FUNDAMENTALS OF FLUID MECHANICS Chapter 8 Pipe Flow 1 MAIN TOPICS General G l Ch Characteristics i i off Pipe Pi Flow Fl Fully Developed Laminar Flow Fully Developed Turbulent Flow Dimensional Analysis of Pipe Flow Pipe Flow Examples Pipe Pi Flowrate Fl Measurement M 2 Introduction Flows completely bounded by solid surfaces are called INTERNAL FLOWS which include flows through pipes (Round cross section), section), ducts (NOT Round cross section) section), nozzles, nozzles diffusers diffusers, sudden contractions and expansions, valves, and fittings. The basic principles involved are independent of the cross cross--sectional shape, although the details of the flow may be dependent on it. The flow regime g (laminar ( or turbulent)) of internal flows is pprimarilyy a function of the Reynolds number (->inertial force/viscous force). Laminar flow: Can be solved analytically. y y Turbulent flow: Rely heavily on semisemi-empirical theories and experimental p data. 3 General Characteristics of Pi Flow Pipe Fl Laminar vs. Turbulent Entrance Region vs. Fully Developed Flow 4 Pipe System A pipe system include the pipes themselves (perhaps (p p of more than one diameter), ), the various fittings, the flowrate control devices valves) , and the pumps or turbines turbines. 5 Pipe Flow vs. vs Open Channel Flow Pipe flow: Flows completely filling the pipe pipe.. (a) g alongg the ppipe p is main driving g force. The ppressure gradient Open channel flow: Flows without completely filling the pipe.. (b) pipe The gravity alone is the driving force. 6 Laminar or Turbulent Flow 1/2 The flow of a fluid in a pipe may be Laminar ? Or Turbulent ? Osborne Reynolds, Reynolds, a British scientist and mathematician, was the first to distinguish g the difference between these classification of flow by using a simple apparatus as shown. 7 Laminar or Turbulent Flow 2/2 For “small enough flowrate” the dye streak will remain as a well--defined line as it flows along, with only slight blurring due well t molecular to l l diffusion diff i off the th dye d into i t the th surrounding di water. t For a somewhat larger “intermediate flowrate” the dye fl fluctuates in i time i andd space, andd intermittent i i bursts b off irregular i l behavior appear along the streak. For F “large l enough h flowrate fl t ” the h dye d streakk almost l immediately become blurred and spreads across the entire pipe in a random fashion. fashion 8 Time Dependence of Fluid Velocity at a Point 9 Indication of Laminar or Turbulent Flow The Th term flowrate fl t should h ld be b replaced l d by b Reynolds R ld number,, Re VL / ,where V is the average velocity in the pipe, number and L is the characteristic dimension of a flow. L is usuallyy D (diameter) in a pipe flow. -> a measure of inertial force to the viscous force. It I iis not only l the h fluid fl id velocity l i that h ddetermines i the h character h off the h flow – its density, viscosity, and the pipe size are of equal importance. p For general engineering purpose, the flow in a round pipe Laminar R e 2100 Transitional Turbulent R e＞4000 10 Entrance Region and Fully Developed Flow 1/5 Any fluid flowing in a pipe had to enter the pipe at some location. The region of flow near where the fluid enters the pipe is termed the entrance (entry) y region g or developing p g fflow region.. region 11 Entrance Region and Fully Developed Flow 2/5 The 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 ppipe p wall (the no slip p boundary y condition)). condition 12 Entrance Region and Fully Developed Flow 3/5 A boundary layer in which viscous effects are important 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, length, section (2), beyond which the velocity profile does not vary with x. x. The boundary layer has grown in thickness to completely fill the pipe. 13 Entrance Region and Fully Developed Flow 4/5 Viscous effects are of considerable importance within the boundary layer. Outside the boundary layer, the viscous effects are negligible. The shape p of the velocity y pprofile in the ppipe p depends p on whether the flow is laminar or turbulent, as does the length of the entrance region, . For laminar flow 0.06R e D For turbulent flow 1/ 6 4.4R e D Dimensionless entrance length 14 Entrance Region and Fully Developed Flow 5/5 Once the fluid reaches the end of the entrance region, section (2), the flow is simpler to describe because the velocity is a function of only the distance from the pipe centerline, r, and independent of x. x. The flow between (2) and (3) is termed fully developed. developed. 15 Pressure Distribution along Pipe In the entrance region of a pipe, pipe the fluid accelerates or decelerates as it flows. There is a balance between pressure, viscous, and inertia (acceleration) force. force. p p constant 0 x u 0 x Th magnitude The i d off the h pressure gradient is larger than that in the fully developed region. There is a balance between pressure and viscous force. force The magnitude of the pressure gradient p g is constant. 