CIVE2400 Fluid Mechanics. Section 1: Fluid Flow in Pipes

Transcription

1 CIVE00 Flid Mechanics Section : Flid Flow in Pipes CIVE00 FLUID MECHNICS... SECTION : FLUID FLOW IN PIPES.... FLUID FLOW IN PIPES.... Pressre loss de to riction in a pipeline..... Pressre loss dring laminar low in a pipe.... Pressre loss dring trblent low in a pipe.... Choice o riction actor The vale o or Laminar low Blasis eqation or Nikradse Colebrook-White eqation or...8. Local Head Losses Losses at Sdden Enlargement Losses at Sdden Contraction..... Other Local Losses....6 Pipeline nalysis....7 Pressre Head, Velocity Head, Potential Head and Total Head in a Pipeline....8 Flow in pipes with losses de to riction Reservoir and Pipe Example Pipes in series Pipes in Series Example...9. Pipes in parallel Pipes in Parallel Example n alternative method.... Branched pipes..... Example o Branched Pipe The Three Reservoir Problem..... Other Pipe Flow Examples dding a parallel pipe example...8 CIVE 00: Flid Mechanics Pipe Flow

2 . Flid Flow in Pipes We will be looking here at the low o real lid in pipes real meaning a lid that possesses viscosity hence looses energy de to riction as lid particles interact with one another and the pipe wall. Recall rom Level that the shear stress indced in a lid lowing near a bondary is given by Newton's law o viscosity: τ d dy This tells s that the shear stress, τ, in a lid is proportional to the velocity gradient - the rate o change o velocity across the lid path. For a Newtonian lid we can write: τ μ d dy where the constant o proportionality, μ, is known as the coeicient o viscosity (or simply viscosity). Recall also that low can be classiied into one o two types, laminar or trblent low (with a small transitional region between these two). The non-dimensional nmber, the Reynolds nmber, Re, is sed to determine which type o low occrs: Re ρ d μ For a pipe Laminar low: Re < 000 Transitional low: 000 < Re < 000 Trblent low: Re > 000 It is important to determine the low type as this governs how the amont o energy lost to riction relates to the velocity o the low. nd hence how mch energy mst be sed to move the lid.. Pressre loss de to riction in a pipeline. Consider a cylindrical element o incompressible lid lowing in the pipe, as shown Figre : Element o lid in a pipe CIVE 00: Flid Mechanics Pipe Flow

3 The pressre at the pstream end,, is p, and at the downstream end,, the pressre has allen by Δp to (p-δp). The driving orce de to pressre (F Pressre x rea) can then be written driving orce Pressre orce at - pressre orce at p ( p Δp) Δp Δp πd The retarding orce is that de to the shear stress by the walls shear stress area over which it acts τw area o pipe wall τwπdl s the low is in eqilibrim, driving orce retarding orce Δp d π τπ w dl τwl Δp d Eqation Giving an expression or pressre loss in a pipe in terms o the pipe diameter and the shear stress at the wall on the pipe. The shear stress will vary with velocity o low and hence with Re. Many experiments have been done with varios lids measring the pressre loss at varios Reynolds nmbers. These reslts plotted to show a graph o the relationship between pressre loss and Re look similar to the igre below: Figre : Relationship between velocity and pressre loss in pipes CIVE 00: Flid Mechanics Pipe Flow

