Lecture notes 02

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Lecture notes 02

  1. 1. Chapter 2 Flows under Pressure in PipesIf the fluid is flowing full in a pipe under pressure with no openings to the atmosphere, itis called “pressured flow”. The typical example of pressured pipe flows is the waterdistribution system of a city.2.1. Equation of MotionLets take the steady flow (du/dt=0) in a pipe with diameter D. (Fig. 2.1). Taking acylindrical body of liquid with diameter r and with the length Δx in the pipe with thesame center, equation of motion can be applied on the flow direction. Flow Δx α F2 r F1 y D γπr 2 Δx x Figure 2.1.The forces acting on the cylindrical body on the flow direction are, a) Pressure force acting to the bottom surface of the body that causes the motion of the fluid upward is, → F1= Pressure force = ( p + Δp )πr 2 1 Prof. Dr. Atıl BULU
  2. 2. b) Pressure force to the top surface of the cylindrical body is, ← F2= Pressure force = pπr 2c) The body weight component on the flow direction is, ← X = γπr 2 Δx sin αd) The resultant frictional (shearing) force that acts on the side of the cylindrical surface due to the viscosity of the fluid is, ← Shearing force = τ 2πrΔxThe equation of motion on the flow direction can be written as, ( p + Δp )πr 2 − pπr 2 − γπr 2 Δx sin α − τ 2πrΔx = Mass × Acceleration (2.1)The velocity will not change on the flow direction since the pipe diameter is keptconstant and also the flow is a steady flow. The acceleration of the flow body will bezero, Equ. (2.1) will take the form of, Δpr 2 − γr 2 Δx sin α − 2τrΔx = 0 1 ⎛ Δp ⎞ τ= ⎜ − γ sin α ⎟r (2.2) 2 ⎝ Δx ⎠The frictional stress on the wall of the pipe τ0 with r = D/2, 1 ⎛ Δp ⎞D τ0 = ⎜ − γ sin α ⎟ (2.3) 2 ⎝ Δx ⎠2We get the variation of shearing stress perpendicular the flow direction from Equs.(2.2) and (2.3) as, 2 Prof. Dr. Atıl BULU
  3. 3. r τ =τ0 (2.4) D2 y D/2 r y τ0 τ Fig. 2.2Since r = D/2 –y, ⎛ y ⎞ τ = τ 0 ⎜1 − ⎜ ⎟ (2.5) ⎝ D 2⎟ ⎠The variation of shearing stress from the wall to the center of the pipe is linear as canbe seen from Equ. (2.5).2.2. Laminar Flow (Hagen-Poiseuille Equation)Shearing stress in a laminar flow is defined by Newton’s Law of Viscosity as, du τ =μ (2.6) dyWhere μ = (Dynamic) Viscosity and du/dy is velocity gradient in the normal directionto the flow. Using Equs. (2.5) and (2.6) together, 3 Prof. Dr. Atıl BULU
  4. 4. ⎛ y ⎞ du τ 0 ⎜1 − ⎜ ⎟=μ ⎟ ⎝ D 2⎠ dy τ0 ⎛ y ⎞ du = ⎜1 − ⎜ D 2 ⎟dy ⎟ μ ⎝ ⎠By taking integral to find the velocity with respect to y, τ0 ⎛ y ⎞ u= μ ∫ ⎜1 − D 2 ⎟dy ⎜ ⎝ ⎟ ⎠ τ0 ⎛ y2 ⎞ u= ⎜y− ⎜ ⎟ + cons (2.7) μ ⎝ D⎟ ⎠Since at the wall of the pipe (y=0) there will no velocity (u=0), cons=0. If the specificmass (density) of the fluid is ρ, Friction Velocity is defined as, τ0 u∗ = (2.8) ρKinematic viscosity is defined by, μ μ υ= →ρ= ρ υ τ 0υ τ 0 u ∗2 u∗ = 2 → = μ μ υThe velocity equation for laminar flows is obtained from Equ. (2.7) as, u∗ ⎛ 2 y2 ⎞ u= ⎜y− ⎟ (2.9) υ ⎜⎝ D⎟ ⎠Using the geometric relation of the pipe diameter (D) with the distance from the pipewall (y) perpendicular to the flow, 4 Prof. Dr. Atıl BULU
