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Power of water pump

Consider a water pump pushing water through a pipe:

We wish to find the energy which the water pump has to expend to keep the water flowing through the pipe. Writing Newton's second law for water flowing out of a cylindrical cross section of it,

$$ F= v \frac{dm}{dt}$$

Now, $$ P = F \cdot v$$

$$ P= v \frac{dm}{dt} v$$

$$ \frac{dm}{dt} = \rho A v$$

$$ P= v^3 \rho A $$

Now the real answer has a factor of half, where in this derivation have I gone wrong?

Answer

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$\begingroup$ niels, that's a cute Darth Vader comment. I'll try to remember it. $\endgroup$
– David White
Commented Jun 10, 2020 at 19:44

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