4 This graph shows that the relationship between pressre loss and Re can be expressed as laminar where.7 < a <.0 trblent Δp Δp a s these are empirical relationships, they help in determining the pressre loss bt not in inding the magnitde o the shear stress at the wall τ w on a particlar lid. I we knew τ w we cold then se it to give a general eqation to predict the pressre loss.. Pressre loss dring laminar low in a pipe In general the shear stress τ w. is almost impossible to measre. Bt or laminar low it is possible to calclate a theoretical vale or a given velocity, lid and pipe dimension. (s this was covered in he Level modle, only the reslt is presented here.) The pressre loss in a pipe with laminar low is given by the Hagen-Poiseille eqation: or in terms o head Where h is known as the head-loss de to riction μl Δ p d μl h ρ gd (Remember the velocity,, is means velocity and is sometimes written.) Eqation. Pressre loss dring trblent low in a pipe In this derivation we will consider a general bonded low - lid lowing in a channel - we will then apply this to pipe low. In general it is most common in engineering to have Re > 000 i.e. trblent low in both closed (pipes and dcts) and open (rivers and channels). However analytical expressions are not available so empirical relationships are reqired (those derived rom experimental measrements). Consider the element o lid, shown in igre below, lowing in a channel, it has length L and with wetted perimeter P. The low is steady and niorm so that acceleration is zero and the low area at sections and is eqal to. Figre : Element o lid in a channel lowing with niorm low CIVE 00: Flid Mechanics Pipe Flow

5 p w p τ LP W sinθ 0 writing the weight term as ρ gl and sin θ Δz/L gives this can be rearranged to give ( p p ) τ LP gδz 0 w ρ [( p p ) Δz] L where the irst term represents the piezometric head loss o the length L or (writing piezometric head p * ) * dp τ o m dx where m /P is known as the hydralic mean depth Writing piezometric head loss as p * h, then shear stress per nit length is expressed as τ o * dp h τ o m m dx L P 0 Eqation So we now have a relationship o shear stress at the wall to the rate o change in piezometric pressre. To make se o this eqation an empirical actor mst be introdced. This is sally in the orm o a riction actor, and written where is the mean low velocity. Hence dp dx * ρ τ o ρ h m L So, or a general bonded low, head loss de to riction can be written L h m More speciically, or a circlar pipe, m /P πd /πd d/ giving Eqation L h gd This is known as the Darcy-Weisbach eqation or head loss in circlar pipes (Oten reerred to as the Darcy eqation) Eqation This eqation is eqivalent to the Hagen-Poiseille eqation or laminar low with the exception o the empirical riction actor introdced. It is sometimes sel to write the Darcy eqation in terms o discharge Q, (sing Q ) CIVE 00: Flid Mechanics Pipe Flow

6 Or with a % error Q πd 6 LQ LQ h gπ d.0d LQ h d Eqation 6 Eqation 7 NOTE On Friction Factor Vale The vale shown above is dierent to that sed in merican practice. Their relationship is merican Sometimes the is replaced by the Greek letter λ. where λ merican Conseqently great care mst be taken when choosing the vale o with attention taken to the sorce o that vale.. Choice o riction actor The vale o mst be chosen with care or else the head loss will not be correct. ssessment o the physics governing the vale o riction in a lid has led to the ollowing relationships. h L. h v. h /d. h depends on srace roghness o pipes. h depends on lid density and viscosity 6. h is independent o pressre Conseqently cannot be a constant i it is to give correct head loss vales rom the Darcy eqation. n expression that gives based on lid properties and the low conditions is reqired. CIVE 00: Flid Mechanics Pipe Flow 6

7 .. The vale o or Laminar low s mentioned above the eqation derived or head loss in trblent low is eqivalent to that derived or laminar low the only dierence being the empirical. Eqation the two eqations or head loss allows s to derive an expression o that allows the Darcy eqation to be applied to laminar low. Eqating the Hagen-Poiseille and Darcy-Weisbach eqations gives: μl d L gd 6μ ρvd 6 Re Eqation 8.. Blasis eqation or Blasis, in 9, was the irst to give an accrate empirical expression or or trblent low in smooth pipes, that is: Re Eqation 9 This expression is airly accrate, giving head losses /- % o actal vales or Re p to Nikradse Nikradse made a great contribtion to the theory o pipe low by dierentiating between rogh and smooth pipes. rogh pipe is one where the mean height o roghness is greater than the thickness o the laminar sb-layer. Nikradse artiicially roghened pipe by coating them with sand. He deined a relative roghness vale k s /d (mean height o roghness over pipe diameter) and prodced graphs o against Re or a range o relative roghness /0 to /0. Figre : Regions on plot o Nikrades s data CIVE 00: Flid Mechanics Pipe Flow 7

