Bernoulli Question

if gate valves are close at 2 & 3, then 1, 2 & 3 will have the same pressure. But if both 2 &3 are open then, for an incompressible liquid using bernoulli's equation

Q=A1V1=A2V2+A3V3, since A1=A2=A3=A & V2&V3 = V;

AV1=2AV, so V1 = 2V, V1>V2 or V3

since Pressure & Velocity is inversely proportional, the P1<P2 or P3

Not clear what do you want to say.

Pressure at inlet is lower than that of the outlet.

? I ...

I could not agree on you answer, really. Total Head discharge is the sum of pressure & velocity heads. So the two will just give up & give in vise versa to compensate with the total head. It does not mean that when pressure is very low, it can not delivery a flow. Water stream along a conduit with out change of pressure, only velocity-This is possible along same cross sectional pipes.

Reversing the flow, arrived the same results, otherwise the bernoulli's equation will not be that acceptable.

Correct me if I'm wrong.

Not difficult. By the way you are using a derived equation from Bernoulli equation for an incompressible fluid.

p1=p2 only when ρ=0 (this only occurs when the incompressible fluid has no density) and as you point out when one includes the added head pressure from the change in cross sectional are the dominant factors that makes Bernoulli's equation a useful relation. But if instead of just using the collection of formulas one finds on a page, one carefully reads the text you will find the third paragraph states:

Bernoulli's principle can be derived from the principle of conservation of energy. This states that, in a steady flow, the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy and potential energy remain constant. If the fluid is flowing out of a reservoir the sum of all forms of energy is the same on all streamlines because in a reservoir the energy per unit mass (the sum of pressure and gravitational potential ρ g h) is the same everywhere.^[4]

Adding friction to any analysis immediately makes any system a non-conservative energy system. Certainly in most cases of liquids running through a pipe, the friction components that happen from cohesion and surface tension are so small that they can be often ignored. Once and awhile though the friction in the system cannot be ignored. Long ago it was found that the sleeve turbulence of a fabric hose reduced the friction sufficiently that more water could be pumped through a cloth fire-hose than a smooth fire-hose.

Please expand that analysis to the 1-inlet, 3-outlet situation in the original post. Eager readers are ready to learn more.

GA from me.

By the way, how can any one get to the conclusion that the inlet pressure can become higher than the outlet? (Outlet means flow out...).

Also, on your other comment (bernouilli...), apart from the conclusion that ρ = 0, the vs could be = 0 for the same conclusion, which brings them back to zero flow as you mentioned.

GA. I always like the fluid flow / electrical flow analogy.

Actually, p0=p1=p2=p3 can occur not only at zero flow but also at a specific forward flow so long as friction is not ignored and the distance between inlet and outlet is sufficient.

As stated by the OP, the inlet and outlet pipes are the same size, so the three outlet pipes total 3 times the cross sectional area of the inlet.

The velocity must be 1/3 at the outlet as it is at the inlet.

Think of an airplane wing, higher velocity = lower pressure.

When the head loss due to frictional losses along the length of pipe equals the pressure gained by the decrease in velocity, p0=p1=p2=p3

BBB

I'm confused, is this not what I said in #4, above?

"Bernoulli's principle can be derived from the principle of conservation of energy. This states that, in a steady flow, the sum of all forms of mechanical energy in a fluid along a streamline is the same at all points on that streamline. This requires that the sum of kinetic energy and potential energy remain constant. If the fluid is flowing out of a reservoir the sum of all forms of energy is the same on all streamlines because in a reservoir the energy per unit mass (the sum of pressure and gravitational potential ρ g h) is the same everywhere.^[4]"

"Not difficult. By the way you are using a derived equation from Bernoulli equation for an incompressible fluid."

To simplify what you said here, do you agree with the insights of PWSlack @#2, sayiing

P1> P2 or P3 to satisfy a flow condition.

"Not difficult. By the way you are using a derived equation from Bernoulli equation for an incompressible fluid."

To simplify what you said here, do you agree with the insights of PWSlack @#2, sayiing

P1> P2 or P3 to satisfy a flow condition.

__________________

Both you and PW are hitting the right concepts. For a frictionless example with no change in elevations, you are correct. PW's is including friction, but I'll resubmit PW's answer with one slight modification/clarification. The confusion may lie in terminology (I've heard people use pressure to refer to the static pressure and others who say pressure to refer to the total head).

The total head is actually slightly higher [at the inlet] than the total head at the outlet during forward flow so as to overcome friction losses in the manifold.

If it were equal, then the flowrate would be zero.

If it were lower, then the flow direction would be sdrawkcab.

It is funny, several people hurry to drop bernoulli's name, but then fail to actually utilize his work.....

