NAS-Unisel -MAC13 1At the end of this course students should be able to: Explain and analyze the basic concept of flow of incompressible fluid in a pipe NAS-Unisel -MAC13 2 At the end of this course students should be able to: identify and understand various characteristics of the flow in pipes discuss the main properties of laminar and turbulent pipe flow and appreciate their differences. calculate losses in straight portions of pipes as well as those in various pipe system components. NAS-Unisel -MAC13 3 Reynolds Number where V avg average flow velocity (in m/s),ρ is density (in kg/m 3 ) D is characteristic length of the geometry (in m), υ is kinematic viscosity of the fluid (in m 2 /s) µ Is the dynamic viscosity of the fluid (in N.s/m 2 or Pa.s or kg/m.s )of the fluid Inertial forces Re Viscous forces avg avg V D V D ρ υ µ = = = Osborne Reynolds Sir George Stokes NAS-Unisel -MAC13 4 NAS-Unisel -MAC13 5 • At small or moderate Reynolds numbers, the viscous forces are large enough to suppress these fluctuations and to keep the fluid “in line” • At large Reynolds numbers, the inertial forces, which are proportional to the fluid density and the square of the fluid velocity, are large relative to the viscous forces, and thus the viscous forces cannot prevent the random and rapid fluctuations of the fluid NAS-Unisel -MAC13 6 NAS-Unisel -MAC13 7 Flow issuing at constant speed from a pipe: (a) High viscosity, low-Reynolds-number, laminar flow; (b) Low viscosity, high-Reynolds-number, turbulent flow. NAS-Unisel -MAC13 8 For flow in tubes/pipes, Re 2300 laminar 2300 Re 4000 transitional Re 4000 turbulent ≤ ≤ ≤ ≥ NAS-Unisel -MAC13 9 For flow through noncircular pipes, the Reynolds number is based on the hydraulic diameter, D h 4 : Cross section area perimeter c h c A D p where A p = − − NAS-Unisel -MAC13 10 Calculate the Reynolds number and state whether the fluid is in laminar, transition or turbulent. i. Oil with a kinematic viscosity of 0.00062 m 2 /s and 25 m/s velocity is being discharged by a 5-mm-diameter horizontal pipe from a storage tank open to the atmosphere. (Re=202, Laminar) ii. Oil with a density of 850 kg/m 3 , viscosity of 0.527 N.s/m 2 and 200 m/s velocity is being discharged by a 5-mm- diameter horizontal pipe from a storage tank open to the atmosphere. (Re=1613, Laminar) NAS-Unisel -MAC13 11 iii. Water with a density of 1kg/m 3 , kinetic viscosity of 1.52x10 -6 m 2 /s being discharged by a 50-mm- diameter horizontal pipe from a storage tank open to the atmosphere with a velocity of 0.1 m/s. (Re=3289, transition) iv. Oil with a Reynolds number of 5000 and density of 850 kg/m 3 being discharged by a 5-mm-diameter horizontal pipe from a storage tank open to the atmosphere. (Re=5000, turbulent) NAS-Unisel -MAC13 12 NAS-Unisel -MAC13 13 Fluid typically enters the pipe with a nearly uniform velocity profile at section 1 As the fluid moves through the pipe, viscous effects cause it to stick to the pipe wall Thus, a boundary layer is produced along the pipe wall such that the initial velocity profile changes with distance along the pipe, x, until