FUNDAMENTALS OF FLUID MECHANICS. Chapter 8 Pipe Flow. Jyh-Cherng. Shieh Department of Bio-Industrial

Chapter 8 Pipe Flow FUNDAMENTALS OF FLUID MECHANICS Jyh-Cherng Shieh Department of Bio-Industrial Mechatronics Engineering National Taiwan University 1/1/009 1

MAIN TOPICS General Characteristics of Pipe Flow Fully Developed Laminar Flow Fully Developed Turbulent Flow Dimensional Analysis of Pipe Flow Pipe Flow Examples Pipe Flowrate Measurement

類關, 但流場的細節分佈則與截面形狀有關分Introduction 定義 Internal flows: 流體完全被 solid surface 包圍的流動依 solid surface 形狀可分成 pipes ducts( 非圓 ) nozzle 等等 Flows completely bounded by solid surfaces are called INTERNAL FLOWS which include flows through pipes (Round cross section), ducts (NOT Round cross section),, nozzles, diffusers, sudden contractions and expansions, valves, and fittings. The basic principles involved are independent of the cross-sectional sectional shape, although the details of the flow may be dependent on it. The flow regime (laminar or turbulent) of internal flows is primarily a function of the Reynolds number. Laminar flow: Can be solved analytically. Turbulent flow: Rely heavily on semi-empirical empirical theories and experimental data. Internal flow 的基本原理與截面形狀無如此分類的意義? 務實, 便於分析分類的依據, 指標? 如何界分? 3

General Characteristics of Pipe Flow 先談談管流的一般特徵 4

Pipe System 號稱管系統, 含括? 管系 : 管本身與管配件 A A pipe system include the pipes themselves (perhaps of more than one diameter), the various fittings, the flowrate control devices valves, and the pumps or turbines. 5

Pipe Flow vs. Open Channel Flow 相對 pipe flow, 有所謂 open-channel flow Pipe flow: Flows completely filling the pipe. (a) The pressure gradient along the pipe is main driving force. Open channel flow: Flows without completely filling the pipe. (b) The gravity alone is the driving force. 沒有填滿管子的管流就非管流 Open channel flow Pipe flow 流體一部分與大氣接觸 6

Laminar or Turbulent Flow 1/ 管流依? 標準如此分類 The flow of a fluid in a pipe may be Laminar? Turbulent? 會有後續的分類, 來自 Reynolds Osborne Reynolds,, a British scientist and mathematician, was the first to distinguish the difference between these classification of flow by using a simple apparatus as shown. Reynolds 先進行觀察調節流率大小, 觀察染料的崩裂情形 7

Laminar or Turbulent Flow / 當 flowrate 很低中度到很大, 染料的崩裂情況不同! For small small enough flowrate For flowrate the dye streak will remain as a well-defined line as it flows along, with only slight blurring due to molecular diffusion of the dye into the surrounding water. intermediate flowrate For a somewhat larger intermediate flowrate the dye fluctuates in time and space, and intermittent bursts of irregular behavior appear along the streak. For large large enough flowrate For flowrate the dye streak almost immediately become blurred and spreads across the entire pipe in a random fashion. 染料分子擴散到附近, 並出現輕微的模糊現象染料隨時空波動, 沿著 streak 出現間歇性的不規則的崩裂染料開始 Random 的崩裂, 並逐漸擴散充滿整個管子 8

Time Dependence of Fluid Velocity at a Point 在某一位置記錄速度變動流率 9

Indication of 剛剛提到分類的標準? 指標? Laminar or Turbulent Flow The term flowrate should be replaced by Reynolds number, R e ρvd / μ,where V is the average velocity in the pipe. 有共識的指標 It is not only the fluid velocity that determines the character of the flow its density, viscosity, and the pipe size are of equal importance. 該指標內涵不只是速度而已, 還包括流體的黏度 For general engineering purpose,, the flow in a round pipe Laminar R e < 100 Transitional 一般工程應用的角度 Turbulent R e >4000 10

Reynolds Number 1/ Re ρvl Vl μ υ In honor of Osborne Reynolds (184~191), the British engineer who first demonstrated that this combination of variables could be used as a criterion to distinguish between laminar and turbulent flow. The Reynolds number is a measure of the ration of the inertia forces to viscous forces. If the Reynolds number is small (Re<<1), this is an indication that t the viscous forces are dominant in the problem, and it may be possible to neglect the inertial effects; that is, the density of the fluid will no be an important variable. 物理意義紀念 Osborne Reynolds 是一個衡量 inertial force / viscous force 的指標當 Re << 1 表示 viscous forces 為主,inertial force 重要性低, 可以忽略, 流體的密度不是一個至重要的變數 11

Reynolds Number / 是一個衡量 inertial force / viscous force 的指標 Flows with very small Reynolds numbers are commonly referred to as creeping flows. Reynolds number 很低的 flow 稱為 creeping flows For large Reynolds number flow, the viscous effects are small relative to inertial effects and for these cases it may be possible to neglect the effect of viscosity and consider the problem as one involving a nonviscous fluid. Flows with large Reynolds number generally are turbulent. Flows in which the inertia forces are small compared with the viscous forces are characteristically laminar flows. Re>>1, 表示 inertial force 為主,viscous force 重要性很低,viscous effect 可以忽略, 此種流體可考慮被歸類為無黏性流體 Reynolds number 大的流體一般為 turbulent flow, inertial force<viscous force 則為 laminar flow 的主要特徵之一 1

Example 8.1 Laminar or Turbulent Flow Water at a temperature of 50 flows through a pipe of diameter D 0.73 in. (a) Determine the minimum time taken to fill a 10-oz glass (volume 0.15ft3) with water if the flow in the pipe is to be laminar. (b) Determine the maximum time taken to fill the glass if the flow is to be turbulent. Repeat the calculation if the water temperature is 140. 13

Example 8.1 Solution If the flow in the pipe is to maintain laminar, the minimum time to fill the glass will occur if the Reynolds number is the maximum allowed for laminar flow, typically Re100. Thus V 100 μ / ρd 0.486 ft / s t V V... Q ( ρπ / 4)D V 8.85s 14

流體在管內的發展歷程? 從進口處開始 How flowing fluid developed within pipe 15

Entrance Region and Fully Developed Flow 1/5 總要有.. 開始 Any fluid flowing in a pipe had to enter the pipe at some location. 由進口端開始說明管流的發展 The region of flow near where the fluid enters the pipe is termed the entrance region. 稱為進口區 16

Entrance Region and Fully Developed Flow /5 The fluid typically enters the pipe with a nearly uniform velocity profile at section (1). The region of flow near where the fluid enters the pipe is termed the entrance region. As the fluid moves through the pipe, viscous effects cause it to stick to the pipe wall (the no slip boundary condition). 一開始幾乎是 uniform flow, 不受管的影響, 但由於 No slip boundary condition 的存在, 管壁開始影響, 即 viscous effect 開始浮現在進口處幾乎可以看成 uniform flows, 但稍微前進之後 17

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, section (), beyond which the velocity profile does not vary with x. The boundary layer has grown in thickness to completely fill the pipe.??? 有時候邊界層不見得厚到填滿整個 pipe! 受到 viscous effect 影響的範圍稱為 boundary layer, 在層內, 流體速度由管壁的 ZERO(no slip condition) 向管中心增加 NOTE: velocity profile 順著管流方向改變, 到某一位置 ( 進口區結束 ) 就維持穩定不再改變 : 換言之,Boundary layer 厚度一直有變化, 到某一位置 ( 進口區結束 ) 就不再改變 18

前一頁提到 BL 不見得填滿整個 PIPE.. 依流況而定回頭說到進口區有多長?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 of the velocity profile in the pipe depends on whether the flow is laminar or turbulent, as does the length of the entrance region, l l. 進口區長度? For laminar flow l l l 6 l D 0.06R e Dimensionless entrance length For turbulent flow 1/ D 4.4R e BL 內外,viscous, effect 重要程度不同! 19

Entrance Region and Fully Developed Flow 5/5 發展過程結束,velocity profiles 不再改變! Once the fluid reaches the end of the entrance region, section (), 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. The flow between () and (3) is termed fully developed. 進口區之後稱為完全發展區, 在完全發展區內 velocity profile 已經定型此階段的 velocity profiles 只是 function of r 0

