[204 Fluid Mechanics ] Half of Z, will be below the inital height of water in the vessel 2 _ 3934 A a But height of paraboloid for closed vessel = CO = CA + AO = (120 ~ 80) + 44.67 em 40 + 44.67 = 84.67 cm. Ans. Problem 5.27 For the data given in Problem 5.26, find the speed of rotation of the vessel, when axial depth of water is zero. Solution, Gen ry ie Ao = 44.67 om Diameter of vessel 20 em +. Radius, R= 10cm Initial height of water 80 cm T ‘so Length of vessel = 120m 120 Let ais the angular speed, when axial depth is zero ao When axial depth is zero, the height of paraboloid is 120 em and +. Using the relation, 129= 2% %980 oP =2 x 980 x 120 = 235200 of) ‘Volume of air before rotation = Volume of air after paraboloid mR? x (120 ~ 180) = Volume of paraboloid Baz enexZ 2 or x10? «40 = =x 120 x10? x40x2_ 8000 or Pe RXNO X40? _ 2000 «65.67 x 120 120 Substituting the value of ? in equation (i), we get @? x 66.67 = 235200 59.4 rads 6667 2nN 2 Speed Nis given by @ = 2. P aiven by @= or N= BO OO%504 _ 567.22 ppm. Ans, 2 aR Problem 5.28 The cylindrical vessel of the problem 5.26 is rotated at 700 r.p.m. about its vertical axis. Find the area uncovered at the bottom of the tank: Solution. Given Initial height of water = 80 cm Diameter of vessel 20 em e Radius, 10 em Length of vessel = 120 em[ Kinematics of Flow and Ideal Flow _205 | Speed, N= 700 rpm. PRN _ 228X700 _ 73.30 rads, 0 0 If the tank is not closed at the top and also is very long, then the height of parabola corresponding to @= 733 will be oP xR? _ 7337 x10? 2xg 2980 From Fig. 5.24, 4) + 120 +x, = 274.12 or xy + y= 274.12 ~ 120 = 154.12 em ...() From the parabola, KOM, we have oe _ B3? x4 120 +x) = 2 = eli cs » 2g 2x 980 Se For the parabola, LON, we have Lal * 208, BR xe , ef 7 2g 2x 980 . Fig. 5.24 Now, volume of air before rotation = Volume of air after rotation Volume of air before rotation = nR* x (120 ~ 80) = x x 10° x 40 = 1256.3 em* =i) Volume of air after rotation = Volume of paraboloid KOM ~ volume of paraboloid LON cnr (BEA) gee AL © Equating (iv) and (0), we get m2 (104%) _ efx mit a) _ alc Substituting the value of r,? from (ii) in (vi), we get (120+ x,) «2980 7 1256.3 = =i) (204%) _ nex 2 2980 x (120 + x (733) 12566.3 = x x or 12566.3 = 0.573 (120 + x)? Substituting the value of x from (iff) in the above equation 33? xr J 2 “2980 12566.3 = 0.573 | 120+ 2980 0.573 (120 + 2.74 133-43 x rx ry? = 0.573 [120° + 2.74? r+ 2 120 x 2.74 re] 43 ry[206 Fluid Mechanics = 0.573 [14400 + 7.506 r,! + 657.6 73°] — 4.3 1 125663 21930 = 14400 + 7.506 r,* + 657.6 r;° 4.3 734 0573, or 13! (7.506 ~ 4.3) + 657.6 r,° + 14400 - 21930 = 0 or 3.206 rf + 657.6 r,? ~ 7530 =0 + 657.6" —4 x(~7530) x (3206) > _ 657. 23206 657.6 + [432A37.16 + 96504.72 6a = 9516S 2732 _ _ 915.98 of 10.87 o4l2 Negative value is not possible 2 rp = 10.87 em? +. Area uncovered at the base = mr,? = x 10.87 = 34.149 em®, Ans. Problem 5.29 A closed cylindrical vessel of diameter 30 cm and height 100 cm contains water upto a depth of 80 cm. The air above the water surface is at a pressure of 5.886 N/cm?. The vessel is rotated at a speed of 250 r.p.m. about its vertical axis. Find the pressure head at the bottom of the vessel : (a) at the centre, and (b) at the edge. Solution, Given : Diameter of vessel = 30cm v. Radius, R= 150m Initial height of water, H/= 80 em Length of cylinder, L= 100em Pressure of air above water = 5.886 Niem? or p= 5.886 x 10'S Head due to pressure, h= ppg = 5886%10" 6 m of water 1000x9381 Speed, N= 250 rpm. 2nN _ 2nx250 =e 26.18 