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raaaid
10-01-2005, 02:44 PM
how would you make 1000 tons of water go all the way up 10 km and then down again with no friction?

have you heard about a tiphon? your toilet has one, you would make use of that principle if you wanted to take gas from a car deposit with a plastic tube

with an ideal fluid you could make the tube go 10 km up and move all the fluid you wanted up always that the end of the tube is lower than the level of water where the other end of the tube is

now imagine that in the top of the 10 km tiphon you trick the system and divide the tube in two, one which goes down again ending one meter below the level of water but the other that lets half of the water up not letting air come through the tube so the tiphon keeps alive

for what i have studied as long as density*gravity*height is higher in one end of the tube the fluid will move but it doesnt talk about mass so the fluid would not be noticing the splitting in two of the tubes at the top keepeng constant the variables that count but not the mass that doesnt count in the equation

the key would be not letting air entering in the stealing top tube what may not be posible

by the way if you havent noticed this is not posible because then you could move 500ton 10 km up by moving 500 tons 1 meter down but why not

raaaid
10-01-2005, 02:44 PM
how would you make 1000 tons of water go all the way up 10 km and then down again with no friction?

have you heard about a tiphon? your toilet has one, you would make use of that principle if you wanted to take gas from a car deposit with a plastic tube

with an ideal fluid you could make the tube go 10 km up and move all the fluid you wanted up always that the end of the tube is lower than the level of water where the other end of the tube is

now imagine that in the top of the 10 km tiphon you trick the system and divide the tube in two, one which goes down again ending one meter below the level of water but the other that lets half of the water up not letting air come through the tube so the tiphon keeps alive

for what i have studied as long as density*gravity*height is higher in one end of the tube the fluid will move but it doesnt talk about mass so the fluid would not be noticing the splitting in two of the tubes at the top keepeng constant the variables that count but not the mass that doesnt count in the equation

the key would be not letting air entering in the stealing top tube what may not be posible

by the way if you havent noticed this is not posible because then you could move 500ton 10 km up by moving 500 tons 1 meter down but why not

AOD_DieHard_X
10-01-2005, 03:19 PM
Tiphon? or Siphon?

Anyway, the thing about the siphon is that it will theoretically (if the tube could support the weight)work but the tube has to be primed. Does this "tiphon" you refer to have a way around this?

Tully__
10-01-2005, 07:32 PM
Siphon limit is approximately 32 feet (about 10 metres) for water at sea level and less @ higher altitudes and it is subject to friction. Even then the outlet must be lower than the intake or it just stops. Go read some more to find out how a siphon really works.

Edit: What you're missing is that the pressure reduction caused by the mass of the water in the tube trying to descend reaches zero (vacuum) when there's 32 feet of water below. At heights above that you'll simply end up with a vacuum in the tube and the water wont lift any further. The lift in the tube is provided by the air pressure above the surface of the source tank and there is only enough air pressure to support a 32 foot column of water at sea level. As altitude increases and air pressure drops the maximum height of the water column is reduced.

GRYPHON_401Cdn
10-01-2005, 10:03 PM
I'm not sure what the fundamentals of fluid flow is doing in this forum...

The only fluid flowing in here is Tequila

raaaid
10-02-2005, 06:18 AM
first those who dont like me writing this here forgive me but i tried many forums and here is the smartest people http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

you are wrong a siphon doenst work by atmospheric pressure but by liquid pressure given by the product density*gravity*height,
liquid will always go downhill or to the lower end (where the h is lower with the same dens and grav)indepently of how much up it went before(neglecting friction of course)

this is confirmed by a experiment ive just done with one thicker tube united to one thinner tube
and shaped in the form of an U being the thick in one side and the thin in the other

if the height is the same in both ends of the tube when you put it upside down independently that the thicker part holds much more water and weights more the thin tube water holds for it because the end is at the same height than the other end, the moment you put any of the ends lower all the water will come out, its interesting

so i take one thick tube up and a thin one down stealing half of the water that pass by the top and only sending half down, as long as the end of the tube is lower than the begining it will be ok as proved is no a matter of mass but of height

Tully__
10-02-2005, 06:52 AM
The water wont come out at the top because the pressure inside the tube is lower than the pressure outside.

What keeps the water in the top of the U shape is the external air pressure. There is only enough external air pressure at sea level to push the water 32 feet up, after that the tube will be empty (vacuum). Ask your fluid mechanics teacher if you don't believe me. Go on, I dare you... http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

Chuck_Older
10-02-2005, 07:10 AM
With no friction? An H20 molecule rubbing against another H20 molecule should produce friction. Each individual molecule would have to be in vacuum, or else they are touching

raaaid
10-02-2005, 07:21 AM
yes ive read some more and you can rise water only 10 m in normal conditions, fluid at 0 velocity once the speed increases the altitude should increase as well though

but well you can still pump 1000 tons of water 10 m up and only send 500 down (the end of the tube one meter below initial water level)

so you have raised 500 tons 10 meters up by lowering 500 tons 1 meter down taking into account friction say you rise it 3 meters

in the top you only have to divide the tube in two one to steal keeping the vaacuum and the other to send the rest down to keep the water chain

seems that nature use siphons in jirafes to pump up blood and even in humans

major_setback
10-02-2005, 07:40 AM
What about capillary action. Trees use this to move sap a few hundred feet upwards! Just get a bunch of (very fine capillary) tubes and wrap them together.

