HARDY CROSS METHOD IN OPEN NETWORKS

 HARDY CROSS METHOD IN OPEN NETWORKS

Example: Determine the flow rates and head loss in each of the sections in the figure.

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Solution:

Step 1:Assume flow rates that go through each pipe complying with the continuity equation. Let's say that the proposed flow rate for section A-D is 26.5 l/s, for section B-D a flow rate of 12 l/s and for section C-D a flow rate of 12 l/s.

Step 2: We find the head losses due to friction with the Hazen and Williams equation.

hf=1741*L/D^4.87*Q^n/C^n   where : n=1.85

R=1741*L/(D^4.87*C^n )

hf=R*Q^n

* First let's calculate R

FIRST PSEUDOCIRCUIT:

For section A-D:

RA-D=1741*4000/(10^4.87*120^1.85 )=0.013377

For section D-B:

RD-B=1741*6000/(8^4.87*120^1.85 )=0.059486

SECOND PSEUDOCIRCUIT:

For section D-C:

RD-C=1741*5000/(6^4.87*120^1.85 )=0.201227

For section D-B: (Common section)

    RD-B=1741*6000/(8^4.87*120^1.85 )=0.059486

*Once R has been calculated we will calculate hf:

FIRST PSEUDOCIRCUIT:

For section A-D:

    hfA-D=-0.013377*26.5^1.85=-5.7462

For section D-B: (Common section)

hfD-B=-0.059486*12.5^1.85=-5.9007

    ZA=100 m          ZB = 91 m  

    ▲Z(A-B)=100-91=9 m

SECOND PSEUDOCIRCUIT:

For section D-C:

hfD-C=-0.013377*26.5^1.85=-19.9605

For section D-B: (Common section)

 hfD-B=-0.059486*12.5^1.85=5.9007

    ▲Z(B-C)=91-80=11 m

Step 3:We find the flow correction for each pseudocircuit.

FIRST PSEUDOCIRCUIT:

∆ Q = -( -11.6468443) + ( 100 - 91 ) / (1.85 x 0.7085595)

∆ Q = 2.01920451388774 l/s

SECOND PSEUDOCIRCUIT:

∆ Q = - ( -14.0598592) + ( 91 - 80 )  / (1.85 x 2.155101)

∆ Q = 0.767471203280844 l/s

QA-D =-24.4807954861123 l/s

QD-B = -10.7482666893931 l/s

QD-C = -11.2325287967192 l/s

QD-B = 10.7482666893931 l/s

Step 4: We do the necessary iterations until the correction is <= 10^-4, the more iterations there will be more precision.
Doing 8 iterations we finally have the result:

Flow rates :

QA-D =-23.2913684395083 l/s

QD-B =-10.3335081505777 l/s

QD-C = -10.4578602889306 l/s

QD-B = 10.3335081505777 l/s

Load losses:

hfA-D =-4.52568390261812 m

hfD-B =-4.47481293121938 m

hfD-C = -15.4759075534977 m

hfD-B = 4.47481293121938 m

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