# CHAPTER 3 BASIC EQUATIONS IN FLUID MECHANICS NOOR ALIZA AHMAD

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1 CHAPTER 3 BASIC EQUATIONS IN FLUID MECHANICS 1

2 INTRODUCTION Flow often referred as an ideal fluid. We presume that such a fluid has no viscosity. However, this is an idealized situation that does not exist. 2

3 INTRODUCTION (cont d) Real fluid, the effects of viscosity are introduced into the problem 3

4 INTRODUCTION (cont d) Classification of types of flow Incompressible fluid flow assumes the fluid has constant density ( = constant), though liquids are slightly compressible we usually assume them to be incompressible 4

5 INTRODUCTION (cont d) Steady flow means steady with respect t time. Thus all properties of the flow at every point remain constant with respect to time. Uniform flow happened when the cross section ( shape and area) through which the flow occurs remains constant 5

6 INTRODUCTION (cont d) Path line is the trace made by a single particle over a period of time. The path line shows the direction of the velocity. 6

7 INTRODUCTION (cont d) Stream line shows the mean direction of a number of particles at the same instant time. Flowrate is known as quantity of fluid flowing per unit time across any section. The flowrate can be expressed in terms of i) Volume flow rate ( discharges) BG cfs (cubic per second), gpm ( gallon per minute, mgd ( million gallon per day) SI m 3 /d 7

8 INTRODUCTION (cont d) ii) mass flow rate BG slugs per second SI kg/s iii) weight flow rate BG pounds per second SI kn/s Incompressible fluid volume flow rate Compressible fluid weight & mass flowrate 8

9 INTRODUCTION (cont d) Mean velocity at Point P = u, volume flow rate passing through the element of area da is 9

10 INTRODUCTION (cont d) dq = u. da = ( u cos ) da = u ( cos da ) = u. da da = projection of da on the plane normal to the direction of U. This indicates that the 10

11 INTRODUCTION (cont d) Volume flow rate = Q = magnitude of the mean velocity multiplied by the flow area at right angle to the direction of the mean velocity. Q uda A AV (4.3) How to calculate mass flow rate( m) and weight flow rate (G)??? 11

12 INTRODUCTION (cont d) V Q A m G A A (4.6) Write the mean velocity,v for circular pipe. 12

13 INTRODUCTION (cont d) Example 1 The velocity of a liquid (s =1.4 m) in a 150 mm pipeline is 0.8 m/s. Calculate the rate of flow in L/s, m 3 /s, kg/s and kn/s. 13

14 INTRODUCTION (cont d) Solution: Q AV m s Q Convert to L/s Known 1000 L = 1 m m L s 3 s 1000L 3 1m 14

15 Solution: m Q kg/ 3 3 m m / s 19.79kg/ m G Q kN / m m / s kN / s INTRODUCTION (cont d) 15

16 EQUATION OF CONTINUITY (PERSAMAAN KESELANJARAN) Figure shows the short length pf stream tube where no fluid can leave or enter the stream tube except at the ends. The fixed volume between two section 1 and 2 is a control volume 16

17 Steady flow m in mout EQUATION OF CONTINUITY( cont d) (PERSAMAAN KESELANJARAN) (control volume) AV A V g G in G out m g 1 m2 1 AV A V (control volume) 17

18 EQUATION OF CONTINUITY( cont d) (PERSAMAAN KESELANJARAN) Incompressible flow ( = constant) for both steady and unsteady flow A 1 V 1 = A 2 V 2 = Q (4.17) 18

19 Example 2 EQUATION OF CONTINUITY( cont d) (PERSAMAAN KESELANJARAN) Water flows in a river at 9.00 am the flow past bridge 1 is 37.2 m 3 /s. At the same instant the flow past bridge 2 is 26.9 m 3 /s. At what rate is water being stored in the river between the two bridges at this instant? 19

20 Solution: 1 EQUATION OF CONTINUITY( cont d) (PERSAMAAN KESELANJARAN) Q Q 2 m 3 s dv dt dv dt dv 10.30m 3 dt s 20

21 BERNOULLI EQUATION Consider frictionless steady flow of an ideal fluid along the streamline as shown below. We shall consider the forces acting in the direction of the streamline on a small element of the fluid in the stream tube and we shall apply Newton s second law, that is F = ma. 21