16 Fully Developed Laminar Flow There are numerous ways to derive important results pertaining to fully developed laminar flow: From F=ma applied directly to a fluid element. From the Navier Navier--Stokes equations of motion From dimensional analysis methods 17 From F=ma 1/8 Considering a fully developed axisymmetric laminar flow in a long, straight, constant diameter section of a pipe. The fluid element is a circular cylinder of fluid of length l and radius r centered on the axis of a horizontal ppipe p of diameter D. 18 From F=ma 2/8 Because the velocity is not uniform across the pipe, the initially flat end of the cylinder of fluid at time t become distorted at time t+t when the fluid element has moved to its new location along the pipe. If the flow is fully developed and steady, the distortion on each end of the fluid element is the same, and no part of the fluid experiences any acceleration as it flows. Steady V 0 t u Fully developed V V u i 0 x 19 From F=ma 3/8 Apply the Newton’s second Law to the cylinder of fluid Fx ma x The force (pressure & friction) balance p1r 2 p1 p r 2 rx 2r 0 p 2 rx p p1 p2 r Basic balance in forces needed to drive each fluid particle along l the h pipe i with i h constant velocity l i Not function of r p 2 rx r Not function of r Independent of r ? rx Cr B C r=0 rx=0 B.C. r=D/2 rx= w < 0 2 w r rx D 20 From F=ma 4/8 Th pressure drop The d andd wall ll shear h stress are related l d by b 2 w r p 2 rx 4 w rx p D r D Valid for both laminar and turbulent flow. flow. Laminar rx du dr 21 From F=ma 5/8 Since du dr L i Laminar p du r dr 2 p 2 rx r p 2 p du 2 rdr u 4 r C1 With the boundary conditions: u=0 at r=D/2 pD 2 C1 16 Velocity distribution p 4 w D 2 2 pD 2r 2r u (r ) 1 VC 1 16 D D 2 w D r u (r ) 1 4 R 2 22 From F=ma 6/8 The shear stress distribution rx du r p dr 2 Volume flowrate Q u dA A D 4 p Q 128 R 0 pD 2 u (r ) 16 2r 2 2r 2 1 VC 1 D D u ( r ) 2rdr ..... R 2VC 2 Poiseuille’s Law Valid for Laminar flow only 23 From F=ma 7/8 Average velocity V average Q Q pD 2 2 A R 32 D 4 p Q 128 Point of maximum velocity du 0 dr u u max at r=0 pD 2 u (r ) 16 R 2 p U 2V average 4 2r 2 2r 2 1 VC 1 D D 24 From F=ma 8/8 Making adjustment to account for nonhorizontal pipes p p sin >0 if the flow is uphill <0 0 if the th flow fl is i downhill d hill pr 2 p p r 2 rx 2r gr 2 sin 0 p g sin 2 rx r Specific weight p sin D 2 p sin D 4 Vaverage , g Q 32 128 25 Example 8.2 8 2 Laminar Pipe Flow An oil with a viscosity of μ= 0.40 N· N·s/m2 and density ρ= 900 kg/m3 flows in a pipe of diameter D= 0.20m . (a) What pressure drop, drop p1-p2, is needed to produce a flowrate of Q=2.0× Q=2.0 Q 2 0× 0×10-5 m3/s if the pipe is horizontal with x1=0 and x2=10 m? (b) How steep a hill, θ,must the pipe be on if the oil is to flow through the pipe at the same rate as in part (a), but with p1=p2? (c) For the conditions of part (b), if p1=200 kPa, what is the pressure at section, x3=5 5m m, where x is measured along the pipe? 26 Example 8 8.2 2 Solution1/2 R e VD / 2 .87 2100 V Q 0 .0637 m / s A The flow is laminar flow 128 Q p p1 p 2 ... 20 .4 kPa 4 D If the pipe is on the hill of angle θ with Δp=0 sin 128Q ... 13.34 4 gD 27 Example 8 8.2 2 Solution2/2 Wi h p1=p2 the With h length l h off the h pipe, i , does d not appear iin the h flowrate fl equation Δp=0 for all p1 p 2 p 3 200 kPa 28 From the Navier Navier--Stokes Equations 1/3 General motion of an incompressible Newtonian fluid is governed by the continuity equation and the momentum equation Mass conservation Navier-Stokes Equation in a cylindrical coordinate Acceleration 29 From the Navier Navier--Stokes Equations 2/3 Simplify the NavierNavier-Stokes equation axial component: z Th flow The fl is i governedd by b a balance b l off pressure, weight, i ht andd viscous i forces in the flow direction. For steady, steady fully developed flow in a pipe, pipe, the velocity contains only an axial component, component , which is a function of onlyy the radial coordinate V u ( r ) i 30 From the Navier Navier--Stokes Equations 3/3 axial component: x V u( r ) i p 1 u g sin r r r r x Function of, at most, only x Function of ,at most, only r p p p const. x x Integrating Velocity profile u(r)= B.C. (1) r = R , u = 0 ; (2) r = 0 , u < ∞ or u/ u/r=0 31 From Dimensional Analysis 1/3 Assume that the pressure drop in the horizontal pie, Δp, is a function of the average velocity of the fluid in the pipe, V, the length of the pipe, , the pipe diameter, D, and the viscosity of the fluid, μ. p F ( V , , D , ) Dp V D Dimensional analysis an unknown function of the length to diameter ratio of the pipe. pipe k-r = 5 (총 변수 변수)) -3 (reference dimension)) =2 dimensionless ggroupp 32 From Dimensional Analysis 2/3 Dp C where C is a constant. V D p CV 2 p 2 ( / 4C )pD 4 Q AV D D 2 D 4 C The value of C must be determined by theory or experiment. experiment. For a round pipe, pipe C=32. C=32. For duct of other crosscross-sectional shapes, the value of C is different. different. ( / 4C ) D 4 1 Q p p Flow resistance 32 V For a round pipe p D2 Analogy: i V R 33 From Dimensional Analysis 3/3 p 32 V / D 2 64 For a round pipe 1 64 2 2 1 VD D Re D 2 V 2 V V 2 p f D 2 D p p f V 2 2 Dynamic pressure -> Characteristic pressure V 2 D 2 * p * For laminar flow f is i termed t d the th friction f i ti factor, factor f t , or sometimes the Darcy friction factor -> dimensionless pressure drop for internal flows. flows. 