If the flow truly is laminar, and there are no changes in elevation (so the density doesn't matter, assuming constant volume flow rate), the pressure drop is going to be proportional to the viscosity. But, aside from polymer melts and solutions, I highly doubt that the flow will be laminar. If...
This is not quite correct, because the friction factor is going to depend on the Reynolds number, which is inversely proportional to the fluid viscosity.
If the expansion is reversible (quasi-static), you can use the ideal gas law to get the pressure of the gas at the piston face where the work is being done (and it is equal to the external pressure of the piston face on the gas). If the expansion is irreversible, an ideal gas does not satisfy...
No problem. Just make sure that the values match at the upper right corner.
No problem. The boundary condition is just that the tangential component of stream function is constant. You don't need to set the value there. The differential equation becomes $$\frac{\partial^2 \psi}{\partial...
You can set those boundaries in motion tangentially in any way you want, and it still is not going to cause any fluid flow within the region. Discontinuities in tangential velocity are perfectly comparable with inviscid flow.
I'm not sure I understand what you are saying. The flow is into the region across the left boundary and out of the region across the right boundary? There is no flow across the upper and lower boundaries? Across the left and right boundaries the velocity in the x-direction is a quadratic...
I have some ideas on how you can proceed to get your CN program to work properly.
Phase 1: To test out your spatial finite difference scheme, solve the problem using forward Euler (rather than CN ) with a small enough step size that the solution remains stable.
Phase 2: Use the CN equation...
I am not going to discuss Problem 1 because I am opposed to using differentials for irreversible processes.
For problem 2, I would add that, if the entropy of state B is higher than state A, then, if you could use an adiabatic irreversible path to go from state B to state A, the entropy would...
At equilibrium, the partial pressure of water vapor in the air is equal to the equilibrium vapor pressure. This is because, in an ideal gas mixture, each gas behaves as if the other gas is not present.
All the water would flow from the upper reservoir to the lower reservoir on the right. There would be no flow from 2 to 1 unless a huge flow were forced by the pump. It would have to provide a pressure of at least 10 psi.
https://www.engineeringtoolbox.com/steel-pipe-schedule-40-friction-loss-diagram-d_1145.html
6 ft of water is about 2.6. psi. 2.6 psi over at 50' pipe is about 0.05 psi/ft. That would result in an initial flow rate of about 7 gpm for a steel pipe.
If the upward motion of the plate is $$h=A\cos{\omega t}$$, then the upward acceleration of all the fluid is $$-A\omega^2\cos{\omega t}$$. So, from the moving frame of reference, the equivalent upward gravity would be $$-g+A\omega^2\cos{\omega t}$$ and the equivalent downward gravity would be...
If you didn't make any mistakes, it would work. Try a calculation in which you use only 21 grid points and choose delta t such that it makes alpha close to 1.