‘and is called co-efficient of lift ao 2 plat plate 1.5m X 1.5 m moves at 50 kahour in stationary car of density 1.15 kgf 44 proviem! - + drag and lft are 0.15 and 0.75 respec ively, determine : ee lift FOrCEs (ii) The drag force, rhe resultant js function of Re If the force, and ii) 7 a te quired to keep the plate in motion. jy) The power Fe ion. Given rea of the plates As15x15=2.25 m ea of the 50 x 1000 vetocity of the plates U = SO kmihr = "5p ms = 13.89 mils pensity of ait p= 1.15 kg/m? Co-eficient of drags Cp = 0-15 Coefficient of Hilt, = Ors ). Using equation (14.4), {y Litt Fore (Fr 2 pentyl mgs areoe ts Ot «in Drag Force (Fp) Using equation (14.3), 2 2 Fp=CpXA xn = 0,15 x 2.25 MSxEe N=37.44. N. Ans. Using equation (14.5), Fe=Viath = 3744? +187.207 N = 1400 + 35025 = 190.85 N. Ans. i) Power Required to keep the Plate in Motion Force in the direction of motion x Velocity. (uj) Resultant Force (Fx) oe 1000 = Fy XU _ 374251389 yw = 0.519 kW. Ans. 1000 1000 conducted in a wind tunnel with a wind speed of 50 km/hour on @ flat Problem 14.2 Experiments were iris 1.15 kg/m, The co-efficients of lift and drag are 0.75 plate of sige 2 m long and 1 m wide, The density of and 0.15 respectively. Determine : (i) the lift force, (ii) the drag force, (iii) the resultant force, (iv) direction of resultant force and (0) power exerted by air on the plate. Solution. Given : Area of plate, Az=2xl=2m Velocity of air, Density of air, Value of[640 Fluid Mechanics (i) Lift force (Fi) : ‘ C,xAxpx U2 Usi vation (14.4), a ae 0.95 x 2x 1.15 x 13.897/2 = 166.404 N. Ans. (ii) Drag force (Fp) Using equation (14.3), Fy= CpXAXpx U2 = 0.15 x2 1.15 x 13.89 Te ea Using equation (14.5), Faz Vf + Fe = 3328" +166404 (iv) The direction of resultant force (8) ‘The direction of resultant force is given by, _ Fr _ 166.38 oe aeUTS 9 =tan! 5 = 78.69". Ans. 2/2 = 33.28 N. Ans. (iii) Resultant force (F) 169.67 N. Ans. (v) Power exerted by air on the plate = Force in the direetion of motion x Velocity = Fy x UN mls = 33.280 x 13.89 W = 462.26 W. Ans. Power Watt = N m/s) Problem 14:3. Find the difference in drag force exerted on a flat plate of size 2 m xm when the plate is Iroving at a speed of 4 m/s normal ots plane in: (i) water, (i) air of density 1.24 Kee Co-efficient of drag is given as 1.15. Solution. Given: Area of plate, 2x2=4m Velocity of plate, 4mis =115 Co-efficient of drag, (@ Drag force when the plate is moving in water. Using equetion((43)) | p= Gye Ace a where p for water = 1000 z = 1.15 4% 1000 5 N=36800 0 (ii) Drag force when the plate is moving in air, 2 Fo=Cpx Axe, where p for air = 1.24 Wi) Fy = 1.15 x 4.0 1.24x i) - i) = 36800 ~ 45.6 = 36754. 4 N. Ans. Problem 14.4. A truck having a projected area of 6.5 square metres travelling at 70 km/hour has a total resistance of 2000 N. Of this 20 per cent is due to rolling friction and 10 per cent is due to surface friction. ‘The rest is due to form drag. Calculate the co-efficient of form drag. Take density of air = 1.25 kg/m’. Difference in drag force Solution. Given : Area of truck, A=65m U= 10 km/hr = F,=2000N 70100 60 x 60 = 19.44. m/s Speed of truck, Total resistance, Rolling friction resistance, Fc = 20% of total resistance = 2 x 2000 = 400 Nrion sistance, = 10% of fot tesisance = x 2000 = 200 N surf : orm dr Fy = 2000 ~ Fe — Fs = 2000 — 400 — 200 = 1400 N tying ewation (143) Fy itfp2 Form drag then Cp = Co-efficient of form drag rere if Fo e 19.44 1400 Co Ce a (p= Density of air = 1.25 kg/m’) eee D> BSwl as RIO Odd vee one problem 44.5 A circular disc 3 min diameter is held normal to a 26.4 m/s wind of density 0.0012 givec. Har force is required £0 hold it at rest? Assume co-efficient of drag of dise = 1.1. olution. Given Diameter of disc =3m roa , Area, at x (3)? = 7.0685 m’ Yelocity of wind, U=264 mis BE 00D eo. ee Density of wind, 9 = 0.0012 gem = 90> kel 00012 ys KB. : = Foo X10 as = 12 kel Co-efficient of drag, Cpe ll coorce required to hold the dise at rest i equal tothe drag exerted by wind on the disc. Drag (Fp) is given by equation (14.3) as 2 2 Fy=Cpxax ee 1x 7.0685 12264" _ 3951.41, Ans. 2 20 |g Aman weighing 90 kgf descends 10 the ground from an aeroplane with the help ofa ce of air. The velocity with which the parachute, which is hemispherical in ‘Find the diameter of the parachute. Assume Cp = 0.5 and density of air Problem 14. parachute against the resistant ‘shape, comes down is 20 m/s. g/m’ Fy=90 kgf Solution. Given : Weight of man, W=90 kgf = 90x 9.81 N= 882.9 N (2 Lkgf=9.81.N) Velocity of parachute, U= 20 mvs Co-efficient of drag, Cp=05 Density of air, p= 1.25 kg/m* Let the diameter of parachute =D. Area, Ask Dim. ‘omes down with a uniform velocity, fal to the weight of man, neglecting the the hemispherical parachute will be When the parachute with the man ci U=20 m/s, the drag resistance will be equ weight of parachute, And projected area of equal to ™ D, 4 Fy = 90 kgf = 90 x 981 = 882.9N 2 Using equation (14.3), Fy=Cpxax Pe Drag,[642 Fluid Mechantes 882. 882.9 4x 2.0, Y= 05x mx 125% 20x20 or D = V89946 = 2.999 m. Ans. Problem 14.7. A man weighing 981 N descends to the ground from an aeroplane with the help of @ parachute against the resistance of air. The shape of the parachute is hemispherical of 2 m diameter. Find the velocity of the parachute with which it comes down. Assume Cg = 0.5 and p for air = 0.00125 gm/ce and = 8.9946 m? 0.015 stoke. Solution. Weight of the man, Drag force, Diameter of the parachute, ven, W=981N Projected area, Co-e t of drag, 0.00125 4 1000 o Ke 3 x 10° = 1.25 kg/m .00125 gm/er 0.00125 1000 Density for Let the velocity of parachute Using equation (14.3), Fy =CpXAX a ee ORI Oper 125% = | 81X20 31.61 mis. Ans. x mx 125 Problem 14.8 A man descends to the ground from an aeroplane with the help of a parachute which is hemispherical having a diameter of 4 m against the resistance of air with a uniform velocity of 25 m/s. Find the weight of the man if the weight of parachute is 9.81 N. Take Cy = 0.6 and density of air = 1.25 kg/m’. Solution. Given: Diameter of parachute, D=4m Projected area, A= a xDP= = x4? = 4mm? Velocity of parachute, U=25 mis Weight of parachute, Co-efficient of drag, Density of air, Let the weight of man ; Then weight of man + Weight of parachute = W> + W, = (W, +9.81) Hence drag force will be equal to the weight of man plus weight of parachute. © Drag force, Fy =(W2 +981) Using equation (14.3), we have Fi = Cp x AX me .25x 257 or (W, + 9.81) = 0.6x4nx ——* 2945.24 N 20. 