從 pressure distribution 來看管流的發展過程 1

Pressure Distribution along Pipe In the entrance region of a pipe, the fluid accelerates or decelerates as it flows. There is a balance between pressure, viscous, and inertia (acceleration) force. 參與角力者眾多從壓力降的觀點來看管流變化 The magnitude of the pressure gradient is constant. The magnitude of the pressure gradient is larger than that in the fully developed region. p x 還在變動中.. p x Δp < 0 l 已經固定此區壓力降梯度大於完全發展區

局限於完全發展區且區內為 Laminar flow 先討論流況被歸類為 Laminar flow 者 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-Stokes equations of motion From dimensional analysis methods 不同的切入方法求解管流的 velocity distribution 3

F ma 切入先討論 Force balance( ( 無關流況 )- 剪力壓力與重力 ma 再面對不能迴避的問題 : 剪應力與速度的關係? Laminar flow 或 Turbulent flow,, 其剪應力與速度關係不同! Laminar flow 者, 剪應力與速度關係比較簡單, 這也就是何以先集中火力討論 Laminar flow 可以想像者,Turbulent, flow 比較複雜! 後頭再討論 4

From Fma 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 pipe of diameter D. 在管流中選一個 fluid element 先忽略重力項 5

From Fma /8 因為 velocity 非 uniform, 因此從 t 發展到 tδt,fluid element 移動新位置時形狀也有改變 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 tδt 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 r r r u 0 Fully developed V V u i 0 t x 因為在完全發展區,fluid element 兩端的形狀改變是相同的, 且在區內的加速度為零 6

From Fma 3/8 左邊與 r 無關 Apply the Newton s s second Law to the cylinder of fluid F x ma x The force balance p πr 1 Δp l Δp τ ( p Δp) πr τl( πr) 0 l r Basic balance in forces needed to drive each fluid particle along the pipe with constant velocity Not function of r τ r Not function of r 力已平衡且加速度為 0 Independent of r, 要如何才能做到? τ? τ Cr B.C. r0 τ0 rd/ τ τw 力平衡維持等速度移動由 B.C. 求常數 τ 先忽略重力項 τ r w D 7

4/8 From Fma 4/8 物理意義 : 在完全發展區內壓力降與管壁剪應力平衡 The pressure drop and wall shear stress are related by τ w r Δp τ τ 4lτw Δp 注意沒有放入 D l r D 因為水平擺放 Valid for both laminar and turbulent flow. 注意沒有放入重力重力到目前為止, 無關流體為 Laminar 或 Turbulent flow 開始假設是 Laminar flow Turbulent flow? 剪應力關係不單純! 後頭再討論 Laminar du τ μ dr 8

9 From Fma From Fma 5/8 5/8 Since Since With the boundary conditions: u0 at rd/ With the boundary conditions: u0 at rd/ dr du μ τ 1 C r 4 p u rdr p du r p dr du μ Δ μ Δ μ Δ l l l μl Δ 16 pd C 1 μ τ μ Δ w C R r 1 4 D u(r) D r 1 V D r 1 16 pd u(r) l Velocity distribution Velocity distribution D 4 p w τ Δ l 切記切記 : 因為有假設 : 因為有假設 Laminar flow Laminar flow 因為存在這個關係讓後續變得可為, 若是 Turbulent flow, 就沒這麼簡單目標達成

From Fma 6/8 The shear stress distribution τ μ du dr rδp l Volume flowrate r R Q u da u(r)πrdr Q A 4 πd Δp 18μl 0... 4 πr V Poiseuille s Law Valid for Laminar flow only C 30

From Fma 7/8 Average velocity V average Q A Q πr ΔpD 3μl Point of maximum velocity du 0 at dr r0 R Δp u u max U 4μl V average 31

From Fma 8/8 當管流不是水平時 Δp Δp γlsin θ Making adjustment to account for nonhorizontal pipes Δp Δp γlsin Δp γlsinθ l V average θ τ r ( Δp γlsin θ) 3μl D 重力項考慮進來 θ>0 if the flow is uphill θ<0 if the flow is downhill Q π ( Δp γlsin θ) 18μl D 4 3

Example 8. Laminar Pipe Flow An oil with a viscosity of μ 0.40 N s/mn and density ρ 900 kg/m 3 flows in a pipe of diameter D 0.0m. (a) What pressure drop, p 1 -p, is needed to produce a flowrate of Q.0 10-5 m 3 /s if the pipe is horizontal with x 1 0 and x 10 m? (b) How steep a hill, θ,must the pipe be on if the oil is to flow through the pipe at the t same rate as in part (a), but with p 1 p? (c) For the conditions of part (b), if p 1 00 kpa,, what is the pressure at section, x 3 5 m, where x is measured along the pipe? 33

1/ Example 8. Solution 1/ R e ρvd / μ.87 < 100 Q V A 0.0637 m / s The flow is laminar flow 18μlQ Δp p1 p... 4 πd 0.4kPa If the pipe is on the hill of angle θ with Δp0 18μlQ sin θ... θ 13. 34 4 πρgd 34

/ Example 8. Solution / With p 1 p the length of the pipe, l,, does not appear in the flowrate equation Δp0 for all l p p p3 1 00 kpa 35

從 Navier Stokes equation 切入透過合理假設, 簡化 36

From the Navier-Stokes Equations 1/3 General motion of an incompressible Newtonian fluid is governed by the continuity equation and the momentum equation V r 不可壓縮的牛頓流體簡化 Navier-Stokes equation r 0 r g gk r V t Steady flow r r V V p ρ r g ν 加速度為 0 在完全發展區內 r For steady, fully developed flow in a pipe, the velocity contains only an axial component, r rwhich is a function of only the radial coordinate V u(r) i V 由簡化的 momentum equation 與連續方程式來描述速度僅與 r 有關 37

Equation of Motion chapter 6 非線性方程式 δ F ρg ρg ρg x y z x δma σ x τ x xx xy τ x xz x τ σ y τ y y τ z τ z σ z v ρ t w ρ t These are the differential equations of motion for any fluid satisfying the continuum assumption. How to solve u,v,w? yx yy yz δf y δma zx zy zz y u ρ t δf z δma z u u u u v w x y z v v v u v w x y z w w w u v w x y z 微分型式的運動方程式 38

Stress-Deformation chapter 6 The stresses must be expressed in terms of the velocity and pressure field. Cartesian coordinates σ σ σ τ τ xx yy zz xy xz r u p μ V μ 3 x r v p μ V μ 3 y r w p μ V μ 3 z v u τ yx μ x y w u τzx μ x z τ yz τ zy μ w y v z 39

40 The The Navier Navier-Stokes Equations Stokes Equations chapter 6 chapter 6 Under Under incompressible flow with constant viscosity incompressible flow with constant viscosity conditions conditions, the the Navier Navier-Stokes equations are reduced to: Stokes equations are reduced to: μ ρ ρ μ ρ ρ μ ρ ρ z y x z w y w x w g z p z w w y w v x w u t w z v y v x v g y p z v w y v v x v u t v z u y u x u g x p z u w y u v x u u t u 再來一步步透過假設, 簡化 Navier-Stokes equations 假設不可壓縮且黏度是 constant

From the Navier-Stokes Equations /3 Simplify the Navier-Stokes equation V r 0 r p ρgk μ 連續方程式 r V 簡化後的 Navier-Stokes equation The flow is governed by a balance of pressure, weight, and viscous forces in the flow direction. 顯示在完全發展區內 pressure weight 與 viscous force 形成平衡 41

From the Navier-Stokes Equations 3/3 r V r u(r)i p x ρgsin θ μ 1 r r r u r Function of, at most, only x Function of,at most, only r p x const. p x Δp l 積分邊界條件 Integrating Velocity profile u(r) B.C. (1) r R, u 0 ; () r 0, u < or r 0 u/ u/ r0 4

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, l,, the pipe diameter, D, and the viscosity of the fluid, μ. Chapter 7 Δp F(V, l, D, μ) Dimensional analysis DΔp l φ an unknown function of the length to μv D diameter ratio of the pipe. 函數關係?UNKNOWN 43

