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As a simplification, you can consider that you have a 2D viscous flow between two boundaries that approach each other. Assuming that the flow is symmetrical about the line (with the line along the Y direction), you can simplify this further to "no flow at x=0". What you are left with is a pressure distribution $p(x,t)$ whose integral in $x$ should equal the ...


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Yes it is possible to have water coming out be colder than the water going in. Imagine a big pipe connecting a big container of ice water to a big container of boiling water. Then one end starts out colder than the other. Now raise the container of boiling water higher up so that water starts to flow from the boiling water to the freezing water. Heat ...


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Take a 1-cm square tube and place it vertically in the container from top to bottom, touching the bottom so that the bottom of the container is the bottom of the tube. The pressure at the bottom of the tube is nothing but the weight of water it is supporting - the water in the tube. Supporting means to keep from falling. (Forget the air pressure - that's ...


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If you remove the bottom you no longer have a container. The pressure is atmospheric and you just have gravity at work.


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By the way, it's customary to use primes or some other indicator inside the integral to be slightly less confusing. I.e. $$W(t) = \int_{t_0}^{t} P(t')Q(t')dt'$$ Be careful about how you think about Q(t). You've described it as a "flow rate", but really, you are saying $Q(t) = \frac{dV}{dt}$, or the rate of change of the volume with time. Now, the term ...



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