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Ideal Fluid with Pressure..

Started by V4viet
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V4viet

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Can you guys help me out with this question?

An ideal fluid with pressure P flows through a horizontal pipe with radius r. If the radius of the pipe increased by a factor of 2, which of the following most likely gives the new pressure?

A. P
B. 4P
C. 16P
D. the new pressure cannot be determined without more informations

the answer is D.. can someone please help me with this question..maybe im reading too much into the question..
 
This may be related to Bernoulli's equation. You cannot determine the pressure because they don't give you the fluid's velocity?
 
maybe I'm wrong, but here's my guess:

it seems like the only direct result of increasing the pipe radius (and thus area) will be to reduce the fluid velocity at a given point in the pipe, but we don't know what magnitude a velocity change will have on pressure, because we don't have a reference point.

though not relevant to the answer, it seems like doubling radius should quadruple pipe area. If volume flow rate remains constant, then new velocity at a given point will be 1/4 of old velocity at that point.

r= density
K= constant

in equations, we have 0.5rv^2 + rgh + P = K
assuming gh is constant, we can substract it from K, which is still K.

thus, if initially P = K - .5rv^2,
then new P = K - .5r (v/4)^2

but what does this tell us - we have P1/P2 = (K+x)/(K+y)

we need to know K.

Suppose this pipe were filled with ideal water and were very long and descended 10,000 metres, had a flat stretch where this problem takes place, and has an extremely slow velocity.

Assuming we could use the approximation of 1 atm / 10 atm, and the pressure were approximately 1000 atm, doesn't it seem counterintuitive that making a change to the radius of one little part of the pipe could increase pressure by thousands of atm?


warning: I haven't taken any science classes in 8 years, so please read my posts with caution.
 
Can you guys help me out with this question?

An ideal fluid with pressure P flows through a horizontal pipe with radius r. If the radius of the pipe increased by a factor of 2, which of the following most likely gives the new pressure?

A. P
B. 4P
C. 16P
D. the new pressure cannot be determined without more informations

the answer is D.. can someone please help me with this question..maybe im reading too much into the question..

To get the best answers I'm going to direct you to the subforum link at the top of the MCAT forum.
 
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Correction. ...the answer is NOT C! lol sorry about that!



The correct answer is C. 16P

This is because the flow Q, and attributes of a tube it is flowing in (radius, length, etc) are related by Pouisselle's equation:

Q = change in pressure x 3.14 x r^4/8nL

This is not an equation we need to memorize, but be familiar with...what it says in terms of concept is:

1. Flow is inverseley proportional to length. When the length is longer (tube)...the flow will be less (Q = flow = can be thought of as velocity of fluid)

2. and it also says Flow is directly proportional to pressure, and if pressure goes up (and it DOES, since you are increasing the radius of the tube...ie, more fluid particles slamming against the walls of the tube), ...then flow goes up. But by how much? From the equation, Q ~ r^4....so if you increase the radius by 2, then you increase the flow by 2 ^4.....which is 16. And remember Q is directly proportional to P....so it makes sense then that the answer is 16P (16 x 1P).





Can you guys help me out with this question?

An ideal fluid with pressure P flows through a horizontal pipe with radius r. If the radius of the pipe increased by a factor of 2, which of the following most likely gives the new pressure?

A. P
B. 4P
C. 16P
D. the new pressure cannot be determined without more informations

the answer is D.. can someone please help me with this question..maybe im reading too much into the question..
 
The correct answer is C. 16P

This is because the flow Q, and attributes of a tube it is flowing in (radius, length, etc) are related by Poiseuille's equation...
No it isn't. The question is about an ideal fluid; Poiseuille's equation applies to real fluids. (Incidentally, your answer isn't right for a real fluid either -- your analysis gives an answer for pressure gradient, not total pressure.)

With an ideal fluid, the problem is straightforward (if not obvious):

- First, by the continuity equation (flow rate stays constant), flow speed will decrease by a factor of four.

- Second, that will decrease the kinetic energy density (the 0.5(rho)v^2 term in the Bernoulli equation) by a factor of 16.

- Third, the effect on P cannot be computed because you don't know what the other two terms (pressure and potential energy density) were to start with -- the effect could be anywhere from negligible to infinite, in terms of the factor by which P is increased.


In the future, ask science questions in the study question subforum (see my sig).
 
OOps...thanks for the correction. Looks like i gotta review that section again.


Why does the flow speed decrease by a factor of 4?
 
flow rate = Q = (Area)(Velocity) = constant

Area for a circle = (Pi)(r)(r)

So according to the flow rate equation, increasing the radius by a factor of 2 (you have to square it to find area) will decrease the velocity by a factor of 4. The flow rate is always constant.

OOps...thanks for the correction. Looks like i gotta review that section again.


Why does the flow speed decrease by a factor of 4?