64 8 w f Re V 2 4 w p D 34 Energy Consideration 1/3 The energy equation for incompressible, steady flow between two locations p2 2 V22 p1 1V12 z1 z2 hL 2g 2g 1V12 2 V22 2g 2g p1 p 1 z1 2 z 2 hL p1 p2 z1 z 2 4 w 1 2 p g sin r D p g sin 2 rx r 2 w r D The head loss in a pipe is a result of the viscous shear stress on the wall. 35 Example 8 8.3 3 Laminar Pipe Flow Properties 1/2 The flowrate, Q, of corn syrup through the horizontal pipe shown in Figure E8.3 is to be monitored by measuring the pressure difference between sections (1) and (2). (2) It is proposed that Q=KΔ Q=KΔp, p where the calibration constant, K, is a function of temperature, T, because of the variation of the syrup’s y p viscosityy and densityy with temperature. p These variations are given in Table E8.3. (a) Plot K(T) versus T for 60˚ 60˚F≤T ≤ 160 160˚˚F. (b) Determine the wall shear stress and the pressure drop, Δp=p1-p2, for Q=0.5 T=100˚F. ft3/s and T=100˚ (c) For the conditions of part (b) (b), determine the nest pressure force.(π force.( force (π (πD2/4) /4)Δ Δp, p and the nest shear force, πDτw, on the fluid within the pipe between the sections (1) and (2). 36 Example 8 8.3 3 Laminar Pipe Flow Properties 1/2 37 Example 8 8.3 3 Solution1/2 If the flow is laminar ((-> should be verified) pD4 Q Kp 128 1 .60 10 5 K For T=100˚ T=100˚F,, μ=3.8 =3.8××10-3 lb·s/ft2, Q Q=0.5ft3/s 128Q 2 ... 119 lb / ft p D4 Q V ... 10.2ft / s A R e VD / ... 1380 2100 4 w pD p w ... 1.24lb / ft 2 D 4 Laminar 38 Example 8 8.3 3 Solution2/2 The new pressure force and viscous force on the fluid within the pipe between sections (1) and (2) is D Fp p ... 5.84lb 4 D Fv 2 w ... 5.84lb 2 2 The values of these two forces are the same. The net force is zero; there is no acceleration. acceleration 39 Fully Developed Turbulent Flow Turbulent pipe flow is actually more likely to occur than laminar flow in practical situations. Turbulent flow is a very complex process. Numerous persons have devoted considerable effort in an attempting to understand the variety of baffling aspects of turbulence. Although a considerable amount if knowledge about the topics has been developed, the field of turbulent flow still remains the least understood area of fluid mechanics. Much remains to be learned about the nature of turbulent flow. flow 40 Transition from Laminar to Turbulent Flow in a Pipe 1/2 For any flow geometry, there is one (or more) dimensionless parameters such as with this parameter value below a particular value the flow is laminar, whereas with the parameter value larger than a certain value the flow is turbulent. The important parameters involved and their critical values depend on the specific flow situation involved. For flow in ppipe p : Re~4000 For flow along a plate Rex~500000 Turbulence initiated initiated. Consider a long section of pipe that is initially filled with a fluid at rest. 41 1/7 7 (1(1-6: option) Energy Considerations 1/ Considering the steady flow through the piping system, including a reducing elbow. The basic equation for conservation of energy – the first law of thermodynamics Q net in e d V e V n dA in nn V n dA CS CV CS t WShaft in e d V e V n dA nn V n dA CS CS t CV W Shaft Q net in Energy equation -p p V2 edV (uˆ gz ) V n dA Q net in W Shaft in CS 2 t CV e uˆ 2 Kinetic energy V gz 2 Internal energy N Note: Th The shear h stress power iis negligibly li ibl smallll on a control surface. 42 2/7 7 Energy Considerations 2/ When the flow is steady CV edV 0 t The integral of p V2 CS û 2 gzV ndA ??? Uniformly distribution p V2 p V2 p V2 û gz m û gz m CS û 2 gzV ndA 2 2 out in Only one stream entering and leaving p V2 CS ûˆ 2 gzV ndA p V2 p V2 outt u ini uû gz m û gz m 2 2 out in 43 3/7 7 Energy Considerations 3/ If shaft h ft workk is i involved…. i l d 2 p p Vout Vin2 uû out uû in m g z out z in 2 out in One-dimensional energy equation OneQ W net in shaft net in for steadysteady-inin-thethe-mean flow p Enthalpy h l ĥ ûˆ h The h energy equation i is i written i in i terms of enthalpy. 