CHAPTER 3 VISCOUS FLOW IN PIPES

Transcription

1 HPTER 3 VISOUS FOW IN PIPES 3. Introduction This chapter will deal almost exclusiely with pipes or conduits that are lowing ull. ree surace will not exist in the pipe itsel, and the Froude number, which is important to open channel lows, will accordingly hae no signiicance. The Reynolds number will be the signiicant dimensionless parameter, but its importance will also diminish at high Reynolds numbers as the low becomes independent o iscous eects. Flow in a circular pipe is undoubtedly the most common closed luid low. It is encountered in the eins and arteries in a body, in a city s water system, in a armer s irrigation system, in the piping systems transporting luids in a actory and in the ink jet o a computer s printer. For a suiciently low Reynolds numbers (Re < 000 in a pipe) a laminar low results, and at suiciently high Reynolds number a turbulent low occurs. For all lows inoled in this hapter, we assume that the pipe is completely illed with the luid being transported as is shown in Figure 3.a Thus, we will not consider a concrete pipe through which rain water lows without completely illing the pipe, as is shown in Figure 3.b. Such lows, called open-channel low will be treated in next chapters. The dierence between the open channel low and the pipe low is in the undamental mechanism that dries the low. For open channel low, graity alone is the driing orce such that the water lows down a hill like in streams and riers. For pipe low, graity may be important (the pipe need not be horizontal), but the main driing orce is likely to be a pressure gradient along the pipe. I the pipe is not ull, it is not possible to maintain this pressure dierence p p. Figure 3. (a) pipe low and; (b) open channel low 3. aminar or Turbulent low The low o a luid in a pipe may be laminar low or it may be turbulent low. Osborne Reynolds (8-9), a ritish scientist and mathematician, was the irst to distinguish the dierence between these two classiications o low by using a simple apparatus as shown in Figure

2 Figure 3. (a) Experiment to illustrate type o low and; (b) typical dye streaks I water runs through a pipe o diameter with an aerage elocity, the ollowing characteristics are obsered by injecting a dye as shown in Figure 3.. For small enough lowrates the dye streakline will remain as a well deined line as it lows along; or a somewhat larger intermediate lowrate the dye streakline luctuates in time and space. On the other hand, or large enough lowrates the dye streakline almost immediately becomes blurred and spreads across the entire pipe in a random ashion. These three characteristics are denoted as laminar, transitional and turbulent low, respectiely. The cures shown in Figure 3.3 represents the x-component o the elocity as a unction o time at a point in the low. Figure 3.3 Time dependence o luid elocity at a point Question 3- The -cm diameter pipe is used to transport water at 0 o. What is the maximum aerage elocity that may exist in the pipe or which laminar low is guaranteed? Solution 3- The Reynolds Number ranges or which laminar, transitional or turbulent pipe lows are obtained cannot be precisely gien. The actual transition rom laminar, transitional or turbulent low may take place at arious Reynolds numbers, depending on how much the low is distributed by ibrations o the pipe, roughness o the entrance region, and the like. In general, or engineering purposes the ollowing alues are appropriate. The low in a round pipe is laminar i Reynolds number is less than approximately

3 The low in a round pipe is turbulent i Reynolds number is greater than approximately 000 The low in a round pipe is in transition i Reynolds number is ranging between 00 and 000 The kinematic iscosity o water at 0 o is 0 6 m / sec. Using a Reynolds number o 000 so that a laminar low is guaranteed, we can simply ind that, U Re () U (0.0m) 000 (0 6 m / sec) U 0.m / sec This aerage elocity is quite small. Such small elocities are not usually encountered in actual situations. Hence laminar low is seldom o engineering interest except or specialized topics such as lubrication. Most internal lows are turbulent lows and thus the study o turbulence gains much attention. 3.3 Entrance Region and Fully eeloped Flow ny luid lowing in a pipe had to enter the pipe at some location. The region o low near where the luid enters the pipe is termed the entrance region and is illustrated in the Figure 3.. s it is shown in the igure, the luid typically enters the pipe with a nearly uniorm elocity proile. s the luid moes through the pipe, iscous eects cause it to stick to the pipe wall. Thus, a boundary layer in which iscous eects are important is produced along the pipe wall such that the initial elocity proile changes with distance along the pipe, until the luid reaches the end o the entrance length beyond which the elocity proile does not ary with distance x. Figure 3. Entrance region o a laminar low in a pipe or a wide rectangular channel Question 3- laminar low is to occur in an experimental acility with 0 o water lowing through a cm diameter pipe. alculate the aerage elocity and the entrance length i the Reynolds number is 8000? Solution 3- The shape o the elocity proile in the pipe depends on whether the low is laminar or turbulent, as does the length o the entrance region, E. s with many other properties o pipe low, the dimensionless entrance length, E /, correlates quite well with Reynolds number. Typical entrance lengths are gien by - 3 -