rads oO 60 Height of paraboloid formed, if the vessel is assumed open at the top and it is very long. y°R? _ 26.18" x15 2g 2x 981 Let r; is the radius of the actual parabola of height x2 Then we have a = 78.60 cm oo)[ Kinematics of Flow and Ideal Flow 207 | o'r? _ 2618? x4 Then x 35, di 23g 7 a qog1 O57 @ ‘The volume of sir before rotation RR? (100 ~ 80) = wx 15? x 20 = 14137 em? Volume of air after rotation = Volume of paraboloid EOF 1 =p xan But volume of air before and after rotation is same. : 14137 = bx are But from (i), 43 £ 0.35 7)? 14137 = 4 x nr? 035 1? pte RAMS Sosa #x035 r= 25714)" = 12.66 cm Substituting the value of r, in (i), we get 2) = 0.35 x 12.66? = 56.1 em Pressure head at the bottom of the vessel (2) At the centre. The pressure head at the centre, -e., at H = Pressure head due to air + OH .0 + (HL~ LO) (OH = LH- Lo} + HL=1000m=1m = 60 + (1.0 0.561 soeda-assy fore Scien sétal (b) At the edge, i.e., at G = Pressure head due to air + height of water above G soracesos (GM Ht) =60 + G03) om it~ o00s6 { risemceasn | = 6.0 + 0.439 + 0.786 = 7.225 m of water, Ans. Problem 5.30 A closed cylinder of radius R and height H is completely filled with water. It is rotated about its vertical axis with a speed of @ radians/s. Determine the total pressure exerted by water on the top and bottom of the cylinder. Solution. Given Radius of cylinder =R Height of eylinder =H ‘Angular speed =o[208 Fluid Mechanics ] AS the cylinder is closed and completely filled with water, the rise of water level at the ends and depression of water at the centre due to rotation of the vessel, will be prevented. Thus the water will exert force on the complete top of the vessel. Also the pressure will be exerted at the bottom of the cylinder. te Rf ‘Total Pressure exerted on the top of cylinder. The top of cylinder is in contact with water and is in horizontal plane. The pressure variation at any radius in horizontal plane is given by equation (5.21) D or ae Integrating, we get Jap =Spotrdr or p= Consider an elementary circular ring of radius r and width dr on the top of the cylinder as shown in Fig. 5.26. ‘Area of circular ring = 2nrdr Force on the elementary ring = Intensity of pressure x Area of ring Fig. 5.26 = px 2nrdr x 2nrdr. ® wr} & a'r? x amr { e Total force on the top of the cylinder is obtained by integrating the above equation between the limits 0 and R. Toa fore or Fy= [ 207s? xanntr= Be? x2K f" Par 6.25) Total pressure force on the bottom of cylinder, Fy = Weight of water in cylinder + total force on the top of cylinder 1g x MR x H+ Fy (3.26) pgx nex H+ ob x er p= Density of water. Problem 5.31 A closed cylinder of diameter 200 mm and height 150 mm is completely flled with water. Calculate the total pressure force exerted by water on the top and bottom of the cylinder, if it is rotated about its vertical axis at 200 r-p.m. Solution. Given Dia. of cylinder = 200 mm = 0.20 m Radius, R=01m Height of cylinder, H= 150 mm = 0.15 m Speed, N= 200 rp.m, 2nN _ 2nx200 Angular speed, ea) ues oo 60[ Kinematics of Flow and Ideal Flow 209 | Total pressure force on the top of the cylinder is given by equation (5.25) y= Bat cca « 120 52908 52% 0." SAME N An 4 Now total pressure force on the bottom of the cylinder is given by equation (5.26) as F y= pg x mk? x H+ Fy 1000 x 9.81 x mx (OI)? x 0.15 + 34.44 = 46.22 + 34.44 = 80.66 N. Ans. 5.10.6 Equation of Free Vortex Flow. For the free vortex, from equation (5.20), we have vx r= Constant = say ¢ Substituting the value of v in equation (5.23), we get ap = pdr ~ pg de = px dr~pgde=px 5 dr~pg dz Pxr Consider two points 1 and 2 in the fluid having radius r, and r, from the central axis respectively as shown in Fig. 5.27. The heights of the points from bottom of the vessel is z, and z. Integrating the above equation for the points 1 and 2, we get fio or PoP far fog [ Pearmpe fae [Ef mete 3-vi]-psle-a] v? v3] ps [22-2] (5.21) Equation (5.27) is Bernoulli's equation, Hence in case of free vortex flow, Bernoulli's equation is applicable. Fig. 5.27[210 Fluid Mechanics ] Problem 5.32 In a free cylindrical vortex flow, at a point in the fluid at a radius of 200 mm and at a height of 100 mm, the velocity and pressures are 10 m/s and 117.72 kN/m? absolute. Find the pressure at a radius of 400 mm and at a height of 200 mm. The fluid is air having density equal to 1.24 kg/m’. Solution. At Point 1 : Given Radius, = 200 mm = 0.20 m Height, z= 100 mm = 0.10 m Velocity, 0 mis Pressure, 17.72. kNN/m? = 117.72 x 10° Nim? At Point 2: 400 mm = 0.4 m 2 = 200 mm = 0.2 m Py = pressure at point 2 p= 1.24 kg/m? For the free vortex from equation (5.20), we have vx r= constant oF 7 10x02 ofa mis But p= 1.24 kg/m’ Uu772x10" 10g my aot 124x981" 2x981 pe 2x981 Py MII, 10? yy) 5? _ 9g pe 124x981 2x98” 2x93 9677.4 + 5.096 + 0.1 ~ 1.274 ~ 0.2 = 9676.22 1676.22 x py = 9676.22 x 1.24 x 9.81 17705 N/m? = 117.705 x 10° Nim? 17.705 kN/m? (abs.) = 117.705 kN/m?, Ans. (B) IDEAL FLOW (POTENTIAL FLOW) > 5.11 INTRODUCTION [Wal fluid is a fluid which is incompressible and inviscid. Incompressible fluid is a fluid for which density (p)remains constant. Inviscid fluid is a fluid for which viscosity (1) is zero. Hence a fluid for which density is constant and viscosity is zero, is known as an ideal fluid. ‘The shearstons is given by, t= 1 SE, Hence for idea id the shear ston wll e210 a8 = 0 for ideal fluid. Also the shear force (which is equal to shear stress multiplied by area) will be zero in[ Kinematics of Flow and Ideal Flow 211 | case of ideal or potential flow. The ideal fluids will be moving with uniform velocity. All the fluid particles will be moving with the same velocity. ‘The concept of ideal fluid simplifies the typical mathematical analysis. Fluids such as water and air have low viscosity. Also when the speed of air is appreciably lower than that of sound in it, the compressibility is so low that air is assumed to be incompressible, Hence under certain conditions, certain real fluids such as water and air may be treated like ideal fluids. > 5.12 IMPORTANT CASES OF POTENTIAL FLOW ‘The following are the important cases of potential flow : (i) Uniform flow, (ii) Source flow, (iii) Sink flow, (iv) Free-vortex flow, (v) Superimposed flow. > 5.13 UNIFORM FLOW In a uniform flow, the velocity remains constant. All the fluid particles are moving with the same velocity. The uniform flow may be : (i) Parallel to x-axis Gi) Parallel to y-axis. 5.13.1. Uniform Flow Parallel to x-Axis. Fig. 5.27 (a) shows the uniform flow parallel to a-axis. In a uniform flow, the velocity remains constant. All the fluid particles are moving with the same velocity. Fig. 5.27 (a) Le U = Velocity which is uniform or constant along x-axis uw and y = Components of uniform velocity U along x and y-axis. For the uniform flow, parallel t0 x-axis, the velocity components u and v are given as u=Uandy=0 (5.28) But the velocity uw in terms of stream function is given by, ay and in terms of velocity potential the velocity u is given by, ay _ a ua VL 5.29) a (5.29) 5 av _% Similarly, it can be shown that y = ~ 2¥ = 2% 5.294 ae te oe ay oe) But w= U from equation (5.28). Substituting u = U in equation (5.29), we have[212 Fluid Mechanics ] Le 5.3 u (5.30) & sand also u = 2 or = and atso u = First part gives dy = U dy whereas second part gives d@ = U dx. Integration of these parts gives as We Uy+C, and = Uxt Cy where C, and Cy are constant of integration, Now let us plot the stream lines and potential lines for uniform flow parallel to x-axis. Plotting of Stream lines. For stream lines, the equation is waUxy+G Let y= 0, where y = 0. Substituting these values in the above equation, we get 0=Ux04C, or C,=0 Hence the equation of stream lines becomes as y weU.y (5.31) ‘The stream lines are straight lines parallel to x-axis and at a distance y from the x-axis as shown in Fig. 5.28. In equation (5.31), U. y represents the volume flow rate (i. m/s) be- tween x-axis and that stream line at a distance y. Note. The thickness of the fluid stream perpendicular to the plane is assumed to be unity. Then y x I or y represents the area of flow. And Uy represents the produet of velocity and area. Hence U. y represents the volume flow rate. Stream lines ° Plotting of potential lines. For potential lines, the equation is O=U.x+C, Let @=0, where x= 0. Substituting these values in the above equation, we get C, Hence equation of potential lines becomes as oe Ux ‘The above equation shows that potential lines are straight lines parallel to y-axis and at a distance of x from y-axis as shown in Fig. 5.29. Fig. 5.30 shows the plot of stream lines and potential lines for uniform flow parallel to x-axis. The stream lines and potential lines intersect each other at right angles. be y cle Potental ines esis Potanal sys ines J rts fe 4 tC ers * We vs chee ease wo Fig. 5.29 Fig. 5.30[ Kinematics of Flow and Ideal Flow 213] 5.13.2 Uniform Potential Flow Parallel to y-Axis. Fig. 5.31 shows the uniform potential flow parallel to y-axis in which U is the uniform velocity along y-axis. Fig. 531 The velocity components 1, v along x-axis and yraxs are given by u=0andv=U (533) ‘These velocity components in tems of steam function (y) and velocity potential function (9) are piven as av _ 2% = A534 dye ae ay _ a penta (5.35) - ee 635) But from equation (5.33), v = U. Substituting v = U in equation (5.35), we get oy _ 4% oy 29 ya oe ya and aso = % a ay By Firs part gives dy = — U ce whereas second part gives p= U dy Integration ofthese pars gives as, W=-U.x+C, and 0=U.y+Cy (5.36) where C, and C, are constant of integration, Let us now plot the stream lines and potential lines. Plotting of Stream lines. For stream lines, the equation is y= U.x + C, Let w= 0, where x= 0. Then C, = 0. Hence the equation of stream lines becomes as y = ~ U.x (3.37) ‘The above equation shows that stream lines are straight lines parallel to y-axis and at adistance of x from. the y-axis as shown in Fig. 5.32. The ve sign shows that the str:am lines are in the downward direction. stream lines I 5 +7 Pojentatinos Ys al ‘[214 Fluid Mechanics ] Plotting of Potential lines. For potential lines, the equation is 6 = U.y + C, Let = 0, where y = 0. Then C; = 0. Hence equation of potential lines becomes as 6 = U.y (5.38) ‘The above equation shows that potential lines are straight lines parallel to x-axis and at a distance of 1y from the x-axis as shown in Fig. 5.32. > 5.14 SOURCE FLOW ‘The source flow is the flow coming from a point (source) and, moving out radially in all directions of a plane at uniform rate. Fig. 