If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

On Friday, May 26, 2017 at 2:18:45 PM UTC-4, wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

I think it depends on how much the flow is restricted. If you try to measure at the output of the pump and the water is just going into the air right at the output of the pump, then the pressure would be zero.

If you block the output of the pump I think it would be 90. So I think you could get zero to 90 , depending on how constricted the output is.

This is my guess.

Dan

On Friday, May 26, 2017 at 2:48:20 PM UTC-4, wrote:
On Friday, May 26, 2017 at 2:18:45 PM UTC-4, wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

I think it depends on how much the flow is restricted. If you try to measure at the output of the pump and the water is just going into the air right at the output of the pump, then the pressure would be zero.

If you block the output of the pump I think it would be 90. So I think you could get zero to 90 , depending on how constricted the output is.

This is my guess.

Dan

My guess is "no." The pump creates pressure from slinging a *mass* of water. I don't think it matters what the pressure is entering the pump.

--
Ed Huntress

On Friday, May 26, 2017 at 3:39:34 PM UTC-4, wrote:
On Friday, May 26, 2017 at 2:48:20 PM UTC-4, wrote:
On Friday, May 26, 2017 at 2:18:45 PM UTC-4, wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

I think it depends on how much the flow is restricted. If you try to measure at the output of the pump and the water is just going into the air right at the output of the pump, then the pressure would be zero.

If you block the output of the pump I think it would be 90. So I think you could get zero to 90 , depending on how constricted the output is.

This is my guess.

Dan

My guess is "no." The pump creates pressure from slinging a *mass* of water. I don't think it matters what the pressure is entering the pump.

--
Ed Huntress

To clarify that, I think that a typical 10 psi centrifugal pump that has 80 psi at the inlet will have something like 80 psi at the outlet.

--
Ed huntress

wrote in message
...
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

http://www.engineersedge.com/pumps/multi_stage_pump.htm

Check Harbor Freight for a pump that boosts water pressure.

Hul

wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

On Fri, 26 May 2017 22:10:04 +0000 (UTC), Hul Tytus
wrote:

Check Harbor Freight for a pump that boosts water pressure.

Hul

wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric
I guess i could do that and see what kind of pump it is.

On 05/26/2017 01:25 PM, wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?

It would, but unless that pump is designed as a pressure-boosting pump, the likelihood of the shaft seal being able to contain such pressures is not high. On a cheap pump, even the housing might fail.

--
Bob Nichols AT comcast.net I am "RNichols42"

wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

This is a common question in the fire service. Now if the pump is rated
for a higher pressure and flow it could boost the pressure. That is how
a 2 stage high rise pump operates. Those may pull in 90 psi hydrant
water through a 5", feed that through the first stage and boost it to
125 psi. That is then fed into the second stage which can boost it to a
higher pressure. Our old two stage could put out 300 psi through a 4"
line. with 70 psi at the 5" inlet. BUT if the inlet pressure dropped the
outlet dropped much farther.

Currently our biggest pump can lift water 20' and put out 2,075 gpm at
150 psi and keep pumping that way until she runs out of fuel. That's
with 2 6" suction lines

--
Steve W.

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

--
Ed Huntress


This is a common question in the fire service. Now if the pump is rated
for a higher pressure and flow it could boost the pressure. That is how
a 2 stage high rise pump operates. Those may pull in 90 psi hydrant
water through a 5", feed that through the first stage and boost it to
125 psi. That is then fed into the second stage which can boost it to a
higher pressure. Our old two stage could put out 300 psi through a 4"
line. with 70 psi at the 5" inlet. BUT if the inlet pressure dropped the
outlet dropped much farther.

Currently our biggest pump can lift water 20' and put out 2,075 gpm at
150 psi and keep pumping that way until she runs out of fuel. That's
with 2 6" suction lines

--
Steve W.

wrote in message
...
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out
of
the pump be 90 psi? I think the pressure will be 90 psi. Am I
wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out
will be
at most 80 psi. if the pump is designed to produce 10 psi. It may
be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can
produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add
any
pressure because it cannot pump faster than the water is already
flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage
pumps 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.

--
Ed Huntress

That reference showed it could be done, but not how. I was hoping it
would lead someone who knows more than I do about fluid dynamics to a
better description.

On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

Consider connecting the pump to the bottom of a 180 foot tall (approx
80 psi head) tank. Will the pump be capable of pumping, at the
specified flow, to a height of 23 feet (10 psi head) or 203 feet?

Practical matters of seal design aside.



--
Ned Simmons

On Sat, 27 May 2017 18:57:07 -0400, Ned Simmons
wrote:

On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

Consider connecting the pump to the bottom of a 180 foot tall (approx
80 psi head) tank. Will the pump be capable of pumping, at the
specified flow, to a height of 23 feet (10 psi head) or 203 feet?

Practical matters of seal design aside.
Thanks Ned and everyone else who posted. I am going to put in a hot
water recirculating setup in my house. I was going to buy the pump,
valves, and pressure switch but I got lucky when a friend gave me a
new Grundfos pump that was slated for in floor heating. I was pretty
sure the pump would work but the Grundfos web site didn't mention what
type of hot water circulating system it would work for.
Eric

wrote on 5/26/2017 4:18 PM:
On Friday, May 26, 2017 at 3:39:34 PM UTC-4, wrote:
On Friday, May 26, 2017 at 2:48:20 PM UTC-4, wrote:
On Friday, May 26, 2017 at 2:18:45 PM UTC-4, wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

I think it depends on how much the flow is restricted. If you try to measure at the output of the pump and the water is just going into the air right at the output of the pump, then the pressure would be zero.

If you block the output of the pump I think it would be 90. So I think you could get zero to 90 , depending on how constricted the output is.

This is my guess.

Dan

My guess is "no." The pump creates pressure from slinging a *mass* of water. I don't think it matters what the pressure is entering the pump.

--
Ed Huntress

To clarify that, I think that a typical 10 psi centrifugal pump that has 80 psi at the inlet will have something like 80 psi at the outlet.



Why do you think there is something called "two-stage" or "dual-stage"
air compressor?

The second stage will use the output of the first stage as its intake to
add more pressure to it.

wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law. It's much harder to imagine it with liquids.

--
Ed Huntress

On Saturday, May 27, 2017 at 8:17:29 PM UTC-4, rIOdEš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›UCAXF wrote:
wrote on 5/26/2017 4:18 PM:
On Friday, May 26, 2017 at 3:39:34 PM UTC-4, wrote:
On Friday, May 26, 2017 at 2:48:20 PM UTC-4, wrote:
On Friday, May 26, 2017 at 2:18:45 PM UTC-4, wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

I think it depends on how much the flow is restricted. If you try to measure at the output of the pump and the water is just going into the air right at the output of the pump, then the pressure would be zero.