The outlet has 3 times the cross section of the inlet, so those claiming that 'pressure at the inlet must be greater than at the outlet for flow in the forward direction' are as wrong as they were confident

pressure at the inlet is lower than the outlet in many scenarios fitting the ops conditions....ignoring or allowing for friction, and at a constant elevation.

It is kind o

Ever stand in the threshold between one room that is at a higher pressure (maybe a pressurized lab) and one room at a lower pressure? Which way is the air blowing?

You have to use the conservation of mass principle here too buddy.

Cingold,

I think some of these folks might be stuck, my friend.

Anyone who would argues that fluid flows from low to high pressure must not be argued with. By doing so you are distracting them from solving the world's energy problems through the development of a permanent magnet motor.

-A-

Ahh my deepest apologies.

I was really just hoping I didn't step in troll dung after I posted my comment.

Consider a venturi, my friend. You'll find that the fluid flows from the low-pressure zone at the centre of the venturi to the high-pressure zone after the venturi.

By the way, all of this only applies if there is flow through the manifold. If there is no flow then the pressure is constant at every point.

Now that's funny.

Cingold, it may seem, at first glance, counter intuitive that a point upstream in a flow may have less pressure than one downstream, but bernoulli's equation makes sense when you consider conservation of energy.

Consider a closed system, wherein a pipe that changes in cross section along the length, going from 1 square foot ID out to 3 square feet ID, and then tapers down to 2 square feet ID afterwhich a pump returns the fluid back to the 1sq ft ID inlet.

As the fluid moves from the 1 square foot cross section to the 3 square foot cross section, the velocity slows. Agreed?

Where did the kinetic energy go the fluid had when it was traveling 3 times as fast?

Bernouilli will explain that the kinetic energy is not lost, but converted into pressure.

Likewise when the fluid moves from the 3 sqtf cross section to the 2 squarefoot cross section, the speed increases. Where did the energy come from to accelerate the fluid? It came from, and can be measured as a decrease in pressure.

. How else would you account for the energy required to accelerate the fluid?

BBB

Yes, that is the classic explanation that Bernoulli provides to explain a Venturi tube pressure drop. However, in considering the whole system there must be a net movement from a higher pressure to a lower pressure or the fluid will not move. The best demonstration of this (that may or may not also include a venturi pinch off) is the simple action of a siphon. At the apex of the siphon the pressure will be less than at the siphon entry point. But if the siphon exit point is not below the elevation of the entry point (at a higher pressure) then the siphon action does not happen.

I'm sure you can think of several examples where there is a higher pressure downstream than up....

Consider from the thickest section of an airfoil to the very end of airfoil. Pressure is lowest where the airfoil is thickest....are you suggesting the air refuses to flow over the later part of the airfoil, protesting aboud an increase in pressure?

...or consider an eductor pump. With no blades nor pistons, these pumps can draw a high vacuum. The outlet pressure is much higher than the suction section just up stream.

We aren't talking about initiating flow, we are talking about steady state.

I can think of many many flows that parts of the fluid flow go from a low pressure zone to a high pressure zone. All of these conditions require either momentum to be a component or an external force to be applied to the fluid. I cannot think of a single scenario, including the ones you have cited, where the net flow of a fluid in the whole system (not a small region) goes from a low pressure region to a high pressure region without an external force. This seems to be so fundamental to me that I'm sure that we are not talking about the same thing.

Who specified flow without a external power source.... the op certainly didn't. It is weird so maky people deny the pressure measured downstream in a wide slower section could be higher than that upstream in a smaller faster section. Pitot tube, anyone? Venturi flow meter? Absurd.

I think it depends on how the pitot tube is oriented, and whether static pressure or total pressure is being measured. (The terms "total dynamic head" and "total energy" are also used.)

Well la de freakin da. If you're going to nit pick that the OP didn't say that an external power source didn't exist, he also didn't say that it did exist. I said that you have to consider the whole system and not just a small part.

The op specified a portion of a system, being a manifold. The manifold has flow, so an external source of power is implied. ....at least from the perspective of those who understand that steady state, nonzero flow must be driven by some external power. You make it a point to note that the op did not specifically condition the absence of something external driving the flow. Always eager to learn new things, I am intrigued by your inference of steady state flow that is not driven by an outside source, with respect to the to a scenario similar to the ops manifold. Enthrall me with your acumen!

Flattery may take you far and wide, but never far enough to make your concepts/comments correct on this issue.

While I didn't have any self doubt about being a capable being, i none the less appreciate the complement and olive branch you extended in attempting to reassuring of such, since you are conversing with me.

I'll have to decline having any of the Troll, even if you do decide to feed it to beings, as i hear they are high in trans-facts.

I will reply with a (real) complement. I refuse to have a battle of wits with an unarmed person....so the fact that i am engaging you is evidence I don't consider you unarmed.

I would ask you to leave aside the question of what is causing the flow, and consider the gauge pressure upstream in a high flow small cross section as compared to downstream at a low flow large cross section.

