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fluid speed vs pressure 2
- Thread starter Hagar1
- Start date
The pressure in a system is constant if we ignore losses (i.e. a perfect world)..conservation of energy and all that.
Total pressure =static +velocity (or dynamic) pressure
so if SP=10 and VP=10 TP=20
if you speed up the system, (and have no losses) then the VP goes up and the SP goes down but the total is the same
voila
SP=5 VP=15 TP=20
etc
Friar Tuck of Sherwood
Total pressure =static +velocity (or dynamic) pressure
so if SP=10 and VP=10 TP=20
if you speed up the system, (and have no losses) then the VP goes up and the SP goes down but the total is the same
voila
SP=5 VP=15 TP=20
etc
Friar Tuck of Sherwood
Hagar1,
The two responses above while correct, are not really on the mark.
In Oil & Gas operations we often talk about velocities being greater at low pressures. What is left out of that statement is "for a constant volume flow rate at STP, actual velocity will be higher at lower pressure".
We're always talking about volume flow rate at STP (and we assume that everyone has the same defination of STP). So when someone says "the critical flow rate in 2-3/8 tubing at 10 psig is 100 MSCF/d and at 500 psig it is 900 MSCF/d", they are probably using a program like Perform or Prosper to calculate a critical flow rate behind the scenes. The terminal velocity at flowing conditions underlying the programs is pretty similar between the two cases, but when you do the conversion to standard conditions the numbers are far different.
David Simpson, PE
MuleShoe Engineering
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
The Plural of "anecdote" is not "data"
The two responses above while correct, are not really on the mark.
In Oil & Gas operations we often talk about velocities being greater at low pressures. What is left out of that statement is "for a constant volume flow rate at STP, actual velocity will be higher at lower pressure".
We're always talking about volume flow rate at STP (and we assume that everyone has the same defination of STP). So when someone says "the critical flow rate in 2-3/8 tubing at 10 psig is 100 MSCF/d and at 500 psig it is 900 MSCF/d", they are probably using a program like Perform or Prosper to calculate a critical flow rate behind the scenes. The terminal velocity at flowing conditions underlying the programs is pretty similar between the two cases, but when you do the conversion to standard conditions the numbers are far different.
David Simpson, PE
MuleShoe Engineering
Please see FAQ731-376 for tips on how to make the best use of Eng-Tips Fora.
The Plural of "anecdote" is not "data"
Visualize a short length of pipe with a large fluid filled pressure vessel on one end and the other end of the pipe is open to the atmosphere. The higher the pressure inside the pressure vessel, the greater the flow will be through the pipe. At first glance, this seems to contradict the statement, “The Speed of a liquid in a pipe is slow when the pressure is high and the speed is fast where the pressure is low.” But it doesn’t. When you consider that the pressure vessel is part of the fluid distribution system (AKA very big pipe with caps on the end) and the velocity of the fluid within the large vessel is slow due to its large diameter, you have high pressure and low velocity within the vessel. The other end of the short pipe is open to the atmosphere and has low pressure (AKA atmospheric pressure) and also has high velocity. Something happened to energy involved in the pressure drop between the vessel and the end of the pipe. The change in static fluid pressure accelerated the fluid through the pipe. The mathematical relationship between the change in pressure and the change in velocity is accounted for with Bernoulli's equation.
All answering posts are right. It is just a matter of conservation of energy: an exchange between kinetic energy (velocity) and pressure energy. To zdas04: since Hagar1 refers to a liquid (~incompressible fluid), the answer by Latexman seems the most appropriate one.
The flow work when the liquid is displaced through a volumetric space 1/[ρ] against the restraint of pressure p is p/[ρ].
The kinetic energy per unit mass is v2/2.
The potential energy per unit mass with the horizontal reference level 0-0 is zg.
Bernoulli's equation for frictionless flow is:
p1/[ρ] + V12/2 + z1g = p2/[ρ] + V22/2 + z2g
Dividing by g all the dimensions in the equation are in units of length, e.g., m.
Multiplying by [ρ] all the dimensions are in units of force per unit area, or pressure, e.g., N/m2.![[pipe]](/data/assets/smilies/pipe.gif "[pipe] [pipe]")
The flow work when the liquid is displaced through a volumetric space 1/[ρ] against the restraint of pressure p is p/[ρ].
The kinetic energy per unit mass is v2/2.
The potential energy per unit mass with the horizontal reference level 0-0 is zg.