  5. 5. D D r= − y → y = −r 2 2 2 ⎡ ⎞ ⎤ 2 u D 1 ⎛D u = ∗ ⎢ − y − ⎜ −r⎟ ⎥ υ ⎢2 ⎣ D⎝ 2 ⎠ ⎥⎦ u∗ ⎛ D 2 2 2 ⎞ u= ⎜ ⎜ 4 −r ⎟⎟ (2.10) υD ⎝ ⎠Equ. (2.10) shows that velocity distribution in a laminar flow is to be a paraboliccurve.The mean velocity of the flow is, Q ∫AudA V= = A APlacing velocity equation (Equ. 2.10) gives us the mean velocity for laminar flows as, 2 Du ∗ V= (2.11) 8υSince τ0 u∗ = 2 ρAnd according to the Equ. (2.3), 1 ⎛ Δp ⎞D τ0 = ⎜ − γ sin α ⎟ 2 ⎝ Δx ⎠2 5 Prof. Dr. Atıl BULU
  6. 6. 1 ⎛ Δp ⎞D u∗ = 2 ⎜ − γ sin α ⎟ 2 ρ ⎝ Δx ⎠2Placing this to the mean velocity Equation (2.11), D 2 ⎛ Δp ⎞ V= ⎜ − γ sin α ⎟ (2.11) 32μ ⎝ Δx ⎠We find the mean velocity equation for laminar flows. This equation shows thatvelocity increases as the pressure drop along the flow increases. The discharge of theflow is, πD 2 Q = AV = V 4 πD 4 ⎛ Δp ⎞ Q= ⎜ − γ sin α ⎟ (2.12) 128μ ⎝ Δx ⎠If the pipe is horizontal, πD 4 Δp Q= (2.13) 128μ ΔxThis is known as Hagen-Poiseuille Equation.2.3. Turbulent FlowThe flow in a pipe is Laminar in low velocities and Turbulent in high velocities.Since the velocity on the wall of the pipe flow should be zero, there is a thin layerwith laminar flow on the wall of the pipe. This layer is called Viscous Sub Layer andthe rest part in that cross-section is known as Center Zone. (Fig. 2.3) 6 Prof. Dr. Atıl BULU
  7. 7. δ Viscos Sublayer y Center Zone τ τ0 Fig. 2.3.2.3.1. Viscous Sub LayerSince this layer is thin enough to take the shearing stress as, τ ≈ τ0 and since the flowis laminar, du τ =μ = τ 0 = ρu ∗ 2 dy ρ 2 du = u ∗ dy μBy taking the integral, ρ 2 μ ∫ u= u ∗ dy ρ 2 u = u ∗ y + cons μSince for y =0 → u = 0, the integration constant will be equal to zero. Substitutingυ=μ/ρ gives, 2 u∗ u= y (2.14) υ 7 Prof. Dr. Atıl BULU
  8. 8. The variation of velocity with y is linear in the viscous sub layer. The thickness of thesub layer (δ) has been obtained by laboratory experiments and this empirical equationhas been given, υ δ = 11.6 (2.15) u∗Example 2.1. The friction velocity u*= 1 cm/sec has been found in a pipe flow withdiameter D = 10 cm and discharge Q = 2 lt/sec. If the kinematic viscosity of the liquidis υ = 10-2 cm2/sec, calculate the viscous sub layer thickness. υ δ = 11.6 u∗ 10 − 2 δ = 11.6 1 δ = 0.12cm = 1.2mm2.3.2 Smooth PipesThe flow will be turbulent in the center zone and the shearing stress is, + (− ρ u ′ v ′) du τ =μ (2.16) dyThe first term of Equ. (2.16) is the result of viscous effect and the second term is theresult of turbulence effect. In turbulent flow the numerical value of Reynolds Stress(− ρ u ′ v ′) is generally several times greater than that of (μ du dy ) . Therefore, theviscosity term (μ du dy ) may be neglected in case of turbulent flow.Shearing stress caused by turbulence effect in Equ. (2.16) can be written in the similarform as the viscous affect shearing stress as, du τ = −ρ u ′v ′ = μ T (2.17) dy 8 Prof. Dr. Atıl BULU
