pipe a sits 1 meter under the river surface its 20cm in diameter and rises vertical 40 meters. pipe a joins pipe b which is also 20cm in diameter. pipe b descends 40 meters downwards. pipe b 1 meter down opens up to 2 meters in diameter and the rest of pipe b is 2 meters in diameter. pipe a is full of water the water weighs 1250kg and is held in position by check valves. at the top of pipe b where pipe b widens to 2 meters diameter is a piston that is 2 meters in diameter. the piston weighs 3000kg considering everything is airtight if the piston is lowered slowly would it be able to draw water up pipe a into pipe b. so if the piston is lowered at a constant rate would the water in pipe a be constantly drawn. the bottom of pipe b is open so there is no pressure under the piston
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pipe a sits 1 meter under the river 1
- Thread starter p kansik
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Sorry, no.
Pressure/gamma IS head due to static height of fluid.
And no again.
Pressure/gamma plus velocity head equals total head.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
Pressure/gamma IS head due to static height of fluid.
And no again.
Pressure/gamma plus velocity head equals total head.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
No there are three terms in the Bernouli equation for total head. P/gamma + V2/2g + H, which without friction is constant along any length of pipe. Friction reduces the total head. P/gamma is the pressure at any point in the system which may be produced by a pump, V2/2g is the velocity head, H is the static head developed or subtracted by the height of the liquid as you transverse along the length of the pipe.
Please don't get off on tangents that don't matter. Just respond to my original post as to why I am wrong as to what I said. Either you can argue against what I said or you can't but don't completely change the subject and obfusicate the issues.
P/gamma + V2/2g + H
H is not a head due to fluid state. Actually in my text books "H" is denoted as "Z", probably so that it is not confused with static head. "Z" is a height above some reference elevation, like mean sea level, or any other arbitrary elevation reference point, from which total head at any point in the system is calculated to ensure that total heads at each point are relative to that same reference elevation Z.
If you lay a pipeline on the earth's surface and set its reference elevation as mean sea level, then each point on the pipeline will have a Z value in a Bernoulli expression that equals its elevation above, or below MSL.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
H is not a head due to fluid state. Actually in my text books "H" is denoted as "Z", probably so that it is not confused with static head. "Z" is a height above some reference elevation, like mean sea level, or any other arbitrary elevation reference point, from which total head at any point in the system is calculated to ensure that total heads at each point are relative to that same reference elevation Z.
If you lay a pipeline on the earth's surface and set its reference elevation as mean sea level, then each point on the pipeline will have a Z value in a Bernoulli expression that equals its elevation above, or below MSL.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
Correct but you still haven't addressed the points I raised in my original post to your analysis.
We have a total head in the pipe connection at the tank of 2.459m.
Head at the tank connection is not created by the pump but by the static liquid height in the tank. Total head of the flowing fluid at the tank connection point is the static height of liquid in the tank plus the vlocity head of the fluid as it enters the tank.
If total head in the tank is less, we know we can get water in the pipe into the tank.
Water will flow into the tank as long as the static liquid head of the tank is less than the pump no-flow dead head pressure period.
So we need to know what the head in the tank is at the nozzle connection point. It might help to think of a double swing door at the tank entry point. We can open the door by just pushing on it, or by running at it and slaming into it, so we will try to do both at the same time. The Pressure Head will push the door and the velocity head will do the slamming. Note that this is potential energy of pressure and kinetic energy from the velocity. Added together we have 2.459 m of energy to open the door into the tank. But the tank may have some energy keeping it closed. What has the tank got?
There is no imbalance of forces at the tank connection between the tank side and the pumps side as the pressure on the pump/pipe side is developed by the pressure on the tank side (which is simply the static pressure due to height of liquid in the tank). The pump output total pressure minus the friction drop will always equal the tank static pressure at the connection plus the kinetic energy of the flowing fluid into the tank. (Static pressures in tank and pipe are equal but pipe flow has additional flowing kinetic energy).
The tank has potential energy from any fluid inside it. Does it have kinetic energy? No. The velocity of fluid in a tank is 0. No kinetic energy. So, the tank only has fluid level to hold that door closed. We have 2.459m, so iif the tank has less than 2.459 m of water in it (above the height of the nozzle), we're getting in, at least with some of our water.
