snipped-for-privacy@gmail.com wrote on 5/29/2017 5:11 PM:
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"Turbomachinery, in mechanical engineering, describes machines that transfer energy between a rotor and a fluid"
You are confusing yourself by reading too much and overloading your brain, again.
The rotor in a turbine functions much like an electric fan or a ship's propeller. It is designed to propel fluid in a direction perpendicular to its plane of rotation.
On the other hand, the rotor (impeller) in a centrifugal pump is throwing fluid radially outward parallel to its plane of rotation.
They are different in the principle they work. That's why one is called a "turbo pump", the other is called "centrifugal pump", Ed.
I'm not overloaded. I'm trying to figure out the fluid dynamics within the involutes of a pump.
No shit?
If you're comfortable with your understanding of it, that's fine. I'm not c omfortable with it. Something in the explanations is being oversimplified t o the point where they make no sense.
They're both turbomachinery. In the context in which I used the term, it's correct. I was quite careful about how I used it.
How did you ever dream up that requirement? 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).
On May 29, 2017, snipped-for-privacy@gmail.com wrote (in article):
Model isn?t quite right - water is incompessible, which means that the volume of a package of water is constant - mass does not change either. And Bernoulli?s equation is a direct consequence of the conservation of energy applied to the flow of an incompressible fluid.
Pumps in series do work, but the stages must be correctly matched for the cascade to be effective. A good example is an axial-flow turbine - each fan disk stage gives the passing fluid a kick (increases its velocity). If this flow is impeded, the pressure will rise, until the stagnation (max) pressure is reached.
If you look at the performance curve for any fan, it will be max flow at zero head, and max head at zero flow, and max delivered aerodynamic power somewhere between.
I?d dig up a college intro to physics textbook and read the chapter about Bernoulli?s equation. This will clarify the issue.
snipped-for-privacy@gmail.com wrote on 5/29/2017 6:14 PM:
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the involutes of a pump.
ot comfortable with it. Something in the explanations is being oversimpli fied to the point where they make no sense.
t's correct. I was quite careful about how I used it.
You are hopelessly thick.
Please look at this impeller of a centrifugal water pump. The fins are acting like pedals, not blades. It is designed to throw fluid outward, instead of pushing fluid forward.
Would you do us a favour? Please beat your head against the wall 10 times to clear your mind and then think again before you get back to us.
Because if they are not completely filled, there is no physical way to tran sit any positive pressure at the inlet to the outlet. My further reading su ggests they are filled, although some illustrations show them partly filled . Photos taken through transparent windows show them partly filled, but tho se are illustrations of cavitation. I'm reaching the conclusion that they'r e completely filled in normal operation.
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.
First some centrifugal pumps do not have involutes.
Second are you thinking there could be air in the pump? THa woud be bad as you would have cavitation.
The pump would be completely filled. How could it not be completely filled? And why would the velocity have to be the same from input port to the periphery?
The flow would be the same, but the passages vary in cross section, so there is no way the velocity could be the same.
If you have a column of water, the pressure at the bottom of the column in higher than the pressure at the top. So there is an example of no flow with a pressure differential.
If you take something like say a large nut and tie a string to it and whirl it around your head, then you have a force on the string and a velocity of the nut. But the nut does not go flying off , unless you let go of the string.
In the same way with the outlet blocked off on a centrifugal pump. the water has a velocity , but it just goes around and around. Produces a pressure, but no flow from the pump.
Think of a centrifuge with test tubes in it. The material in the test tubes are subjected to force, but there is no flow.
d stage? Is it completely full when it's operating?
f so, how is that possible unless the velocity is the same from the input p ort to the periphery of the wheel?
Most do, but anything that will accelerate the flow to the periphery will d o.
as you would have cavitation.
Not necessarily. The air could come from partial filling, which almost cert ainly is the case at start-up. Once the pump is running, apparently the inv olute spaces fill.
Easy. Low input pressure; no back pressure; vanes sling the small amount of water to the periphery, with no resistance.
