# Question No.5

#### Topic : Solving Pipe Networks using Quantity Balance Method

Determine the flows in the network illustrated in the below figure; minor losses are given by Cm v2/2g.

Data :

 Pipe Length (m) Diameter (mm) Roughness (mm) Minor Loss Coefficient Cm AB 20,000 500 0.3 20 BC 5,000 350 0.3 10 BD1 6,000 300 0.3 10 BD2 6,000 250 0.06 10

Given :

Steps of solution :

Step (1) - Estimate ZB (pressure head elevation at B) = 80 m (Note that the elevation of the pipe junction itself does not affect the solution.)

Step (2) - Allocate estimated flow velocities in the pipes.

The friction factor f may be obtained from the Moody diagram, or using Barr's equation, using an initially estimated velocity in each pipe. Subsequently f can be based on the previously calculated discharges. However, unless there is a serious error in the initial velocity estimates, much effort is saved by retaining the initial f values until perhaps the final correction.

Step (3) – Evaluate the head loss coefficient K for each pipe. The friction coefficient f is obtained through Moody chart or using Barr's equation. The calculations proceed in tabular form. Note that Q is written in liter/sec simply for convenience; all computations are based on Q in m3/sec.

To calculate the velocity at each pipe :

$v=QA=Qπ4D2$

To calculate Reynolds number :

$Re=vDν,ν=1.011∗10−6m2/sec$

To calculate the friction coefficient, Barr's equation is used :

$1f=−2logk/D3.7+5.1286Re0.89$

Barr's equation is based partly on an approximation to the logarithmic smooth turbulent element in the Colebrook-White equation.

To calculate friction head loss coefficient Kf for each pipe  (using Darcy-Weisbach equation) :

$Kf=8fLπ2gD5$

To calculate minor head loss coefficient Km for each pipe (representing minor losses) :

$→$ $Km=Cm2gA2$

To calculate head loss coefficient K for each pipe (representing both friction & minor losses) :

$K=Kf+Km=8fLπ2gD5+Cm2gA2$

To calculate the correction required in ZB value :

$ΔZB=2(ΣQIB−FB)ΣQIBhL,IB$

For this problem, no outflow occurs at junction B, therefore :

$FB=0$

Therefore :

$ΔZB=2ΣQIBΣQIBhL,IB$

For the first iteration :

Initial estimate for velocity is 2 m/sec for all pipes !

Correction to ZB :

For the second iteration :

The initial estimates for velocity is updated according to the previously calculated value !

Correction to ZB :

For the third iteration :

The initial estimates for velocity is updated according to the previously calculated value !

Correction to ZB :

For the fourth iteration :

Correction to ZB :

Final Discharges in the given pipe network :

Final discharges (after 4 iterations) :

 Pipe Q (l/sec) Flow Direction AB 156.513 A to B BC 56.518 B to C BD1 58.979 B to D BD2 41.016 B to D