Also, how would you account for conservation of energy from the slowing fluid?

Ok, first I apologize if my rant offended you or anyone else.

Now for the point I really thought I had made clear but that obviously eludes a few. Bernoulli does very clearly explain why an apparent loss of energy happens in a fluid flow when one just considers the pressure involved in the fluid. As Bernoulli and you accurately point out the pressure is not the sole metric for the energy of the fluid, the velocity must also be taken into account. It is when one considers both the pressure and velocity that one has the true metric (energy) to determine the dynamics of a fluid flow. So let me start using the single metric of energy for now instead of the two of pressure and velocity.

In any flow of material going from point A to point Z there must be a decrease in energy if the energy robbing component of friction exists in any form. (Second Law of Thermodynamics) Now along the way (D) there maybe an energy source to the chain of material that can make Z at a higher energy level than A but Z will still be at a lower energy level than the original energy level A plus the added energy applied at D. So to reiterate my point here, in a flowing non-conservative system downstream will always be at a lower energy than all energy applied to the system.

Now to get back to the flow of a fluid with pressure and velocity considered separately. Bernoulli's analysis works quite well with fluid flow that has reached an equilibrium. (Yes, that is an oxymoron. I cannot think of a better way to phrase this idea.) What I mean is that the moving fluid has reached in its downstream travel a pressure and velocity that is relatively uniform. But Bernoulli alone does not explain the dynamics that explain what happens when a sharp transition occurs. In a tangible real world thought experiment example lets compare the dynamics that happen in a horizontal venturi column with a horizontal conical section like a nozzle.

The only thing different between the two transitions is their shape. Both transitions have the same orifice area and shape. The upstream velocity and pressure of the identical fluids are the same. The upstream and downstream areas are identical, too. In the venturi column the pressure and velocity exactly match what Bernoulli predicts, even when taking small snippets of regions to measure. However in the case of the nozzle right after the restriction transitions to the larger exit area the velocity and pressures are not uniform in this region. In this turbulent transition region, near the axis that coincides with the orifice's center the fluid velocity is close to the velocity in the orifice. In contrast the velocity in the turbulent region near the exit diameter approaches zero. Downstream with both orifices Bernoulli's relationships hold true.

Now that I've gone through the trouble of setting up this thought experiment, you maybe wondering what is my point of this lengthy exercise. In both orifices the dynamics of the streams are identical (given) and agree with Bernoulli's equation. In both cases the far downstream velocity and pressures agree with Bernoulli's equations. In the nozzle case an initial central stream moves with the kinetic energy and momentum that existed in the orifice and gets diffused with the turbulence that happens after transition. This makes sense when you consider what happens with a steam adjusted nozzle on a garden hose that is spraying water. The fluid continues through the low resistant air with a similar velocity as that flowing through the orifice. There is my point. For Bernoulli's equations to be valid there must be some non-conservative diffusing interactions occurring. So with the venturi shape the energy losses after the orifice must be so gradual and likely minimize these losses that Bernoulli's equations agree in all points.

Since Bernoulli's relationships requires a non-conservative system to be true, the energy downstream must be less than the energy upstream and all added energy.

I completely agree that the total energy in the fluid will be less downstream than upstream as losses present in any real system occur. That was never disputed.

. .

The OP's question and the difference in viewpoints related to pressure (not total energy). Specifically, I disagree with statements (made by several respectable minds) that claimed (paraphrasing) 'inlet pressure must be greater than outlet pressure for flow to occur from inlet to outlet'.

. .

I also disagree with the assertion that (paraphrasing again) 'if inlet pressure equals outlet pressure, then flow must be zero', since in a real system at a certain rate of flow headloss can offset any pressure that would be gained by the decrease in velocity.

. .

Your last comment suggests you are now in agreement. Which puts you in the ridiculed minority. I'm greatful for the company.

. .

:-)

My friend Cingold is correct in thinking there is something wrong with fluid flowing from low pressure to high pressure. It is counter intuitive because it is wrong.

Potential energy and kinetic energy are both components of the total energy of a system. In Bernoulli's equation he treats these as "static" pressure and "dynamic" pressure.

It is when people and professors start to refer to "static" pressure as Pressure that things become counter intuitive.

Total Pressure is the combination of static pressure and dynamic pressure (and head). When velocity equals zero, dynamic pressure equal zero. If there is then a velocity component, and the total pressure on the system is the same, the static pressure will be less (all other things being equal).

There is nothing counter intuitive about it, IF you use the entire equation.

-A-

If the total cross section of the outlet pipes was less than the area of the inlet pipe, the velocity would be greater and the pressure lower ~ conservation of energy. The opposite would be true if the inlet was the smaller.

