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wrote on 5/29/2017 7:19 PM:
On Monday, May 29, 2017 at 6:21:59 PM UTC-4, Jim Wilkins wrote:
wrote in message
...
On Monday, May 29, 2017 at 5:13:41 PM UTC-4, wrote:
On Monday, May 29, 2017 at 3:34:50 PM UTC-4,
wrote:
sw

Right. So what is the condition inside of one involute in the
second stage? Is it completely full when it's operating?

--
Ed Huntress

It is exactly the same as the first stage.

Dan

And what is that condition? Are the involutes completely filled?
And, if so, how is that possible unless the velocity is the same
from the input port to the periphery of the wheel?

--
Ed Huntress

How did you ever dream up that requirement?

Because if they are not completely filled, there is no physical way to transit any positive pressure at the inlet to the outlet. My further reading suggests they are filled, although some illustrations show them partly filled. Photos taken through transparent windows show them partly filled, but those are illustrations of cavitation. I'm reaching the conclusion that they're completely filled in normal operation.

It isn't a positive
displacement pump. If the outlet valve is closed the input and output
velocities will be zero, yet the periphery is still spinning. The
pressure differential (p) from inlet to outlet will be what the
discharge curve shows for zero flow (q).
-jsw

If there is no flow, there will be no pressure differential. Pressure will be the same throughout the volume of liquid from inlet to outlet.

You need to use your brain to think, Ed.

When the outlet valve is shut off, the impeller is spinning the fluid as
a disk inside the housing.

The impeller is a solid piece (balanced weight) so it does not have the
tendency to fly away from the center when it is spinning at high speed.
The fluid in the disk will try to fly away from the center due to
centrifugal force. The fluid will exert highest pressure on the wall at
the outermost part of the housing while there is no pressure at the
center of the rotating disk.

If the outlet valve is opened at this time, the fluid will be expelled,
and the center of the rotating disk will form a negative pressure to
suck in fluid from the intake valve.

So, if there is no flow, there will definitely be pressure differential
if the impeller is spinning the fluid at high speed.

Another way to prove that there will be pressure differential is by
inspecting the equation for centrifugal force:

https://en.wikipedia.org/wiki/Centrifugal_force#Force

Centrifugal force of a rotating object is proportional to its radius
from the center of rotation. It is obvious that centrifugal force is
zero at the center, and maximum at the farthest point from the center.

centrifugal force = m × ω × ω × r
http://keisan.casio.com/keisan/lib/real/system/2006/1271292951/%B1%F3%BF%B4%CE%CF.gif

You should use angular velocity (ω) instead of linear velocity (v)
because all the fluid molecules in the disk spin at the same angular
velocity regardless of its distance from the center.