Bernoulli's equation for frictionless flow is:
p1/[ρ] + V12/2 + z1g = p2/[ρ] + V22/2 + z2g
Dividing by g all the dimensions in the equation are in units of length, e.g., m.
Multiplying by [ρ] all the dimensions are in units of force per unit area, or pressure, e.g., N/m2.
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2
- #10
Keith,
As a former Naval Instructor, I had to find a way to relate pressure and velocity, besides stating Bernoulli's Theory.
Here is a simple example I used and that all the sailors I taught thoroughly understood;
(1) You and your buddies are sitting in a crowded bar.
(2) A fire breaks out.
(3) There is only one door.
(4) Everyone rushes to the door.
How long does it take to get to the door is your speed. How many people standing, pushing, and shoving to get through the door can actually be measured if you had a scale on your back and everyone behind you is pushing on it. That is pressure or force. Or in simpler terms count the number of people within a 10' x 10' square near the door. That gives you the bodies per square foot. It will be alot.
Now hear is the ringer? How many people can go through the door at one time is determined by pressure or how many people are pushing on those at the door. Keep in mind that as you ran from the bar stool to the door, your speed dramatically decreased. Actually came to a screaching halt.
As you finally exit the door with your shirt almost on fire, what do you do? You run because there are very few people around you. Thus, as you exit the door your speed will increase but "bodily" pressure decreases.
Todd
As a former Naval Instructor, I had to find a way to relate pressure and velocity, besides stating Bernoulli's Theory.
Here is a simple example I used and that all the sailors I taught thoroughly understood;
(1) You and your buddies are sitting in a crowded bar.
(2) A fire breaks out.
(3) There is only one door.
(4) Everyone rushes to the door.
How long does it take to get to the door is your speed. How many people standing, pushing, and shoving to get through the door can actually be measured if you had a scale on your back and everyone behind you is pushing on it. That is pressure or force. Or in simpler terms count the number of people within a 10' x 10' square near the door. That gives you the bodies per square foot. It will be alot.
Now hear is the ringer? How many people can go through the door at one time is determined by pressure or how many people are pushing on those at the door. Keep in mind that as you ran from the bar stool to the door, your speed dramatically decreased. Actually came to a screaching halt.
As you finally exit the door with your shirt almost on fire, what do you do? You run because there are very few people around you. Thus, as you exit the door your speed will increase but "bodily" pressure decreases.
Todd
oxilume
That is one of the best explanations I've heard in all my life. The fact that it is so simple that any everyday person can understand it underlies its genius and therefor the depth of your understanding of the subject.
Well done!
remove.marius@mailbox.co.za
That is one of the best explanations I've heard in all my life. The fact that it is so simple that any everyday person can understand it underlies its genius and therefor the depth of your understanding of the subject.
Well done!
remove.marius@mailbox.co.za
If you have a horizontal pipe with two different diameters in which water is running (full) and install a pressure gage on each section, chances are that the gage in the larger pipe will show a higher pressure (lower velocity) even if it is located downstream !.
This is exactly why an air plane can take off and land, this is also why a centrifugal pump can deliver fluid past the impeller. Velocity increases at the expense of the pressure orthogal to the direction of this velocity, in other words the product of velocity and pressure is a constant at any section of a pipe or a control volume. This is the reduced form of the general energy equation, also called the Bernoulli equation in thermofluid mechanics.
We deliver steam as if your life depends on it.
We deliver steam as if your life depends on it.
I think 25362 owes me an explanation, atlast, for the following reason.
A pressure gauge on a liquid line shows the pressure drop occuring after the point where it was installed. Suppose, I have two horizontal pipelines(different sizes) kept side by side and with the ends open to the atmosphere. Let us also assume the pipes are full and velocity in the bigger pipe is lower than that of the smaller pipe(let us further assume flow is constant in both the pipes).
Neglecting any other losses except friction,
Hf = flv2/2gd, increased d and reduced [/b]v[/b] should reduce Hf
PS: I a speaking about constant cross section pipelines here.
Regards,
A pressure gauge on a liquid line shows the pressure drop occuring after the point where it was installed. Suppose, I have two horizontal pipelines(different sizes) kept side by side and with the ends open to the atmosphere. Let us also assume the pipes are full and velocity in the bigger pipe is lower than that of the smaller pipe(let us further assume flow is constant in both the pipes).
Neglecting any other losses except friction,
Hf = flv2/2gd, increased d and reduced [/b]v[/b] should reduce Hf
PS: I a speaking about constant cross section pipelines here.