1945.24 ~ 9.81 = 2935.43 N. Ans,t oneean Problem 14.9 Calculate 1p, that the maximum terminal ve| be treated as hemispherical 1g Solution. Given Mass of object, ¢ diameter o| locity of 1) 4 Parachute 19 dropping is. 5 pe ve used for drop bect of freee tt. The trae ct of mass re Mensity of air is 1 216 bees etlcient for the parachuie wo Weight of obje OO ke Drag force E 981N Velocity of object, Drag co-effi Density of ai Pp Let the diameter of parachute’ = Projected area, As es Dm Using equation (14.3), p= Cp x A x 2U* 2 a 8151.3 x Epty 1216 x5? : pi Pee Sica Z 13Xmx1216xsx5 = 63.21 D= 6321 =7.95 m, Ane. #8 3.924 N assumes an angle of 12°19 the horizontal. The string horizontal. The pull on the ‘String is 24.525 N when the wind is @ffcient of drag and lift. Density of air is siven atiached 10 the kite makes an angle of 45°16 the flowing ata speed of 30 km/hour. Find the corresponding co- as 1.25 kg/m Solution. Given: Area of kite, A=08 x 0.8 = 0.64 m? Weight of kite, W=3.924N Angle made by kite with horizontal, 6, = 12° Angle made by string with horizontal, , Pull on the string, P=24.525N Speed of wind, U = 30 km/hr = exten m/s = 8.333 mis * Density of air, P= 1.25 ke/m? Kite may be assumed to be a flat plate. ora Drag force, Fp = Porce exerted by wind in the direction of motion (ie, in the X-X direction) e ae And lift force, F = Force exerted by wind on the kite perpendicular to the direction of motion (ée., along ¥-¥ direction) ; The kite is in equilibrium and, therefore, lift and drag must be balanced by the string tension and weight of the kite as shown in Fig. 14.4. aoe Resolving forces into horizontal aie i ea * 224.525 x 00s 45° =17.34N =0,F,=P sin 45° + W 24.525 sin 45° + 3.924 = 17.34 43.924 = 21.264NFig. 144 (i) Drag Co-efficient (Cp). Using equation (14.3), we have 2x Fp 2x 1754 = 0.624, Ans, “Ap? 064 x1.25%8333° efficient (C,)s Using equation (14.4), we have : put cxaxee (i) Litt Co- 221264 “AxpxU? 0,64 x 1.258. Notes. 1 Here full area ofthe kite has been used for calculating Cp and Cy, 2 Here full area of kite has been used in calculating co-efficient of drag and lift beeause the projected dea varies with angle made by kite with horizontal. Ths is also the accepted practice in the case of aero or Gs 0.768. Ans. foils. Problem 14.11 A kite weighing 7.848 N has an effective area of 0.8 m?, It is maintained in air at an angle of 10° to the horizontal. The string attac shed to the kite makes an angle of 45° to the horizontal and at this position the value of co-efficient of drag and lift are 0.6 and 0.8 respectively. Find the speed of the wind and the tension in the string’ Take the density of air as 1.25 kg/m’. Solution. Given: Weight of kite, 848 N Effective area, 1.8m ‘Angle made by kite with horizontal = 10° ‘Angle made by string with horizontal = 45° Density of air, Let the speed of the wind and tension in string The free body diagram for the kite is shown in Fig. 14.5. ‘The kite is in equilibrium and, therefore, the lift and drag must be balanced by the string tension and weight, of kite. Resolving forces into horizontal and vertical components, Fig. 145Component of T along X-X = Tcos 45° 0 Component of 7 vertically downward + Weight of kite n 45° 47.848 Mii) and But drag force Fy is also 13 UP (iti) pxAx PU" _06x08x125xU" 2 2 F, = CLXAXPU? _ 08x 08% 1250" 5 2 20, Equating the two values of Fp, given by equations (i) and (ii) Teos 45° = 0,3 U2 ) Bquating the two values of F, given by equations (if) and (it) T sin 45° + 7848 = 04 U2 = sin 45° = 04076 U>_ 7,848 (vi) From equations (v) and (0) we get, as T cos 45° = Tsin 45°, 0.3 U? =0.4 U>~ 7.848 fe 7.848 = 0.4 U?—03 U’ 2 1848 Also lift force, aie Civ) On 8.48 RES pre SOOO re ie Substituting the value of U? = 78.48 given by equation (iii) into equation (i), we get T cos 45° = 0,3 x 78.48 = 23.544 SH -333N. aes ate Problem 14.12 The air is flowing over a cylinder of diameter 50 mm and infinite length with a velocity of 0.1 m/s. Find the total drag, shear drag and pressure drag on I'm length of the cylinder if the total drag co-efficient is equal to 1.5 and shear drag co-efficient equal to 0.2. Take density of air = 1.25 kg/cm’. Solution. Given : Diameter of cylinder, Length of cylinder, Projected Area, Velocity of air, Total drag co-efficient, Shear drag co-efficient, Density of a Total drag is given by, (0.1)? =15% 05% 25500" .000468 N. Ans. a put Shear drag is given by, Fps= Cys Ax P 25% (0)? = 0.2.x.05 x 2% (00 0000625 N. Ans. From equation (14.1), Total drag, fpr = Pressure drag + Shear drag = Total drag ~ Shear drag Pressure drag, = 0,000468 ~ 0.000625646 Fluid Mechanics as a projected area 1.5m normal to the d Problem 14.13. A body of length 2.0 m has a projected area 1.5m? normal to th direction of, The body is moving through water, which is having viscosity = 0.01 poise. Find the drag on the bo, " mon, a drag co-efficient 0.5 for a Reynold number of 8 X 10°. dy ify 7 Solution. Given Length of body, Projected Area, OUI 10 Viscosity of water, 11 = 0.01 poi a Drag co-effici C,=0.5 Reynold number, R,=8x 10° Let the drag force on body > 1 the velocity with which body is moving in water. Itis calculated from the given Reynoig uy Ibe, irstf R= aE where p for water = 1000 1000 U x20 or = ———___— = 2x 10° U Bad 0.001 < B2c0Ssan arn 5 v= 5x 10° vu 4 Using equation (14.3), Fy = Cy xX AX ae = 05 x 1.5 1000“ Problem 14.14 A sub-marine which may be supposed to approximate a cylinder 4 m in diameter andy long travels submerged a 1.3 mds in sea-water. Find the drag exerted om it ifthe drag co-efficient for Rew number greater than 10° may be taken as 0.75. The density of sea-water is given as 1035 ke/nt and kinemayie = 6049 N. Ans, viscosity as 015 stokes. Solution. Given Dia. of cylinder, D=4m Length of cylinder, L=20m Velocity of cylinder, U=13 mls Density of sea-water, p = 1035 kg/m? Kinematic viscosity V =0,015 stokes = .015 cm*/s = .015 x 104 m/s Let the drag force =i, UxD Reynold number, R= 3.466 x 108 Since R, > 10°, hence Cp=0.75 Drag force is given by equation (14.3) as where A = projected area of cylinder = L x D = 20 x 4.0 = 80.0 m? 2 ‘ Fy =0.75 80x 1521S" po = 52472.2 N. Ans. Problem 14.15 A jet plane which weighs 29.43 kN and having a wing area of 20 m flies at a velocity 950 km/hour, when the engine delivers 7357.5 kW power. 