8 nmber o distinct regions can be identiied on the diagram. The regions which can be identiied are:. Laminar low ( 6/Re). Transition rom laminar to trblent n nstable region between Re 000 and 000. Pipe low normally lies otside this region. Smooth trblent The limiting line o trblent low. ll vales o relative roghness tend toward this as Re decreases.. Transitional trblent The region which varies with both Re and relative roghness. Most pipes lie in this region.. Rogh trblent. remains constant or a given relative roghness. It is independent o Re... Colebrook-White eqation or Colebrook and White did a large nmber o experiments on commercial pipes and they also broght together some important theoretical work by von Karman and Prandtl. This work reslted in an eqation attribted to them as the Colebrook-White eqation: log 0 k s.7d.6 Re Eqation 0 It is applicable to the whole o the trblent region or commercial pipes and ses an eective roghness vale (k s ) obtained experimentally or all commercial pipes. Note a particlar diiclty with this eqation. appears on both sides in a sqare root term and so cannot be calclated easily. Trial and error methods mst be sed to get once k s Re and d are known. (In the 90s when calclations were done by slide rle this was a time consming task.) Nowadays it is relatively trivial to solve the eqation on a programmable calclator or spreadsheet. Moody made a sel contribtion to help, he plotted against Re or commercial pipes see the igre below. This igre has become known as the Moody Diagram. [Note that this igre ses λ ( ) or riction actor rather than. The shape o the diagram will not change i were sed instead.] CIVE 00: Flid Mechanics Pipe Flow 8

9 Figre : Moody Diagram. He also developed an eqation based on the Colebrook-White eqation that made it simpler to calclate : 6 / 00k s d Re Eqation This eqation o Moody gives correct to /- % or 0 < Re < 0 7 and or k s /d < 0.0. Barr presented an alternative explicit eqation or in 97 or ks.86 log d Re k.86 log Re d s 0 Eqation Eqation Here the last term o the Colebrook-White eqation has been replaced with.86/re 0.89 which provides more accrate reslts or Re > 0. The problem with these ormlas still remains that these contain a dependence on k s. What vale o k s shold be sed or any particlar pipe? Fortnately pipe manactres provide vales and typical vales can oten be taken similar to those in table below. CIVE 00: Flid Mechanics Pipe Flow 9

10 Pipe Material k s (mm) Brass, copper, glass, Perspex 0.00 sbestos cement 0.0 Wroght iron 0.06 Galvanised iron 0. Plastic 0.0 Bitmen-lined dctile iron 0.0 Spn concrete lined dctile 0.0 iron Slimed concrete sewer 6.0 Table : Typical k s vales. Local Head Losses In addition to head loss de to riction there are always head losses in pipe lines de to bends, jnctions, valves etc. (See notes rom Level, Section - Real Flids or a discssion o energy losses in lowing lids.) For completeness o analysis these shold be taken into accont. In practice, in long pipe lines o several kilometres their eect may be negligible or short pipeline the losses may be greater than those or riction. general theory or local losses is not possible, however rogh trblent low is sally assmed which gives the simple ormla hl kl g Eqation Where h L is the local head loss and k L is a constant or a particlar itting (valve or jnction etc.) For the cases o sdden contraction (e.g. lowing ot o a tank into a pipe) o a sdden enlargement (e.g. lowing rom a pipe into a tank) then a theoretical vale o k L can be derived. For jnctions bend etc. k L mst be obtained experimentally... Losses at Sdden Enlargement Consider the low in the sdden enlargement, shown in igre 6 below, lid lows rom section to section. The velocity mst redce and so the pressre increases (this ollows rom Bernolli). t position ' trblent eddies occr which give rise to the local head loss. Figre 6: Sdden Expansion CIVE 00: Flid Mechanics Pipe Flow 0

