EK 1001 num 569

[What up doc]

Advertisement - Members don't see this ad

When comparing two points of fluid flowing through the same horizaontal pipe, if the fluid velocity is greater, then:

a. temp is less
b. temp is greater
c. temp is unchanged
d. pressure is greater


answer is A....wtf? can someone explain...

 

[iA-MD2013]

I don't know if you can say this definitively for any fluid...because it depends on the fluid's viscosity. But generally for less viscous fluids and gases, increasing the temp decreases the velocity because it increases the number of collisions between the molecules. Because the collisions aren't exaclty elastic, this causes loss of kinetic energy/velocity and the fluid moves slower within a pipe.
So, lower temperatures causes greater velocity.
EDIT: Somebody please verify this...I'm not completely sure. I am second guessing myself.

 

[tncekm]

This is a pretty out there question... really.

Okay, so, when this fluid in motion speeds up it loses pressure because it loses its "random motion" kinetic energy to a translational kinetic energy. Random motion kinetic energy is the one related to temperature.

So, we can look at it this way:

ΔP + ρgΔy + ½ρΔ(v&#178😉 = 0 ; rewritten in terms of the energy values being conserved here and nothing there is no ρgΔy (no change in gravitational potential energy of the fluid):

&#916;P = -&#189;&#961;&#916;(v&#178😉 < 0 (because v_final > v_initial)
- This is proof we lost "pressure" (or kinetic energy that is NOT translational) to translational kinetic energy which does not exert any pressure.

Their explanation calls on KE = 3/2 RT, so we can rewrite this as &#916;KE = 3/2 R&#916;T

&#916;T = 2/3&#916;KE this tells us that since &#916;KE < 0, that &#916;T< 0, so the temperature had to go down.

This is according to their explanation; personally, I would have thought that temperature would have increased -- but maybe it wouldn't in the absence of friction?

 

[What up doc]

This is a pretty out there question... really.



&#916;P = -½&#961;&#916;(v²) < 0 (because v_final > v_initial)
- This is proof we lost "pressure" (or kinetic energy that is NOT translational) to translational kinetic energy which does not exert any pressure.



explain this again...?

 

[Kaustikos]

This is a pretty out there question... really.



&#916;P = -&#189;&#961;&#916;(v&#178😉 < 0 (because v_final > v_initial)
- This is proof we lost "pressure" (or kinetic energy that is NOT translational) to translational kinetic energy which does not exert any pressure.



explain this again...?

This is using bernoulis equation and saying that it equals to zero for that point. We have, therefore, the final equation: &#916;pressure = -1/2density * &#916;velocity^2.

Derived from: &#916;P+ dg&#916;y(which is = 0 since no change in height) + 1/2d&#916;v^2 = 0

&#916;P + 1/2d&#916;v^2 = 0. For simplicity's sake, we're taking out the 1/2 and density

&#916;P +&#916;v = 0. Subtract the v to get &#916;P = -&#916;V. We know that we get an increase in velocity, which means a positive &#916;, which makes the &#916;P a negative increase, which means the pressure decreases.

I don't know how that correlates to KE = 3/2RT, though.

Couldn't you just use PV = nRT? Pressure decreases and therefore the temperature decreases since volume remains?

Actually, I'm an idiot. Temperaute is average kinetic energy of a particle. It all comes together!

 

[tncekm]

&#916;P + 1/2d&#916;v^2 = 0. For simplicity's sake, we're taking out the 1/2 and density

&#916;P +&#916;v = 0. Subtract the v to get &#916;P = -&#916;V. We know that we get an increase in velocity, which means a positive &#916;, which makes the &#916;P a negative increase, which means the pressure decreases.

I don't know how that correlates to KE = 3/2RT, though.

Couldn't you just use PV = nRT?

The 'v' in Bernoulli's equation (which is invaluable for fluids in motion) is the velocity, not the volume 🙂

 

[tncekm]

This is a pretty out there question... really.



&#916;P = -½&#961;&#916;(v²) < 0 (because v_final > v_initial)
- This is proof we lost "pressure" (or kinetic energy that is NOT translational) to translational kinetic energy which does not exert any pressure.



explain this again...?

That statement above says that because the velocity increased, the pressure decreased.

There are two types of kinetic energy that I'm familiar with: 1) the kinetic energy of a molecule associated with its high speed random motion (which is also associated with its pressure and thermal energy), and 2) the translational kinetic energy of a large moving body (i.e. a fluid mass flowing in one direction).

Pressure decreasing will cause a reduction in type 1 kinetic energy.

Since KE = 3/2 RT, KE is proportional to T, so if KE goes down, T must go down.

And, its important to note that you cannot use PV=nRT here because PV=nRT is the "ideal gas" law and we're working with ideal fluids in motion.

What personally gets me is that if you take some water and run it through a very long tubing it will increase in energy. But, I think that's from friction increasing its internal energy, so in the case where there is no friction, I guess that wouldn't happen and we'd get the result I described.

 

[Kaustikos]

The 'v' in Bernoulli's equation (which is invaluable for fluids in motion) is the velocity, not the volume 🙂

No, I know that. God, if I don't know that by now, I've been getting remarkably lucky on questions.