65% of the power is used to overcome the dras resistance of the wing. Calculate the co-efficients of lift and drag for the wing. The density of the atmospheri air is 1.21 kg/m’.gotution. ciiven weight of plane, W = 29.43 KN = 29.43 1000 N = 29430 N wing are A= 20m? 950 x 1000 of plane, Us a speed of plane 950 km/hr 569 2653:88 mis fingine powers P= 7357.5 kW rower used 10 overcome drag resistance = 65% of 7357.5 = 2 pow 199 % 7357.5 = 4782375 kW 21 kg/m’ efficient of drag and C, = Co-efficient of lift. Density of at, p= OF Let Now power used in kW to overcome drug resistance = 2*Y. oy 4749 475 Lo % 26388 1000 7, ww 4782375 x 1000 Dy é 263.88 2 put from equation (14,3), we have Fy= Cp .A, LUO 2 4782.375 x 1000 263.88? pesgg COX 20% 1.21 x SE Cy = 4782375 1000 x2 Te CALA Ara "20 xi2ix2638@ = 040215: Ans the lift force should be equal to weight of the plane F F, = W=29430N a 2 But Fy2c,.A. Pe or 29430 = C20 1.21 x 2258 . 29430x2 9 = ar = 0.0349, +" 29x 12126388 = 084 Ams 14.3.2 Pressure Drag and Friction Drag. ‘The total drag on a body is given by equation (14.1) as Total drag, Fy = p cos OdA + J t sin OdA A) where {pcos OdA = Pressure drag or form drag, and J 1 sin 0dA = Friction drag or skin drag or shear drag The relative contribution of the pressure drag and friction drag to the total drag depends on = (i) Shape of the immersed body, (ii) Position of the body immersed in the fluid, and (iii) Fluid characteristics. Consider the flow of a fluid over a flat plate when the plate is placed parallel to the direction of the flow as shown in Fig. 14.6. In this cos 0, which is the angle made by pressure with the direction motion, will be 90°. Thus the term J p cos OdA will be zero and hence total drag will be equal to friction drag (or shear drag). If the plate is placed perpendicular to the flow as shown in Fig. 14.7, the angle 0, made by the pressure with the direction of motion will be zero. Hence the term J ty sin @dA will become equal to zero and hence total drag will be due to the pressure difference between the upstream and downstream side of the plate. If the plate 9 isheld at an angle with the direction of flow, both the —§ ————+ ee eee terms J p cos @dA and f ty sin @dA will exist and total =§ ————> eS drag will be equal to the sum of pressure drag and = ———— friction drag Fig. 146 Flat plate parallel to flow.oe ‘SEPARATION POINT Sy 48 Fluid Mechanics au PLATE DOWNSTREAW, rae ae . WAKE sig. 147 Palate perpendicular to flow. A strefined body is defined a that body whose surface yo ln that cas the separation of Ow wl ae nthe boundary ayer will start at tela 14.3.3 Stream-lined Body. the streamlines, when the body is place wart of the body). Though the trailing edge (or rearmost p' ‘ome turbulent from laminar, yet it does not separate jined body. Thus behind. astream-lined body, wé the pressure drag will be very small, Then the total drag on the (shear) only. A body may be stream-lined at low velocities but may not be so at higher velocities. 2. when placed in a particular position in the flow but may not be so when placed in another pi 'A bluff body is defined as that body whose surface does mot coincide aflow. Then the flow is separated from the surface of the body mucha trailing edge with the result of avery lage wake formation 206, “Then the drag due to pressure wil Tage as compared tothe drag due to fiction onthe Body. Thus the bodies of such a shape in i pressure dea is very large as compared to fiction drag are called bluff bodies. Ply, ¢ upto the fearmost part of the body ing formation zone will be very small ang a 1e stream-lined body will veda stream 14.3.4 Bluff Bod: streamlines, when placed in > 14.4 DRAG ON A SPHERE Consider the flow of areal fluid past a sphere. U= Velocity of the flow of fluid over sphere, D=Diameter of sphere, p =Mass density of fluid, and = Viscosity of fluid Let ae <0}, the viscous ra ; asin this case viscous forces are much more predomi the inertial forces, which may be assumed negligible. G.G. Stokes, developed a mathematical equtotl total drag on a sphere immersed in a flowing fluid for which Reynolds number is upto. 0.2, so th forces may be assumed negligible. According to his solution, total drag is. Fy = 3miDU. 5 He further observed that out of the total drag given by equation (14.8), two-third is contribuel friction and the remaining one-third by pressure difference. Thus, Ifthe Reynolds number of the flow is very small less than 0.2 (: eR, much more important than the inertial forces ‘Skin friction drag, Feats 2 3m.DU = 2nuDU es e il and pressure drag, Fy, = 3 Fo 4 x 3mDU = DU. = (i) Expression of Cy for Sphere when Reynolds Number is less tham 0.2. From equation ce ae drag is given by2 Fy CoxAx PE Fy =3muDU A = Projected area of the sphere = For sphere, 3muDU Cox % pp Bue 2 SSO a 2a G Ea ~1499( sta") qx Equation (14.9) is called ‘Stoke's law’, (ii) Value of Cp for Sphere when R, is between 0.2 and 5. With the increase of Reynolds number, the inertia forces increase and must be taken into account. When R, lies between 0.2 and 5, Oseen, a Swedish physicist, improved Stoke’s law as 2 -2(14.10) Equation (14.10) is called Oseen formulae and is valid for R, between 0.2 and 5. (ii) Value of Cp for R, from 5.0 to 1000. The drag co-efficient for the Reynolds number from 5 to 1000 is equal to 0.4 (j) Value of Cp for R, from 1000 to 100,000. In this range, C,, is independent of the Reynolds number and its value is approximately equal to 0.5. (v) Value of Cy for R, more than 10°. The value of C, is approximately equal to 0.2 for the Reynolds number more than 10°. Problem 14.16 Calculate the weight of a ball of diameter 80 mm which is just supported in a vertical air stream which is flowing at a velocity of 7 m/s. The density of air is given as 1.25 kg/m’. The kinematic viscosity of air = 1.5 stokes. Solution. Given : Dia. of ball, D=80 mm = 0.08 m Velocity of air, U=7 mis Density of air, p= 1.25 kg/m* 5 stokes = 1.5 x 10% m/s UxD__17x.08 Se Ven LS ge lORs ‘Thus the value of R, lies between 1000 and 100,000 and hence Cy =0.5. When the ball is supported in a vertical air stream, the weight of ball is ‘equal to the drag force as shown in Fig. 14.8. Kinematic viscosity, Reynold number, R, = 0.373 x 10° = 3730. 