22 BERNOULLI EQUATION (cont d) Element moving along streamline( ideal fluid) The crosssectional area of the element at right angles to the streamline may have any shape and varies from A to A + da 22

23 BERNOULLI EQUATION( cont d) In steady flow the velocity does not vary at a point (local acceleration = 0). but that it may vary with position (convective acceleration 0). One-dimensional Euler Equation: dp dz d V 2 2g 0 (5.6) 23

24 BERNOULLI EQUATION (cont d) Incompressible fluid For the case of an incompressible fluid ( = constant), we can integrate Eq. (5.6) to give 24

25 BERNOULLI EQUATION (cont d) Incompressible fluid If we multiply each term first by g and then by, we obtain the following alternate forms: 25

26 BERNOULLI EQUATION (cont d) Example 3 Glycerin (specific gravity 1.26) in a processing plant flows in a pipe at a rate of 700 L/s. At a point where the pipe diameter is 600 mm, the pressure is 300 kpa. Find the pressure at a second point where the pipe diameter is 300mm if the second point is 1.0 m lower than the first point. Neglect head loss. 26

27 BERNOULLI EQUATION (cont d) 27

28 BERNOULLI EQUATION (cont d) For an incompressible fluid ( = constant), we can integrate directly. Integrating from some point 1 to another point 2 on the same streamline, where the distance between them is L,we get for an incompressible real fluid 28

29 BERNOULLI EQUATION (cont d) Element moving along streamline( real fluid) 29

30 BERNOULLI EQUATION (cont d) 30

31 BERNOULLI EQUATION (cont d) Example 4 If h = 10.5 m in Fig. X5.3.4 and the pressures at A and B are 170 and 275 kpa respectively, find the direction of flow and the pipe friction head loss in meters of liquid. Assume the liquid has a specific gravity of

32 BERNOULLI EQUATION (cont d) Solution: 32

33 APPLICATION OF BERNOULLI EQUATION 1.Water Discharge From An Orifice (Example 5, 6 ) 2.Velocity Measurement by A Pitot Tube (Example 7) 3. Flow Measurement Using Meter Venturi (Example 8) 33

34 Example 5 A large tank open to the atmosphere is filled with water to a height of 5 m from the outlet tap. A tap near the bottom of the tank is now opened, and water flows out from the smooth and rounded outlet. Determine the water velocity at the outlet. APPLICATION OF BERNOULLI EQUATION (cont d) 34

35 APPLICATION OF BERNOULLI EQUATION (cont d) Solution: We take point 1 to be at the free surface of water so that P 1 = P atm (open to the atmosphere), V 1 = 0 (the tank is large relative to the outlet), and z 1 = 5 m and z 2 = 0 (we take the reference level at the center of the outlet). Also, P 2 = P atm (water discharges into the atmosphere). Then the Bernoulli equation simplifies to 35

36 APPLICATION OF BERNOULLI EQUATION (cont d) The relation V is called the Toricelli equation. 36

37 APPLICATION OF BERNOULLI EQUATION (cont d) Example 6 A closed tank has an orifice 0.025m diameter in one of its vertical sides. The tank contains oil to a depth of 0.61m above the centre of the orifice and the pressure in the air space above the oil is maintained at N/m 2 above atmospheric. Determine the discharge from the orifice. (Coefficient of discharge of the orifice is 0.61, relative density of oil is 0.9). 37

38 APPLICATION OF BERNOULLI EQUATION (cont d) Solution: Apply Bernoulli, 38

39 APPLICATION OF BERNOULLI EQUATION (cont d) Take atmospheric pressure as 0, 39

40 APPLICATION OF BERNOULLI EQUATION (cont d) Example 7 A piezometer and a Pitot tube are tapped into a horizontal water pipe, as shown, to measure static and stagnation (static dynamic) pressures. For the indicated water column heights, determine the velocity at the center of the pipe. 40

