Bernouli and Specific Gravity Problem

[Cramster]

What would happen to the pressure in a pipe with a flowing liquid if I were to double the specific gravity of the fluid?

Intuitively I would think the pressure at any point in the pipe would increase by a factor of 2 [this is the answer given in EK] but I don't see why this is true mathematically.

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[sangria1986]

What would happen to the pressure in a pipe with a flowing liquid if I were to double the specific gravity of the fluid?

Intuitively I would think the pressure at any point in the pipe would increase by a factor of 2 [this is the answer given in EK] but I don't see why this is true mathematically.

I'm gonn give this a shot...Im not entirely sure but I think this works...

pressure is the pressure of the liquid exerted on the pipes. As a result, it is the force F of the liquid over an area A on the pipe. Thus pressure of liquid on pipe equals F/A. Since F=ma and density times volume equals mass, F is directly proportional to the density of the liquid. Think about it, the denser the liquid is the more it is able to strike the pipes, increasing its force. because pressure is directly related to force whiich is directly related to density, increasing any one of these variables by a factor x will increase the rest by the same factor.

I bring up denisty because density is merely 10^3 Times specific gravity....since specific gravity is defined by the density of water!

 

[RogueUnicorn]

are we assuming no resistance? if so i don't understand how increasing the density would increase lateral pressure.. according to B's principle, pgh + 0.5pv^2 + P is constant, which means an increase in specific gravity (i.e. density) of a fluid while maintaining all else contant should decrease the lateral pressure.

 

[Cramster]

are we assuming no resistance? if so i don't understand how increasing the density would increase lateral pressure.. according to B's principle, pgh + 0.5pv^2 + P is constant, which means an increase in specific gravity (i.e. density) of a fluid while maintaining all else contant should decrease the lateral pressure.

Exactly!!! Mathematically it should decrease, though not necesarily by a factor of 2. The EK explanation states the pressure should increase by a factor of 2 anywhere along the pipe. Ugggh. If we assume there IS resistance is there some equation that would allow me to justify that the pressure increases by a factor of 2? (i'm thinking poiselles law but its related to pipe diameter by exponent of 4 i think) Heres a picture of the pipe, the vertical branches are sealed off.

 

[RogueUnicorn]

the only thin i can think of is the Reynolds number equation that deals with turbulent flow, but a high density would serve to increase resistance further dropping the lateral pressure... Poiseuille's law uses viscosity, not density.

i'm quite puzzled myself, perhaps someone could help us here

 

[sangria1986]

are we assuming no resistance? if so i don't understand how increasing the density would increase lateral pressure.. according to B's principle, pgh + 0.5pv^2 + P is constant, which means an increase in specific gravity (i.e. density) of a fluid while maintaining all else contant should decrease the lateral pressure.

Lol....are you guys just ignoring what I said?

there's a flaw in your thinking in using Bernouill's theorem...

his theorem is for a single liquid. that means, that the equation can only be used for a particular liquid with that density. therefore changing the density of the liquid means all of its parameters will change in that equation you mentioned.

for example, say you have a liquid with density X traveling at velocity V and at constant height throughout (so we can get rid of the potential energy part).....the velocity V it travels at is only for the density that is CONSTANT in that equation. changing the density does not mean plug and chug in this situation. you have to understand the context...changing density means the fluid is now different...i.e. you cannot use B's equation anymore. as a result, you cannot properly determine velocity or pressure or anything else that comes from the equation. BTW....where are you getting pressure from B's equation anyway? B's equation has P in it but it is ATMOSPHERIC PRESSURE. not the pressure of the fluid (just making sure no one is confused about that)

this is why I went back to the basics of pressure to discuss P=F/A

 

[RogueUnicorn]

Lol....are you guys just ignoring what I said?

there's a flaw in your thinking in using Bernouill's theorem...

his theorem is for a single liquid. that means, that the equation can only be used for a particular liquid with that density. therefore changing the density of the liquid means all of its parameters will change in that equation you mentioned.

for example, say you have a liquid with density X traveling at velocity V and at constant height throughout (so we can get rid of the potential energy part).....the velocity V it travels at is only for the density that is CONSTANT in that equation. changing the density does not mean plug and chug in this situation. you have to understand the context...changing density means the fluid is now different...i.e. you cannot use B's equation anymore. as a result, you cannot properly determine velocity or pressure or anything else that comes from the equation. BTW....where are you getting pressure from B's equation anyway? B's equation has P in it but it is ATMOSPHERIC PRESSURE. not the pressure of the fluid (just making sure no one is confused about that)

this is why I went back to the basics of pressure to discuss P=F/A

no you're incorrect on at least two counts. 1) the P is lateral pressure exerted by the fluid. 2) density has no effect on velocity when turbulence is ignored. i read your "basics" and i ignored it because it was miguided, or at least your understanding of it is.

 

[Phantastic]

Poiseuille's law gives the volume flow rate as being equal to the change in pressure * (pi*r^(4)/(8*viscosity*L).

If the volume flow rate is to remain constant, if we increase the viscosity, deltaP also must increase. Therefore, final pressure must be greater than initial.