2 But drag force, Fy= Coxax oo where A = projected area of ball male rw) Rio _k 2 2 aes 4B? =F (0.08) = 0.005026 mi =008m 7 Drag force, Fy = 0.5 x 0.005026 x Pare .07696 N Weight of ball "p = 0.07696 N. Ans. Fig. 14.8Lose fluid Mechanics ay > 14.5 TERMINAL VELOCITY OF A BODY Terminal velocity is defined as the maximum constant velocity of a falling body (such ay phe Composite body such as parachute together with man) with which the body will be travelling. Whenga ra 's allowed to fal from rest in the atmosphere, the velocity ofthe body increases due to acceleration ype® My With the inerease of the velocity, the drag force, opposing the motion of body also increases Bi reached when the upward drag force acting on the body will be equal to the weight of the body Then gee extemal free acting on the body willbe zero and the body will be traveling at Constant speed. Ths ca speed is called terminal velocity of the falling body. — IE the body drops in aud, at the instant it has acquired terminal velocity the net force acting on ty ied will be zero, The forces acting on the body at this state will be ty 1. Weight of body (WW), acting downward, 2. Drag force (F jing vertically upward, and 3. Buoyant force (F,), acting vertically up. ‘The net force on the body should be zero, i.e., W= Fot+Fg Problem 14.17 A metallic sphere of sp. gr 7.0 falls in an oil of density 800 ke/m?. The diameter of te sphere ts 8 mm and it attains a terminal velocity of 40 mm/s. Find the viscosity of the oil in ‘poise. Solution. Given Sp. gr. of metallic sphere 0 Density of metallic sphere,p, = 7 x 1000 = 7000 kg/m? Density of oil, ~(4.1) Dia. of sphere, Terminal velocity, Let the viscosity of oil Weight of sphere, W =p, xg x Volume of sphere =7000%9.81 x * D? = 1000 x 9.81 x A x (8x 107) = 18.4 x 10° = 0.0184 N Buoyant force on sphere, Fy = Density of oil x g x Volume of sphere 00 9.81 x F (8x 10)" N= 0,002103 N. Drag force, F, on the sphere is given by equation (14.8) as Fp = 3m.DU = 3p x 8 x 10 x .04 = .003015 1 Using equation (14.11), W= Fy + Fy or 0.0184 = .003015 41 + 0.002103 or (003015 jx = 0.0184 — 0.002103 = 0.016297 0.016297 or = aie - 4 # 0.003015. m 4X 10 = 54.0 poise. Ans. ‘The expression for drag given by equation (14.11) is valid only upto Reynolds number less than 0.2. Hence itis necessary to calculate Reynold number for the flow. uD Reynold number, y= rad where p for oil = 800 kg/m? 04x 8x10? R= 800 x EX = 0474 54 Hence R, < 0.2 and so the expression Fr = 3muDU is valid.problem 14.18 A sphericay steel batt op Ee Tatiawaey ihe bale L228 of eb pee ve Kgl and Cy a G4 Peet ain, Fina the increase in ten, 8 O45. Fi Solution. Given Diameter of steel ball, Density of ball, Cp for ball in water D=40 ™M = 0.04 m Ps= 8500 kgm? 0.45 =U, ‘al ball are t We Density Of ball x. &* Volur Let the terminal velocity in water ‘The forces acting on the spheric, 1, Weight, 2. Buoyant force, = 1000 x 9,81 x. e (0.04)? = 0.3086, 3. Drag force, =Cyx Ax DU? 2 See an where A = projected area = ™ py 4 694)", 9 = 1000 for water 2 Fp =0.45 x fe X (0.04)? x 1000 Ui : Using equation (14.11), we get w= Fp+F, i 2.794 = 0.2825 U)? +0.3286 2.794 ~ 0.3086 uP? = e zi O85 a aes U, = V8725 = 2.953 mis. Ans, When ball is dropped in air. Let the terminal velocity = U, Weight, W=2.794 | _Buoyant force, Fy= Density of air x g x Volume of ball = 1259.81 x © (0.08) = 0.000811 N 2 Drag force, Fo=Cyxax Pu where p for air = 1.25 2 Fp =0.1 x © (04)? x 1.25 =0.0000785 Up. S 4 The buoyant force in air is 0.000411, while weight of the ball is 2.794 N. Hence buoyant force is negligible. For equilibrium of the ball in air, Fy = Weight of ball 2.794 eo 0.0000785 U,? = 2.794 or Up = a = = 185.717 m/s. Ans. Increase in terminal velocity in air = U; ~ Uj = 188.67 2.9533 188.67 m/seae Problem 14.19 A metallic ball of diameter2 x 10 m drops in a fluid of sp. The density of the metallic ball is 12000 kg/m’. Find () The drag force exerted by fluid on metallic ball, (i) The pressure drag and skin friction drag, (41) The terminal velocity of ball in fluid. Solution, Giv n Diameter of metallic ball, D =2x 10 m Sp. gr. of fluid, Sy = 0.95 Density of fluid, Po = 0.95 x 1000 = 950 kg/m? Viscosity of fluid, H = 15 poise 2 1535 Density of ball, P, = 12000 kg/m* The forces acting on the ball are Weight of ball, W = Density of ball x g x Volume of ball = 12000 x 9.81 xt D = 12000 x 9.81 xE xx 10°)3 N = 0.000493 N Buoyant force Fy = Density of fluid x g x Volume of ball = 950 x 9.81 x . (2x 10)? N = 0.000039 N When the metalli w .000454 N. Ans. .000454 N (i) Drag force, om (i) Pressure drag 2 Skin friction drag My (iii) Let the terminal velocity = U Then drag force (F'p) is given by equation (14.8) as F, But Fp = 0.000454 N Equating the two values of Fp, we have 3muDU: 3mUDU = 0.000454 or sax x2x 107 x U= 0.000454 10 x 0.000454 or U= 0.016 m/s. Ans. 3nx15x2x10> Let us check for Reynold number, R, UD R, am + Where p = 950 kg/m? 0.016 x2 x10 7 Ls =950 x = 0.02 Hence the Reynolds number is less than 0.2 and so the expression F;, = 3m.DU for calculating termini velocity is valid. 87. 0.95 and vis, Cosity 15 ball reaches the terminal velocity, equation (14.11) is applicable D+F, or Fp=W— Fy =0.000493 — 0.000039 * 0.000454 = 0.0001513 N. Ans. x 0.000454 = 0.0003028 N. Ans.— 7m 14.20 Determine the ver See proble he velocity o on Submerged Bod sere on d eae acsae mer les 653 | and kinematic viscosity 0.15 eme/, fat Frain drops solution. Given; oF x10 om dicy “ameter density 0.0012 amen Diameter of rain drops, =, Volu +* Volume of particle Density of rain drops, : Kinematic viscosity, Using the relation, wae pO ONG ee 0.0012 m = 0-15 x.0012 = 0.00913 —gm_ Now weight of rain drop =PX8x Volume ofr; Ae in drop =pxexE ps Seer nee 6 (Rain drop is a sphere) Drag foree, Fp, on rain drop is given by equation (14,8) When rain drop is falling with a uniform velocity U, the d Hence equating te two values, we get vine Grag force must be equal to the weight of rain drop, Weight of rain drop = Drag force X ™ Pxgx= xp? or PX 8x — D> =3mupu or. Us 2 age pet 3m 18 = 2.0012 981 (3010-3)? TS ROONTH = 0.327 emis. Ans. = UD _0327x30x1073 oe v Osis = As the Reynolds number is less than 02, the expression Fy = Smu.DU is vai. Problem 14.20 (a) Determine: (i) terminal velocity, (it) Reynolds number corresponding to terminal yelocity