2 Vin2 Vout ĥ out ĥ in m g z out z in Q net / in W shaft net / in 2 44 4/7 7 Energy Considerations 4/ For steady, steady incompressible flow…One flow… flow One One--dimensional energy equation 2 p out pin Vout Vin2 û out û in m gz out z in Q net in 2 m 2 pout Vout pin Vin2 gz out gz in û out û in q net in 2 2 where q net in Q net in / m For steady, incompressible, frictionless flow… flow… p out Examples? 2 2 Vout V t i in z out p in z in Bernoulli equation 2 2 uˆ out uˆ in q net in 0 Frictionless flow… flow… 45 5/7 7 Energy Considerations 5/ For steady, steady incompressible, incompressible frictional flow… flow… û out û in q net in 0 Frictional flow… flow… p V2 Defining “useful or available energy”… gz 2 Defining “loss of useful or available energy”… û out û in q net in loss 2 p out Vout p in Vin2 gz out gz in loss 2 2 46 6/7 7 Energy Considerations 6/ F steady, For t d incompressible i ibl flow fl with ith friction f i ti andd shaft h ft workk… 2 p out p in Vout Vin2 m uû out u û ini gz out z ini Q net ini W shaft h f net in i 2 m 2 p out Vout pin Vin2 gz g out gz g in w shaft net in ( uˆ out uˆ in q net in ) 2 2 2 2 p outt Vout p V t gzout ini ini gzin w shaft net in loss l 2 2 g Shaft head 2 pout Vout pin Vin2 z out z in h s h L 2g 2g hS w shaft net in g W shaft net in g m W shaft net in Q Head loss h L lloss g 47 7/7 7 Energy Considerations 7/ 2 pout Vout pin Vin2 z out z in h s h L 2g 2g Total head loss , hL, is regarded as the sum of major losses, hL major, due d tto frictional f i ti l effects ff t in i fully f ll developed d l d flow fl in constant area tubes, tubes, and minor losses, hL minor, resulting from entrance, fitting, area changes, changes, and so on. hL hLmajor hLminor 48 Major Losses: Friction Factor The energy equation for steady and incompressible flow with zero shaft work V2 p1 1V1 p 2 2V 2 g 2 g z1 g 2 g z 2 hL 2 2 A V ndA 2 1 V2 m 2 For fully developed flow through a constant area pipe, α1= α2, V1= V2 , where is the kinetic energy coefficient and V is the average velocity. For a uniform velocity, α=1. p1 p 2 ( z 2 z1 ) hL g For hori horizontal ontal pipe, pipe z2=z1 p1 p 2 p hL g g 49 Major Losses: Laminar Flow In fully developed laminar flow in a horizontal pipe pipe,, the pressure drop 128 Q 128 V D 2 / 4 p 32 D 4 D 4 D p 64 64 1 VD D Re D V 2 2 V 2 V V2 p f h L 32 D 2 D D D 2 Friction Factor V D 64 V 2 64 VD R e D 2 f pD / / V 2 / 2 f laminar l i 64 Re 50 Major Losses: Turbulent Flow 1/3 In turbulent flow, we cannot evaluate the pressure drop analytically; we must resort to experimental results and use dimensional analysis to correlate the experimental data. data In fully developed turbulent flow the ppressure drop, p, △p , caused byy friction in a horizontal constantconstant-area pipe is known to depend on pipe diameter,D, pipe length, , pipe roughness,e, average flow velocity, V, fluid densityρ densityρ, andd fluid fl id viscosity,μ. viscosity, i it μ. p FV , D, , , , 51 Major Losses: Turbulent Flow 2/3 Applying dimensional analysis, the result were a correlation of the form VD p , , 2 1 D D 2 V Experiments E i show h that h the h nondimensional di i l head h d loss l is i directly di l proportional to /D. Hence we can write p Re, 2 1 D D 2 V f Re, D V 2 p f D 2 Darcy--Weisbach equation Darcy V2 h L major f D 2g 52 Roughness for Pipes 53 Moody chart Depending on the specific circumstances involved. 54 About Moody Chart For F laminar l i fl flow, f 64/R which f=64/Re, hi h iis iindependent d d t off the th relative roughness. roughness. For very large Reynolds numbers, numbers f= f=Φ Φ(ε/D), /D) which is independent of the Reynolds numbers. numbers. For flows with very large value of Re, Re, commonly termed completely turbulent flow (or wholly turbulent flow), the laminar sublayer is so thin (its thickness decrease with i increasing i Re) R ) that th t the th surface f roughness h completely l t l dominates the character of the flow near the wall. For flows with moderate value of Re, Re, the friction factor f=Φ f= Φ(Re, (Re,εε/D). 55 Major Losses: Turbulent Flow 3/3 Colebrook – To avoid having to use a graphical method for obtaining f for turbulent flows. Valid for the entire nonlaminar h Moody M d chart. h 1 2 .51 range off the / D 2 .0 log Colebrook formula -> needs 3 . 7 f Re f iteration. Miler suggests that a single iteration will produce a result within 1 percent if the initial estimate is calculated calc lated from / D 5 .74 f 0 0 .25 log l 0 .9 3 . 7 Re 2 56 Example 8.5 Comparison of Laminar or Turbulent pressure Drop Air under standard conditions flows through a 4.04.0-mmmm-diameter drawn tubing with an average velocity of V = 50 m/s. For such conditions the flow would normally be turbulent. turbulent However, However if precautions are taken to eliminate disturbances to the flow (the entrance to the tube is veryy smooth,, the air is dust free,, the tube does not vibrate, etc.), it may be possible to maintain laminar flow. (a) Determine the pressure drop in a 0.10.1-m section of the tube if the flow is l i laminar. (b) Repeat the calculations if the flow is turbulent. 