This why I love to look thru this site. Good discussion and good answers with lots of questions. Way to Go Folks!

ok. from the above discussion. I will try to summarize based on my scenario below:

Inlet ----->manifold----->outlet 1 + outlet 2 + outlet 3

& assuming:-

a) Same cross sectional area for all hoses. Same length for all hoses. Short length. Same height.

b) Water pressure 100bar supplied by water pump going into inlet through hoses to manifold. Constant flowrate.

c) Hoses of same type as b) used by outlet 1,2 &3

So:-

1) If all outlets are shut. P = 100bar = P1 = P2 = P3

2) If all outlets are open, each pressure shooting out of outlet 1,2 & 3 would be simply ~100/3. If we ignore the friction. Correct?

3)If friction is consider, each pressure at outlet 1,2 & 3 would be lower than 100/3. Correct?

No.

For all the hoses the same size, the inlet velocity would be three times the outlet velocity. Ke is proportional to the square of the velocity.

9Ke+Pin = Ke +Pout

Don't get confused with PWSlack @ #2 I will explain his assumptions with his answer P1> P2 or P3. Try to look at the configuration of your problem below.

PWSlack answer is correct P1>P2 or P3 when his point of reference will be moved for 2 & 3 to 4 & 5.

Total Head = Head @ 1 = Head2 +Head3 this Total head is not caused actually by the point 1 to 2 or 3 (Take note this is just equating inlet and outlet mass or mass balance)

The head is caused by the difference of states between system & surrounding.

Consider this analysis, we put the head in pressure so that you will understand, if the gate valve is closed both 2 &3, the total Head will head caused by the Pressure with in that system.

Total Head = Hp1+Hv1=(Hp2+Hv2)+(Hp3+Hv3), since V=0 this will be simplified to

Total Head = Hp or Pressure Head (this is at shut off condition)

The question is would there be a flow, if the surrounding pressure equals to that system pressure? Obviously NO.

Are state at one 2 & 3 not included in your system? Yes it is. Total Head1 = Total Head @ (2+3) because you just did a mass balance or energy balance there, then 2 & 3 does not cause the flow. Therefore, the flow is caused by condition other than 2 & 3.

I think we need to know more about the configuration of your manifold. Is it like in #5 where Bernoulli would apply, or is it more like a fire department wye?

You said all 4 pipes/hoses are the same size, so I suspect it is more of a wye.

This is going to depend on flow rate, and manifold configuration. At on extreme flow rate = 0,, pressures are all equal,,,

I going against the crowd on this one.

The outlet pressure will be the greater than or equal to the inlet pressure ....

...if incompresible frictionless flow is assumed, and in many cases, even if friction is considered.

As stated by the OP, the sizes of the inlet and outlet pipes are the same.

Given that there are 3 outlet pipes and only one inlet pipe, velocity at the inlet pipe will be 3 times that at any outlet.

Assuming level incompressible flow and all other conditions being equal... For non-zero flow rates, in either direction, pressure at any outlet pipe will be greater than at the inlet pipe, except for cases in which headloss from inlet to the outlet exceeds pressure gained (via a conversion from kinetic energy) by the fluid slowing into the increase cross section.

BBB

Correct me if I'm wrong on my thinking here.

Let's assume that the outlet areas are all equal to the inlet area (cross sectional). Conservation of mass says that total mass inlet = total mass outlet.

So we're flowing 6 apples through this pipe. The pressure at the inlet would be 6 apples per square inch. Flow would divide evenly for each outlet which would mean each outlet would take 2 apples. So now the outlet pressure on each pipe would be 2 apples square inch.

Inlet pressure > Outlet Pressure.

You are mixing up flow and pressure. What you say would apply to flow, but not to pressure.

The OP says the outlet pipes are the same size as the inlet pipe, so with three of them there is three times the cross-sectional area. Mass flow is the same at inlet and outlet, therefore only one third of the velocity at each outlet compared to the inlet. According to Bernoulli if the velocity decreases then the pressure increases.

So we would have a higher outlet pressure than inlet pressure?

I think what we are mixing up here is pressure and pressure. Bernoulli said the SUM of pressures was equal. Herein he was talking about static AND dynamic pressures. Heretofore we have ONLY referred to pressure. That would infer the SUM of static and dynamic. Everyone arguing that pressure would be greater at the outlet of a fluid system is only arguing half of the equation. True, the static pressure increases through a nozzle or wye or whatever, but the difference is offset by the dramatic loss of dynamic pressure, which is (obviously) a function of velocity. The difference in TOTAL pressure is friction.

When it is all said and done, water flows downhill. From high to low. Fini.

OK, now that I have stuck my neck out, I am prepared for ridicule.

-A-

I think right now your head is a high pressure system. All that hot air is just flowing out.

Just kidding, I think the only instance where the outlet pressure might be higher than inlet is when the momentum of the fluid is enough to overcome the increasing outlet pressure. To my knowledge, though, this doesn't happen by simply flowing water through a valve or manifold.

