PROPERTIES OF FLUIDS

Unit - I Chapter - PROPERTIES OF FLUIDS Solutions of Examples for Practice Example.9 : Given data : u = y y, = 8 Poise = 0.8 Pa-s To find : Shear stress. Step - : Calculate the shear stress at various points It is given that, u = y y du dy = y As per Newton's law of viscosity, = du dy = 0.8 y At y = 9 cm = 0.09 m = 0.8 0.09 = 0.84 N m Example.0 : Given data : d =6mm=6 0 m, d =mm= 0 m, Ans. = 0.075 N/m, = 0º. To find : Difference of water levels. Step - : Calculate the difference of levels in the tubes Capillary effect in smaller limb is, 4 cos 4 0.075 cos(0) h = = d 980 6 0 h = 4.969 0 m = 4.969 mm Capillary effect in larger limb is, 4 cos 4 0.075 cos (0) h = = d 980 0 - Fluid Mechanics

Properties of Fluids h =.464 0 m =.464 mm Difference in the levels of capillary tubes is, h = h h = 4.969.464 =.465 mm Ans. Example. Given data : Side of cube b = 0.5 m, m = 50 kg, 45º, =.5 0 N-s m, t = 0.05 mm = 0.05 0 m 0. m Contact area W sin θ0. m G θ W cos θ t = 0.05 mm θ = 45º W Step - : Calculate the shear force on the bottom surface of cube Weight of cube W = mg = 50 9.8 = 490.5 N Component of weight W acting along the plate is equal to the shear force on the bottom surface of cube. Shear force F = W sin = 490.5 sin (45) F = 46.858 N Step - : Calculate the speed of the block As per Newton's law of viscosity, = du dy = F A F = du dy A = U t A Fig.. (A = Contact area) 46.858 =.5 0 U (0.5 0.5) 0.05 0 U = 7.7468 m/sec Ans. - Fluid Mechanics

Properties of Fluids Exmaple. : Given data : d = 0.5 m, t = 0.05 m, N = 80 rpm, T = 6 0 4 N-m. To find : Viscosity () Step - : Calculate the viscosity of oil Refer Fig... Velocity of the disc is, dn V = 60 V = 0.5 80 60 V = 0.68 m/sec. We know that, T = Force Radius = F 6 0 4 0.5 = F F=8 0 N As per Newton's law of viscosity, = du dy = F A d 0.5 m 80 rpm Fig.. t = 0.05 m F = du dy U A = d t t 8 0 = 0.68 0.5 0.05 = 0.540 Pa-s Ans. Example. Given data : r d =.5 cm, d d =.5 = cm, r j = 0.8 cm d j = 0.8 =.6 cm, p atm = 00 kn m, = 0.05 N/m To find : p gauge and p abs Step - : Calculate the gauge pressure and absolute pressure for a droplet and for a jet of water - Fluid Mechanics

Properties of Fluids i) For a droplet Gauge pressure is, p = 4 d = 4 0.05 0 = 66.6667 N/m Ans. Absolute pressure is, p abs = p p atm = 66.6667 + 00 0 p abs = 00.0667 kn/m Ans. ii) For a jet of water Gauge pressure is, p = d = 0.05.6 0 = 6.5 N/m Ans. Absolute pressure is, p abs = p p atm = 6.5 + 00 0 p abs = 00.065 kn/m Ans. Properties of Fluids ends... - 4 Fluid Mechanics

Unit - I Chapter - Fluid pressure and its measurement Solutions of Examples for Practice Example. Given data : S m = 0.8, h = 00 cm = m, h = 0 cm = 0. m, h =50cm=0.5m To find : Pressure difference (p p ) Step - : Calculate the pressure difference between M and N We know that, p p = h h h p M M N M N m = (980 ) (980 0.8 0.) (980 0.5) p = 550.6 kn m Ans. N Example. We know that, Pressure in left column = Pressure in right column p h h = 0 p = [(980 0.) (980.6 0.)] p M = 8.645 0 N m = 8.645 kn m Ans. Example. Given data : Z = 0.4 m To find : Pressure p Step - : Calculate the pressure due to 0.4 m column i) For water, p = g Z = 000 9.8 0.4 - Fluid Mechanics

Fluid Pressure and its Measurement p = 94 N/m Ans. ii) For oil, Pressure is given by, p = g Z = (0.9 000) 9.8 0.4 p = 5.6 N/m Ans. iii) For mercury, p = g Z = (.6 000) 9.8 0.4 p = 5.664 0 N/m Ans. Example.4 Pressure in left column above datum = Pressure in right column above datum pa h = h h p A = h h h p A = 980 0.6.6 980 0.45 0.88 980 0. p A = 56.740 0 N/m Ans. Fluid Pressure and its Measurement ends - Fluid Mechanics

Unit - I Chapter - Hydrostatic forces Solutions of Examples for Practice Example. Given data : d =. m, Distance of C.P. from liquid surface = m, 90º To find : Total pressure (P) and center of pressure (h) Step : Calculate the total pressure and center of pressure Refer Fig... CG CP x h CG CP m. m Fig.. Total pressure on the plate is, P = A x = 4 d x P = 980 4. P =.896 0 N =.896 kn... Ans. Depth of center of pressure vertically below liquid surface is, h = x + I d 4 G =x+ 64 Ax d x 4 - Fluid Mechanics