when a particle of diameter I mm and of density 1 Hi m 100 kg/nt! falls freely from rest in an oil of density 900 kgint’ and of viscosity 0.03 N s/m?., Find also the time taken by the particle to attain 99% of terminal 1.0654 Solution. Given: Diameter of particle, d= 1 mm = 1x 10° m Density of particle, p, = 1100 ke/m* Density of oil, 5 = 900 kg/m? Viscosity of oil, 1 = 0,03 Ns/m? Stokes' law is applicable, Find : (i) Terminal velocity of particle, (ii) Reynolds number corresponding to the terminal velocity, and (iii) Time taken by the particle to attain in a velocity of 99% of terminal velocity, ()) Terminal velocity: Let U = terminal velocityPenh nmin anemia A ey de he 4 «OTe she oon a a ge i om Pt Fe Wh pd Ses, a Toelanbee N98 E eases Ed eaenene wet s980. sinaraie 3000 ya 98x02 yp SAM 66689 1 a Sore era at tne epee ‘Cinewnathperenemee ont pi sriahibenmansane seme ~iahirerncantnetoam ey ventana eg + aad Mean vegck ot 2 NH jecameay oe nate sen yeaa TST 9sbem pa Nn é aw -woay) “aaa oping te sonora. et Tay ‘Tenia wl 6 9 i ely (099a34he ey Po toncanorm oo[656 Fluid Mechanics " 1 og, (1.783 ~ 490.9 VI hs yoo x ait sce 1 fog, (1.783 ~ 490.9 3.596 x 10°] — 1 5 (loge 4905 "8 Lag [log, (1.783 ~ 1.765) —log, 1.783] 309 | fog, 0.018 log, 1.783] 490.9 t a | —|__ jog, (99.05) zo05 | (are | 4909 = | 4.505 = 0.00936 seconds. Ans. 490.9 RAG ON A CYLINDER le aS real fluid flowing over al circular cylinder of diameter Pig oe When the oyn 4 aced in the fluid such that its length is perpendicular to the direction ‘eynolds Dumberat ge : f Uxd flow is less than 0.2 { ie, — saa er wil be symmetrical. As the Reynolds number is increase ae a a ae ain Secreaieh eee areca aca over cylinder. Wine | increase of the Reynolds number, the flow pattern becomes unsymmetrical with respect 4p a perpendicular to the direction of flow. The drag force, ie, the force exerted by the flowing fig a cylinder in the direction of flow depends upon the Reynolds number of the flow. From exper been observed that : (®) When Reynolds number (R,) < 1, the drag force is directly proportional to velocity and co-efficient (Cp) is inversely proportional to Reynolds number. : (a) With the inteae ofthe Reynolds umber from 102000, the drag co-efficient decreases and 4 minimum value of 0.95 at R, = 2000. (iii) With the further increase of the Reynolds number from 2000 to 3 x 10°, the co-efficient of dog increases and attains maximum value of 1.2 at R, = 3 x 10%. (iv) The value of co-efficient of drag decreases if the Reynolds number is increased from 3 x 10*to3 xi AtR, =3x 10° the value of Cy = 0.3. (») Ifthe Reynolds number is increased beyond 3 x 10°, the value of C, increases and it becomes equalio 0.7 in the end. < 02), the neta ere eaiebly smal s compared Viscous yg iments ity NCE the dag > 14.7 DEVELOPMENT OF LIFT ON A CIRCULAR CYLINDER ‘When « body is placed ina fluid in such a way that its axis is parallel to the direction of fluid flow and bed 's symmetrical, the resultant force acting on the body is inthe direction of flow. There is no force ‘component on the body perpendicular to the direction of flow. But the component to the force on the body perpendicular to the direction of flow, is known as ‘Lift. Hence in this case lift will be zero, {he lift willbe acting on the body when the axis ofthe symmetrical body is inclined tothe direction off or body is unsymmetrical. In the case of circular cylinder, the body is symmetrical and the axis is parallel the direction of flow when cylinder s stationary. Hence the lift will be zero, But if the cylinder is rotated te4 lectin Se ae a ov eo xt, where velocity is zero. For a rotating cylinder as Snow” cat points on the surface of the cylinder sats a7 hg resultant velocity is given by equation (14.15) as mig. 12> i: u=2Usino+ 2nR ror stagnation points u=0 es J sin @ + —— = 0 or 2U sin® 2U sin O + si nose oe 14.19) ov 4nUR -qhe solution of equation (14.19) gives the location of stagnation points on the surface of the cylinder. ‘There -vrjues of 8, which satisfy equation (14.19). As sin @ is negative in equation (14-19), it means @ is more 2H ess than 360°. The two values of @ are such that one value is between 180° and 270° and other than Ts between 270° and 360°. value Ofpale stagnation point, ©= 270° and then equation (14.19) becomes as sin 270° = - —— or c ( sin270°=- 1) 4nUR 4nUR (14.20) 3 P= 4nUR. ac1.7 Magnus Effect. When a cylinder is rotated in a uniform flow, a lift force is produced on the i menon of the lift force produced by a rotating cylinder in a uniform flow is known as cylinder. This pheno kno Stagnus Effect. Ths fact was investigated by a German physicist H.G. Magnus and hence the name is 8 Magnus Effect. problem 14.21. A cylinder rotates at 150 rp.m. with its axis perpendicular in an air streans which is prone uniform velocity of 25 més. The cylinder is 1.5.m in diameter and 10 m long. Assuming ideal fluid vars find (i) the circulation, (i) lift force, and (iii) position of stagnation points. Take density of air as 1.25 kg/m. iven as Solution. Given: 2 Speed of cylinder, Velocity of air, Diameter of cylinder, N=150rp.m. Radius of cylinder, s; Length of cylinder, Density of air, Tangential velocity of cylinder is en as ee eae EDN _ RXIOXO* = 11.78 mis. Coe ry (i) Circulation (I) is obtained from equation (14.14), as ug = aR T= 2nR X up = 2X 0.75 x 11.78 = 55.51 m/s. Ans. (ii) Lift foree, Fis given by equation (14.16) as F, LUT = 1.25 x 10 x 25 x 55.51 17344.N. Ans. (iii) Position of stagnation points are given by equation (14.19) as P 5551 4nUR ”~ 4nx2sx07s °° sin =oa hag Sctstion ce Sesser ma Am Thee it per ei eg = = 995.77 Nm. Ans. os exmx07 sear 07 ss 45175 36" (287 anda 25) ho 2 128 a 3835-Am. Al Org an act Rt Foe Now aa toc peaby Fux! pte, 2 + 1x2@ximaPsaee nein beim’ For acta drop and if. sake ie : eS nde hich lg ny ig he s rs = S gett atin ng, Serer tame ta Peden tesa0 1281s volte ee, BD gry STA asap Am cy i. fm ene ut ft ink rer ra ede ene Sing miei rm fn ne en nd oa ‘Sra an ar ne tte PR kd ———— ee acon ee tment, ea ani i iniatign? Foe 40. wee tw pt 0 -tasea POs {ee rt oreo min 36 Wied ae "thn sana po ey epi 9) sem Am—————S— Ole _ 93643 =—sin 21.36" : Fin (190° + 21.36") and sit [360° - 21.36°) 901.36° and sin. 338.64" sin 201.36" and 338.64". Ans: ee 664 Fluid Mechanics fs wou 6 NT OF LIFT ON AI » 14.8 