You and several others seem to be getting pressure (lb/sq in) and flow rate (gallons/minute) mixed up. Ignoring friction, elevation changes, etc, the inlet and outlet pressure will be the same, but the flow rate, volume, will be roughly 1/3 in the outlets.

You and several others are ignoring the fact that TOTAL pressure is comprised of static AND dynamic pressure.

Everyone thus far has only talked about static pressure. Static pressure is only one part of total pressure.

p_tot=p_stat+½ρV²+ρgz

It's like everyone is pretending static pressure IS total pressure. And then arguing about it!

p_tot in = p_tot out - frictional losses

Therefore, because we live on Earth, p_tot in > p_tot out.

If it wasn't, the flow would not be in that direction.

PLEASE, someone, tell me I'm wrong.

-A-

I don't know if you're right or wrong because I do not understand the unidentified variables and therefore your point at all. I assume that your small case rho (ρ) is the density of the fluid but we should all remember the definition of assume. I'm not sure if you intend V to be the Velocity, Viscosity, Volume or another variable all together. If you cite where this equation comes from, I'm sure that the web page will identify what each of these are.

You are absolutely correct. My deepest apologies.

ρ = density

g = gravity

p_stat = static pressure

p_tot = total Pressure

V = velocity

z = height (change)

Citation:

(1) Fundamentals of Fluid Mechanics, sixth edition. Munson, Young, Okiishi, Huebsch, Wiley. ISBN 978-0470-26284-9

(2) Pump Characteristics and Applications, second edition. Michael Volk, P.E. Taylor & Francis CRC. ISBN0-8247-2755-X

I agree with you, we have the same basis, all these guyz seem to entangled with Pressure Head as to Total Head.

Total Energy at a point of reference in the pipe is the sum of the Pressure Head + Velocity head + Static head.

that is why the formula shows

Total Head/energy =p_stat+½ρV²+ρgz

this is basic engineering, i wonder why people are not so well configured with it.

Thanks -A- ... I was trying to point that out earlier, but you did a much better and more thorough job at explaining it.

No problem. Just doing my part. It doesn't change anything, but I keep doing it. I'll let someone else with more time take the glory. My comments are mostly just filler.

-A-

Had to go back over 30 years to basic fluid mechanics course.

In a fluid network, the pressure at each node is a constant and the sum of the flows is equal to 0.

The pressure in the manifold has only 1 value. If the 3 outlets go to the same node at a distance from this node, the pressure drop in each pipe has to be equal.

Please be clear what you are discussing. Flow, velocity pressure, static pressure and pressure drop are entirely different entities.

@ v=0 answer = yes

since flow (inlet/outlet) indicates v>0 answer = No

Given the information that you have provided and assuming that you are located somewhere on this planet then the answer is:

Inlet pressure >= outlet pressure. Given the information that you gave us there is no way that the outlet pressure can be larger than the inlet pressure - if it was what would be causing the pressure rise? Imagine if you could then connect this higher outlet pressure back to the inlet then you would somehow have managed to create an over unit device - and most of us here know that this is just horse poo.

People bang on about Bernoulli all the time but forget to use a bit of common sense.

In a static system the multiple outlets would have the same pressure as the inlet. This is I believe what is confusing the OP. It would appear that we now have turned a single inlet into multiple outlets with the same pressure, therefore "creating" additional pressure sources. A static system is however a useless system. As soon as we go dynamic, everything changes.

Hi WJMFIRE:

Those arguing that the inlet pressure is less than that at the outlets are wrong, it has to be equal; or higher if we are discharging to open atmosphere.

It looks like somewhere in their minds they can't stop thinking of a piston, which inlet pressure gets multiplied by the cross sectional area of the bore, therefore giving hell of a lot of times the inlet "pressure" (actually force), so you can rise your car and replace a tire. Obviously, this is not the case but they're stuck to it. Others confuse pressure and flow etc. Oh boy !

Yahlasit

The common sense really there, is try to study (mv2/2) a.k.a kinetic energy component.

Question: At real world, is flow possible with out change in pressure? Answer only Yes or No

Let us see how far you can go.

Are you making a point or asking a question? I really can't make out what you actually want - ok now I realise: 34.6- is that OK

Yes, if there is no flow! But the moment line 1, line 2 and line 3 are fully opened, the line with the least resistance (line losses) wil have the highest pressure at the outlet, which also means the highest flow. If the line losses for the three lines are the same, the pressure on each of the line will be the same (but seldom this happens in actual installations). What you can do to equalize the pressure on each line is to provide a throttling valve for each line, so you can adjust the outlet pressure on each line. Hope this answers your question.

I think I, like everyone else, got lost is side arguements about Bernoulli's equation instead of focusing on the question.

Did anyone actually answer your question?

