fluids

[Sonyfan08]

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I don't understand fluids!! So, this is a really basic question.. but exactly what is the bouyant force? How is it related to the the force due to the weight of the fluid (gauge pressure)?

So, if an object was floating on a fluid.. there is greater pressure on the submerged part (i assume because of gauge pressure) than the portion that is above. And it says the bouyant force is what pushes it up.. but I guess what I'm trying to get at is.. wouldn't the force be pushing down due to the weight of the fluid?

 

[Rabolisk]

Buoyant force is the force that opposes gravity in objects in water, and thus acts upwards. A solid object would normally free fall in vacuum, but the liquid applies a pressure that counteracts gravity. The buoyant force is equal to the weight of the fluid displaced by the solid object. The reason that some objects float is because only a percentage of that object's volume is submerged. The weight of any object tends to bring it down, and it is equal to mg. The weight of the fluid is then equal to ρVg, where ρV = mass of the fluid displaced. This makes sense because density is mass/volume. The important thing to remember is that the volume of the fluid can vary from infinitely small to the volume of the object, but not any larger. The weight of the solid object stays constant (more or less) no matter how much of it is submerged. If the buoyant force cannot fully negate gravity, then the object sinks, because F = ρsolidVsolidg - ρfluidVfluidg. Since Vfluid cannot be larger than Vsolid, that means that ρfluid must be equal to or larger than ρsolid.

As far as your second question, the greater pressure would push it up, not down. There is greater pressure, or force on the bottom of an object pushing it up, than there is pressure, or force, on the top of an object pushing it down. You derive the buoyant force from pressure, noting that pressure is force/area.

 

[Sonyfan08]

Thanks!

Why is the volume never greater than the volume of the object?

I was working on a problem in the TPR workbook, it gives you the buoyant force of an object of being 100 N, making the object 100 N lighter than in air. It then asked what was the volume of the object. I looked at the solution and it equates 100N = density of fluid (v)(g); I guess my question is - how can you assume what the V is? For instance, if the object was floating, the V in Fb would be equal to the volume submerged. but if it's submerged, is it always equal to the volume of the object?

 

[fizzgig]

when you push an object into a basin of water, the water level goes up, right? you've moved some water out of the way, to make room for the object, and that water you moved out of the way is sitting at a higher gravity potential than it was previously. that water you moved out of the way weighs something, and it wants to go back down and fill the container - obviously, since when you remove the object the water level goes back down and fills the space the object took up.

so if you put a rock into the water, it will push water out of the way until it is fully submerged. it CANNOT push any more water out of the way than it has volume. once it is fully submerged, the rock may have more weight (mg) than the weight of water it pushed out of the way (Fb, mg of water that you displaced). unless it changes shape it can't create more buoyant force. you have rock of a higher density than water taking up the space water used to occupy. since Fb < mg_rock, the two forces to NOT balance out, and the rock sinks vs just hanging out right below the surface.

 

[Sonyfan08]

Another question regarding fluids:

In the TPRH workbook passage 20 #3

In a cylindrical tank, if the hole was made towards the bottom of the tank... what can you expect the velocity emerging from the hole to be as the water in the level in the tank drops to the level of the hole?

The answer says that the speed will increase.

I thought it would decrease because I thought with high pressure there would be decrease in speed, while low pressure has a higher speed. As more water drains out, the pressure will decrease at the hole, correct? Then why is it that the speed increases??

 

[Rabolisk]

I think you answered your question. As you said, if lower pressure has a higher speed, then the speed increases because the pressure drops as water drains out...

P + &#961;gh + 1/2&#961;v^2 is constant. P is the ambient pressure, and h approaches 0 as the water drains out, thus velocity must go up.

 

[fizzgig]

ok someone come with me on this - doesn't make sense to me...

you have a tank, hole at bottom. hole area is << tank top's surface area

(P+rhogh+.5rhov^2)tanktopopening = (P+rhogh+.5rhov^2)hole

P for both is Patm =0
v for tank top opening is approximately zero because compared to what's getting out of the hole this fluid is moving really slow
h at the hole is set to be 0

you're left with

rhogh tank = 0.5rhov^2 hole

h tank is proportional to v^2 hole

as the height of the water drops, the velocity of water leaving the hole drops.
if you drain a bucket of water it does not start flowing slow and then rocket out the last few seconds...

the speed vs pressure thing is related to P i believe when you are drawing a fluid in a pipe. the amount of pressure besides that does to velocity or height, due to random fluid movement, exerted on the pipe walls.

 

[soundnin]

i too came to the same equations as above for the hole in bucket question... can anyone explain why the speed increases as the height drops??

 

[fizzgig]

yeah i'd be up for an explanation from anyone. this doesn't intuitively even make sense. i've watched water troughs drain. i'm wrong a LOT but i'd go so far as to say that unless i'm missing part of the question, this answer from the book was incorrect.

 

[vsl5]

Any of you guys ever learned about the classic fluids problem involving opening a spigot in a tank of water? The solution is ALWAYS v=sqrt(2gh)

P1+pgh1 +1/2pv1^2 = P2+pgh2+1/2pv2^2 where 2 is the position at the spigot or hole

hole is open to atm so P2 = 1 atm, P1 is at the top with no fluid above it so P1 = atm. Take h2 to be ground level so h2=0 and v1= zero because fluid doesn't move at this point

You get:

pgh1=1/2 pv2^2
gh1=v2^2
sqrt(gh1) = v

v decreases as h increases.

Edit: Just read the above post saying the same thing. My guess is that in the question, the bottom isnt open to the air and is connected to some other pipe or something.

 

[soundnin]

Any of you guys ever learned about the classic fluids problem involving opening a spigot in a tank of water? The solution is ALWAYS v=sqrt(2gh)

P1+pgh1 +1/2pv1^2 = P2+pgh2+1/2pv2^2 where 2 is the position at the spigot or hole

hole is open to atm so P2 = 1 atm, P1 is at the top with no fluid above it so P1 = atm. Take h2 to be ground level so h2=0 and v1= zero because fluid doesn't move at this point

You get:

pgh1=1/2 pv2^2
gh1=v2^2
sqrt(gh1) = v

v decreases as h increases.

Edit: Just read the above post saying the same thing. My guess is that in the question, the bottom isnt open to the air and is connected to some other pipe or something.

um, you just derived the exact set of equations as fizzgig above. and again, v is proportional to square root of h, or h is proportional to v^2. either way, v does NOT decrease as h increases like you wrote.

edit: lol didn't read your edit. so i guess we still don't have an explanation for this.

 

[Prateek]

V = sqrt (2gh) is the correct formula

Therefore, as the water flows out, the height will decrease, thus the speed of the water out of the hole will also decrease

 

[sangria1986]

dd

 

[Rabolisk]

..That's clearly not what the question was asking. It's saying that the velocity increases as fluid drains out, lowering its height.

 

[sangria1986]

no calculus needed...

Think about it this way... the change in the amount of mass going through the hole must equal the change in the amount of mass entering directly above it. regardless if whether the height of water is decreasing, the above principle holds. Even as the height decreases, the same amount of water leaves the tank per second. Thus, delta mass1/time1= delta mass2/time2. This equation directly translates into m1v1=m2v2. Thus this equation holds regardless if whether the amount of liquid above is decreasing in height-the same amount at any point in time has to enter the hole (until there is none left).

Then you just use bernouill's theorem.