# Question No.5

#### Topic : Pipe Network Analysis (Hardy-Cross Method)

In the system shown in the following figure, a pump is installed in BC to boost the flow to C. Neglecting minor losses, determine the flow distribution and head elevations at the junctions of the pump delivers a head of 15 m. All pipes have the same roughness size of 0.03 mm. The outflows at the junctions are shown in l/sec.

Data :

 Pipe AB BC CD DE EA BE Length (m) 500 600 200 600 600 200 Diameter (mm) 200 150 100 150 200 100

Pressure head elevation at A = 60 m

Given :

Steps of solution :

Step (1) - Identify loops. When using hand calculations, the simplest way is to employ adjacent loops.

Loop 1 : ABEA

Loop 2 : BCDEB

The inlet flow at junction A is equal to the sum of the outlet flow at the other junctions, therefore the inlet flow at junction A is equal to :

= 60 + 50 + 50 + 40 = 200 l/sec

Step (2) - Allocate estimated flows in the pipes. Only one estimated flow in each loop is required; the remaining flows follow automatically from the continuity condition at the nodes.

Assume that water is flowing in pipe AB with a flowrate of 120 lit/sec (from junction A to junction B) and flowrate in pipe BC is 40 lit/sec (from junction B to junction C).

Therefore, the initial estimates of flowrates through the network are :

 Pipe Q (l/sec) AB 120 BC 40 CD -10 DE -50 EA -80 BE +20/-20

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 :

$Re=vDν$

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 K for each pipe (using Darcy-Weisbach equation) :

$K=8fLπ2gD5$

To calculate the correction required in the discharge value :

$ΔQ=−ΣKQQ2ΣKQ$

For the 1st iteration :

For the 2nd iteration :

For the 3rd iteration :

For the 4th iteration :

For the 5th iteration :

For the 6th iteration :

Final discharges (after 6 iterations) :

 Pipe Q (l/sec) Head loss (m) AB 106.373 21.405 BC 52.558 13.337 CD 2.558 0.258 DE 37.442 14.876 EA 93.627 20.123 BE 6.185 1.282

To calculate the pressure head elevations at different junctions, use the energy equation between point A and different points ,total energy at any point can be calculated through :

$T.E=Z+Pγ+v22g$

Since the minor losses can be neglected, the velocity head term is negligible (equal to zero), therefore the total energy at point A is equal to :

$T.EA=ZA+PAγ+vA22g$ $→$ $T.EA=60+0=60m$

To calculate the pressure head elevation at point B, apply Bernoulli's equation between points A & B :

$T.EA=T.EB+hlA→B$

$T.EB=ZB+PBγ+vB22g$ $→$ $T.EB=ZB+PBγ+0$

$60=ZB+PBγ+21.405$ $→$ $ZB+PBγ=38.595m$

Similarly, the pressure head elevations at different points is summarized in the following table :

 Junction T.E (m) A 60 B 38.595 C 25.258 D 25.001 E 39.877

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