If you block the output of the pump I think it would be 90. So I think you could get zero to 90 , depending on how constricted the output is.

This is my guess.

Dan

My guess is "no." The pump creates pressure from slinging a *mass* of water. I don't think it matters what the pressure is entering the pump.

--
Ed Huntress

To clarify that, I think that a typical 10 psi centrifugal pump that has 80 psi at the inlet will have something like 80 psi at the outlet.



Why do you think there is something called "two-stage" or "dual-stage"
air compressor?

Because it has two stages.


The second stage will use the output of the first stage as its intake to
add more pressure to it.

With a gas, no problem. With a liquid, the problem is more complex.

--
Ed Huntress

wrote on 5/27/2017 11:08 PM:
On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law. It's much harder to imagine it with liquids.


Water does compress. It takes a lot of pressure to compress water just a
little bit, but that little bit reduction in volume is what gives the
compressed water the "stored pressure".

If you daisy-chain two power-washers, theoretically you should get a
more powerful water jet than using just one power-washer.

wrote on 5/27/2017 11:08 PM:
On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law. It's much harder to imagine it with liquids.


The idea is the same. You use pressure to reduce the volume of a fluid.
The gas/water restores it original volume after losing the pressure.

The graph in the link below shows that 200-bar of pressure (2900 psi)
will compress water by about 0.7% at 4°C (under STP, water is 1g/cc at 4°C):

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

On Sunday, May 28, 2017 at 2:17:58 AM UTC-4, HJARTš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›lBYWJ wrote:
wrote on 5/27/2017 11:08 PM:
On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law.. It's much harder to imagine it with liquids.


The idea is the same. You use pressure to reduce the volume of a fluid.
The gas/water restores it original volume after losing the pressure.

I don't think so. Centrifugal pumps are very lossy machines. They couldn't hold pressure that way.

From a physics point of view, I think the answer lies in sorting out the kinetic aspects of a turbo pump (velocity) and the potential aspects (pressure). A turbine pump that's pumping a liquid must be producing potential energy from kinetic energy.

I'd need to see a good, expert explanation to understand it. I see no way that an ordinary turbine pump can hold the pressure generated by a previous stage, unless the entire thing is kinetic, which we're then measuring as potential energy (pressure).


The graph in the link below shows that 200-bar of pressure (2900 psi)
will compress water by about 0.7% at 4°C (under STP, water is 1g/cc at 4°C):

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

No turbine pump could hold that 0.7% compression.

--
Ed Huntress

wrote on 5/28/2017 9:23 AM:
On Sunday, May 28, 2017 at 2:17:58 AM UTC-4, HJARTš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›lBYWJ wrote:
wrote on 5/27/2017 11:08 PM:
On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law. It's much harder to imagine it with liquids.


The idea is the same. You use pressure to reduce the volume of a fluid.
The gas/water restores it original volume after losing the pressure.

I don't think so. Centrifugal pumps are very lossy machines. They couldn't hold pressure that way.

From a physics point of view, I think the answer lies in sorting out the kinetic aspects of a turbo pump (velocity) and the potential aspects (pressure). A turbine pump that's pumping a liquid must be producing potential energy from kinetic energy.

I'd need to see a good, expert explanation to understand it. I see no way that an ordinary turbine pump can hold the pressure generated by a previous stage, unless the entire thing is kinetic, which we're then measuring as potential energy (pressure).


The graph in the link below shows that 200-bar of pressure (2900 psi)
will compress water by about 0.7% at 4°C (under STP, water is 1g/cc at 4°C):

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

No turbine pump could hold that 0.7% compression.

According to you, turbo fan aircraft engines are also very lossy
machines, but most turbo fan aircraft engines use two-stage air
compression.

https://www.youtube.com/watch?v=_LaKlE2h3Jw

Both water and air are fluids. I am sure you can modify the aircraft
turbo fan engine design to propel a submarine underwater (without using
fuel and combustion chamber, of course).

If a "very lossy" turbo fan engine can use two stages to compress air or
water, then why wouldn't a two-stage centrifugal pump work?

On Sunday, May 28, 2017 at 9:54:50 AM UTC-4, SbzWrš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›AJFeU wrote:
wrote on 5/28/2017 9:23 AM:
On Sunday, May 28, 2017 at 2:17:58 AM UTC-4, HJARTš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›lBYWJ wrote:
wrote on 5/27/2017 11:08 PM:
On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law. It's much harder to imagine it with liquids.


The idea is the same. You use pressure to reduce the volume of a fluid..
The gas/water restores it original volume after losing the pressure.

I don't think so. Centrifugal pumps are very lossy machines. They couldn't hold pressure that way.

From a physics point of view, I think the answer lies in sorting out the kinetic aspects of a turbo pump (velocity) and the potential aspects (pressure). A turbine pump that's pumping a liquid must be producing potential energy from kinetic energy.

I'd need to see a good, expert explanation to understand it. I see no way that an ordinary turbine pump can hold the pressure generated by a previous stage, unless the entire thing is kinetic, which we're then measuring as potential energy (pressure).


The graph in the link below shows that 200-bar of pressure (2900 psi)
will compress water by about 0.7% at 4°C (under STP, water is 1g/cc at 4°C):

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

No turbine pump could hold that 0.7% compression.

According to you, turbo fan aircraft engines are also very lossy
machines, but most turbo fan aircraft engines use two-stage air
compression.

They are very lossy machines. The part that's of interest here is compressor efficiency, which ranges from 0.70 to 0.85 in the best turbo machinery, including stationary and aircraft gas turbines. At the high end, 85% compressor efficiency, they're losing 15% to gas friction.

But again, you're talking about a gas turbine. Gas has a close ratio between pressure and volume (Boyle's law). Compressing gas with a machine, whether it's positive or non-positive displacement, like a turbo compressor, is not a problem. An example of positive displacement types is a vane-type supercharger. An example of non-positive-displacement types is a turbocharger. Or the compressor stage of a gas turbine engine, such as an aircraft jet engine.

Liquids don't behave according to Boyle's law.


https://www.youtube.com/watch?v=_LaKlE2h3Jw

Both water and air are fluids. I am sure you can modify the aircraft
turbo fan engine design to propel a submarine underwater (without using
fuel and combustion chamber, of course).

If a "very lossy" turbo fan engine can use two stages to compress air or
water...

How do you know that a turbo fan engine can compress water?

, then why wouldn't a two-stage centrifugal pump work?

Obviously, it *does* work. The question is "how."

--
Ed Huntress

wrote on 5/28/2017 10:27 AM:
On Sunday, May 28, 2017 at 9:54:50 AM UTC-4, SbzWrš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›AJFeU wrote:
wrote on 5/28/2017 9:23 AM:
On Sunday, May 28, 2017 at 2:17:58 AM UTC-4, HJARTš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›lBYWJ wrote:
wrote on 5/27/2017 11:08 PM:
On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law. It's much harder to imagine it with liquids.