From Dimensional Analysis /3 DΔp μv Cl D Δp l where C is a constant. CμV D Q AV ( π/ 4C) ΔpD μl The value of C must be determined by theory or experiment. For a round pipe, C3. For duct of other cross-sectional sectional shapes, the value of C is different. For a round pipe 假設是一個線性關係 Δp 對圓型管而言,C 3 3μlV D V 4 常數 C 可由理論或實驗來推算 average Q A Q πr ΔpD 3μl 44

3/3 From Dimensional Analysis 3/3 Δp 3μlV D For a round pipe Δp f l D D Δp l ρv f 1 Δp ρv ρv 3μlV / D ρv 1 μ l 64 ρvd D 64 Re f is termed the friction factor, or sometimes the Darcy friction factor. For laminar flow 物理意義 : 在完全發展區內壓力降與管壁剪應力平衡 f 64 Re 8τ ρv Δp w l D 4lτ D w 45

Example 8.3 Laminar Pipe Flow Properties 1/ 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 (). It is proposed that QKΔp, where the calibration constant, K, is a function of temperature, T, because e of the variation of the syrup s s viscosity and density with temperature. These variations are given in Table E8.3. (a) Plot K(T) versus T for 60 F T 160 F. (b) Determine the wall shear stress and the pressure drop, Δpp 1 -p, for Q0.5 ft 3 /s and T100 F. (c) For the conditions of part (b), determine the nest pressure force.(πd /4)Δp, and the nest shear force, πdlτ w, on the fluid within the pipe between the sections (1) and (). 46

Example 8.3 Laminar Pipe Flow Properties 1/ 47

1/ Example 8.3 Solution 1/ If the flow is laminar Q πδpd 18μl 4 KΔp 1.60 10 K μ For T100 F, μ3.8 10-3 lb s/ft, Q0.5ft 3 /s 18μlQ Δp... 119lb / ft 4 πd Q V... 10.ft /s A Δ 4lτ D ΔpD 4l w p τw... 5 R e ρvd / μ... 1380 < 1.4lb / ft 100 48

/ Example 8.3 Solution / The new pressure force and viscous force on the fluid within the pipe between sections (1) and () is F F p v πd Δp 4 D π lτ... 5.84lb w... 5.84lb The values of these two forces are the same. The net force is zero; there is no acceleration. 49

當管流是 Turbulent flow 50

Fully Developed Turbulent Flow 實務上, 管流為 turbulent pipe flow 的機率遠大於 Laminar pipe 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. 即便有很多投入, 但對於 turbulent pipe flow 的了解還是有限 Much remains to be learned about the nature of turbulent flow. Turbulent flow 的特性還是最吸引人去關注的議題紊流, 一個複雜的過程 51

Transition from Laminar to Turbulent Flow in a Pipe 1/ 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. 之前所使用的指標, 小於某一值歸屬 Laminar flow, 大於某一值歸屬 Turbulent flow The important parameters involved and their critical values depend on the specific flow situation involved. 因 flow situation 不同 For flow in pipe : 100<Re..Re>4000 For flow along a plate Re x ~5000 討論 Laminar turbulent Consider a long section of pipe that is 為了了解 transition initially filled with a fluid at rest. 管內先塞滿流體 turbulent flow 的過程似乎很難一刀畫下, 存在一個過渡的階段 5

Transition from Laminar to Turbulent Flow in a Pipe / 閥打開, 管內流體開始流動,Re 由 0 增加 As the valve is opened to start the flow, the flow velocity and, hence, the Reynolds number increase from zero (no flow) to their maximum steady flow values. For the initial time period the Reynolds number is small enough for laminar flow to occur. 速度出現隨機波動 At some time the Reynolds number reaches 100, and the Re > 100 崩裂 flow begins its transition to turbulent conditions. 速度 Intermittent spots or burst appear.. Re>100 開始由 transition 觀察不同時間的速度變動, 並計算 Re 53

Description for Turbulent Flow 1/4 紊流下, 流體參數也都出現隨機波動的特性在特定點追蹤速度的軸向分量 A typical trace of the axial component of velocity measured at a given location in the flow, uu(t). 注意其中的速度定義 Turbulent flows involve randomly fluctuating parameters. The character of many of the important properties of the flow (pressure drop, heat transfer, etc.) depends strongly on the existence and nature of the turbulent fluctuations or randomness. 流體的參數與紊流的隨機波動特性有很強烈的關聯進入 turbulent status The time-averaged, ū,, and fluctuating, ú description of a parameter for tubular flow. 54

Description for Turbulent Flow /4 Turbulent flows are characterized by random, three- dimensional vorticity. 也可以如此描述 Turbulent flows can be described in terms of their mean values on which are superimposed the fluctuations. 放大描述 turbulent flow 以隨機的, 三維的漩渦來描述紊流 u 其他參數也有這種現象 u' 1 T to T u t u O ( x, y,z,t) u dt u u u' 55

Description for Turbulent Flow 3/4 The time average of the fluctuations is zero. T 1 u' T t t 1 T O ( u u) dt ( Tu Tu) 0 O The square of a fluctuation quantity is positive. (u') 1 t T O t O T ( u' ) dt > 0 Turbulence intensity or the level of the turbulence 紊流強度 I (u') u 1 T to T t O u ( u' ) dt Turbulence intensity 越大, 速度變動越大 The larger the turbulence intensity, the larger the fluctuations of the velocity. Well- designed wind tunnels have typical value of I0.01, although with extreme care, values as low as I0.000 have been obtained. 56

Description for Turbulent Flow 4/4 In some situations, turbulent flow characteristics are advantages. In other situations, laminar flow is desirable. Turbulence: mixing of fluids. Laminar: pressure drop in pipe, aerodynamic drag on airplane. Turbulent flow 有其優點, 當然也有其缺點 ; Laminar flow 亦是有優點, 也有缺點 57

不能迴避的問題, 剪力與速度的關係 1/ Shear Stress for Laminar Flow 1/ Laminar flow 的剪應力 Laminar flow is modeled as fluid particles that flow smoothly along in layers, gliding past the slightly slower or faster ones on either side. The fluid actually consists of numerous molecules darting about in an almost random fashion. The motion is not entirely random a slight bias in one direction. As the molecules dart across a given plane (plane A-A, for example), the ones moving upward have come from an area of smaller average x component of velocity than the ones moving downward, which have come from an area of large velocity. 58

Shear Stress for Laminar Flow / The momentum flux in the x direction across plane A-A give rise to a drag of the lower fluid on the upper fluid and an equal but opposite effect of the upper fluid on the lower fluid. The sluggish molecules moving upward across plane A-A must accelerated by the fluid above this plane. The rate of change of momentum in this process produces a shear force. Similarly, the more energetic molecules moving down across plane A-A must be slowed down by the fluid below that plane. BY combining these effects, we obtain the well-known Newton 如果流況歸類為 Laminar flow, 則 viscosity law du Shear stress is present only if there is a τ yx μ gradient in u u(y). dy 59

Shear Stress for Turbulent Flow 1/ Turbulent flow 的剪應力 The turbulent flow is thought as a series of random, three- dimensional eddy type motions. These eddies range in size from very small diameter to fairly large diameter. This eddy structure greatly promotes mixing within the fluid. 一連串 random 3D eddy 型態的運動簡言之, 紊流的剪力與速度關係無法像 Laminar flow 一樣 60

Shear Stress for Turbulent Flow / The flow is represented by u (time-mean mean velocity ) plus u u and v v (time randomly fluctuating velocity components in the x and y direction). The shear stress on the plane A-AA τ μ ρu' du dy v' ρu' v' 在 wall 附近 ρu' v' 0 τ la min ar τ turbulent The shear stress is not merely proportional to the gradient of the time-averaged velocity, u(y). 剪力非與速度梯度成正比關係 is called Reynolds stress introduced by Osborne Reynolds. 稱為 Reynolds stress As we approach wall, and is zero at the wall (the wall tends to suppress the fluctuations.) 不像 Laminar flow 的剪應力與速度梯度存在簡單的關係 61

怎麼辦? 分靠近牆壁與遠離牆壁者兩區 Structure of Turbulent Flow in a Pipe 1/ 管內紊流結構 Near the wall (the viscous sublayer), the laminar shear stress τ lam is dominant. 管壁附近 laminar shear stress 為主 Away from the wall (in the outer layer), the turbulent shear stress τ turb is dominant. The transition between these two regions occurs in the overlap layer. 近管壁處 Laminar flow 主導 ; 近管心處 Turbulent flow 主導剪力比 transition 越靠近管中心 τ turb > τ lam 速度曲線 6