41 APPLICATION OF BERNOULLI EQUATION (cont d) Solution: We take points 1 and 2 along the centerline of the pipe, with point 1 directly under the piezometer and point 2 at the tip of the Pitot tube. This is a steady flow with straight and parallel streamlines, and the gage pressures at points 1 and 2 can be expressed as 41

42 APPLICATION OF BERNOULLI EQUATION (cont d) Noting that point 2 is a stagnation point and thus v 2 =0 and z 1 =z 2, the application of the Bernoulli equation between points 1 and 2 gives 42

43 APPLICATION OF BERNOULLI EQUATION (cont d) Example 8 A Venturimeter of throat diameter 0.076m is fitted in a 0.152m diameter vertical pipe in which liquid of relative density 0.8 flows downwards. Pressure gauges are fitted to the inlet and to the throat sections. The throat being 0.914m below the inlet. Taking the coefficient of the meter as 0.97 find the discharge when the pressure gauges read the same. 43

44 APPLICATION OF BERNOULLI EQUATION (cont d) Solution: 44

45 APPLICATION OF BERNOULLI EQUATION (cont d) By continuity: 45

46 APPLICATION OF BERNOULLI EQUATION (cont d) 46

47 MOMENTUM EQUATIONS 47

48 MOMENTUM EQUATIONS (cont d) But since the flow we are considering is steady. from continuity,m 1 = m 2 = 1 Q 1 = 2 Q 2 = Q. Also. using the vector relations of Fig. 6.lb, let us for convenience write V = V 2 - V 1. Using these. Eq. (6.5) becomes Steady flow m(av) = Q( V) = Q(V 2 V 1 ) (6.6) 48

49 MOMENTUM EQUATIONS (cont d) 49

50 APPLICATION OF MOMENTUM EQUATIONS a) Structure open to the atmosphere (Example 9 & 10) b) Force of Pressure Conduits and Bend ( Example 11 ) c) Force of A Stationary Vane or Blade (Example 12,13 &14) d) Force of Nozzle ( Example 15) 50

51 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Example 9 The water passage in Fig. S6.1 is 3 m wide normal to the plane of the figure. Determine the horizontal force acting on the shaded structure. Assume ideal flow. 51

52 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Solution: Free Body Diagram 52

53 APPLICATION OF MOMENTUM EQUATIONS (Cont d) 53

54 APPLICATION OF MOMENTUM EQUATIONS (Cont d) 54

55 Example 10 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Flow occurs over the spillway of constant section as shown in Fig. X Given that y 1, = 4.2 m and y 2 = 0.7 m, determine the horizontal force on the spillway per meter of spillway width (perpendicular to the spillway section). Assume ideal flow. 55

56 APPLICATION OF MOMENTUM EQUATIONS (Cont d) 56

57 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Energy: V 12 /(2x9.81) = V 22 /(2x9.81) (1) Continuity per m: 4.2 V 1 = 0.71V 2 (2) Substituting (2) into (1) yields: V 1 = m/s, V 2 = 8.40 m/s ; Q = A 1 V 1 = 5.88 m 3 /s per meter 57

58 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Eq. 6.7a: F = F 1 -F 2 - F x = Q(V 2x -V 1x ) 9810(2.1) (0.35)0.7 F x = 1000(5.88)( ) = N/m to the left. Water (F w/s ) x acts on spillway to the right with 42.9 kn/m. 58

59 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Force of Pressure Conduits and Bend Consider the case of horizontal flow to the right through the reducer of Fig. 6.3a. A free-body diagram of the forces acting on the fluid mass contained in the reducer (the control volume (CV) is shown in Fig. 6.3b. We shall apply Eq. (6.7a) to this fluid mass to examine the forces that are acting in the x direction. 59

60 APPLICATION OF MOMENTUM EQUATIONS (Cont d) The forces P 1 A 1 and P 2 A 2 represent pressure forces that fluid located just upstream and just downstream exerts on the control volume. The forces Fx represents the force exerted by the reducer on the fluid ( CV) in the x direction. 60