the fluid reaches the end of the entrance length, section 2. Beyond section 2, the velocity profile does not vary with x. NAS-Unisel -MAC13 14 The boundary layer has grown in thickness to completely fill the pipe. Viscous effects are of considerable importance within the boundary layer. For fluid outside the boundary layer [inviscid core surrounding the centerline from 1 to 2], viscous effects are negligible. NAS-Unisel -MAC13 15 Inviscid Flow? = nonviscous = frictionless flow Flow with small (or no) viscosity Can use Bernoulli equation 2 Constant 2 dP V gz ρ + + = ∫ 2 Constant 2 P V gz ρ + + = General (valid for compressible and incompressible flow) Only valid for incompressible flow) NAS-Unisel -MAC13 16 Once the fluid reaches the end of the entrance region, section 2 or when inviscid core terminates and the flow is all viscous, the velocity is a function of only the distance from the pipe centerline, r, and independent of x. Starting from section 2 the flow is now called a Fully Developed Flow The velocity profile becomes parabolic NAS-Unisel -MAC13 17 The length of the pipe between the start and the point where the fully developed flow begins is called the Entrance Length, le The entry length is much shorter in turbulent flow *- the are many formulae to calculate the entrance length suggested by different scholars. To make it simple we will use the one written here ( ) 16 0.06Re laminar flow 4.4 Re turbulent flow e e D D = = NAS-Unisel -MAC13 18 Blue and yellow streams of paint at (each with a density of 825 kg/m 3 and a viscosity 1000 times greater than water) enter a pipe with an average velocity of 1.2 m/s as shown in Figure. Would you expect the paint to exit the pipe as green paint or separate streams of blue and yellow paint? Explain. Repeat the problem if the paint were “thinned” so that it is only 10 times more viscous than water. Assume the density remains the same. (viscosity of water, 1.12x10 -3 N.s/m 2 ) NAS-Unisel -MAC13 19 3 Re 825 1.2 0.05 Re 1000 1.12 10 Re 44.2 44.2 2300 this is laminar flow avg V D x x x x ρ µ − = = = < 3 Re 825 1.2 0.05 Re 10 1.12 10 Re 4419.6 4419.6 4000 this is turbulent flow avg V D x x x x ρ µ − = = = > The paint will exits separately The paint will exits in green (mixed- turbulent) NAS-Unisel -MAC13 20 for laminar flow 0.06Re 0.06Re 0.06 45.5 0.05 0.1365 e e e e D D x x m = = = = ( ) ( ) ( ) 16 16 16 for turbulent flow 4.4 Re 4.4 Re 4.4 0.05 4553.6 0.89 e e e e D D x x = = = = NAS-Unisel -MAC13 21 NAS-Unisel -MAC13 22 ∆p = p 2 -p 1 ∂ p/ ∂ x is larger at the entrance region compared to fully developed region. ∂ p/ ∂ x is constant at the fully developed region. ∂ p/ ∂ x = ∆p/ l < 0 NAS-Unisel -MAC13 23 Free body diagram of a cylinder of fluid NAS-Unisel -MAC13 P 1 -P 2 =∆P P 2 =P 1 -∆P 24 A. For shear stress dx 2πr 2r dx , 2 Area A rdx π = τ τ 2 F PxA F rdx τ π = = NAS-Unisel -MAC13 25 B. For Pressure 2r dx 2 , Area A r π = 2r P P- ∆P 1 2 1 P F PxA F r π = = Point 1 Point 2 ( ) 2 2 2 F PxA F P p r π = = − ∆ NAS-Unisel -MAC13 26 Force equilibrium/balance We can see that ∆p and l are independent to r, so 