration due to gravity g = 10 m/s’.4.26 A hydrometer weighs 0.03 N and has a stem at the upper end which is cylindrical and 3 mm in diameter. It will float deeper in oil of specific gravity 0.75, than in alcohol of specific gravity 0.8 by how much amount? (a) 10.7mm (b) 43.3 mm (c) 33mm (d) 36mm [ESE : 2007] 4.27 Awooden rectangular block of length of L is made to float in water with its axis vertical. The centre of gravity of the floating body is 0.15 L above the centre of buoyancy. What is the specific gravity of the wooden block? (a) 0.6 (b) 0.65 (c) 0.7 (d) 0.75 [ESE : 2007]3.17 A container with two circular vertical tubes of diameters d, = 39.5 mm and d= 12.7 mm is partially filled with mer- cury. The equilibrium level of the liquid is shown in the left diagram. A cylindrical object made from solid brass is placed in the larger tube so that it floats, as shown in the right dia- gram. The object is D =37.5 mm in diameter and H =76.2 mm high. Calculate the pressure at the lower surface needed to float the object. Determine the new equilibrium level, h, of the mercury with the brass cylinder in place. _ i Brass d, l—d; “| |Vortex Motion Fer free Works Ve & ¥ we Tsxtot o4 lots C= 14 ag: (at gS > Tsxetx wt 20-03 24 —___@ Ml oO <free Ue eS Fe fe Wor 1S Tastete ce Catins ae Seated Cm Taxon ot 2608 Uermemetines de Wher = $350" a fre Surber pe flaPign ace fast ~ fitar ae efict 2 30 ES, witie (AL A] Se Shape 2 hype batts. a-n- £{ ha) est. PAL. 2 (4-)Ls = 9028 ™ fe Zee Ml Oo <3.65 The gate shown is 3 m wide and for analysis ean be considered massless. For what depth of water will hi re: {angular gate be in equilibrium as shown? PRS A Ge Ae 3 Cc fe Ra fied py F ro COs ow 4! fh, fe Fo slomgrdyd 03 ] as=0 (a) EM=0 ys = FAB <0 5 d sf *s = if Ss d= LSKS 22 > Sinbo ESKS 43 2? lox & (aetese|ES CTEAPRY Use of the Pressure Prism Concept GIVEN A prostze unt crt ol (SG — 090) nt tas» FIND What he magic an can fe rea fore Square, my OG pe bot oH sie a hed stn teach le? Fig. E2a The presue ge on he top ef te kre 0, tcl han eter pes. 86 ete 2 Fld Stat: Sowution ‘The res dro wg te se sua fe phos shotn in Fig E28h The pressure agen pot onthe plate Guetta pres, at cl surace andthe resi oe he ol wich vars near With dep shown in eM. “Themslat eceon th plate hviganare As othe com ponent, Fad Fehr F, nd reo he ecg od ‘anglais oF he pres daraton, espe Ths Ro (atyma [50% 10 Nat " (090K081 % 10°Ny2m) (038 7) = 0c0N ao(85 a apa seus mm oe ‘The mage of he esl fo, Fy trfore Fea Fit R= 254X10N=2541N (Ans) ‘The ve! aetna Fy cm eatin by sein m- mets rand a as gh pot O50 at Fayo™ F(0m) + F02m) jpn G4 ENC m) + (0964 x 1ENIO 2A 1N 0206 (Ans) “Tha he foc acts at sane of 0.206 m above te bata of ‘he plate cong i vere ax syne COMMENT Note he a presen tbe cain he foce we ge rie Atrspere ree dost: ec the sat fre (naga ola) see 28 08 ‘thsi of he phen, ey canon10nd Mss oF Machines Bxample 218 Desrmine he get of ie sure of water nk 4 og. moving with ensue acetone the shown aig. 28. fn.242) igidins an nde norm aceletaton Solution Lets cometh pres aa point nto of and 2 2» tere, p= a. asm From Bs 2.465) sn 2.85. orp a, aie -pera) where a4 ete ieelern Ina detonate ew. a= Ose “These (29) Rees, ap= -plOSede+eds) Po -polosrss)ee sy vege ss ecmin Coming eri of conta eat fe sri =P ctmenper ese) x4 Oaed 9 “hase (29) bers pelosPsat Baty ation 08 “The ego fe sac ecb by Png P= Pan SES 29) mpese+2)= 0 Bxample 219 A retnglr nko eng = 10m, neigh b= 2m and wie pedir a {he pene Fg 243) 1m bly bal il ath wer Te tk aeny| tcc ang the horn ecb with a seen elation topany) Wala rrp tea?Buoyancy Sagan t 1000 PM A block of wood (SG = 0.6) floats in fluid X in Fig. P2.126 such that 75 percent of its volume is sub- merged in fluid X. Estimate the vacuum pressure of the air in the tank. Po + 4&4) <0 Paz Se geen —@ Lrorden Qady 10 {loa airy colt 15% Submerged. Woody = Favoyont tog fa My otk fe he O15) Sa = 06 KO _ Rp -Gi) . os Byer © — Pas — Boo x91 xoy = — 3194P2.47 The cylindrical tank in Fig. P2.47 is being filled with water at 20°C by a pump developing an exit pressure of 175 kPa. At the instant shown, the air pressure is 110 kPa and H = 35 cm. The pump stops when it can no longer raise the water pressure. For isothermal air compression, estimate Hat that time, pe t 7 db fax Fy u-3som (t “Pp = Wo 1eha b hay | Ea erm Wu) + Ros For loo trey mal Compres Sim, hiv AEROS Ay, =PVe Maw) Fes (Ilo tefa) x05 ah@ Rost @= (Memb) x08 4 Qin)Re Wo Kay xe? iy Qi-¥)> full - After aCempressim ht Pgh = VisKh Pr= Wis tcbe) — Garon) —G) Byeqr da 8p qa (11s xy? — Garon) = Woxte® xo7's Qi-#) (1s = 9-84) ( (1-H) = Wovors 192-5 —1sw — 10-744 + 4-811 = Hoxo-Is q-a1H® — (9574 H + Mo =o Br 1843 + WLI=0 Hy=o-6na%m = !8-31 mM the Morimum Neigh H, Me sunie i HSNote 23-Apr-2019 23-hoe 2019 910513 PA » Fons Jane qh Fo he J Pane = feqnsa.n Fe rae ey 1 Fae Toe Fhe $9 To0 FoRe $9( Tan + HAY he bgTen Me™ Fe DRE he Qt a ft u Al - Aye ho®S Aye hyXS Fle 4 For inclined Susfyce Tz Ht AD Fa sind Fy3.70 The plane rectangular gate can pivot about the support at B, For the conditions given, is it stable or unstable? Neglect the ‘weight of the gate. Justify your answer with calculations. ‘cate a Ye A PROBLEM 3.70 S41 72 8 Sinus?” Anopencircula eyinder 12 mighis ited with ‘igi ois top. The hquidis vena rigid body rotaton about the axe of tho oyinder and the pressure atthe centre Ene atthe bot surface ound obo 0 6m iqud. Wha 's the raioot volume ofiquc spilled out te eye t he ‘xiginalvolumo? we © 38 ow (a ae [ese : 20071 Inacyineicalvorex maton abouta vertical ax, 1,and ae he radial cistancesoltwo ports o9 the novzontal plane ( f > f) I for a given tangential iui velocity at, the pressure peonposdes aq Aew 4009 S14) jo WE ON IYI MON ‘GUORENIGNG ASV 30el Public ' Atuid particle can accelerate & ina steady nonuniform flow field ‘only if the flow field is both unsteady and (©) only in an unsteady flow fold (4) ina steady uniform flow field it the viscous forces are large enough 5 The acceleration components of afuid particle are denoted as 1. Localtangential acceleration 2. Convective tangential acceleration 3. Local normal acceleration 4. Convective normal acceleration Inacurved nozzie fitted to the endofa straight Pipeline carrying water under variable head, 2 ‘acceleration components thal are present woud Include (a) tana2 (©) 1.2ana4 (o) Sands (@) 1.2,3and4 DAS: 1997] 16 A flow field which has only convective acceleration is (@) a steady uniform flow (©) an unsteady uniform tlow (©) asteady non-uniform tow (9) an unsteady non-uniterm tow [GATE : 2014] 17 Ata point on a steaming, the velocity is 3 ms and the radius of curvature is 91m. 1 the rate of Increase of velocity along the streaming at this point is 1/3 ms, then the total acceleration at this point would be (b) mvs? (a) 1s? ( pms? ( Sams? 8 Fora tuid fiow through a divergent pipe of length ‘L having inlet and outlet radii of R, and Ri, respectively and a constant flow rate of Q, ‘assuming the velocity to be axial and uniform fatany cross-section, Ue accoleration athe ext is 201R, - FR) nL 207(R,-F,) () Re CETL WRD BON THRENTS VE FOYNTI IRD DOANE HOMIE 3 3 Z § 204R,-F) 207 Fy -R © ane 0 ae IGATE : 2004 0.19 Given tat the temperature Tlx, 7) does ‘change along a sireamling in a steady 21 incompressible flow, the equation ofthe stroat line is obtained trom ar we {3 “Gl ar ar o% He @ 2-H 0.20 Wnichot the tolewing equations willbe satisti by itrataional low of an incompressible tic! av au au_ aw av_ aw de” ay’ a2” Oe aE” By ‘Select the correct answer from the codes giv below: (@) Sand (© \and3 (b) vand2 (9) Vanc4 WAS : 19 For acertain 2D steady incompressible tow, horizontal and vertical velocity components given by u= 6y, v= 0, where yis the vert Gistance, The angular velocity and rate ost a2 strain respectively are (@) -3and3 ——(b) Sand-3 (©) 3and-6 —(@) - Gand 3SGATE Previous Years Solved Papers: II | ‘ 3.17 Atwo-dimensional flow feld has velocitgg the x and y directions given by u = ga v= -2xyt respectively; where is time, th equation of streamline is (@) x°y= constant 2 Marks] (qd) not possible to determine [2006 : 2 Marks] 3.18 Ina two-dimensional velocity field with velocities uand valong the x and y directions respectively, the convective acceleration along the x-direction is given by 2 Marks] av au ju av seit fet mur OU, equation au au au | au (1,2)i8 Oley, OMe [2006 : 1 Mark] 1 Mark] 3.19 Ina steady fow through anozzle, the flow velocity on the nozzle axis is given by v= up (1 + 3x), where «is the distance along the axis of the nozzle fromitsinlet plane and Lis the length ofthe nozzle. The tim required fora uid particle on the axis to travel rom the inlet tothe exit plane of the nozzle is firections wand v, L Oa © 3, L L © aa © 355 [2007 : 1 Mark] fh combination of the following statements steady incompressible forcad vortex flow rect? ; stress is 2010 at al points inthe flow. ticity i at all ints in the flow. icity is 2er0-at al POmnIS ius pictwot Eh 50 ote Re wn va i 8 am ®) te) 020 Anoamorsora resrprsabetnctenen tov io oven by B= s7. 