well wait is this what u meant...
since total volume has to equal .027 m^3 i can write equations in terms of x and y as
.027 = (.02)x + (.01)y
then use pressures equal so that
rho(g)x +Ma(g)/Aa = Rho(g)y + Mb(g)/Ab
from these two equations solve?
I still not sure how this will work, But if i understand correctly...
Make tank A height x and Tank B height Y. Then Tank B = y = x - .3. is that what your talking about?
then Pressure tank A = x and Pressure tank B = y per say then x = y + difference between two pressures due to the pistons...
Once valve is opened the two tanks should go to equilibrium pressure wise. Intuitively the volume from tank a would rush into tank B till equilibrium is reached.
Yes, that area would be the area of the respective cylinders and the pistons form upper bound of the tank so they add to pressure. I think it might help to say that the bases of each cylinder are on the same level if that helps.
before the tank valve is open the following equations determine...
Two tanks filled with oil with density 860 kg/m3 are connected by a valve. Tank A has piston with 2.2 kg mass, cross sectional area of .02 m^3 and is suspended 1 m from the base of the tank. Tank B has a pistion mass of 1 kg, with cross sectional area .01 m^2 and is suspended .7m. when the...