2τ/r must be dependent in the radius, r ( ) 2 2 2 2 2 2 2 P 2 0 P P 2 0 2 0 2 2 r P p r rl r r p r rl p r rl p r rl p l r π π τ π π π π τ π π τ π π τ π τ − − ∆ − = − + ∆ − = ∆ − = ∆ = ∆ = Cr τ = NAS-Unisel -MAC13 27 From here we can have, Into the pressure difference equation we will get, when 0, 0, 0 when 2, , 2 w w w r C r D C D r τ τ τ τ τ = = = = = = = 2 w r D τ τ = 4 w l p D τ ∆ = NAS-Unisel -MAC13 28 If we still remember that, To make the τ always in positive, it will be easier to change above equation to, This is because τ >0 with du/dr < 0. (du/dr < 0 mean that the velocity decreases from the pipe centerline to the pipe wall) du dr τ µ = du dr τ µ = − NAS-Unisel -MAC13 29 Using this equation into the pressure different equation we will get Because we want to know the velocity profile in term of radius, r, we need to integrate above equation ( ) 4 2 4 4 2 l du p D dr p r du pD pr dr l l l µ µ µ µ | | ∆ = − | \ . ∆ | | | | | | ∆ ∆ = − = − = − | | | \ . \ . \ . NAS-Unisel -MAC13 30 To solve this equation we need to have the boundary condition We can see that, u=0 at r=D/2. Using this we can get the value of C 1 2 1 2 2 2 pr du dr l p r u C l µ µ | | ∆ = − | \ . ∆ = − + ∫ ∫ 2 2 1 16 4 pD pR C l l µ µ ∆ ∆ = = NAS-Unisel -MAC13 31 The equation we can simplify becoming, 2 2 2 2 2 2 2 4 16 4 1 16 2 1 16 pr pD u l l pD r u l D pD r u l D µ µ µ µ ∆ ∆ = − + | | ∆ = − | \ . | | ∆ | | = − | | | \ . \ . NAS-Unisel -MAC13 32 When we plot the equation in R vs. U, we will this about this graph -10 -8 -6 -4 -2 0 2 4 6 8 10 -4000 -3000 -2000 -1000 0 1000 2000 NAS-Unisel -MAC13 33 We can see that at R = 0, the velocity is maximum. So we introduce a new term that we call as, centerline velocity, V c 2 16 c pD V l µ ∆ = NAS-Unisel -MAC13 34 Previous equation can be simplify as Previously we know that, By re arranging the equation we can get, Velocity equation can again re write as, 2 2 2 1 1 , 2 c c r r u V V D R where R D | | | | | | | | = − = − | | | | | | \ . \ . \ . \ . = 4 w l p D τ ∆ = 4 w pD l τ ∆ = 2 2 2 1 1 4 4 w w D D r r u D R τ τ µ µ | | | | | | | | = − = − | | | | | | \ . \ . \ . \ . NAS-Unisel -MAC13 35 The flow rate, is given by By using previous equation, we can re write above equation Q udA = ∫ 2 2 1 1 2 c r R c r o r Q V dA R r Q V rdr R π = = | | | | = − | | | \ . \ . | | | | = − | | | \ . \ . ∫ ∫ NAS-Unisel -MAC13 36 We can solve above equation 2 2 3 2 2 2 4 2 2 0 1 2 2 2 2 2 2 4 2 c r R r R c c c c r o r o R c c c r Q V dA R V r V r Q V rdr V r dr R R V r V r V R Q R π π π π π π = = = = | | | | = − | | | \ . \ . | | | | = − = − | | \ . \ . ( = − = ( ¸ ¸ ∫ ∫ ∫ NAS-Unisel -MAC13 37 If we still remember that, From this equation we can re write above equation becoming where, V is the average velocity From previous equation, So the velocity, V can been wrote as, 2 2 2 2 2 c c V Q R V R V V R π π π = = = Q VA = 2 16 c pD V l µ ∆ = 2 32 pD V l µ ∆ = NAS-Unisel -MAC13 38 As we seen before, most of the equation are in term of the radius. If we want to get the equation in term of the diameter, 2 2 32 4 Q VA pD D Q x l π µ = ∆ = NAS-Unisel -MAC13 39 This