Regards,
To quark:
This is not as complicated as it seems. In my example I totally disregarded any difference in friction drop, by locating both gages at a short distance one from the other. That's also the reason I said: "chances are...".
If, on the other hand, the gages were located far apart, the friction drop may have "eaten up" the corresponding pressure "gain" due to velocity "reductions" and the downstream readings would probably become lower than those upstream.
![[smile]](/data/assets/smilies/smile.gif "[smile] [smile]")
This is not as complicated as it seems. In my example I totally disregarded any difference in friction drop, by locating both gages at a short distance one from the other. That's also the reason I said: "chances are...".
If, on the other hand, the gages were located far apart, the friction drop may have "eaten up" the corresponding pressure "gain" due to velocity "reductions" and the downstream readings would probably become lower than those upstream.
Guys,
I'm new to the field and I have a pretty simple question I think...How do you calculate operating pressure in your system. I have a Chilled water closed loop...40 ton chiller to cased chilled water coils. ewt=45 lwt=55. Is it as simple as converting my head on my pump to psi?
I'm new to the field and I have a pretty simple question I think...How do you calculate operating pressure in your system. I have a Chilled water closed loop...40 ton chiller to cased chilled water coils. ewt=45 lwt=55. Is it as simple as converting my head on my pump to psi?
venkatParam
Mechanical
Hello Guys,
Regarding Bernoulli's equation. Will a flow with high static pressure, low velocity have the same impact on a blockage in path as a flow with low static pressure and high velocity.
Thank you
Regarding Bernoulli's equation. Will a flow with high static pressure, low velocity have the same impact on a blockage in path as a flow with low static pressure and high velocity.
Thank you
venkatParam
A short answer would be: no.
The force of impact of a fluid stream on a perpendicular stationary plate is estimated from the law of conservation of momentum as:
Bernoulli's equation is used, for example, to estimate the velocity of a fluid using Pitot tubes by measuring the stagnation pressure from the kinetic energy component converted into head, corrected by a factor that depends on the geometry of the Pitot tube.
When a liquid of density [ρ], moving at a speed V in a rigid pipe, is suddenly stopped, the resulting pressure (water hammer) wave travels at the speed of sound c in the liquid, and the pressure increase from that stoppage is:
This pressure must be added to the prevailing static pressure to verify that the pipe is not subjected to excessive tensile stresses.
A short answer would be: no.
The force of impact of a fluid stream on a perpendicular stationary plate is estimated from the law of conservation of momentum as:
F = ([ρ])(q)(v)/g
Bernoulli's equation is used, for example, to estimate the velocity of a fluid using Pitot tubes by measuring the stagnation pressure from the kinetic energy component converted into head, corrected by a factor that depends on the geometry of the Pitot tube.
When a liquid of density [ρ], moving at a speed V in a rigid pipe, is suddenly stopped, the resulting pressure (water hammer) wave travels at the speed of sound c in the liquid, and the pressure increase from that stoppage is:
([ρ])(c)(V)/g
This pressure must be added to the prevailing static pressure to verify that the pipe is not subjected to excessive tensile stresses.
sailoday28
Mechanical
25362 (Chemical) Sep 25, 2004 writes
When a liquid of density ?, moving at a speed V in a rigid pipe, is suddenly stopped, the resulting pressure (water hammer) wave travels at the speed of sound c in the liquid, and the pressure increase from that stoppage is:
(?)(c)(V)/g
This pressure must be added to the prevailing static pressure .........
THIS IS A RESPONSE TO FLOW WITH BERNOULLI'S EQUATION which in original form does not include friction. AND is valid when the local velocity is negligible with respect to the sound speed.
My experience is that normally the local velocity IS small with regard to sound speed. However, if friction is not negligible, then another type of analysis (method of characteristics) must be used to determine the resulting pressure and velocity distribution in the pipe.
When a liquid of density ?, moving at a speed V in a rigid pipe, is suddenly stopped, the resulting pressure (water hammer) wave travels at the speed of sound c in the liquid, and the pressure increase from that stoppage is:
(?)(c)(V)/g
This pressure must be added to the prevailing static pressure .........
THIS IS A RESPONSE TO FLOW WITH BERNOULLI'S EQUATION which in original form does not include friction. AND is valid when the local velocity is negligible with respect to the sound speed.
My experience is that normally the local velocity IS small with regard to sound speed. However, if friction is not negligible, then another type of analysis (method of characteristics) must be used to determine the resulting pressure and velocity distribution in the pipe.
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