Bernouli's eqn applies to ideal fluid flow I believe. Because they mentioned doubling the specific gravity, I assumed this wasn't an ideal fluid due to viscosity concerns.

Also, Bernouli's eqn should only apply when energy is not being added to the system (initial energy of fluid = final energy of fluid). By doubling the specific gravity, we've increased the momentum of each particle, and will therefore, in my mind, double the force exerted, as noted by the previous poster's response.

 

[RogueUnicorn]

we didn't increase viscosity, viscosity and density are separate terms and not necessarily correlated.

additionally, even if we accept the two are interchangeable, the rise in resistance means more pressure lost meaning even less lateral pressure exerted.

what's clear to me is that the question needs to give us more info about the assumptions

 

[surag]

a

 

[surag]

a

 

[sangria1986]

no you're incorrect on at least two counts. 1) the P is lateral pressure exerted by the fluid. 2) density has no effect on velocity when turbulence is ignored. i read your "basics" and i ignored it because it was miguided, or at least your understanding of it is.

Lets make it even simpler...I'll give you an example

lets say you have one tube with fluid A with density p(A) flowing at velocity A through a pipe with points X and Y at atmospheric pressure on both ends such that Bernouilli's equation is reflected as such:

P1+ p(A)*g*h1+ 1/2*v1*p(A)= P2+p(A)*g*h2+1/2*v2p(A).

where v1 h1 are velocity and height at point X and h2 and v2 are height and velocity of fluid at point Y
Because P1 and P2 equal atmosphere pressure you can ignore P1 and P2 (they cancel)....however, what you WILL notice is that DENSITY IS CONSTANT....you cannot change density for this equation to be usable!

However, you can compare two equations SEPARATELY with DIFFERENT densities.

Thus... we create TWO pipes. The first pipe with the same fluid as above and same equation and a SECOND pipe with a DIFFERENT density B.


Now that we have made that distinction lets compare the two pipes forces agains the walls of the pipe (ie definition of lateral pressure)

in pipe A you will have a pressure equal to F/A where F is the force of the fluid with density pA and area is the area of the pipe the force is acting on.

Since Force of the fluid is equal to the mass of the fluid times its acceleration or F=ma the Pressure on a given area of the pipe is also equal to P=ma/A

because density is equal to m/V and the change in force is dependent on change in mass, for a constant volume we can see that the forces is ALSO dependent on the DENSITY.

Thus, if the density of fluid B is twice that of the density of fluid A then the Force and resulting PRESSURE of fluid B is also TWICE THAT OF fluid A.

Jesus....its that simple.

 

[sangria1986]

basically you cant use bernouilli's principle in this example.

And to clarify....I WAS NOT WRONG...

ONE i never said that velocity is related to density...i merely stated that you CANNOT used density to calculate these values, it is NOT plug and chug. I was merely using velocity as an example.

Also TWO the P is NOT lateral pressure...lateral pressure, the pressure exerted by THE FLUID is actuall pgh which is derived from P=F/A as I had SAID BEFORE....before you, for absolutely no reason, condescendingly insult someone trying to help you maybe you should reread the "basics" of physics. Thanks for your underhand insult.

 

[RogueUnicorn]

i seriously can't continue you this because you clearly have no idea what you're talking about. i stopped reading as soon as you kept insisting P is atmospheric pressure. i didn't insult you, i said you are misguided. that being said, you're clearly an idiot. go back to the kids' table.

force is dependent on mass AND acceleration, get it right and hit some books or something. christ you're dense.

 

[kentavr]

May be I missed something but the question in the form stated by the topic-starter is undermined.

To Author: Can you, please post the full text of the problem?

The reason why it is not determined is that changing the density from the problem context something should be kept as constant:

1. It can be flow rate = const. (Mass per second through pipe section). Then given the same rate and twice density the speed will decrease (twice etc).

2. It can be energy flow = const, then Bernouli can be used. (assuming pump with the same power)

3. It can be speed=const, (If continuity is stated in the problem)

Some other conditions in the problem may be a clue what is left as constant.

All of these additional assumptions will give different answer. Without full specification of the problem it is useless to look for the right answer...

 

[sangria1986]

i seriously can't continue you this because you clearly have no idea what you're talking about. i stopped reading as soon as you kept insisting P is atmospheric pressure. i didn't insult you, i said you are misguided. that being said, you're clearly an idiot. go back to the kids' table.


force is dependent on mass AND acceleration, get it right and hit some books or something. christ you're dense.

For the record dude... I took the MCAT in 2005 and got a 14 on physical sciences, a 10 on verbal and a 11 on bio. I let my MCAT expire and I'm retaking it now (low GPA issues etc.) seriously...I'm trying to help you out.

You seriously have issues my man. I'm not going to insult you or anything but I'm really trying to help you out. And now, I want to make sure anyone who comes across this thread is not confused by you. You are SIMPLY WRONG.