9 8 dato ha dh exten er ude gus etal ay okt fim id ohn wo alatssi) weve) © alt) (a elt) © Tor ep rh srg aa Bekok + aro: 9 - O+0j+-ak snail 2,30 sent = len Rewiti seen = ofstseniclysoe0] xp ofas0i] = oti] see ls esi] mi of off) leAccelevahm Companentn in polar (cylindrical) Systerra \hy = $0, 80, z,+) Up= 407, ¥89,2,4) ay: DY Ye y= - De, Uap uo Poa = £( 7,140,2,*) O,= DV dt 5 Qa 2 V- eg: a, = [P55 + Wort be 2 a ay » Vp Qe 4 Ue] 4 De “Pett a6 roa a a wy Cowvec ive LocalThe components of acceleration in cylindrical coordinates may be obtained from Eq. 5.10 by expressing the velocity, V, in cylindrical coordinates (Section 5.1) and utilizing the appropriate expression (Eq. 3.19, on the Web) for the vector operator V. Thus,* OV, , Vo OV, OOF +78 = vy, Va Ve 4 = Veo + a6 * OV, | Ve OV, fe Vg i oe v3 Ov; Bee ae (5.12a) ViVo , y Ve = aa (5.12b) OV, , OV, e Vives oe (5.12c)([E RIE! 4 wo dimensional flow is desribed inthe Lagrangian system as weet +yo(1-e) and =e Find (a) the equation of path ine ofthe particle and (b) the velocity components in Eulerian system. Sotution (a) Path line ofthe particle is found by eliminating from the equaions describing its motion as follows: viv Hence, xebioly) + yd ~ 67) which finally gives after some rearrangement (-y0b? —ayy97 +30 This isthe required equation of path Tne, (0) (the x component of velocity) de a Slne +y0-<%)] = She 4 2 bye Hk ley e142 ky e™ mie thy (te) niet ete (the ycomponentof velocity)> ‘ need 4 Mechanical Engineering ® Fluid Mechanics 2.14 A fluid particle can accelerate (@) ina steady nonuniform flow field (0) only if the flow field is both unsteady and nonuniform (©) only in an unsteady flow field (d) ina steady uniform flow field if the viscous forces are large enough Q.15 The acceleration components of a fluid particle are denoted as: 1. Local tangential acceleration 2. Convective tangential acceleration 3. Local normal acceleration 4. Convective normal acceleration Ina curved nozzle fitted to the end of a straight pipeline carrying water under variable head, the acceleration components that are present would include (a) tand2 (b) 3and4 (©) 1,2and4 —d) 1,2, 3and4 DIAS : 1997] flow field which has only convective acceleration is (a) a steady uniform flow )) an unsteady uniform flow steady non-uniform flow (2) an unsteady non-uniform flow [GATE : 2014] QW Ata point on a streamline, the velocity is 3 m/s and the radius of curvature is 9 m. If the rate of increase of velocity along the streamiine at this. 20 peorpoide/ 69 Fa 9000 Si 0 ued oN OG MON BARU SGA GVH 0 OHRID BSTanG TNBURESD @ee 3.8 A velocity field is given as V=3x?yi ~6xyzk where x, Oeste 8 mand Vin ms, Determine i represents an incompressible flow (ii) The flow is irrotational (ii) The flow is steady [1993 : 2 Marks) Streamlines, Path lines and streak lines are Virtually identical for 3.7 (@) Uniformfiow —_(b) Flow of ideal fluids (c) Steady flow (d) Nonuniform flow [1994 : 1 Mark] 3.8. Ina flow field the stream lines and equipotential lines (@) Are parallel (b) Cutatany angle (0) Are orthogonal every where inthe field (d) Cutorthogonal exceptat the stagnation points [1994 : 1 Mark] 3.9. Foratluid element in atwo dimensional low field ( yplane), if twill undergo (@) Translation only (6) Translation and rotation (¢) Translation and deformation (d)_ Deformation only [1994 : 1 Mark] 3.10 Existence of velocity potential implies that (a) Fluid isin continuum (b) Fluidis rotations! (c) Fluid is ideal 5 Fluid is compressible [1994 : 1 Mark] das line integral of tangential eee (fill in the ation is 3.11 Circuit of velocity about @ ———" _amponent ofViscous incomoresile Flows mo 8.2. GENERAL VISCOSITY LAW ‘Thewelhknown Newton's vssosty law i we E an ‘where 1s the coordinate direction normal tothe solidi interface, J is the coefficient of viscosity and V is velocity, This law is valid for parallel flows ‘There are more generalized relations which can relate sress field and velocity field for any kind of Now. Such relations are called constitutive equations, We shal consider her the Stokes viscosity la. ‘According to Stokes’ law of viscosity, shear stress is proportional toratof shear strain so that 2a) (820) (8.20) ‘The frst subscript of denotes the direction ofthe normal to the plane on hich the sires acts, while the second subscript denotes dretion ofthe free ‘hich causes the sess. "The expressions of Stokes’ w of viscosity for normal stresses are 2 2 5 ye [9H 20 9 . au=-nmu St 44/330] aay a a) 24 | se whee fin sect cy yt ip Woe 4 Gen my Me hve sdy we tt Deoodjate om Now,ifwe addthe three Eqs(8.3a),(8.38)and (836), soa [4 224 2) aye [He ele] [Be 3p +(2u+ 3H) IE 2 +] ax" ay* az[OI roduction to Fluid Mechanics and Flud_ Machines For incompressible uid, 3% 2 Reve 0 Sop {is satisfied in the same manner. For compressible Aids, Stokes" hypotess is y’=-2 evoking this tq, (84), wil ally conclude that p =~ (@at0a*09) Generally, uids obeying the ideal gas uation follow tis hypothesis and they ae ealledStokesian Mids, I mny alo ‘bementiond that he second coefficient of acon, has been verified be ‘elisa. Now, we ean write ha ae as asf afte] an In deriving the above strese-stain rate relationship, twas assumedthata fluid has the flowing properties 1. Fluid is homogencous and isotropic, Le. the relation between components of stress and those of rte of siran isthe same in all directions. 2, Stes i linear funtion of stain rate. 3. The stress-strain relationship will hold good inespective of the ‘orientation of the reference coordinate system. 4. The stress components must reduce 10 the hydrostatic pressure (typically thermodynamic pressure = hydrostatic pressure) p when all the gradients of velocities are zero. 8.3 NAVIER-STOKES EQUATIONS Generalized equations of motion ofa real flow are named after the inventors of| them and they are known as Navier-Stokes equations. However, theyare derived fiom the Newton's second law which states that the product of mass and acceleration is equal to sum of the extemal forees acting on a body. Extemal forces are of two kinds—one ats throughout the mass ofthe Body and another ‘ets on the boundary. “The first one is known as body force (gravitational fore, electromagneticsr 1 sna one meg Pr Ti, ster penning =—2 ting i 8 ly cena £22) ey. a sig esto i syphilis Pid may aso ‘ementoned hatte sean pfiffi oe eerie 4/6202] ay de a,]21, 30,35 aren a2] ay 6,2 —pory Mn2,[% 4 20 20) 1830) uP De aH Lan ay ae, nds the sve stroman elaine ta ld TT Rleld Te Manogenous a upl, ie.he elaion bawsen Capone of wre a th fate of a hs sue all 2, Stetson of sain me 2 Pe tteaearinweltonhip wil ol good lspci of the ier of tie tere crate 4, Tag tue compensa rece te the hosintic ponte (spilly hamaiynanie psa = hrs pesu)7eaun aw {es as aa jn th x ae __ Ti angle betwen te ot dani he one. eetcoga tee els necator ene erates by eri tne in then pe ‘See sratn a0, ily pepo ns inte xan ee Mowe Y(t) oo at tren tga 25 ce oly si te ees Te en Le wena) t=2(S+E) on l(Se8) ae rally urs ot ha we cn mtemtily combine linear in at and shoar stain ate ito ae lone symmetric secre ord tereor ead the ees ond strain rate tensor sich 3 coniaion of Eq #25 one 26 eshearstrain Garaiertne StainratetemarinCartaion cones ingle Between 7 Maines Cine iy a) alyta) ok | The sain rate tensor obeys all hw laws of methomatia esos, such as ters invariants, asformation law, and erp ses Figure 4-29 shows 2 goeral (although two-dimensional) shuaton in 2 compresib fluid flew it which all posible motors and dfermaions ae present simuRanecusly. In poricula there's Tarsation, raion, linea Strain, end snes strain. Because ofthe compresible ntu of te fd there i a30 volume stain (latation) You should row have 9 bee appreciation ofthe inert comple of fluid dytamics, and the mate ‘matical sophistication Yequrd to fully describe ltd motion, EXAMPLE 4-6 openies ag ansation, fstab consider the saad, twoimensona valor td of Example 4-1 V= ao} = 5 + 00x) + (15 ~ 089j o fre gt ae nis cm, oan wien mT 8 Zaputon pln ‘at (-0.625, 1.875) as shown in Fig, 4-40. Streamlines of Ferment le Fg 4-0, en he os erate roe letenyt tel tet, toon rt ea ea See data: San a, ny at sto ereHence, +E Finally, we can write Gu The transverse displacement of B with respect to-4 and the lateral displacem of D with respect to A (Fig. 3.7) can be considered as the rotations of the lit segments AB and AD about A and brings the concept of rotation in a flow fi The rotation ata point is defined as the arithmetic mean of the angular veloci ‘of two perpendicular linear segments meeting at that point. The angular veloci oF 4B and AD about A are 9% ang £P respectively, bu in the opposite se Considering the anticlockwise direction as positive, the rotation at A car written as, \ j(aat a8) 2a ae 1f av _ au pa 8 24 3. or, %, (% ax) r ‘The suffix z in «represents the rotation about z-axis, ‘Therefore, itis observed that when w= u(x, y) and v= v(x, ») the rot ‘angular deformation ofa fluid clement exist simultaneously Ina special ease, when av ox . fy = 0 (from Ea, 3.19) ei = a= %-- - (from Eq. 3.20) @: ‘This implies that the linear segments 4B and AD move with the same angcoer \ Flui¢ ee 3.1. The velocity potential funcisn for & source varies withthe distance ras @ w (uF we ine (1987 : 1 Mark) 3.2. A-streamiined body is defined as a body about whieh @) Thetlow laminar (©) The low is along the strearsines {6} Tho ow sepatationis suppressed (6) Tha dragis 200 [1987 : 1 Mark) 3.2. The Newtonian fi haste folweng elected yi van yek The rate shear doformatin ¢, at ie point gn, y= and z= 2 fer he gen ow is a6 2 @ -2 was [1988 : 2 Marks} ” nis ntyed testo 24 mace svat paint 1 1) @2 (0) a2 ‘ oe © {1989 : 2 Marks} jaretrveepipe sections trough which 3.5 Showbaae hour. opton ato below prin watt Tow tomoraltzeonte poe seen soir tw lat. ove etguom tesa Yas (© fortwo vex and v0 ys ©) dee tobot ve rand ve yates {6} 45"toboth-ve and yes Ste which ote options indicate the corect tector fcefr pips 1 poe 2and pipe 3 =. 