Can head loss be added in parallel like resistors in circuits?

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SUMMARY

The discussion centers on the addition of head loss in parallel valve/coil systems, specifically questioning if it can be treated similarly to resistors in electrical circuits. The consensus is that head loss can be combined in parallel only if the head loss is identical across all systems. The relevant equation for this scenario is 1/h = 1/h1 + 1/h2 + 1/h3. If the head loss varies among the systems, additional information about the resistance of each leg is necessary for accurate calculations.

PREREQUISITES
  • Understanding of fluid dynamics principles, specifically head loss.
  • Familiarity with electrical circuit theory, particularly the behavior of resistors in parallel.
  • Knowledge of hydraulic resistance calculations.
  • Ability to interpret and analyze system diagrams.
NEXT STEPS
  • Research "fluid dynamics head loss calculations" for a deeper understanding of head loss in systems.
  • Study "hydraulic resistance in parallel systems" to learn how to analyze multiple pathways.
  • Explore "equivalent resistance in electrical circuits" to draw parallels between fluid and electrical systems.
  • Investigate "flow rate and pressure drop relationships" to understand how they affect head loss.
USEFUL FOR

Engineers, fluid mechanics students, and anyone involved in the design or analysis of hydraulic systems will benefit from this discussion.

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Homework Statement


I have three valve/coil systems connected in parallel as shown in the attached figure. If I know the head loss of each valve/coil is 40 feet, can I add them in parallel like you do with resistors in circuits?
Thanks!

Homework Equations



1/h=1/h1+1/h2+1/h3 (possibly)

The Attempt at a Solution

 
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You forgot the diagram.

Not my field but I believe you are incorrect. Isn't head drop equivalent to "voltage drop"? In which case consider what happens when you connect three resistors in parallel... the current flowing in each changes until they end up sharing a common voltage drop (head drop).

Only _if_ the head drop was the same on each leg could you easily add them in parallel. In that I believe case you get h = h1 = h2 = h3. This is equivalent to adding three identical resistors in parallel.

If the head loss is different for each leg... Do you know the resistance of each leg ? eg the head loss per unit of flow?
 

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