The idea is the same. You use pressure to reduce the volume of a fluid.
The gas/water restores it original volume after losing the pressure.

I don't think so. Centrifugal pumps are very lossy machines. They couldn't hold pressure that way.

From a physics point of view, I think the answer lies in sorting out the kinetic aspects of a turbo pump (velocity) and the potential aspects (pressure). A turbine pump that's pumping a liquid must be producing potential energy from kinetic energy.

I'd need to see a good, expert explanation to understand it. I see no way that an ordinary turbine pump can hold the pressure generated by a previous stage, unless the entire thing is kinetic, which we're then measuring as potential energy (pressure).


The graph in the link below shows that 200-bar of pressure (2900 psi)
will compress water by about 0.7% at 4°C (under STP, water is 1g/cc at 4°C):

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

No turbine pump could hold that 0.7% compression.

According to you, turbo fan aircraft engines are also very lossy
machines, but most turbo fan aircraft engines use two-stage air
compression.

They are very lossy machines. The part that's of interest here is compressor efficiency, which ranges from 0.70 to 0.85 in the best turbo machinery, including stationary and aircraft gas turbines. At the high end, 85% compressor efficiency, they're losing 15% to gas friction.

But again, you're talking about a gas turbine. Gas has a close ratio between pressure and volume (Boyle's law). Compressing gas with a machine, whether it's positive or non-positive displacement, like a turbo compressor, is not a problem. An example of positive displacement types is a vane-type supercharger. An example of non-positive-displacement types is a turbocharger. Or the compressor stage of a gas turbine engine, such as an aircraft jet engine.

Liquids don't behave according to Boyle's law.


https://www.youtube.com/watch?v=_LaKlE2h3Jw

Both water and air are fluids. I am sure you can modify the aircraft
turbo fan engine design to propel a submarine underwater (without using
fuel and combustion chamber, of course).

If a "very lossy" turbo fan engine can use two stages to compress air or
water...

How do you know that a turbo fan engine can compress water?

, then why wouldn't a two-stage centrifugal pump work?

Obviously, it *does* work. The question is "how."


If you daisy-chain two centrifugal pumps together, I am sure you will
get a lot higher pressure output than using just one (too optimistic to
expect 2 times the pressure).

On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. 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 stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Pressure, thus, is converted to velocity. Unless the machine *compounds* the velocity at each stage, I don't see how it works. And, in order to compound velocity by a factor of, say, three, either the shaft driving the stage either has to be turning at (square root of 3) times that of the first stage, or the the second stage has to have a completely different scroll design..

But you can carry that only so far. Go to three stages, or four, and the shaft rotational speeds become outrageous, or the scroll design does.

Obviously, I'm missing something here, but I haven't yet seen what it is.

Ed Huntress


Consider connecting the pump to the bottom of a 180 foot tall (approx
80 psi head) tank. Will the pump be capable of pumping, at the
specified flow, to a height of 23 feet (10 psi head) or 203 feet?

Practical matters of seal design aside.



--
Ned Simmons

wrote in message
...
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT),
wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi
is
supplied with water at 80 psi will the water pressure coming
out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I
wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out
will be
at most 80 psi. if the pump is designed to produce 10 psi. It may
be
lower depending on the size of the housing and the restriction
the
impeller creates. Say your input side is 2" and the pump can
produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't
add any
pressure because it cannot pump faster than the water is already
flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage
pumps 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.

The pumps in series business is confusing the issue. The output of
the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as
long
as the flow is constant for both inlet conditions, the delta P
across
the pump will be the same. In other words, the pump will increase
the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular centrifugal
turbines, the outlet of the first stage is fed into the axis of the
second stage. 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 stage, the outlet volume of which is LARGER than
the inlet volume between any two blades.

Pressure, thus, is converted to velocity. Unless the machine
*compounds* the velocity at each stage, I don't see how it works. And,
in order to compound velocity by a factor of, say, three, either the
shaft driving the stage either has to be turning at (square root of 3)
times that of the first stage, or the the second stage has to have a
completely different scroll design.

But you can carry that only so far. Go to three stages, or four, and
the shaft rotational speeds become outrageous, or the scroll design
does.

Obviously, I'm missing something here, but I haven't yet seen what it
is.

Ed Huntress


Consider connecting the pump to the bottom of a 180 foot tall
(approx
80 psi head) tank. Will the pump be capable of pumping, at the
specified flow, to a height of 23 feet (10 psi head) or 203 feet?

Practical matters of seal design aside.
--
Ned Simmons

===========================
http://net.grundfos.com/doc/webnet/m...p-handbook.pdf
The pressure of a 10 meter head of water is close to one atmosphere.
Section 3.2 describes pumps in series. Fig 3.2.5 and 3.2.6 show how
the pressures add when the pumps are of equal or mismatched sizes.
-jsw

wrote on 5/28/2017 11:35 AM:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. 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 stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Pressure, thus, is converted to velocity. Unless the machine *compounds* the velocity at each stage, I don't see how it works. And, in order to compound velocity by a factor of, say, three, either the shaft driving the stage either has to be turning at (square root of 3) times that of the first stage, or the the second stage has to have a completely different scroll design.

But you can carry that only so far. Go to three stages, or four, and the shaft rotational speeds become outrageous, or the scroll design does.

Obviously, I'm missing something here, but I haven't yet seen what it is.

Ed Huntress


A centrifuge pump creates pressure by spinning fluid away from the
center (hence "centrifuge").

You are confusing yourself by of processing too much information at one
time. To make it easy for you, let's consider the output valve is closed
(the system is creating pressure but not expelling anything).

The first pump supplies the pressurized fluid to the second pump, the
second pump spins the pressurized fluid away from the center to add more
pressure to the housing wall. The pressure gauge in the second stage
should register more pressure than the first stage.

Does this help you understand better now, Ed?

On Sunday, May 28, 2017 at 12:43:23 PM UTC-4, lvCCLš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›IvVyG wrote:
wrote on 5/28/2017 11:35 AM:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. 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 stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Pressure, thus, is converted to velocity. Unless the machine *compounds* the velocity at each stage, I don't see how it works. And, in order to compound velocity by a factor of, say, three, either the shaft driving the stage either has to be turning at (square root of 3) times that of the first stage, or the the second stage has to have a completely different scroll design.

But you can carry that only so far. Go to three stages, or four, and the shaft rotational speeds become outrageous, or the scroll design does.

Obviously, I'm missing something here, but I haven't yet seen what it is.

Ed Huntress


A centrifuge pump creates pressure by spinning fluid away from the
center (hence "centrifuge").