Hydrostatic Forces (.) 4 h = + 64 (.) 4 h =.045 m below free liquid surface. Ans. Example.4 Given data : b = m, Distance of top edge from liquid surface =.5 m. To find : Total pressure (P) and centre of pressure Step - : Calculate the total pressure and position of centre of pressure Refer Fig.. Total pressure is, P = Ax = BC AD x P = 0.9 980 cos 0 (.5 +.5) A 0º CG CP Fig.. C D B.5 m m P = 0.8 0 N = 0.8 kn Ans. The position of centre of pressure is, h = x+ I G = x+ Ax But I G = bh BC (AD) = 6 6 h = (.5.5) I G BC AD x = ( cos (0)) 6.464 cos (0) (.5.5) =.464 m 4 h =.5 m below free liquid surface. Ans. Example.5 4 Given data : Diagonal AC = 4 m Side of square b =.884 m To find : P and h Step - : Calculate the hydrostatic force and the depth of centre of pressure - Fluid Mechanics

Hydrostatic Forces Refer Fig... A m h b b x=+=m 4m B O CG D P CP Hydrostatic force is, P = Ax = 0.85 980 (.884. 884) ( ) P = 00.0 0 N 00.0 kn Ans. Depth of centre of pressure is, h = I G x Ax 4 Fig.. But, I G = b 4 (.884) = = 5. m 4 5. h = ( ) (.884.884) ( ) h =. m below free liquid surface Ans. C Example.6 Given data : b=5m, d=6m, 45º To find : Total pressure and centre of pressure. Step - : Calculate the total pressure and the position of centre of pressure Refer Fig..4. x =5+OAsin45=5+sin45=7. m Total pressure is, P = Ax = 980 (5 6) 7. P =.0958 0 6 N Position of centre of pressure is, Ans. - Fluid Mechanics

Hydrostatic Forces 45º h x 5m B 45º A Upper edge B cos45 45º A O 5m sin45 m CG O CP 6m I h = x G sin Ax = x Fig..4 bd sin (b d) x = 7. 5 6 (sin 45) 5 6 7. h = 7.9 m below free liquid surface Ans. Example.7 Given data : d=4m,b=m, = 90º To find : Total pressure (P) and centre of pressure Step - : Calculate the total pressure and position of centre of pressure Refer Fig..5. Total pressure is, P = Ax But, A = 6 m and x =.5 m.5 m P = 980 6.5 P = 06.0 0 N Ans. The position of centre of pressure is, I G h = x Ax But, I G = bd 4 6 m 4 CG CP m Fig..5 h P 4m - 4 Fluid Mechanics

Hydrostatic Forces h = 6.5 6.5 h = 4.69 m below free liquid surface Ans. Hydrostatic Forces ends... - 5 Fluid Mechanics

Unit - II Chapter - 5 Kinematics of fluid motion Solutions of Examples for Practice Example 5.7 u=6x, v = 8xyz, w = 4xz xz Now, differentiate the velocity components with respect to x, y and z u x = x u y = 0 u z = 0 v x = 8yz v y = 8 xz v z = 8xy w x = 4z z w y = 0 w z = 8xz x i) To satisfy continuity equation (for continuous flow) u v w = 0 x y z x + 8xz 8xz x = 0 Hence, the flow is continuous. Ans. ii) For rotational flow, the rotational component about z-axis should not be zero. The component of rotation about z-axis is, z = v u x y z = (8yz 0) =4yz0 As z 0, the flow is rotational Ans. iii) Rotation components at point (,, ), x = w v y z x = (0 8 xy) (0 8 ) 5 - Fluid Mechanics

Kinematics of Fluid Motion x = 4 Ans. y = u w z x y = (0 z)) ( 4 z y = (0 ) ( 4 y = 8 Ans. z = v x u y z = (8 yz ) 0 = (8 0 ) z = 4 Ans. Example 5.8 To satisfy the continuity equation, u v w = 0 (i) x y z u = x y z u x = x v = xy xy yz = v y x x z Substituting these values in equation (i), w x x x z z = 0 w z = x x z w = (x x z) z Integrating on both sides, w = x z xz + z C But, constant of integration cannot be a function of z. Hence, it is a function of x and y i.e. C = f(x, y). w = x z xz + z f(x,y) Ans. 5 - Fluid Mechanics

Example 5.9 Stream function = xy i) As per defination of stream function, x = v=y and y = u = x u = x and v = y We know that, V = u v at point (, ), u = x = = 4 v = y = =6 V = ( 4) 6 = 7. m/s Ans. ii) As per defintion of potential function, = u = ( x) = x (i) x and y = v = (y) = y (ii) From equation (i), = x x Integrating on both sides, = x C = x C But, constant of integration is a function of y i.e. C = f (y). Kinematics of Fluid Motion = x f(y) (iii) To find f (y) we will use equation (ii). Initially differentiate equation (iii) with respect to y i.e. y = 0 f(y) But, y = y [From equation (ii)] f(y) = y or f(y) = y Constant 5 - Fluid Mechanics

f(y) = y constant Kinematics of Fluid Motion Substituting this value in equation (iii), = x y Constant Ans. Example 5.0 Velocity potential = xy x As per definition of velocity potential, = u=(y ) u= y x and y = v= x v= x Now, as per definition of stream function, and x y From equation (i), x =v= x = x x Integrating on both sides, = v= x (i) = u=y (ii) = x = x C C But, the constant of integration cannot be a function of x. Hence, it is a function of y i.e. C = f(y). = x f(y) (iii) Initially differentiate equation (iii) with respect to y i.e. y = 0 f (y) But, y = y y = f (y) 5-4 Fluid Mechanics