Structure of Turbulent Flow in a Pipe / The relative magnitude of τ lam compared to τ turb is a complex function dependent on the specific flow involved. Typically the value of τ turb is 100 to 1000 times greater than τ lam in the outer region. 兩者間的相對大小, 依流況而定, 一般是 100~1000 倍 63

Alternative Form of Shear Stress 1/ 另一種表達 shear stress 的方式 τ turb turb : requiring an accurate knowledge of the fluctuations u and v, v, or ρu' v' The shear stress for turbulent flow is given in terms of the eddy viscosity η. τ turb η du dy 強調 : 只是另一種表達方式而已 This extension of of laminar flow terminology was introduced by J. Boussubesq,, a French scientist, in 1877. η? A semiempirical theory was proposed by L. Prandtl to determine the value of η 64

Alternative Form of Shear Stress / η ρl m du dy 不管那一種方式, 都一樣不簡單 du τturb ρl m dy mixing length, is not constant throughout the flow field. There is no general, all-encompassing, useful model that can accurately predict the shear stress throughout a general incompressible, viscous turbulent flow. 沒通用, 可全面涵蓋的 model,, 可以精確預估 viscous turbulent flow 的 shear stress 65

Turbulent Velocity Profile 1/5 Shear stress 與 velocity 關係不簡單之下, 還是得面對如何寫出 velocity profiles 前面已講過完全發展管流內分成三區 Fully developed turbulent flow in a pipe can be broken into three region: the viscous sublayer,, the overlap region, and the outer turbulent sublayer. 說明不同 sublayer, 主導之 stress 不同 Within the viscous sublayer the shear stress is dominant compared with the turbulent stress, and the random, eddying nature of the flow is essentially absent. In the outer turbulent layer the Reynolds stress is dominant,, and there is considerable mixing and randomness to the flow. Within the viscous sublayer the fluid viscosity is an important parameter; the density is unimportant. In the outer layer the opposite posite is true. 在 viscous sublayer 中黏度重要密度不重要因為沒有一個簡單的 shear stress vs. velocity gradient, 所以 66

式取自半經驗公式Turbulent Velocity Profile /5 探討 velocity profiles 的方法 : 因次分析與半經驗公式 Considerable information concerning turbulent velocity profiles has been obtained through the use of dimensional analysis, and semi- empirical theoretical efforts. 不同 layer 有不同的公式 In the viscous sublayer the velocity profile can be written in dimensionless form as 管壁附近 * u yu u y * u ν Law of the wall Where y is the distance measured from the wall yr-r. u * ( τ / ρ) 1/ w Kinematic viscosity is called the friction velocity. * Is valid very near the smooth wall, for yu 0 5 ν 67

Turbulent Velocity Profile 3/5 In the outer region the velocity should vary as the In the logarithm of y u u.5 ln yu y 5.0 yu * for > 30 ν In transition region or buffer layer U u u.5 ln R y 管中心附近 Determined experimentally for 5-7 yu ν * 30 68

Turbulent Velocity Profile 4/5 u u * yu ν * 管中心 u u.5ln yu y 5.0 靠近管壁 69

5/5 Turbulent Velocity Profile 5/5 另一種公式 The velocity profile for turbulent flow through a smooth pipe may also be approximated by the empirical power-law equation u U y R 1/ n 1 r R 1/ n Where the exponent, n, varies with the Reynolds number. The power-law profile is not applicable close to the wall. 可其限制 - 近管壁處不管用 70

Example 8.4 Turbulent Pipe Flow Properties Water at 0 (ρ998kg/m 3 and ν1.004 1010-6 m /s) flows through a horizontal pipe of 0.1-m m diameter with a flowrate of Q4 10 - m 3 /s and a pressure gradient of.59 kpa/m.. (a) Determine the approximate thickness of the viscous sublayer.. (b) Determine the approximate centerline velocity, V c. (c) Determine the ration of the turbulent to laminar shear stress, τ turb /τ lam at a point midway between the centerline and the pipe wall (i.e., at r0.05m) 71

1/3 Example 8.4 Solution 1/3 The thickness of viscous sublayer, δ s, is approximately δ ν u * s DΔp 4l 5 s 5 ν τw... 64.8N / m δ * * u u ( τ / ) 1/ w ρ... 0.55m / s ν 5 δs 5... 1.97 10 m 0.0 mm * u The centerline velocity can be obtained from the average velocity y and the assumption of a power-law velocity profile Q A 3 0.04m π(0.1m) V / s / 4 5.09m / s R e VD / ν... 5.07 10 5 7

/3 Example 8.4 Solution /3 Q u U AV y R 1/ n uda 1... r R 1/ n πr V c (n n 1)(n 1) πr V V V c (n n 1)(n 1) V c... 6.04m / s τ τ w r D τwr τ D τ lam Valid for laminar or turbulent flow (64.8N / m )(0.05m) (0.1m) τ turb 3.4N / m n8.4 R e VD / υ... 5.07 10 73 5

3/3 Example 8.4 Solution 3/3 τ τwr D (64.8N / m )(0.05m) (0.1m) τ lam τ turb 3.4N / m τ lam μ du dr μ Vc nr 1 r R (1 n) / n 0.066N / m τ τ lam 3.4 0.066 0.066 turb lam lam τ τ τ 10 74

Dimensional Analysis of Pipe Flow 75

Energy Considerations 1/8 Page 1/8~7/8 出現在 Chapter5 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& t net in Q& CV net in W& Shaft W& eρdv Shaft in CS in (û CS Energy equation e u σ nn t p ρ V r r V nda CV V eρdv gz t CS CV r eρv r r r gz) ρv nda eρdv r nda Q& net CS in r eρv CS σ W& r nda nn Shaft r r V nda Work done by normal stresses at the CS in 76

Rate of Work done by CV W & W& W& W& Shaft Shaft work W & Shaft : the rate of work transferred into through the CS by the shaft work ( negative for work transferred out, positive for work input required) 藉由 shaft 傳遞的功 Work done by normal stresses at the CS: W& normal r δf normal Work done by shear stresses at the CS: W& shear Other work t v V r r τv nda CS cv eρdv normal CS CS σ nn r r V nda r r eρv nda shear Q& net CS Negligibly small in W& r r pv nda W& other 輸入系統者,- 輸出系統者 shaft net in CS r r pv nda 77

Energy Considerations /8 When the flow is steady The integral of t CVeρdV 0 CS û p ρ V r r gz ρ V nda??? CS p û ρ Uniformly distribution V r r gz ρ V nda 進一步假設 out p û ρ V gz m& p V r r CS û gz V nda Only one stream ρ ρ entering and leaving p V p û gz m& out û Special & simple case ρ out ρ 單進單出 in p û ρ V V gz in gz m& m& in 78

Energy Considerations 3/8 If shaft work is involved. m& û Q& out net in Enthalpy û in W& ĥ shaft û p ρ net p ρ out in 當 shaft work 包括進來 p ρ in V out V Vout Vin m& ĥout ĥin g & in g ( z z ) out One-dimensional energy equation for steady-in in-the-mean flow 單維能量方程式 The energy equation is written in terms of enthalpy. ( z ) out zin Qnet / in Wshaft net / in & in 79

4/8 Energy Considerations 4/8 p p Vout V in m & ûout ûin g ρ ρ out For steady, incompressible flow One One-dimensional energy equation m p ρ out in out in & ûout ûin g m& p out ρv out p ρ out where γz p ρ V out out q V gz net ( zout zin ) Q& net in W& shaft net in p in in out V Q& p ρ net ρv in in û out û in q net in in 0 ( zout zin ) Q& net in V in / m& γz in gz For steady, incompressible, frictionless flow in ( û û q ) out V in net Bernoulli equation Frictionless flow e 沒有摩擦損失 û gz 沒有 shaft work 有 normal stress 做的功之前得到的 Bernoulli eq. in 80