61 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Neglecting shear forces at the boundary of the reducer, the force Fx is the resultant (integrated) effect of the normal pressure forces that the wall of the reducer exerts on the fluid. 61

62 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Applying Eq. (6.7a) and assuming the fluid is ideal with F directed as shown, since the entry and exit velocities are parallel to the x direction, we get Fx = P 1 A 1 P 2 A 2 Fx = Q(V 2 -V 1 ) (6.10) Fx = P 1 A 1 -P 2 A 2 - Q(V 2 -V 1 ) (6.11) 62

63 APPLICATION OF MOMENTUM EQUATIONS (Cont d) 63

64 APPLICATION OF MOMENTUM EQUATIONS (Cont d) If the fluid undergoes a change in both direction and velocity, as in the reducing pipe bend, with applying Eq (6.7a) by summing up x- forces acting on the fluid in the CV, and equating them to the change in fluid momentum in the x direction, gives Fx = P 1 A 1 P 2 A 2 cos Fx = Q(V 2x -V 1x )(6.12) 64

65 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Known that V 2x = V 2 cos and V 1x = V 1, Fx = P 1 A 1 P 2 A 2 cos - Q(V 2 cos - V 1 ) (6.13) in the y direction, Fy = 0 P 2 A 2 sin + Fy = Q(V 2y - V 1y ) ( 6.14) Known that V 2x = V 2 sin and V 1y = 0, Fx = P 2 A 2 sin + Q(V 2 sin ) (6.15) 65

66 Example 11 APPLICATION OF MOMENTUM EQUATIONS (Cont d) A water flows in a pipe which bend to the horizontal axis at The inlet pipe s diameter is 600mm and reduce to 300mm in the end. Given the inlet s pressure and flow rate are 140kPa and m 3 /s respectively. Neglecting the friction, calculate the resultant force at the bend.. P 1 A 1 1 v 1 2 P 2 A 2 v 2 =

67 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Solution: v 1 = Q/A 1 = 0.425/ ( (0.6) 2 /4) = 1.5 m/s v 2 = Q/A 2 = 0.425/ ( (0.3) 2 /4) = 6.0 m/s A 1 = ( (0.6) 2 /4) = m 2 A 2 = ( (0.3) 2 /4) = m 2 67

68 APPLICATION OF MOMENTUM EQUATIONS (Cont d) 1) Apply Bernoulli Eqn. at Point 1 and 2 (P 1 / g) + (v 12 /2g) + (z 1 ) = (P 2 / g) + (v 22 /2g) + (z 2 ) Placed in the horizontal axis, z 1 = z 2 (140 x 10 3 / 10 3 x 9.81) + (1.5 2 / 2 x 9.81) = (P 2 / 10 3 x 9.81) + (6 2 / 2 x 9.81) P 2 = 1.23 x 10 5 N/m 2 68

69 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Fx = Momentum Changes in x direction P 1 A 1 P 2 A 2 kos - Fx = Q(v 2 kos - v 1 ) (140 x 10 3 )( 0.282) (123x 10 3 ) ( 0.071) ( kos 45) Fx = (10 3 ) ( 0.425) ( 6 kos ) Fx = N Fy = Momentum Changes in y direction P2A2sin + Fy = Q(v2sin - 0) 123x 10 3 ) ( 0.071) ( sin 45) + Fy = (10 3 ) ( 0.425) ( 6 sin 45) Fy = 7970 N 69

70 F = Fx 2 + Fy 2 = (32260) 2 + (7970) 2 = N = 33 kn Direction, = tan-1 Fx /Fy = tan /7970 = (to horizontal) APPLICATION OF MOMENTUM EQUATIONS (Cont d) 70

71 Example 12 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Refer to Fig. P6.15. Assuming ideal flow in a horizontal plane, calculate the magnitude and direction of the resultant force on the stationary blade. Note that the jet (Vj, = 12 mis, D = 150 mm) is divided by the splitter so that one-third of the water is diverted toward A. 71

