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|dw:1324391218466:dw| Ok you have this problem
quick analysis reveals that time is independent on the density rho, so thats one down
We can translate the units into units/m2 -rho*v*g => -rho*h*g m*g => rho*g 1/2*rho*v^2 = p*g + p*g*h So that's initially, but in the end h=0. We are gonna view this as a linear problem (not 100% true, but ok. A better assumption would be to take an integral over it). So at the end the formula is: 1/2*rho*v^2 = p*g To get the average velocity, solve this problem. Because you know the volume of the total box (A1*h) and now know the liquid flow (in m/s which multiplied by A2 becomes volume/s)
The average velocity can be found by combining both formulas: rho*v^2 = 2*p*g + p*g*h
Wher I say p, I mean rho ;P ... I mix em up sometimes
after we solve for v, what can we do to obtain ttime?
You translate V (m/s) into cuubs per second (by multiplying with outlet surface A2). Now you have a box with a fixed volume, that decrease with that rate over time
I don't usually work with inviscid flow, so maybe your teacher wants a more complicated solution, idk that.. this gives at least a relatively accurate solution
im stuck on the very first problem :(
Always draw before you start your analysis... It makes the problem a lot easier.. I think my solution is the one your teacher is searching for... But you might want to open your book to see if it's done there differently.
it was done quite differently :( http://ocw.mit.edu/courses/physics/8-01sc-physics-i-classical-mechanics-fall-2010/units-and-dimensional-analysis/MIT8_01SC_problems01_soln.pdf
They've made more assumptions than I did. Haha, dimensional analysis, I've had that a long while back. I don't think I'll be to good at that. I only use it as a check, not as a way to work ;P
Yeah I don't even understand how they could pull out those assumptions.