DEVELOPME! Fig, 14.15 shows the two shapes of ie a fines bodies which | =F airfoils, which are stream may be symmetrical oF unsymmetrical i a Shapes. The airfoil is characterized by its chord symerrICA fength C, angle of attack a (which is the angle ‘ARFOR between the direction of the fluid flowing and sup chord line) and span Z of thegirfll, TRE lift on = ure created on von tne ee the airfoil is due to negative pre the upper part of the airfoil, The drag force on the airfoil is always small due'to the deste of the shape of the body, which is stream-ined. From the theoretical analysis, the circulation F developed on the edge ofthe airfoil is tangential to the a oil, is given as T=nCU sino where C= Chord length, U = Free stream velocity 0 Lift force Fis given by equation (14.16) as pULT = pUL x nCU sin & noe @ ig. 14.15 Shapes of airfoils, airfoil so that the stream-line atthe ti ily (14) tack airfoil, 0. = Angle of = mC The lift force is also given by equation (14.4), as p= C,xAxee where C, = Co-efficient of lift Zs rojected area = CX L for airfoil A pu? a ae (042) Equating the two value of lift force given by equations (14.22) and (14.23), we get 2 mpCvtt sinaa=C,xCxLx 2 \ 2 2npCU*Lsin & sin (49 aeiascrepUe ‘Thus its clear fron tus itis clear from equation (14.23), that co-efficient of lit depends upon the angle of attack. 14.8.1 Ste ro! in ; Hecan oe ofa Flying Object. When a flying object for example airplane isi iirplane is equal to the lift force and thrust devek i . cae pare leveloped by the engine is equal tothe drag ost pAU? (0429 W= Liflt force = C, pAU? Probl 5 roblem 14.24 Anairfoil of chord I of chord length 2m moving with a a eat a Sera < given as 0.03 and 0.5 re he values of co-efficient of drag and lift weight of the airfoil and 5 respectively ig and lift corresponding to angle of ack where W = Weight of the airplane and C, force. L rolution. Given : Chord length span of airfoil, “Angle of attack, Yelocty of airfoil, 0 rive Density of air, 25k Co-fficient of drag, ‘con Coelficient of lift, 50 From equation (14.25), we know that Weight of airfoil = Lift forces ¢ PAU? 5G, hae BOS (as 7), a0: = 050% 1.25% seta), ROE Now drag force, Fox Gy xax a $0008 x 1.25 218) x80? 2 iced et Foxu — x :. Power required in kW = BAT _ 3600x 80 1000 "1009 =288 KW. Ans, Problem 14.25 A jet plane which weighs 29439 n 250 Wis the engine delivers 7357.5 kW. 63% ae a me open ata velocity of the wing. Calculate the co-efficient of lift and cor gp ome cometh dag resistance of 121 kent, ‘-effcient of drag for the wing. Take density of air equal 10 Solution. Given : Weight of plane, Wing area, Velocity of plane, Power delivered by engine = 7357.5 kW Power required to overcome drag resistance 65% of 7357.5 = 0.65 x 7357.5 = 4782.375 kW. Density of air, 21 kg/m? 2 Now, weight of plane G,xAx oe 2 29430 = C, x 20x 1.21 x 244 2 Seer Ae 20% 1.21x 69.44 Let Fy = Drag force FyxU kw Power required to overcome drag resistance = —!o-— x 69.44 4782,315 = 2X 4782.375 x 1000 _ 699706 N Fo= 69.44[.666__ Fiuld Mechani* i ieee Now drag force, 121 69.447 EgR70IG = Op x 20 mero 70.6 * 2 20 x 121 x 69.44 18. Ans. TON A FLAT PLATE tigeussed in Art. 143.2 030. A thin flat plate may be lag Hae own in Fig. 14.16 (a) and et ) > 14.9 DRAG AND LIF direction of flow as shi te is briefly d Drag and lift on a flat plat to the direction of flow and perpendicular to the Oa p. aan Direction —> oto of low u Thin flat Plate UThin tat plate (b) Plate is 1 to direction of flow (a) Plate is parallel to direction of flow Fig. 14.16 14.9.1 Plate is Parallel to Direction of Flow. [Refer to Fig 14.17 (q). The pressure acting on upperan lower faces will make an angle of 90° with direction of flow. Pressure) Shen stu Hence here 6 = 90°. ae fy (@) The drag force (Fp) acting on upper face is given by —> eae equation (14.1) as ae U i Fy = [pcos @dA+ fr, sin 0dA ovat | Fig. 14.17 (a) ‘As @ = 90° hence cos @ = 0 and sin @ = 1. The above equation becomes as Fy= Jrxoxda+ fro x1xda = fro xdA only due to shear stress acting on the surface of the plate. If ty is constant over Hence, drag force is surface, then Fp = Ty x Area of plate =) xb x L where b = Width and L = Length of plate. (ii) The lift force (F,) is given by equation (14.2) as F,= Jroc0s 0 dA fp sin aA As @ = 90°, cos 0 = 0 and sin @ = 1 F,= fo x0xdA-— [px1xdA = [Sign —ve is due to pressure acting downwards on upper face of plate]. Jpxaa© Hence, lift force is only pressure forees are equal Hence, there will be no lift Force inthis as ase 14.9.2 Plate is Perpendicular to face and back face is actin, | zero. Hence = 0. Dire, ig in the directi () The drag force (Fp) acting on the f (al) as Font face is given bY equation i= Jrcos 84+ fr sin oun AsO=O,bence cos 8= 1 and sin 99,7, *0. The above equation bec, Somes as Fo= Joxixaay Ftox0xda Jraa+o= Joan iE pressure distribution on front face is kn own. Fig. 14.17 (b) Similaty, drag fore F,*for back face win oo Gras force canbe obtaineg, M7) F,* Jos aa where p* is pressure on back face Pressure p* will be nepaMive Hence ii wf \ormal ? ; ting ne (G) The lift foree (Fis given by equation (14.2) = Oe as FL= ft cos @aa— Jrsine aa =0, hence cos @ = 1 and sin 8 = 0. Now, the above equation becomes, FL= Jox1xdA-[pxoxaa = fro aa ‘he shear stress distribution onthe upper half and lower half ofthe front pposit steat forces on the front face will be zero, Also: the Shear MEAT eal and opposite, Hence Sure distribution on the upper half and lower half of back fc will be equal and opposite. Hi lence shear forces on back face will be also zero. Hence lift force will be zero, The lift force is zero on the plate, when itis placed parallel to the direction of flow or perpendicular to direction of flow. ‘The lft will be created if plate is placed inclined to direction of flow as shown in Fig. 14.17 (c). In this case, there would be both a lift and a drag, each of which would be dependent on both shear stress and pressure. Both the shear stress and Pressure distribution would be different for top and bottom Fig. 14.17 (©) surfaces. Problem 14.26 Find the drag and lift forces exerted by air on aflat plate of length 1.2 m and of width 2 m when the plate is placed (i) parallel to the direction of flow and (ii) Perpendicular tothe direction offow. The prepure and shear stress distributions on the surfaces of the plate are shown in Fig. 14.18. The aris lowing with a velocity of 8 mis. Solution, Given: Width of plate, = 2m12m Length of Pl velocity of a aN més p=0 9000) araffel to the di and botton Ist Case : Plate is Pa distribution on (OP p= 1 10 Pressure Shear stress distribution O Drag Foree ‘ e side of the plate is given by equation (14.1) as ‘The drag force (Fp) o” On Foe Jpcosoaa + fo sin OdA But 6 (the angle made by pressure with direction of flow) ence, above equation becomes 2 Fiy= Jpx0xda+ froxixda =0+ froxixaa frou Diag force on both sides (top and bottom) will be. wl =2Fp=2 froas je not constant. It varies with x. To find integral of to dA, €0 sas shown in Fig. 14.18 (D) ) is equal to 90°. Hence, £08 0 Damaiagag sider an elemeatl In the above equation % thickness de at a distance x from y ‘Area of element, dA = bx dr Wall shear stress at xis given as i %9 60x10? 