You are confusing yourself by of processing too much information at one
time. To make it easy for you, let's consider the output valve is closed
(the system is creating pressure but not expelling anything).

The first pump supplies the pressurized fluid to the second pump, the
second pump spins the pressurized fluid away from the center to add more
pressure to the housing wall. The pressure gauge in the second stage
should register more pressure than the first stage.

Does this help you understand better now, Ed?

That's well said, but how is the fluid pressurized once it's fed into the second stage?

Take a look at the pump impellers in these photos:

https://www.google.com/search?q=pump...1 920&bih=974

(or, Tiny URL):

https://tinyurl.com/y9trhytf

Remember that the inlet is at the center of the impeller. The liquid then makes a 90-deg. turn and enters the involutes. As the liquid travels from the center to the periphery, the volume *increases*. Poof! There goes your pressure out the window.

But I think I have it figured out now. The problem starts with the concept of "pressure." Except when you're dealing with gases, that's always problematic. The physical measures that are involved here actually are mass, velocity, and force. Forget about pressure for a moment. Think energy instead.

If the previous stage can supply enough liquid to fill the subsequent involute more than it would be filled without that previous stage, then that subsequent stage increases the energy of the water by increasing its velocity. "Pressure" is irrelevant. The energy going in is the product of mass and force (forget the actual formula for now). The energy coming out is the same thing, but along the way, an increase in velocity has increased the force. Restrict that mass and force at the pump exit, and you get pressure.

So that's what I think is happening.

--
Ed Huntress

On Sunday, May 28, 2017 at 11:35:39 AM UTC-4, wrote:


This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. 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 stage, the outlet volume of which is LARGER than the inlet volume between any two blades.



Obviously, I'm missing something here, but I haven't yet seen what it is.

Ed Huntress

You are over thinking the situation. A multistage pump has a bunch of identical sections all turning at the same speed. Each stage increases the pressure. So you might have a 6 stage pump with each stage increasing the pressure by 10 psi. Which makes for a fairly efficient pump which will supply water at 60 psi. Google it.

In Eric's case it is a little different. He wants to circulate water. The system is pressurized to 80 psi and the pump has 80 psi on the input. The output is at 90 psi into a heating loop. and the friction of the water flow reduces the pressure so that at one end of the loop you have 90 psi and at the other end you have 80 psi. The pumps are not in series.

Dan


Consider connecting the pump to the bottom of a 180 foot tall (approx
80 psi head) tank. Will the pump be capable of pumping, at the
specified flow, to a height of 23 feet (10 psi head) or 203 feet?

Practical matters of seal design aside.



--
Ned Simmons

wrote in message
...
On Sunday, May 28, 2017 at 12:43:23 PM UTC-4, lvCCL?? Mighty ? Wannabe
??IvVyG wrote:
wrote on 5/28/2017 11:35 AM:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT),
wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi
is
supplied with water at 80 psi will the water pressure coming
out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I
wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out
will be
at most 80 psi. if the pump is designed to produce 10 psi. It
may be
lower depending on the size of the housing and the restriction
the
impeller creates. Say your input side is 2" and the pump can
produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't
add any
pressure because it cannot pump faster than the water is
already flowing
through it.

That's exactly what I thought, but Jim's reference to
multi-stage pumps 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.

The pumps in series business is confusing the issue. The output
of the
first (centrifugal) pump in the chain is far from an ideal
pressure
source. The original question was about a pump with a constant
inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as
long
as the flow is constant for both inlet conditions, the delta P
across
the pump will be the same. In other words, the pump will increase
the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular
centrifugal turbines, the outlet of the first stage is fed into
the axis of the second stage. 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 stage, the outlet
volume of which is LARGER than the inlet volume between any two
blades.

Pressure, thus, is converted to velocity. Unless the machine
*compounds* the velocity at each stage, I don't see how it works.
And, in order to compound velocity by a factor of, say, three,
either the shaft driving the stage either has to be turning at
(square root of 3) times that of the first stage, or the the
second stage has to have a completely different scroll design.

But you can carry that only so far. Go to three stages, or four,
and the shaft rotational speeds become outrageous, or the scroll
design does.

Obviously, I'm missing something here, but I haven't yet seen what
it is.

Ed Huntress


A centrifuge pump creates pressure by spinning fluid away from the
center (hence "centrifuge").

You are confusing yourself by of processing too much information at
one
time. To make it easy for you, let's consider the output valve is
closed
(the system is creating pressure but not expelling anything).

The first pump supplies the pressurized fluid to the second pump,
the
second pump spins the pressurized fluid away from the center to add
more
pressure to the housing wall. The pressure gauge in the second stage
should register more pressure than the first stage.

Does this help you understand better now, Ed?

That's well said, but how is the fluid pressurized once it's fed into
the second stage?

Take a look at the pump impellers in these photos:

https://www.google.com/search?q=pump...1 920&bih=974

(or, Tiny URL):

https://tinyurl.com/y9trhytf

Remember that the inlet is at the center of the impeller. The liquid
then makes a 90-deg. turn and enters the involutes. As the liquid
travels from the center to the periphery, the volume *increases*.
Poof! There goes your pressure out the window.

But I think I have it figured out now. The problem starts with the
concept of "pressure." Except when you're dealing with gases, that's
always problematic. The physical measures that are involved here
actually are mass, velocity, and force. Forget about pressure for a
moment. Think energy instead.

If the previous stage can supply enough liquid to fill the subsequent
involute more than it would be filled without that previous stage,
then that subsequent stage increases the energy of the water by
increasing its velocity. "Pressure" is irrelevant. The energy going in
is the product of mass and force (forget the actual formula for now).
The energy coming out is the same thing, but along the way, an
increase in velocity has increased the force. Restrict that mass and
force at the pump exit, and you get pressure.

So that's what I think is happening.

--
Ed Huntress

If you can understand the steam injector then centrifugal pumps are
easy. Both transform velocity into pressure.
https://en.wikipedia.org/wiki/Injector

-jsw

On Sunday, May 28, 2017 at 5:44:40 PM UTC-4, Jim Wilkins wrote:
wrote in message
...
On Sunday, May 28, 2017 at 12:43:23 PM UTC-4, lvCCL?? Mighty ? Wannabe
??IvVyG wrote:
wrote on 5/28/2017 11:35 AM:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT),
wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi
is
supplied with water at 80 psi will the water pressure coming
out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I
wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out
will be
at most 80 psi. if the pump is designed to produce 10 psi. It
may be
lower depending on the size of the housing and the restriction
the
impeller creates. Say your input side is 2" and the pump can
produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't
add any
pressure because it cannot pump faster than the water is
already flowing
through it.

That's exactly what I thought, but Jim's reference to
multi-stage pumps 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.