Kinematics of Fluid Motion f(y) = y y f(y) = y y Substituting this value in equation (iii), = x y y Ans. 5-5 Fluid Mechanics

Unit - III Chapter - 6 FLUID DYNAMICS Solutions of Examples for Practice Example 6. Given data : d =60cm=0.6m a = 4 d = 4 (0.6) = 0.87 m d =0cm=0.m a = 4 d = 4 (0.) = 0.04 m Z = 0 cm, = 0. m, S = 0.9, p = 80 kpa = 80 0 Pa p = 00 kpa = 00 0 Pa, h L =%ofh, C d = 0.97. To find : Discharge Q. Step - : Calculate the flow rate through the venturimeter Applying Bernoulli's equation between inlet (section ) and throat (section ) by considering section as a datum, p V Z g = p V Z h g f (i) Inlet But, for vertical venturimeter differential head is 0 cm h = p p (Z Z ) Throat h = 80 0 00 0 0.9 980 (0. 0) S f w h = 9.60 m (ii) Head lost, h L =%h=0.0 9.60 = 0.85 m Also by continuity equation, Q = a V = a V 0.87 V = 0. 04 V V = 0.0 V Outlet Fig. 6. 6 - Fluid Mechanics

Fluid Dynamics Equation (i) can be written as, p p (Z Z ) = V V g h L 9.60 = V (0.0 V ) 9.8 9.0758 = 0.050 V 0.85 V =.45 m/s Discharge is, Q = a V = 0.04.45 Q = 0.47 m sec = 4.780 lps Ans. Example 6.4 Given data : d 0 = 0 cm = 0.0 m a 0 = 4 d 0 = 4 (0.0) = 0.04 m d =40cm=0.4m a = 4 d = 4 (0.4) = 0.56 m x = 0.8 m, S = 0.9, C d = 0.64 To find : Flow rate of oil Q. Step - : Calculate the flow rate of oil through the orifice meter Pressure head through orifice meter is, h = x Sm = 0.8.6 S 0.9 h =.888 m Discharge through the orifice meter is, Q = C a a d 0 a gh a 0 Q = 0.64 0.04 0.56 9.8.888 (0.56) (0.04 ) Q = 0.088 m sec = 08.885 lps Ans. Example 6.5 Given data : x = 00 mm = 0. m, S m =, air =6Nm, C v = 0.98 6 - Fluid Mechanics

Fluid Dynamics To find : Velocity of aeroplane (V). Step - : Calculate the velocity of aeroplane Pressure head is, h = x S m m = x S air h = 0. 980 =.45 m of air 6 Velocity of aeroplane is, air V = Cv gh = 0.98 9.8.45 V = 48.097 m/sec = 7.9069 km/hr Ans. Example 6.6 Given data : d = 00 mm = 0. m a = 4 d = 4 (0.) =.459 0 m d c = 6.4 cm = 0.64 m a c = 4 d c = 4 (0.64) =.40 0 m h = 5. m, Q = 80 lps = 80 0 m sec To find : Hydraulic coefficients (C,C,C andc ) c d v r Step - : Calculate the values of hydraulic coefficients Coefficient of contraction is, C c = a c = a.40 0.49 0 = 0.67 Ans. Theoretical discharge is, Q th = a V th =.469 0 9.8 5. (Vth gh) Q th = 0.7 m sec Coefficient of discharge is, C d = Q act Q th We know that, = C d = C C c v 80 0 0.7 = 0.567 Ans. C v = C d Cc C v = 0.567 = 0.847 Ans. 0.67 6 - Fluid Mechanics

Coefficient of resistance is, C r = C = = 0.4046 Ans. v (0.847) Example 6.7 Given data : = 0º, S = 0.9, h m = 600 mm = 0.6 m, C d = 0.98 Inlet diameter d = 00 mm = 0. m To find : Flow rate (Q) Area of inlet a = 4 d = 4 (0.) = 0.0706 m Throat diameter d = 00 mm = 0. m Area of throat a = 4 d = 4 (0.) = 0.04 m Fluid Dynamics 0.7 m 0.7 sin 0 = 0.5 m 0º 0.6 m Fig. 6. Mercury Step - : Calculate the flow rate of oil Applying Bernoulli's equation between section and by considering horizontal through section as a datum, p V Z g = p V Z g V V g = p p (Z Z ) =h But, h = x S S m o 6-4 Fluid Mechanics

Fluid Dynamics h = 0.6.6 0.9 h = 8.4666 m Discharge through venturimeter is, Q = C d a a gh a a Q = 0.98 0.0706 0.04 9.8 8.4666 (0.0706) (0.04) Q = 0.448 m sec Ans. Fluid Dynamics ends 6-5 Fluid Mechanics

Unit - IV Chapter - 7 LAMINAR FLOW Solutions of Examples for Practice Example 7.5 Given data : B = 00 mm = 0. m, u max = m/sec, =.5 poise =.5 Pa-s. To find : i) max ii) dp ii) du iv) u dy Step - : Calculate the shear stress at the plates For laminar flow between parallel plates, u avg = u max = =. m/sec But, u avg = B p x. = (0.) p.5 x p x = 000 Pa m Shear stress at the plates is maximum, hence max = p B 0. = 000 x max = 00 Nm or Pa Ans. Step - : Calculate the difference in the pressure and velocity gradient at the plates Pressure difference between two points which are 0 m apart is, p p = dp = u avg L.5. 0 = B (0.) p p = dp = 9.9995 0 N m As per Newton's law of viscosity, = du dy Ans. 7 - Fluid Mechanics