But before I clarify, its painfully obvious you merely skimmed the last sentence I made and then threw yout that I was saying that force is ONLY dependent on mass...I never said that. The reason I mentioned only mass is because in my example with different fluids of different densities if the acceleration of the fluid is constant, IE only the mass changes, then the force change is dependent only on the change in mass. I have since changed it to state the CHANGE in force is dependent on the CHANGE in mass when the acceleration is constant (I left out the acceleration is constant part)

I mentioned only mass because when you change density ONLY the mass changes. I figured you'd be able to make that realization...I'm not some idiot who doesnt know what hes talking about.

Now that I've said that....go ahead, open your text book. You will see that P1 and P2 is the pressure exerted on the pipe at that point from the outside...IE the pressure exerted on the liquid at one opening is P1 and the pressure exerted on the other end is P2. This is NOT the pressure that the fluid exerts in the pipe.

 

[RogueUnicorn]

For the record dude... I took the MCAT in 2005 and got a 14 on physical sciences, a 10 on verbal and a 11 on bio. I let my MCAT expire and I'm retaking it now (low GPA issues etc.) seriously...I'm trying to help you out.

You seriously have issues my man. I'm not going to insult you or anything but I'm really trying to help you out. And now, I want to make sure anyone who comes across this thread is not confused by you. You are SIMPLY WRONG.

But before I clarify, its painfully obvious you merely skimmed the last sentence I made and then threw yout that I was saying that force is ONLY dependent on mass...I never said that. The reason I mentioned only mass is because in my example with different fluids of different densities if the acceleration of the fluid is constant, IE only the mass changes, then the force change is dependent only on the change in mass.

I mentioned only mass because when you change density ONLY the mass changes. I figured you'd be able to make that realization...I'm not some idiot who doesnt know what hes talking about.

Now that I've said that....go ahead, open your text book. You will see that P1 and P2 is the pressure exerted on the pipe at that point from the outside...IE the pressure exerted on the liquid at one opening is P1 and the pressure exerted on the other end is P2. This is NOT the pressure that the fluid exerts in the pipe.

bad assumption.

for the record, i'm not impressed by your mcat score. trust me.

finally, why don't you open it up, and explain to me the basis for reduced lateral pressure with increased fluid velocity. jesus.. it's a sealed pipe. do you know anything?

 

[PiBond]

looks like i missed all the fun

 

[sangria1986]

bad assumption.

for the record, i'm not impressed by your mcat score. trust me.

finally, why don't you open it up, and explain to me the basis for reduced lateral pressure with increased fluid velocity. jesus.. it's a sealed pipe. do you know anything?

Fine. That I can do. First let me clarify. I made it very clear in the beginning that in certain situations P1 equals ATM pressure and =P2. Ie when the pipe is open ended in same environment. I did not say that P1 has to equal ATM pressure.

Oh and btw. Only an idiot would think the whole pipe is closed! For if so. Tell me oh genius what exactly is causing the ffluid to flow? A closed pipe woul necessarily leave no room for the fluid to move....it would by stationary. Have u ever seen the fluid in a CLOSED pipe move? LOL. Even if there was a height difference a closed pipe would stop flowing at some point. Just use common sense. You must have that. So let me ask you...do you know anything? I don't mean that. I'm not trying to be mean. I only say it because you did.
Bernoullis principle does not apply for closed pipes
because for flow to take place non steady flow through a pump must exist. As you know the human circulatory system is closed but the heart pumps blood and pressure varies-systole and diastole. Making bernoullis principle useless for the system as a whole. The poster arguing however does not understand this. Please ignore him.
Second. It seems to me that your reading comprehension is poor. Repeatedly you have made comments that were patently false based on misreading what I've wrote. I assumed you just ddnt care about what I wrote. That's fine since obviously you don't respect me. But you also FAILED to read the OPs description. The pipe is open ended ONLY the vertical columns are closed!!

Please read again!

That being said I will explain why veloicity is related to inverse of pressure and why you are wrong. The velocity is inversly related to pressure.

suppose we take the beginning portion of a pipe and a pressure P is exerting on it To the right so that the fluid moves at velocity v to the right. Since pressure is related to force( before you bite my head off yes I know area is involved but it's not important here regarding my explanation) since pressure is related to force the fluid must also be accelerating. Thus at point two along the pipe in the direction of the fluid flow the velocity must be greater. As a result the pressure at this point on the fluid must be LOWER. Now the pressure measured and as described as P2 in the equation is in a direction OPPOSITE to P1!!!!! That is the way pressure is measured. Not in the same direction as P1. Why is it not in the same direction? Imagine the fluid at p1 pushing the fluid at p2 to the right. Because of inertia the fluid at p2 will resist. As a result it will exert a pressure opposite to the pressure of p1. If the fluid moves against p2. Ie to the right that means that p1 exerts greater pressure than p2. Thus the fluid at p2 has a velocity greater than at p1 due to acceleration! If u dont believe me u can look it up in YOUR textbook. Lol. Seems like you don't rmmbr much about physics

Anyway that being said why is this related to all this? The point is that bernoullis principle CANNOT be applied strictly here. This is because P1 and P2 reflect the pressure exerted on the fluid from EXTERNAL area. As I said before P1 is the force exerted on the water To somewhere and p2 is external force from opposite end(think about it the inertia at point 2 that resists motion of the flow from p1 is in part due to external pressure from opposite end.