4 and E 0.06 Re or laminar low () E / 6 or turbulent low (3).(Re) Once the luid reaches the end o the entrance region, the low is simpler to describe because the elocity is a unction o only the distance rom the pipe centreline, r and independent o x. U Re U (0.0m) 8000 (0 6 m / sec) and since the low in the pipe is turbulent low, U 0.m / sec E E 0.0 / 6.(Re) / 6.(8000) E 0.0. (8000) / m. 3. Pressure change and shear stresses Fully deeloped steady low in a constant diameter pipe is drien by pressure orces. For horizontal pipe low, graity has no eect except or a hydrostatic pressure ariation across the pipe,, that is usually negligible. It is the pressure dierence, p p p between one section o the horizontal pipe and another which orces the luid through the pipe. Viscous eects proide the restraining orce that exactly balances the pressure orce, thereby allowing the luid to low through the pipe with no acceleration. I iscous eects were absent in such lows, the pressure would be constant throughout the pipe, except or the hydrostatic ariation. The need or the pressure drop can be iewed rom two dierent standpoints. In terms o a orce balance, the pressure is needed to oercome the iscous orces generated. In terms o an energy balance, the work done by the pressure orce is needed to oercome the iscous dissipation o energy throughout the luid. Figure 3.5: Element o luid in a pipe - -

5 I graitational eects are neglected, the pressure is constant across any ertical cross-section o the pipe, although it aries along the pipe rom one section to the next. Thus, i the pressure is p p at section (), it is p p p at section (). We can oreseen the act that the pressure decreases in the direction o low so that, p 0. The driing orce due to pressure (F = Pressure x rea) can then be written as = Pressure orce at - pressure orce at ( p)( ) ( p p)( ) ( p)( ) ( p)( ) ( p)( ) Since the pipe area at section () and at section () are same, = =, the equation reduces to, ( p)( ) ( p)( ) ( p)( ( p)( ) ( p)( r The retarding orce is that due to the shear stress by the walls and can be written as = shear stress area oer which it acts = τ w area o pipe wall ( w ) rl s the low is in equilibrium, the driing orces must be equal to the retarding orces. which can be simpliied to gie p w l r ( p) r ( w) rl the aboe equation represents the basic balance in orces needed to drie each luid particle along the pipe with constant elocity. It is clear that at r 0 (the centerline o the pipe) there is no shear stress ( 0 ). t r (at the pipe wall) the shear stress is maximum, denoted w, the wall shear stress. The pressure drop and wall shear stress are related by ) ) () l w p (5) it is important to note that small shear stress can produce a large pressure dierences in the closed conduits i the pipe is relatiely long ( l ). The shear stress dependence or turbulent low is ery complex. Howeer, or laminar low o a Newtonian luid, the shear stress is simply proportional to the elocity gradient du du (6) dy dy the negatie sign is included to gie 0 with du dr 0 (the elocity decreases rom the pipe centerline to the pipe wall). There are two goerning laws or ully deeloped laminar low o a Newtonian luid within a horizontal pipe. The one is Newton s second law o motion and the other is the deinition o a Newtonian luid. y combining these two equations we can obtain, - 5 -

6 - 6 - r l p dr du (7) which can be integrated to gie the elocity proile as 6 ) ( r V r l p r u c (8) where V c is the centerline elocity. n alternatie expression can be written by using the relationship between the wall shear stress and the pressure gradient to gie ) ( R r r u w (9) where R is the pipe radius. This elocity proile plotted in Figure 3.5 is parabolic in the radial coordinate, r, has a maximum elocity, V c, at the pipe centerline, and a minimum elocity at the pipe wall. The olume lowrate through the pipe can be obtained by integrating the elocity proile across the pipe. rdr R r V rdr r u ud Q R c R r r 0 0 ) ( or V c Q R (0) y deinition, the aerage elocity is the lowrate diided by the cross-sectional area and can be attained as the one-hal o the maximum elocity. l p R V R R Q V V c c 3 l p Q 8 () Figure 3.5 Shear stress distributions within the luid in a pipe and typical elocity proiles. The aboe results conirm the ollowing properties o laminar pipe low. For a horizontal pipe the lowrate is:

7 a. irectly proportional to the pressure drop b. Inersely proportional to the iscosity c. Inersely proportional to the pipe length d. Proportional to the pipe diameter to the ourth power I we introduce the riction actor, a quantity o substantial interest in pipe low, a dimensionless wall shear, deined by () ( / 8) V we see that the change in pressure per unit weight in pipe can be written as p h V g (3) where h is the head loss with dimension o length. This equation is oten reerred to as the arcy-weisbach equation, named ater Henri P.G. arcy ( ) and Julius Weisbach (806-87). ombining aboe equations: 6 () Re or laminar low in a pipe. Substituting this back into aboe equation we see that p h 3 the head loss is directly proportional to the aerage elocity (and hence the discharge also) to the irst power, a result that generally is applied to deeloped, laminar lows in conduits, o shape other than circular. (5) Question 3-3 n oil with a iscosity o =0. Ns/m and density = 900 kg/m 3 lows in a pipe o diameter =0.0m. What pressure drop, p p p is needed to produce a lowrate o x0-5 m 3 /s i the pipe is horizontal with x =0 and x =0m. Solution 3-3 The question mentions that the Reynolds number is less than 00.The aerage elocity in the pipe can be calculated as: Q V The Reynolds number is; 5 0 V (0.0) V Re in which m/sec

8 (0.0637) (0.0 ) Re m (.0 m / sec) Hence, the low is laminar. Thereore the pressure dierence can be deducted by the ollowing equation with l x x 0m..87 8l Q p 5 5 8(0.)(0)(0 ) 00 p 000N / m 0. kpa 7 (0.0) 5.00 Question 3- small diameter horizontal tube is connected to a supply reseroir as shown in the ollowing Figure. I 6600 mm 3 is captured at the outlet in 0 seconds estimate the iscosity o the water. Solution 3- The tube is ery small, so we expect iscous eects to limit the elocity to a small alue. Using ernoulli s equation rom the surace to the entrance to the tube, and neglecting the elocity head, we hae, letting 0 be a point on the surace: Ptan k tan k Ptube tube ztan k ztube g g where we hae used gage pressure with p o 0 and thus ptan k 0 and the elocity head in the reseroir is accepted as negligible. The equation becomes P tube P tube P tube N / m 9. 6kPa at the exit o the tube the pressure is zero; hence, p kPa / m l. The aerage elocity is ound to be 9 Q (66000 ) /0 V 0.80m / s (0.00) - 8 -

9 heck to make sure the elocity head is negligible: ( /g)=0.036m compared with pressure head. So the assumption o negligible elocity head is alid. The iscosity o the luid can be ound by: p V 3l where p 3Vl (6300) Ns / m 3(0.80) we should check the Reynolds number to determine i our assumption o a laminar low was acceptable. It is V Re in which (000)(0.8m / sec) (0.00m) Re 38.6 (6.060 m / sec) Hence, the low is laminar. Thereore the pressure dierence can be deducted by the ollowing equation with l x x 0m.This is obiously a laminar low since Re<00, so the calculations are alid proiding the entrance length is not too long. It is E 0.06Re E 0.06 Re m this is approximately 8% o the total length, a suiciently small quantity, hence calculations are assumed reliable. 3.5 Turbulent Flow in a pipe The study o turbulent low in a pipe is o substantial interest in actual lows since most lows encountered in practical applications are turbulent lows in pipes. In a turbulent low all three elocity components are nonzero. I we measure the components as a unction o time, the graphs similar to those shown in Figure 3.6 would result or a low in a pipe where u, and w are in the x-, r- and θ-directions. Howeer, there is seldom interest to the ciil engineers in the details o the randomly luctuating elocity components; hence, here, only the notation o a timeaeraged quantity will be introduced. The elocity components (u,,w) are written as ' u u u (6) ' (7) w where a bar oer a quantity denotes a time aerage and a prime denotes the luctuating part. Using the component u as an example, the time aerage is deined as, ' w w (8) T u u( t) dt (9) T 0 Where T is a time increment large enough to eliminate all time dependence rom u. s the low in the pipe becomes deeloped it is obsered that the horizontal elocity parameter u would be - 9 -