The pumps in series business is confusing the issue. The output
of the
first (centrifugal) pump in the chain is far from an ideal
pressure
source. The original question was about a pump with a constant
inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as
long
as the flow is constant for both inlet conditions, the delta P
across
the pump will be the same. In other words, the pump will increase
the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular
centrifugal turbines, the outlet of the first stage is fed into
the axis of the second stage. 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 stage, the outlet
volume of which is LARGER than the inlet volume between any two
blades.

Pressure, thus, is converted to velocity. Unless the machine
*compounds* the velocity at each stage, I don't see how it works.
And, in order to compound velocity by a factor of, say, three,
either the shaft driving the stage either has to be turning at
(square root of 3) times that of the first stage, or the the
second stage has to have a completely different scroll design.

But you can carry that only so far. Go to three stages, or four,
and the shaft rotational speeds become outrageous, or the scroll
design does.

Obviously, I'm missing something here, but I haven't yet seen what
it is.

Ed Huntress


A centrifuge pump creates pressure by spinning fluid away from the
center (hence "centrifuge").

You are confusing yourself by of processing too much information at
one
time. To make it easy for you, let's consider the output valve is
closed
(the system is creating pressure but not expelling anything).

The first pump supplies the pressurized fluid to the second pump,
the
second pump spins the pressurized fluid away from the center to add
more
pressure to the housing wall. The pressure gauge in the second stage
should register more pressure than the first stage.

Does this help you understand better now, Ed?

That's well said, but how is the fluid pressurized once it's fed into
the second stage?

Take a look at the pump impellers in these photos:

https://www.google.com/search?q=pump...1 920&bih=974

(or, Tiny URL):

https://tinyurl.com/y9trhytf

Remember that the inlet is at the center of the impeller. The liquid
then makes a 90-deg. turn and enters the involutes. As the liquid
travels from the center to the periphery, the volume *increases*.
Poof! There goes your pressure out the window.

But I think I have it figured out now. The problem starts with the
concept of "pressure." Except when you're dealing with gases, that's
always problematic. The physical measures that are involved here
actually are mass, velocity, and force. Forget about pressure for a
moment. Think energy instead.

If the previous stage can supply enough liquid to fill the subsequent
involute more than it would be filled without that previous stage,
then that subsequent stage increases the energy of the water by
increasing its velocity. "Pressure" is irrelevant. The energy going in
is the product of mass and force (forget the actual formula for now).
The energy coming out is the same thing, but along the way, an
increase in velocity has increased the force. Restrict that mass and
force at the pump exit, and you get pressure.

So that's what I think is happening.

--
Ed Huntress

If you can understand the steam injector then centrifugal pumps are
easy. Both transform velocity into pressure.
https://en.wikipedia.org/wiki/Injector

-jsw

Right. I have no problem with steam injectors or with centrifugal pumps. The problem I was having is with compounded centrifugal fluid pumps.

After brushing up on impulse and momentum, and remembering that "pressure" is a frequently misleading term, I'm pretty sure I see what's happening.

--
Ed Huntress

On 5/28/2017 3:59 PM, wrote:
After brushing up on impulse and momentum, and remembering that "pressure" is a frequently misleading term, I'm pretty sure I see what's happening.

-- Ed Huntress

Absolutely, your head is lodged in the toilet trap again, Crazy Eddy.

wrote on 5/28/2017 2:50 PM:
On Sunday, May 28, 2017 at 12:43:23 PM UTC-4, lvCCLš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›IvVyG wrote:
wrote on 5/28/2017 11:35 AM:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. 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 stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Pressure, thus, is converted to velocity. Unless the machine *compounds* the velocity at each stage, I don't see how it works. And, in order to compound velocity by a factor of, say, three, either the shaft driving the stage either has to be turning at (square root of 3) times that of the first stage, or the the second stage has to have a completely different scroll design.

But you can carry that only so far. Go to three stages, or four, and the shaft rotational speeds become outrageous, or the scroll design does.

Obviously, I'm missing something here, but I haven't yet seen what it is.

Ed Huntress


A centrifuge pump creates pressure by spinning fluid away from the
center (hence "centrifuge").

You are confusing yourself by of processing too much information at one
time. To make it easy for you, let's consider the output valve is closed
(the system is creating pressure but not expelling anything).

The first pump supplies the pressurized fluid to the second pump, the
second pump spins the pressurized fluid away from the center to add more
pressure to the housing wall. The pressure gauge in the second stage
should register more pressure than the first stage.

Does this help you understand better now, Ed?

That's well said, but how is the fluid pressurized once it's fed into the second stage?

Take a look at the pump impellers in these photos:

https://www.google.com/search?q=pump...1 920&bih=974

(or, Tiny URL):

https://tinyurl.com/y9trhytf

Remember that the inlet is at the center of the impeller. The liquid then makes a 90-deg. turn and enters the involutes. As the liquid travels from the center to the periphery, the volume *increases*. Poof! There goes your pressure out the window.

But I think I have it figured out now. The problem starts with the concept of "pressure." Except when you're dealing with gases, that's always problematic. The physical measures that are involved here actually are mass, velocity, and force. Forget about pressure for a moment. Think energy instead.

If the previous stage can supply enough liquid to fill the subsequent involute more than it would be filled without that previous stage, then that subsequent stage increases the energy of the water by increasing its velocity. "Pressure" is irrelevant. The energy going in is the product of mass and force (forget the actual formula for now). The energy coming out is the same thing, but along the way, an increase in velocity has increased the force. Restrict that mass and force at the pump exit, and you get pressure.

So that's what I think is happening.


No, that's no what's happening.

You are confusing yourself by making things complicated. You don't need
force/energy equations to figure this out.

A centrifugal pump is a "centrifuge". It is spinning the fluid
(gas/liquid) outward. The fluid is continuously being thrown against the
peripheral wall of the housing which results in fluid pressure against
the retaining wall. It is just like you use your hands to push against
the retaining wall to create pressure on the retaining wall, except that
the fluid is being spun around continuously to create the radially
outward pressure all around in 360 degrees against the retaining wall
that is preventing it from flying out.

Of course, the center of the centrifuge will be a partial vacuum. That
is where the input valve is usually placed so it can suck in fluid if
the pump is properly primed.

On Fri, 26 May 2017 11:25:07 -0700, wrote:

If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

Depends entirely on the particular pump. Some pumps can suck from
atmosphere once primed. Other pumps require a Net Positive Suction
Head (pressure on the inlet). How much pressure the pump will add
depends on the pump characteristics and how much flow you're
demanding. A data sheet on the pump will give you this information.

John
John DeArmond
http://www.neon-john.com
http://www.tnduction.com
Tellico Plains, Occupied TN
See website for email address

On Sunday, May 28, 2017 at 6:06:28 PM UTC-4, tyre biter wrote:
On 5/28/2017 3:59 PM, wrote:
After brushing up on impulse and momentum, and remembering that "pressure" is a frequently misleading term, I'm pretty sure I see what's happening.