Equation is commonly referred as Poiseuille’s Law and the flow is termed as Hagen-Poiseuille flow Gotthilf Heinrich Ludwig Hagen J ean Louis Marie Poiseuille 4 128 p D Q l π µ ∆ = NAS-Unisel -MAC13 40 Take note that this equation only valid for laminar flow only! From equation before, we know that, By arranging the equation we will get, Previously, the dynamic pressure is given by Dividing the equation to the dynamics pressure, we will get, If we still remember, the definition of Reynolds number is given by 2 32 pD V l µ ∆ = 2 32l V p D µ ∆ = 2 2 V ρ 2 2 2 2 32 64 2 2 p l V l V D V D V µ µ ρ ρ ρ ∆ = = Re avg V D ρ µ = NAS-Unisel -MAC13 41 Using the Reynolds number, we can simplify previous equation becoming In term of pressure drop, we can write it as, 2 64 2 Re p l V D ρ ∆ = 2 64 Re 2 l V p D ρ ∆ = NAS-Unisel -MAC13 42 As we know, for certain velocity, the Reynolds number are constants. Above equation can be simplify by(only valid for horizontal pipes) Where f is what we call friction factor or Darcy Friction Factor We can also write this friction factor in term of shear stress. By recalling that The equation can be written as, 2 2 l V p f D ρ ∆ = Henry Philibert Gaspard Darcy 2 64 64 64 2 Re pD f VD VD V l µ υ ρ ρ ∆ = = = = 4 w pD l τ ∆ = 2 8 w f V τ ρ = NAS-Unisel -MAC13 43 Do not confuse with Fanning Friction Factor, f/4, which is less used in fluid mechanics, unless those who follow British conversion or chemical engineer. NAS-Unisel -MAC13 44 Head loss, h L something that commonly used can be extract from Using the previous pressure drop equation, We can create new equation This equation is well known as Darcy-Weisbach equation and is valid for any fully developed, steady, incompressible, horizontal or inclined pipe p g h ρ ∆ = ∆ 2 2 l V p f D ρ ∆ = 2 2 L l V h f D g = J ulius Ludwig Weisbach NAS-Unisel -MAC13 45 Take note that only valid for horizontal pipe only but valid for turbulent or laminar flow Head loss h L represents the additional height that the fluid needs to be raised by a pump in order to overcome the frictional losses in the pipe The required pumping power to overcome the pressure loss is obtained from 2 2 l V p f D ρ ∆ = NAS-Unisel -MAC13 46 L pump pump pump Q gh Q p W ρ η η ∆ = = For fully developed flow the kinetic energy is the same at any section and the energy equation becomes Sometimes Head loss is refereed as Major Loss 1 2 1 2 L p p z z h g g ρ ρ | | | | + − + = | | \ . \ . NAS-Unisel -MAC13 47 The basic equation is written as; Remember that the V is the average velocity in the pipe α – kinetic energy coefficient NAS-Unisel -MAC13 2 2 1 1 2 2 1 1 2 2 2 2 pump turbine L p V p V z h z h h g g g g α α ρ ρ + + + = + + + + 48 Uphill flow, θ> 0 Downhill flow, θ< 0 NAS-Unisel -MAC13 49 Using the same approach before, we can get these equations: Using the same approach we can get these equations: Centerline velocity, Average velocity, 2 2 2 sin 1 16 D p r u g l D ρ θ µ | | ∆ | | | | = − − | | | | \ . \ . \ . 