10 nonzero and =0, w =0 will be attained. Figure 3.6 Velocity components in a turbulent pipe low ll materials are rough when iewed with suicient magniication, although glass and plastic assumed to be smooth with e=0 (sometimes mentioned as k s or δ). s noted in the preceding sections shear stresses are signiicant only near the wall in the iscous wall layer with thickness δ. I the thickness δ is suiciently large, it submerges the wall roughness elements so that they hae negligible eect on the low; it is as i the wall were smooth. Such a condition is oten reerred to as being hydraulically smooth. I the iscous wall layer is relatiely thin, the roughness elements protrude out o this layer and the wall is rough. The relatie roughness e/ and the Reynolds number can be used to determine i a pipe is smooth or rough. (See Figure 3.7) Figure 3.7: (a) smooth wall and (b) a rough wall Question 3-5 Water at 0 o lows in a 0 cm diameter pipe at an aerage elocity o.6 m/sec. I the riction actor, o the pipe is 0.08, calculate the wall shear stress and riction (shear) elocity o the pipe. Solution 3-5 In the wall region the characteristic elocity and the length are the shear elocity u τ is deined by u o / (0) and the wall shear is calculated rom o V () 8 Thereore the wall shear stress occurring on the walls o pipe is calculated as: Pa 8-0 -

11 and the riction elocity is m/sec Question 3-6 The -cm diameter smooth, horizontal pipe as shown in the Figure transports 0.00 m 3 /sec o water at 0 o. alculate (a) the wall shear, (b) the pressure drop oer a 0-m length, o the pipe is Solution 3-6 The aerage low elocity in the pipe can be obtained by using the discharge relation, Q V Q 0.00 V 3.8m / sec 0.0 s it was same in the preious question, the wall shear stress occurring on the walls o pipe can be calculated as: o Pa 8 s it was deined beore, the pressure drop and wall shear stress are related by l o p 0 3 p 3000 Pa osses in eeloped Pipe Flow Perhaps the most calculated quantity in pipe low is the head loss. I the head loss is known in a deeloped low, the pressure change can be calculated; The head loss that results rom the wall shear in a deeloped low is related to the riction actor by the arcy-weisbach equation, namely, p h V g () onsequently, i the riction actor is known, we can ind the head loss and then the pressure drop. The riction actor, depends on the arious quantities that aect the low, written as, (,, V,, e) (3) where the aerage wall roughness height e accounts or the inluence o the wall roughness elements. dimensional analysis proides us with - -

12 V e (, ) () where e/ is the relatie roughness. Experimental data that relate the riction actor to the Reynolds number hae been obtained or ully deeloped pipe low oer a wide range o wall roughnesses. The results o this data are presented in Figure 3.8, which is commonly reerred to as the Moody diagram, named ater ewis F. Moody ( ). There are seeral eatures o the Moody diagram that should be noted. For a gien wall roughness, measured by the relatie roughness e/, there is a suiciently large alue o Re aboe which the riction actor is constant, thereby deining the completely turbulent regime. The aerage roughness element size e is substantially greater than the iscous wall layer thickness δ, so that iscous eects are not signiicant. For the smaller relatie roughness e/ alues it is obsered that, as Re decreases, the riction actor increases in the transition zone and eentually becomes the same as that o a smooth pipe. The roughness elements become submerged in the iscous wall layer so that they produce little eect on the main low. For Reynolds numbers less than 000, the riction actor o laminar low is shown. The critical zone couples the turbulent low to the laminar low and may represent an oscillatory low that alternately exists between turbulent and laminar low. The ollowing empirical equations represent the Moody diagram or Re>000: Smooth pipe low: Transition zone: ompletely turbulent zone: 0.86 ln Re 0.8 e ln 3.7 Re (5) (6) e 0.86ln (7) 3. 7 The transition zone equation that couples the smooth pipe equation to the completely turbulent regime equation is known as the olebrook Equation. Smooth pipe low is the olebrook equation with e=0, and completely turbulent zone equation is the olebrook equation with Re= Three categories o problems can be identiied or deeloped turbulent low in a pipe o length. atagory Known Unknown Q,,e,ν h h,,e,ν Q 3 Q,e,ν,h category problem is straightorward and requires no iteration procedure when using the - -