  9. 9. Here μT is known as turbulence viscosity and defined by, du μ T = ρl 2 (2.18) dyHere l is the mixing length. It has found by laboratory experiments that l = 0.4y forτ≈τ0 zone and this 0.4 coefficient is known as Von Karman Coefficient.Substituting this value to the Equ. (2.18), du μ T = 0.16 ρy 2 dy 2 du ⎛ du ⎞ τ 0 = μT = 0.16 ρy 2 ⎜ ⎟ ⎜ dy ⎟ dy ⎝ ⎠ 2 τ0 ⎛ du ⎞ = u ∗ = 0.16 ρy 2 ⎜ ⎟ 2 ⎜ dy ⎟ ρ ⎝ ⎠ du u ∗ = 0.4 y dy dy du = 0.4 y u∗ dy du = 2.5u ∗ yTaking the integral of the last equation, dy u = 2.5u ∗ ∫ y (2.19) u = 2.5u ∗ Lny + consThe velocity on the surface of the viscous sub layer is calculated by using Equs.(2.14) and (2.15), 9 Prof. Dr. Atıl BULU
  10. 10. 2 u∗ u= y υ υ y = δ → δ = 11.6 u∗ u = 11.6u ∗Substituting this to the Equ. (2.19) will give us the integration constant as, cons = u − 2.5u ∗ Lny ⎛ υ ⎞ cons = 11.6u ∗ − 2.5u ∗ Ln⎜11.6 ⎜ ⎟ ⎟ ⎝ u∗ ⎠ ⎛ υ ⎞ cons = 11.6u ∗ − 2.5u ∗ ⎜ Ln11.6 + Ln ⎜ ⎟ ⎟ ⎝ u∗ ⎠ υ cons = 5.5u ∗ − 2.5u ∗ Ln u∗Substituting the constant to the Equ. (2.19), υ u = 2.5u ∗ Lny + 5.5u ∗ − 2.5u ∗ Ln u∗ yu ∗ u = 2.5u ∗ Ln + 5.5u ∗ υ (2.20) u yu = 2.5 Ln ∗ + 5.5 u∗ υEqu. (2.20) is the velocity equation in turbulent flow in a cross section with respect toy from the wall of the pipe and valid for the pipes with smooth wall.The mean velocity at a cross-section is found by the integration of Equ. (2.20) for areA, 10 Prof. Dr. Atıl BULU
  11. 11. Q ∫ udA V= = A A A ⎛ u D ⎞ V = ⎜ 2.5Ln ∗ + 1.75 ⎟u ∗ ⎝ 2υ ⎠ (2.21) V Du ∗ = 2.5 Ln + 1.75 u∗ υ2.3.3. Definition of Smoothness and RoughnessThe uniform roughness size on the wall of the pipe can be e as roughness depth. Mostof the commercial pipes have roughness. The above derived equations are for smoothpipes. The definition of smoothness and roughness basically depends upon the size ofthe roughness relative to the thickness of the viscous sub layer. If the roughnesess aresubmerged in the viscous sub layer so the pipe is a smooth one, and resistance andhead loss are entirely unaffected by roughness up to this size. Pipe Center Flow e Pipe wall Fig. 2.4 υSince the viscous sub layer thickness (δ) is given by, δ = 11.6 pipe roughness size u∗e is compared with δ to define if the pipe will be examined as smooth or rough pipe. 11 Prof. Dr. Atıl BULU
  12. 12. υa) e < δ = 11.6 u∗The roughness of the pipe e will be submerged in viscous sub layer. The flow in thecenter zone of the pipe can be treated as smooth flow which is given Chap. 2.3.2. υb) e > 70 u∗The height of the roughness e is higher than viscous sub layer. The flow in the centerzone will be affected by the roughness of the pipe. This flow is named as WhollyRough Flow. υ υc) 11.6 < e < 70 u∗ u∗This flow is named as Transition Flow.2.3.4. Wholly Rough PipesPipe friction in rough pipes will be governed primarily by the size and pattern of theroughness. The velocity equation in a cross section will be the same as Equ. (2.19). u = 2.5u ∗ Lny + cons (2.19)Since there will be no sub layer left because of the roughness of the pipe, theintegration constant needs to found out. It has been found by laboratory experimentsthat, e u=o→ y= 30The integration is calculated as, 12 Prof. Dr. Atıl BULU
