- #1
maverick280857
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Hi everyone
I have a relatively simple fluid dynamics problem I need some help with:
We have two cylinders L and R of crossectional areas A and 2A respectively. Initially, the level of water in X is H and Y is empty. At t = 0, the two cylinders are joined at the bottom by a tube of cross-section a (after a hole of the same cross-section is opened in each cylinder). Find the time at which the water level is equal in both cylinders.
This is what I've done so far.
Denote the water levels in the left (L) and right (R) cylinders by y_{L} and y_{R} respectively. So y_{L}(t=0) = H and y_{R}(t=0) = 0.
Mass conservation (or continuity equation) leads to AH = Ay_{L} + 2Ay_{R} or equivalently H = y_{L} + 2y_{R}. This gives 0 = \dot{y_{L}} + 2\dot{y_{R}}.
Applying Bernoulli's Theorem to two points at the surface of each meniscus, we get
P_{atm} + \frac{1}{2}{\rho v_{1}^2} + \rho g h_{1} = P_{atm} + \frac{1}{2}{\rho v_{2}^2} + \rho g h_{2}
where P_{atm} is the atmospheric pressure, \rho is the density of water, v_{1} = -\dot{y_{L}}, v_{2} = \dot{y_{R}}, h_{1} = y_{L}, h_{2} = y_{R}.
Hence,
\frac{1}{2}{\rho \dot{y_{L}}^2} + \rho g y_{L} = \frac{1}{2}{\rho \dot{y_{R}}^2} + \rho g y_{R}
After simplifying a bit, this gives a differential equation in y_{R} with the boundary conditions y_{R}(t=0) = 0 and y_{R}(t = T) = \frac{H}{3} where H/3 is the equilibrium height of water level (in each cylinder--this follows from the mass conservation equation above) and T is the time when this happens.
Now, if you try this out you get an imaginary (and therefore ridiculous) solution for y_R. As far as I think, my algebra is okay so there must be a conceptual fault somewhere. I would be very grateful if someone could offer some advice.
PLEASE NOTE: This is not a homework problem but I couldn't think of a better place to post it on PF.
Thanks and cheers,
Vivek
EDIT: The answer should involve a, the cross-section of the orifice, but it doesn't in my case.
I have a relatively simple fluid dynamics problem I need some help with:
We have two cylinders L and R of crossectional areas A and 2A respectively. Initially, the level of water in X is H and Y is empty. At t = 0, the two cylinders are joined at the bottom by a tube of cross-section a (after a hole of the same cross-section is opened in each cylinder). Find the time at which the water level is equal in both cylinders.
This is what I've done so far.
Denote the water levels in the left (L) and right (R) cylinders by y_{L} and y_{R} respectively. So y_{L}(t=0) = H and y_{R}(t=0) = 0.
Mass conservation (or continuity equation) leads to AH = Ay_{L} + 2Ay_{R} or equivalently H = y_{L} + 2y_{R}. This gives 0 = \dot{y_{L}} + 2\dot{y_{R}}.
Applying Bernoulli's Theorem to two points at the surface of each meniscus, we get
P_{atm} + \frac{1}{2}{\rho v_{1}^2} + \rho g h_{1} = P_{atm} + \frac{1}{2}{\rho v_{2}^2} + \rho g h_{2}
where P_{atm} is the atmospheric pressure, \rho is the density of water, v_{1} = -\dot{y_{L}}, v_{2} = \dot{y_{R}}, h_{1} = y_{L}, h_{2} = y_{R}.
Hence,
\frac{1}{2}{\rho \dot{y_{L}}^2} + \rho g y_{L} = \frac{1}{2}{\rho \dot{y_{R}}^2} + \rho g y_{R}
After simplifying a bit, this gives a differential equation in y_{R} with the boundary conditions y_{R}(t=0) = 0 and y_{R}(t = T) = \frac{H}{3} where H/3 is the equilibrium height of water level (in each cylinder--this follows from the mass conservation equation above) and T is the time when this happens.
Now, if you try this out you get an imaginary (and therefore ridiculous) solution for y_R. As far as I think, my algebra is okay so there must be a conceptual fault somewhere. I would be very grateful if someone could offer some advice.
PLEASE NOTE: This is not a homework problem but I couldn't think of a better place to post it on PF.
Thanks and cheers,
Vivek
EDIT: The answer should involve a, the cross-section of the orifice, but it doesn't in my case.
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