5.33 shows a source flow in which the point 0 is the source from which the fluid moves radially outward. The strength of a ol source is defined as the volume flow rate per unit depth, The unit 3 ‘ of strength of source is m/s. It is represented by 4. oer Let u, = radial velocity of flow at a radius r from the source O “(sy volume flow rate per unit depth dius ‘The radial velocity u, at any radius r is given by, Fig. 5.33 Source low (Flow atway (5.39) from source) ‘The above equation shows that with the increase of r, the radial velocity decreases. And at a large distance away from the source, the velocity will be approximately equal to zero, The flow is in radial direction, hence the tangential velocity ug = 0. Let us now find the equation of stream function and velocity potential function for the source flow. As in this case, up = 0, the equation of stream function and velocity potential function will be obtained from u, Equation of Stream Function By definition, the radial velocity and tangential velocity components in terms of stream function are given by [See equation (5.124)] But [See equation (5.39)] p24 or A w= hal nr? Qn a Integrating the above equation wat. 8, we gt w= dh x0 6 Cy, whee C, i constant of isp Let y=0, when 0 =0, then C, =0. Hence the equation of stream function becomes asKinematics of Flow and Ideal Flow 215 4 at 4 vege (5.40) In the above equation, q is constant. ‘The above equation shows that stream function is a function of 8. For a given value of , the stream function will be constant, And this will be a radial line. The stream lines ean be plotted by having different values of @. Here 0 is taken in radians. Plotting of stream lines sa When @=0,y=0 Stream ines 0= 45° = = radians, y=. = =4 units ‘are radal 4 m4 ok q m4 vo 0= 90° = radians, y=. =4 units 4 2 me oe 24 a Equation of Potent wai By definition, the radial and tangential components in Fig. 5.34 Stream line for terms of velocity function are given by source flow. [See equation (5.94)] ar But from equation (5.39), u,= nr Equating the two values of u, we get og 4 2 or aye ar ar wr Or Integrating the above equation, we get ja-[fa Potent ines 2nr mere afl 4 or Ap dar fis Lisaconstan tem ante | ae 4 A top,r St Qn Be (S41) In the above equation, q is constant. ‘The above equation shows, that the velocity potential function is ‘a function of r. For a given value of r, the velocity function 6 will be constant. Hence it will be a circle with origin at the source, The velocity potential lines will be circles with origin at the source as shown in Fig. 5.35. Let us now find an expression for the pressure in terms of radius. Fig. 5.35. Potential lines for source[216 Fluid Mechanics ] Pressure istribution in a plane source flow ‘The pressure distribution in a plane source flow can be obtained with the help of Bernoulli's equa- tion. Let us assume that the plane of the flow is horizontal. In that case the datum head will be same for two points of flow. Let p= pressure at a point 1 which is at a radius r from the source at point 1 1, = velocity at point 1 Po = Pressure at point 2, which is at a large distance away from the source. The velocity will be zero at point 2. [Refer to equation (5.39)] Applying Bernoulli's equation, we get Pt Po gy gq (PAB) ps 28 pg Pg or (pp) =-= Bat fom enusion(S.3), 4, = 52 (6) 2) \ ar, In the above equation, p and q are constants. ‘The above equation shows that the pressure is inversely proportional to the square of the radius from the source. (5.42) > 5.15 SINK