It is important to note that, y is the distance measured from the plates and shear stress () at the plates is maximum. max = du (At the plate y = 0) dy y 0 00 =.5 du dy y 0 Laminar Flow du dy y 0 = 40 sec Step - : Calculate the velocity at 0 mm from the plate Velocity for the fixed parallel plates is, u = p x (By y ) Ans. u = 000 (0. 0.0 0.0 ).5 u =.0 m/sec Ans. Example 7.6 Given data : = 0.08 N-s m, Width Z=m, B=0mm=0.0 m At midway u = u max = 6 m/sec p To find : i) ii) u x avg iii) Q Step - : Calculate the pressure gradient along the flow For laminar flow through fixed parallel plates, u max = p 8 x B 6= 8 0.08 p (0.0) x p x = 9600 p or = 900Nm Pa 6 m or Ans. x m Step - : Calculate the average velocity and discharge of an oil Average or mean velocity is, u avg = u max = 6 = 4 m/sec Ans. Discharge of an oil is, Q = Area Mean velocity = (Z B) u avg Q = ( 0.0) 4 = 0.6 m sec Ans. 7 - Fluid Mechanics

Example 7.7 Given data : D = 400 mm = 0.4 m R= D To find : i) u mean or u avg ii) Radius at u avg 0.4 iii) u Step - : Calculate the mean velocity and its radius For laminar flow through a pipe, = 0. m, u max = m/sec. Laminar Flow u avg = u max = u avg = m/sec Ans. The radius for mean velocity is, r = 0.707 R = 0.707 0. r = 0.44 m Ans. Step - : Calculate the velocity at 60 mm from the pipe wall Velocity for the pipe flow is, r u = umax R [From equation (7.5)] But, r = R y = 0. 0.06 = 0.4 m [y is measured from the pipe wall] u = 0.4 0. u =.0 m/sec Ans. Example 7.8 Given data : =.5 poise = 0.5 Pa-s, S = 0.9, D = 50 mm = 0.05 m R = 0.05 m, p = 0 kn m = 0 0 Nm x To find : i) Mass flow rate m ii) iii) Re iv) Power P Step - : Calculate the mass flow rate and shear stress at pipe wall For laminar flow through a pipe, mass flow rate is, m = A u avg = R u avg (i) oil But, pressure drop is, p p = u L D avg 7 - Fluid Mechanics

Laminar Flow or p = u x D 0 0 = avg 0.5 u (0.05) avg ( L = dx) u avg = 5.08 m/sec Substituting in equation (i), m = 0.9 000 (0.05) 5.08 ( =S ) m = 9.08 kg/sec m = 9.08 60 = 55.95 kg/min Ans. Shear stress is maximum at the pipe wall i.e. p R max = x = (0 0 ) 0.05 max = 5 N m Ans. Step - : Calculate the Reynolds number of flow and the required power Reynold's number of flow is, Re = uavg D 0.9 000 5.08 0.05 = 0.5 Re = 56.49 Ans. Re < 000, it means flow is laminar. Power required to maintain the flow is, p p P = Qh f = Area Mean velocity P = R u (p p ) avg oil water P = R P = u uavg D avg (0.05) 5.08 0.5 5.08 60 (0.05) L ( L=60m) P = 65.8446 W = 6.58 kw Ans. Example 7.9 : Given data : D = 80 mm = 0.08 m, R= D 0.08 = 0.04 m, L = 500 m, 7-4 Fluid Mechanics

Slope = in 50, = 0.8 N-s/m, S = 0.9, Q = 6 lps = 6 0 m s To find : i) Flow is laminar or not? ii) P pump iii) u and du dy Step - : Calculate the Reynolds number to find the type of flow Discharge through the pipe is, Q = A u avg = 4 D u 6 0 = 4 (0.08) u avg u avg =.96 m/sec Reynolds number is, Re = f u avg D S w u avg D Re = avg 0.9 000.96 0.08 0.8 Re = 07.44 < 000 Flow is laminar Ans. Step - : Calculate the required power of pump Loss of head due to friction is, h f = u avg L 0.8.96 500 = D 0.9 980 (0.08) f h f = 70.86 m A slope of in 50 indicates that, for 50 m length there is m height (inclined pipe) of the pipe. It means, for 500 m length of pipe there is height of 0 m. Hence, the pump has to overcome frictional resistance in the pipe and the extra 0 m height of the pipe. Total head against which the pump works is, H = h f + Head due to slope = 70.86 + 0 = 80.86 m Power required by the pump is, P = f QH = 0.9 9806 0 80.86 P = 4. 859 0 W Ans. Step - : Calculate the centre line velocity and velocity gradient at the pipe wall Centre line (maximum) velocity is, u max = u avg.96 =.87 m/sec Ans. Pressure drop is given as, p p = 8 QL D 4 p p L p x 8 Q D4 Laminar Flow 7-5 Fluid Mechanics

Laminar Flow p = x 8 0.8 60 ( 0.08) 4 47746. 0 N m m Negative sign indicates pressure drop. As per Newton's law of viscosity, = du dy du dy = But at the pipe wall, y = 0 and shear stress is maximum, du dy y 0 = max = p x R du dy y 0 = 4.7746 0 0.8 0.04 du dy y 0 = 9.66 sec Ans. Example 7.0 Given data : = 6 poise =.6 P a-s, S = 0.9, D = 60 mm = 0.06 m R = 0.0 m p p =0kN/m =00 N/m, L =.5 m To find : i) Q in lpm iii) u max iii) F D Step - : Calculate the rate of flow of oil For laminar flow through pipe, p p = 8 QL D 4 0 0 = 8. 6 Q.5 006. 4 Q =.5 0 m /s... Ans. Step - : Calculate the center line velocity Discharge through pipe is, Q = Area Mean velocity = R u avg. 5 0 = (. 00) u avg u avg = 0.4687 m/s Maximum velocity in the pipe is, 7-6 Fluid Mechanics