3.6 A vboiy fod is ¢ Goa heres. y. ae in (represents 2 (i) Thetiow isirot, Gi) The fowis ste 3.7 Streamines, patt ‘ally identical fe {@) Uniform ow (©) Stoads fon 3.8 inallowteld the: lines (a) Aveparalet (0) Cutat any any (@) Aveorthogone (@} Cutortrogone 3.9 Forathidelemer (o- yplane), its (@) Tansiatono (6) Wnsiationa {) Transition (¢) Delormatior 3.10 Existence of ve (@) Fludis inc (0) Fluidisiro (6) Fuuidis ide (@) Fhidisco. 8.11 Circulation is component o! blanks)the free shigher In) are is the 1) are ithe false true. 014] rain dby ress 15] ate, 6] Baal 1.62 1.63 nae 0 se = up inal paion coefficient of water in air is Ea ym, The work required in spiiting up the is fay 0.96 x 10-° (b) 0.69 x 10 J (c) 0.32 x 40%) (d) 0.23 x 10%) {ESE : 2017] A 150 mm diameter shaft rotates at 1500 rpm \within a 200 mm long journal bearing with 150.5 mm internal diameter. The uniform annular space between the shaft and the bearing is filled with oil of dynamic viscosity 0.8 poise, The shear stress on the shaft will be (a) 1.77 KN/m? (b) 2.77 KN/m? (c) 3.77 KN/m? (d) 4.77 KN/m? [ESE : 2018] The normal stresses within an isotropic Newtonian fluid are related to 1.Pressure 2.Viscosity of fluid 3.Velocity gradient Which of the above are correct? (a) 1and2only (b) 1 and 3only () 2and3only (d) 1,2and3 [ESE : 2018]AT ~~ ape 5,53 For an irrotational motion (a) The fluid element does not undergo any she (6) The fluid particles do not undergo g stevia motion r (c) The circulation around any path is zero (d) The viscosity may not be zero [ESE : 2011] | Bora Pes tote 2A Consider the velocity field given by u= # and v= - 2xy. Find the cious ) ok around an area bounded by A(1, 1), 8 (2, 1), C (2, 2) and D(1, 2). Solution : r= flude +vdy+ waz) = ine jude +f vay fe 7 udv if” vdy jxatya2plications the line AB fude = the line BC Along the line DA Judy = bo Ta Lary —_fuidrinemac | 494 ro bid a a2 1 ' On2 can 3 ts = gL)-()]= +2333 | (8) - -2)] = -6.00 (FT. 7 y=2, x= 2 - 1 . bye bh {(-1) - (-4)] = + 3.00 T = 2.333 - 6.0 - 2.333 + 3.00 = - 3.0 units {s \ - 210-@)] = 23334.39 Water enters a two-dimensional, square channel of constant width, A=75.5 mm, with uniform velocity, U. The channel makes 90° bend that distorts the flow to produce the linear velocity profile shown at the exit, With Ug =2 Vpia- Evaluate Unio, if U=75 mis, 4.39, 4.80. 4.80 Assume the bend of Problem 4.39 is a segment of a larger channel and lies in a horizontal plane. The inlet pressure is 170 kPa (abs), and the outlet pressure is 130 kPa (abs). Find the force required to hold the bend in Vinay FF 3 Umiv 9, 434 wt vu atl 6: For Steady {lov ul Oo <LW _ Wels % 28 ‘Qe ¥ oe “32 4 f dou) ae am Tegi The velocity potential function for asourcevaries with the distance ras (a) Wr (b) 4/7? () e" (d) Inr [1987 : 1 Mark] Laake alnnuitDiocharge Q= Uy. 2NTK Vdedty Pratite y= ea Posem tial junction 24 a i oe 7 5 eal 2) a0, 24 eb s ay - 5 26 a wv. AQ wv” and g = & rs i) -@ UR ge. ors@s jw > ‘ Up = oF HO) forYpash tp) = 0 Rey= comst By eh ® Inv +c a NYA closer look at the proof of the Pascal’s law shows that the implication of the fluid being at rest was the absence of shear stresses. But then, there are other cases of fluid flow where shear stresses do not exist. Pascal's law of equal pressure in-all directions in space is, therefore, valid TT (a) for a fluid a rest, whether the fluid is viscous or inviscid, compressible or —~(b) for a liquid in solid-body n either at constant linear acceleration or at constant rotational vel ina j here again, there is no relative motion between different fluid ‘Tayers and (c) for the flow of an inviscid fluid and for ideal fluid flow where 4 = O corresponds to absence of shear stresses. ,Tyhe 4, flas Convective acc! Wav BS aw ag * Ot i) Steady & Unifermfin! 0 ° Ny a0 di Sieady & Non Grif |g yigt ° Ue low Kil) Unsteady & Uniform 0 Spink Wo Fino at v_ 9g7o WM Unsteady & Nom ‘ BW yy © Uniberm | Exlot Evid a Flasa ‘Thus it may be concluded that if the flow is irrotational then applicable to all the. in the flow field, thatis for all the stream| the value of the constant is same. Hows Totational flow, the Bernoulli's equation applicable only to particular streamline that is the value of the constant is different for differ streamlines. SSS a el wsy2 & nee eenQ.40 4 frictionless circular piston of area 10? m? and mass 100 kg sinks into a cylindrical container of the same area filled with water of density 1000 kg/m? as shown in the figure. The container has a hole of area 10~* m? at the bottom that is open to the atmosphere. Assuming there is no leakage from the edges of the piston and considering water to be incompressible, the magnitude of the piston velocity (in nv/s) at the instant shown is (correct to three decimal places). | e=10me 0.5mProblem 6.12 A siphon is shown in Fig P 6.12, Point Aix me above the water lee. India by point 1. The bottom ofthe sphan i 8x below Io! A. Aawumingfrition to be ealgbe determine the peed of the Jet at outlet and also the prewure af A Sol" Gn, Uses Bernal eatin, tween and 2 ‘ ororly ay = Ls xyxte ran ort v tt Tigre PE Prion mods son +4 Bez, P,= P= atmospheric pressure Consider level 1 as datum. The velosity of water at the surface is zero, ve o+0= M7 2¢ s Vor JTXOXOBI = 11.72 mis = Vy Considering surface 1 and level A. As flow is the same, Pa vi flee Considering Py'y= 10.3 m of water, Pa Ay YE lio9-1-7 yoy) Re =2.8 m of water column (absolute) A P04082.46 Statement (I) : There exists a positive pressure difference between the inlet and throat of a venturi meter. Statement (II) : The coefficient of discharge of a venturimeter accounts for the non-uniformity of flow at both inlet and throat. (a) Both Statement (I) and Statement (II) are individually true and Statement (Il) is the correct explanation of Statement (1). (b) Both Statement (1) and Statement (II) are individually true but Statement (II) is not the correct explanation of Statement (I). (c) Statement (I) is true but Statement (II) is false. (d) Statement (|) is false but Statement (II) is true. [ESE : 2018)Q.36 If the coefficient of discharge of a venturimeter is 0.96, the venturimeter constant is 0.3 m2-5/sec and venturi headi is 0.2m, the discharge through the venturimeter isGATE- 2019 Gvechamical) Q33 Atank open atthe top witha water level of Im, as shown inthe figure, has a hole at a height (00.5 mA ee jet leaves horizontally from the smooth hole, The distance X (in) where the jet strikes the lor is — aos Lo © 20 ©) 40 ~%o> For Smoom nate? (PrictionJen) Vasa * Now Cyat56. A vertical cylindrical tank, 2 m diameter has, at the bottom, a 5cm diameter, sharp-edged orifice, for which C,= 0-6. Water enters the tank at a constant rate of 9l/sec. At what depth above the orifice will the level _in the tank become steady? — (a) 2-95 m ab) 275m hes atc) 2-60 m 4 -b en A (92-50 m ee “or fe oe a eae “\ y }totational 2. raters to energy balance, (6), Nor-uniorm iron m (gfAtor-unform rotational 3. develops into Bernout’s equation under = appropriate conditions 4. is applicable to rotational well 85 6.39 An open cylindrical tank of 75 om dame iotatonal ows 15 mnigh contans water upto 1.2m ¢ Wich of these statements are correct? the ylinder 1 foaled abouts verte (@) 1,2. 3.and 4 (O) 1 and 2 only seat stne maximum anguiatvelocty (ey (©) tnd 8 only) and 4 only por socond) hat can be attained wih (ESE : 2019] any water? (@) 755 (o) 808 6.26 Ata distance of 20 cm tomthe axis fa whiipoo be tang in an ideal baud, the velocity is 1Omvs. At a © a radius of 50 cm, wat will be the depression of the tree surace ol he qua below thal at avery g.34 A horizontal venturimeter with inlet dia large aistance? 