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wrote on 5/29/2017 7:43 PM:
On Monday, May 29, 2017 at 7:23:55 PM UTC-4, wrote:
On Monday, May 29, 2017 at 5:37:34 PM UTC-4, wrote:
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

First some centrifugal pumps do not have involutes.

Most do, but anything that will accelerate the flow to the periphery will do.

Second are you thinking there could be air in the pump? THa woud be bad as you would have cavitation.

Not necessarily. The air could come from partial filling, which almost certainly is the case at start-up. Once the pump is running, apparently the involute spaces fill.

The pump would be completely filled. How could it not be completely filled?

Easy. Low input pressure; no back pressure; vanes sling the small amount of water to the periphery, with no resistance.

And why would the velocity have to be the same from input port to the periphery?

The velocity doesn't have to be the same. But if it's higher at the periphery, the involute space doesn't fill. If it's higher at the inlet, you have a possible energy-conservation dilemma, unless the volume in each space from center to periphery increases as to the square of distance from the center. I haven't done the math on that but it's easy to tell.

This is the outward radial velocity we're talking about here. The tangential velocity increases by definition.

The flow would be the same, but the passages vary in cross section, so there is no way the velocity could be the same.

Again, it's a matter of which "velocity" you're talking about -- radial or tangential.

When you are dealing with centrifugal pump, it is easier to think
"angular velocity" (ω) rather than "linear velocity" (v).

Please let me repeat myself from my other post:

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.