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

AI Thread Summary
Head loss in parallel systems cannot be directly added like resistors unless the head loss is identical across all legs. The relationship for head loss in parallel systems is similar to that of resistors, where the total head loss can be calculated using the formula 1/h = 1/h1 + 1/h2 + 1/h3. If the head loss varies for each valve/coil, additional information about the resistance or head loss per unit of flow for each leg is necessary for accurate calculations. Understanding the differences in head loss is crucial for determining the overall system performance. Proper analysis of each component's head loss is essential for effective system design.
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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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