30 em and throat diameter of 15 cm is (Take 17g = 0102 sm) measure the flow of water, The readi @ Gem @ osi2m diferential manometer connected to the @ Gsm (0) 1.224m the throat is 20 cm of mercury. It C,, (ese 2019] tate o flow is nearly 6.27 The Bernoulli's equation is applicable to (a) 1251s (©) 25 is (a) Both stoady and unsteady flows () 125 Us (0) 250 us (©) Real fvids lest (6) All fuids and flows along a stream tube 6.32A mercury water manometer has : (6) Steady flow of ideal fds along a stream i ditferonce of 0.8 m. The difference in ad measured in metres of waters iesen2013) (@ 08 (©) 1.06 {6.28 Statement (1): The integration of diferent form (©) 1005 (a) 802 ‘oF Eulo"'s equation of motion yields to Bernoul's les ‘equation ‘Statement (I): Evlers equation is based on the 6.33 An orilice is located in the side of a sation 3 Conaar ain ol Tas its centre 10 cm above the base of the (@) Both Statement (1) and Statement (Il) are Constant water level is 1.0m above | individually tue and Statement (It) is the of otice, The coetticient of velocity i Correct explanation of Statement (}) the issuing jet, the horizontal distanc (©) Both Statement () and Statement (I) are vena-contracta to where the jet is 10 ‘ndivigually true but Statement (l) is NOT the vena-contractais Correct explanation of Statement (1) (a) 162m () 100m (6) Statement I) rue but Statement (I) is false (©) 062m (@) 032m (4) Statement (is fase but Statement (Il) is true fe [ESE ; 2013] 6.34 The velocity of water at the outer edgs ‘measurements with a Prandtl-Pilot tube diameter whirlpool, where the wat ‘rowed ha hop reacngs varied on across ——hzonal s Bais. re wong Be tow att side opening eudngs vated ——_—lcaeotedaneto olen aw the direction of flow. The type of flow is (a) 1 mvs. (0) 5 mis tomioaong (0) Uniform rotational cen (o) 10 mi>Q Uniferm los _ spt fegie= Pa Deo For 1D Adit - Wepot = Foxit ™ Mm Ye u U = ye 20) f. uw 93 -aP_ pra ae P [Se we Vat my Ip th cara x 2-196 Pasm qe25 | seietoltwmeni © Mamet A PIstin: RE bo @ £@ 0.20 ina Siphon the sunmitis Sm above the water lovl the tank rom which te ow is being] 4641, = Be av + Ls thy discharged. Il the head lss frm the ine 0 Y 34 the summits 25 m and veloc heed atthe ‘summit is 0.5 m Take (@ = 10000 N/m?) the pressucheadal sunt UB -80 KPa (b) -3 mol water (abs) {© Simot water abs) & Yy (9 tomotwatatam) Civ |p Mechtmicss coonebar 2111 R= Iomoex ( — 8) i1.29 kin!) in wdc of ‘gn Fw ton ooh) cals en econ 2 Ly) OMe Te tw wen aves Gate -t08 reste ilewncstetwcen beans ST egg Science ‘The dametr mati, ‘Nisa of oe ieee astm or] y= 250mm Ae fy> ors toes ot ae sat Bee st g Re De Aa Ag, Ar hot aa Ay oe aail = aut es h = 2 1 ae 2 d, @) sim Loss sidearm ann p By mh + Ney ose = + ad 7 aie ? 4 RP < ch Sk 4) = eagle aye ( & ce ek 1, 1 = xi a | {ie asst} [350-14 | By Py =~ [298-76 treSd: Ye uted Stagnation point V=0, U=0, Yo 0-5+0-0% =0 7 1-5 -08ys0 hes \s 8 a a (#8) Ore paint.Civil Work boo, Bermattinen fur adiabatic. 4lno for Compressible fig et Vdv + 9dz=0 a Race VP spe afvav tghz<0 bephe | h i ape reg agc it a) VP e+ wha gree ‘ nie ake ™ % aha y pM dpc BF Yr 6 ' ity ges C ae vp ateD : Fluid Mechanics Useful Data: See jesattn nisl 2ita ve 23 [fm -Jy Density of water p, = 1000 kgm? SE Q.1-Q.6 carry one mark each. "Aime (085 -\eMns /Nowto [822 -i]o mo" Pitot tube 530m; Q.1 A projection’ manometer is to be used for measuring the dynamic pressure of airsream (= 1.2 kg/m’). The least count of the manometer scale is 0.1 mm, The ‘manometric liquid is alcohol of specific gravity 0.8. The lowest velocity that can be ‘measured is A Bi2ms rims © Sims (D) 2ms 2 Thevelosity of sound torees UA isa thermodynamic state variable (B) is constant fora particular fluid (C) depends on the velocity field forthe flow (D) depends on whether the flow is laminar or turbulent Qs ©. The most appropriste set of conditions for this equation to hold good is: Tveo Trcompressite (A) steady or unsteady, and compressible or incompressible (B) steady or unsteady, and compressible steady or unsteady, and incompressible (D) steady, and compressible or incompressible Promat) Numbers 1 Goa ctv Q4 The non-dimensional number obtained from specific heat (c, (dandviconiy Gis Meet ‘vue om cot me . Meh Wiiein ial, = Rule, ©) ue, th Q5 For a two-dimensional turbulent boundary layer, the wall shear stress can be expressed as r, = 45] where m isthe velocity paral tothe wall andy is he ‘coordinate perpendicular to the wall. In the above expression, denotes Vi) the molecular visosiy ofthe uid ( Loma yar oF Sutface) {B) the tubutent eddy viscosity (C) an effective viscosity which is greater than the molecular viscosity ye (D)aneffective vost) Ua ses than the molecular saclQ.23 A 30 cm diameter, 90° elbow has one limb vertical. Average velocity of flow of water through the elbow is 5 m/s and the pressure intensity is 4 kPa. The vertical component of force to keep the elbow in'position will be (a) 0.28 kN (b) 1.49 kN | (c) 1.77KN (d) 2.05 kN ) [ESE : 2001]Civil Gate L019 (48 A rough pipe of 0.5 m diameter, 300 m length and roughnes height of 0.25 mm, cates water (kinematic viscosity = 0910" mls) with velocity of 3 ms. Friction factor ) for laminar low is given’ by J 64lRy and for hxbulent flaw it is gen” by Freel jets, where, Re Reynolds number, r= radius of pipe, k= roughness ‘height and g = 9.81 m/s’, The head loss (in m, up to three decimal places) inthe pipe due to fiction is i K+ 0-Lsmm V2 3myy dz05 d Letom Re= Pus _ vd _ 3x05 MiP oq xtors Re= 166x057 1200 Turple Aas i 14 4 Lope (BA. a) \ = let 2x3 = PUNY o-0161x300x3" 2d LxGe KOS, hye 454m Pao0te7 *Sdution 4, GS dy FN qorcbove tere Covet: MIS me dincharrge in akon aut ntfermly dang M fergh =, ‘Dioct F; ~ a ont tag OE x dhy= fang ag or TET ae Qe a aK Se Ml o AQ.18 A triangular pipe network is shown in the figure ‘The head loss in each pipe is given by h, = rQ"®, with the variables expressed in consistent set of units. The value of r for the pipe AB is 1 and for the pipe BC is ¢ If the discharge supplied at the point A (‘.e., 100) is equally divided between the pipe ABand AC, the value of r(up to 2 decimal places) for the pipe AC should be (0.62) Given h, = r- Q'® Because of the given condition of equal discharge distribution in pipe AB & AC, th discharge in AB and AC will be 50 and 50. Now satisfy continuity at point B and ( For close Loop ABCA Eh,=0 =91(50)"8 - 2(20)"® - (50)"# = 0 => — 1143.26 - 439.42 = (50)! r= 0.62UrBy=2om Conventional Questions Sept, Rave Diochange 3 al 2.20 Vir tova tm ares ough ere Lose, cnt inetd. Bios aredas vw a Sue OF042,= OFF Te Lesson 12002 and avalabieload = 20m, Keg = 1 (Center tine of pipes shown in tigure are eral at care ive) Loo = Lr yee @ @ s § we -o a Seam Ww& wet * Rot Fue 8 Qeaanic> | vs Kr es “® em == aa Sepa: 94 ad Minor Lassen one Wegleaea- ho + het he=20 4-4 se = xt Shah , Fab | Bhat 7, crea teas tee eras TTD 00 WU fap hah yh - ‘a Li as + Bete = S30 Qy= 14. 1x 13 Ws4) | (@ — The 1/7 power law for turbulent boundary layer is not applicable very neaglto fM solid | voundary. Do you agree? If yes, why ? Justify your answer. “~*~Q.51 Consider a laminar flow in the x-direction between two infinite parallel plates, ic ‘The lower plate is stationary and the upper plate is moving with a velocity of | cm/s in the direction. The distance between the plates is 5 mm and the dynamic viscosity of the fluid is 0.01 Nesim’. Ifthe shear stress on the lower plate is zero, the pressure gradient, 22 (in Nn? per m, round offto I decimal place) is Sol: h | 1d For Covet Flos ue My Gote td Ta Ud ley Tues eB]Flow Through Pipes ptt ater via vey V gue stan bet) anderen cf cre secton , entra steam” bring valerie of erea A end vay i The wo teams rir nh he ome pra Fvecmonugiiyaeaighibepbe teeter eupe aw angle area wih veloc Vs and presse Pi there a no lesos in he om, Seti (P,P) for =20 mis Y= 10 mie 4,= 001 m2. A, = 0.02m?, density of water p= 1000 kg/m. seo LA, oft = =" /{==.. |e o ® Gate (1990: 5 Marks} Arwrepsions : @ Cid Steady —p Cito stovage) tou Va Cit) Trcempsessive E =. Cun 1D cy Fux? Sas * ay ett Z her, = Sa. - FQVin PAate- MA, = Sars - PL aws@n-Ad i) L (Rha = Gadi — PL AAE+ Reve) ms Swe - SCE cwewd + vz)PB = Pyr- sf wane a Continyity brween © 2@ Min=Mout fh +@r-A) 2) = Kris Bucs Rh apis (Lette) = 2V3 Vg 2 SSC By eqhcir © rans th emt EH) =-C1s¥ leo) = “1s KPa AgatAyChaper—6 ototly 542, Q.26 Two reservoirs are connected through @ 930 m_ Nude long, 0.3 m diameter pipe, which has a gate _1 valve. The pipe entrance is sharp (loss 2, | ‘coefficient = 0.5) and the valve is half-open | C (loss coetfcient = 5.5). The head difference jy, t betwen the two reservoir is 201m. Assume the 3-430 hegix Iriction factor for the pipe as 0.03 and Wp de03 9 = 10 mis2, The discharge in the pipe Sigg OH mor anc meer ose i Kensomce = 0°S IGATE : 2015] Kuawe = 5° ZL. = Lom f=0-08 E, = E, + Losseo gato ac OAT, = OFF TA henscomee + hg hvawe Feel U-h= osvty fut 4 ssity we A ye Ute Thy guvt ys 74 ref + Qotuase | vt 2a0 O° BeLH0 «& 4 VeL be = ovum es Rdyy= Fabs eensn 12. 