13 Moody diagram. ategory and 3 problems are more like problems encountered in engineering design situations and require an iteratie trial and-error process when using the Moody diagram. Each o these types will be illustrated with ollowing examples. Figure 3.8: Moody diagram (Note that i e/ or k s /=0.0 and Re=0, the reads =0.03 Question 3-7 Water at 0 o is transported or 500m in a cm-diameter wrought iron (commercial steel) horizontal pipe with a low rate o 3/sec. alculate the pressure drop oer the 500-m length o pipe, using Moody diagram. Solution 3-7 The aerage low elocity in the pipe can be obtained by using the discharge relation, Q V V 3 Q m /sec The Reynolds number is U Re Obtaining e or k s rom the Moody diagram we hae, using =/00 e 0.06mm mm The riction actor is read rom the Moody diagram to be 0.03 The head loss is than calculated as - 3 -

14 p h The pressure drop can also be calculated as V g h m p =830N/m or h p 8.3kPa Question 3-8 pressure drop o 700kPa is measured oer a 300-m length o horizontal, 0-cm diameter commercial pipe that transports oil (S=0.9, ν=0-5 m /s). alculate the low rate using Moody diagram. Solution 3-8 The relatie roughness is e 0.06mm mm ssuming that the low is completely turbulent (Re is not needed), the Moody diagram reads to be The head loss is than calculated as p h V g h p oil m note that S oil water The elocity can be calculated rom the head loss equation o arcy Weisbach V h g V gh V 5.6m /sec This proides us with a Reynolds number o U Re using this Reynolds number and e/=0.0006, the Moody diagram gies the riction actor as

15 0.03 This corrects the original alue or. the elocity is recalculated to be V m /sec This proides us with a Reynolds number o U Re From the Moody diagram =0.03 appears to be satisactory. Thus the low rate is Q V m 3 /sec Question 3-9 rawn tubing o what diameter should be selected to transport 0.00 m 3 /sec o 0 o water oer a 00-m length so that the head loss does not exceed 30-m? Use Moody diagram to sole the question Solution 3-9 In this problem we do not know. Thus, a trial and error solution is anticipated. The aerage is related to by Q V The riction actor and are related as ollows Thereore, The Reynolds number is 30 V h g 00 ( / ) U Re 6 0 Now, let us simply guess a alue or and check with the relations aboe and the Moody diagram. The irst guess is =0.03 and the correction is listed in the ollowing table. Note: the second guess is the alue or ound rom the calculations o the irst guess. (m) Re e/ (rom Moody) x x

16 The alue o =0.0 is acceptable, yielding a diameter o 3.88 cm. Since this diameter would undoubtedly not be a standard a diameter o cm Would be the tube size selected. This tube would hae a head loss less than the limit o 30-m imposed in the problem statement. 3.6 osses in eeloped noncircular conduits. Good approximations can be made or the head loss in circuits with noncircular cross sections by using the hydraulic radius R, deined by: R (8) P Where is the cross-sectional area and P is the wetted perimeter where the luid is in contact with the solid boundary. Question 3-0 ir at standard conditions (ν=.5x0-5 ) is to be transported through a smooth, horizontal, 30 cm x 0 cm rectangular duct whose length is 500-m. The low rate is 0. m 3 /sec. alculate the pressure drop. Solution 3-0 For a circular pipe lowing ull, the hydraulic radius is; r0 r0 R r0 Hence we simply replace the radius with double R and use the Moody diagram with U U (R) Re nd relatie roughness as e e R The head loss thereore becomes, h V V g R g The pressure drop in the duct can be calculated by p V h g Thereore it is necessary to calculate the head loss in the duct which is directly related with the riction coeicient o arcy. On the other hand, since the duct is rectangular it is necessary to ind the hydraulic radius instead o. The hydraulic radius is - 6 -