Ok so now what? Basically when u change density the effects of p1 and p2 which are due to EXTERNAL(the pressure has To come initially from somewhere right) pressure are NOT affected. The change in density only affects the internal molecular activity of the fluid. Ie the mass change duetodensity causes a change in force and change in pressure. To better understand this read my explanations in the posts above where I describe how P=F/A is the only way you can go about it. I really wishi could draw to explain but I can't

you do not need anymore information than what is present. The guy arguing with me either needs to reread his textbook and not insult others. Dude if you want to be a doctor condescending to idiots(assuming I am one which I am not) isn't a great way to go about it. Shame on you. I didn't mention mcat to impress. Only to show that I'm not an idiot. Please reread the OPs posts and your physics text before you blather

 

[RogueUnicorn]

Fine. That I can do. First let me clarify. I made it very clear in the beginning that in certain situations P1 equals ATM pressure and =P2. Ie when the pipe is open ended in same environment. I did not say that P1 has to equal ATM pressure.

no you didn't say that in the beginning at all.

and btw. Only an idiot would think the whole pipe is closed! For if so. Tell me oh genius what exactly is causing the ffluid to flow? A closed pipe woul necessarily leave no room for the fluid to move....it would by stationary. Have u ever seen the fluid in a CLOSED pipe move? LOL.

so your blood is stationary??! lololol!!! twit. rest of your post is a tl;dr situation.

 

[sangria1986]

no you didn't say that in the beginning at all.



so your blood is stationary??! lololol!!! twit. rest of your post is a tl;dr situation.

The only type of pump that follows the principle is a centrifugal pump which utilizes conversion of kinetic and potential energy. As a result the principle can be applied

http://en.wikipedia.org/wiki/Centrifugal_pump

unfortunately our heart is not one such pump. Lol.

Are u seriously going to continue insulting me? Clearly your background in physics is subpar. BLOOD DOES NOT FOLLOW BERNOULLIS PRICIPAL.

It is PUMPED. Meaning that the pressure is constantly varied for forces outside of potential and kinetic energy. WOW ok. You're just stupid man. Seriously. Wow. I can't believe your out giving advice. Bernoulis principal is not useful for closed objects because for pressure and flow to exist a pump or energy not related to kinetic and potential must be used. Ie non conserved work. Bernoullis does not work for such reasons in human blood. Wow dude. At first I thought maybe you weren't paying attention. Now Im thinking you never took physics.

For the principle to work. Flow must be STEADY. Wow. Even ek will let u know bernoullis doesn't work. Go away man. Seriously. Just leave and study on your own

http://forums.studentdoctor.net/showthread.php?t=526219

 

[sangria1986]

And yes. I made it very clear for atmosperic pressure. Your reading comprehension skills are poor, you're extremely arrogant which means ur incapable of reasoning rationally, and you have knowledge issues considering you don't even know when bernoullis principle can be used. I see u post 5000 times. This is probably why. I'm through proving anything to you. You're an idiot. Anyone following this will know I'm right. Itruly hope for your sake on the mcat you study instead of wasting everyones time on this forum

 

[RogueUnicorn]

Fine. That I can do. First let me clarify. I made it very clear in the beginning that in certain situations P1 equals ATM pressure and =P2. Ie when the pipe is open ended in same environment. I did not say that P1 has to equal ATM pressure.

let's see again, all your posts previous.

For the record dude... I took the MCAT in 2005 and got a 14 on physical sciences, a 10 on verbal and a 11 on bio. I let my MCAT expire and I'm retaking it now (low GPA issues etc.) seriously...I'm trying to help you out.

You seriously have issues my man. I'm not going to insult you or anything but I'm really trying to help you out. And now, I want to make sure anyone who comes across this thread is not confused by you. You are SIMPLY WRONG.

But before I clarify, its painfully obvious you merely skimmed the last sentence I made and then threw yout that I was saying that force is ONLY dependent on mass...I never said that. The reason I mentioned only mass is because in my example with different fluids of different densities if the acceleration of the fluid is constant, IE only the mass changes, then the force change is dependent only on the change in mass. I have since changed it to state the CHANGE in force is dependent on the CHANGE in mass when the acceleration is constant (I left out the acceleration is constant part)

I mentioned only mass because when you change density ONLY the mass changes. I figured you'd be able to make that realization...I'm not some idiot who doesnt know what hes talking about.

Now that I've said that....go ahead, open your text book. You will see that P1 and P2 is the pressure exerted on the pipe at that point from the outside...IE the pressure exerted on the liquid at one opening is P1 and the pressure exerted on the other end is P2. This is NOT the pressure that the fluid exerts in the pipe.

basically you cant use bernouilli's principle in this example.

And to clarify....I WAS NOT WRONG...

ONE i never said that velocity is related to density...i merely stated that you CANNOT used density to calculate these values, it is NOT plug and chug. I was merely using velocity as an example.

Also TWO the P is NOT lateral pressure...lateral pressure, the pressure exerted by THE FLUID is actuall pgh which is derived from P=F/A as I had SAID BEFORE....before you, for absolutely no reason, condescendingly insult someone trying to help you maybe you should reread the "basics" of physics. Thanks for your underhand insult.