13, 3 ‘A pipe of 300 mm diameter and 1000 m ength Cconnects two reservoirs kept al 15 m level iference. Find the ow through the pipe. an ‘additional ength of 800 m of 300 mm diameter pipe ts connected athe lator hal othe exsing pipe, find the inerease in discharge. Neglect frinor losses, Take f= 0.02 [ESE : 2001] [Ans. 26.47%) | 7 when a suciden contraction rom 60.em dato ‘30 emdia siniroducedinahorzortaippelee, the pressure drops trom 100 KPa at the upstream of the contraction to 80 «Pa on the downstream, Assuming a Co = 065. Find. (i) How rate, i) head loss due to sudden contraction. Ans. 0.404 m/s, 0.482 m] Water lows through a 90" bend in ahotzcrtal plane as depicted a he gue. 14 [A pressure of 140 KPa is measured at section 1-1. The inet diameter marked at section t-15 ar J ca. Wine ha nozze hareter markedas 14 section 22s cm. Assume te folowng () Acceleration due to grauty = 10 mis? {i Weights of both the bent pipe segment as well as water are negligible (i) Fricton across the bend is negligible “Tne magnitude ofthe force (in kN uptotwo ‘ecimal places) that would be required to hold the pipe section ‘Apptotwaterissues fom anozze with avelocty Poms and itimpinges normally on aft pate moving away from it at 10 mis, The cross: sectional atea ofthe jet is 0.01 m?. and the density of water = 1000 Kgl. The force developed on the plates t - Loe wet0ms (a) 1000 (b) 100N (©) 10N (o) 2000 [1990 ; 2 Marks}TABLE6.3; RECOMMENDED FRICTION FACTORS IN DARCY-WEISBACH FORMULA [SINo Pipe Material Diameter(mm) Friction Factor For Design From | To New Period of 30 Years . | 100 | 2000 | 001 to 0.02 | O01 to 0.02 100 | 600 [001 to 002 | 001 to 002 ~ HDPE/PVE 20 i100 | 001 to 002 | 001 to 002 SGSW 100 | 600 | 0.01 to 0.02 | 001 to 0.02 rer 100 | 1000 | 0.01 to 0.02 | 0.053%0 0.03 (for corrosive waters) 6 Ci 100 [1000 | 001 to 002 | 0034t0 0.07 (for non-corrosive waters) ~ Cement Mortar or | 100 | 2000 | 001 to 002 | O01 to 0.02 xy Lined metallic pipes (Cast Iron, Ductile Iron, Steel) GI. 15 |” 100 [oor to 0.03 | 0.0315 to 0.06950 Tw we rere mmc pon ing fl of a eg 0 nd Samet of 010m, show blo ar a an Civil Gare erat a sorceat Setenscaireeeer es ga: Capone ae es gr0m ao Apblying BED LG fyetiny © 1 #0 Y= 0t2; %am +042, = + att, ‘7 as Pee lo-3 +042, = ak ay, pus 78-ut ty ts 1 33 & = T:8-0.mS - QoLxtomrows ol 1g, = SSM Lg oeusoeo {é) ~ [ESE : 2003] 39 em diameter 500 long water pipe with action factor f= 0.025, leads from a COnetani- head reservoir and terminates at the delivery end nto a nezzie discharging into air. (Neglect all nergy losses other than those due to pipe con). What is the approximate diameter of the ye! for maximum power? A S8imm (b) 598mm C) 667mm (d) 59.8 ilies 7.30 440mm diame [ESE : 2004] fer 2mlonn ates. aw8.8 LUBRICATION MECHANICS ‘Whenever there is rubbing or sliding between two metallic surfaces due to relative linear or rotational motion between the two, friction and corresponding loss of ‘energy, mechanical wearing and generation of heat take place. This is avoided by providing different types of bearings. The hydrodynamical analysis of bearings is based on the theory of laminar flow. The common types of bearings are slipper or sliding bearing, journal bearing, collar bearing, conical bearing, etc. Their analysis is explained through illustrative examples and problems. 89 TRANSITION FROM LAMINAR TO TURBULENT FLOW [Laminar flow takes place only at small values of Reynolds number. For pipe flow UDpAL must be less than 2100; corresponding value of Reynolds umber for open channels, defined as URp/i, is $00, Here R is the hydraulic radius which is the ratio of area to the wetted perimeter. The Reynolds {at which flow ceases to be laminar is known as the critical Reynolds number. Characteristics of laminar flow and its transition to turbulent flow in the case of pipes were studied by Osborne Reynolds through his well-known experiments. Laminar flow changes into turbulent flow through repeated breakdown of ‘external disturbances imposed on the flow in the form of velocity or pressure fluctuations. These are caused by faulty inlet conditions, vibrations of pump and Pipe, etc. Local instability of flow occurs when the parameter x (chi) defined as xo] exceeds a value of 500. ILLUSTRATIVE EXAMPLES: 8.1 Consider steady uniform laminar flow of an incompressible fluid between two parallel plates which are kept at distance B apart. Assume plates to be sufficiently ‘wide in z direction. (Fig. 8.3). Reduce the N.S. equations to appropriate differential ‘equation and obtain expression for velocity distribution and average velocity. ‘Assume plates to be horizontal. Because the flow is steady 2.50; since flow is uniform -2-m0 chowever, 2 in no sero as explsined below); further 20 since Now is two dimensional ‘Also it is obvious that v= w= 0. Therefore continuity equation gives cs which on integration yields u=/(y) since it cannot be a function of x,z and 1. SRemther exhetinaton Gi chrue condiicns Be. 01er. In a laminar flow through pipe, the point 1 of maximum instability exists at a distance of y from the wall which is (a) 3 of pipe radius R Pr 2 of pipe radius R 0 (o) $ of pipe radius R A 1 (d) 4 of pipe radius R vf es \ ~~ 2019-8-7 OFZUGH Be CWY L ah ae 248” TWO R-vey Te Ry v4, ARG) e xX > Fyre _ a ke Prices Ju Xe fab [N- sy aK, P (a2) [th sy" od ay Bao any -aytes 2R- 34% a 3 on> wo = arse a wae Geax (2004: 1 Marky spe om gy ond 4 inetd cro a ifei is BBE) oe tre re seam velo is 2s, ard ai has Ie pic wacentyot 15x 10" nd donsty weep ugh, he wal shoar sress atx = 16 Gh ae tONi? (0) 486% 102 Nem? {9 £28 x102N fe () 2.18 109 [2004 : 2 Marks) «521 Revolt rele tuly developed laminar How pp pipe of diameter D is given by ee igtte 42/0), where ris he radia distance one cent the viscosity o he tid i fe pressure crop across alengh Lofthe pe is oF owe Ki oye BH (2006 : 2 Marks), rtad Annwor Questions 5.22 and 6.23: A smooth ‘ute win sharp leasing edge is placed along a assteu Novng at U= 10 ms. The thickness of tho ‘eat aye at gocton ris 10 me, the breadth of ‘ples 1 m (rt the paper) and the densty othe (5 p= 10 gh Assure thatthe boundary layer is ‘hn, wodimansional, and follows @ linear velocity srt, u = Uy), athe section 1-5, where ys ‘ohegh tom pate tet e Palpite a ass flow rato (ngs) acrous the eoction (0 40 (0) 006 or (6018 [2006 : 2 Marks) ‘sy |_ Po ech nd dae Machines | 15.29 Tho inngrated cag force (nN) on te pal, betweon ps8 ¢ (9 o6r 03s (ow (8 ew {2006 : 2 Marks} 5.24 Consist anhccrorenabi rina boundary aye ow over ala pate of ongth L ignad win tro rection af anincoming uniform re seam. F ‘sho rato olthe crag toren onthe ont hat lhe plate tothe drag force on he rear hal. en (a) Fete (Fa 12 (ret (ret (2007 1 Mark) 15.25 Consider steady laminar incompressible ax Symmaire tly developed wacous ow hough ‘| vaigh crouar pipeot constant cross-ectorat ‘reat aReynoiss ruber of5. The ato oferta force to viecous force on aid parce as (0) 15 @o @- (2007 : 1 Mark LUnked Anewar Questions 5.26 and 6.27: Consider a ‘steady compressible How twougnachannel as shown below The velocity proile i utr wih avai of Uy at the inlet section A. The velocity protie at secton 8 ‘sounstoam 8 mgs OSSS uel, BaysHs H-y =¥, H-ssysit a ttSY, H-8y 5.26 The ratio Vl 1 omm 1 \ L ocum mim elm) seen) _28 Air flows in a square duct of side 10cm. At the entrance, te velocity is uniform at 10 m/s and the boundary layer thickness is negligible. At the exit, the displacement thickness is 5 mm (on each wall) The velocity outside the boundary layers atthe exits Wi235ms 108m) Om (©) 9mvs So 98: v. e ‘ = 3 Uo x 1 a, Uj= toms Lom Foy luis Fow Ay = Ave x U Ly ve= Les Th 3 @txeH) xto= [o-t- 240-005) x Yo Us= 1234™IsQ.21 An automobile with projected area 2.6 m? is running on a road with a speed of 120 km per hour, The mass density and the kinematic viscosity of air are 1.2 kg/m? and 1.5 x 10°§ m%/s, respectively. The drag coefficient is 0.30. The drag force on the automobile is (a) 620N (b) 600N (c) 580N (d) 520N [GATE : 2008]400q S14} JO Wed ON “1Y18Q Many ‘sUOTeOK Ww) v2 Ww Wve [ESE : 2007] Q.18 Asolid sphere (diameter 6 mm) is rising through oil (mass density 900 kg/m, dynamic viscosity 0.7 kg/m-s) at a constant velocity of 1 cm/s. What is the specific weight of the material from which the sphere is made? (Take g = 9.81m/s?) (a) 4.3 kN/m3 (b) 5.3 KN/m$ (c) 8.7 kN/m$ (d) 12.3 kN/m? [GATE : 2003]Got ta8 (026 Ae fowing overs snot yn of anc in ind tee! pers woe ‘ea ond ner tit see ewe pn: Terres ot nde Seen hn prota Wy rsa Ths (0) pea down doping on he eines ‘Sale: Mamontven 4 turbulent fas in mae tram Lamiver 4a D0 i detox epra. tions Awovoble — Adverte: sity ine sn vin ic i yee ing om a wu ie ee ee fe, 2 ow I taf acy oe pe ia | _, Seroeatneecinescamersce | weit se data ey pt els ‘Son Zien Sun he cay al me a ‘nary tet ofa ini han san fo Sec SES aed pt Cm nes ac ‘he toy espe fa if he ce vc show ae iti ty i ese flr sy a or ‘Sp pu ine nei gh pce of a bh be ee ‘neat eg mal Rize ae 22 oe Fig- VG “he locon of epron,te widh of he wake region bind be nde rear das one mi oped on i tt ft buyer ow Cor Ford wit laminar bondary hye ate! Boda Hye Now as mae Htc hg) th momen amocatd hit eee (I) at ead a He ee Sg poe for me neat Te the al wif poi nd 2) ere cm be ona el cgy secede sting uno copes of th ey tet et tepcr nt inesvcaged comport of coi To ince it Fi ‘Grr tony lpr can Ao fan ound he Cader (rer pte freeia Se holo ie ames Dede iechewtoher ince,Lominay Yar Do is deloys Depration. favovable — Adverse we | ee oe 2S? | | Physically, inthe absence of vioous effet, lid patie tein fom the font tothe back of the eylinder casts down Ue esse hil” from 8 = O10 9 = 97 Com point to Cin Fig. and hen backup the hil 0 ~ 190" from pont CtoF) wile ut any lou of energy. There isan eichange between Knctc and premers energy, bt tere fue no energy loses. The same fresue dtibton is imposed onthe viscous Mui within ‘he boundary layer. Te decreas in pressure inthe dieton of low along the front half of the eylndr is ened favorable pressure gradient The incest in peste inthe