72 APPLICATION OF MOMENTUM EQUATIONS (Cont d) 72

73 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Solution: 73

74 Example 13 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Refer to Fig. X Assume that friction is negligible, that =115, and that the water jet has a velocity of 25 m/s and a diameter of 40 mm. Find (a)the component of the force acting on the vane in the direction of the jet; (b)the force component normal to the jet; and (c) the magnitude and direction of the resultant force exerted on the blade. 74

75 APPLICATION OF MOMENTUM EQUATIONS (Cont d) 75

76 Solution: APPLICATION OF MOMENTUM EQUATIONS (Cont d) 76

77 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Example 14 A water jet strikes on a vane at If water flows and velocity are 0.68 kg/s and 24 m/s respectively, calculate: a) Resultant force at stationary vane b) Resultant force at vane if the vane moving at velocity 8m/s in jet direction c) Output Power if (b) replaced by the set of vanes 77

78 APPLICATION OF MOMENTUM EQUATIONS (Cont d) y Solution: a) Jika bilah pegun v 1 v x Direction x, v 1x = 24 m/s v 2x = - v 1x kos 30 =- 24 m/s kos 30 v = v 2x- v 1x = (-24 m/s kos 30) 24 m/s = m/s 78

79 Given, Flow =0.68kg/s Fx = Flow x v =0.68kg/s x-44.8 m/s = N ( ) APPLICATION OF MOMENTUM EQUATIONS (Cont d) Direction y,v 1y = 0 m/s v 2y = - v sin 30 v = v 2y - v 1y = (-24 m/s sin 30) 0 m/s = m/s Given, Flow =0.68kg/s Fy = Flow x v =0.68kg/s x-12 m/s = N ( ) 79

80 APPLICATION OF MOMENTUM EQUATIONS (Cont d) F = Fx 2 + Fy 2 = (30.45) 2 + (8.16) 2 = 31.5 N ( magnitude) Direction, = tan-1 Fy /Fx = tan /30.45 = 15 0 x 15 0 y 80

81 APPLICATION OF MOMENTUM EQUATIONS (Cont d) b) Resultant force at vane if the vane moving at velocity 8m/s in jet direction Given, Flow =0.68kg/s= av 0.68 kg/s = 10 3 x a x v a = 0.68kg/s / 10 3 x v Moving vane, = 2.83 x 10-5 m 2 flow = a(v-u) = 10 3 x 2.83 x 10-5 ( 24-8) = kg/s 81

82 APPLICATION OF MOMENTUM EQUATIONS (Cont d) Direction x, Initial velocity, v 1x = v-u = 24-8 = 16 m/s Final Velocity, v 2x = - v 1x kos 30 =- 16 m/s kos 30 v = v 2x - v 1x = (- 16 m/s kos 30) 16 m/s = m/s Resultant force, Fx = flow x v =0.453kg/s x m/s = N ( ) 82

83 Direction y, Initial velocity, v 1y = 0 m/s Final Velocity v 2y = - (v-u ) sin 30 =- 16 m/s sin 30 v = v 2x - v 1x = (- 16 m/s sin 30) 0 m/s = -8 m/s APPLICATION OF MOMENTUM EQUATIONS (Cont d) Resultant force, Fy = flow x v =0.453kg/s x-8 m/s = -3.6 N ( ) S2S2008/

84 F = Fx 2 + Fy 2 = (13.5) 2 + (3.6) 2 = 14 N APPLICATION OF MOMENTUM EQUATIONS (Cont d) Direction, = tan-1 Fy /Fx = tan /13.5 = 15 0 x 15 0 y S2S2008/

85 APPLICATION OF MOMENTUM EQUATIONS (Cont d) c) Output Power if (b) replaced by the set of vanes Notes: Vanes move at 8 m/s. Water jet strikes in x direction Direction x, Initial velocity, v1x = v-u = 24-8 = 16 m/s Final velocity v2x = - v1x kos 30 =- 16 m/s kos 30= S2S2008/

86 v = v 2x - v 1x = (- 16 m/s kos 30) 16 m/s = m/s Fx = flow x v =0.68kg/s x m/s = N ( ) Output Power = Fx x u = 20.3 N x 8m/s = Nm/s = watt APPLICATION OF MOMENTUM EQUATIONS (Cont d) S2S2008/

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