  13. 13. e 0 = 2.5u ∗ Ln + const 30 e const = −2.5u ∗ Ln 30 e u = 2.5u ∗ Lny − 2.5u ∗ Ln 30The velocity distribution at a cross section for wholly rough pipes is, 30 y u = 2.5u ∗ Ln e (2.22) u 30 y = 2.5Ln u∗ eThe mean velocity at that cross section is, ⎛ D ⎞ V = u ∗ ⎜ 2.5Ln + 4.73 ⎟ ⎝ 2e ⎠ (2.23) V D = 2.5Ln + 4.73 u∗ 2e2.4. Head (Energy) Loss in Pipe FlowsThe Bernoulli equation for the fluid motion along the flow direction between points(1) and (2) is, p + Δp V12 p V2 z1 + + = z 2 + + 2 + hL (2.24) γ 2g γ 2gIf the pipe is constant along the flow, V1 = V2, Δp hL = − ( z 2 − z1 ) (2.25) γ 13 Prof. Dr. Atıl BULU
  14. 14. V2 hL 2g Energy Line p + Δp H.G.L p γ γ Flow 2 Δx 1 Z2 α Z1 Horizontal Datum Figure 2.5If we define energy line (hydraulic) slope J as energy loss for unit weight of fluidfor unit length, hL J= (2.26) ΔxWhere Δx is the length of the pipe between points (1) and (2), and using Equs. (2.25)and (2.26) gives, Δp z 2 − z 1 J= − γΔx Δx (2.27) Δp J= − sin α γΔx 14 Prof. Dr. Atıl BULU
  15. 15. Using Equ. (2.3), 1 ⎛ Δp ⎞D τ0 = ⎜ − γ sin α ⎟ 2 ⎝ Δx ⎠2 γD ⎛ Δp ⎞ τ0 = ⎜ γΔx − sin α ⎟ ⎜ ⎟ (2.28) 4 ⎝ ⎠ γD τ0 = J 4Using the friction velocity Equ. (2.8), τ0 u∗ = ρ τ 0 = ρu ∗2 γD ρgD ρu ∗2 = J= J 4 4 2 4u ∗ J= (2.29) gDEnergy line slope equation has been derived for pipe flows with respect to frictionvelocity u*. Mean velocity of the cross section is used in practical applications insteadof frictional velocity. The overall summary of equational relations was given inTable. (2.1) between frictional velocity u* and the mean velocity V of the crosssection. 15 Prof. Dr. Atıl BULU
  16. 16. Table 2.1. Mathematical Relations between u* and V Laminar Flow (Re<2000) 2 Du ∗ V= 8υ Smooth Flow Turbulent υ Flow e < 11.6 u∗ ⎛ Du ∗ ⎞ (Re>2000) V = u ∗ ⎜ 2.5Ln + 1.75 ⎟ ⎝ 2υ ⎠ Wholly Rough Flow ⎛ D ⎞ υ V = u ∗ ⎜ 2.5 Ln + 4.73 ⎟ e > 70 ⎝ 2e ⎠ u∗After calculating the mean velocity V of the cross-section and finding the type of low,frictional velocity u* is found out from the equations given in Table (2.1). The energyline (hydraulic) slope J of the flow is calculated by Equ. (2.29). Darcy-Weisbachequation is used in practical applications which is based on the mean velocity V tocalculate the hydraulic slope J. f V2 J= (2.30) D 2gWhere f is named as the friction coefficient or Darcy-Weisbach coefficient. Frictioncoefficient f is calculated from table (2.2) depending upon the type of flow where ρVD VDRe = = . μ υ 16 Prof. Dr. Atıl BULU
  17. 17. Table 2.2. Friction Coefficient Equations Laminar Flow 64 (Re<200 f = Re 0) Smooth Flow ⎛ υ ⎞ ⎜ e < 11.6 ⎟ ⎜ ⎝ u∗ ⎟ ⎠ 8 ⎛ = 2.5Ln⎜ Re f ⎞ ⎟ + 1.75 f ⎜ 32 ⎟ Turbulent ⎝ ⎠ Flow (Re>2000) Wholly Rough Flow ⎛ υ ⎞ 8 ⎛D⎞ ⎜ e > 70 ⎟ ⎜ = 2.5Ln⎜ ⎟ + 4.73 ⎝ u∗ ⎟ ⎠ f ⎝ 2e ⎠ Transition Flow ⎛ ⎜ ⎛ υ υ ⎜ ⎜11.6 < e < 70 8 ⎛D⎞ 9.2 ⎜ = 2.5Ln⎜ ⎟ + 4.73 − 2.5Ln⎜1 + ⎝ u∗ u∗ ⎝ 2e ⎠ f ⎜ Re ⎜ ⎝ DeThe physical explanation of the equations in Table (2.2) gives us the following results. a) For laminar flows (Re<2000), friction factor f depends only to the Reynolds number of the flow. f = f (Re ) b) For turbulent flows (Re>2000), 1. For smooth flows, friction factor f is a function of Reynolds number of the flow. f = f (Re ) 17 Prof. Dr. Atıl BULU