Lets make it even simpler...I'll give you an example

lets say you have one tube with fluid A with density p(A) flowing at velocity A through a pipe with points X and Y at atmospheric pressure on both ends such that Bernouilli's equation is reflected as such:

P1+ p(A)*g*h1+ 1/2*v1*p(A)= P2+p(A)*g*h2+1/2*v2p(A).

where v1 h1 are velocity and height at point X and h2 and v2 are height and velocity of fluid at point Y
Because P1 and P2 equal atmosphere pressure you can ignore P1 and P2 (they cancel)....however, what you WILL notice is that DENSITY IS CONSTANT....you cannot change density for this equation to be usable!

However, you can compare two equations SEPARATELY with DIFFERENT densities.

Thus... we create TWO pipes. The first pipe with the same fluid as above and same equation and a SECOND pipe with a DIFFERENT density B.


Now that we have made that distinction lets compare the two pipes forces agains the walls of the pipe (ie definition of lateral pressure)

in pipe A you will have a pressure equal to F/A where F is the force of the fluid with density pA and area is the area of the pipe the force is acting on.

Since Force of the fluid is equal to the mass of the fluid times its acceleration or F=ma the Pressure on a given area of the pipe is also equal to P=ma/A

because density is equal to m/V and the change in force is dependent on change in mass, for a constant volume we can see that the forces is ALSO dependent on the DENSITY.

Thus, if the density of fluid B is twice that of the density of fluid A then the Force and resulting PRESSURE of fluid B is also TWICE THAT OF fluid A.

Jesus....its that simple.

Lol....are you guys just ignoring what I said?

there's a flaw in your thinking in using Bernouill's theorem...

his theorem is for a single liquid. that means, that the equation can only be used for a particular liquid with that density. therefore changing the density of the liquid means all of its parameters will change in that equation you mentioned.

for example, say you have a liquid with density X traveling at velocity V and at constant height throughout (so we can get rid of the potential energy part).....the velocity V it travels at is only for the density that is CONSTANT in that equation. changing the density does not mean plug and chug in this situation. you have to understand the context...changing density means the fluid is now different...i.e. you cannot use B's equation anymore. as a result, you cannot properly determine velocity or pressure or anything else that comes from the equation. BTW....where are you getting pressure from B's equation anyway? B's equation has P in it but it is ATMOSPHERIC PRESSURE. not the pressure of the fluid (just making sure no one is confused about that)

this is why I went back to the basics of pressure to discuss P=F/A

nope, never stated at all. go back into your cave trolltard and work on your mcat score. come talk to me when it starts with a 4. and seriously. just look up the principle on google. it's right there in google. P is lateral pressure exerted by the fluid. it's only necessarily equal to atmospheric P at the open end of a pipe if it's flowing out.

 

[RogueUnicorn]

here i even did it for you.

http://www.scienceclarified.com/eve...ol-3-Physics-Vol-1/Bernoulli-s-Principle.html

..a slow-moving fluid exerts more pressure than a fast-moving fluid..

 

[sangria1986]

here i even did it for you.

http://www.scienceclarified.com/eve...ol-3-Physics-Vol-1/Bernoulli-s-Principle.html

Dude post # 16 bottom of paragraph
Now that I've said that....go ahead, open your text book. You will see that P1 and P2 is the pressure exerted on the pipe at that point from the outside...IE the pressure exerted on the liquid at one opening is P1 and the pressure exerted on the other end is P2. This is NOT the pressure that the fluid exerts in the pipe.

Ohhhhhh great your going to use a blogs wording as scientifically accurate language. Bravo. Wait. Oh right. It's the Internet. Is that how we study for the mcats now? Hmmm.

Let me find you some articles from ACTUAL sources


Hmm ohh here's from an UNIVERSITY
If Bernoulli's Equation and the model of ideal fluids applied to the human circulatory system, then the heart would have to beat but a few times to generate the required blood pressure in the closed circulatory system in order to circulate blood and then never have to beat again. Blood is viscous and lacks laminar flow so both Bernoulli's Equation and the Equation of Continuity do not directly apply to the human circulatory system. Still, some arguement may be made for the continuity equation and we use the continuity equation to describe some of the physics of a heart attack. To actually describe the physics of a heart attack would require some incredible effort of the author that is beyond his skill in physics. So in the following section, we take a more qualitative approach to the physics of a heart attack.

http://www.uaf.edu/pair/physics/heart.html

please ask your "physics professor" at the "med" school your at to clarify thus for you. Perhaps you were asleep in class?

One of the most important contexts for Fluid Mechanics reasoning for future doctors is the human circulatory system. However, Bernoulli's Law can only be applied in very special and localized circumstances in the cardiovascular system because blood is not an ideal fluid
http://www.wikipremed.com/mcat_course.php?code=0101080402000000

second link that it only works in certain areas like the aorta....

 

[RogueUnicorn]

again with the tl;dr. when you finally realize P is pressure exerted by the fluid, you'll feel pretty dumb. once you've finally gotten around to it, please return, then we can have an actual discussion. kthxbye, mr. 35.