11 pply the momentm eqation between positions and to give: p ( ) p ρ Q Now se the continity eqation to remove Q. (i.e. sbstitte Q ) Rearranging gives p ( ) p ρ p p g ( ) Eqation 7 Now apply the Bernolli eqation rom point to, with the head loss term h L nd rearranging gives Combining Eqations 7 and 8 gives p p g g h p p h L g ρ g L g ( ) g h L ( ) Eqation 8 hl g Eqation 9 Sbstitting again or the continity eqation to get an expression involving the two areas, (i.e. / ) gives Comparing this with Eqation gives k L h L g Eqation 0 k L Eqation When a pipe expands in to a large tank << i.e. / 0 so k L. That is, the head loss is eqal to the velocity head jst beore the expansion into the tank. CIVE 00: Flid Mechanics Pipe Flow

12 .. Losses at Sdden Contraction Figre 7: Sdden Contraction In a sdden contraction, low contracts rom point to point ', orming a vena contraction. From experiment it has been shown that this contraction is abot 0% (i.e. ' 0.6 ). It is possible to assme that energy losses rom to ' are negligible (no separation occrs in contracting low) bt that major losses occr between ' and as the low expands again. In this case Eqation 6 can be sed rom point ' to to give: (by continity / /0.6 /0.6) h 0.6 ( / 0.6 ) L g i.e. t a sdden contraction k L Other Local Losses h L 0. g Eqation Large losses in energy in energy sally occr only where low expands. The mechanism at work in these sitations is that as velocity decreases (by continity) so pressre mst increase (by Bernolli). When the pressre increases in the direction o lid otside the bondary layer has enogh momentm to overcome this pressre that is trying to psh it backwards. The lid within the bondary layer has so little momentm that it will very qickly be broght to rest, and possibly reversed in direction. I this reversal occrs it lits the bondary layer away rom the srace as shown in Figre 8. This phenomenon is known as bondary layer separation. Figre 8: Bondary layer separation CIVE 00: Flid Mechanics Pipe Flow

13 t the edge o the separated bondary layer, where the velocities change direction, a line o vortices occr (known as a vortex sheet). This happens becase lid to either side is moving in the opposite direction. This bondary layer separation and increase in the trblence becase o the vortices reslts in very large energy losses in the low. These separating / divergent lows are inherently nstable and ar more energy is lost than in parallel or convergent low. Some common sitation where signiicant head losses occr in pipe are shown in igre 9 divergent dct or diser Tee-Jnctions Y-Jnctions Figre 9: Local losses in pipe low Bends The vales o k L or these common sitations are shown in Table. It gives vale that are sed in practice. k L vale Practice Bellmoth entry 0.0 Sharp entry 0. Sharp exit bend tees In-line low 0. Branch to line. Gate vale 0. (open) Table : k L vales CIVE 00: Flid Mechanics Pipe Flow

14 .6 Pipeline nalysis To analyses the low in a pipe line we will se Bernolli s eqation. The Bernolli eqation was introdced in the Level modle, and as a reminder it is presented again here. Bernolli s eqation is a statement o conservation o energy along a streamline, by this principle the total energy in the system does not change, Ths the total head does not change. So the Bernolli eqation can be written or p z H constant g Pressre Kinetic Potential Total energy per energy per energy per energy per nit weight nit weight nit weight nit weight s all o these elements o the eqation have nits o length, they are oten reerred to as the ollowing: pressre head p velocity head g potential head z total head H In this orm Bernolli s eqation has some restrictions in its applicability, they are: Flow is steady; Density is constant (i.e. lid is incompressible); Friction losses are negligible. The eqation relates the states at two points along a single streamline. CIVE 00: Flid Mechanics Pipe Flow