  18. 18. 2. For trannsition flow f depend on Reyno number (Re) of the flow and relative ws, ds olds r e r roughn of the pipe (e/D), f = f (Re, e D) ness p 3. For wh holly rough flows, f is a fun h nction of t the relative roughness (e/D) e f = f (e D )Fric ction coeffic cient f is c calculated fr from the eq quations giv in Tabl (2.2). In case of ven leturb bulent flow, the calcul , lation of f will always be done by trial and error method. Adiag gram has be prepare to overco een ed fficulty. It i prepared by Nikura ome this dif is adse andshow the func ws ctional relati ions betwee f and Re, e/D as curv (Figure 2.6) en , ves. e Figure 2.6Sum mmary a) Energy loss for uni length of pipe is calc it culated by D Darcy-Weisb bach equati ion, f V2 J= D 2g For a pipe w length L, the energy loss will be, with l 18 Prof. Dr. Atıl BULU A
  19. 19. h L = JL (2.31)b) The friction coefficient f will either be calculated from the equations given in Table (2.2) or from the Nikuradse diagram. (Figure 2.6)2.5. Head Loss for Non-Circular PipesPipes are generally circular. But a general equation can be derived if the cross-sectionof the pipe is not circular. Let’s write equation of motion for a non-circular prismaticpipe with an angle of α to the horizontal datum in a steady flow. Fig. (2.7). Flow τ0 α p Δx p+Δp A W = γAΔx P=Wetted Perimeter Figure 2.7 ( p + Δp )A − pA − τ 0 PΔx − γAΔx sin α = Mass × accelerationWhere P is the wetted perimeter and since the flow is steady, the acceleration of theflow will be zero. The above equation is then, ΔpA − τ 0 PΔx − γAΔx sin α = 0 A ⎛ Δp ⎞ (2.32) τ0 = ⎜ − γ sin α ⎟ P ⎝ Δx ⎠ 19 Prof. Dr. Atıl BULU
  20. 20. Where, A R= = Hydraulic Radius (2.33) PHydraulic radius is the ratio of wetted area to the wetted perimeter. Substituting thisto the Equ. (2.32), ⎛ Δp ⎞ τ 0 = R⎜ − γ sin α ⎟ ⎝ Δx ⎠ ⎛ Δp ⎞ τ 0 = γR⎜ ⎜ γΔx − sin α ⎟ ⎟ ⎝ ⎠Since by Equ. (2.27), Δp J= − sin α γΔxShearing stress on the wall of the non-circular pipe, τ 0 = γRJ (2.34)For circular pipes, A πD 2 4 D R= = = (2.35) P πD 4 D = 4RThis result is substituted (D=4R) to the all equations derived for the circular pipes toobtain the equations for non-circular pipes. Table (2.3) is prepared for the equationsas, 20 Prof. Dr. Atıl BULU
  21. 21. Table 2.3. Circular Pipes Non-Circular pipes D τ0 =γ J 4 τ 0 = γRJ f V2 J= × D 2g f V2 J= × 4R 2 g f = f (Re, D e ) f = f (Re, 4 R e ) VD Re = υ V 4R Re = υ2.6. Hydraulic and Energy Grade LinesThe terms of energy equation have a dimension of length [L ] ; thus we can attach auseful relationship to them. p1 V12 p V2 + + z1 = 2 + 2 + z 2 + h L (2.36) γ 2g γ 2gIf we were to tap a piezometer tube into the pipe, the liquid in the pipe would rise inthe tube to a height p/γ (pressure head), hence that is the reason for the name ⎛ p V2 ⎞hydraulic grade line (HGL). The total head ⎜ + ⎜ γ 2g + z ⎟ in the system is greater ⎟ ⎝ ⎠ ⎛p ⎞ V2than ⎜ + z ⎟ by an amount ⎜γ ⎟ (velocity head), thus the energy (grade) line (EGL) ⎝ ⎠ 2g V2is above the HGL with a distance . 2gSome hints for drawing hydraulic grade lines and energy lines are as follows. 21 Prof. Dr. Atıl BULU