 

[sangria1986]

Ok man...here's an example that should shut you up.

Do you agree that for open ended pipes that Pressure P1 and P2 must equal the outside pressure? You've agreed before and even if you dont you'll come across a dunce because theres ample evidence on your Internet and in YOUR TEXTBOOK that P1 and P2 equal atmospheric pressure for open ended pipes on surface....

suppose this is the case. Then P1 and P2 can be ignored for a pipe(with uneven height, ie P1 is higher than P2) that is on earths surface.... In this case, the height of the pipe is not so great that atmospheric pressure changes enough to be a difference between P1 and P2. This means that you can IGNORE P1 and P2.

But suppose YOU are correct in that P1 and P2 reflect the pressure of the fluid....LOL....THAT WOULD MEAN THAT there is NO pressure of the fluid since you just said that P1 and P2 describe the pressure of the fluid itself....this means that if the fluid is not exerting a pressure it is not moving. Yet we know that it must be if the pipes heights are not the same...WHICH MEANS YOU ARE WRONG.

Game. Set. Match

Hook. Sink. Line

Check. And. Mate

Read. A. Physics. Book.


Haha. We man. U clearly lack any real background in physics. I'm done arguing w u idiot. Lol. Have fun plug n chugging mcat. I'm sure ull do great.

If u PM me I'll show u my mcat.. 35 isn't spectacular although my 14 is.

 

[RogueUnicorn]

lol i already took my mcat thanks. glad you know your 35 from 5 years ago is in fact nothing spectacular. at least you got something right.

 

[sangria1986]

lol i already took my mcat thanks. glad you know your 35 from 5 years ago is in fact nothing spectacular. at least you got something right.

you're a ******...I took the paper version...and yes that was MUCH harder.

oh and btw...genius, you have no comment for what I said above...or anything else. other than personal attacks. Basically....you're just wrong. GO AWAY

 

[RogueUnicorn]

you're a ******...I took the paper version...and yes that was MUCH harder.

oh and btw...genius, you have no comment for what I said above...or anything else. other than personal attacks. Basically....you're just wrong. GO AWAY

awww you want a kiss on the booboo? i have no comment on what you have to say because you've shown yourself to be clueless. check again buck-o, you're the ones who came in with the guns blazing, although it was cute of you to edit it out after the fact.

 

[Cramster]

May be I missed something but the question in the form stated by the topic-starter is undermined.

To Author: Can you, please post the full text of the problem?

The reason why it is not determined is that changing the density from the problem context something should be kept as constant:

1. It can be flow rate = const. (Mass per second through pipe section). Then given the same rate and twice density the speed will decrease (twice etc).

2. It can be energy flow = const, then Bernouli can be used. (assuming pump with the same power)

3. It can be speed=const, (If continuity is stated in the problem)

Some other conditions in the problem may be a clue what is left as constant.

All of these additional assumptions will give different answer. Without full specification of the problem it is useless to look for the right answer...

The volume flow rate is kept constant and the specific gravity is doubled.

I should have mentioned this before, my bad

 

[sangria1986]

awww you want a kiss on the booboo? i have no comment on what you have to say because you've shown yourself to be clueless. check again buck-o, you're the ones who came in with the guns blazing, although it was cute of you to edit it out after the fact.

what are you talking about man? the only things I edited out were clarifications that for anybody else would be obvious.

you have issues.... oh and check out what OP says above...flow rate is constant...no pumping, no external forces....conservation of energy maintained....IE NOT A CLOSED PIPING...GO AWAY.

 

[RogueUnicorn]

what are you talking about man? the only things I edited out were clarifications that for anybody else would be obvious.

you have issues.... oh and check out what OP says above...flow rate is constant...no pumping, no external forces....conservation of energy maintained....IE NOT A CLOSED PIPING...GO AWAY.

you can have a completely open system - e.g. wings of an airplane in air - and P is STILL pressure exerted by the fluid, nimrod. god you're pathetic

 

[kentavr]

The volume flow rate is kept constant and the specific gravity is doubled.

I should have mentioned this before, my bad

Thanks, Cramster.

That clarifies the conditions. May be it has been already answered above, I didn't check. Too much words, so I will share my view on this paradox.

1. This question is not simple. Since it includes some internal paradox which was not cleanly mentioned in standard hydrodynamics textbooks.

Let first simplify the question. Let pipe be straight, with constant radius and at the same level. Then Bernouli said:

P+rho*g*h+0.5*rho*v^2=CONST.

What it tells us? if delta h = 0, cross-section does not change and volume per second = const, then we mast conclude that P=const. at any pipe point.
It looks right, because if the speed is const, then sum of external forces should be zero.

Here is two consequences:

a) The rho does not affect pressure; Since I can apply any additional external pressure to both ends of the pipe, but movement will not change.

b) We need to avoid a mistake, to equate two bernoulli equations. For different rho we don't know if the CONST is the same. (it is not)

All in all for simple pipe Pressure does not depend on density. One more very important point that I will be using later: No work is done. It is just inertia.