In reality, you will have pressure drop across the manifold so the answer is that the outlet pressures are less than the inlet. By how much, I don't know. Sometimes pressure drops across a valve are insignificant.

Remember, the system determines the pressure as much as anything else. The reason I say this is that you can probably treat the whole system as one pressure.

Don't get lost in these arguements about higher outlet pressure. That's possible by redefining the boudaries of your system to make it true but not very useful. Remember, velocity is squared in Bernoulli's equation, so it takes more v's than p's to move a fluid.

For goodness sake. The use of the Bernoulli equation cannot be a sidetrack! It is the title of the thread!

A manifold is a chamber with tubes sticking out of it. The question did not specify that they were not on the same level so we must assume that the manifold is horizontal and all of the tubes are horizontal and on the same level (no change of head). The question specified that there is one inlet and three outlets and that they are all the same size.

The total energy (T_e) of the fluid is equal to the sum of it's potential energy (P_e) and its kinetic energy (K_e). T_e = P_e + K_e.

Under the rules of energy conservation, if we neglect friction losses, the T_e of the fluid at the inlet must equal the total T_e at the outlets. Because the total area of the outlets is three times the area of the inlet, the fluid is moving at one third of the velocity. Let the outlet velocity be V and the inlet velocity be 3V. Let m be the unit mass of the fluid.

T_ein = P_ein + m(3V)²/2. = T_eout = P_eout + mV².

and P_eout - P_ein = 4.5mV² - 0.5mv² = 4mV²

Pressure is higher at the outlets by 4mV².

I don't know how to make plainer than that.

Excuse me again but when all three outlets are open the total energy available at each opening cannot equal the energy available at the inlet. So if an even sharing of the energy occurs between the three outlets and if we neglect both friction losses and any effects from an elevation I get instead:

T_ein=3T_eout

P_ein + m(3V)²/2=3(P_eout + m(V)²/2)

P_ein +9/2mV² = 3P_eout + 3/2 mV²

P_ein +3mV² = 3P_eout

1/3 P_ein + mV2 = P_eout <or if you prefer>

P_ein = 3P_eout - 3mV²

Either way P_ein > P_eout by almost a factor of three. You should recognize that I was using your choice of unit symbols for velocity and mass.

I'll have to give it some thought but my initial response is to stick with my version because the total mass leaving must equal the total entering and I think your version has three times the mass leaving. I'll kick it around in the morning when I'm more able to reason.

I thought that you had just replaced the symbol m for rho because the character set CR4 uses looks too much like p. The above citation (#43 & #46)for the energy equations has rho as the density, not the mass itself. Clearly one third of the mass will be going through each of the three openings but the density will remain the same. It dawns on me now that it is also easy in these types of problems to have people inadvertently calculating the pressure at close to each other but in different locations. I believe my calculations are correct for the pressure just inside one of the three manifold exits. I calculated for this point because we really do not know the area of the manifold where all three exit manifolds join. This area has to be greater than the sum of the other three pipes. In that region I believe your analysis is correct if you assume the necessary added area (that reduces the velocity even further) and thus the pressure is higher in this common manifold part that is wider than the entrance.

To clarify, my calculations are just downstream of the manifold to each of the three pipes. The question was what is the outlet pressure not the manifold pressure. I believe you, passingontogreen, and others who insist that on a higher pressure have calculated the manifold pressure that is just upstream of the outlet but before the inlet. I agree that the manifold must be a higher pressure.

thx for all the replies...though it have been a bit confusing :P

Regarding the manifold it really just quite small, can hold it by hand; rectangular shape with 4 openings ...the hose size is 3/8".

frankly i am not very sure bernoulli principle applies here. just a suggestion from a friend and since it's a fluid dynamic.

anyway the system here is actually a water jet cutting system and its pressure is actually 300~500bar create by water jet pump. I just put 100bar earlier to simplify it.

the 3 outlets size where the water goes out to for cutting is a lot smaller than 3/8". I forget the size.

Within this system, i am planning to put valves to divert the flow before the outlets. That's why i would like to find out if outlet pressure is lower than inlet pressure, so i can safely get a (500 x S.F.)bar valves.

I will provide the layout later.

Are you talking about the outlet of the hose or the outlet of the manifold? The reason I ask is that you'll have some pressure drop across the manifold (may or may not be negligible) but coming out of the hose the pressure is atmosphere and you've traded all of your pressure energy for velocity.

And Bernoulli's equation is the conversation of energy so it always applies.

Hopefully my writing is understandable . I notice a lot replies state pressure would be lesser but how much? is it Poutlet = 500/3 = 167bar? Or it is less than that due to other factors.

First of all, Pout will definitely not be 500/3. You could divide the flowrate like this, but not the pressure.

