Hardy Cross method correction problem

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SUMMARY

The discussion centers on the Hardy Cross method for correcting flow rates in a network of loops, specifically addressing a junction where two loops converge. The user queries the flow rates (Q) in the left and right loops, questioning the values of 21 clockwise and 21 anticlockwise, respectively. The user proposes alternative calculations of 12 for the left loop and -19 for the right loop, indicating a misunderstanding of the correction process. The Hardy Cross method requires iterative adjustments to achieve flow balance, which the user is attempting to grasp.

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  • Basic knowledge of fluid dynamics and flow rate calculations
  • Familiarity with loop analysis in hydraulic systems
  • Ability to perform iterative calculations for convergence
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Engineering students, hydraulic engineers, and professionals involved in fluid dynamics who are looking to understand the Hardy Cross method for flow corrections in network systems.

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Homework Statement


I have problem with the flow rate (Q) in junction that join 2 loops together , where k = 4 ,

Why in the second correction , the corrected Q in first and second loop = 21 clcockwise for left loop , and for 21 anticlockiwise for right loop ?

Homework Equations

The Attempt at a Solution


IMO, it should be 10+2 = 12 for the left loop , and for the right loop , it should be -10-9= -19 , am I right ?[/B]
 

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