17 R 0.06 m P (0.3 0.) The aerage elocity is Q 0. V m/sec This gies a Reynolds number o U (R) 0.06 Re Using the smooth pipe cure o the Moody diagram, there results Hence, V 500 h m R g Thereore, the pressure drop is p gh Pa The arcy-weisbach equation is alid or the laminar or turbulent pipe low o any Newtonian luid. number o additional equations are also aailable or ealuating the head loss in a pipe due to riction. These tend to be more empirical in nature and limited to the turbulent low o water. This limitation is rarely restrictie in hydraulic engineering. Their main adantage is generally their ease o application. lthough they are less accurate than the arcy-weisbach equation, many hydraulic engineering applications do not require or justiy the higher accuracy anyway. One such empirical equation is the Hazen-Williams equation: or V V H R S (9) H R S (30) For use with U.S. customary or SI units, respectiely. In either case, H is a discharge coeicient with typical alues gien in the ollowing table Hazen Williams coeicient or pipes Pipe description or material H Extremely smooth and straight 0 sbestos-cement 0 New, unlined Steel 0-50 New, rieted Steel 0 Old, rieted Steel 95 Very smooth Steel 30 New,unlined ast iron 30 5-year-old ast iron 0 0-year-old ast iron 0 0-year-old ast iron 00 0-year-old ast iron Smooth wood or masonary 0 Vitriied clay 0 Ordinary brick 00 Old pipe in bad condition

18 R is the hydraulic radius (equal to / in a ull pipe), and S is the slope o the total head or energy grade line, that is, the ratio o the head loss to the pipe length. There are both nomograph and tabular solutions o the Hazen-Williams equation that proide rapid solutions to head loss/discharge problems. 3.7 Minor osses in Pipe Flow We now know how to calculate the losses due to a deeloped low in a pipe. Pipe systems do, howeer, include ales, elbows, enlargements, contractions, inlets, outlets, bends etc. that cause additional losses. Each o these deices causes a change in the magnitude and/or the direction o the elocity ectors and hence results in a loss. In general, i the low is gradually accelerated by a deice, the losses are ery small; relatiely large losses are associated with sudden enlargements or contractions. minor loss is expressed in terms o a loss coeicient K, deined by h m K (3) g the alues o K hae been determined experimentally or the arious ittings and geometry changes o interest in piping systems. One exception is the sudden expansion rom rea to area or which the loss can be calculated as h m (3) Question 3- I the low rate through a 0-cm diameter wrought iron pipe is 0.0m 3 /sec, ind the dierence in eleation H o the two reseroirs. Solution 3- The energy equation written or a control olume that contains the two reseroir suraces can be gien as P P z h hm z g g where elocities and pressures at point and point are equialent and equal to zero. z h hm z Thus letting z -z =H, we hae - 8 -

19 Figure 3.9: Minor oss coeicients K - 9 -

20 H ( K entrance K ale K elbow K exit ) g g The aerage elocity, Reynolds number and relatie roughness are Q 0.0 V m/sec This gies a Reynolds number o U ( ) Re e 0.06mm mm Using the Moody diagram we ind that Using the loss coeicient rom the gien table, or an entrance, a globe ale, screwed 0-cm diameter standard elbows, and an exit there results H ( (0.6) ) H... 6m The minor losses are about equal to the rictional losses as expected, since there are ie minor loss elements in 500 diameters o pipe length. 3.8 Hydraulic and Energy Grade ines When the energy equation is written in the orm o P z g h h m P z g The terms hae dimensions o length. This has led to the conentional use o the hydraulic grade line and the energy grade line. The hydraulic grade line (HG), the dashed line in Figure 3.0, in a piping system is ormed by the locus o points located a distance p/γ aboe the center o the pipe, or (p/γ) + z aboe a preselected datum; the liquid in a piezometer tube would rise to the HG. The energy grade line (EG), the solid line in Figure 3.0 is ormed by the locus o points a distance /g aboe the HG, or the distance ( /g) + (p/γ) + z aboe a preselected datum; the liquid in a pitot tube would rise to the EG. The ollowing points are noted relating to the HG and the EG. s the elocity approaches to zero, the HG and the EG approach each other. Thus, in a reseroir, they are identical and lie on the surace. The EG and HG slope downward in the direction o the low due to the head loss in the pipe. The greater the loss per unit length, the greater the slope; s the aerage elocity in the pipe increases the loss per unit length increases sudden change occurs in the HG and EG wheneer a loss occurs due to a sudden geometry change as represented by the ale or the sudden enlargement