ce- ‘ion af fiw along te rer ba ofthe cylinder is teed an adverse pressure gradient. ‘Consider Tid particle within the boundary ayer inde in Fig 912 Ins atempt 10 flow from Ato Fit expences the sme pressure distibution the prises in the ree stam immediately outside the boundary layer—the inviscid ow field pressure. However. because ofthe viscous effects involved the price inthe boundary layer experiences a ost sfenengy ait Hows long. This les mean thatthe parle docs ae ane enough energy to coast ll ofthe way up the pressure hil (fom C > F) and reach plat Fat tho mat ofthe cylinder. This Kinetic energy dtc is sen inthe velocity profile deal a point C ‘Showa in Fig 9.128, Because of friction, the boundary layer Aas canse tel rr the from tothe rear of the cylinder (This conclusion ean also be obtained fom the concept that do o viscous effects the particle a C doesnot have enough morentam to allow it to cet up the presse Bll 16 F) “Ths, the id lows against the increasing pessre as fr es it can, at which pont the boundary Taye sepaaes from ifs of the srface, as indicated by te gues inthe ‘margin. Ts Boundary layer separation i licted in Fig 9.12, Tplel veloc profiles ‘trepresentative locations along the surface are shown in Fig.» Ath separation locae Yin (pole Dy, the velocity gradient atthe wall andthe wal shar ses are zero. Beyond that ction (fom D 0 £) there i reverse ow inthe Boundary het ‘Asis inlieiad in Fig,‘ Boome of tomsiny yr sya th wings pr tho cylinder is considerably less thn at othe rot bal. Ths. SEE agit oe soe gene ett = the viscous shear drag may be quite smal, ae aQin Civil & Lonchaoe CRA ICY -8) on Ao per Stoker low the Drog (eve om Sprew io Given by pe 3m vd a Ww 5 &é _y® Vols Lyd 3h0 Ane 3) Far ining tsite Constomt weiney 20 yey =p? o- Cy 4 Mg) =O My > Fa & mng= SruiyQiog - 3x ved Vpoag VO! = VORA KGaahag - imei 3 AGO! K0°00h \oaiy =49 CHSal Airtel 4G < 1 CHAPTER-| CONDUCTION Wa wc no | 6 [Son ne 2.4 28 wa [46 1) E Re Fm pp) me [areza Be | ma aD | A [a8 85) 1.8 mc ie we [96 isp | ae a1 [20.679 C a a ae) Bc] [st 368) [inp BA 3A anc] 32.4229 I isp | 260 asa | ae 153.00) se) (CHAPTER-2 HEAT EXCHANGER. TT sa] ep] nem i Tio.p Tw ci103) | ic [15.645 [Tc [ eine) | 9.555) ____CHAPTER-3 RADIATION Le sc | 22 me [58.105 (=o mo Bo 3o.c_ br eas aa WA] tea 3.€ 38 STW 4c 1a] 28¢ [Reavy sa Be A BOM ee rb (Os 7A A 38 CHAPTER-4 CONVECTION Ta so] ue | eo [A [ewan] 20 | 20 IBA aioe ic BA ixD A Tega] [40 a es ef > 00243 | 5c we ib) wu | 2s ey fw eonresyDimensimal fests ( simmtavse t =trwad a Jeet teasne ee GE pa For dynamic Similarity Je Ge ote. fr ot © bse da Ue Dem = Se te “ap jen in madd & prototype Se fan unkev © an as oh. For viocao 0 Reps ‘Re o oh Oo 2 BP elo "(Sao Ve 5 ee SS, (Gra bo 6 YY, ose To Used Te asd . lh Ml oO < yp Bee ES Ue© OEea < a Ia OE HEHE Comd: Gp model Yuk Go Ou Seis Von Dm So Vp ome ate Be as :- oe) ee eu Be) = (UREx10 og, ub Bop > Ur oe = (bre (Paty acuget6 OP enter Connery 4% Cl) BO) @Pdar , LENSE Geman De Ml fa} <2 Problem 3. The prfirmane of aeroplane yo 210m hight at ped of 250, Inuph cb elated by 18 tl ed in pas dtd ttn onlay The conto the fig altace a temperate =, pee = TS AN = #74 130g, Te tet condone re 2100 BN and 18. = 181 x 10%hg/ me, Te dag ree on ml mam 1 a 27 swe sthvan 968 Determine the dag the preter AL gen Bea ena, Voy anand ie (oa » SFTTRERTTE= 390 = 180 mph eae Rey nar sino yee ere Dena tet cndonn = 210 1 aa) = 901 Ket rely a hm Te MeN 8 = 88 gs tng ayer igh ‘okey iinet” = enon = ee Ossr0He Ly 280 Syne SR ‘hie ae neat isd i et ie A Rk Take? BAe 7 1 ioumotition vg ey of drag at 25.498 wi le ohne stam. Dragon potatype= 78524185 = 302 N J este ates [Problem 8.12 sonar anode inthe shape a a ophrof 300 mm dame wed (a betta be noel at 26 mote water at 20. To doerine the da onthe ana wl of 10 rameter i fre 8 wa tee th ng of 30 Te drag fost ‘itor ae {3 Detrewthcaped of afr the eat. Batiede the dag on he pray ‘A to fally ered ype of fam, Reo umber sir soak be ait ‘Toe dnsty and neat ieosty vale a 1800x106 th 016 10 ate Dt a3 2 O10 96 wy 100 * 0050 oe | 28200 Ml oO <a< 7-59 Measaremens of rg force ae made om 2 del ato. ‘moble ia stg tank led ith Gah at. The adel Tenth seus that of the prtype Sate the eons rue oegire dn lary betwee he del 394 proto, DETTE te raion ofthe poten spe inst {rich the snd etal be made in ter fo exe ‘analy sini condlom, Mosscomeat made tis Speeds show tat th cmensonle ore rao Recor cor ‘Man at vel tt yet above Voy tt, hg foc smeared dings tet i peed Fy = HON. Cae {hegre he rte vee pring Mae (By « le) Bye) + (BO 2 esas 5 Cag Cop o> Gre Fane | Sei foe sa i Faye Hox 18 > Lac Ml oO <eor-aar ComO HN) geconsx Lo) OA pilin cing water tax muice raghaee of acme bight {tom ie str se gd 36 wane tet (2) The ana of ow x aly carded 0 em pipe i ou to areas {7 pow wo ned fom «pr fom ie allt 20 eh ‘Some all dasssg pipe ae s0up ip, etatte Dl of sm a sa Gy Ke sm ae 86 EB soa twrnctar Saclarer he Veen nipoie e tv 2073 Pa~ See Ve” Ue tot 1 $ “ = MOH gays stort 36 lew Ag esas? H=C0L 7G Rar pipe 8 las 33 to" Une 120 Be Tom Gi tom For Rovyn pipe Os 4 Sts be DY Ml oO <Loco For Req pe i Bs Sas lyf yelodty ) be Os ters tu (88) a) YL Ost Sas Lego (2) -ay 8 Ne is. Uacth _ s-releye( Me) - s7steq (| Ue Lah 5 ss bl) 506 on, a = Se « Soy (%) 4 S18 at Ly = O-3Mas has Seah (i o-3UCts, ee Bye CAME = 95+ 515 lope oe kz 0-028 Ml oO <2008 ENGINEERING SCIENCES. XE Common Data Questions Assuming Common Data for Questions 29 and 30: ™ Ly 2 = (9 The Darcy-Weisbach equation for head loss through a pipe is given as hy, = ID - A reservoir, as shown in the figure, stores water to a height of 8 m. The entrance from the reservoir to the pipe (length 50 'm, diameter 10 em) is sharp, with a loss coefficient of 0.5, and the friction factor for the pipe is 0.017. Oto rot, = OOF + OSV ELE, yt &-z, 9 2gs 7 A) = P54 ome yy yt mo Ve 4m aE iw t @ h Teal 3 O+O42y ogy’ jiameter = 10 em ‘ cnc ben Q.29 What would be the dischargé through the pipe? zy 4 YI0.0314 mis (B) 0.0322 m’/s (C) 0.0331 m/s (D) 0.0341 m’s Q30 Ifit is desired to increase the discharge, the following four options are available: 1 Increase the pipe length, keeping everything else the same 2 Inerease the pipe diameter, keeping everything else the same “9 Migr 3. Add a valve atthe end of the pipe, keeping everything else the same 44 Replace the sharp entrance by «rounded entrance, keeping everything else the same Teond Enramce decrtanco eddy lessea Only two of these options serve our purpose. Which are they? A) 1.2 (B13 W248 342 plugin oom ofa esd ak ison tape, sh NP F287 Thea ess KP te qi 1 ; Ym 4 oS ‘hn a et wh 2 AN Demi he a V tentoe T tierce acing eee t ete oo oe win et 0 etiga Trost st Metned : eur Gams apf sa- snr) she - Yond mY edd x KKB eRe 3 a i en t Lay Gane by Gey® NT we or Methed Cetaym +, Feet (Py Gorn + wolmrdsa - trite] = Fret (end + a7 xi) Wo-s Th 4 11 Ki (ON ost] ater 3 Fee = INT tH Sent4:52 Water flows steadily past « porous fat plate. Constant ‘suetion is applied along the porous section, The velocity profile at section ed de -a6 28" Evaluate the mas flow rate across section be, Sealy fed ing tiny Bae” Pap {Ipc HBqs 48a} =° $78 Gay Has Pes 2(5H4)) Uw UCew)— fudge’) alstsdte- Ofer - vs fepte- vad, a Qype = 165x103 is< 2ufrep_ME-GATE-2019-Session-2. [T] MADE EASY (nian Best Fetcute for 168, GATE & PSUs Q.23 it xis the mean of data 3, x, 2 and 4, then the mode is Ans. (3) B+x+2e4 Mean, x = —7=—= dye9en Br=9 xa3 So the given numbers are 3, 3, 2, 4 Mode = 3 Mode = Maximum times repeated value. 0.24 Water entors a circular pipe of length L = 5.0 m and diameter D = 0.20 m with Reynolds number Re, = 500. The velocity profile at the inlet of the pipe is uniform while it is parabolic at the exit. The Reynolds number at the exit of the pipe is (500) cy 2¥, 1} Re = PE tor pipo tow V= mean velocity Exit mean velocity V, In = Mog PAM, = PAN, ts Water is incompressible fluid, so density will remain same} Gen, AtA : Yaw So, Reynolds number is constant i.e. 500 at ex= 1000 kent iscosty of water, p= 10° kph) Dre vist of ap, = 18 104g, . ‘one markeach Tia om inp =Ovaid? * 1 Sesdy incompressible low (@:Uastendy incompressible ow “> i Steady compresibleflow °° 8: Unsteady compressible flow @RQR, Mars Vo RRS POS seam function CANNOT be defied for {@) wo dimensional incompressible flow (0) two dimensional compressible daw, WA three dimensional incompressible ow “ asigymietric incompressible flow 'k Which ene efthe followings a irrtatinal ow? ‘(US ree'vartex ow (@) Forced vortex flow (©) Couetteow ” "(Wake fow Inder strong ‘wind conditions, electrical canbe rete to wind inden oclations. ‘one ofthe following non-dimensional relevant o this problem? 