57 Example 8 8.5 5 Solution1/2 Under standard temperature and pressure conditions Ρ=1.23kg/m3, μ=1.79 μ=1.79 10-5Ns/m The Reynolds number R e VD / ... 13,700 Turbulent flow If the flow were laminar f=64/Re=…=0.0467 f 64/R 0 0467 1 p f V 2 ... 0 .179 kPa D2 58 Example 8 8.5 5 Solution2/2 If the flow were turbulent From Moody chart f= f=Φ Φ(Re, (Re,εε/D) =…0.028 1 p f V 2 ... 1 .076 kPa D2 59 Minor Losses 1/5 Most pipe systems consist of considerably more than straight pipes.. These additional pipes components (valves, bends, tees, and the like) add to the overall head loss of the system. Such losses are termed MINOR LOSS. But, it is not minor at all and it may be larger than the major losses. The flow pattern through a valve 60 Minor Losses 2/5 The theoretical analysis to predict the details of flow pattern (through these additional components) is not, as yet, possible. p is The head loss information for essentiallyy all components given in dimensionless form and based on experimental data.. The most common method used to determine these data head losses or pressure drops is to specify the loss coefficient,, KL. coefficient 61 Minor Losses 3/5 KL h Lmin or 2 V / 2g p 1 p K L V 2 1 2 2 V 2 Minor losses are sometimes gi en in terms of an equivalent given eq i alent length eq Large K : large pressure drops for given velocities. eq V 2 V2 h Lmin or K L f 2g D 2g D eq K L f The actual value of KL is strongly dependent on the geometry of the component considered. It may also dependent on the fluid properties. That is K L (geometry, Re) 62 Minor Losses 4/5 For F many practical i l applications li i the h Reynolds R ld number b is i large enough so that the flow through the component is dominated by inertial effects, effects, with viscous effects being of secondary importance. In a flow that is dominated by inertia effects rather than viscous effects, it is usually found that pressure drops and head losses correlate directly with the dynamic pressure. pressure. This is the reason why the friction factor for very large Reynolds number, number fully developed pipe flow is independent of the Reynolds number. number. 63 Minor Losses 5/5 This is true for flow through pipe components. Thus, in most cases of ppractical interest the loss coefficients for components are a function of geometry only, only y, K L ( geometry t ) 64 Minor Losses Coefficient Entrance flow 1/3 Entrance flow condition and loss coefficient (a) Reentrant, KL = 0.8 (b) sharpsharpp-edged, g , KL = 0.5 (c) slightly rounded, KL = 0.2 (d) wellwell-rounded,, KL = 0.04 KL = function of rounding of the inlet edge. 65 Minor Losses Coefficient Entrance flow 2/3 A vena contract region may result because the fluid cannot turn a sharp right right--angle corner. The flow is said to separate from the sharp corner. The maximum velocityy velocityy at section ((2)) is ggreater than that in the pipe section (3), and the pressure there is lower. If this high speed fluid could slow down efficiently, the kinetic energy could be converted into pressure. 66 Minor Losses Coefficient Entrance flow 3/3 Such is not the case. Although the fluid may be accelerated very efficiently efficiently, it is very difficult to slow down ((decelerate)) the fluid efficently. y (2) (2) (3) The extra kinetic energy of the fluid is partially lost because of viscous dissipation, so that the pressure d does nott return t to t the th ideal id l value. Flow pattern and pressure distribution f a sharpfor sharp h -edged d d entrance 67 Minor Losses Coefficient Exit flow Exit flow condition and loss coefficient (a) Reentrant, KL = 1.0 (b) sharpsharpp-edged, g , KL = 1.0 (c) slightly rounded, KL = 1.0 (d) wellwell-rounded,, KL = 1.0 68 Minor Losses Coefficient varied diameter Loss coefficient for sudden contraction, expansion,typical conical diffuser. diffuser A K L 1 1 A2 2 69 Minor Losses Coefficient Bend Character of the flow in bend and the associated loss coefficient. coefficient Carefully designed guide vanes help direct the flow with less unwanted swirl and disturbances. 70 Internal Structure of Valves (a) globe valve (b) gate valve (c) swing check valve (d) stop check valve 71 Loss Coefficients for Pipe Components 72 Example 8.6 8 6 Minor Loss 1/2 Air at standard conditions is to flow through the test section [between sections (5) and (6)] of the closedclosed-circuit wind tunnel shown if Figure E8 E8.6 6 with a velocity of 200 ft/s. ft/s The flow is driven by a fan that essentially increase the static pressure by the amount p1-p9 that is needed to overcome the head losses experienced p byy the fluid as it flows around the circuit. Estimate the value of p1-p9 and the horsepower supplied to the fluid by the fan. 73 Example 8.6 8 6 Minor Loss 2/2 74 Example 8 8.6 6 Solution1/3 The maximum velocity within the wind tunnel occurs in the test section (smallest area). Thus, the maximum Mach number of the flow is Ma5=V5/c5 V5 200ft / s c 5 ( KRT5 )1 / 2 1117ft / s The energy equation between points (1) and (9) p 9 V9 2 p1 V12 z1 z 9 h L19 2g 2g h L19 p1 p 9 The total head loss from (1) to (9). 