Question

The 3 tanks shown in the figure are connected to each other by pipes with a friction factor of 0.0187. Calculate the speed of water in each pipe according to the table and picture below.

can bernoulli energy equation be used in this question?

z+9 z+3 (2) - (1) 2-5 D(mm) L (cm) (3) 400 2500 Z 1 pipe 2 pipe 3 pipe 500 3918 650 5670 х

Answer 1

Applying Bernoulli's equation at different different points we can solve this problem as below-
Sol :- The data provided in the question are as below - Pipe ! 400 mm = 0.4 m; L= 2.5 mi Pire 2 :- D = 0.5 m ;L=91918 m. sol:- The data Provided in the question are as below Pipe I :- D = 0.4 ; L= 25 m. Pipe 2 :- D = 0.5 m ; L = 39.18 m. Pipe 3:- D = 0.65 m; L = 56.7 m. Friction factor , f = 0.0187 (for all Pipes) We have to find out -a)Speed of water in each Pipe Bernoulli's equation be used in this questions 6) can Z + 9 B (2) #243 D (1) C X 2-5 (3) b) we can use Bernoull's energy equation in this Problem. The basic equations used for solving flow through branched pipes question are

1. continuity equation . 2. Bernoulli's lu 3. Darcy Weisbach equation a ) دیبا . flow is steady, Now, applying Bernoullies s energy equation by w reservoir A and function Assumptions Po ZA + gg along the streamline. : fluid is Incompressi- or Z+g f L₂ V2² ble. 2 g D₂ Minor losses are 0-0187x39.18 V2 = Z.+ 9 negrected 2x 9.81xors D +20) = % 2 fed = Zt 0.074685612? 0 applying Bernoulli's equation and reservoir B b/w junction we get

Po sg + az tha, = ZB Po or + Z + 3 fs on 0.0187 X 25% V, 2 2x9-81% 0.4 = E + 3 0.059569 V, 2 2 applying Bemouli's equation b/w junction ® f reservoir c we get- Pp eg +30) = Zc t - hla of bf Or (°2 + 0.0187 456.7 XV² zo = Z-5 t 2 x 9.8140.65 =Z-5 + 0-0831404Vz? 3 from f Zt9 0.0746856522 = Z +3 -0.059569V, 2 0.0595691,2 + 6 = 0.074685612? 1.2538 V2 100.723 or, va 4

from 4 3 = :2+3 -0.059569 V,? Z-sto-0831404 3² or, V² = 96.2228 - 0.71657,2 = 96.2228-0.7165( 112538 V2?–100.723) [from 0.89835 V2² - applying continuity equation - Q3 - 927 Qi or, 0.65 x V 0.5% ₂ + 0.4²V = 158 39 5 NOCJ) = or V₃ - 0.5917 V2 +0.3787V, 6 or, √168.39-0089835V2² = 0.5917V2 +0:3787 X 1.2.538 V₂ -100.723 or, 6.5917V; (168.3 8.39-0-8983512) 0.7574 X 51168.39-0.8983512") (1 25386,1 1001723) 0.179812 V₂² - 14:4450 + as onsole or, (168639-0-8983512") (1-2538V2 - 100.723) 1.0686 5V2 - 153.94512 0.7574 = v.-1531915 19

or (1.411 V3? - 203, 254) = 211,127 V-112635V24 1.6960.75 +90.4845v, 1 2 301.6115V2 or, 1.99112"573583V2 + 41312:188 - 1112635124 16960.75 4 2 or, 3.11735 V2 875.1995 V. +58272.94 = 0, 4. 2 Or, V 280.75V2 + 18693:10 = 0 2 280.75 + J 280.75?- 4x 18693-10 - 2 2 280.75 + 63.625 = 2 - 172,1875 ) 108.5625

then + If V = 108.5625 V2 = 10.42 m/sec 5.951 m/sec ; V3=81419 m/sec- * If "= 172.1875 then vz = 13:122 meseca isVi = 10:731 m/sec 5 from V3 = 3.702 m/sec from 11.83 m/sec. Two Different values at the same time is not w possible so, correct Velocities in the Pipes are Vi - 5.951 m/sec = 10.42 m/sec. 8.419 m/sec Ans V₂ = These three velocities satisfies all eqh s 46 Ans