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$\begingroup$ On topic point: This Q discusses the same point as this one $\endgroup$– Cathartic EncephalopathyCommented Mar 27, 2022 at 5:18
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$\begingroup$ Why did my original method fail? $\endgroup$– Cathartic EncephalopathyCommented Mar 27, 2022 at 12:17
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$\begingroup$ You need to consider the change in kinetic energy for the special case of: a horizontal pipe with no friction loss and negligible inlet velocity, with a pump that only increases pressure and the increase in pressure is constant regardless of the flow rate. $\endgroup$– John DarbyCommented Mar 27, 2022 at 19:59
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$\begingroup$ For the special case summarized in my above comment, pump power is ${1 \over 2} \rho A V^3$. I address this special case after a general discussion. It took me a while to see the special conditions where this is the pump power. So I agree with other responders, but I want to provide the more general approach, then apply it to this special case, to emphasize this special result is not in general the pump work. Sorry it took me a while to understand the special case. Answer updated. $\endgroup$– John DarbyCommented Mar 27, 2022 at 20:01
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$\begingroup$ Hi, we've noticed that you have made a large number of minor edits to this post. Please be mindful that every edit bumps the post in the "active" tab of the site and try to make your edits substantial. If you foresee improving this post repeatedly, maybe collect several edits and make them in one go instead of submitting them individually. $\endgroup$– Chris ♦Commented Mar 28, 2022 at 3:12
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