4 a 92 mee NS ° ; 7 ‘Substituting the values of 4A and tp in equation (i), we get ee —- 12m Drag on both sides = 2 ft 4A Fig. 14.18 ig. 14.18 () _, !2{oox1o" alt ( x (bd) Limits of integration wart. x are 010 1.2 (betel 2 3 a _, p12 (60x10 rag on both sides =2 (a x2xde*P*(0% 109) 1 L “bh 2 ae sue 2 ent =4X00 103 |peee cs T +1) lo 12 L =4x60x 103 | 2 1 2 =4x 60x 107 x2 [viz] =0526N. Ans, Lift Force Tie if fore on one side of the pati given by equation (142) L2) as Fe= fia cos aA~ fosino ay For 8 = 90°, cos @ = 0 and sin 8=1 Fs Jrox0xda~ frxircan =0- Ioxixan pan =~ [oxaa Total lift on both sides an =- foxda+ J pxai=o af pw &p tesa pada ‘There is no lift in this case. 2nd Case : Plate is perpendicular to the direction of flow as shown in Fig. 14.18 (c) Pressure distribution on front face, N p=40 (1-2 ( m and on back face p* =~ 42 N/m? (constant) Shear stress distribution on upper half of front face is equal and opposite on lower half of front face. Hence shear force on front face = 0 Similar is the case with back face. Hence shear force on back face is also zero, Hence in this case, the drag force is due to difference of pressure on front face and back face. Drag Force Drag force is given by equation (14.1) as Fp= Jpcos0.d 0+ fry sin da cali) Fig. 14.18 (c)670 Fluid Mechanics But 0 0 (Angle made by press t qual (0 0° for frome 4 _ 88, Hey, Hence drag f irag force on front face becomes a we Fy= Joxtxdh + fry Oa Jraa +0 J pan force on both sides (front and back) J pans f vran Total drag where p = pressure distribution on front face = 40 (1 ~ 2-) Nim (given) (varies with y) p* = pressure distribution on back 42. N/m? (constant) fe and away from surface] ww, for back face of plate will be 180° Hence for besht [This pressure will be acting 1.10 stl Je by pressure with direction of flo 1 and sin @ = 0. alues in equation (ii), We B = if eeCnrdas fro xoxo The angle ma @ = 180°. Then cos 8 = these va et drag force for back face as Substituting Fp* ee |e f 42) ae? ju ck bck mot Joni [ spotal drag force = Drag fore? SP Tront face + Drag mee a AE face. _ J PA +P pt = feo | da + 100.8 (ii) j} ickness dy at @ an element of thi To find integral of pdA (iz. of y from x-@ distance ment, dA Area of ele! ure at y is ive > + 0.6m ea fpta = Li Limits of integrasio® with respect 0 y’) x2xdy fous =*9 ty 0.6 me areAl = 80 | (06-¢ oo)-{%5 c =a 2 93.12 Nim? total drag force rave lift foree on On side (front face) # = Jpsin 044 = fovea [Pron6. ‘The drag on a sphere for Reynolds nu 2 e is total drag, Skin friction dag, = 3: x SRHDU = 2nuDU Out of th 1 + 4 3m DU = muDU. and pressure drag 37am 4, Values of C, for sphere for different Reynold number is a4 Cp = when R, < 0.2 = N14 3 | when 02-<2, <5.0 R| 16R = 0c sn when R, ies between 5 and 1000 umber less than 0.2 is given by Fp = 3mupy (0.5 vin. when R, lies between 100 and 100000 = 0.2 van when R, > 10° 8, Terminal velocity is defined asthe maximum constant velocity ofa falling body with which it iy terminal velocity, the weight of the body is equal to the drag force plus the buoyant force, 49, ‘The velocity of ideal fluid at any point on the surface ofthe cylinder is given by — uy = 2U sin ® ‘where ug = Tangential velocity on the surface of the cylinder U-= Uniform velocity or free stream velocity wee {@= Angle made by any point on the surface of the cylinder with the direction of flow, 10, Circulation is the flow along a closed curve and is obtained when the product of velocity along ‘he laste and length ofthe small elementis integrated around the curve. Circulation for free vortex at any rd ripe by T= 2RR x uy 14, The resultant velocity on a circular cylinder which is rotated at 2onstant speed in uniform flow-feld 1" U sin 0+ —— u=2U sin + > 12, When a circular eylinder is rotated ina uniform flow, a lift force is produced on the cylinder. The magni the lift force Fis given by F,=pLur. where L = Length of cylinder, U = Free stream velocity, T= Circulation. 13, ‘The expression for lift co-efficient for a rotating cylinder in a uniform flow is given by r oe “= Ry 2p U i 14, The location of stagnation points is given by sin @ = - > a ee “4x0 where R = Radius of cylinder, U= Free stream velocity. 15. For a single stagnation point, the condition is. T=4nur or y= 2U 16, Circulation developed onthe airfoil is given by P=nCUsina where C = Chord length, U = Velocity of airfoil, = Angle of attack. 17, The expression for co-efficient of lift for an airfoil is C, = 2n sin o. 18, When an airplane is in steady-state, Weight of plane = Lift foree ‘Thrust by engine = Drag force, Cin term of eee ~-Gin term of tangential ge) Cin terms of ciredlalag (in terms of tangential vel)EXERCISE (A) THEORETICAL PROBLEMS terms: grag and efine the : sch 1. pe he er pderatand by : Total drag on a body, resultant force on a body, co-efficient of drag and co-efficien 3, What doy : of lift i if te ‘ane ate between (i) stream-lines body and bluff body, (é) Friction drag and pressure drag. 4, Differentiate Be f the flow is upto 0.2 ? Hene ssion for the drag on a sphere, when the Reynolds number o! nt of drag for sphere for this range of the Reynolds number is given by (a) What isthe €xP! Cp= = where R, = Reynolds number. why Cp suddenly drops at R, = 3 * 10°. raw Cp versus R, diagram for a sphere and exp oo () Draw &p .re when fluid has no viscosity {C) Draw pressure distribution diagrams in dimensionless form for flow past spher When R, = 10" and when R,= 10% ; be g, What do you mean by “Terinal velocity of body” 2 What isthe elation between the weight of the body, drag + Wipe on the body and buoyant force when the body has acquired terminal velocity ? 