2 sin 16 c p gl D V l ρ θ µ ∆ − = 2 sin 32 p gl D V l ρ θ µ ∆ − = sin 2 p g l r ρ θ τ ∆ − = NAS-Unisel -MAC13 50 Flow rate ( ) 4 sin 128 p gl D Q l ρ θ π µ ∆ − = NAS-Unisel -MAC13 51 NAS-Unisel -MAC13 52 A large artery in a person’s body can be approximated by a tube of diameter 9 mm and length 0.35 m. Also assume that blood has a viscosity of approximately 4x10 -3 N.s/m 2 , a specific gravity of 1.0, and that the pressure at the beginning of the artery is equivalent to 120 mm Hg. The specific gravity of mercury is 13.3. If the flow were steady (it is not) with V=0.2m/s, determine the pressure at the end of the artery if it is oriented a)vertically up (flow up) or b)horizontal. NAS-Unisel -MAC13 53 P 2 ?? P 1 P 2 D=0.009m l=0.35m ρ=1000SG µ= 4x10 -3 N.s/m 2 V=0.2m/s 1 1 1 1 120 13.3 1000 9.81 0.12 15.96 P mmHg P gH P x x x P kPa ρ = = = = 3 1000 0.2 0.009 Re Re 450 450 4 10 2300 laminar x x x − = = < NAS-Unisel -MAC13 54 this equation can be simplify 1 2 1 2 L p p z z h g g ρ ρ | | | | + − + = | | \ . \ . ( ) ( ) 1 2 1 2 2 1 1 2 2 2 1 1 2 2 L L p p z z h g g p p g z z h l V p p g z z f D ρ ρ ρ ρ ρ | | | | + − + = | | \ . \ . = + − − = + − − NAS-Unisel -MAC13 55 For vertical 64 Re 64 0.1422 450 f f = = = ( ) ( ) 2 2 1 1 2 2 2 2 2 2 1000 0.350.2 15.96 1000 9.81 0.35 0.1422 0.009 2 15.96 3.433 0.1106 12.52 l V p p g z z f D x p x p kPa kPa kPa p kPa ρ ρ = + − − = + − − = − − = NAS-Unisel -MAC13 56 For horizontal Or can just use ( ) ( ) 2 2 1 1 2 2 2 2 2 2 1000 0.350.2 15.96 1000 9.81 0 0.1422 0.009 2 15.96 0 0.1106 15.95 l V p p g z z f D x p x p kPa kPa p kPa ρ ρ = + − − = + − = − − = 2 1 2 2 l V p p p f D ρ ∆ = − = NAS-Unisel -MAC13 57 Can we use Bernoulli equation? For Vertical flow For Horizontal We can use, only and ONLY if we ignore the friction effects NAS-Unisel -MAC13 ( ) ( ) 2 1 1 2 2 2 2 15.96 1000 9.81 0.35 15.96 3.433 12.527 p p g z z p x p kPa kPa p kPa ρ = + − = + − = − = ( ) 2 1 1 2 2 15.96 p p g z z p ρ = + − = 58 Oil with ρ= 876 kg/m 3 and µ=0.24 kg/m·s is flowing through a 1.5-cm-diameter pipe that discharges into the atmosphere at 88 kPa. The absolute pressure 15 m before the exit is measured to be 135 kPa. Determine the flow rate of oil through the pipe if the pipe is a) horizontal, b) inclined 8° upward from the horizontal, and c)inclined 8° downward from the horizontal. Assume that the flow is in laminar NAS-Unisel -MAC13 59 NAS-Unisel -MAC13 60 A) horizontal pipe 1 2 135 88 47 p p p kPa ∆ = − = − = 4 128 p D Q l π µ ∆ = 4 5 3 47 0.015 1.622 10 / 128 15 0.24 k Q x m s x x π − = = NAS-Unisel -MAC13 61 B) for inclined uphill C) for inclined downhill ( ) ( ) 4 4 5 3 sin 128 47 876 9.81 15 sin8 0.015 128 15 0.24 1 10 / p gl D Q l k x x x Q x x Q x m s ρ θ π µ π − ∆ − = − = = ( ) ( ) ( ) 4 4 5 3 sin 128 47 876 9.81 15 sin 8 0.015 128 15 0.24 2.24 10 / p gl D Q l k x x x Q x x Q x m s ρ θ π µ π − ∆ − = − − = = NAS-Unisel -MAC13 62 As we can see the highest flow rate is the downhill pipe. Is this flow really laminar? 