Energy Considerations 5/8 For steady, incompressible, frictional flow û out û in q net in > Defining useful or available energy 0 Frictional flow Defining loss of useful or available energy û out û in q net in p ρ 考量能量損失 V gz Loss 發生在 in out 過程中 loss p ρ out V out 下游 gz out 因為 1 有摩擦損失, p V gz ρ p ρ in V in 上游 gz in loss 1 p1 V gz ρ 自然而然就低於 1 81

Energy Considerations 6/8 For steady, incompressible flow with friction and shaft work m p ρ p ρ V V out in out in & ûout ûin g m& p ρ out p ρ out V out p γ V out gz V g gz out out p ρ in p ρ in V V in ( zout zin ) Q& net in W& shatf net in in gz gz V g in in w w shaft shaft g out out in in zout zin hs hl Shaft head h S w shaft g net / in W& shaft mg & p γ net / in W& shaft γq net / in 有摩擦損失有軸功進來在 in out 注入 net net in in (û out loss û 在 in out 注入 Head loss h L in loss g q net in 8 )

Energy Considerations 7/8 p γ out For turbine h s h T (h T > 0) For pump h s h P hp is pump head The actual head drop across the turbine h T (h s h L ) T The actual head drop across the pump h V g out z out p γ in (h h ) p s L p Vin g z in h s h L in out 輸出想像 : 讓 loss 擴大想像 : 讓 loss 減緩 ht is turbine head in out 輸入 83

總結 1/8~7/8 p γ V p V z 1 1 z h h 1 s L g γ g Pipe system 內 LOCATION 1 LOCATION h h L h L major L min or 84

Energy Considerations 8/8 Total head loss, h L, is regarded as the sum of major losses, h L major,, due to frictional effects in fully developed flow in constant area tubes, and minor losses, h L minor,, resulting from entrance, fitting, area changes, and so on. Head loss 可以分成 major loss 與 minor loss h h L h L major L min or 85

Major Losses: Friction Factor The energy equation for steady and incompressible flow with zero shaft work p ρg >>> 1 α1v1 p α V z 1 z h L p g p ρg ρg 1 (z z h 1) L g For fully developed flow through a constant area pipe 簡化一 For horizontal pipe, z z z1z 簡化二 >>> p p ρg α V g 1 1 α Δp ρg 1 h L V g 86

From Fma 6/8 Q 4 πd Δp 18 μl Poiseuille s Law Valid for Laminar flow only V average Q A Q πr ΔpD 3μl 87

Major Losses: Laminar Flow 先探討 Laminar flow 的 major loss In fully developed laminar flow in a horizontal pipe, the pressure drop 水平完全發展的 laminar flow l ρ Δp f 18μlQ 18μlV( πd / 4) l μv Δp Δp 1 ρv Δp f 4 πd μ 64 ρvd l ρv D >> l D h L πd 64 Re 4 l D l μv 3 D ρd 3 l D D V D μ 64 ρvd Friction Factor f Δp(D / l) /( ρv / ) 壓力降可以解出, 然 turbulent flow 可就不是那麼容易 QV AV AV πd /4 64 R e f l D V Δp ρg h L f la min ar D h L l D V V g 64 Re 88

Major Losses: Turbulent Flow 1/3 再探討 Turbulent flow 的 major loss In turbulent flow we cannot evaluate the pressure drop analytically; lly; we must resort to experimental results and use dimensional analysis to correlate the experimental data. In fully developed turbulent flow the pressure drop, p, caused by friction in a horizontal constant-area area pipe is known to depend on pipe diameter,d, pipe length, l,, pipe roughness,e, 只得訴諸實驗, 訴諸 dimensional analysis average flow velocity, V, fluid densityρ,, and fluid viscosity,μ. ( V, D, l, ε, μ ρ) Δ p F, 列出影響壓力降的因子困難的關鍵在無法解析解出壓力降管壁粗糙度 89

Major Losses: Turbulent Flow /3 Applying dimensional analysis, the result were a correlation of the form Δp ρvd l ε φ,, 諸多驗實驗顯示 : head 1 ρv μ D D loss 與 l/d 成正比來自累積的實驗結果 Experiments show that the nondimensional head loss is directly proportional to l/d. Hence we can write f 1 Δp ρv 透過 dimensional analysis φ Re, l φ Re, D ε D 把 l/d 拿出來是沒有問題的 ε D l ρv Δp f D Darcy-Weisbach equation l V h L major f D g Δp ρg h L Frictional factor? 90

Roughness for Pipes 影響 frictional factor 的因子之一 91

Friction Factor by L. F. Moody Depending on the specific circumstances involved. 9

About Moody Chart 從 Moody chart 看出什麼 For laminar flow, f64/re, which is independent of the relative roughness. 在 Laminar flow 中,f 與粗糙度無關 For very large Reynolds numbers, fφ(ε/d), which is independent of the Reynolds numbers. For flows with very large value of Re,, commonly termed completely turbulent flow (or wholly turbulent flow), the laminar sublayer is so thin (its thickness decrease with increasing Re) that the surface roughness completely dominates the character of the flow near the wall. For flows with moderate value of Re, the friction factor fφ(re, (Re,ε/D). 中度的 Reynolds number 在 Reynolds number 很大時,f 僅與粗糙度有關, 原因邊界層很薄很薄 93

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 1 ε / D.51 range of the Moody chart..0 log f 僅適用於非 Laminar flow 範圍 3.7 Re f Colebrook formula Miler suggests that a single iteration will produce a result within 1 percent if the initial estimate is calculated from f 0 有沒有可以不用 Moody chart 的管道猜 f 的初始值 0.5 log ε / D 3.7 5.74 Re 0.9 94

Example 8.5 Comparison of Laminar or Turbulent pressure Drop Air under standard conditions flows through a 4.0-mm mm-diameter drawn tubing with an average velocity of V 50 m/s. For such conditions the flow would normally be turbulent. However, if precautions are taken to eliminate disturbances to the flow (the entrance to the tube is very smooth, the air is dust free, the tube t does not vibrate, etc.), it may be possible to maintain laminar flow. (a) Determine the pressure drop in a 0.1-m m section of the tube if the flow is laminar. (b) Repeat the calculations if the flow is turbulent. ulent. 95

1/ Example 8.5 Solution 1/ Under standard temperature and pressure conditions Ρ1.3kg/m 3, μ1.79 1010-5 N s/m The Reynolds number R e ρvd / μ... 13,700 Turbulent flow If the flow were laminar f64/re 0.0467 0.0467 Δp f l D 1 ρv... 0.179 kpa 96

/ Example 8.5 Solution / If the flow were turbulent From Moody chart fφ(re, (Re,ε/D) 0.08 Δp f l D 1 ρv... 1.076 kpa 97

Minor Losses 1/5 Pipe system 不是只有直管而已, 其他管元件所導致的損失 Most pipe systems consist of considerably more than straight pipes. These additional components (valves, bends, tees, and the like) add to the overall head loss of the system. Such losses are termed MINOR LOSS. Additional components 所導致的 loss 稱為 MINOR LOSS The flow pattern through a valve 98

Minor Losses /5 The theoretical analysis to predict the details of flow pattern (through these additional components) is not, as yet, possible. The head loss information for essentially all components is given in dimensionless form and based on experimental data. The most common method used to determine these head losses or pressure drops is to specify the loss coefficient, K L 理論預測流體在元件內的流況根本做不到以無因次化型式呈現 head loss information 資料來自實驗最常用的方法連結 Loss coefficient 與 head loss 或壓力降 99

Minor Losses 3/5 K 如何連結? L h V L min or / g Δp 1 ρv Δp Minor losses are sometimes given in terms of an equivalent length l eq K L 1 ρv 把 minor loss 視同另類 major loss h l L eq min or K L K L D f V g f l D eq V g 相當於多長的管損失 The actual value of K L is strongly dependent on the geometry of the component considered. It may also dependent on the fluid properties. That is K φ(geometry, L Re) 100

Minor Losses 4/5 流體在元件內的 Reynolds number 很大, 主導力來自 inertial effect,viscous effect 可略過 For many practical applications the Reynolds number is large enough so that the flow through the component is dominated by inertial effects, with viscous effects being of secondary importance. Viscous effect 相對不重要 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. This is the reason why the friction factor for very large Reynolds number, fully developed pipe flow is independent of the Reynolds number. Minor loss Friction factor 與 Re 無關壓力降及 head loss 與 dynamic pressure 有關 101