15 .7 Pressre Head, Velocity Head, Potential Head and Total Head in a Pipeline. By looking at the example o the reservoir with which eeds a pipe we will see how these dierent heads relate to each other. Consider the reservoir below eeding a pipe that changes diameter and rises (in reality it may have to pass over a hill) beore alling to its inal level. Figre 0: Reservoir eeding a pipe To analyses the low in the pipe we apply the Bernolli eqation along a streamline rom point on the srace o the reservoir to point at the otlet nozzle o the pipe. nd we know that the total energy per nit weight or the total head does not change - it is constant - along a streamline. Bt what is this vale o this constant? We have the Bernolli eqation p g g z H p g g z ρ ρ We can calclate the total head, H, at the reservoir, p 0 as this is atmospheric and atmospheric gage pressre is zero, the srace is moving very slowly compared to that in the pipe so 0, so all we are let with is total head H z the elevation o the reservoir. sel method o analysing the low is to show the pressres graphically on the same diagram as the pipe and reservoir. In the igre above the total head line is shown. I we attached piezometers at points along the pipe, what wold be their levels when the pipe nozzle was closed? (Piezometers, as yo will remember, are simply open ended vertical tbes illed with the same liqid whose pressre they are measring). pressre head elevation Total head line Figre : Piezometer levels with zero velocity s yo can see in the above igre, with zero velocity all o the levels in the piezometers are eqal and the same as the total head line. t each point on the line, when 0 H CIVE 00: Flid Mechanics Pipe Flow

16 p z H ρ g The level in the piezometer is the pressre head and its vale is given by p ρ g. What wold happen to the levels in the piezometers (pressre heads) i the water was lowing with velocity? We know rom earlier examples that as velocity increases so pressre alls velocity head pressre head elevation Total head line hydralic grade line Figre : Piezometer levels when lid is lowing p ρ g g z H H We see in this igre that the levels have redced by an amont eqal to the velocity head,. Now as g the pipe is o constant diameter we know that the velocity is constant along the pipe so the velocity head is constant and represented graphically by the horizontal line shown. (this line is known as the hydralic grade line). What wold happen i the pipe were not o constant diameter? Look at the igre below where the pipe rom the example above is replaced by a pipe o three sections with the middle section o larger diameter pressre head elevation velocity head Total head line hydralic grade line Figre : Piezometer levels and velocity heads with lid lowing in varying diameter pipes The velocity head at each point is now dierent. This is becase the velocity is dierent at each point. By considering continity we know that the velocity is dierent becase the diameter o the pipe is dierent. Which pipe has the greatest diameter? H CIVE 00: Flid Mechanics Pipe Flow 6

17 Pipe, becase the velocity, and hence the velocity head, is the smallest. This graphical representation has the advantage that we can see at a glance the pressres in the system. For example, where along the whole line is the lowest pressre head? It is where the hydralic grade line is nearest to the pipe elevation i.e. at the highest point o the pipe..8 Flow in pipes with losses de to riction. In a real pipe line there are energy losses de to riction - these mst be taken into accont as they can be very signiicant. How wold the pressre and hydralic grade lines change with riction? Going back to the constant diameter pipe, we wold have a pressre sitation like this shown below velocity head pressre head elevation Total head line hydralic grade line H h Figre : Hydralic Grade line and Total head lines or a constant diameter pipe with riction How can the total head be changing? We have said that the total head - or total energy per nit weight - is constant. We are considering energy conservation, so i we allow or an amont o energy to be lost de to riction the total head will change. Eqation 9 is the Bernolli eqation as applied to a pipe line with the energy loss de to riction written as a head and given the symbol h (the head loss de to riction) and the local energy losses written as a head, hl (the local head loss). p p z z h h L g g Eqation.9 Reservoir and Pipe Example Consider the example o a reservoir eeding a pipe, as shown in igre. Figre : Reservoir eeding a pipe CIVE 00: Flid Mechanics Pipe Flow 7

