Solving the Hardy Cross Problem

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Homework Help Overview

The discussion revolves around the Hardy Cross method, specifically addressing the relationship between flow rate (Q) and head loss (hf) in a fluid dynamics context. Participants are examining the implications of the equation hf = k(Q^2) and questioning the signs of the values involved.

Discussion Character

  • Conceptual clarification, Assumption checking

Approaches and Questions Raised

  • The original poster attempts to understand why hf, derived from the equation hf = k(Q^2), should be positive given that Q has a power of 2. They explore the implications of negative values in their calculations and question the sign of the product (554)(-40^2).
  • Another participant questions the reasoning behind the assertion that (554)(-40^2) would yield a negative result.

Discussion Status

The discussion is ongoing, with participants seeking clarification on the signs of the values in the equations. There is an exploration of different interpretations regarding the calculations, but no consensus has been reached yet.

Contextual Notes

Participants are navigating the complexities of the Hardy Cross method and its assumptions, particularly regarding the signs of flow rates and head loss. The original poster's inquiry suggests a potential misunderstanding of the mathematical relationships involved.

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Homework Statement


in this problem , we know that Q (flow rate ) has power of 2 , hf = k(Q^2) , shouldn't the hf = k(Q^2) has positive value ?

Homework Equations

The Attempt at a Solution


since (554)(-40^2) will become positive , same thing goes to hf / Q , when the hf = positive , the hf / Q will become hf / Q divided by Q(has negative value ) , so the hf / Q will have negative value , am i right ?
 

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why will (554)(-40^2) will become negative ?
 
anybody can answer ?
 
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