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Gyroscope
Jan31-07, 01:03 PM
1. The problem statement, all variables and given/known data
This is my creation.
I have a reservoir at an height h and another one on the floor. I connect them through a pipe with constant diameter. If the water start flowing from the higher reservoir to the lowest with will flow with constant velocity. This is true, isn't it?
By Bernoulli, I can say that potential energy per volume is being converted in pressure energy as the water flows down. Am I right?

denverdoc
Jan31-07, 01:07 PM
1. The problem statement, all variables and given/known data
This is my creation.
I have a reservoir at an height h and another one on the floor. I connect them through a pipe with constant diameter. If the water start flowing from the higher reservoir to the lowest with will flow with constant velocity. This is true, isn't it?
By Bernoulli, I can say that potential energy per volume is being converted in pressure energy as the water flows down. Am I right?

What are the relevant equations? Forget the second tank for a second, assume the tank is simply allowed to drain, is the rate of drainage constant? Hint: Imagine it nearly full and then nearly empty.

John

Gyroscope
Jan31-07, 01:09 PM
No. As the tank drains, the velocity the water leaves the tank is decreasing.

I created this problem. The relevant equations is just Bernoulli's and continuity.

Gyroscope
Jan31-07, 01:40 PM
But imagine that the higher tank is being filled with water so that the level remains constant. In this case, the velocity of the water flowing through the pipe is constant in space and time. Right?

denverdoc
Jan31-07, 01:56 PM
I suppose, it depends on the configuration, if the tubing connecting the two is under water at any point then the flow will change as a function of the growing backpressure.

Gyroscope
Jan31-07, 01:57 PM
It is not underwater. So am I right?

denverdoc
Jan31-07, 02:05 PM
p+1/2rhoV^2 + rho*g*h=constant. But this neglects the frictional losses in the pipe. Not clear what is being asked really at this point? Sorry I can't be of more assistance,
John