2. Now, the interesting part. Let say our pipe has two segment: one with cross-section S, another S'.

Should we perform work in order to keep speed steady in this system? Surprise: We should. Why? Because we changing the Kinetic energy for the same small mass of liquid. Let me show that,

Let's apply force F to the entrance of the pipe using piston. Since we already proved that P=const in the pipe of the same radius, then we will apply force equal P*S for some distance X and perform the work P*S*X.
Are we done with work. Not yet: We need to remove the work which was done from another side of the pipe, or P'S'X', Total work o this segment of liquid is : W=(P-P')*Volume. (Volume=S*X=S'*X')

Where this work is going? To change the kinetic energy of the volume that we pushed to the second segment. Speed there is faster. Change in kinetic energy will be:
0.5*M*V^2=0.5*rho*Volume* (delta V)^2.

Work is equal to the change in energy for ideal liquid and we get:

(P-P')*Volume = 0.5*rho*Volume*(delta V)^2 or

P-P' = 0.5 * rho * (delta V)^2.

3. Last step.
Since by the problem conditions Volume per second = const, then (delta V)^2 is the same for any liquid. Then(collecting the constants)

P = P' + k*rho

Linear equation.... P grows with rho. The EK's authors may assume that P'=0 and get direct proportionality as in the answer.

4. Conclusions:
a) To keep constant speed in the pipe with changing diameter, we must do the work: Not against the friction or viscosity, but to speed up or slow down the mass that we pushing throw. This point is missed in basic textbooks.
b) If we are not keeping up with this work, the speed will drop or we will get not stationary flow.
c) We can use calculus to generalize this model for pipe of any shape.

Sorry, if it gets too long, but I hope this will clear the misunderstanding. I think it is not very fair from EK, to ask this question and expect a solution in 90 seconds. Best of luck.

 

[sangria1986]

you can have a completely open system - e.g. wings of an airplane in air - and P is STILL pressure exerted by the fluid, nimrod. god you're pathetic

WTF. This has absolutely nothing to do w what I said. It's like your posting about another topic. Lol. U are I think just confused

 

[sangria1986]

Thanks, Cramster.

That clarifies the conditions. May be it has been already answered above, I didn't check. Too much words, so I will share my view on this paradox.

1. This question is not simple. Since it includes some internal paradox which was not cleanly mentioned in standard hydrodynamics textbooks.

Let first simplify the question. Let pipe be straight, with constant radius and at the same level. Then Bernouli said:

P+rho*g*h+0.5*rho*v^2=CONST.

What it tells us? if delta h = 0, cross-section does not change and volume per second = const, then we mast conclude that P=const. at any pipe point.
It looks right, because if the speed is const, then sum of external forces should be zero.

Here is two consequences:

a) The rho does not affect pressure; Since I can apply any additional external pressure to both ends of the pipe, but movement will not change.

b) We need to avoid a mistake, to equate two bernoulli equations. For different rho we don't know if the CONST is the same. (it is not)

All in all for simple pipe Pressure does not depend on density. One more very important point that I will be using later: No work is done. It is just inertia.

2. Now, the interesting part. Let say our pipe has two segment: one with cross-section S, another S'.

Should we perform work in order to keep speed steady in this system? Surprise: We should. Why? Because we changing the Kinetic energy for the same small mass of liquid. Let me show that,

Let's apply force F to the entrance of the pipe using piston. Since we already proved that P=const in the pipe of the same radius, then we will apply force equal P*S for some distance X and perform the work P*S*X.
Are we done with work. Not yet: We need to remove the work which was done from another side of the pipe, or P'S'X', Total work o this segment of liquid is : W=(P-P')*Volume. (Volume=S*X=S'*X')

Where this work is going? To change the kinetic energy of the volume that we pushed to the second segment. Speed there is faster. Change in kinetic energy will be:
0.5*M*V^2=0.5*rho*Volume* (delta V)^2.

Work is equal to the change in energy for ideal liquid and we get:

(P-P')*Volume = 0.5*rho*Volume*(delta V)^2 or

P-P' = 0.5 * rho * (delta V)^2.

3. Last step.
Since by the problem conditions Volume per second = const, then (delta V)^2 is the same for any liquid. Then(collecting the constants)

P = P' + k*rho

Linear equation.... P grows with rho. The EK's authors may assume that P'=0 and get direct proportionality as in the answer.

4. Conclusions:
a) To keep constant speed in the pipe with changing diameter, we must do the work: Not against the friction or viscosity, but to speed up or slow down the mass that we pushing throw. This point is missed in basic textbooks.
b) If we are not keeping up with this work, the speed will drop or we will get not stationary flow.
c) We can use calculus to generalize this model for pipe of any shape.

Sorry, if it gets too long, but I hope this will clear the misunderstanding. I think it is not very fair from EK, to ask this question and expect a solution in 90 seconds. Best of luck.

Thank you. This is along the lines of what I said. Finally someone else to confirm!!!!!

 

[RogueUnicorn]

Thank you. This is along the lines of what I said. Finally someone else to confirm!!!!!

delusion lol

 

[kentavr]

Thank you. This is along the lines of what I said. Finally someone else to confirm!!!!!