Laminar Flow u avg = u max 0.4687 = u max u max = 0.975 m/s... Ans. Step - : Calculate the total friction drag over km of the pipe Total friction drag for 500 m length is, F D = F D = max Contact area = max DL p x R 0 0 DL = 000 00. 006..5 0 F D = 4.45 N... Ans. Laminar Flow ends 7-7 Fluid Mechanics

Unit - IV Chapter - 8 TURBULENT FLOW Solutions of Examples for Practice Example 8. Given data : D = 0. m, R = 0. = 0. m, u max = 5 m/sec, At 60 mm from the centre, y = 00 60 = 40 mm = 0.04 m, u = 4.4 m/sec To find : 0 Step : Calculate the shear stress at the wall For turbulent flow, umax u = 5.75 log u * 0 R y 5 4.4 = 5.75 log 0. u 0 * 0.04 u * = 0.6 m/sec Shear velocity is, u * = 0 0.6 = 0 000 0 = 68.759 N m Ans. Example 8. Given data : f = 0.04, u avg = 0.6 m/sec, Radial distance = 0.5 r r = Pipe radius y =r 0.5 r = 0.5 r To find : i) u ii) u max Step : Calculate the local velocity at a radial distance of 0.5 r 8 - Fluid Mechanics

Turbulent Flow For rough pipe, f But, R = log 0.74 k 0.04 = log R 0.74 k R log 0 =.6 k u u avg * R = 5.75 log 0 475. k 0.6 = 5.75.64.75 u * u * = 0.044 m/sec For smooth and rough pipe, u uavg y = 5.75 log.75 u 0 R * u 0.6 0.044 = 5.75 log 0 0.5r r.75 (R=r) u = 0.6857 m/sec Ans. Step : Calculate the velocity at the centre of pipe At the centre of pipe, y = D =randu=u max For smooth and rough pipe, u uavg = 5.75 log u * 0 y.75 R u max 0.6 = 5.75 log r 0.044 0 75 r. u max = 0.759 m/sec Ans. Example 8. Given data : D = 0.5 m, R = 0.5 = 0.5 m, u max =.5 m/sec. u = m/sec, y = 0.5 0.5 = 0. m To find : i) Q ii) f iii) k Step : Calculate the discharge through the pipe For turbulent flow, 8 - Fluid Mechanics

Turbulent Flow u max u * u.5 u * = 5.75 log 0 R y 0.5 = 5.75 log 0 0. u * = 0.85 m/sec For smooth and rough pipe, umax uavg =.8 u *.5 u avg 0.85 =.8 u avg =.6696 m/sec Discharge through the pipe is, Q = Area Velocity = 4 D u Q = 0.5. 6696 = 0.54 m 4 /sec Ans. Step : Calculate the friction factor and height of roughness projection We know that, u * = 0.85 = f u 8 avg f.6696 8 f = 0.055 Ans. Height of roughness projection for turbulent flow, = log R 0.74 f k 0.055 = log 0.5 0.74 k 0.5 log 0 =.898 k 0.5 k = 9.4908 k =.865 0 m =.865 mm Ans. avg Turbulent Flow ends 8 - Fluid Mechanics

Unit - V Chapter - 9 FLOW THROUGH PIPES Solutions of Examples for Practice Example 9.0 Given data : i) Pipes are connected in parallel : d =0cm=0.m, d =40cm=0.4m, Q = 00 lps = 0. m sec, L L = 50 m, f f = 0.0 ii) Pipes are connected in series : L = 600 m, Q = 0. m sec, f f = 0.0 To find : i) Q and Q ii) H Step - : Calculate the discharge through the each pipe when pipes are connected in parallel Refer Fig. 9. (a) For pipes in parallel, Q = Q Q = 0. (i) and h f = h φ 0. m f flq.d 5 = f L Q.d 5 Q Q Q But, f = f and L = L Q 5 (0.) = Q 5 (0.4) Q = 0.7 Q Q φ 0.4 m L = 50 m Fig. 9. (a) Q = 0.487 Q Substituting this value in equation (i), 0. = 0.487 Q Q 0. =.487 Q Q = 0.445 m sec = 4.48 lps Ans. 9 - Fluid Mechanics

Flow Through Pipes and Q = 0.487 Q = 0.487 0.45 Q = 0.0655 m sec = 65.5 lps Ans. Step - : Calculate the water level difference between the tanks Refer Fig. 9. (b). For pipes in series, Q = Q =Q=0.m sec Total head is given by, H = h h H = flq.d 5 + f L Q.d 5 f f H φ 0. m φ 0. m 50 m 50 m Fig. 9. (b) H = 0.0 50 (0.). (0.) 5 0.0 50 (0.). (0. 4) 5 H = 9.58 +.598 H =.78 m Ans. Example 9. Given data : H = 8 m, d = 00 mm = 0. m, L = 00 m, h c =4m, L = 0 m, f = 0.0. To find : i) Discharge Q ii) Pressure at summit (p c ) Step - : Calculate the discharge through the siphon If minor losses are neglected, then the loss of head is, h f =H = flq. d 5 8 = 0.000Q. (0.) 5 Q = 0.09 m sec Ans. 9 - Fluid Mechanics