-- Ed Huntress

Absolutely, your head is lodged in the toilet trap again, Crazy Eddy.

Poor biter. He must have really been stung when he found out he missed his IP geolocation "target" by over 20 miles. And he even had detailed latitude and longitude values for it, too! ggg

Now he can't get over it. He thought he was a clever troll, however cowardly. Now he's exposed that he's stupid, too.

--
Ed Huntress

On Sunday, May 28, 2017 at 6:14:05 PM UTC-4, izlsbš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›TSwdF wrote:
wrote on 5/28/2017 2:50 PM:
On Sunday, May 28, 2017 at 12:43:23 PM UTC-4, lvCCLš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›IvVyG wrote:
wrote on 5/28/2017 11:35 AM:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. 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 stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Pressure, thus, is converted to velocity. Unless the machine *compounds* the velocity at each stage, I don't see how it works. And, in order to compound velocity by a factor of, say, three, either the shaft driving the stage either has to be turning at (square root of 3) times that of the first stage, or the the second stage has to have a completely different scroll design.

But you can carry that only so far. Go to three stages, or four, and the shaft rotational speeds become outrageous, or the scroll design does.

Obviously, I'm missing something here, but I haven't yet seen what it is.

Ed Huntress


A centrifuge pump creates pressure by spinning fluid away from the
center (hence "centrifuge").

You are confusing yourself by of processing too much information at one
time. To make it easy for you, let's consider the output valve is closed
(the system is creating pressure but not expelling anything).

The first pump supplies the pressurized fluid to the second pump, the
second pump spins the pressurized fluid away from the center to add more
pressure to the housing wall. The pressure gauge in the second stage
should register more pressure than the first stage.

Does this help you understand better now, Ed?

That's well said, but how is the fluid pressurized once it's fed into the second stage?

Take a look at the pump impellers in these photos:

https://www.google.com/search?q=pump...1 920&bih=974

(or, Tiny URL):

https://tinyurl.com/y9trhytf

Remember that the inlet is at the center of the impeller. The liquid then makes a 90-deg. turn and enters the involutes. As the liquid travels from the center to the periphery, the volume *increases*. Poof! There goes your pressure out the window.

But I think I have it figured out now. The problem starts with the concept of "pressure." Except when you're dealing with gases, that's always problematic. The physical measures that are involved here actually are mass, velocity, and force. Forget about pressure for a moment. Think energy instead.

If the previous stage can supply enough liquid to fill the subsequent involute more than it would be filled without that previous stage, then that subsequent stage increases the energy of the water by increasing its velocity. "Pressure" is irrelevant. The energy going in is the product of mass and force (forget the actual formula for now). The energy coming out is the same thing, but along the way, an increase in velocity has increased the force. Restrict that mass and force at the pump exit, and you get pressure.

So that's what I think is happening.


No, that's no what's happening.

You are confusing yourself by making things complicated. You don't need
force/energy equations to figure this out.

A centrifugal pump is a "centrifuge". It is spinning the fluid
(gas/liquid) outward. The fluid is continuously being thrown against the
peripheral wall of the housing which results in fluid pressure against
the retaining wall. It is just like you use your hands to push against
the retaining wall to create pressure on the retaining wall, except that
the fluid is being spun around continuously to create the radially
outward pressure all around in 360 degrees against the retaining wall
that is preventing it from flying out.

That doesn't explain how it adds to the pressure at the inlet.


Of course, the center of the centrifuge will be a partial vacuum.

No, it isn't. In fact, in this case, it's 80 psi.

That
is where the input valve is usually placed so it can suck in fluid if
the pump is properly primed.

wrote on 5/28/2017 10:27 AM:
On Sunday, May 28, 2017 at 9:54:50 AM UTC-4, SbzWrš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›AJFeU wrote:
wrote on 5/28/2017 9:23 AM:
On Sunday, May 28, 2017 at 2:17:58 AM UTC-4, HJARTš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›lBYWJ wrote:
wrote on 5/27/2017 11:08 PM:
On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law. It's much harder to imagine it with liquids.


The idea is the same. You use pressure to reduce the volume of a fluid.
The gas/water restores it original volume after losing the pressure.

I don't think so. Centrifugal pumps are very lossy machines. They couldn't hold pressure that way.

From a physics point of view, I think the answer lies in sorting out the kinetic aspects of a turbo pump (velocity) and the potential aspects (pressure). A turbine pump that's pumping a liquid must be producing potential energy from kinetic energy.

I'd need to see a good, expert explanation to understand it. I see no way that an ordinary turbine pump can hold the pressure generated by a previous stage, unless the entire thing is kinetic, which we're then measuring as potential energy (pressure).


The graph in the link below shows that 200-bar of pressure (2900 psi)
will compress water by about 0.7% at 4°C (under STP, water is 1g/cc at 4°C):

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

No turbine pump could hold that 0.7% compression.

According to you, turbo fan aircraft engines are also very lossy
machines, but most turbo fan aircraft engines use two-stage air
compression.

They are very lossy machines. The part that's of interest here is compressor efficiency, which ranges from 0.70 to 0.85 in the best turbo machinery, including stationary and aircraft gas turbines. At the high end, 85% compressor efficiency, they're losing 15% to gas friction.

But again, you're talking about a gas turbine. Gas has a close ratio between pressure and volume (Boyle's law). Compressing gas with a machine, whether it's positive or non-positive displacement, like a turbo compressor, is not a problem. An example of positive displacement types is a vane-type supercharger. An example of non-positive-displacement types is a turbocharger. Or the compressor stage of a gas turbine engine, such as an aircraft jet engine.

Liquids don't behave according to Boyle's law.


https://www.youtube.com/watch?v=_LaKlE2h3Jw

Both water and air are fluids. I am sure you can modify the aircraft
turbo fan engine design to propel a submarine underwater (without using
fuel and combustion chamber, of course).

If a "very lossy" turbo fan engine can use two stages to compress air or
water...

How do you know that a turbo fan engine can compress water?

, then why wouldn't a two-stage centrifugal pump work?

Obviously, it *does* work. The question is "how."


I suggest you watch this centrifugal pump video to clear up your brain fog:

http://www.wermac.org/video/how-does-a-centrifugal-pump-work.html

On 5/28/2017 5:01 PM, wrote:
On Sunday, May 28, 2017 at 6:06:28 PM UTC-4, tyre biter wrote:
On 5/28/2017 3:59 PM, wrote:
After brushing up on impulse and momentum, and remembering that "pressure" is a frequently misleading term, I'm pretty sure I see what's happening.