Again there is no discontinuity of pressure between pipe side and tank side as the pressure at the tank connection is due to the static height of the liquid in the tank. The flowrate will balance in any case so that the static head output of the pump minus the friction loss equals the static height of liquid at the tank connection. The pump output head is not constant but is determined by the intersection of the system curve with the pump curve.
If the water level in the tank is below our nozzle door, we get in with full flow. Whatever our pressure and velocity heads combined will push into it. That's (V^2)/(2g) and we have both Pressure and Velocity to play with. The total of 2.459m
If the water level is below door then pressure at the tank connection will then be 0 psig so you are operating the pump on a different part of the pump curve at a different output head such that the pump output static head minus friction loss now = 0 psig, but with a difference in head of the fluid equal to the velocity head of the fluid entering the tank.
Let's solve for Velocity. 2.459m = V^2/2/g
So,
V = sqrt(2.459m * 2 * 9.81m/s2)
V=6.98 m/s
We're going to blow the door off it's hinges, at least when we start.
Pressure head plus velocity head (total head) in the pipe does not convert to all velocity head in the tank. Pressure head in pipe at pipe exist cannot convert into anything because it is in equillibrium with the static head in the tank so there is no extra pressure head available to convert to any additional velocity.
Since we are outside the pipe, the cross-sectional area of the flow depends of the "jet stream" shape. Our inside pipe flow rate of 0.0246 m3/s continues into the tank, but now at a velocity of 6.98 m/s.
So we recalculate the cross-sectional area of Flow to be (from V= Q/A), A = Q/V
A = 0.0246 m3/s / 7m/s = 0.00351 m2
A jet stream can never decrease in diameter when it exits the pipe. For water jet in water it will increase due to slowing down by the fluid in the tank and at the same time some of the fluid in the tank being caught up in the jet causing it to expand. For water in air (like a fire hose nozzle) the jet will maintain its jet diameter longer since there is nothing causing it to slow down and any air picked up into the stream is of little consequence.
We have a total head in the pipe connection at the tank of 2.459m.
Head at the tank connection is not created by the pump but by the static liquid height in the tank. Total head of the flowing fluid at the tank connection point is the static height of liquid in the tank plus the vlocity head of the fluid as it enters the tank.
If total head in the tank is less, we know we can get water in the pipe into the tank.
Water will flow into the tank as long as the static liquid head of the tank is less than the pump no-flow dead head pressure period.
So we need to know what the head in the tank is at the nozzle connection point. It might help to think of a double swing door at the tank entry point. We can open the door by just pushing on it, or by running at it and slaming into it, so we will try to do both at the same time. The Pressure Head will push the door and the velocity head will do the slamming. Note that this is potential energy of pressure and kinetic energy from the velocity. Added together we have 2.459 m of energy to open the door into the tank. But the tank may have some energy keeping it closed. What has the tank got?
There is no imbalance of forces at the tank connection between the tank side and the pumps side as the pressure on the pump/pipe side is developed by the pressure on the tank side (which is simply the static pressure due to height of liquid in the tank). The pump output total pressure minus the friction drop will always equal the tank static pressure at the connection plus the kinetic energy of the flowing fluid into the tank. (Static pressures in tank and pipe are equal but pipe flow has additional flowing kinetic energy).
The tank has potential energy from any fluid inside it. Does it have kinetic energy? No. The velocity of fluid in a tank is 0. No kinetic energy. So, the tank only has fluid level to hold that door closed. We have 2.459m, so iif the tank has less than 2.459 m of water in it (above the height of the nozzle), we're getting in, at least with some of our water.
Again there is no discontinuity of pressure between pipe side and tank side as the pressure at the tank connection is due to the static height of the liquid in the tank. The flowrate will balance in any case so that the static head output of the pump minus the friction loss equals the static height of liquid at the tank connection. The pump output head is not constant but is determined by the intersection of the system curve with the pump curve.
