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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I should wait for your sketch, but I think you have an inverted "U" and are trying to draw up water 40m into pipe A from a river by lowering a piston in pipe B.
No, it will not draw up water to 40m, but with perfect seals, you could draw up water in pipe A to a height of 10m +/-. Once the pressure anywhere in the water reaches its vapour pressure, no more water can be drawn upwards. In a case where water is already filled to the top of the U and held there by check valves, the water in pipe A will form a vapor pocket and any vapor bubbles will collect at the high point. Any downward motion of the piston will just create a larger vapor pocket at the top of the inverted "U".
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
No, it will not draw up water to 40m, but with perfect seals, you could draw up water in pipe A to a height of 10m +/-. Once the pressure anywhere in the water reaches its vapour pressure, no more water can be drawn upwards. In a case where water is already filled to the top of the U and held there by check valves, the water in pipe A will form a vapor pocket and any vapor bubbles will collect at the high point. Any downward motion of the piston will just create a larger vapor pocket at the top of the inverted "U".
--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
Sounds like an exam question??
There is a student forum.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
There is a student forum.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
I don't think it's an exam question, unless it's written in a a true/false, or essay context, as there is no definitive answer to it.
--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
Simple answer is no, assuming that pipe a is open to the river with on other pump or thing providing pressure/head forcing the water up pipe a.
The max "lift" is between 8 to 9 metres at sea level before you create a vacuum.
You can fill pipe a with water from the top with it sitting on check valves at the bottom, but you will not be able to lift that water.
Forget about mass and "weight" and use pressure.
Basically you can't make water flow uphill.
So unless there is a pump or other thing creating at least 3 bar pressure at the base of pipe a, you won't lift anymore water.
A sketch would definitely help....
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
The max "lift" is between 8 to 9 metres at sea level before you create a vacuum.
You can fill pipe a with water from the top with it sitting on check valves at the bottom, but you will not be able to lift that water.
Forget about mass and "weight" and use pressure.
Basically you can't make water flow uphill.
So unless there is a pump or other thing creating at least 3 bar pressure at the base of pipe a, you won't lift anymore water.
A sketch would definitely help....
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
When the pump is running, at any given point along the pipe, add the discharge pressure of the pump to that point's static pressure, then subtract from it the friction loss between pump and that point.
At a point in a pipeline near a tank bottom, the pressure head will be equal to the height of water in the tank, whether the pump runs or not, however total head is not the same. To find total head, add (v^2)/2/g.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
At a point in a pipeline near a tank bottom, the pressure head will be equal to the height of water in the tank, whether the pump runs or not, however total head is not the same. To find total head, add (v^2)/2/g.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
The gage pressure at the fluid surface in an unpressurized tank is 0. Its absolute pressure is equal to atmospheric pressure.
The pressure at the discharge from nozzle into the tank is equal to the pressure equivalent of the height of fluid in the tank above the nozzle.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
The pressure at the discharge from nozzle into the tank is equal to the pressure equivalent of the height of fluid in the tank above the nozzle.
--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
BB4 - are you the OP now?
If not please don't hijack someone else's thread.
The velocity right at the tip of the exit will be the same as the pipe velocity, or maybe 0.5 to 1D inside it. After that the water jet starts to disperse and slow down at the edges, but if it's high enough it can continue and hit the far side of the tank if it's small enough.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
If not please don't hijack someone else's thread.
The velocity right at the tip of the exit will be the same as the pipe velocity, or maybe 0.5 to 1D inside it. After that the water jet starts to disperse and slow down at the edges, but if it's high enough it can continue and hit the far side of the tank if it's small enough.
Remember - More details = better answers
Also: If you get a response it's polite to respond to it.
Well it's already jacked ... so I'll continue here.
BB4 If you start a new thread, I'll delete this and put it into your new thread.
Pipe exit velocity
The velocity inside the pipe is not necessarily the same as outside the pipe. Outside the pipe we cannot make the same assumptions that we do inside the pipe. Normally inside the pipe we assume that the pipe is flowing full and that velocity is equal to the flow rate divided by cross-sectional area. But outside the pipe, we are not sure what the cross-sectional area is.
There is one case where the velocity inside the pipe is equal to the velocity outside the pipe. Let's look at that case in detail.
We will work out an example problem of a whole system.
Pump
Let's say there is a pump driving flow through a 50m long 102mm inside diameter pipe that fills a tank. Additionally we will buy a pump able to discharge at a flow rate of Q=0.0246m3/s
at a Total discharge head (pressure head+velocity head) of 6.459m
Pipe
ID= 102mm
cross-sectional area is A=0.008219m2
I chose those dimensions because the water's velocity in the pipe is then Q/A = 3m/s
Pipe Friction Loss
Head loss to friction over the length of the pipe.