0) (1 Froude number 1G) Weber nurmber |G Faradayaumber a Strouhal number [ Dinples are ide 9 golf alla for which tho the equ to make the flow over the ball turbulent to make the flow over the ball laminar 8 :tocrete a separated boundary layer flow over ‘De ball "UAP,Q 1 ut @RS onc ORR oot 02D boundary layer flow, x and y.are the tramvise and wil-normal coordinates, ‘eapectively Ira denotes the velocity along the x which one ofthe following represents ‘Gecondition at the point of low separation? ie ‘iloving reasons? — ‘Sg Pstomskethe ball travel a longer distance j i e y pts due to gravity. g = 10m" Sctemce, 1 Which one among the following boundary layer Awan the LEAST suscep tofoweparation? (©) Turbulent boundary layer in favourable ‘ressre gradient ©) Laminar boundary layer tna favourable Socet ‘Tate wy er i verse remeron (@ Laminar boundary layer in an adverse pressure gradient 8. Airfom the blower ofa hairdryer lows between ‘roidential liptical elinders eunpended freely, ep. ees Bi i Seetaccs oe Blogs ona “tS ‘eachother for Case 2 AS towards cach other for Case 2 and away from each other for Case 2 (6 aveay from each other for Case 1 and away from each other for Cose 2° ' + towards uch oer for Cove Y and tans (0 each ether for Case 2 @9-24 marks exch ‘8A 40 cm cubical block slides on oll (viscosity '=0.80Paa),overa large plane horizontal surface, Ifthe oil film between the block and the surface ‘haga uniform thickness of 04 mam, what willbe the force required to drag the block at 4 mvs? Tguore the end effects and treat the flow as two dimensional. @) 128081 (© 1920 2B ates 80 nd Mr nein in ghiravity. centre ofbuayaney, and the metacentre, 11 reepectively. Th by willbe table if. @) Gisleeated shove B a) mae UuktQue Use Ue ‘at Bre x0 Oye 4 | leCamScanr eld Qatari oar asbestos =O Sb titpun te Reeapaeiay bergen mat Sian eee oa Ox (00-6 g time 80 Bayt 4yvOF te ® q “= 9 []Gee ye go “ie @ The mot sual relatin fo Oow Y Acie al wept rn sig ° Shgermanactonaoetpenty onan ‘teal cet sy at at en Seger nemecnesta en a ~Eeciaaamareesagertas 2 ae he te ve Ine loin ine Tada eran nee Si OH SSW AGEL Stash Comite NitO11) Ifks02im snd therenltant of etal acslerations ° lerady-drenimnstecelyysioa io, Semana wine oan & g 8 i F Scanned with CamS¥: Sony pertain usa) ok zo Ue baer Us Pinay YS we . g. . a Tl. Se Pacey o~ 8 (alg) @d any ae Ta, usb = *8 uo ap COor ° “ Re Um ve opine tyEomp + ty Com) « Fares per on he, Poe ran denon ea.ese Oyen winnie” Gee hig Morente lina te bay) ve Yatgie oma The Lietieatenadiyetecamet eee ‘SSfeinaeawae-nonrcae “aac ioe Semin tne reap aapSE b-8ece iste Sitetenene dan Soma sta, Aesd Bh hie Ment 2 Bedwikinsrtctey Termes hun Ft Yo Srodentcricdsa® Wee time Nenana Vie Mumetocmranee nist irieted 9. peyman Sa ee | eammecepaatie ne Gaston, Heylrasiatte | Se Acer eine heme enn, Batad ex | 281 arth munemet fait tones dirs o08 hen Se nm Ke apiece iam 1 tt na romp] = te ages 4 Ase lips gay EERE ws Gaagiagon Renney aoe Meat ane Gone 1 Geren pt ene arise haha) tira Belson ioe gee op det eosin tain yy enn teenae q 4 Sept aad a ecurstavaliviee S20 ca se nee er SS TS Sines bey “heh, eros s Ghovacorintic longth (2) A rotoant nex: A 46m W490 ee Mies same @ pow? at ate 12, When large tank vaaining water i placed on ca | ‘ree nee ers Soteteeaedmnsncnom pede Paced fo punde ie et Tears eieaatecdiwtcrings | ema Nim tevire mond ewer Se a ee tsemiageraa \ wearin Tranter tae of el im Brine pent ng RUE tothe waghng sae tthe stat Ui ae ee ee pened ed he we je eae ot Dent of (©) Te degon he iinet Teese (© Bothene drags 1 Adnan sete bedoo Uo vey sby i) Bette nee meres | a aan ew Bite © ye fos renomination im va! Cel ER Sas aaa eae ak aL fon “J Wea [See Sy BB oan ate 2 Sst ea on ventana nee Geet 3 tne 2 Na Anon oes Bt Be BEL, Vrmorsioatios ‘A str steam of meld th veicity U ad pes pfs ps 2 ‘eel inkr sows nites Woe Ve Gx [-Ssints é5 O=™ G=-3 ali en by = 2in@. The pesre The minimum tals of Con the urfce of yh web teem th of a OY ore aaa Rego Rey ons Soy oat wd by ed cnecin eo ress wm bya hy t=) * Ue Bele “ em 41004Useful Data: Q2 Qs as D : Fluid Mechanics ‘Acceleration due to gravity g = 10 m/s? Density of water p,= 1000 kg/m? 0.1-0.6cary one markeact, Pitot tube ion manometer i o be wid for measring the dynamic psu ot aiateam (7) = 12 kya), The least count of the manomeet sae i 01 mm. The ‘manometric liquid is alcohol of specific gravity 0.8. The lowest velocity that can be ‘measured is A Bi2ms rims © Sims (D) 2ms ‘The velocity of sound 1A) is a thermodynamic state variable (is constant fora paricular Nid (C) depends on the velocity field forthe flow (D) depends on whether the flow is laminar or turbulent ©. The most appropriste set of conditions for this equation to hold good is: (A) steady or unsteady, and compressible or incompressible {(B) steady or unsteady, and compressible steady or unsteady, and incompressible (D) steady, and compressible or incompressible ‘The non-dimensional number obtained from specific heat (c, 1 Phal canuctiviy Norte (222 -fome? (es -{]o mos $4 (and viconiy Gis Meet ‘ve nomi coh i +. Meh, W fiin Viale, Bale, ©) ue, tb For a two-dimensional turbulent boundary layer, the wall shear stress can be expressed as r, = 45] where m isthe velocity paral tothe wall andy is he ‘coordinate perpendicular to the wall. In the above expression, denotes WA) the molecular viscosity ofthe uid ( Laminar ius of Surface) (8) the turbulent eddy viscosity (Chan effective viscosity which is greater than the molecular viscosity (Dyan effective viscosity thai less than the molecular viseosity Levee Tmcompressitte Promatl Number12, Consider a ‘vodimensional velocity field given Goste Pals by V=nyi-nj, where j and j are the unit vectors in the directions of the rectangular Cartesian coordinates x and y, respectively. A fluid particle is located initially at the Point (— 1, 1). Its position after unit time is (a) (-2,-2) @® a,-0 OGD (@ @,-D z So Ye Ry eT vt Strewmine eqtr ee eI * lestert _4 = Ry Ba -ke he Joe = Fay Time toteen by dre Ponticle fer one Tevstorion 2, E-Bee ts BE. BE unin Pays 20 ae OS n=, Yor Fox luvit time iF Cover holy Tevdietion, Cay . a) FW T=AT Greose -13 13. The stream function for a potential flow field is ven by y= 2'~ 7 The corresponding potential on, assuming zero potential at the origin, is pore O29 Ate (@ x-v A yony . Ghere —ve Showy las rom af a. TArs0 high potem-tiod Yo Lao ppotentit i as “Hey. Woy u A MM cs) Tea 96. _dy Fivot CR: $0 dg 24 (== | Ga tay my we =U ae 4 ™ = ony + HQ) OF 0, Yea, F=0 O)=0Katt 201s 7. Consider a combined forced-free vortex. Tt central region with radius R and angulit velocity w is the forced vortex and the restit the free vortex. The pressure at the edge of the combined vortex is p,. If the density of the fluid isp, the pressure at the center of the combinel vorteris aye @)p,-parR? sb) =doatrt © Pt doa Rs @) p+ pan gah yer? 4 war Ned, ot tu 8 aie Sot Nov R (=> Wer Res Im free Vorep Zone: ° & ut. foal, nut bs 4Z eye lk B43 8 fim Wg cn Tm fated vortey 2. 3 he hots Boag? & fe fo este {Wires gyeae 7 wong a Byeandy ea my wets sand 9 tre - - Palgt ee ey tu Pa toy - fot c corm daiauee, «yD > a . atthe Gk Neo | Y<0 Peo, C<6 Sener Ea-sbayhglly p80, owt saya Ce — Lpat( rey -8 fete) ee ~ Sfomeo +6 ay) ee |-y 3f Coss} - S99 ) us 6) su () 5-564 (0) Chapter Ww a-50 3 (8)Sno na es (sere — 200) ~~ ie ACFE Hae 1802.) ae % & eee ew on. eee oS 18. ®2e we ne wom aweae ee ao mw © 2 © a 0 0 @ BF ® pRS sO onwaoe a 6. 68 ooo n aan me ar no mw mw a © &@ © 8 © 9 oO ow or. @ 98 OB. toh wh Me) HOD 1S TH wo ee Tor, Hw) ton TO TG ong aa 7 om om ee mS ee em me ty we Rem e mo me we me 8 we ee me me me OO BO ee me wow ee 1a owe we me we we mE SO ee so wee wo we ee ee we OS we 7 we mw me m7 me me we mew ee Oo me ee me TEElbow Meter (or Pipe-bend Meter). An elbow meter (or pipe- bend meter) consists of a simple 90° pipe bend provided with two pressure taps, one each at the inside and the outside of the bend, as shown in Fig. 7.10. Its operation is based on the fact that as liquid flows round a pipe bend its pressure increases with the radius, due to approximately free vortex conditions being developed in the bend. As such a pressure difference is produced on the inside and outside of the bend which is used asa measure of the discharge. The pressure taps are connected toa differential manometer to measure the differential pressure head h. The discharge Q may then be computed by the following expression 1 and 2 are pressure tappings Q=CA\Pgh Figure 7.10 Elbow meter where C, isthe coefficient of discharge of the elbow meter and A isits cross-sectional area. The coefficient of discharge C, depends mainly on the ratio R/c (where Ris the radius of the axis of the bend and c is the radius of the pipe), and its value can be obtained by calibration. The main advantage associated with an elbow meter is that it entails no additions or alterations to an existing pipe system, except forthe drilling of pressure taps, and ifsuitably calibrated iéean be used or precision measmemenixCS 5. If O(x, y) is velocity potential and y(x, y) is stream function for a 2-D, steady, incompressible and irrotational flow, which one of the followings is incorrect? aeenrecey dy 1 ol (a) (Z)_ == MXM, ® artZrn9 + ivvo tutional CB8aundary Condivia\ @ +L =0 © —» Ina pressileub GATE 2014 22 Water is flowing through a venturimeter having a diameter of 0.25 m atthe eotrance (St through ) and 0.125 m at the throat (Staion 2) as shown in the figure. A mercury manometer measures the piezometric head difference between Stations | and 235 1.3805 m. The loss of head between these stations, is 1/7 times the velocity head atthe Station 2. Assume the acceleration due to gravity two tobe 10 m/s". The velocity of water at the throat is__ mvs. diameter 0.25m Ruy, ceo 0125 Woser nd 9 cebu J vot ae ON PAPER 5 hoe a a Sa 95" a ic rie 1-3 Sos Vy = 44s WS\ «for turbulent flow, where Re is the Reynolds —_ 0,2 m? and the velocity across sect iin number based on the diameter. For fully developed density of water may be taken ai flow of a fluid of density 1000 kg/m* and dynamic _ viscous effects and elevation change: viscosity 0.001 Pa.s through a smooth pipe of 5 is 0 wo AeA, diameter 10 mm with a velocity of 1 m/s, determine Ve Say, sng Pheu ns «Ate oR tity the Darcy friction factor. 