The velocity doesn't have to be the same. But if it's higher at the periphe ry, 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 fro m center to periphery increases as to the square of distance from the cente r. 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 tangentia l velocity increases by definition.
ere is no way the velocity could be the same.
Again, it's a matter of which "velocity" you're talking about -- radial or tangential.
Oh, right. I was thinking of externally applied pressure. Gravity or centrifugal force would result in a differential.
Any pressure applied from the inlet, as in the case of a multi-stage centrifugal pump, would be the same throughout the volume within one involute. But the centrifugal force added by the spinning rotor would be greatest at the periphery.
Again, the velocity is the tangential velocity. There is no radial velocity.
transit any positive pressure at the inlet to the outlet. My further read ing suggests they are filled, although some illustrations show them partl y filled. Photos taken through transparent windows show them partly fille d, but those are illustrations of cavitation. I'm reaching the conclusion that they're completely filled in normal operation.
ill 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:
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.
You should use angular velocity (?) 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.
d stage? Is it completely full when it's operating?
if so, how is that possible unless the velocity is the same from the inp ut port to the periphery of the wheel?
ad as you would have cavitation.
certainly is the case at start-up. Once the pump is running, apparently t he involute spaces fill.
t of water to the periphery, with no resistance.
iphery, 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 s pace 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.
ntial velocity increases by definition.
there is no way the velocity could be the same.
or tangential.
When you are dealing with centrifugal pump, it is easier to think "angular velocity" (?) 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:
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.
You should use angular velocity (?) 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.
In 16 years you still haven't figured out how to make water drain into the ground. Yet now you think there must be someone who'll take advice from you about fluid dynamics, based on some of your wildly exaggerated "experience." LOL
That's close. A pump on atmospheric air (1 bar) that makes a 10 bar output , will be different, because the density of inlet air determines the force on the motor to spin it up. Feeding in 10bar air, you have higher density , and the motor torque and work done will rise (because an AC motor operates at constant speed). Water, being nearly incompressible, doesn't create any such scaling issues with inlet pressure.
s threw me. Since water isn't compressible, I don't see how the multi-stage pumps work. For gas, no problem, but I don't get it for liquids.
For liquids, the multistage pumps I've used have pistons, and the low-press ure pistons stall out when backpressure is high, then the same motor moves smaller high
-pressure pistons to achieve higher fluid pressures with lower volume displacement (b ut nearly equal motor power). I suspect there's a differential gear driving all the pistons with different mechanical advantage, and a reverse-inhibiting clutch on the low pressure pistons.
ines, the outlet of the first stage is fed into the axis of the second stag e. The pressure from the first-stage outlet is retained at the second-stage inlet, but from there it feeds into the whirling blades of the second stag e, the outlet volume of which is LARGER than the inlet volume between any t wo blades.
Why? Several stages of centrifugal pump in series for a liquid would have same volume flow, and same work per shaft would deliver the same head, at each stage. So, t hey'd be the same size rotors, too. Maybe different shaft seals, though.
mps threw me. Since water isn't compressible, I don't see how the multi-sta ge pumps work. For gas, no problem, but I don't get it for liquids.
ssure pistons
he pistons with
w pressure pistons.
rbines, the outlet of the first stage is fed into the axis of the second st age. The pressure from the first-stage outlet is retained at the second-sta ge inlet, but from there it feeds into the whirling blades of the second st age, the outlet volume of which is LARGER than the inlet volume between any two blades.
same volume flow,
they'd be the same
The issue I was trying to resolve was the question of whether the pump(s) i n series were, each, completely filled. Illustrations often show centrifuga l pumps as being less than completely filled. But I realized that they have to be completely filled, especially for a multi-stage pump.
If those spaces aren't completely filled, the outlet pressure from one stag e, which would be the inlet pressure of the next stage, wouldn't communicat e to the output of that following stage. It would be like pushing on the en d of a rope.
All my posts in this thread should be a good read. The impeller of a centrifugal pump forms compartments inside the tight fitting casing to trap incoming fluid so that when the impeller spins, it is forcing the fluid in all the compartments to go in a circular motion. The centrifugal force from the rotating fluid generates the pressure radially outward against the outer wall of the pump.
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