If the water level in the tank is below our nozzle door, we get in with full flow. Whatever our pressure and velocity heads combined will push into it. That's (V^2)/(2g) and we have both Pressure and Velocity to play with. The total of 2.459m
If the water level is below door then pressure at the tank connection will then be 0 psig so you are operating the pump on a different part of the pump curve at a different output head such that the pump output static head minus friction loss now = 0 psig, but with a difference in head of the fluid equal to the velocity head of the fluid entering the tank.
Let's solve for Velocity. 2.459m = V^2/2/g
So,
V = sqrt(2.459m * 2 * 9.81m/s2)
V=6.98 m/s
We're going to blow the door off it's hinges, at least when we start.
Pressure head plus velocity head (total head) in the pipe does not convert to all velocity head in the tank. Pressure head in pipe at pipe exist cannot convert into anything because it is in equillibrium with the static head in the tank so there is no extra pressure head available to convert to any additional velocity.
Since we are outside the pipe, the cross-sectional area of the flow depends of the "jet stream" shape. Our inside pipe flow rate of 0.0246 m3/s continues into the tank, but now at a velocity of 6.98 m/s.
So we recalculate the cross-sectional area of Flow to be (from V= Q/A), A = Q/V
A = 0.0246 m3/s / 7m/s = 0.00351 m2
A jet stream can never decrease in diameter when it exits the pipe. For water jet in water it will increase due to slowing down by the fluid in the tank and at the same time some of the fluid in the tank being caught up in the jet causing it to expand. For water in air (like a fire hose nozzle) the jet will maintain its jet diameter longer since there is nothing causing it to slow down and any air picked up into the stream is of little consequence.
Snickster, I agree.
I was going to correct the forked thread, but deleted the entire thread instead of only deleting and rewriting my post. Sorry. In any case its all here anyway.
As Katmar pointed out in the forked thread, If I want to separate potential and kinetic at the end of the pipe, I have to do that at the beginning. I didn't and got too much head at the end.
Katmar, thanks for weighing on that one.
I must learn to turn off this tablet after 12pm.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
I was going to correct the forked thread, but deleted the entire thread instead of only deleting and rewriting my post. Sorry. In any case its all here anyway.
As Katmar pointed out in the forked thread, If I want to separate potential and kinetic at the end of the pipe, I have to do that at the beginning. I didn't and got too much head at the end.
Katmar, thanks for weighing on that one.
I must learn to turn off this tablet after 12pm.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
If we are splitting hairs - 12pm is neither noon nor midnight. There is no time that would correctly be written as 12pm. PM stands for the Latin "Post Meridiem" which translates as "After Midday". On the other hand, AM stands for the Latin "Ante Meridiem" or "Before Midday".
Any time in the morning before noon is AM and after midday is PM. 12 o'clock midday should be written as "12 noon" because it is neither before nor after midday - it is midday. But nobody follows this correctly and it is only pedantic old gits like me that worry about it.
Katmar Software - AioFlo Pipe Hydraulics
"An undefined problem has an infinite number of solutions"
Any time in the morning before noon is AM and after midday is PM. 12 o'clock midday should be written as "12 noon" because it is neither before nor after midday - it is midday. But nobody follows this correctly and it is only pedantic old gits like me that worry about it.
Katmar Software - AioFlo Pipe Hydraulics
"An undefined problem has an infinite number of solutions"
This is really strange. If you look closely at the face of the clock at St. Peter's Basilica in Rome, you notice that it only has 6h. Its one hand revolves 4x per day. Less gears needed?
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
LittleInch
Petroleum
Now this thread has gone off the rails, it's why no one timetables things at 00.00 or 24.00 as no one knows what it means.
All planes, trains etc leave at 2355 or 0005.
Though one place I used to travel from confused everyone when due to daylight saving, the plane used to leave at 23.55, but then moved to 0055. Problem was every one used to refer to it as "the e.g. Thursday night flight". Too many missed the plane by 23 hours as the Thursday flight had left at 00.55.....
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
All planes, trains etc leave at 2355 or 0005.
Though one place I used to travel from confused everyone when due to daylight saving, the plane used to leave at 23.55, but then moved to 0055. Problem was every one used to refer to it as "the e.g. Thursday night flight". Too many missed the plane by 23 hours as the Thursday flight had left at 00.55.....
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
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