Physics Help-Fluids

[johnjohn320]

I'm just not a scientific thinker I guess. I'm ok with a lot of stuff, but fluids has me stumped.

23) (see diagram) A water tower (spherical reservoir) contains 5.25x10^5 kg of water when full. The reservoir is vented to the atmosphere at the top. For a full reservoir, find the gauge pressure that the water has at the faucet at house A and house B. Ignore the diameter of the delivery pipes.

I have absolutely no idea on that one. I know p=m/v, other than that, I'm really clueless (sorry, I was sick the day he explained this stuff). Any help?

There's another problem that's givine me trouble, but rather than spelling the whole thing out, I'll just ask a specific question: I'm trying to find the torque of something. Something is applying pressure to rotate something else. I have a distance from the axis of rotation to where the pressure is being applied. I have the pressure. I have the radius of the tube the pressure is being applied through (it's a hydraulic oil pump). I know T=Fd in this case. I have the d, so I need the force. How do I use the information about pressure to find the force? Where does the radius of the tube come in?

Thanks guys...that'll teach me to get sick.

 

[DXM]

Check your book, you'll find that pressure in fluids is simply a funcion of the ( density of fluid ) * ( gravity ) * ( height from reference point ). It's quite surprising but the gauge pressure at faucet B is the same pressure if the faucet were under the water tower 7.3m off of the ground.


EDIT: fixed some spelling mistakes.

 

[Alex]

dude you dont go to guelph or anything eh?
cuz theres physiscs tomorrow and everybodys freakin out... this is prolly just a coincidence tho... had to ask anyway

 

[DXM]

Oh, and about the second part of your question, you should know that F = (Pressure)*(Area).

 

[johnjohn320]

Originally posted by: DXM
Oh, and about the second part of your question, you should know that F = (Pressure)*(Area).

but which area?

 

[DXM]

Originally posted by: johnjohn320

Originally posted by: DXM
Oh, and about the second part of your question, you should know that F = (Pressure)*(Area).

but which area?

Well, I'm not too sure without a diagram to help me, but I believe the area would be the cross sectional area of the pipe.

 

[johnjohn320]

Originally posted by: DXM

Originally posted by: johnjohn320

Originally posted by: DXM
Oh, and about the second part of your question, you should know that F = (Pressure)*(Area).

but which area?

Well, I'm not too sure without a diagram to help me, but I believe the area would be the cross sectional area of the pipe.

Yeah, I guess that would make sense lol. Thanks for the help.

 

[johnjohn320]

Originally posted by: DXM
Check your book, you'll find that pressure in fluids is simply a funcion of the ( density of fluid ) * ( gravity ) * ( height from reference point ). It's quite surprising but the gauge pressure at faucet B is the same pressure if the faucet were under the water tower 7.3m off of the ground.


EDIT: fixed some spelling mistakes.

Crap. I did all that. According to my book, density of water is 1000. Gravity is 9.8 . The height is 15m for faucet A. The Volume is 5.25x10^5kg.

pgv=5,145,000,000
pgh=147,000

You would add those together, right? Well, the book's answer is 2.45x10^5 for faucet. I can't make heads or tails of how they got that...

 

[Evadman]

Question A = (height of tower from gound to top of tower ) * 9.81^2 * (density of water)
Question B = (height of tower from gound to top of tower - 7.3) * 9.81^2 * (density of water)

Question C
Force = pressure * area.
in this instance:
Force = Pressure of fluid * (pi * (radius of pipe^2))

 

[johnjohn320]

Originally posted by: Evadman
Question A = (height of tower from gound to top of tower ) * 9.81^2 * (density of water)
Question B = (height of tower from gound to top of tower - 7.3) * 9.81^2 * (density of water)

Question C
Force = pressure * area.
in this instance:
Force = Pressure of fluid * (pi * (radius of pipe^2))

Why did you square gravity? And, that still doesn't get the right answer...

 

[johnjohn320]

bump

 

[GoodToGo]

Give me the radius.

 

[johnjohn320]

Originally posted by: GoodToGo
Give me the radius.

Of what? The water tower? isn't given. Everything they give I've given.

 

[GoodToGo]

Is the answer for the force 5.145 *10^6? I know I havent calculated the pressure but I just wanted to try something else.

 

[johnjohn320]

Originally posted by: GoodToGo
Is the answer for the force 5.145 *10^6? I know I havent calculated the pressure but I just wanted to try something else.

No, that's what I got the first time. The book, as I said before, got 2.45x10^5 for faucet A.

 

[GoodToGo]

I think I got it. The sphere contains 5.25*10^5 kg of water. Divide that by the density and you will get the volume of the sphere. Use that to find the radius which you will see is 5m. So the actual height is 10 (the diameter) +15 = 25m. Use pgh and get the answer. Similar for B.

 

[johnjohn320]

Originally posted by: GoodToGo
I think I got it. The sphere contains 5.25*10^5 kg of water. Divide that by the density and you will get the volume of the sphere. Use that to find the radius which you will see is 5m. So the actual height is 10 (the diameter) +15 = 25m. Use pgh and get the answer. Similar for B.

Yep, that's it. I'm still confused as to how pgV never came into play, but whatever. Thanks for the help!

 

[GoodToGo]

Originally posted by: johnjohn320

Originally posted by: GoodToGo
I think I got it. The sphere contains 5.25*10^5 kg of water. Divide that by the density and you will get the volume of the sphere. Use that to find the radius which you will see is 5m. So the actual height is 10 (the diameter) +15 = 25m. Use pgh and get the answer. Similar for B.

Yep, that's it. I'm still confused as to how pgV never came into play, but whatever. Thanks for the help!

You got it, now get some sleep