Flow Through Pipes c Summit A h c Reservoir Inlet leg Outlet leg B H Z A Z c Reservoir ZB Datum Step - : Calculate the pressure at the summit Pressure at the summit is, p c V = h g + fl c d But, Q = AV V= Q A = 0.09 =.49 m/sec (0.) 4 p c (.49) = 4 9.8 Fig. 9. 0.0 0 0. p c =.05 m of water Ans. or p c =.05 980 = 8.4 0 Pa (Vacuum) Ans. Example 9. Given data : d = 5 cm = 0.5 m, L = 4 km = 4 0 L =km= 0 m, f = 0.05 m, H = 6 m To find : Increases in discharge. Step - : Calculate the discharge through the single pipeline 9 - Fluid Mechanics

Flow Through Pipes Refer Fig. 9. (a). Discharge through the single pipe line is, H=h f = flq. d 5 6 = 0.054 0 Q. (0.5) 5 φ 5 cm 4km Fig. 9. (a) 6 m Q = 0.05 m sec Step - : Calculate the increase in discharge due to installation of new pipe Refer Fig. 9. (b). As pipes and are in parallel, h f = h f 5 = fl Q. d 5 fl Q. d But, d = d, L = L km km φ 5 cm Fig. 9. (b) 6 m Q = Q Total discharge is, Q = Q Q = Q For pipes in series (pipes and ), 6 = H = h h f f = fl Q. d 0.050 Q. (0.5) 5 5 + 5 fl Q. d 0.050 (Q ). (0.5) 5 6 = 54.460 Q + 7.6648 0 Q = 7.080 0 Q Q = 9.7754 0 m sec Q = Q = 9.7754 0 = 0.0955 m sec Increase in discharge is Q Q = 0.0955 0.05 Q Q = 4.55 0 m sec Ans. 9-4 Fluid Mechanics

Example 9. Given data : d = 7 cm = 0.7 m, L =8m,d =.5 cm = 0.5 m, L = 8 m, H = 8 m, Coefficient of friction = 0.04, To find : Discharge (Q) and draw EGL Step - : Calculate the discharge through the pipe line Refer Fig. 9.4 (a). Friction factor, f = 4 Coefficient of friction f = 4 0.04 = 0.6 As per continuity equation, Q = A V = A V V = A A V V = d d V = (0.5) (0.7) V =.88 V Total head loss through the pipe is, Loss of head Loss of head H = at entrance due to friction Loss of head due to friction H = hen hf he hf hex H = 0.5 V fl V (V V ) g gd g Loss of head at exit fl V gd Loss of head due to sudden expansion V g 8 = 0.5(.88 V ) 0.6 8 (.88 V ) (.88 V V ) 9.8 9.8 0.7 9.8 0.6 6 V V + 9.8 0.5 9.8 8 = 0.86 V + 5.77 V + 0.47 V + 0.895 V + 0.05 V 8 = 7.068 V V =.0677 m/sec and V =.88 V =.88.0677 = 4.46 m/sec d = 0.7 m 8m 8 m Flow Through Pipes d = 0.5 m Fig. 9.4 (a) 8m 9-5 Fluid Mechanics

Flow Through Pipes Discharge through the pipe line is, Q = A V = 4 d V = (0.5).0677 4 Q = 0.094 m sec Ans. h en = 0.86 V = 0.86 (.0677) = 0.47 m h f = 5.7 V = 5.7 (.0677) = 5.897 m h e = 0.47 V = 0.47 (.0677) h f = 0.895 V = 0.895 (.0677) = 0.488 m = 0.4440 m h ex = 0.05 V = 0.05 (.0677) = 0.058 m Refer Fig. 9.4 (b) for Energy Gradient Line (EGL). A 0.058 m 0.4096 m EGL 5.897 m HGL 0.488 m H=8m EGL 0.4440 m HGL C 0.058 m B Fig. 9.4 (b) Example 9.4 Given data : L = 800 m, d = 50 mm = 0.5 m, H = 8 m, f = 0.0 To find : Discharge and length of new pipe. Step - : Calculate the discharge through the single pipe Refer Fig. 9.5 (a). Ø 0.5 m 8 m Neglecting minor losses, 800 m H = h f flq. d 5 Fig. 9.5 (a) 9-6 Fluid Mechanics

Flow Through Pipes 8 = 0.0 800 Q. (0.5) 5 Q = 0.78 m sec Step - : Calculate the length of new pipe Refer Fig. 9.5 (b). Increase in discharge = 5 % 0.5 = Q Q = Q 0.78 Q 0.78 Q = 0.05 m sec For parallel pipes, h f = h f fl Q. d 5 = fl Q. d L Q = L Q 5 900 m 900 m Ø 0.5 m L Fig. 9.5 (b) 6 m Also, Q = Q Q Total loss of head is, flq H = hf hf. d 5 8 = 0.0 900 (0.05). (0.5) 5 Q = 0.467 m sec fl Q 5. d 0.0 900 Q. (0.5) 5 (i) But, Q = Q Q 0.05 = 0.467 + Q Q = 0.058 m sec Substituting these values in equation (i), 900 (0.467) = L (0.058) L = 5698.5665 m = 5.6985 km... Ans. Flow Through Pipes ends 9-7 Fluid Mechanics

