Bernoulli's equation

[fashafosho]

So I hate it.
Am I royally disadvantaged if I completely ignore it?

---

Members don't see this ad.

 

[BloodySurgeon]

umm... my momma always says, "just sit ur A S S down and study it, you lazy bum!"

(she doesn't really say that but it sound better that way).

 

[RSAgator]

royally disadvantaged? Well that all depends on whether you get a question or not. I can't count how many times I've told myself "oh i don't need to know this" only to sit down and take a test and think "man I'm an idiot for not learning this."

Bernoulli's equation is an important one and fluid mechanics I think is one of the more important topics of physics because it has significant biological impact (ala blood flow, etc). Not only can it not be skipped, it may be one of the sections that you should know the best =)

 

[Kaustikos]

royally disadvantaged? Well that all depends on whether you get a question or not. I can't count how many times I've told myself "oh i don't need to know this" only to sit down and take a test and think "man I'm an idiot for not learning this."

Bernoulli's equation is an important one and fluid mechanics I think is one of the more important topics of physics because it has significant biological impact (ala blood flow, etc). Not only can it not be skipped, it may be one of the sections that you should know the best =)

Yeah, seriously. Bernouli is a must, even if it's not focused that much on the mcat, it'll be important in the long run.


Question:

Could someone elaborate on the height component of the equation? Kaplan offered a terrible explanation and I never quite figured out what the height component meant in terms of flow.

 

[RSAgator]

Yeah, seriously. Bernouli is a must, even if it's not focused that much on the mcat, it'll be important in the long run.


Question:

Could someone elaborate on the height component of the equation? Kaplan offered a terrible explanation and I never quite figured out what the height component meant in terms of flow.

well you can draw an analogy between the height and potential energy. Think of it in reverse. You have a hose and you connect it to a bucket at the bottom of the bucket. The hose fills the bucket up to the top. In order to fill up the bucket you need to do work on the water. Once the water level rises to a certain level you disconnect the hose and plug a hole in the bucket. You now have a water of some height in the bucket that is essentially potential energy. You open up the hole at the bottom and this potential energy gets converted to kinetic energy as your water flows out of the bucket.

Now in this example the height obviously decreases so eventually your potential energy is 0 and you have no flow.


Here's a good working example of bernoulli's equation between 2 points.

Lets assume you have a pool or some other large body of water with a drain at the bottom that empties out somewhere. The water level we'll assume doesn't change (so the pgh is constant) and we'll assume the velocity at the top is 0. Also since the top is open to the atmosphere and so is the drain, the pressure terms are equal. The original equation would be:


pgh1 + P1 + .5*pv1^2 = pgh2 + P2 + .5*pv2^2

where 1 is at the top and 2 is at the drain.

P1 = P2, and v1 = 0, so pgh1 = pgh2 + .5pv2^2

you can set h2, the height of your drain, as your reference point and make it equal to 0 (this is fair because the water below the drain doesn't affect pressure, it is only the water above the drain that does). This results in

pgh1 = .5pv2^2

As you can see from this worked out example, the velocity leaving the drain is a direct result of the height of the pool. That is to say that the height is the driving force for the velocity. In order for you to have a velocity moving the fluid through the drain you need some sort of driving force. This could either be potential energy (from the height of the fluid), a pressure difference (if say one end was a vacuum, your pressure difference would drive the fluid through), or an initial velocity (say you were comparing 2 different cross sections of a pipe). Hope that gives a decent explanation.

 

[tncekm]

Bernoullis equation should be looked at no differently than the conservation of energy equations you've used previously, just applied to fluids.

ΔUg, or mgΔy (change in gravitational potential energy) is analogous to ρgΔy (the gravitational component of Bernoulli's equation), and ΔKE, or ½mΔ(v&#178 is analogous to ½ρΔ(v&#178.

So, restated for a fluid, the change in gravitational potential energy of a fluid + the change in kinetic energy + the change in pressure (which is an energy density) will always be equal to zero. Its conservation of energy, no more no less. The only difference is that you may need to know the continuity equation I=A₁v₁=A₂v₂ in order to solve a problem.

ΔP + ρgΔy + ½ρΔ(v&#178 = 0

mgΔy + ½mΔ(v&#178 = 0

A 100kg skier skies down an extremely slippery 10m slope of unknown angle. What is his speed after he arrives at the bottom of the slope if he twists his ankle and falls ¼ second after he starts his way down the slope?

A mercury dam is let loose from an initial velocity of 0m/s ( 0.001m deep and 15m wide, specific gravity = 13.6) and flows down a 10m frictionless cliff. What is the speed of the cross section immediately in contact with the dam reservoir as it hits y = 0m?

Both equations, which are statements of conservation of energy, will yield Toricellis results (only called that for fluids) v=√(2gD) which can be used to solve both problems, which have the same answer-- In other words, the fluid and the skier are both moving about 14m/s (or 10√2 m/s). Conservation of energy, it works the same way as you thought it did

 

[Kaustikos]

Bernoullis equation should be looked at no differently than the conservation of energy equations you've used previously, just applied to fluids.