75 Example 8 8.6 6 Solution2/3 The energy across the fan, from (9) to (1) Hp is the actual head rise supplied pp p 9 V9 2 p1 V12 z9 hp z1 by the pump (fan) to the air. 2g 2g p1 p 9 hp h L19 The actual power supplied to the air (horsepower, Pa) is obtained from the fan head by Pa Qh p A 5 V5 h p A 5 V5 h L19 76 Example 8 8.6 6 Solution3/3 The total head loss h L19 h Lcorner 7 h Lcorner 8 h Lcorner 2 h Lcorner 3 h Ldif h L noz h Lscr V2 V2 h Lcorner K L 0. 2 2g 2g K L noz 0.2 K Lscr 4.0 h Ldif K Ldif V2 V2 0 .6 2g 2g p1 p 9 h L19 (0.765lb / ft 2 )(560ft ) ... 0.298psi Pa ... 34300ft lb / s 62.3hp 77 Noncircular Ducts 1/4 The Th empirical i i l correlations l i for f pipe i flow fl may be b usedd for f computations involving noncircular ducts, provided their cross sections are not too exaggerated exaggerated. The correlation for turbulent pipe flow are extended for use with noncircular geometries by introducing the hydraulic diameter, diameter, defined as 4A Dh P where A is cross cross-sectional area, and P i wetted is tt d perimeter. i t 78 Noncircular Ducts 2/4 For F a circular i l duct d t 4 A 4R 2 Dh D P 2R For a rectangular duct of width b and height h 4A 4bh 2h Dh P 2( b h ) 1 ar ar h / b h b The hydraulic diameter concept can be applied in the approximate range ¼<ar<4. So the correlations for pipe flow give acceptably accurate results for rectangular ducts. 79 Noncircular Ducts 3/4 The friction factor can be written as f=C/Reh, where the constant C depends on the particular shape of the duct, and Reh is the Reynolds number based on the hydraulic diameter.. Note: C = 64 for a circular tube. diameter The hydraulic diameter is also used in the definition of the friction factor, h L f ( / D h )( V 2 / 2g) , and the relative roughness /Dh. 80 Noncircular Ducts 4/4 For Laminar flow, the value of C=f·Reh have been obtained from theory and/or experiment for various shapes. For turbulent flow in ducts of noncircular cross section, calculations are carried out byy usingg the Moodyy chart data for round pipes with the diameter replaced by the hydraulic diameter and the Reynolds number based on the hydraulic diameter. The Moody chart, developed for round pipes, can also be used for noncircular ducts. ducts. 81 Friction Factor for Laminar Flow in Noncircular Ducts Note: C = 64 for a circular tube tube. 82 Example 8 8.7 7 Noncircular Duct Air at temperature of 120˚ 120˚F and standard pressure flows from a furnace through an 88-in.in.-diameter pipe with an average velocity of 10ft/s It then passes through a transition section and into a square 10ft/s. duct whose side is of length a. The pipe and duct surfaces are smooth ((εε=0). ) Determine the duct size,, a,, if the head loss per p foot is to be the same for the pipe and the duct. 83 Example 8 8.7 7 Solution1/3 The head loss per foot for the pipe h L f V2 D 2g For given pressure and temperature ν=11.89 1.89 8910-4ft2/s 89 VD Re 35300 hL f Vs2 0.0512 For the square duct D h 2g Dh 4A a P Vs Q 3.49 2 A a 84 Example 8 8.7 7 Solution2/3 hL f Vs2 f (3.49 / a 2 ) 2 0.0512 a 1.30f 1 / 5 (1) D h 2g a 2(32.2) The Reynolds number based on the hydraulic diameter Vs D h (3.49 / a 2 )a 1.89 104 Re h 4 1.89 10 a (2) Have three unknown (a (a, f, f and Reh) and three equation – Eqs. 1, 2, and either in graphical form the Moody chart or the Colebrook equation Find a 85 Example 8 8.7 7 Solution3/3 Use the Moody chart A Assume th the friction f i ti factor f t for f the th duct d t is i the th same as for f the th pipe. i That is, assume f=0.022. F From E Eq. 1 we obtain bt i a=0.606 0 606 ft. ft From Eq. 2 we have Reh=3.05 =3.05 104 F From M Moody d chart h we fi findd f=0.023, f 0 023 which hi h does d not quite i agree the h assumed value of f. Try again, using the latest calculated value of f=0.023 as our guess. =3.05 104, and a=0.611ft. …… The final result is f=0.023, Reh=3.05 86 Pipe Flow Examples 1/2 The energy equation, relating the conditions at any two points 1 and 2 for a single single--path pipe system, for steady and incompressible flow with zero shaft work is given by p1 1V12 p 2 2 V2 2 g 2 g z1 g 2 g z 2 h L h L major h L min or By judicious choice of points 1 and 22, we can analyze not only the entire pipe system, but also just a certain section of it that we may be interested in. in 2 Major loss h Lmajor V f D 2g 2 Minor loss h Lmin or V KL 2g 87 Pipe Flow Examples 2/2 Single pipe whose length may be interrupted by various components. Multiple pipes in different configuration Parallel Series Network N t k 88 Single--Path Systems 1/2 Single Pipe flow problems can be categorized by what parameters are given and what is to be calculated calculated.. 