Hardy Cross method correction problem

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The discussion centers on a Hardy Cross method problem involving flow rates at a junction connecting two loops. The user is confused about the corrected flow rates, questioning why both loops have a flow rate of 21, with one clockwise and the other counterclockwise. They believe the calculations should yield different values, specifically 12 for the left loop and -19 for the right loop. The conversation highlights the importance of understanding the correction process in the Hardy Cross method. Clarification on the flow rate corrections is sought to resolve this discrepancy.
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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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