ΔUg, or mgΔy (change in gravitational potential energy) is analogous to ρgΔy (the gravitational component of Bernoulli's equation), and ΔKE, or ½mΔ(v²) is analogous to ½ρΔ(v²).

So, restated for a fluid, the change in gravitational potential energy of a fluid + the change in kinetic energy + the change in pressure (which is an energy density) will always be equal to zero. Its conservation of energy, no more no less. The only difference is that you may need to know the continuity equation I=A₁v₁=A₂v₂ in order to solve a problem.

ΔP + ρgΔy + ½ρΔ(v²) = 0

mgΔy + ½mΔ(v²) = 0

A 100kg skier skies down an extremely slippery 10m slope of unknown angle. What is his speed after he arrives at the bottom of the slope if he twists his ankle and falls ¼ second after he starts his way down the slope?

A mercury dam is let loose from an initial velocity of 0m/s ( 0.001m deep and 15m wide, specific gravity = 13.6) and flows down a 10m frictionless cliff. What is the speed of the cross section immediately in contact with the dam reservoir as it hits y = 0m?

Both equations, which are statements of conservation of energy, will yield Toricellis results (only called that for fluids) v=√(2gD) which can be used to solve both problems, which have the same answer-- In other words, the fluid and the skier are both moving about 14m/s (or 10√2 m/s). Conservation of energy, it works the same way as you thought it did

Minus the two questions, great advice so far. Honestly, I don't see how you could solve the skier question without additional info, considering you only have the mass and distance traveled. But the equation you added definitely made it easier. thanks

 

[tncekm]

I meant to intentionally use confusing questions that were very simple to answer. I think I may not have described the question for the skier well, the 10m slope is 10m tall, not in distance. (My bad, I came up with those pretty quick.)

In that case, all you need is distance traveled in the y direction, the mass and all other factors are useless.

You know where that equation is derived from, right? It comes from the conservation of energy equations above.

From y=10m to y=0m all potential energy in the y direction has been converted into kinetic energy. So, you can say:

ΔKE = ΔU
½mΔ(v&#178 =mgy
½Δ(v&#178=gy; mass is useless--goodbye mass!
Δ(v&#178=2gy; v_initial is 0 ∴
v_final = √(2gy) [or v=√(2gD)] -- the derivation works the same way with bernoulli's equation.

 

[Kaustikos]

I meant to intentionally use confusing questions that were very simple to answer. I think I may not have described the question for the skier well, the 10m slope is 10m tall, not in distance. (My bad, I came up with those pretty quick.)

In that case, all you need is distance traveled in the y direction, the mass and all other factors are useless.

You know where that equation is derived from, right? It comes from the conservation of energy equations above.

From y=10m to y=0m all potential energy has been converted into kinetic energy. So, you can say:

ΔKE = ΔU
½mΔ(v²) =mgy
½Δ(v²)=gy
Δ(v²)=2gy; v_initial is 0 ∴
v_final = √(2gy) [or v=√(2gD)] -- the derivation works the same way with bernoulli's equation.

That did it.

 

[tncekm]

Cool

 

[unsung]

Yeah, seriously. Bernouli is a must, even if it's not focused that much on the mcat, it'll be important in the long run.


Question:

Could someone elaborate on the height component of the equation? Kaplan offered a terrible explanation and I never quite figured out what the height component meant in terms of flow.

You can think of it in terms of electric potential too.

 

[fashafosho]

You should tutor!
I understand this so much more now- none of my study materials defined bernoulli's equation to be a conservation equation; I will not forget it now! I was just getting discouraged by all the variables and figured it was not worth the investment to work out a problem that requires all of them on the MCAT- but it's a lot simpler now and makes a lot more sense- thank you!

I meant to intentionally use confusing questions that were very simple to answer. I think I may not have described the question for the skier well, the 10m slope is 10m tall, not in distance. (My bad, I came up with those pretty quick.)

In that case, all you need is distance traveled in the y direction, the mass and all other factors are useless.

You know where that equation is derived from, right? It comes from the conservation of energy equations above.

From y=10m to y=0m all potential energy in the y direction has been converted into kinetic energy. So, you can say:

ΔKE = ΔU
½mΔ(v²) =mgy
½Δ(v²)=gy; mass is useless--goodbye mass!
Δ(v²)=2gy; v_initial is 0 ∴
v_final = √(2gy) [or v=√(2gD)] -- the derivation works the same way with bernoulli's equation.

 

[tncekm]

You should tutor!
I understand this so much more now- none of my study materials defined bernoulli's equation to be a conservation equation; I will not forget it now! I was just getting discouraged by all the variables and figured it was not worth the investment to work out a problem that requires all of them on the MCAT- but it's a lot simpler now and makes a lot more sense- thank you!

Thanks, and I'm glad I can help.

I do tutor-- unofficially I was offered an SI position, but didn't have time, but I usually spent quite a bit of time helping friends / classmates out before tests. I enjoy helping other people out, and it helps cement the materials