-- Ed Huntress

Absolutely, your head is lodged in the toilet trap again, Crazy Eddy.

Poor biter. He must have really been stung when he found out he missed his IP geolocation "target" by over 20 miles.


LOL!

Oh eddy, crazy eddy...say his to the Tidi-bowl man for me woncha?

wrote on 5/28/2017 7:04 PM:
On Sunday, May 28, 2017 at 6:14:05 PM UTC-4, izlsbš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›TSwdF wrote:
wrote on 5/28/2017 2:50 PM:
On Sunday, May 28, 2017 at 12:43:23 PM UTC-4, lvCCLš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›IvVyG wrote:
wrote on 5/28/2017 11:35 AM:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. 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 stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Pressure, thus, is converted to velocity. Unless the machine *compounds* the velocity at each stage, I don't see how it works. And, in order to compound velocity by a factor of, say, three, either the shaft driving the stage either has to be turning at (square root of 3) times that of the first stage, or the the second stage has to have a completely different scroll design.

But you can carry that only so far. Go to three stages, or four, and the shaft rotational speeds become outrageous, or the scroll design does.

Obviously, I'm missing something here, but I haven't yet seen what it is.

Ed Huntress


A centrifuge pump creates pressure by spinning fluid away from the
center (hence "centrifuge").

You are confusing yourself by of processing too much information at one
time. To make it easy for you, let's consider the output valve is closed
(the system is creating pressure but not expelling anything).

The first pump supplies the pressurized fluid to the second pump, the
second pump spins the pressurized fluid away from the center to add more
pressure to the housing wall. The pressure gauge in the second stage
should register more pressure than the first stage.

Does this help you understand better now, Ed?

That's well said, but how is the fluid pressurized once it's fed into the second stage?

Take a look at the pump impellers in these photos:

https://www.google.com/search?q=pump...1 920&bih=974

(or, Tiny URL):

https://tinyurl.com/y9trhytf

Remember that the inlet is at the center of the impeller. The liquid then makes a 90-deg. turn and enters the involutes. As the liquid travels from the center to the periphery, the volume *increases*. Poof! There goes your pressure out the window.

But I think I have it figured out now. The problem starts with the concept of "pressure." Except when you're dealing with gases, that's always problematic. The physical measures that are involved here actually are mass, velocity, and force. Forget about pressure for a moment. Think energy instead.

If the previous stage can supply enough liquid to fill the subsequent involute more than it would be filled without that previous stage, then that subsequent stage increases the energy of the water by increasing its velocity. "Pressure" is irrelevant. The energy going in is the product of mass and force (forget the actual formula for now). The energy coming out is the same thing, but along the way, an increase in velocity has increased the force. Restrict that mass and force at the pump exit, and you get pressure.

So that's what I think is happening.


No, that's no what's happening.

You are confusing yourself by making things complicated. You don't need
force/energy equations to figure this out.

A centrifugal pump is a "centrifuge". It is spinning the fluid
(gas/liquid) outward. The fluid is continuously being thrown against the
peripheral wall of the housing which results in fluid pressure against
the retaining wall. It is just like you use your hands to push against
the retaining wall to create pressure on the retaining wall, except that
the fluid is being spun around continuously to create the radially
outward pressure all around in 360 degrees against the retaining wall
that is preventing it from flying out.

That doesn't explain how it adds to the pressure at the inlet.

A centrifugal pump creates a partial vacuum at the inlet (usually at the
center for a centrifugal pump)



Of course, the center of the centrifuge will be a partial vacuum.

No, it isn't. In fact, in this case, it's 80 psi.

The partial vacuum is filled in by the incoming fluid because the outlet
is shut off. Once the tap at the outlet is turned on, the partial vacuum
will suck in additional fluid to fill the void.

In both cases (either the tap at the outlet is turned on or turned off),
a lot more than 80 psi will be at the casing radially outward farthest
from the center.

Just watch this video, then you will understand:

http://www.wermac.org/video/how-does-a-centrifugal-pump-work.html

On Sunday, May 28, 2017 at 7:05:42 PM UTC-4, ZScPbš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›tLolr wrote:
wrote on 5/28/2017 10:27 AM:
On Sunday, May 28, 2017 at 9:54:50 AM UTC-4, SbzWrš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›AJFeU wrote:
wrote on 5/28/2017 9:23 AM:
On Sunday, May 28, 2017 at 2:17:58 AM UTC-4, HJARTš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›lBYWJ wrote:
wrote on 5/27/2017 11:08 PM:
On Saturday, May 27, 2017 at 8:28:04 PM UTC-4, dWuVxš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›DiDrO wrote:
wrote on 5/27/2017 3:20 PM:
On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.


Water is actually compressible. The compressed volume doesn't change as
much as gas would.

"The low compressibility of non-gases, and of water in particular, leads
to their often being assumed as incompressible. The low compressibility
of water means that even in the deep oceans at 4 km depth, where
pressures are 40 MPa, there is only a 1.8% decrease in volume."

https://en.wikipedia.org/wiki/Properties_of_water#Compressibility

Thanks. I suspect that most of the people here know that. The pressure/volume relationship, though, isn't in agreement with Boyle's law. Gases approximate it. It's easy to imagine a multi-stage non-positive-displacement compressor that keeps building pressure in a material that obeys Boyle's law. It's much harder to imagine it with liquids.


The idea is the same. You use pressure to reduce the volume of a fluid.
The gas/water restores it original volume after losing the pressure.

I don't think so. Centrifugal pumps are very lossy machines. They couldn't hold pressure that way.

From a physics point of view, I think the answer lies in sorting out the kinetic aspects of a turbo pump (velocity) and the potential aspects (pressure). A turbine pump that's pumping a liquid must be producing potential energy from kinetic energy.

I'd need to see a good, expert explanation to understand it. I see no way that an ordinary turbine pump can hold the pressure generated by a previous stage, unless the entire thing is kinetic, which we're then measuring as potential energy (pressure).


The graph in the link below shows that 200-bar of pressure (2900 psi)
will compress water by about 0.7% at 4°C (under STP, water is 1g/cc at 4°C):

http://www.engineeringtoolbox.com/fluid-density-temperature-pressure-d_309.html

No turbine pump could hold that 0.7% compression.

According to you, turbo fan aircraft engines are also very lossy
machines, but most turbo fan aircraft engines use two-stage air
compression.

They are very lossy machines. The part that's of interest here is compressor efficiency, which ranges from 0.70 to 0.85 in the best turbo machinery, including stationary and aircraft gas turbines. At the high end, 85% compressor efficiency, they're losing 15% to gas friction.

But again, you're talking about a gas turbine. Gas has a close ratio between pressure and volume (Boyle's law). Compressing gas with a machine, whether it's positive or non-positive displacement, like a turbo compressor, is not a problem. An example of positive displacement types is a vane-type supercharger. An example of non-positive-displacement types is a turbocharger. Or the compressor stage of a gas turbine engine, such as an aircraft jet engine.