21 jump occurs in the HG and the EG wheneer useul energy is added to the luid as occurs with a pump, and a drop occurs i useul energy is extracted rom the low as occurs with a turbine. t points where the HG passes through the centerline o the pipe, the pressure is zero. I the pipe lies aboe the HG, there is a acuum in the pipe, a condition that is oten aoided. n exception to this condition is the design o a siphon. Figure 3.0: Hydraulic Grade ine (HG) and Energy Grade ine (EG) or a piping system. Question 3- Water at 0 o lows between two reseroirs at the rate o 0.06m 3 /sec, Sketch the HG and EG What is the minimum diameter allowed to aoid the occurrence o caitation? Solution 3- The EG and the HG will be sketched on the igure including sudden changes at the entrance, contraction, enlargement and the exit. Note the large elocity head (the dierence between the EG and the HG) in the smaller pipe because o the high elocity. The elocity, Reynolds number, and relatie roughness in the 0-cm diameter pipe are calculated to be Q 0.06 V.9 0. m/sec This gies a Reynolds number o U ( ) Re

22 e 0.6mm mm Using the Moody diagram we ind that 0.0 The elocity, Reynolds number, and relatie roughness in the smaller pipe are Q V This gies a Reynolds number o U ( ) Re 6 0 e The minimum possible diameter is established by recognizing that the water apor pressure at 0 o is the minimum allowable pressure. Since the distance between the pipe and the HG is an indication o the pressure in the pipe, we can conclude that the minimum pressure will occur at section. Hence the energy equation applied between section, the reseroir surace and section gies P z g ( K entrance K g cont g g Where the subscript reers to the 0-cm diameter pipe. This simpliies to P ) z g g Which results (0.076 ) 0 0 ( (0.076 ) 50 (0.076 ) ) Where we assume that K cont =0.5. This requires a trial and error solution. The ollowing illustrates the procedure et =0. m Then (e/) =0.006 and Re 7.60 Thereore ? et =0.09 m - -

23 Then (e/) =0.009 and Re 8. 0 Thereore ? We can easily obsere that 0.m is too large and 0.09 m is too small. In act, the alue o 0.09 m is only slightly too small. onsequently, to be sae we must select the next larger pipe size, o 0.meter diameter. I there were a pipe size o 9.5-cm diameter, that could be selected. ssuming that such a size is not aailable we could select =0cm 3.9 Pipeline Systems Pipes in series One type o pipe system has already been considered in Example 3- and 3-, namely pipes in series. Inspection o that example would indicate the two basic premises associated with pipes in series: irst, continuity requires that the discharge is the same in all pipes, and second, the total head loss or the system equals the sum o the indiidual head losses o the respectie pipes. These principles can be generalized or any number o pipes that are connected in series. It should be noted that i minor losses are included, they must be reerenced to the appropriate pipe size and elocity head. Similarly, the head loss due to riction must be ealuated separately or each pipe size. When the arcy-weisbach equation is used in the analysis o any type o pipe system, the resistance coeicients are requently treated as constants. 5 Figure 3.: Three pipes in series system. Thereore, according to aboe deinition in the pipes connected in series; The total head loss is the sum o head losses in each pipe: ( h ) h h h The low rate, in each o the pipes connected is the same. Q Q Q Q total total a a b b c c Pipes in Parallel second pipe system is composed o a number o pipes connected in parallel (Fig. 3-). lthough any number o pipes could be so connected, the deelopment o the concepts will be based on just three pipes. In passing, note that the pipe sections upstream and downstream o the parallel pipes could be considered to be in series with the parallel pipe system o our immediate interest, and once the parallel pipe system is analyzed the remaining system could be treated as a number o pipes in series. Returning to the three parallel pipes, continuity can be applied with the result - 3 -

24 Q total Q a Q b Q c Since each o the pipes must share common alues o piezometric head at the junctions, it ollows that the head loss must be identical in each o the parallel pipes. This gies a second relationship that may be expressed as ( h ) h h h total a b c Figure 3.: Three pipes in parallel system. ranching Pipe System nother type o pipe system may be introducing by considering the three interconnected reseroirs o Fig Subsequently, we will generalize the analysis to include additional reseroirs and/or additional junctions. For the present, consider the problem o estimating the discharge in each o the pipe sections. Figure 3.3: Three-branch reseroir system. uring the solution o branch systems the necessary numbers o equations are obtained by utilizing energy and continuity concepts. Howeer, it is diicult to incorporate the elocity head at the junction, a problem that is circumented by assuming that it is a small quantity relatie to the other energy terms and may justiiably be ignored. It is usual to make the same assumption with respect to possible minor losses. second problem is that the low direction in pipe is not known at the outset. Our procedure will be to ormulate the energy equation so as to determine the actual low direction in pipe and then proceed on the basis o the outcome o this irst calculation. I the piezometric head (since the elocity head is neglected) at the junction is assumed equal to the eleation o reseroir surace, then no low would occur in pipe, and the continuity equation becomes Q =Q. In addition, the ollowing energy equations can be written (neglecting the elocity heads): - -