6, Whats circulation ? Find an expression for circulation fora free-vortex of radius R. 5. Min an expression forthe lift produced on a rotating cylinder placed ina uniform flow field such that the axis + oie eylinder is perpendicular to the direction of flow. §, What is Magnus effect ? Why is it known as Magnus effect ? efficient of lift for a rotating cylinder placed in a uniform flow is given by 9, Prove that the c« where = Circulation, & = Radius of cylinder, U = Free-stream velocity. 10, Define stagnation points. How the position of the stagnation points for a rotating cylinder in a uniform flow determined ? What is the condition for single stagnation point ? 11, Define the terms ; Airfoil, chord length, angle of attack, span of an airfoil. 12, Ifthe circulation developed on an airfoil is equal to %CU sin ar, then prove that co-efficient of lift for airfoil is given by C, = 2 sin a, where ot = angle of attack. 13, Explain the terms : (i) Friction drag, (ii) Pressure drag and profile drag. 14, (@) How are drag and lift forces caused on a body immersed in a moving fluid ? (b) What is the drag force on a sphere in the stoke range ? (B) NUMERICAL PROBLEMS {A flat plate 2m x2 m moves at 40 km/hour in stationary air of density 1.25 kg/m’. If the co-efficient of drag and lift are 0.2 and 0.8 respectively, find : (é) the lift force, (i?) the drag force, (iii) the resultant force, and (iv) the power required to keep the plate in motion, _[Ans. (i) 246.86 N, (ii) 61.715 N, (iii) 254.4 N, (iv) 0.684 kW] Find the drag force difference on a flat plate of size 1.5 mx 1.5 m when the plate is moving at a speed of 5 m/s normal to its plate first in water and second in air of density 1.24 kg/m’. Co-efficient of drag is given as 1.10. [Ans. 30899 N) 3. A truck having a projected area of 12 square metres travelling at 60 km/hr has a total resistance o} 2943 N. Of this 25% is due to rolling friction and 15% is due to surface fraction, The rest is due to form drag. 1, 2, Calculate the coefficient of form drag if the density of air = 1.25 kg/m’. [Ans. 0.847) 4, A circular dise 4 m in diameter is held normal to 30 m/s wind of density 1.25 kg/m’. If the co-efficient of drag o} dise = 1.1, what force is required to hold the dise at rest ? [Ans. 7775.4 N} , Find the diameter of a parachute with which a man of mass 80 kg descends to the ground from an aeroplane against the resistance of air, with a velocity of 25 m/s. ‘Take C,= 0.5 and density of air = 1.25 kg/m’. [Ans. 2.26 ny 6. A man descends to the ground with the help of a parachute from an aeroplane against the resistance of air with : uniform velocity of 10 m/s. The parachute is hemispherical in shape and is having diameter of 5 m. Find th Weight of man if C, = 0.5 and density of air = 1.25 kg/m’. [Ans. 613.52 Nwe tebe t 10. Md. 16. 17. . A body of length 2.5 m has a projected A metallic ball of diameter 5 mm drops in a fluid of sp. gr. ° tal. The string attached to the ki angle of 10° to the horizont h ite sm x 60 em weighing 2.943 N assumes an ang! en pee sie of 45° vo the horizontal, Ifthe pull on the sing 28. be et ovine wa pg akes an angle al te sing ‘of 40 km/hr. Find the corresponding co-efficient to drag al pono a velocity of 150 mmis, Fj and of infinite length with a velocity of A Fie e 0" over a cylinder of diameter 100 mm and of ae a 3 Ae tia P Ueetae sna sxsure drag on 1m length of the exner ifthe total drag co-efficient = 1.5 m ang cae ae r ven as = lve kKerme ffi 2 el n as shear drag co-efficient = 0.25. The density of air is give he omen ce the direction ofits motion. The body is moving 6 x 10° and the drag co-efficient = 0.5. Find the area 1.8 m? normal to Reynold number = through water with a velocity such that the a ea drag on the body. Take viscosity of water poise. : 2592 Calculate the weight ofa ball of diameter 50 mm which is just supported in a vertical ies ve flowin, ata velocity of 10 mvs. The density of air = 1.25 kg/m? and kinematic viscosity = - [Ans. 0.0613 Ny ve fs: i Dae “ {A metallic sphere of sp. gr 8.0 falls in an oil of density 800 kg/m’. The diameter of the oes 10mm andi attains a terminal velocity of 50 mnvs. Find the viscosity of the oil in poise. 8. 78.48 poise] 0.8 and viscosity 30 poise. The specific gravity of the ‘netallic bal, is 9.0. Find : (i) the drag force exerted by fluid on metallic ball, (ii) the pressure drag and skin cl eI jal vel ball in fluid. friction drag, and (iii) terminal velocity of a in UE Ooo Oe Ce aa ‘A cylinder rotates at 200 r.p.m. with its axis perpendicular in an air stream which is having uniform velocity of 20 ims. The eylinderis 2m in diameter and 8 m long, Assuming ideal fluid theory, find (i) the circulation, (i iy force, and (iii) position of stagnation points. Take density of air as 1.25 kg/m’ {Ans. (i) 131.57 m’/s, (ii) 26309.4 N, (iii) @ = 211.56° and 328.449) . For the problem 13, find the speed of rotation of the cylinder which will give only a single stagnation point, (Ans. 381.97 p.m) The air having a velocity of 30 m/s is flowing over a cylinder of diameter 1.4 m and length 10 m, when the axis ofthe cylinder is perpendicular to the air stream. The cylinder is rotated about its axis and a total lift of 58860 N is produced. Find the speed of rotation and location of the stagnation points. The density of air is given as 1.25 kg/m*. [Ans. N = 486.87 r-p.m., 6 = 216.5° and 323.5%) A jet plane which weighs 19620 N has a wing area of 25m? Itis flying at a speed of 200 km per hour. When the engine develops 588.6 kW, 70% of this power is used to overcome the drag resistance of the wing. Calculate the co-efficient of lift and co-efficient of drag for the wing. Taken density of air as 1.25 kg/m, : (Ans. C, = 0.407, Cp =0. Experiments were conducted in a wind tunnel with a wind speed of 50 km/hour on a flat plate of size 2 mr i : and te me is density of air is 1.15 kg/m®. The plate is kept at such an angle that co-efficients of lift and, dae are 0.75 and 0.15 respectively. Determine : (i lift force, (ii), drag force, (iti) resultant alitsraneee (©) power exerted by the air stream on the plate. : eee [Ans. (i) 166.4 N. ON (aN Tie CARR JER De ee