6 BRANCHING PIPES. One tank to another 1- Q 1 = Q 2 +Q 3. One tank to two or more tanks 1- Q 1 = Q 2 + Q 3 or Q 3 =Q 1 + Q 2

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1 6 RNHING PIPES Oe tak to aother - Q Q Q - pply eroulli s equatio betwee & through s & as well as & Oe tak to two or more taks - Q Q Q or Q Q Q - pply eroulli s equatio betwee each two taks 6. ischarge from Oe Tak to other Example : Two taks are coected by a lie km log ad ad 0 cm diameter. Fid the discharge through the. If the last.5 km of the is replaced by s of.5 km log ad 0 cm diameter each. etermie the icrease i Q. Take f faig Give: Z m, Z 0 m Required: Q, ΔQ Solutio:

2 st case : pply eroulli s equatio betwee & kowig that P P atm., vv ~0, hp 0 Z Z h L h L 4 flv g 4x0.005x000xv x9.8x0. π Q x0. v 4 v d case: Q Q Q Q L v i.e., Q L, v Q Q, h Q L h L For flow through s &, apply eroulli s equatio betwee & kowig that P P atm., v v ~0, hp 0 h L but Z Z h L 4 fl v g v substitute i h v 4 f Lv g L v get v v 4 Q Example : Three s are itercoected. The s characteristics are as follows: (i.) L(ft) F(Moody) ft P 0 ft 50 ft Fid the rate at which water will flow i each fid also the pressure at poit P. all legths are much greater tha 000 diameter, therefore mior losses may be eglected. Give: Required: Q, Q, Q

3 Solutio: pply eroulli s betwee & through kowig that P P atmospheric press., v v ~0, hp 0 Z Z 50 h L h L 0.0x000xv 50 g(6 /) apply cotiuity equatio, Q 6v lso, h 80v 4 flv g Q 6v L Q h c 5.6v 4 f Lv g 64v Solvig equatios (), () & () v 7.78 ft/s Q.5 ft /s v 5.6 ft/s Q 0.49 ft /s v 5.88 ft/s Q.0 ft /s l 0.04x4000xv g(8 /) c...()...()...() s a check, Q Q Q To fid pressure at poit P, apply eroulli s betwee ad P through or. eroulli s through ; kowig that P gage 0, v ~ 0 P p 4.75 ft s a check, apply eroulli s equatio betwee & P through P p 4.75 ft

4 6. ischarge from Oe Tak to Two or more Taks Example : Three reservoirs are coected as show: - take f faig for all s - L, km, cm ? - Q Q Q Q 4- Z 5 m & Z.5 m Required: Solutio: pply eroulli s betwee & through - 4 flv.5 g but v v 4 flv g put v v...() i () Q 0.9 m /s Q Q 0.45 m /s pply eroulli s betwee & through - 5 get 4 flv g 0.4 m 4 flv g...()

5 7 NETWORKS OF PIPES Itercoected s through which flow to a give outlet may come from differet circuits are called etworks of s. Problems o these require trial solutios The followig coditios must be satisfied i a etwork of s: F G E - The pressure drop betwee ay two poits i the circuit, for example & G must be the same whether through the -G or through FEG. i.e., ΣhL 0; hl KQ ; sum of all head losses must equal zero - Flow ito each juctio must be equal flow out of juctio (cotiuity equatio) - arcy s equatio or ay equivalet expoetial frictio formula must be satisfied for each. Hardy-ross method is a method i which successive approximatios are utilized. For ay Q o is a assumed iitial discharge. The expoetial equatio may be arcy's or h f rq ; r RL/ m ; L is legth of ad is iside diameter i SI uits. R 0.675/ ; R is resistace coefficiet,.85 ad m

6 f( roughess), values are listed i the table below: 40 Extremely smooth, straight, asbestos- cemet 0 Very smooth, cocrete, ew cast iro 0 Wood stave, ew welded steel 0 Vitrified clay, ew riveted steel 00 ast iro (.I.) after years of use 95 riveted steel after years of use Old s i bad coditios Steps of solutio: ) ssume best distributio of flow that satisfies the cotiuity ) ompute h f i each ) ompute for each circuit 4) Evaluate 5) ompute Q revised Q ΔQ 6) Repeat steps ) 5) begiig with revised Q util desired accuracy is obtaied. (ΔQ<± ) ΔQ is applied to each i a circuit ΔQ is added to Q i the clockwise flow ΔQ is subtracted i the aticlockwise flow

7 Example: Fid the best distributio of the give discharges at juctios,, ad. 0 r ΔQ r r r ΔQ 00 0 r4 0 For the left side circuit ΔQ ca be estimated as follows: x70 xx70 x5 xx5-4x0 x4x0 Sum ΔQ - For the right side circuit ΔQ ca be estimated as follows: 5x5 x5x5 -x5 xx5 -x5 xx5 Sum ΔQ Hece the corrected flows i the idividual s are: Pipe orrected flow 70(-)57 5(-)-57 0-(-)

8 Repeatig the same steps to calculate ΔQ ad ΔQ : x57 xx57 x7 xx7-4x4 x4x4 Sum ΔQ - 5x0 x5x0 -x0 xx0 -x7 xx7 Sum 8 94 ΔQ - Hece the corrected flows i the idividual s are: Pipe orrected flow (-) 4-4 0(-)7 0-(-) Repeatig the same steps to calculate ΔQ ad ΔQ : x58 xx58 x xx -4x4 x4x4 Sum 60 - ΔQ 0 5x7 x5x7 -x xx -x xx Sum

9 ΔQ So, the last corrected flows ca be take as the correct distributio of the discharges i the etwork. REFERENES est wishes - bd El-ziz, H. Prof. of hemical Eg. airo Uiversity. "Lectures i Fluid Mechaics" - bo El-Hasa, M. Prof. of hemical Eg. airo Uiversity. "Lectures i Fluid Mechaics" - Geakoplis,. J., Trasport Processes ad Uit Operatios. d editio. lly ad aco, Ic., Streeter, V.L. ad Wylie, E.. Fluid Mechaics. rd editio. New York: McGraw-Hill,990