The pipe friction factor is 0.018 (from a friction factor iteration I did previously for this problem)
Friction loss in meters for the 50m length is (f * L * V^2)/(D * 2 *g)
= (0.018 * 50m * (3m/s)^2)/(0.102m * 2 * 9.81m/s2)
F = 4m
Velocity Head Inside the Pipe
Velocity head is (V^2)/2/g
Hv =3^2/2/9.81
Hv = 0.459m
I said in another response above that the pressure, or head at any point along the pipe is equal to pump discharge head- friction loss from pump to that point. At 50m, just before entering the tank, the total head (pressure head +velocity head) is
Pump total discharge head of 6.459m - pipe friction of 4m
Hp -F = 2.459m
So we have 2.459m of total head just before entering the tank (still inside the pipe). Just for fun, let's see how much of that is velocity head and how much of that is Pressure Head?
Total head -velocity head = pressure head.
2.459 -0.459
= 2m of Pressure Head
OK, so now let's see what happens at the entry to the tank. As you would expect, it depends on how full the tank is. If the tank is empty, we would expect that we can fill it, at least with some water coming from the pipe. Well that's if we bought a big enough pump with enough head capacity to push it in there.
We have a total head in the pipe connection at the tank of 2.459m. If total head in the tank is less, we know we can get water in the pipe into the tank. 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?
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. 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
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.
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
Jet stream diameter is now (from A = 1/4 * pi D^2)
D = sqrt(4 * A / pi) = 0.0669 m or 66.9 mm, down from 102mm
But filling the tank will stop, if the tank ever gets to a level of 2.459m above our nozzle. The energy on both sides of the door will be equal and the door closes. Or, if there isn't a door, the flow just stops. There is no longer an energy difference remaining to drive flow into the tank..
Now the process of filling while the door is closing subjects the pipe and pump to some dynamic effects. As flow comes to a halt, your pump may increase pressure, depending on its output head vs flow curve. The pipe friction loss will reduce considerably all the way to zero. If it does, then you will have 4 more meters of head to reinstate flow into the pipe and tank. The door will open rapidly, but at low flow rates. Depending in how all those effects work together, surging will likely occur. But that's a subject for another thread, for sure.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
BB4 If you start a new thread, I'll delete this and put it into your new thread.
Pipe exit velocity
The velocity inside the pipe is not necessarily the same as outside the pipe. Outside the pipe we cannot make the same assumptions that we do inside the pipe. Normally inside the pipe we assume that the pipe is flowing full and that velocity is equal to the flow rate divided by cross-sectional area. But outside the pipe, we are not sure what the cross-sectional area is.
There is one case where the velocity inside the pipe is equal to the velocity outside the pipe. Let's look at that case in detail.
We will work out an example problem of a whole system.
Pump
Let's say there is a pump driving flow through a 50m long 102mm inside diameter pipe that fills a tank. Additionally we will buy a pump able to discharge at a flow rate of Q=0.0246m3/s
at a Total discharge head (pressure head+velocity head) of 6.459m
Pipe
ID= 102mm
cross-sectional area is A=0.008219m2
I chose those dimensions because the water's velocity in the pipe is then Q/A = 3m/s
Pipe Friction Loss
Head loss to friction over the length of the pipe.
The pipe friction factor is 0.018 (from a friction factor iteration I did previously for this problem)
Friction loss in meters for the 50m length is (f * L * V^2)/(D * 2 *g)
= (0.018 * 50m * (3m/s)^2)/(0.102m * 2 * 9.81m/s2)
F = 4m
Velocity Head Inside the Pipe
Velocity head is (V^2)/2/g
Hv =3^2/2/9.81
Hv = 0.459m
I said in another response above that the pressure, or head at any point along the pipe is equal to pump discharge head- friction loss from pump to that point. At 50m, just before entering the tank, the total head (pressure head +velocity head) is
Pump total discharge head of 6.459m - pipe friction of 4m
Hp -F = 2.459m
So we have 2.459m of total head just before entering the tank (still inside the pipe). Just for fun, let's see how much of that is velocity head and how much of that is Pressure Head?
Total head -velocity head = pressure head.
2.459 -0.459
= 2m of Pressure Head
OK, so now let's see what happens at the entry to the tank. As you would expect, it depends on how full the tank is. If the tank is empty, we would expect that we can fill it, at least with some water coming from the pipe. Well that's if we bought a big enough pump with enough head capacity to push it in there.
We have a total head in the pipe connection at the tank of 2.459m. If total head in the tank is less, we know we can get water in the pipe into the tank. 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?
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. 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
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.
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
Jet stream diameter is now (from A = 1/4 * pi D^2)
D = sqrt(4 * A / pi) = 0.0669 m or 66.9 mm, down from 102mm
But filling the tank will stop, if the tank ever gets to a level of 2.459m above our nozzle. The energy on both sides of the door will be equal and the door closes. Or, if there isn't a door, the flow just stops. There is no longer an energy difference remaining to drive flow into the tank..
Now the process of filling while the door is closing subjects the pipe and pump to some dynamic effects. As flow comes to a halt, your pump may increase pressure, depending on its output head vs flow curve. The pipe friction loss will reduce considerably all the way to zero. If it does, then you will have 4 more meters of head to reinstate flow into the pipe and tank. The door will open rapidly, but at low flow rates. Depending in how all those effects work together, surging will likely occur. But that's a subject for another thread, for sure.
--Einstein gave the same test to students every year. When asked why he would do something like that, "Because the answers had changed."
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1
- #18
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.
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.
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