I am sorry Sir, I did not confirm what you said. Because I did not read after your first post. The reason I didn't read, that you took strictly dynamic approach based on Newton 2nd law. (If I clear all you have said related to the problem)
This approach is a valid one, but I am not sure that you can finish what you have started. Because, your main assumption is the stationarity of the field for acceleration vector. It is hidden in your(F=ma formula). Why you so sure that acceleration field will not change for different density? For speed in stationary cross-section we are guarantied by continuity equation.
If I say that this has to be proven, you have no other choice, but to deduce the Navier–Stokes equation from the field theory. Let's say you'll do it, and even will be able to solve it. Unfortunately, the border condition for this differential equation as shown by the author of the topic does not have continuous first derivative. And that moves the problem to the rocket science area. (And, I mean literally). Since in dynamic of the gas flowing from the engine the derivative of the rocket wall is also not continuous. There are some solution to that, but would you really go there on MCAT? Sorry, I don't have time to comment anymore.

 

[RogueUnicorn]

I am sorry Sir, I did not confirm what you said. Because I did not read after your first post. The reason I didn't read, that you took strictly dynamic approach based on Newton 2nd law. (If I clear all you have said related to the problem)
This approach is a valid one, but I am not sure that you can finish what you have started. Because, your main assumption is the stationarity of the field for acceleration vector. It is hidden in your(F=ma formula). Why you so sure that acceleration field will not change for different density? For speed in stationary cross-section we are guarantied by continuity equation.
If I say that this has to be proven, you have no other choice, but to deduce the Navier–Stokes equation from the field theory. Let's say you'll do it, and even will be able to solve it. Unfortunately, the border condition for this differential equation as shown by the author of the topic does not have continuous first derivative. And that moves the problem to the rocket science area. (And, I mean literally). Since in dynamic of the gas flowing from the engine the derivative of the rocket wall is also not continuous. There are some solution to that, but would you really go there on MCAT? Sorry, I don't have time to comment anymore.

 

[sangria1986]

I am sorry Sir, I did not confirm what you said. Because I did not read after your first post. The reason I didn't read, that you took strictly dynamic approach based on Newton 2nd law. (If I clear all you have said related to the problem)
This approach is a valid one, but I am not sure that you can finish what you have started. Because, your main assumption is the stationarity of the field for acceleration vector. It is hidden in your(F=ma formula). Why you so sure that acceleration field will not change for different density? For speed in stationary cross-section we are guarantied by continuity equation.
If I say that this has to be proven, you have no other choice, but to deduce the Navier–Stokes equation from the field theory. Let's say you'll do it, and even will be able to solve it. Unfortunately, the border condition for this differential equation as shown by the author of the topic does not have continuous first derivative. And that moves the problem to the rocket science area. (And, I mean literally). Since in dynamic of the gas flowing from the engine the derivative of the rocket wall is also not continuous. There are some solution to that, but would you really go there on MCAT? Sorry, I don't have time to comment anymore.

You are making this unnecessarily complicated. There is no reason to bring up navier stokes into this. This fluid follows newtonian mechanics because it is non compressible and laminar-straight line flow.*
You ask how do I knw that acceleration field does not change. The acceleration field only changes in inviscid or compressible fluids and this is where the navier stokes theorem is used. *

However just because you have two different fluids of different densities does not mean the acceleration fields change. *Indeed if this were the case then bernouillis principle based on conservation of energy and the use of rho *g(which is ACCELERATION btw) *h could not be used. *

Perhaps an example will help. Imagine two pipes one w fluid w density d and in the other pipe fluid w density 2d. If
Both pipes are the same (placement height etc) and both fluids flow unidirectionally then the acceleration field will not change (the acceleration of a linear body will not change direction unless acted upon by another force). **

The only thing changing between the two pipes is the amount forced necessary to move the heavier fluid not the direction !!

And of course we can keep the actual acceleration value the same. We merely compensate for the increase in density(mass) by increasing the force.*

Ie for an object with mass moving in a direction with acceleration a then if you change the mass by factor 2 and move it again w accelleration a to maintain the same acceleration you only need to increase the force by factor 2. *This is what f=ma means

If the fluid ITSELF were to change density in the pipe then we would have a problem.
*
*However nowhere do I say this. Indeed your mention of the continuity equation necessarily requires a constant density rho throughout the fluid. *Otherwise the continuity equation cannot be used. I am not debating that the density is the same throughout the SAME pipe *

Having agreed to this I merely compared two separate pipes w different fluids varying by density rho(say one fluid has density twice the other). In this case assuming that the laminar flow is present then we can compare the force and hence pressure *caused by the fluid in one pipe to the force and pressure *by the other fluid (separate pipe) *

 

[sangria1986]

Shutup idiot

 

[sangria1986]

FYI. While it's true that rho g h is from
the potential energy we can derive it using what I said as a check withe necessary assmptions obviously.

P=F/A. F=ma. So P=ma/A since d(density)*V= m then
P= d*V*a/A
since V=A*h then
P=d*A*h*a/A. The As cancel out and you are left with
P=d*h*a. Substituting g(acceleration due to gravity) for a
P=d*h*g. Or rho g h. Same as the pressure due to potential. And we are using f=ma here. TY