Liquids don't behave according to Boyle's law.


https://www.youtube.com/watch?v=_LaKlE2h3Jw

Both water and air are fluids. I am sure you can modify the aircraft
turbo fan engine design to propel a submarine underwater (without using
fuel and combustion chamber, of course).

If a "very lossy" turbo fan engine can use two stages to compress air or
water...

How do you know that a turbo fan engine can compress water?

, then why wouldn't a two-stage centrifugal pump work?

Obviously, it *does* work. The question is "how."


I suggest you watch this centrifugal pump video to clear up your brain fog:

http://www.wermac.org/video/how-does-a-centrifugal-pump-work.html

I know how a centrifugal pump works. That video doesn't address the issue in question: What happens when the input pressure is higher than the example in your video? And how does it work?

Notice that you did not address the issue of the involute volume increasing as the liquid flows from the center to the periphery, and the effect that has on pressure.

--
Ed Huntress

On Sunday, May 28, 2017 at 7:22:40 PM UTC-4, Pdoihš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›XSEKB wrote:
wrote on 5/28/2017 7:04 PM:
On Sunday, May 28, 2017 at 6:14:05 PM UTC-4, izlsbš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›TSwdF wrote:
wrote on 5/28/2017 2:50 PM:
On Sunday, May 28, 2017 at 12:43:23 PM UTC-4, lvCCLš›†� ï¼*ｉｇｈｔｙ •¬ ï¼·ï½�ｎｎï½�ｂｅ †’š›IvVyG wrote:
wrote on 5/28/2017 11:35 AM:
On Saturday, May 27, 2017 at 6:57:12 PM UTC-4, Ned Simmons wrote:
On Sat, 27 May 2017 12:20:00 -0700 (PDT), wrote:

On Saturday, May 27, 2017 at 2:26:40 PM UTC-4, Steve W. wrote:
wrote:
If a centrifugal pump with a maximum pressure of, say, 10 psi is
supplied with water at 80 psi will the water pressure coming out of
the pump be 90 psi? I think the pressure will be 90 psi. Am I wrong?
Thanks,
Eric

IF the volume of water remains constant the pressure coming out will be
at most 80 psi. if the pump is designed to produce 10 psi. It may be
lower depending on the size of the housing and the restriction the
impeller creates. Say your input side is 2" and the pump can produce 10
psi. at zero head pressure out of a 1.5" outlet.

Feed that pump with an 80 psi head pressure and the pump won't add any
pressure because it cannot pump faster than the water is already flowing
through it.

That's exactly what I thought, but Jim's reference to multi-stage pumps 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.

The pumps in series business is confusing the issue. The output of the
first (centrifugal) pump in the chain is far from an ideal pressure
source. The original question was about a pump with a constant inlet
pressure, either 0 psig or 80 psig. Flow wasn't specified, but as long
as the flow is constant for both inlet conditions, the delta P across
the pump will be the same. In other words, the pump will increase the
pressure by 10 psi in both cases.

This is where I have trouble. Assuming these are regular centrifugal turbines, the outlet of the first stage is fed into the axis of the second stage. 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 stage, the outlet volume of which is LARGER than the inlet volume between any two blades.

Pressure, thus, is converted to velocity. Unless the machine *compounds* the velocity at each stage, I don't see how it works. And, in order to compound velocity by a factor of, say, three, either the shaft driving the stage either has to be turning at (square root of 3) times that of the first stage, or the the second stage has to have a completely different scroll design.

But you can carry that only so far. Go to three stages, or four, and the shaft rotational speeds become outrageous, or the scroll design does.

Obviously, I'm missing something here, but I haven't yet seen what it is.

Ed Huntress


A centrifuge pump creates pressure by spinning fluid away from the
center (hence "centrifuge").

You are confusing yourself by of processing too much information at one
time. To make it easy for you, let's consider the output valve is closed
(the system is creating pressure but not expelling anything).

The first pump supplies the pressurized fluid to the second pump, the
second pump spins the pressurized fluid away from the center to add more
pressure to the housing wall. The pressure gauge in the second stage
should register more pressure than the first stage.

Does this help you understand better now, Ed?

That's well said, but how is the fluid pressurized once it's fed into the second stage?

Take a look at the pump impellers in these photos:

https://www.google.com/search?q=pump...1 920&bih=974

(or, Tiny URL):

https://tinyurl.com/y9trhytf

Remember that the inlet is at the center of the impeller. The liquid then makes a 90-deg. turn and enters the involutes. As the liquid travels from the center to the periphery, the volume *increases*. Poof! There goes your pressure out the window.

But I think I have it figured out now. The problem starts with the concept of "pressure." Except when you're dealing with gases, that's always problematic. The physical measures that are involved here actually are mass, velocity, and force. Forget about pressure for a moment. Think energy instead.

If the previous stage can supply enough liquid to fill the subsequent involute more than it would be filled without that previous stage, then that subsequent stage increases the energy of the water by increasing its velocity. "Pressure" is irrelevant. The energy going in is the product of mass and force (forget the actual formula for now). The energy coming out is the same thing, but along the way, an increase in velocity has increased the force. Restrict that mass and force at the pump exit, and you get pressure.

So that's what I think is happening.


No, that's no what's happening.

You are confusing yourself by making things complicated. You don't need
force/energy equations to figure this out.

A centrifugal pump is a "centrifuge". It is spinning the fluid
(gas/liquid) outward. The fluid is continuously being thrown against the
peripheral wall of the housing which results in fluid pressure against
the retaining wall. It is just like you use your hands to push against
the retaining wall to create pressure on the retaining wall, except that
the fluid is being spun around continuously to create the radially
outward pressure all around in 360 degrees against the retaining wall
that is preventing it from flying out.

That doesn't explain how it adds to the pressure at the inlet.

A centrifugal pump creates a partial vacuum at the inlet (usually at the
center for a centrifugal pump)



Of course, the center of the centrifuge will be a partial vacuum.

No, it isn't. In fact, in this case, it's 80 psi.

The partial vacuum is filled in by the incoming fluid because the outlet
is shut off. Once the tap at the outlet is turned on, the partial vacuum
will suck in additional fluid to fill the void.

In both cases (either the tap at the outlet is turned on or turned off),
a lot more than 80 psi will be at the casing radially outward farthest
from the center.

Just watch this video, then you will understand:

http://www.wermac.org/video/how-does-a-centrifugal-pump-work.html

That's the same video you posted the link to in another message. It still doesn't deal with compound or pressure-inlet conditions.

There is no "partial vacuum" if the inlet is at 80 psi of positive pressure..

--
Ed Huntress