Unit - V Chapter - 0 DIMENSIONAL ANALYSIS Solutions of Examples for Practice Example 0.6 The functional relationship between the dependent and indepenedent variables is given by, T = f(d, N,,,V) or f(t, D, N,,,V ) = 0 or constant Dimensions of different variables are, T D N V [MLT ] [L] [T ] [ML ] [ML T ] [LT ] Total number of variables n = 6 Number of primary dimensions m = Number of terms =n m=6 =(,, ) f(,, ) = 0 or constant (i) As m =, the repeating variables are also. As T is dependent variable, it cannot be taken as repeating variable. The repeating variables are selected such that, i) Geometric property (D) ii) Flow property (N) iii) Fluid property () As per Buckingham theorem the terms can be written as, = Dx Ny z T (ii) = D x y z (iii) = D x y z V (iv) ) Consider equation (ii) and substitute the dimensions, [M0 L0 T 0 ] = [L] x [T ] y [ML ] z [ML T ] Equating the powers of MLT on both sides, 0 - Fluid Mechanics

Dimensional Analysis For M : 0 = z z = For L : 0 = x z x = 5 For T : 0 = y y = Substituting these values in equation (ii), = D 5 N T = D 5 ) Consider equation (iii) and substitute the dimensions, [M0 L0 T 0 ] = [L] x [T ] y [ML ] z [ML T ] Equating the powers of MLT on both sides, For M : 0 = z z = For L : 0 = x z x = For T : 0 = y y = Substitute these values in equation (iii), = D N = ND ) Consider equation (iv) and substitute the dimensions, [M0 L0 T 0 ] = [L] x [T ] y [ML ] z [LT ] Equating the powers of MLT on both sides, For M : 0 = z z = 0 For L : 0 = x z x = For T : 0 = y y = Substitute these values in equation (iv), = D N 0 V = V DN Now, put the values of, in equation (i), or T f N D, ND, V 5 DN = 0 or constant T N D 5 = ND, V DN 0 - Fluid Mechanics

Dimensional Analysis T = N D 5 ND, DN V Ans. Example 0.7 The functional relationship between the dependent and indepenedent variables is given by, = f(,v,d,,k) or f(,, V, D,,K ) = 0 or constant Dimensions of different variables are, V D K [ML T ] [ML ] [LT ] [L] [ML T ] [L] Total number of variables n = 6 Number of primary dimensions m = Number of terms =n m=6 =(,, ) f(,, ) = 0 or constant (i) As the number of primary dimensions m =, the repeating variables are also. As is a dependent variable, it cannot be taken as repeating variable. The repeating variables are selected such that, i) Geometric property (D) ii) Flow property (V) iii) Fluid property () As per Buckingham theorem the terms can be written as, = Dx Vy z (ii) = Dx Vy z (iii) = Dx Vy z K (iv) ) Consider equation (ii) and substitute the dimensions, [M0 L0 T 0 ] = [L] x [LT ] y [ML ] z [ML T ] Equating the powers of MLT on both sides, For M : 0 = z z = For L : 0 = x y z x +y = For T : 0 = y y = and x = 0 Substituting the values of x,y and z in equation (ii), = D0 V 0 - Fluid Mechanics

Dimensional Analysis = V ) Consider equation (iii) and substitute the dimensions, [M0 L0 T 0 ] = [L] x [LT ] y [ML ] z [ML T ] Equating the powers of MLT on both sides, For M : 0 = z z = For L : 0 = x y z x y = For T : 0 = y y = and x = Substituting the values of x,y,z equation (iii), = D V = VD ) Consider equation (iv) and substitute the dimensions, [M0 L0 T 0 ] = [L] x [LT ] y [ML ] z [L] In this case, as the dimensions of D and K are same, We can write by observation, = K D Now, put the values of,, in equation (i), f V, VD, K = 0 D or V = VD K, D Example 0.8 Given data : Scale ratio L m Lp To find : i) H P ii) Q P or constant = 0, H m = 0.4 m, Q m = 0. m /sec As it is a spillway model, Froude law must be applicable.. L r = L m = L Lp 0 P =0L m Ans. H P = 0 H m =0 0.4 =.4 m Ans. 0-4 Fluid Mechanics

Dimensional Analysis. Q r = A r V r = L L r r = L5/ r Q r = 0 5/ = Q m Q p = 0. Q p Q P =.67 m /sec Ans. Example 0.9 Given data : d p =.5 cm = 5 mm, d m = 6 mm, V m =m/s, p = 85 kg/m To find : V p m = 995.7 kg/m, p = 0.0 Pa-S, For flow through pipe, Reynolds law is applicable. As VD = p VD m m = 0.80 Pa-S 85 V 5 = 995.7 6 0.0 0.80 V P = 0.06 m/s Ans. Example 0.0 Given data : Scale ratio L m = Lp 6, H P = 7. m, L p = 50 m, Q p = 50 m /s, h p =4m As it is a spillway model, Froude law must be applicable.. L r = L m = Lp 6 L m = 50 6. Q r = L5/ r Q r = (/6) 5/ = Q m Q P L m = L p 6 = 9.75 m Ans. = Q m 50 Q m =.0996 m /sec Ans.. h m h p = L r = 6 h = 6 h m = 4 = 0.5 m 6 Ans 0-5 Fluid Mechanics

Dimensional Analysis 4. L r = 6 = L m Lp L p =6L m H p = 6 H m 7. = 6 H m H m = 0.45 m Ans Dimensional Analysis ends 0-6 Fluid Mechanics