I meant; couldn't you use the PV = nRT and ignore the volume/moles/R and just have P = T relationship to better explain why the decrease in pressure from bernouli shows a decrease in temperature? But I don't know if that works in this case.

edit - GOD, I need to learn to read. Yeah, it's the ideal gas law, not the ideal particle law. lol. Disregard that statement, despite how it may seemingly work in this example.

 

[What up doc]

ahhh makes sense now! would you say this is beyound the scope of the mcat....well at least, level of difficulty...and i forgot that it said it is an idea fluid, which i think means no friction.....

 

[tncekm]

Not sure if its beyond the scope... I find it "possible", but I haven't seen many questions that tricky on my few practices.

 

[unsung]

This is a pretty out there question... really.

Okay, so, when this fluid in motion speeds up it loses pressure because it loses its "random motion" kinetic energy to a translational kinetic energy. Random motion kinetic energy is the one related to temperature.

So, we can look at it this way:

&#916;P + &#961;g&#916;y + ½&#961;&#916;(v²) = 0 ; rewritten in terms of the energy values being conserved here and nothing there is no &#961;g&#916;y (no change in gravitational potential energy of the fluid):

&#916;P = -½&#961;&#916;(v²) < 0 (because v_final > v_initial)
- This is proof we lost "pressure" (or kinetic energy that is NOT translational) to translational kinetic energy which does not exert any pressure.

Their explanation calls on KE = 3/2 RT, so we can rewrite this as &#916;KE = 3/2 R&#916;T

&#916;T = 2/3&#916;KE this tells us that since &#916;KE < 0, that &#916;T< 0, so the temperature had to go down.

This is according to their explanation; personally, I would have thought that temperature would have increased -- but maybe it wouldn't in the absence of friction?

Yeah, you've got it. I know, it's one of those "concept" Qs that's so typically EK... probably beyond scope of MCAT, but you never know.

I don't remember what page this was on, but EK gave the example of running through this area with bees chasing you... ring a bell? Anyway, their whole point is that there are two different types of velocity:

"Random" motion which is just like Brownian motion (I think of it as just vibrating molecules), and "translational" motion which is actual displacement in the direction of your motion. So random motion is responsible for temp, and translational motion is what we usually associate with KE.

Then it's just a matter of energy conservation: higher temperature fluid uses up more of its energy on "random" motion, so less of it is freed to actually move forward as a collective entity (as opposed to random motions of individual molecules). A lower temperature fluid uses up less of its energy just vibrating around, so has more energy available to it to move forward as a collective entity.

Anyway, that's how I think about it... sounds like you've got a good handle on it now, but it helped me just to type that out again.. heh.

 

[Kaustikos]

Yeah, you've got it. I know, it's one of those "concept" Qs that's so typically EK... probably beyond scope of MCAT, but you never know.

I don't remember what page this was on, but EK gave the example of running through this area with bees chasing you... ring a bell? Anyway, their whole point is that there are two different types of velocity:

"Random" motion which is just like Brownian motion (I think of it as just vibrating molecules), and "translational" motion which is actual displacement in the direction of your motion. So random motion is responsible for temp, and translational motion is what we usually associate with KE.

Then it's just a matter of energy conservation: higher temperature fluid uses up more of its energy on "random" motion, so less of it is freed to actually move forward as a collective entity (as opposed to random motions of individual molecules). A lower temperature fluid uses up less of its energy just vibrating around, so has more energy available to it to move forward as a collective entity.

Anyway, that's how I think about it... sounds like you've got a good handle on it now, but it helped me just to type that out again.. heh.

Best summation EVER👍

 

[tncekm]

Update: I saw a question "slightly similar" to this on a Princeton Review test (got it right 😀). It was a bit more simple, and more based off of PV=nRT because it was for gases, but, I guess its possible to get these types of abstract questions.

 

[generator]

ok. I'm still not getting this one. From what I'm reading on this thread, changeKE = 3/2 RchangeT explains everything. Fine. I can accept this.

BUT, on audio osmosis disc 3 track 12 (fluids in motion) Jordan says, "when the molecules of a fluid move faster, the fluid gets hotter."

Why does ^ the above not apply for this question?

 

[generator]

please help!

 

[generator]

nevermind. ignore all this, i'll just accept it as fact for this particular situation :]

 

[MT Headed]

BUT, on audio osmosis disc 3 track 12 (fluids in motion) Jordan says, "when the molecules of a fluid move faster, the fluid gets hotter."

You need to listen to the rest of the track (it's only 75 seconds long!). Jon explains the difference between random translational motion which creates temperature, and uniform translational motion which does not.

 

[generator]

You need to listen to the rest of the track (it's only 75 seconds long!). Jon explains the difference between random translational motion which creates temperature, and uniform translational motion which does not.

yeah I originally thought #569 would be an example random translational... but I guess its actually not.

Thank you very much MT Headed. I feel like you always answer my mcat questions 😍

btw... just as a note for anyone who may be confused in the future, in the lecture on Bernoulli's equation they say flat out that pressure and temperature decrease as velocity increases :] yay