Minor Losses 5/5 This is true for flow through pipe components. Thus, in most cases of practical interest the loss coefficients for components are a function of geometry only, K φ(geometry) L 就像 major loss 在 Re 很大時,frictional, factor 僅與 roughness 有關 10

Minor Losses Coefficient Entrance flow 1/3 Entrance flow condition and loss coefficient (a)) Reentrant, K L 0.8 (b)) sharp-edged, K L 0.5 (c)) slightly rounded, K L 0. (d)) well-rounded, K L 0.04 依進口條件不同而異 K L function of rounding of the inlet edge. 103

Minor Losses Coefficient Entrance flow /3 轉角處出現 vena contract A A vena contracta region may result because the fluid cannot turn a sharp right-angle corner. The flow is said to separate from the sharp corner. The maximum velocity velocity at section () is greater than that in the pipe section (3), and the pressure there is lower. () 處速度高於 (3) 處速度, 壓力回升 If this high speed fluid could slow down efficiently, the kinetic energy could be converted into pressure. 無法直角轉彎, 自然由轉角處分離如果高速可以有效率緩下來,kinetic energy 當然可以轉回壓力壓力回不來 104

Minor Losses Coefficient 情況並非如此加速度過程是 OK 的, 但減速過程卻沒有效率 Such is not the case. Although the fluid may be accelerated very efficiently, it is very difficult to slow down (decelerate) the fluid efficiently.(1) () () () (3) (3) The extra kinetic energy of the fluid is partially lost because of viscous dissipation, so that the pressure does not return to the ideal value. Kinetic energy 在 () (3) 部分損失, 導致壓力回不到理想位置 Entrance flow 3/3 減速過程出現來自 viscous dissipation 的 loss Flow pattern and pressure distribution for a sharp-edged entrance 105

Entrance HEAD LOSS 流體的 inertial effects 主要是被流體內部的 shear stress 給損失掉, 少部分是肇因於 wall shear stress 106

Minor Losses Coefficient Exit flow Exit flow condition and loss coefficient (a)) Reentrant, KL 1.0 (b)) sharp-edged, KL 1.0 (c)) slightly rounded, KL 1.0 (d)) well-rounded, KL 1.0 107

Minor Losses Coefficient varied diameter 查表找 K L Loss coefficient for sudden contraction, expansion,typical conical diffuser. K L 1 A A 1 108

Minor Losses Coefficient Bend Character of the flow in bend and the associated loss coefficient. 彎管的 K L Carefully designed guide vanes help direct the flow with less unwanted swirl and disturbances. 109

Internal Structure of Valves Valves 內部構造 (a)) globe valve (b)) gate valve (c)) swing check valve (d)) stop check valve 110

Loss Coefficients for Pipe Components 常見 pipe components 的 K L 111

Example 8.6 Minor Loss 1/ Air at standard conditions is to flow through the test section [between sections (5) and (6)] of the closed-circuit circuit wind tunnel shown if Figure E8.6 with a velocity of 00 ft/s. The flow is driven by a fan that essentially increase the static pressure by the amount p 1 -p 9 that is needed to overcome the head losses experienced by the fluid as it flows around the circuit. Estimate the value of p 1 -p 9 and the horsepower supplied to the fluid by the fan. 11

Example 8.6 Minor Loss / 113

1/3 Example 8.6 Solution 1/3 The maximum velocity within the wind tunnel occurs in the test section (smallest area). Thus, the maximum Mach number of the flow is Ma 5 V 5 /c 5 p 1 V1 p9 V9 z1 z9 hl1 9 γ V5 00ft /s c5 (KRT5 ) g γ g 1/ 1117ft /s The energy equation between points (1) and (9) p 1 p9 hl1 9 The total head loss from (1) to (9). γ γ 114

/3 Example 8.6 Solution /3 The energy across the fan, from (9) to (1) p P 9 V9 p1 V z 1 9 hp z1 γ a g γqh h p p p γ γa 1 5 V h 5 p γ 9 p γ h γa L1 9 5 g V 5 h L1 9 H p is the actual head rise supplied by the pump (fan) to the air. The actual power supplied to the air (horsepower, P a ) is obtained from the fan head by 115

3/3 Example 8.6 Solution 3/3 The total head loss h h h h h h h L1 9 L corner 7 L corner 8 L corner L corner 3 L dif L noz h L scr h K L L corner noz K 0. L V g V 0. g K L scr 4.0 h L dif K L dif V g 0.6 V g p P 1 a p 9... γh L1 9 (0.765lb / ft 34300ft lb / s 6.3hp )(560ft)... 0.98psi 116

Noncircular Ducts 1/4 The empirical correlations for pipe flow may be used for computations involving noncircular ducts, provided their cross sections are not too exaggerated. The correlation for turbulent pipe flow are extended for use with noncircular geometries by introducing the hydraulic diameter,, defined as A 為截面積 4A D h P Where A is cross- sectional area, and P is wetted perimeter. 非圓管, 視同圓管 D D h 被流體潤濕的長度要像圓才說得通, 不可以將太誇張的截面硬納入! P: 與流體接觸的周長度 117

Noncircular Ducts /4 For a circular duct 4A D h P D For a rectangular duct of width b and height h 4A 4bh h D h ar P (b h) 1 ar h / b 適用範圍 ¼<ar<4 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. 非範圍內者若硬套, 可稱之為太誇張! 118

Noncircular Ducts 3/4 The friction factor can be written as f C/Re h, where the constant C depends on the particular shape of the duct, and Re h is the Reynolds number based on the hydraulic diameter. 利用 Hydraulic diameter 計算 Reynolds number The hydraulic diameter is also used in the definition of the friction factor, hl f ( l / Dh)(V / g), and the relative roughness ε/d/d h. 利用 hydraulic diameter 來定義 friction factor h L major f l D V g 119

Noncircular Ducts 4/4 For Laminar flow, the value of C f Re h have been obtained from theory and/or experiment for various shapes. For turbulent flow in ducts of noncircular cross section, calculations are carried out by using the Moody chart data for round pipes with the diameter replaced by the hydraulic diameter and the Reynolds number based on the hydraulic diameter. 利用 Moody chart 處理非圓管 The Moody chart, developed for round pipes, can also be used for noncircular ducts. 10

Friction Factor for Laminar Flow in Noncircular Ducts f C/Re h C? 11

Example 8.7 Noncircular Duct Air at temperature of 10 F F and standard pressure flows from a furnace through an 8-in. 8 in.-diameter pipe with an average velocity of 10ft/s. It then passes through a transition section and into a square s duct whose side is of length a. The pipe and duct surfaces are smooth (ε0).( 0). Determine the duct size, a, if the head loss per foot is to be the same for the pipe and the duct. 1

1/3 Example 8.7 Solution 1/3 The head loss per foot for the pipe h L l f D V g For given pressure and temperature ν1.89 1010-4 ft /s VD Re ν 35300 For the square duct h l f D Vs g L h 0.051 4A Q D h a Vs P A 3.49 a 13

/3 Example 8.7 Solution /3 The Reynolds number based on the hydraulic diameter Re h h l L Vs D ν h f D h Vs g (3.49 / a )a 4 1.89 10 0.051 f a 1.89 10 a (3.49 / a ) (3.) Have three unknown (a,f( a,f,, and Re h ) and three equation Eqs.. 1,, and either in graphical form the Moody chart or the Colebrook equation 4 () a Find a 1.30f 1/5 (1) 14

3/3 Example 8.7 Solution 3/3 Use the Moody chart Assume the friction factor for the duct is the same as for the pipe. p That is, assume f0.0. From Eq. 1 we obtain a0.606 ft. From Eq. we have Re h 3.05 10 4 From Moody chart we find f0.03, which does not quite agree the assumed value of f. Try again, using the latest calculated value of f0.03 as our guess. g The final result is f0.03, Re h 3.05 10 4, and a0.611ft. 15

Pipe Flow Examples 1/ The energy equation, relating the conditions at any two points 1 and for a single-path pipe system p α α 1V1 p V 1 z1 z h L h ρg g ρg g h by judicious choice of points 1 and we can analyze not only the entire pipe system, but also just a certain section of it that we may be interested in. Major loss h Single-path pipe system 任兩點間的能量方程式 L major f l D V g Minor loss h L Lmajor L min or min or K L V g 16