I tend to agree with Tornado, there will be very little difference in pressure between the inlet and each outlet. This is mainly due to the fact that pressure is relatively high and flowrate relatively low (I make an assumption here about this type of system, as flowrate has not been given). Any changes in pressure will be due to friction or to the 'venturi effect' (changes in fluid velocity due to changes in cross-sectional area of the fluid path). It is impossible to calculate this without knowing flowrate and nozzle sizes. However, I think that we can be pretty certain that if there were any change in pressure here, it might be significant for a 10bar system, but would be insignificant for a 500bar system.

The main point is, outlet pressure will not be 500/3, so use 500bar valves.

Nope I can understand it and it looks like what I had pictured in my head. Unfortunately for you, I'm not going to solve this. Parallel systems like this need some iterative process to solve (like the Hardy-Cross method).

I will, however, try to point you in the right direction. For pressure drop across a manifold, contact a manufacturer and try to speak to their technical department. If they can't help then you're probably SOL. Also, look at the Crane Technical Paper 410 (I think that's what it's called). Everyone that deals with fluid mechanics needs one of these manuals. Yea, it's that good.

From what I can tell from your picture, I don't think you'll see the same pressure and flow in each outlet.

oops, it's not a hypothetical. Time for me to back away, I don't have the skills for a real problem.

I would point out that the pressure is only lower while there is flow through the tubes. When all three jets are closed, the pressure is equalized through the system.

Dang Redfred, you didn't let me defend myself lol.

He's right. Your calculation is total energy into the manifold is equal to total energy exiting one pipe. That total energy is divided by 3 since there are 3 pipe exiting.

Also, can we really neglect friction here? I know I didn't throw any numbers or variables into my argument but my concept was total energy in equals total energy out (counting losses). That's why I mentioned pressure drop.

I think the OP was asking an idealized question so it is fine to ignore anything not addressed in the question. By the same token, nobody has queried the inlet and outlets being horizontal and at the same elevation. Deviations from these simplifying assumptions just detract from the point of the question.

Agreed that the energy leaving is divided into three, but so is the mass. I think Redfred has three times as much mass leaving than mass entering. When I did my calculation, I mentally combined the three exit pipes into one pipe with three times the cross sectional area. I will return to it for another try. I was thinking of the pressure in the inlet and outlet tubes; the body of the manifold is not described so it cannot be included.

It is interesting that there can be such disagreement between us, even when we eliminate the troll-like contributions. I really do appreciate the civilized give and take.

I thought you had used "m" to represent the density of the fluid because the CR4 Greek character set for rho looks too much like the letter p for pressure. I followed this implied lead and the "m" in my equations is density not mass. Bernoulli's equation uses density instead of mass to simplify the calculations. I am now certain that your derivation is accurate for the manifold pressure. It does not matter how large the manifold area is, the volume of fluid transferred is set by the exit area. Your calculation shows a critical aspect that many can forget. The highest pressure level in a system may not be at the output of any pump.

I also appreciate the give and take, my friend.

Passingtongreen has it EXACTLY right. GA. And although it is exactly right, there is still confusion.

Maybe to help people visualize this we can provide a slightly different example how to properly deal with energy.

Imagine a very large vented tank (large enough to maintain a constant water surface elevation regardless of anything we take out). Now poke a hole in the bottom of it. The available energy at the bottom is equal to the pressure of the water column (rho * g * h). The water exiting the tank will have this same energy, but in kinetic form (1/2 rho v^2). Now move over a few feet and poke another hole in the bottom. The available energy at that hole is STILL (rho * g * h) and it will still exit at the same velocity as the first. The available energy is NOT divided by 2 because there are two holes (or 3 if we had 3 holes, or 4 if we had 4 holes, etc), the available energy is constant because the tank level hasn't changed.

Ok, let me start this reply by saying that passingtongreen does have it exactly right, a theoretical problem like this is fun to debate but when this comes to implementing a real world problem it is best to consult with people who do this daily. It is easy to become confused and mix units in a problem outside of one's field of expertise.

Thank you again redfred but once again I'm a second too late in defending myself lol.

Yes, we've oversimplified this problem to the point that it is not practical and now we're arguing about something that can't exist.

Let's expand on his manifold using the theory that everyone is telling me I'm wrong on. Let's say the manifold splits in 2. Without running calculations to get this exact, let's say velocity decreases by a factor of 2 and pressure increases by a factor of 2. Now, we split again just downstream to get another decrease in velocity (still a factor of 2) and increase in pressure (still a factor of 2). Then we split a third time...notice a pattern? Why bother purchasing a pump when we can simply add a few more manifolds?

I said this before but maybe I worded it wrong. We can't ignore friction (or minor losses) here. If we swage up on a line from 3" to 4", the velocity decreases but the expansion coefficient of the fitting keeps the pressure from changing.

I think we've oversimplified this problem and set the boundaries so narrow that we can't get a good solution in a short time frame.