Unit - VI Chapter - BOUNDARY LAYER THEORY Solutions of Examples for Practice Example. Momentum thickenss is, = = = = = 0 0 0 0 u u U U dy = 0 u u U U dy y y y y y y 9y y y y 6 4 y y 9y y 4 y 6 4 4 6 y y 4 9 y 4 4 y5 4 5 4 5 7 6 7 y 4 7 = 9 = 9 4 8 0 4 8 6 4 8 0 8 = 9 80 0 Ans. Example. u i) U = y y u = U y U y - Fluid Mechanics

Boundary Layer Theory Differentiating above equation with respect to y, du 4Uy Uy = dy at y = 0 du = 0 dy y 0 du As ii) dy y 0 has zero value, the flow is just on the verge of separartion. U u = y y Ans. Uy Uy u = Differentiating above equation with respect to y, du Uy U = dy at y = 0 du U = dy y 0 du As dy y 0 has negative value, the flow is detached or separated. Ans. Example. Given data : L=4m, d=.5m, U=4.4 km/hr = 4 m/sec. air =. kg/m, =.8 0 4 poise =.8 0 5 Pa-s To find : Drag force on both sides of plate. Step - : Calculate the drag force when boundary layer is laminar Reynold's number over the complete plate is, Re L = UL. 4 4 =.8 0 5 =.067 0 6 Average coefficient of drag for laminar boundary layer is, C Da =.8.8 = =.858 0 Re L.067 0 6 - Fluid Mechanics

Boundary Layer Theory Drag force on both sides of plate is, F D = F D = C Da AU.858 0. 4.5 (4) (Both sides) F D = 0.48 N Ans. Step - : Calculate the drag force when boundary layer is turbulent Average coefficient of drag for turbulent boundary layer is, 0.074 C Da = ( Re L ) 5 = 0.074 (. 067 0 6 ) 5 = 4.6089 0 Drag force on both sides of plate is, F D = F D = C Da AU 4.6089 0. 4 5. ( 4) (Both sides) F D = 0.509 N Ans. Example.4 Given data : L=m, b=m, =.75 05Pa - s Re = 5 0 5,U=m/sec., =. kg/m To find : Drag force F D Step - : Calculate the drag force on both sides of the plate. 0.074 700 C Da = (Re ) Re L 5 L But, Re L = UL. =.75 0 5 = 4.4 0 0.074 700 C Da = (4.4 0 ) 5 (4.4 0 ) C Da =.4446 0 Drag force on both sides of the plate is, F D = C Da AU (Both sides) F D =.4446 0. F D = 0.008 N Ans. Boundary Layer Theory ends - Fluid Mechanics

Unit - VI Chapter - FORCES ON IMMERSED BODIES Solutions of Examples for Practice Example. Given data : A = 0.5 m,u=45km/hr=.5 m/sec, W =.7 N, = 8º, T=4N, To find : C L and C D. air =.5 kg/m Step - : Calculate the coefficients of lift and drag From Fig.., F y = 0 = FL W T cos (8) F L = W T cos (8) U = CL A.74 cos (8) = C 0.5.5 (.5) L C L = 0.750 Ans. Now, F x = 0 = FD T sin (8) F D = T sin (8) But, F D U = CD A 4 sin (8) =.5.5 CD 0.5 C D = 0.09 Ans. U T=4N 8º Fig.. F L W =.7 N Kite F D - Fluid Mechanics

Example.4 Given data : W = 8.6 kn = 8.6 0 N, U = 960 km/hr = 66.67 m/sec, Span C = 6 m, A = m, CD 0.0 air =. kg/m To find : i) C L ii) P iii) Step - : Calculate the coefficient of lift We know that, Lift force = Weight of plane i.e. F L = W U CL A = W. (66.67) CL = 8.6 0 C L = 0.069 Ans. Step - : Calculate the required power and boundary layer circulation Power given is, P = FD U U D U U P = CD A. (66.67) = 0.0 P =.45 0 6 W=.45 0 kw Ans. Theoretical boundary layer circulation is, = CUsin As plane is flying in horizontal direction, = 0º. = 5 66.67 sin0 = 0 Ans. Forces on Immersed Bodies Example.5 Given data : D=6cm=0.06 m, S =. kn/m =. 0 N/m, U = m/sec To find : Tension T and its inclination. Step - : Calculate the tension in the string and its inclination Drag force is, - Fluid Mechanics

Forces on Immersed Bodies F D = U CD A = C D D 4 F D =.5 0 0.6 0.06 4 F D = 0.44 N Weight of sphere (W) = S Volume of sphere 4 0.06 W =. 0 U θ Sphere W =.546 N From Fig.., Weight W = T cos But, tan = FD = 0.44 W.546 = 9.559º Ans..546 = T cos (9.559) T =.5599 N Ans. Air flow θ W Fig.. F D Example.6 Given data : L=b=m, air =.4 kg/m,u=60kmph=6.67 m/sec, C D = 0., C L = 0.8 To find : i) F L ii) F D iii) F R iv) v) P Step - : Calculate the lift force and drag force on the plate U Lift force is, F L = CL A.4 6.67 = 0.8 F L = 400.56 N Ans. Drag force is, F D = U CD A.4 6.67 = 0. F D = 50.4 N Ans. Step - : Calculate the resultant force and power required to keep the plate in motion Resultant force is, F R = F F D L = 50.4 400.56 - Fluid Mechanics

F R = 44.6 N Ans. Direction of resultant force is, tan = F L FD tan = 400.56 50.4 = 75.96º Ans. Power required to keep the plate in motion, Forces on Immersed Bodies P = FD U = 50.4 6.67 P = 586.8 W Ans. Forces on Immersed Bodies ends - 4 Fluid Mechanics