Pipe Flow Examples / Single pipe whose length may be interrupted by various components. Multiple pipes in different configuration Parallel Series Network 17

Single-Path Systems 1/ Pipe flow problems can be categorized by what parameters are given and what is to be calculated. 分成幾種型態 18

Single-Path Systems / 重點 Given pipe (L and D), and flow rate, and Q, find pressure drop Δp Given Δp, D, and Q, find L. Given Δp, L, and D, find Q. Given Δp,, L, and Q, find D. 已知, 求 19

CASE 1 Given L, D, and Q, find Δp The energy equation 已知 L D Q, 求壓力降 p α α 1V1 p V 1 z1 z hl h ρg g ρg g h Lmajor L min or The flow rate leads to the Reynolds number and hence the friction factor for the flow. Q D Reynolds number f Tabulated data can be used for minor loss coefficients and equivalent lengths. 求 Minor loss coefficients 或 L eq The energy equation can then be used to directly to obtain the pressure drop. 利用能量方程式求壓力降 130

CASE Given Δp, D, and Q, find L The energy equation p α α 1V1 p V 1 z1 z hl h ρg g ρg g h Lmajor L min or The flow rate leads to the Reynolds number and hence the friction factor for the flow. Q D Reynolds number f Tabulated data can be used for minor loss coefficients and equivalent lengths. 求 Minor loss coefficients 或 L eq The energy equation can then be rearranged and solved directly for the pipe length. 利用能量方程式求 Pipe length 131

Example 8.8 Type I Determine Pressure Drop Water at 60 F F flows from the basement to the second floor through the 0.75-in. (0.065-ft) ft)-diameter copper pipe (a drawn tubing) at a rate of Q 1.0 gal/min 0.067 ft 3 /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 included. 已知管直徑與流率 13

1/4 Example 8.8 Solution 1/4 V 1 Q A 1 μ.34 10... 8.70ft / s 5 lb s / ft The energy equation p ρg ρ 1.94slug / ft Re ρvd / μ 45000 1 α1v1 p αv z1 z hl g p ρg g 3 The flow is turbulent 從已知資訊判定 pipe flow 為紊流 γz 1 ρ(v V ) γh 1 1 L 計算 h L z 1 V 0,z Q / A 0ft, p 0(free... 19.6ft / s jet) Head loss is different for each of the three cases. 133

/4 Example 8.8 Solution /4 (a) If all losses are neglected (h( L 0) 1 p1 γz ρ(v V1 )... 1547lb / ft 10.7psi ε 0.000005 ε / D (b) If the only losses included are the major losses, the head loss l is h L l V1 f 由 Moody chart 求 f Moody chart D g 8 10 5 Re 45000 f0.015 l( 60ft) V 1 1 p1 γz ρ(v V1 ) ρf... 306lb / ft D 1.3psi 134

3/4 Example 8.8 Solution 3/4 (c) If major and minor losses are included p 1 γz 1 ρ(v V 1 ) fγ l D V1 g ρk L V p 1 1.3psi 1.3psi ρk L V (1.94slugs / ft Minor loss 納入 3 (8.70ft / s) ) [10 4(1.5) ] p 1 1.3psi 9.17psi 30.5psi 135

4/4 Example 8.8 Solution 4/4 136

Example 8.9 Type I, Determine Head Loss Crude oil at 140 F F with γ53.7 lb/ft 3 and μ 8 108 10-55 lb s/ft (about four times the viscosity of water) is pumped across Alaska through the Alaska pipeline, a 799-mile mile-along, 4-ft4 ft-diameter steel pipe, at a maximum rate of Q.4 million barrel/day 117ft 3 /s, or VQ/A9.31 ft/s. Determine the horsepower needed for the pumps that drive this large system. 137

1/ Example 8.9 Solution 1/ The energy equation between points (1) and () p γ 1 V1 p V z1 hp z hl g γ Assume that z 1 z, p 1 p V 1 V 0 (large, open tank) g l V hl hp f... 17700ft D g h P is the head provided to the oil by the pump. Minor losses are negligible because of the large length-to to- diameter ratio of the relatively straight, uninterrupted pipe. Minor loss 被忽略 Horse power 克服 head loss f0.014 from Moody chart ε/d(0.00015ft)/(4ft), Re.. 138

/ Example 8.9 Solution / The actual power supplied to the fluid. 1hp Pa γqhp... 550ft lb /s 0000hp 139

CASE 3 1/ Given Δp, L, and D, find Q 1/ 需要 iteration These types of problems required either manual iteration or use of a computer application. The unknown flow rate or velocity is needed before the Reynolds number and hence the friction factor can be found. 因為 flowrate 未知, 無法知道速度, 無法知道 Reynolds number, 所以就不知道 frictional factor 先猜測 f Repeat the iteration process f V Re f until convergence 140

Given Δp,, L, and D, find Q / 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. Repeat the iteration process f V Re f until convergence 先猜測 f 值, 透過 energy equation 計算 velocity, 然後以 Reynolds number 去查詢 f, 再檢驗 f 是否與猜測的 f 一樣? 141

CASE 4 Given Δp, L, and Q, find D 1/ find D 1/ These types of problems required either manual iteration or use of a computer application. The unknown diameter is needed before the Reynolds number and relative roughness, and hence the friction factor can be found. 因為 D 未知, 就無從由 Q 計算 velocity 與 Reynolds number 需要 iteration 14

Given Δp, L, and Q, find D / 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. 由預測的 f 與 energy equation 計算 D 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 143

Example 8.10 Type III, Determine Flowrate According to an appliance manufacturer, the 4-in4 in-diameter galvanized iron vent on a clothes dryer is not to contain more than t 0 ft of pipe and four 90 elbows. Under these conditions determine the air flowrate if the pressure within the dryer is 0.0 inches of water. Assume a temperature of 100 and standard pressure. 直徑 4-in 鍍鋅鐵排氣孔 0 ft pipe 與四個 90 elbows 進口處壓力 0. inches of water, 求 flowrate 144

1/ Example 8.10 Solution 1/ 先把 energy equation 寫出來 Application of the energy equation between the inside of the dryer, point (1), and the exit of the vent pipe, point () gives p γ 1 l z1 z f V1 g p γ V g Assume that z 1 z, p 0, V 1 0 p γ 0.in p (0.in.) V D g 1ft K L V g 1 3 1 (6.4lb / ft ) 1.04lb / ft H O 1in. 能量方程式 With γ0.0709lb/ft 3, V V, and ν1.79 1010-4 ft /s. 945 (7.5 60f )V (1) (1) 代入 f 即可算出 velocity f is dependent on Re, which is dependent on V, and unknown. 145

/ Example 8.10 Solution / VD Re... 1860V () 有 V 即可算出 Reynolds number ν We have three relationships (Eq. 1,, and the ε/d0.0015 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 f0.0 V10.4ft/s (Eq. 1) Re19,300 (Eq.) f0.09 Assume f0.09 V10.1ft/s Re18,800 f0.09 Q AV... 0.881ft 3 /s 146

Example 8.11 Type III, Determine Flowrate The turbine shown in Figure E8.11 extracts 50 hp from the water flowing through it. The 1-ft1 ft-diameter, 300-ft ft-long pipe is assumed to have a friction factor of 0.0. Minor losses are negligible. Determine ermine the flowrate through the pipe and turbine. Friction factor f 已知忽略 minor loess Turbine 敗 50 hp 147

1/ Example 8.11 Solution 1/ 先把 energy equation 寫出來 The energy equation can be applied between the surface of the lake and the outlet of the pipe as p γ V1 p V 1 z h 1 z hl T g γ g Where p 1 V 1 p z 0, z 1 90ft, and V V, the fluid velocity in the pipe h V f l Pa 561 0.093V ft ht... ft D g γq V L 3 0.107V 90V 561 There are two real, positive roots: V6.58 ft/s or V4.9 ft/s. The third root is negative (V-31.4ft/s) and has no physical meaning for this flow. 0 能量方程式 148

/ Example 8.11 Solution / Two acceptable flowrates are π Q D 4 π Q D 4 V... 5.17ft V... 19.6ft 3 3 /s /s 149

