Aim is to transmit water from the source to the end point. Option is to use grav

Question

Aim is to transmit water from the source to the end point. Option is to use gravity pipeline system (may use tools i.e. pump, air valve, pressure reducer, etc if required) Problem The following system transmits water from Reservoir A to Reservoir B by means of a gravity pipeline where, L 2000m, f 0.02 and D 0.8m for all pipes. This pipe line has two high points at C and E, where small-orifice air valves are installed. Assume that the valves require 5m of line pressure head for proper operation. Checking the design specifications [(0.5m/s

Explanation / Answer

SOLUTION

Where

f =0.02

L =2000m

D = 0.8m

Z1 = 200m

Z2 = 170m

Applying Bernoulli’s equation between the Reservoir A and B.

PA/g + VA2/2g + z1 = PB/g + VB2/2g + z2 +hL

0 + 0 + 200 = 0 + VB22/19.62 + 170 +hL

(200 - 170) = VB2/19.62+hL

0.0509 VB2 +hL = 30m Eq. (1)

hf = f ( L/D ) ( V2/2g)

hL = hf,A + hf,B + hf,C+hf,D+hf,E+hf,F

hf,A = 0.02 ( 2000/0.8) ( VA2/2g ) = 2.55VA2

hf,A = hf,B=hf,C=hf,D=hf,E=hf,F

hL = 2.55VA2 + 2.55VB2  +2.55VC2 +2.55VD2 +2.55VE2 +2.55VF2

Substitute the above with Equ 1

0.0509 VB2 +hL = 30m Eq. (1)

2.55VA2 + 2.55VB2  +2.55VC2 +2.55VD2 +2.55VE2 +2.55VF2 = 30m Eq. (2)

we can find other relation between VA,VB,VC,VD,VE and VF

Continuity Equation

Qa + Qb +Qc +Qd +Qe +Qf = 0

/4 × 0.8VA + /4 × 0.8VB + /4 × 0.8VC + /4 × 0.8VD + /4 × 0.8VE + /4 × 0.8VF =0 - Equ 3

VB = VD = VF=VE

VA = VC

3 x (0.785*VA2) = 4  (0.785*VB2) - Equ 4

From above VB= VD=VE=VF=0.75 VA

substitute in Equ 3

/4 × 0.8VA + /4 × 0.8*0.75VA + /4 × 0.8VC + /4 × 0.8*0.75VA+ /4 × 0.8*0.75VA+ /4 × 0.8*0.75VA =0

/4 × 0.8VA + /4 × 0.8VC = -1.884 VA

2.512VA = -0.785 VC

VC = 3.2VA

Apply in Equ 2

2.55VA2 + (2.55*0.75VA2)*4  +2.55(3.2VA)2 = 30m

VA = 1.3 m/s

VC = 3.2 *1.3 = 4.16 m/s

VB= VD=VE=VF=0.75 *1.3 = 0.975 m/s

QA = /4 × 0.8*1.3 = 0.82m3/s

QB = /4 × 0.8*0.975 =  0.61m3/s

QC = /4 × 0.8*4.16 =  2.62m3/s

QD = /4 × 0.8*0.975 =  0.61m3/s

QE = /4 × 0.8*0.975 =  0.61m3/s

QF = /4 × 0.8*0.975 =  0.61m3/s

The Maximum Discharge of System = 2.62m3/s

HEAD LOSS

0.0509 (0.975)2 +hL = 30

hL = 29.95

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