18 The pipe diameter is 00mm and has length m and eeds directly into the atmosphere at point C m below the srace o the reservoir (i.e. z a z c.0m). The highest point on the pipe is a B which is.m above the srace o the reservoir (i.e. z b z a.m) and m along the pipe measred rom the reservoir. ssme the entrance and exit to the pipe to be sharp and the vale o riction actor to be Calclate a) velocity o water leaving the pipe at point C, b) pressre in the pipe at point B. a) We se the Bernolli eqation with appropriate losses rom point to C and or entry loss k L 0. and exit loss k L.0. For the local losses rom Table or a sharp entry k L 0. and or the sharp exit as it opens in to the atmosphere with no contraction there are no losses, so h L 0. g Friction losses are given by the Darcy eqation L h gd Pressre at and C are both atmospheric, is very small so can be set to zero, giving L z zc 0. g gd g L z zc 0. g d Sbstitte in the nmbers rom the qestion m / s b) To ind the pressre at B apply Bernolli rom point to B sing the velocity calclated above. The length o the pipe is L m: p B L z zb 0. g gd g z z That is 8.8 kn/m below atmospheric. pb L 0. g d pb p B B N / m.0 Pipes in series When pipes o dierent diameters are connected end to end to orm a pipe line, they are said to be in series. The total loss o energy (or head) will be the sm o the losses in each pipe pls local losses at connections. CIVE 00: Flid Mechanics Pipe Flow 8

19 .0. Pipes in Series Example Consider the two reservoirs shown in igre 6, connected by a single pipe that changes diameter over its length. The sraces o the two reservoirs have a dierence in level o 9m. The pipe has a diameter o 00mm or the irst m (rom to C) then a diameter o 0mm or the remaining m (rom C to B). Figre 6: For the entrance se k L 0. and the exit k L.0. The join at C is sdden. For both pipes se 0.0. Total head loss or the system H height dierence o reservoirs h head loss or 00mm diameter section o pipe h head loss or 0mm diameter section o pipe h L entry head loss at entry point h L join head loss at join o the two pipes h L exit head loss at exit point So H h h h L entry h L join h L exit 9m ll losses are, in terms o Q: L Q h d L Q h d Q Q Q h Lentry 0. g g d π d d Q Q h Lexit.0 g d d h Ljoin ( ) Q d d Q g π g d d Sbstitte these into h h h L entry h L join h L exit 9 and solve or Q, to give Q 0.8 m /s CIVE 00: Flid Mechanics Pipe Flow 9

20 . Pipes in parallel When two or more pipes in parallel connect two reservoirs, as shown in Figre 7, or example, then the lid may low down any o the available pipes at possible dierent rates. Bt the head dierence over each pipe will always be the same. The total volme low rate will be the sm o the low in each pipe. The analysis can be carried ot by simply treating each pipe individally and smming low rates at the end... Pipes in Parallel Example Figre 7: Pipes in Parallel Two pipes connect two reservoirs ( and B) which have a height dierence o 0m. Pipe has diameter 0mm and length 00m. Pipe has diameter 00mm and length 00m. Both have entry loss k L 0. and exit loss k L.0 and Darcy o Calclate: a) rate o low or each pipe b) the diameter D o a pipe 00m long that cold replace the two pipes and provide the same low. a) pply Bernolli to each pipe separately. For pipe : p pb B z z g g B 0. g l gd.0 g p and p B B are atmospheric, and as the reservoir srace move s slowly and BB are negligible, so l z zb 0..0 d g m / s nd low rate is given by πd Q 0.00 m / s For pipe : p z g pb B zb 0. g g l gd.0 g gain p and p B B are atmospheric, and as the reservoir srace move s slowly and BB are negligible, so CIVE 00: Flid Mechanics Pipe Flow 0