2 2 4 2 5 3 5 4 0.015 1.767 10 4 4 we take the highest flow rate, 2.24 10 / 2.24 10 0.127 / 1.767 10 C C C Q VA Q V A D A x m Q x m s x V m s x π π − − − − = = = = = = = = Re avg V D ρ µ = Re 876 0.127 0.015 Re Re 6.94(the highest Reynolds number) 6 0.2 .94 2300 lam 4 inar avg V D x x ρ µ = = = < NAS-Unisel -MAC13 63 Can we use Bernoulli equation? NAS-Unisel -MAC13 64 time-averaged velocity fluctuating component u u u′ = + v v v ′ = + P P P′ = + T T T ′ = + u u ′ NAS-Unisel -MAC13 65 Shear stress in a pipe Reynolds Stresses lam turb lam turb du dy u v u v τ τ τ τ µ τ ρ ρ = + = ′ ′ = − ′ ′ − → NAS-Unisel -MAC13 66 Turbulent shear stress in term of eddy viscosity,η (or sometime referred also as µ t ) This was proposed by Boussinesq in 1877 In laminar, the constant µ is constant. But for turbulent flow the η is in function of fluid type and flow type turb du u v dy τ ρ η ′ ′ = − = J oseph Valentin Boussinesq NAS-Unisel -MAC13 67 In early 1900, L. Prandtl introduce a way to determine approximate values of η. The method is called mixing length. He said that, turbulent process could be viewed as the random transport of bundles of fluid particles over a certain distance, the mixing length, l m He proposed the eddy viscosity is given by 2 m du l dy η ρ = NAS-Unisel -MAC13 68 By that we can get But the new problem arise is to determining the value of l m This value is not a constant and depend on the distance form a surface 2 2 2 turb m m du du du l l dy dy dy τ ρ ρ | | = = | \ . NAS-Unisel -MAC13 69 Average velocity in a pipe Three region viscous sublayer very near the pipe wall, the overlap region, the outer turbulent layer throughout the center portion of the flow NAS-Unisel -MAC13 70 Viscous sublayer commonly called the law of the wall is valid very near the smooth wall, for u yu u υ ∗ ∗ = 0 5 yu υ ∗ ≤ ≤ NAS-Unisel -MAC13 71 Overlap region 2.5ln 5 u yu u υ ∗ ∗ | | = + | \ . NAS-Unisel -MAC13 72 Outer turbulent Region Sometimes, the well known, Power Law Velocity Profile equation were used. ( ) 2.5ln c V u R u y ∗ − | | = | \ . 1 1 : constant that depend on Reynold Number n c u r V R where n | | = − | \ . − NAS-Unisel -MAC13 73 NAS-Unisel -MAC13 74 NAS-Unisel -MAC13 75 Friction factor in fully developed turbulent pipe flow depends on the Reynolds number and the relative roughness ε/D. But there are no theoretical analysis about these relationship. Cyril F. Colebrook combined available data and developed an implicit relation called Colebrook Equation which valid for turbulent flow. 1 2.51 2log 3.7 Re D f f ε | | = − + | \ . NAS-Unisel -MAC13 76 Lewis Ferry Moody capcuring the equation into a diagram which known as Moody Diagram S. E. Haaland recontruct Colebrook equation becoming This equation can solve the value of f explicitly. 1.11 1 6.9 1.8log 3.7 Re D f ε | | | | = − + | | | \ . \ . NAS-Unisel -MAC13 77 NAS-Unisel -MAC13 78 NAS-Unisel -MAC13 79 Human Target Season 1 episode 1 NAS-Unisel -MAC13 80 Air enters a 10-m-long section of a rectangular duct of cross section 15 cm x 20 cm made of commercial steel at 1 atm and 35°C at an average velocity of 7 m/s. Disregarding the entrance effects, determine the fan power needed to overcome the pressure losses in this section of the duct. At 1 atm and 35°C are ρ = 1.145 kg/m 3 , μ = 1.895×10 -5 kg/m⋅s, and ν = 1.655×10 -5 m 2 /s. NAS-Unisel -MAC13 81 ( ) 4 4 0.15 0.2 0.15 2 0.2 2 0.17 c h h h A D p X D X X D = = + = 5 1.145 7 1.895 1 0.17 Re 72506 72506 2 0 300 avg V D turbulent x x ρ µ − = = = > ⇒ × NAS-Unisel -MAC13 82 From table, ε for commercial table is 0.045mm = 0.000045m Using the Haaland equation, 4 0.000045/ 0.17 2.65 10 D x ε − = = 1.11 1.11 4 1 6.9 1.8log 3.7 Re 1 2.65 10 6.9 1.8log 3.7 72506 1 7.056 0.02008 D f x f f f ε − | | | | = − + | | | \ . \ . | | | | | = − + | | \ . \ . = = NAS-Unisel -MAC13 83 NAS-Unisel -MAC13 84 Or from moody diagram, f =0.021 2 1 2 10 1.145 7 0.02008 0.17 2 33.14 x p p p p Pa ∆ = − = ∆ = 7 0.15 0.2 0.21 Q VA Q x x = = = fan L W Q p Q gh ρ = ∆ = 0.21 33.14 6.9594 fan L fan fan W Q p Q gh W x W W ρ = ∆ = = = NAS-Unisel -MAC13 85 4 128 p D Q l π µ ∆ = Can we use this??? NAS-Unisel -MAC13 86 Loss coefficient, K L Form above equation, we can write minor loss into Minor losses are sometimes given in term of an equivalent length, l eq minor 2 2 1 2 2 L L h p K V g V ρ ∆ = = L eq K D l f = 2 minor 2 L L V h K g = NAS-Unisel -MAC13 87 NAS-Unisel -MAC13 88 Figure 8.6 NAS-Unisel -MAC13 89 NAS-Unisel -MAC13 90 NAS-Unisel -MAC13 91 NAS-Unisel -MAC13 92 NAS-Unisel -MAC13 93 Typical commercial valve geometries: (a) gate valve; (b) globe valve; (c) angle valve; (d) swing-check valve; (e) disk type gate valve. NAS-Unisel -MAC13 94 NAS-Unisel -MAC13 95 2 2 2 2 total minor losses = 2 total major losses = 2 total losses = 2 2 L L V K g L V f D g L V V f K D g g + ∑ ∑ ∑ ∑ 2 2 2 2 1 1 2 2 1 1 2 2 2 2 2 2 L p V p V L V V z z f K g g g g D g g α α ρ ρ + + = + + + + ∑ ∑ NAS-Unisel -MAC13 96 Water flows steadily through the 2-cm diameter galvanized iron pipe system shown in figure below at a rate of 5.6x10 -4 m 3 /s. Your boss suggests that friction losses in the straight pipe sections are negligible compared to losses in the threaded elbows and fittings of the system. Do you agree or disagree with your boss? Support your answer with appropriate calculations. Take the kinematic viscocity of water is 1.12x10 -6 m 2 /s NAS-Unisel -MAC13 97 NAS-Unisel -MAC13 98 Major losses due to straight pipe To find the value of f, we need to whether the flow is laminar or turbulent. To calculate the Re number we must find the value of V 2 2 L l V h f D g = 4 2 , 5.6 10 1.78 / 0.02 4 Q VA Q x V m s A x π − = = = = NAS-Unisel -MAC13 99 To find f for turbulent we need to have the value of ε/D From table we can find for galvanized iron , ε=0.15mm 6 1.12 10 1.78 0.02 Re 31785.7 31785.7 2300 avg avg V D V D x turbulent ρ µ ν − = = = = > ⇒ × 3 2 3 0.15 10 2 10 7.5 10 D x x x ε − − − = = NAS-Unisel -MAC13 100 NAS-Unisel -MAC13 101 From moody diagram f=0.036 Or by using the Haaland equation 1.11 1.11 3 1 6.9 1.8log 3.7 Re 1 7.5 10 6.9 1.8log 3.7 31785.7 1 5.19 0.037 D f x f f f ε − | | | | = − + | | | \ . \ . | | | | | = − + | | \ . \ . = = NAS-Unisel -MAC13 102 Major losses due to straight pipe 2 0.15 0.15 0.1 0.0251.78 0.036 0.02 2 0.1234 L L h g h m + + + = = NAS-Unisel -MAC13 103 Minor losses For reducer, we need to use the graph 2 2 L V K g ∑ for threaded 90° elbow, 1.5 for Tee, 2(threaded,branch