89 Single--Path Systems 2/2 Single Type 1: Given pipe (L and D), and flow rate, and Q, find pressure drop Δp Type 1: Given Δp, D, and Q, find L. Type 2: Given Δp, L, and D, find Q. Type 3: Given Δp, L, and Q, find D. 90 Given L , D, D and Q, Q find Δp The Th energy equation i p1 1V12 p2 2 V2 2 z1 z 2 h L h Lmajor h L min or g 2g 2g g The flow rate leads to the Reynolds number and hence the friction factor for the flow. Tabulated data can be used for minor loss coefficients and equivalent lengths. The energy equation can then be used to directly to obtain the pressure drop. 91 Given Δp, p D, D and Q, Q find L The Th energy equation i p1 1V12 p2 2 V2 2 g 2g z1 g 2g z 2 h L h Lmajor h L min or The flow rate leads to the Reynolds number and hence the friction factor for the flow. Tabulated data can be used for minor loss coefficients and equivalent lengths. The energy equation can then be rearranged and solved directly for the pipe length. 92 Given ∆p ∆p, L L, and D, D find Q These types yp of problems p required q either manual iteration or use of a computer application. The unknown flow rate or velocityy is needed before the Reynolds number and hence the friction factor can be found. First, we make a guess for f* and solve the energy equation for V in terms of known quantities and the guessed friction factor f*. Then we can compute a Reynolds number and hence obtain a new value for f. f. Repeat the iteration process f*→ V→ Re→ f until convergence (f*=f) 93 Given Δp, p L, L and Q, Q find D These types yp of problems p required q either manual iteration or use of a computer application. y The unknown diameter is needed before the Reynolds number and relative roughness, and hence the friction factor can be found. First, we make a guess for f* and solve the energy equation for D in terms of known quantities and the guessed friction factor f*. Then we can compute a Reynolds number and hence obtain a new value for f. Repeat the iteration process f*→ D→ (Re and ε/D) → f until convergence (f*=f) 94 Example 8.8 Type I Determine Pressure Drop Water at 60˚ 60˚F flows from the basement to the second floor through the 0.75--in. (0.06250.75 (0.0625-fy)fy)-diameter copper pipe (a drawn tubing) at a rate of Q = 12 12.0 0 gal/min = 0.0267 0 0267 ft3/s and exits through a faucet of diameter 0.50 in. as shown in Figure E8.8. Determine the pressure at point (1) if: (a) all losses are neglected, (b) the only losses included are major losses, or (c) all losses are i l d d included. 95 Example 8 8.8 8 Solution1/4 Q V1 ... 8.70ft / s A1 1.94slug l / ft 3 2.34 105 lb s / ft 2 Re VD / 45000 The flow is turbulent Nearly uniform velocity profile The energy equation 2 2 p1 1V1 p2 2 V2 z1 z2 hL g 2g g 2g 1 2 1 p1 z 2 (V2 V1 ) hL 1 2 jet ) z1 0, z 2 20 ft f , p2 0( free f V2 Q / A2 ... 19.6 ft / s V1 Q / A1 8.70 ft / s 2 2 Head loss is different for each of the three cases. cases 96 Example 8 8.8 8 Solution2/4 (a) If all losses are neglected (hL=0) 2 2 1 p1 z 2 ( V2 V ) ... 1547lb / ft 2 10.7psi 1 2 (b) If the only losses included are the major losses, the head loss is V12 hL f D 2g Moody chart 0.000005 , / D 8 10 5 , Re R 45000 2 2 p1 z 2 12 ( V2 V1 ) f f=0 0215 f=0.0215 2 1 ( 60ft ) V ... 3062lb / ft 2 21.3psi D 2 97 Example 8 8.8 8 Solution3/4 (c) If major and minor losses are included 2 2 V V 2 2 1 p1 z 2 12 ( V2 V1 ) f K L D 2g 2 V2 p1 21.3psii K L 2 2 Valve (pp (pp. 72)) Faucet ( 8 . 70 ft / s ) 21.3psi (1.94slugs / ft 3 ) [10 4(1.5) 2] 2 Elbow p1 21.3psi 9.17psi 30.5psi 98 Example 8 8.9 9 Type II, Determine Head Loss Crude oil at 140˚ 140˚F with γ=53.7 lb/ft3 and μ= 8 810 10--5 lb·s/ft2 (about four times the viscosity of water) is pumped across Alaska through the Alaska pipeline pipeline,, a 799799-mile mile--along, along 44-ft ft--diameter steel pipe pipe,, at a maximum rate of Q = 2.4 million barrel/day = 117ft3/s, or Q ft/s. Determine the horsepower p needed for the ppumps p V=Q/A=9.31 that drive this large system. 99 Example 8 8.9 9 Solution1/2 The energy equation between points (1) and (2) p1 1V12 2g z1 hP p2 2V2 2 2g z 2 hL hP is the head p provided to the oil by the pump. Assume that z1=zz2, p1=pp2=V V1=V V2=00 (<(<- large, large open tank) tank) Minor losses are negligible because of the large lengthlength-totodiameter ratio of the relatively straight, straight uninterrupted pipe. pipe V2 hL hP f ... 17700ft D 2g Table 8.1 Turbulent (α (α=1) f=0.0124 f 0 0124 from f Moody M d chart h t ε/D=(0.00015ft)/(4ft), /D (0 00015ft)/(4ft) Re=7.76 R 7 76 x 105 For steel pipe 100 Example 8 8.9 9 Solution2/2 The actual power supplied to the fluid. 1hp 202000hp Pa QhP ... 