16. Air flows steadily through a channel. The We Wilibet tt stagnation and static pressures at a point in the SaN"y f + flow are measured by a Pitot tube and a wall Pressure tap, respectively. The pressure difference \\> 3 is found to be 20 mm Hg. The densities of air, water * Din Aas Ayla reenter, are 1.18, 1000 and 13600, 72, The €auge pressure at section-1 respectively, The gravitational acceleration is 981 06) oy ys B18, 11: Me" Determine the air speed in , (9 138 oo. Common Data Questions 8. Tha mamtints nw arent ! co Cie“4 6. Flow past a circular cylinder can be erate Superposition of the following elem , potential flows: . os of Ut Uniform flow, doublet, (5) Uniform flow, vortex |. (© Source, vortex a (d) Sink. vortex ier5. Separation is said to occur at a wall when e at the wall becomes zero. (a) internalenergy —(b) pressure UO shear stress (d) density‘4, Match the devices in Column I with the characteristics in Column II. dds 2 sama Column w ; P. orifice meter) high head loss and low cost. venturi meter, . 2. high head loss and high \ 5; cost.;, ey.j 23. low head loss - high cost \,4. low head loss and low cost a Oy apad rages rhb TEsQ. (10 - 22) carry two marks each. 10. In the following equations, u and v are the velocities in the x- and y- directions, respectively and ¢ is time. The flow field that CANNOT be termed as incompressible is hu exsexy, vey tye (6) u = 10xt,n =- 10 yt . (© u=(yis)",v=0(S=constant) : @) u=2y,v = 2* 6, In the case of a fully developed flow through a ies in | Pipe, the shear stress at the centerline is . ively. ‘* @) a function of the axial distance. ! ving (6) a function of the centerline velocity’ \eY zero @) infinite8. If is the stream function, the Laplace’s equation _ “v2 = 0s true when the flow is ‘Streanm fj . mMcHtin 10 (a) incompressible a > Mn (b incompressible and irotational ag aa by “.(¢) irrotational =! oe 2D “ ow (d) compressibleg, Ina potential flow, the superposition of the stream - functions of a uniform flow and a line source gives + pise to'a dividing streamline representing " oak Rankine’s half-body mm) infinite circular cylinder. - ...-$ © “© infinite rotating circular cylinder = ~ _@ infinite elliptical cour wes Gud é eile aeo-dimensional flow field, : . , th iti i and y- directions are u and ti Ascii . tress for a Newtonian fluid aes ; E ; s oo Ina tw the - The shear § dynamic viscosity p is given by a) Hl x ay, = ! ox oy (b) Qn Be ; du , 6u)- q (© ay wf (2% : | ox i p ax ay, Pf, autegie i pee als(@)Tusbalence cours o> 2105) FIGURE 7-18 ‘Turbulence delays flow separation, Flow separation occurs at about @ = 80° (measured from the stagnation point) when the boundary layer is laminar and at about @ = 140° when iis turbulent (Fig. 7-18). The delay of separation in turbulent flow is caused by the rapid fluctuations ofthe fluid in the transverse direction, which enables the turbulent boundary layer to travel further along the surface before separation ‘occurs, resulting in a narrower wake and a smaller pressure drag. In the range of Reynolds numbers where the flow changes from laminar to turbulent, even the drag force F, decreases as the velocity (and thus Reynolds number) creases. This results in a sudden decrease in drag ofa flying body and insta- bilities in flight. Effect of Surface Roughness We mentioned earlier that surface roughness, in general increases the drag coefficient in turbulent flow. This is especially the case for streamlined bodies. For blunt bodies such as circular cylinder or sphere, however, an increase in the surface roughness may actually decrease the drag coefficient, as shown in Figure 7-19 for a sphere. This is done by tripping the flow into turbulence at ‘lower Reynolds number, and thus causing the fluid to close in behind the body, narrowing the wake and reducing pressure drag considerably. This results in a much smaller drag coefficient and thus drag force for a rough- surfaced cylinder or sphere in a certain range of Reynolds number compared toa smooth one of identical siz atthe same velocity. At Re = 10°, for exam- ple, Cp = 0.1 for a rough sphere with e/D = 0.0015, whereas Cp = 0.5 for ‘asmooth one. Therefore, the drag coefficient inthis case is reduced by a fac tor of 5 by simply roughening the surface. Note, however, that at Re = 108, Co = 0.4 for the rough sphere while Cp = 0.1 forthe smooth one. Obviously, roughening the sphere in this case will increase the drag by a factor of 4 ig. 7-20). ‘The discussion above shows that roughening the surface can be used to ‘great advantage in reducing drag, but it can also backfire on us if we are not ‘areful—specifcally, if we do not operate in the right range of Reynolds num- ber. With this consideration, golf balls are intentionally roughened to induce turbulence ata lower Reynolds number to take advantage ofthe sharp drop in the drag coefficient atthe onset of turbulence inthe boundary layer (the typi- cal velocity range of golf balls is 15 to 150 mis, and the Reynolds number is less than 4 10°). The critical Reynolds number of dimpled golf balls is021 Asati site oto stacomfltian tna i (om ay “Pu (eos deity “Peet “ee r Fp Fon ac 4 Spt “pte Fo Ge Siu Suv a _ .* i Saivvat Sa Vo Wee) [fe Vw Saix MAO ae sien are or ay UF full wire Contam Speed dem in both the Caney nd {ore om the Ole io Jor ale Gugyon gy203 ENGINEERING SCIENCES -XE Linked Answer Questions ‘Statement for Linked Answer Questions 21 and 23 ‘Water enters a symmetric forked pipe and discharges into atmosphere through the two branches as shown in the Figure. The cross-sectional area of section-1 is 0.2 m? and the velocity across section-I is 3 m/s. The density of water may be taken as 1000 kg/m’. The viscous effects and elevation changes may be neglected, 2, Aa=Aue Q.21 The gauge pressure at section-1, in kPa, is (A)0.6 (Byl3.5 (135 (600 Q.22. The magnitude of the force, in KN, required to hold the pipe in place, is (Ay27 B54 or 27sear Gate 2015 aac eee Sete 08 wis cous ome (022 The mgs ofthe fe kN, eure ne pe pce Be es Sq) ho Moby Gs Aywy= O-2x3 = ons Gras.) oars Vz 2 03 4 = 6s forty Oe OLB & 9 he a+ O+ Le 55 ypc h = Vat = * ect ig a sama: a oe Le Fvany =o ee Fre, = PONp, - LANG, ca aT Pity -Fy-0 = (Fay ee" \ Sv \ ( ah, ts ae Sau ‘ayy iole: Fie RAPA Vy cnte’ + fou, Eye (3 5p0%— 2 mend 46 00le's Loa .0% KS Fee tia” 217 Velocity potatial for an incompressible Mi ow i given as: = 2(x+2y—y*). Assume the ‘ale of ream function a the righ tobe zo, The value of steam function wt I(x.) m(2.2)] Som = be wy 24 aw ae” Oy jy a Yo any t £09 -& Be. 4yt HY ae — 20. 4ys$'h N F = [oto staat) = 894409 — 4 +9 = Mr #& fiw = fog =e Y 2 Wy — te Gt ne 0 Yxo, =o arn) Yo dyn = Yxos2 — r= QGate Loty 12 Velaciy in a two-dimensional Mow at time 1 and location (x,y) is desribed as F = 30 1+(x—1) 7. The equation for the path ine of particle passing through the pint (1,0) at 120% (ay xt-4y'=0 (8) (x-1) = (© (x-1)'-64y’ =0 (D) (x+1)'-16y' =0 SoM Ue Be Ve Xt Be Lat aul ii at rh Ch Gueandy Jon =f 36 at dy = (3 -s) dt 4 ne BU yc y= Bag t 4 OF t20 Bel Cet [pe 4 Hi by ch tdi (asQateLotT Correct : 1 Wrong : -0.33 Question Number : 12 Ina given flow field, the velocity vector in Cartesian coordinate system is given as: Va(P ty? +2)i+(ytycty’) J+(-2) Fk ‘What is the volume dilation ree of the fluid at « point where x= 1, y~2 nad 2= 37 ®) 5 © 10 @)0 SoM 6 _ Wy Wy de = Ey - av 34 toL Let (Wtt42y) t+ 1D) SE Ut e494 LUE CI TAD fy = 5 (A) 6 tlhate Dole : 18 An riie mes, vig ote diameter of 32mm ic placed in wate pipeline having Now rate, Quer = 3 x 10~* m¥/s. The ratio of orifice diameter to pipe diameter is 0.6. The contraction ‘coefficient is also 0.6. The density of water is 1000 Ay/in’. If the pressure drop across the orifice pia Path nrg co-Ticentaf theeerrtrasfowReyos manor NP incl toon Sot i Ee Gace= 3004 Ay z wb ge Ce fe awe ase § Mh Tak AP= ygskla \P a ast Bag, <4 Post tan i Re AL foie: Gsser [23s : titel aeGATE-2016 Q.20 Consider 2-D, steady, incompressible, fully developed flow of viscous, Newtonian fluid through to estan ple gar in Cure ord ye) goes Aa pees ve “in-rdirection, wide in z-direction (also there is no variation of velocity in z direction) and distance 2) Thera Os ocd the caer wer phe I = ane SO Between values of stream functions passing through y_ is. mm‘is. Sats qh dW L el? hie a 2 rma [I~ + ) Q J by "Goa (Rehm ta faa J B= Umno (2 - wo = Wye =I Voorn fe oe o bE 2 LoD tt sFaz (a = pgm A | CM =bue = loon g x15 x(3ox1) aw Wyam = 63-Val ; oe £9 * (Pram!) ‘Donn = 420+) x30+(Ga6)