FLOW The sink flow is the flow in which fluid moves radially inwards towards a point where it disappears at a constant rate. This flow is just opposite t0 the source flow. Fig. 5.36 shows 4 sink flow in which the fluid moves radially inwards towards Point O, where it disappears at a constant rate. The pattern of stream lines and equipotential lines of a sink flow is the same as that of a source flow. All the equations derived for a source Sink ra flow shall hold to good for sink flow also except that in sink flow equations, q is to be replaced by (~ q). Problem 5.33 Plot the stream lines for a uniform flow of (5 m/s parallel to the positive direction of the x-axis and Fig.5.36 Sink flow (ii) 10 ms parallel to the positive direction of the y-axis. (Flow toward centre) Solution. (i) The stream function for a uniform flow parallel to the positive direction of the avaxis is given by equation (5.31) as w=Uxy The above equation shows that stream lines are straight lines parallel to the x-axis at a distance y from the x-axis. Here U = 5 mis and hence above equation becomes asKinematics of Flow and Ideal Flow 217 w= sy For y=0, stream function y= 0 For y= 02, stream function y= 5 %0.2=1 unit For y= 044, stream function y= 5 x0.4 = 2 unit The other values of stream function can be obtained by substituting the different values of y. The stream lines are horizontal as shown in Fig. 5.36 (a). ¥ 0g; + y=08 06 ee y=08 04, + y=0a 02} + y=02 ° yo Fig. 8.36 (a) (Gi The stream function fora uniform flow parallel tothe positive direction ofthe y-axis is given by equation (5.37) as, wo-Uxx The above equation shows that stream lines are straight lines parallel to the y-axis at a distance x from the y-axis. Here U = 10 mvs and hence the above equation becomes as ye-l0xx ‘The negative sign shows that the stream lines are in the downward direction. For x=0, — the stream function y= 0 For x= 0.1, the stream function y =~ 10 x 0.1 =~ 1.0 unit For x= 0.2, the stream function y =~ 10x 0.2 =~ 2.0 unit For x= 0.3, the stream function y =~ 10 x 0.3 =~ 3.0 unit ‘The other values of stream function can be obtained by substituting the different values of x. The stream lines are vertical as shown in Fig. 5.36 (b).. 0 01 02 03 Fig. 5.36 (b)[218 Fluid Mechanics ] Problem 5.34 Determine the velocity of flow at radii of 0.2 m, 0.4 m and 0.8 m, when the water is flowing radially outward in a horizontal plane from a source at a strength of 12 m/s. Solution. Given : Strength of source, q = 12 m’is ‘The radial velocity wat any radius r is given by equation (5.39) as, sz nr When r= 0.2 m, Z 9.55 m/s. Ans. In x02 When r= 04 m, = 4.77 mis. Ans. Inxoa When r= 0.8 m, a 2.38 m/s. Ans. 2m x08 Problem §.35 Two discs are placed in a horizontal plane, one over the other. The water enters at the centre of the lower disc and flows radially outward from a source of strength 0.628 m*/s. The pressure, at a radius 50 mm, is 200 kN/m*, Find : (i) pressure in kN/m’ at a radius of 500 mm and (ii) stream function at angles of 30° and 60° if y= 0 at 8 = 0°. Solution. Given : Source strength, q= 0.628 m/s Pressure at radius 50 mm, p, = 200 kN/m’ (@ Pressure at a radius $00 mm Let p= pressure at radius 500 mm (u,), = velocity at radius 50 mm (u,), = velocity at radius 500 mm The radial velocity at any radius r is given by equation (5.39) as 100 x 10° Nim? When r= 50mm =0.05m, (u,), = 288 = 1.998 mis = 2 mis 2m x005 When r= 500mm =0.5m, (u,),= 28 = 0.2 mis 2x05 Applying Bernoulli's equation at radius 0.05 m and at radius 0.5 m, or