550 ft lb / s -> It requires many pump stations to be set up. 101 Example 8.10 8 10 Type II, II Determine Flowrate According to an appliance manufacturer, the 44-inin-diameter galvanized iron vent on a clothes dryer is not to contain more than 20 ft of pipe and four 90 90˚˚ elbows. elbows Under these conditions determine the air flowrate if the pressure within the dryer is 0.20 inches of p of 100℉ 100℉ and standard ppressure. KL = water. Assume a temperature 0.5 for an entrance and 1.5 for elbows. 102 Example 8 8.10 10 Solution1/2 Application of the energy equation between the inside of the dryer, dryer, point (1), and the exit of the vent pipe pipe,, point (2) gives 1V12 2V2 2 V2 V2 z1 z2 f KL 2g 2g D 2g 2g p1 p2 A Assume th thatt z1=z2, p2=0, 0 V1=00 p1 H 2O 1V12 2V2 2 V2 V2 z1 z2 f KL 2g 2g D 2g 2g p1 p2 1ft 3 2 0.2in p1 (0.2in.)) (62.4lb / ft ) 1.04lb / ft i . 12in With γ=0.0709lb/ft3, V2=V, and ν=1.79 =1.79××10-4 ft2/s. 1.04 lb ft 2 20ft V2 1 f 0.5 4 1.5 3 2 0.0709 lb ft 4 12ft 232.2 ft s 945 (7.5 60 f )V 2 (1) Assume turbulent flow (α=1) f is dependent on Re, which is dependent on V, and unknown. 103 Example 8 8.10 10 Solution2/2 Re VD ... 1860V (2) For galvanized iron We have three relationships (Eq. (Eq 1, 1 2, 2 and the ε/D=0.0015 ε/D=0 0015 curve of the Moody chart) from which we can solve for the three unknowns f, Re,, and V. This is done easily by iterative scheme as follows. Assume f=0 f=0.022→V=10.4ft/s 022→V=10 4ft/s (Eq (Eq. 1)→Re=19,300 1)→ 1)→Re=19 300 (Eq (Eq.2)→f=0.029 2)→f=0 →f=0 029 →f=0.029 Assume f=0.029 →V10.1ft/s→Re=18,800 →f=0.029 (Final value) Q AV ... 0.881ft /s 3 Turbulent 104 Example 8.11 8 11 Type II, II Determine Flowrate The turbine shown in Figure E8.11 extracts 50 hp from the water flowing through it. The 11-ftft-diameter, 300 300--ft ft--long pipe is assumed to have a friction factor of 0.02. 0 02 Minor losses are negligible. negligible Determine the flowrate through the pipe and turbine. 105 Example 8 8.11 11 Solution1/2 The energy equation can be applied between the surface of the lake and the outlet of the pipe as p1 1V12 2g z1 p2 2V2 2 2g z 2 hL hT p1 1V12 2g z1 p2 2V2 2 2g z 2 hL hT Where p1=V1= p2=z2=0, =0 z1=90ft, =90ft and V2=V, =V the fluid velocity in the pipe Pa 561 V2 2 ... ft Assume turbulent flow hL f 0.0932V ft h T (α=1) Q V D 2g V2 561 2 3 90ft 0 . 0932 V ft ft 0 . 107 V 90V 561 0 2 2 32.2 ft s V There are two real, p positive roots: V=6.58 ft/s or V=24.9 ft/s. The third root is negative (V=-31.4ft/s) and has no physical meaning for this flow. 106 Example 8 8.11 11 Solution2/2 T acceptable Two bl flowrates fl are 2 Q D V ... 5.17ft 3 / s 4 2 Q D V ... 19 9.6ftt 3 / s 4 At 60oF, F 11.21 21 x 10-5 (ft2/s) Re ≈ 105 107 Example 8.12 Type III Without Minor Losses, Determine Diameter Air at standard temperature and pressure flows through a horizontal horizontal,, galvanized iron pipe (ε=0.0005 ft) at a rate of 2.0ft3/s. Determine the minimum pipe diameter if the pressure drop is to be no more than 0.50 psi per 100 ft of pipe. pipe. 108 Example 8 8.12 12 Solution1/2 Assume the flow to be incompressible with ρ=0.00238 slugs/ft3 and μ=3.74××10-7 lb μ=3.74 lb． ．s/ft2. If the pipe were too long, the pressure drop from one end to the other, p1-p2, would not be small relative to the pressure at the beginning, and compressible flow considerations would be required. 1V12 2V2 2 V2 V2 z1 z2 f KL 2g 2g D 2g 2g p1 p2 V 2 p1 p 2 f With i h z1=z2, V1=V2 , The h energy equation i becomes b D g 2 ( 100 ft ) V p1 p 2 (0.5)(144)lb / ft 2 f (0.00238slugs l / ft 3 ) D 2g Q 2.55 V 2 A D D 0.404f 0 404f 1/5 (1) 109 Example 8 8.12 12 Solution2/2 1.62 10 4 VD Re ... D 0.0005 D D (2) (3) We have four equations (Eq. 1, 2, 3, and either the Moody chart or the Colebrook equation) and four unknowns (f (f, D D, ε/D, ε/D and Re) from which the solution can be obtained by trialtrial-andand-error methods. Repeat the iteration process f*→ D→ Re and ε/D→ f until convergence (1) (2) (3) 110 Example 8.13 Type III With Minor Losses, Determine Diameter Water at 60˚ 60˚F (ν (ν=1.21 =1.21××10-5 ft2/s) is to flow from reservoir A to reservoir B through a pipe of length 1700 ft and roughness 0.0005 ft at a rate of Q= 26 ft3/s as shown in Figure E8.13. E8 13 The system contains a sharpsharp-edged entrance and four flanged 45˚ 45˚ elbow elbow.. Determine the p pipe p diameter needed. 111 Example 8 8.13 13 Solution1/2 The energy gy equation q can be applied pp between two ppoints on the surfaces of the reservoirs (p1=V1= p2=z2=V2=0) 2 2 p1 1V1 p2 2V2 z1 z 2 hL 2g 2g V2 z1 f KL 2g D Q 33.1 V 2 KLent=00.5, 5 KLelbow 0.2, 2 and KLexiti =11 lb =0 A D V2 1700 44ft f [ 4 ( 0 . 2 ) 0 . 5 1 ] 2 2(32.2ft / s ) D f 0.00152 D5 0.00135D (1) 112 Example 8 8.13 13 Solution2/2 VD 2.74 106 ... Re D 0.0005 (3) D D (2) We have four equations (Eq. 1, 2, 3, and either the Moody chart or the Colebrook equation) and four unknowns (f (f, D D, ε/D, ε/D and Re) from which the solution can be obtained by trialtrial-andand-error methods. Repeat the iteration process D→ f* →Re and ε/D→ f until convergence (1) (2) (3) 113