25 P z z h P z z h P z z h Where the head losses are deined using the arcy-weisbach equation. V h g the continuity equation is h h V g V g Q Q or V V V y substituting the headloss expressions respectiely into equations, the three energy equations hae our unknowns P /γ, V, V and V. The continuity equation proides the ourth equation to sole or the our unknowns. Q Question 3-3 Water lows at a rate o m 3 /sec rom reseroir to reseroir through three pipes connected in series (=0.05) as shown in the ollowing igure. Neglecting minor losses, determine the dierence in water surace eleation. Solution 3-3 Write down the energy equation rom reseroir to reseroir

26 P g z h z h z where h z z Writing the total head loss in expanded orm results in: Which results in: Using the arcy-weisbach equation h h h h h h h Thereore the energy equation is turn out to be z z /000 z h h P z g h z z V g V g V 0.05 (9.8) V g /000 V 0.05 (9.8) z 6.37V 0.6V. 58V /000 V (9.8) Use continuity equation to determine the elocities; Q 0.03 V (00/000) V V Q Q 0.03 (80/000) 0.03 (0/000) 0.955m/sec.80 m/sec 0.79 m/sec z z 6.37(0.955) 0.6(.80).58(0.790) z z z z 7.8 m Question 3- The three pipe system shown in Figure below has the ollowing characteristics pipe (cm) (m) Find the lowrate o water in each pipe and the pressure at point 3. Neglect minor losses - 6 -

27 Solution 3- Write down the energy equation rom to. P g z h P z g h 0 80 g Where V is V c and h h h and using the arcy Weisbach equation, we get 0 g g g ecause the pipes and are parallel, the headloss in is equal to the headloss in, so the headloss or can also be used in the aboe energy equation instead o or. This energy equation has two unknowns V and V, so that continuity can be used as a second equation. Q Q Q Which introduces a third unknown V. Since h h the third equation is g g (9.8) 0.30 (9.8) So now, we hae three equations and three unknowns. Using the continuity equation, we get (0.5) (0.30) (0.35) Substituting (0.859) Substituting into energy equation - 7 -

28 (9.8) (0.87 ) (9.8) (9.8) nd sole or V c The low rate is then, 0 ( ) 6.7 (9.8) (9.8).3m/sec 0.35 Q.3 0. m 3 /sec The pressure at 3 can be computed using the energy equation rom to 3 or rom 3 to. Using P3 3 P z3 h (3) z g g Since V 3 =V, the elocity head terms cancel out: P3 0 h (3) 80 and P3 60 P 60 3 g 000 (.3) (9.8) P P 3.5 m P P Pa P kpa (kn/m ) Question 3-5 The pipe system shown in Figure below connects two reseroirs that hae an eleation dierence o 0m. This pipe system consists o 00m o 50-cm concrete pipe (pipe ), that branches into 00-m o 0-cm pipe (pipe ) and 00-m o 0-cm pipe (pipe ) in parallel. Pipes and join into a single 50-cm pipe that is 500m long (pipe ). For =0.030 in all pipes, what is the low rate in each pipe o the system? - 8 -

29 Solution 3-5 The objectie is to compute the elocity in each pipe. We know that V =V because they are the same diameter pipe, h h because pipes and are in parallel, and Express Since Using Q Q h h in term o the elocities, and = Q Q, we get substituting (50 00) yields Q g Q Q g or or or (0 00) (0 00) or Next conert the parallel pipes to a single equialent = =50-cm diameter pipe, with a length E. E g g E 00 E E E 88 m Write the energy equation rom reseroir surace to reseroir surace h 0 m. 0 0 ( ( E ) g ) m/sec

30 also (50 00) Q (.033) m 3 /sec. and Q Q 0.939(.033).909 m/sec (0 00) Q (.909) m 3 /sec. Q Q Q m 3 /sec