flow) L L L K K K = = ∑ ( ) ( ) 2 2 2 1 1.5 2 0.5625 D D = = NAS-Unisel -MAC13 104 K L = 0.18 NAS-Unisel -MAC13 105 Total minor losses Disagree with boss, because as we can see, the total losses about 14.7% less if we neglect the major losses. ( ) 2 1.78 1.5 2 2 0.18 0.8365 2 X m g + + = ∑ major loss 0.1234 0.147 14.7% minor loss 0.8365 = = = NAS-Unisel -MAC13 106 Water at 5 °C flows through the coils of the heat exchanger as shown in Figure at a rate of 3.5L/min. Determine the pressure drop between the inlet and outlet of the horizontal device. Take the kinematic viscocity of water is 1.519x10 -6 m 2 /s NAS-Unisel -MAC13 107 To find the value of f, we need to whether the flow is laminar or turbulent. 2 2 2 2 1 1 2 2 1 2 1 2 1 2 2 2 1 2 2 2 1 2 2 2 2 2 assuming and 2 2 2 2 L L L p V p V L V V z z f K g g g g D g g V V z z L V V p p g f g K D g g L V V p p f K D ρ ρ ρ ρ ρ ρ + + = + + + + = = − = + − = + ∑ ∑ ∑ ∑ ∑ NAS-Unisel -MAC13 108 To calculate the Re number we must find the value of V 6 0.44 1.519 10 0.013 Re 3765.6 3765.6 2300 avg avg V D V D turbulent x ρ µ ν − × = = = > ⇒ 2 , 3.5 1 0.44 / 1000 60 0.013 4 Q VA Q V x m s A x x π = = = = NAS-Unisel -MAC13 109 To find f for turbulent we need to have the value of ε/D From table we can find for drawn tubing, ε=0.0015mm 4 0.001513 1.15 10 D x ε − = = NAS-Unisel -MAC13 110 NAS-Unisel -MAC13 111 From moody diagram f=0.042 Or by using the Haaland equation 1.11 1.11 4 1 6.9 1.8log 3.7 Re 1 1.15 10 6.9 1.8log 3.7 3765.6 1 0.2032 0.0413 D f x f f f ε − | | | | = − + | | | \ . \ . | | | | | = − + | | \ . \ . = = NAS-Unisel -MAC13 112 for threaded 180° return bend, 1.5 1.5 7 10.5 L L L K K x K = = ∑ ∑ ∑ 2 2 1 2 2 2 1 2 1 2 2 2 0.45 81000 0.44 1000 0.44 0.042 10.5 0.013 2 2 2142.3 2.142 L L V V p p f K D x x x p p p p Pa kPa ρ ρ − = + − = + − = = ∑ NAS-Unisel -MAC13 113 A horizontal pipe has an abrupt expansion from D 1 = 8 cm to D 2 = 16 cm. The water velocity in the smaller section is 10 m/s and the flow is turbulent. The pressure in the smaller section is P 1 = 410 kPa. determine the downstream pressure P 2 , and estimate the error that would have occurred if Bernoulli’s equation had been used. NAS-Unisel -MAC13 114 For expansion, we need to use the graph NAS-Unisel -MAC13 ( ) 2 2 1 1 2 2 1 2 , ,min 1 2 2 2 1 1 2 2 ,min 2 2 1 2 2 1 ,min 2 2 2 2 2 L major L or L or L or p V p V z z h h g g g g z z p V p V h g g g g V V p p gh ρ ρ ρ ρ ρ ρ + + = + + + + = + = + + − = + − ∑ ∑ ( ) ( ) 2 2 1 2 0.08 0.16 0.25 D D = = 115 K L = 0.57 NAS-Unisel -MAC13 116 NAS-Unisel -MAC13 2 2 2 10 0.58 2 9.81 2.96 L L L L V h K g h x h = = = ( ) 1 1 2 2 1 1 2 2 2 2 2 2 10 0.08 0.16 2.5 / V A V A V A V A V V m s = = = = ( ) ( ) 2 2 1 2 2 1 ,min 2 2 2 2 2 1000 10 2.5 410 1000 9.81 2.96 2 427.84 L or V V p p gh p k x x p kPa ρ ρ − = + − − = + − = 117 If we just ignore the friction effect (using Bernoulli equation) NAS-Unisel -MAC13 ( ) ( ) 2 2 1 1 2 2 1 2 1 2 2 2 1 1 2 2 2 2 1 2 2 1 2 2 2 2 2 2 2 2 2 1000 10 2.5 410 2 456.875 p V p V z z g g g g z z p V p V g g g g V V p p p k p kPa ρ ρ ρ ρ ρ + + = + + = + = + − = + − = + = 118