21 z z B l 0..0 d g m / s nd low rate is given by πd Q m / s b) Replacing the pipe, we need Q Q Q m /s For this pipe, diameter D, velocity, and making the same assmptions abot entry/exit losses, we have: p p l B B z zb 0..0 g g g gd g z z l 0..0 D g D D The velocity can be obtained rom Q i.e. πd Q Q 0.08 πd D So which mst be solved iteratively B D D 0 D.D. n approximate answer can be obtained by dropping the second term: 0 D. Writing the nction D. D 0.08m ( D) D (0.08) 0.6 So increase D slightly, try 0.07m ( 0.07) 0.0.D. i.e. the soltion is between 0.07m and 0.08m bt 0.07 i siciently accrate. CIVE 00: Flid Mechanics Pipe Flow

22 .. n alternative method n alternative method (althogh based on the same theory) is shown below sing the Darcy eqation in terms o Q LQ h d nd the loss eqations in terms o Q: Q Q Q h L k k k k g g gπ d d For Pipe 0 h Lentry h h Q Q 0.00 m / s Q.litres / s Lexit Q 0.0 Q For Pipe 0 h Lentry h h Q Q 0.088m / s Q 8.8litres / s Lexit Q 0. Q Branched pipes I pipes connect three reservoirs, as shown in Figre 7, then the problem becomes more complex. One o the problems is that it is sometimes diiclt to decide which direction lid will low. In practice soltions are now done by compter techniqes that can determine low direction, however it is sel to examine the techniqes necessary to solve this problem. B D C Figre 7: The three reservoir problem For these problems it is best to se the Darcy eqation expressed in terms o discharge i.e. eqation 7. LQ h d When three or more pipes meet at a jnction then the ollowing basic principles apply: CIVE 00: Flid Mechanics Pipe Flow

23 . The continity eqation mst be obeyed i.e. total low into the jnction mst eqal total low ot o the jnction;. at any one point there can only be one vale o head, and. Darcy s eqation mst be satisied or each pipe. It is sal to ignore minor losses (entry and exit losses) as practical hand calclations become impossible ortnately they are oten negligible. One problem still to be resolved is that however we calclate riction it will always prodce a positive drop when in reality head loss is in the direction o low. The direction o low is oten obvios, bt when it is not a direction has to be assmed. I the wrong assmption is made then no physically possible soltion will be obtained. In the igre above the heads at the reservoir are known bt the head at the jnction D is not. Neither are any o the pipe lows known. The low in pipes and are obviosly rom to D and D to C respectively. I one assmes that the low in pipe is rom D to B then the ollowing relationships cold be written: z h h h a D h Q D D z b h zc h Q Q The h expressions are nctions o Q, so we have eqations with or nknowns, h D, Q, Q and Q which we mst solve simltaneosly. CIVE 00: Flid Mechanics Pipe Flow

24 The algebraic soltion is rather tedios so a trial and error method is sally recommended. For example this procedre sally converges to a soltion qickly:. estimate a vale o the head at the jnction, h D. sbstitte this into the irst three eqations to get an estimate or Q or each pipe.. check to see i continity is (or is not) satisied rom the orth eqation. i the low into the jnction is too high choose a larger h D and vice versa.. retrn to step I the direction o the low in pipe was wrongly assmed then no soltion will be ond. I yo have made this mistake then switch the direction to obtain these or eqations z h h z a b h D D h hd zc h Q Q Q Looking at these two sets o eqations we can see that they are identical i h D z b. This sggests that a good starting vale or the iteration is z b then the direction o low will become clear at the irst iteration... Example o Branched Pipe The Three Reservoir Problem Water lows rom reservoir throgh pipe, diameter d 0mm, length L 0m, to jnction D rom which the two pipes leave, pipe, diameter d 7mm, length L 60m goes to reservoir B, and pipe, diameter d 60mm, length L 0m goes to reservoir C. Reservoir B is 6m below reservoir, and reservoir C is m below reservoir. ll pipes have 0.0. (Ignore and entry and exit losses.) We know the low is rom to D and rom D to C bt are never qite sre which way the low is along the other pipe either D to B or B to D. We irst mst assme one direction. I that is not correct there will not be a sensible soltion. To keep the notation rom above we can write z a, z b 6 and z c 0. For low to D ssme low is D to B z a h D h L Q h 607Q h h D D D z b h d L Q 8 880Q d For low is D to C h h D D z c h L Q 0 768Q d The inal eqation is continity, which or this chosen direction D to B is Q Q Q CIVE 00: Flid Mechanics Pipe Flow