My suggestion to the OP is contact the manifold manufacturer and get a resistance coefficient (or whatever they might call it) for the manifold and then apply Bernoulli's equation using the minor and major loss terms.

I didn't intend to step on your or anyones toes, but it does seem that you and I are in the same camp here. So I took this as defending our perspective. With the ambiguous jumbling of terms like energy per unit mass, total energy, kinetic and potential energies along with the myriad of different poorly defined locations for the OP original query of "outlet pressure" I'm certain that more than one of us is correct in implementing our own perspective.

This problem though has given me a lot more respect for the people who edit engineering textbooks.

Actually, the pressure will not vary by much throughout this manifold. Venturi considerations are only part of the picture. Pipe friction is another part, and absolute rather than relative velocities must be considered.

If you get some gauges, tees, and adapters, you could perform an experiment. If the inlet pressure is say 300 bar, the outlet pressures would probably be between 295 and 300 bar.

(The vast bulk of pressure drop occurs at the washer nozzles, unless you have a very unusual situation.)

But if you ignore friction then the pressure (force per unit area) remains constant at all points in a venturi, manifold, and even after the water leaves the outlet. We just spent this entire thread saying that at all points of a water flow the pressure must remain constant. The pressure must remain constant, even when it's diverted into three equal distributions. It must I tell you.

_{Somewhere there has to be a vector component. Changing the direction of a moving mass must involve a vector somewhere. Oh hello Victor, where did you come from? Who's this with you, glad to meet you Daniel. You say you know something about statistics? Great, lets chat. Crawling back under rock, with some new friends.}

Sorry, I don't follow this:

"But if you ignore friction then the pressure (force per unit area) remains constant at all points in a venturi"

If this were the case, then venturis wouldn't work, and neither would aircraft wings (help!). The principle is that a change in velocity results in a change in pressure.

"We just spent this entire thread saying that at all points of a water flow the pressure must remain constant. "

I haven't been saying this. Pressure is only constant if there is no change in velocity, or if there is zero flow.

I was pretty soundly disciplined that Bernoulli points out that one must consider the total pressure as the pressure of a system. The total pressure must always be preserved by the conservation of pressure rule. So no matter how many different directions an incompressible fluid flows in an N dimensioned frictionless manifold the total pressure of each exiting fluid flow has the same total pressure.

I know when I've been properly tutored here.

Ignorance is Strength.

Freedom is Slavery.

War is Peace.

I love Big Brother.

I think you are misremembering. It is total energy that is conserved. If the kinetic energy is changed because of changed velocity, the potential energy must be changed by an equal and opposite amount. If not, all hell breaks out, or perhaps hell freezes over, I'm not sure which.

One of the interesting things I've noted from this thread is that the final answer can be entirely different depending on whether you view the manifold as 'a system' (in isolation) or as 'part of a system'.

I thought it was conservation of energy, not pressure. In a venturi the increased velocity at the constriction results in a pressure drop. Energy remains the same. This is how a carburettor draws fuel in. So the pressure is not constant at all points in a venturi (nor at all points on an aircraft wing.

I seemed to be quite in sympathy with you, until your last sentence re. Big Brother (OMG!)

I brought Mr. Huxley's seminal work to this brawl because this is what I feel the discussion has devolved into and I have had to give up Julia.

I've certainly tried to bring (most of this time) a professional presentation of my understandings and analysis on how to use Bernoulli's equation. I too believed that the pressure in the middle of a flowing Venturi was less. Instead I've been shown that the total pressure must stay the same. I tried to derive the static pressure in each individual manifold pipe but was rebuked that one must include also the kinetic pressure for a result. Instead of any explanation on correcting my methodology, I feel that scripted dogma have been returned. Dogma can only be blindly accepted or rejected. I give up. Do it to Julia!

Oh, you mean THAT Big Brother! Respect restored.

I think most everyone here has felt the frustration with the terminology and definitions. I tried to clear it up at the beginning and did a poor job. -A- did a better job. It first appeared to me that people had some wild conceptual errors, which might be the case, but because it appears so many people are operating with different definitions for the same term I'm not sure anymore.

So I'm going to try something... define my terms so everyone is on the same page and work the problem with explanation where appropriate... Hopefully we'll all end up on the same page...

more to come...

Here lies one of my difficulties, the use of the term, "Kinetic Pressure". To me, kinetic pressure is felt by solid objects in air, it is the bombardment by randomly moving atoms and molecules. in the case of the balloon, there are more molecules per unit of space inside so the bombardment is greater on the inside and keeps the balloon inflated.

I have not heard it used with regard to fluid flowing through a system.

Related to that and my earlier capitulation, the term "kinetic" implies to me a vector component and yet it is used in the equation as a scalar value.

The Ministry of Truth will explain this when we need it.

The horrible thing about the Two Minutes Hate was not that one was obliged to act a part, but that it was impossible to avoid joining in.