  22. 22. Figure 2.8 e1. By definition, the EGL is positio oned above the HGL a amount equal to an e the velocity head. Thu if the vel y us locity is zer as in lak or reserv ro, ke voir, the HGL and E EGL will cooincide with the liquid s h surface. (Figgure 2.8)2. Head loss f flow in a pipe or c for channel alw ways means the EGL will lope w downward in the direc ction of flow The only exception to this rule occurs w. y n e when a pum supplies energy (an pressure to the flo Then an abrupt mp s nd e) ow. n rise in the EGL occur from the upstream si to the d rs ide downstream side of m the pump.3. If energy is abruptly t s taken out of the flow b for exam f by, mple, a turbbine, the EGL and H HGL will dro abruptly as in Fig… op y ….4. In a pipe or channel w r where the pre essure is zer the HGL is coincide with ro, L ent the water in the system because p γ = 0 at these point This fact can be n m ts. t used to locate the HGL at certain points in th physical system, su as at L n he l uch the outlet e of a pipe where the liquid cha end e, e arges into th atmosphe or at he ere, the upstream end, whe the press m ere sure is zero in the reser rvoir. (Fig.2 2.8)5. For steady flow in a pipe tha has unif y at form physi ical characcteristics (diameter, rroughness, shape, and so on) alon its length the head loss per ng h, unit of leng will be constant; th the slop (Δh L ΔL ) of the E gth hus pe EGL and HGL will b constant and parallel along the l be l length of pi ipe.6. If a flow pa assage chan nges diamete such as i a nozzle or a change in pipe er, in e size, the ve elocity there in will also change; h e o hence the diistance betwween the EGL and H HGL will ch hange. Mor reover, the sslope on the EGL will change e l because the head per unit length will be la e h arger in the conduit with the e w larger veloccity.7. If the HGL falls below the pipe, p γ is neg L w gative, there indicati sub- eby ing atmospheri pressure ( ic (Fig.2.8). 22 Prof. Dr. Atıl BULU A
  23. 23. Figure 2.9If the pressure head of water is less than the vapor pressure head of thewater ( -97 kPa or -950 cm water head at standard atmospheric pressure),cavitation will occur. Generally, cavitation in conduits is undesirable. Itincreases the head loss and cause structural damage to the pipe fromexcessive vibration and pitting of pipe walls. If the pressure at a section inthe pipe decreases to the vapor pressure and stays that low, a large vaporcavity can form leaving a gap of water vapor with columns of water oneither side of cavity. As the cavity grows in size, the columns of watermove away from each other. Often these of columns of water rejoin later,and when they do, a very high dynamic pressure (water hammer) can begenerated, possibly rupturing the pipe. Furthermore, if the pipe is thinwalled, such as thin-walled steel pipe, sub-atmospheric pressure can causethe pipe wall to collapse. Therefore, the design engineer should beextremely cautious about negative pressure heads in the pipe. 23 Prof. Dr. Atıl BULU
  24. 24. 24 Prof. Dr. Atıl BULU

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