25 Now it is a matter o systematically qesting vales o h D ntil continity is satisied. This is best done in a table. nd it is sally best to initially gess h D z a then redce its vale (ntil the error in continity is small): hj Q Q Q QQQ err So the soltion is that the head at the jnction is 7. m, which gives Q 0.00m /s, Q 0.007m /s and Q 0.000m /s. Had we gessed that the low was rom B to D, the second eqation wold have been z h h b D L Q 8 hd 880Q d Q Q and continity wold have been Q. I yo then attempted to solve this yo wold soon see that there is no soltion. CIVE 00: Flid Mechanics Pipe Flow

26 n alternative method to solve the above problem is shown below. It does not solve the head at the jnction, instead directly solves or a velocity (it may be easily amended to solve or discharge Q) [For this particlar qestion the method shown above is easier to apply bt the method shown below cold be seen as more general as it prodces a nction that cold be solved by a nmerical method and so may prove more convenient or other similar sitations.] gain or this we will assme the low will be rom reservoir to jnction D then rom D to reservoirs B and C. There are three nknowns, and the three eqations we need to solve are obtained rom to B then to C and rom continity at the jnction D. Flow rom to B p p L L B B z zb g g gd gd Ptting p p B and taking and B as negligible, gives L L z zb gd gd Pt in the nmbers rom the qestion g0. g (eqation i) Flow rom to C p p L L C C z zc g g gd gd Ptting p p c and taking and c as negligible, gives L L z zc gd gd Pt in the nmbers rom the qestion g0. g (eqation ii) Fro continity at the jnction Flow to D Flow D to B Flow D to C with nmbers rom the qestion πd Q Q πd d d Q πd d d (eqation iii) CIVE 00: Flid Mechanics Pipe Flow 6

27 the vales o, and mst be ond by solving the simltaneos eqation i, ii and iii. The techniqe to do this is to sbstitte or eqations i, and ii in to eqation iii, then solve this expression. It is sally done by a trial and error approach. i.e. rom i, 9.8. rom ii, sbstitted in iii gives This table shows some trial and error soltions Giving.8 m/s, so.8 m/s,.69 m/s Flow rates are πd Q m / s πd Q 0.00m / s πd Q 0.00m / s Check or continity at the jnction Q Q Q () ( ) CIVE 00: Flid Mechanics Pipe Flow 7

28 .. Other Pipe Flow Examples... dding a parallel pipe example pipe joins two reservoirs whose head dierence is 0m. The pipe is 0. m diameter, 000m in length and has a vale o a) What is the low in the pipeline? b) It is reqired to increase the low to the downstream reservoir by 0%. This is to be done adding a second pipe o the same diameter that connects at some point along the old pipe and rns down to the lower reservoir. ssming the diameter and the riction actor are the same as the old pipe, how long shold the new pipe be? 0m Original pipe New pipe 000m a) LQ h d Q 0 0. Q 0.06m / s Q.6litres / s b) H h 0 h h Q h h L L Q d d as the pipes and are the same, same length and the same diameter then Q Q. h By continity Q Q Q Q Q So and Then Q Q L 000 -L CIVE 00: Flid Mechanics Pipe Flow 8

29 ( )( ) / d Q L d L Q d Q L d L Q h h s, d d ( ) L L d Q The new Q is to be 0% greater than beore so Q m /s Solve or L to give L.6m L m CIVE 00: Flid Mechanics Pipe Flow 9