The problem is
How much work is done by pressure in forcing $1.4 m^3$ of
water through a pipe having an internal diameter of $13 mm$ if the
difference in pressure at the two ends of the pipe is $1.0 atm$?
I've seen the following answer:
$$ W = Fd = \left(\frac{F}{A}\right)Ad = PV $$
where $P$ is the pressure difference. Then plug and chug.
However, this last jump, from identifying $F/A$ with pressure at any point in a given area, to now identifying it with a pressure difference, makes no sense to me. Can anyone explain or suggest an alternate understanding of the problem?
Best Answer
$F$ is the sum of two forces - the inward force at the start and the smaller inward force at the end. Since they act in opposite directions, it's also the difference in the magnitudes of the forces which directly corresponds to the difference in pressures. Eg if there's 5.0 atm at the start and 4.0 atm at the end, then $$ F = F_{start} + F_{end} $$ $$ =({5.0 \space atm \times A}) + - ({4.0 \space atm \times A}) $$ $$ = \Delta P \times A $$ $$ = 1.0 \space atm \times A $$
In the special case where the water is released into a vacuum, then $F_{end} = 0$ and you don't need to use any differences.