Example 8.1 Type IV Without Minor Losses, Determine Diameter Air at standard temperature and pressure flows through a horizontal, galvanized iron pipe (ε0.0005( 0.0005 ft) at a rate of.0ft 3 /s. Determine the minimum pipe diameter if the pressure drop is to be no more than 0.50 psi per 100 ft of pipe. 鍍鋅鐵管, 粗糙度已知 Q 已知壓力降已知求管直徑 150

1/ Example 8.1 Solution 1/ Assume the flow to be incompressible with ρ0.0038 slugs/ft 3 and μ3.74 10-7 lb.s/ft. If the pipe were too long, the pressure drop from one end to the other, p 1 -p, would not be small relative to the pressure at the beginning, and compressible flow considerations would be required. l ρv With z 1 z, V 1 V, The energy equation becomes p p f p1 p (0.5)(144)lb / ft f (100ft) D (0.0038slugs/ ft 1 3 D V ) g g Q V A.55 D 1/5 D 0.404f (1) 151

/ Example 8.1 Solution / 4 ρvd 1.6 10 Re... () μ D ε D 0.0005 D (3) We have four equations (Eq. 1,, 3, and either the Moody chart or the Colebrook equation) and four unknowns (f, D, ε/d, and Re) from which the solution can be obtained by trial-and and-error methods. Repeat the iteration process f D Re and ε/d f until convergence (1) () (3) 15

Example 8.13 Type IV With Minor Losses, Determine Diameter Water at 60 F F (ν1.1( 1.1 1010-5 ft /s) is to flow from reservoir A to reservoir B through a pipe of length 1700 ft and roughness 0.0005 5 ft at a rate of Q 6 ft 3 /s as shown in Figure E8.13. The system contains a sharp-edged entrance and four flanged 45 elbow. Determine the pipe diameter needed. Flowrate 管長度粗糙度已知求管直徑 153

1/ Example 8.13 Solution 1/ The energy equation can be applied between two points on the surfaces of the reservoirs (p 1 V 1 p z V 0) p1 V1 p V z1 z hl γ g γ g V l z1 f K L g D Q 33.1 V K A D Lent 0.5, K Lelbow 0., and K Lexit 1 V 1700 44ft f [4(0.) 0.5 1] (3.ft /s ) D f 0.0015D 5 0.00135D (1) 154

/ Example 8.13 Solution / 6 VD.74 10 Re... () ν D ε D 0.0005 D (3) We have four equations (Eq. 1,, 3, and either the Moody chart or the Colebrook equation) and four unknowns (f, D, ε/d, and Re) from which the solution can be obtained by trial-and and-error methods. Repeat the iteration process D f Re and ε/d f until convergence 先猜測 D (1) () (3) 155

Multiple-Path Systems Series and Parallel Pipe System 流量關係與能量損失關係多管系統 Q 1 Q Q3 hl h A B L h 1 L h L 3 依據管排列架構 Q Q 1 Q Q3 hl h h 1 L L3 156

Multiple-Path Systems Multiple Pipe Loop System 流量關係不同路徑, 寫下不同能量方程式 Q 1 p γ A p γ A h Q L V g A V g A Q h L3 3 z z A A p γ B p γ B V g B V g B z z B B h h L1 L1 h h L L3 ( 1 ) ( 1 3) 兩個貯存槽 157

Multiple-Path Systems Three-Reservoir System If valve (1) was closed, reservoir B reservoir C If valve () was closed, reservoir A reservoir C If valve (3) was closed, reservoir A reservoir B With all valves open. Q Q Q 流量關係不同路徑, 寫下不同能量方程式 1 p γ A p γ B VA g VB g 3 z z A B p γ p γ B C VB g VC g z z B C h h L1 L h h L L3 ( A B) ( B C) 三個貯存槽 158

Example 8.14 Three reservoir, Multiple Pipe System Three reservoirs are connected by three pipes as are shown in Figure E8.14. For simplicity we assume that the diameter of each pipe is 1 ft, the frictional factor for each is 0.0, and because of the large l length-to to-diameter ratio, minor losses are negligible. Determine the flowrate into or out of each reservoir. 159

1/4 Example 8.14 Solution 1/4 The continuity equation requires that Flows out of reservoir B Q 1 Q Q3 V1 V V3 The diameters are the same for each pipe The energy equation for the fluid that flows from A to C in pipes (1) and (3) can be written as pa VA pc VC l1 V1 l 3 V3 za zc f1 f3 γ g γ g D g D g By using the fact that p A V A p C V C z C 0 1 (1) 3 z A f 1 l D 1 1 V1 g f 3 l D 3 3 V3 g 3 V () 1 0.4V 160

/4 Example 8.14 Solution /4 Similarly the energy equation for fluid following from B to C p γ B VB g z B z B p γ C l D VC g V g z C f l D l D V3 g V g f 3 l D 3 3 V3 g (3) 3 f f3 64.4 0.5V 0.4V3 3 No solution to Eqs.. 1,, and 3 with real, positive values of V 1, V, and V Thus, our original assumption of flow out of and V 3. Thus, our original assumption of flow out of reservoir B must be incorrect. 161

3/4 Example 8.14 Solution 3/4 The continuity equation requires that Flows into reservoir B Q 1 Q Q3 V1 V V3 (4) The energy equation between points A and B and A and C z z A A z z B C f f 1 1 l D 1 l D 1 1 1 V1 g V1 g f f 3 l D l D 3 3 V g V3 g 58 V 3 V 1 0.5V 1 0.4V3 (5) (6) Solve V 15.9ft /s V 1.88ft / s 16

4/4 Example 8.14 Solution 4/4 The corresponding flowates are Q Q Q 1 3 A V A Q 1 1 1 V Q 1.5ft 3.6ft /s 3 10.ft /s 3 from A /s int o B int o C 163

1/ Pipe Flowrate Meters 1/ 管狀流量量具 The theoretical flow rate may be related to the pressure differential between section 1 and by applying the continuity and Bernoulli equations. Then empirical correction factors may be applied to obtain the actual flow rate. 先介紹理論值 Basic equation ρdv t CV p ρ 1 1 V gz 1 CS r r ρv da p ρ V 0 gz 164

/ Pipe Flowrate Meters / p 1 p Q ideal ρv 1 A A 1 V Theoretical mass flow rate V 孔的截面積 A A (p ρ 1 1 ρ (p ( 4 (D /D ) ) p 1 [ 1 (A / A ) ] ) 1 依據形狀 Orifice Meter Nozzle Meter Venturi Meter p 理論上可由上下游壓力與管的幾何形狀算出 Q 值 ) 1 Q actual Q Δp?? ideal 實際上?? 165

Pipe Flowrate Meters Orifice Meter (p1 p Q actual C oq ideal C oa o 4 ρ(1 β A C d / 4 Area of the hole in the orifice plate 0 π o C ( β d / D, Re ρvd / μ) Orifice meter discharge coefficient o ) ) Orifice meter discharge coefficient C 0 得由實驗取得 166

Pipe Flowrate Meters Nozzle Meter (p1 p Q actual C nq ideal C na n 4 ρ(1 β A C n n πd / 4 Area of the hole C n ( β d / D, Re ρvd / μ) Nozzle meter discharge coefficient ) ) 167

Pipe Flowrate Meters Venturi Meter (p1 p Q actual C VQ ideal C VA T 4 ρ(1 β A C T V πd / 4 Area of the throat C V ( β d / D, Re ρvd / μ) The Venturi discharge coefficient ) ) 168

Linear Flow Measurement Float-type type Variable-area area Flow Meters The ball or float is carried upward in the tapered clear tube by the flowing fluid until the drag force and float weight are in equilibrium. Float meters often called rotameters. Rotameter-type type flow meter 169

Linear Flow Measurement Turbine Flow Meter A A small, freely rotating propeller or turbine within the turbine meter rotates with an angular velocity that is function of the average fluid velocity in the pipe. This angular velocity is picked up magnetically and calibrated to provide a very accurate measure of the flowrate through the meter. Turbine-type flow meter 170

Volume Flow Meters 1/ Measuring the amount (volume or mass) of fluid that has passed through a pipe during a given time period, rather than the instantaneous flowrate. Nutating disk flow meters is widely used to measurement the net amount of water used in domestic and commercial water systems as well as the amount of gasoline delivered to your gas tank. Nutating disk flow meter 171