BTW what are you going to do with the water when you get it up there?
You could let it fall over a water wheel - then you would have perpetual motion! http://forums.ubi.com/images/smilies/53.gif

http://forums.ubi.com/images/smilies/1072.gif http://forums.ubi.com/images/smilies/1072.gif and http://forums.ubi.com/images/smilies/1072.gif

Maybe I shouldn't give you ideas raaid!
http://forums.ubi.com/images/smilies/16x16_smiley-very-happy.gif

PlaneEater
10-02-2005, 01:10 PM
Raaaid, I'm getting this hunch that you'd have an absolute blast in college doing a physics major. Molecular physics, quantam physics, thermodynamics, nuclear physics, ergonomical physics (physics of how muscles and organic structural machines work), any of that stuff--you should look into it. I think you'd find something you can sink your teeth into like nothing you've ever dreamed of.

Tully__
10-02-2005, 09:09 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">but well you can still pump 1000 tons of water 10 m up and only send 500 down (the end of the tube one meter below initial water level)

so you have raised 500 tons 10 meters up by lowering 500 tons 1 meter down taking into account friction say you rise it 3 meters

in the top you only have to divide the tube in two one to steal keeping the vaacuum and the other to send the rest down to keep the water chain </div></BLOCKQUOTE>

Not using siphoning you can't because the pressure inside the tube is lower than the pressure outside!!! Even with some means for the water to leave the tube, it wont as the pressure gradient goes the wrong way. If you want any of the water to come out of the tube at the top you must use some sort of pressure pump upstream of the point where you want to split the stream. At that point it ceases to be a syphoning process.

The capillary action method suggested above might work, but the flow rate would be so slow as to be almost useless for anything other than what trees use it for if you need any quantity of water.

Freelancer-1
10-02-2005, 09:53 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by raaaid:
how would you make 1000 tons of water go all the way up 10 km and then down again with no friction?

have you heard about a tiphon? your toilet has one, you would make use of that principle if you wanted to take gas from a car deposit with a plastic tube

with an ideal fluid you could make the tube go 10 km up and move all the fluid you wanted up always that the end of the tube is lower than the level of water where the other end of the tube is

now imagine that in the top of the 10 km tiphon you trick the system and divide the tube in two, one which goes down again ending one meter below the level of water but the other that lets half of the water up not letting air come through the tube so the tiphon keeps alive

for what i have studied as long as density*gravity*height is higher in one end of the tube the fluid will move but it doesnt talk about mass so the fluid would not be noticing the splitting in two of the tubes at the top keepeng constant the variables that count but not the mass that doesnt count in the equation

the key would be not letting air entering in the stealing top tube what may not be posible

by the way if you havent noticed this is not posible because then you could move 500ton 10 km up by moving 500 tons 1 meter down but why not </div></BLOCKQUOTE>

Hi raaaid,

First off, you can not siphon water above aprox. 32 feet at sea level before the weight of the water simply overcomes the forces applied to it.

Two, any break in the system to divert some of the water will also break the lift.

What your describing is a really really big barometer. Handy for weather prediction but not much use as a perpetual motion device.

Try your experiments by replacing the water with mercury. You'll save a bundle in tubing costs as you will only need a meter or so. Uhh, I should mention it's not a good idea to suck mercury through a hose. Best use a vaccuum pump for that.

Freelancer

Waldo.Pepper
10-02-2005, 10:06 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">I should mention it's not a good idea to suck mercury through a hose </div></BLOCKQUOTE>

Drat! If only you'd have let him suck the mercury, it could have all been over. http://forums.ubi.com/images/smilies/51.gif

raaaid
10-03-2005, 04:56 AM
tully just what my teacher said which has convinced me for two hours, in vengeance i asked him how could he explain vectorially why a leaned balance recovers de horizontality http://forums.ubi.com/images/smilies/784.gif

i agree that to pump out water in the top of the 10 meter loop aparently requires to beat a presure of 760 mm which is the equivalent force to rise the same water 10 meters but just with the water stopped

but consider that if i expand the top of the loop to increase the volume ill be pumping up water but the water is moving so im expanding the top of the loop as the water fills it so im exercing no force because im not pumping im expanding as it fills

in fact the water im recovering at the top has a pressure of 0 mm not 760

with a simple plastic ballon and minimal effort i could fill it as water passes by and the pressure inside would be 1mm (but fill of water that doesnt contract nor expand with the variation of presure) and 760 mm outside the ballon

Tully__
10-03-2005, 06:29 AM
I give up http://forums.ubi.com/groupee_common/emoticons/icon_rolleyes.gif

Think about it.... pressure on the outside of the balloon &gt; pressure inside, the balloon will not fill!!

ruf9ii
10-03-2005, 09:59 AM
i think the most important question is 'why?'

what would you need water 10km in the sky for?

and as alluded to before, imagine the support you'd need for it...

WWSensei
10-03-2005, 10:21 AM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Tully__:
I give up http://forums.ubi.com/groupee_common/emoticons/icon_rolleyes.gif

Think about it.... pressure on the outside of the balloon &gt; pressure inside, the balloon will not fill!! </div></BLOCKQUOTE>

The real question is why do you bother. In every question he's ever posted he always ignores the science and facts that interferes with his ideas. Being innovative with new ideas does get incredibly easy when you simply ignore the data that doesn't fit.

danger13
10-03-2005, 11:16 AM
so, are you actually trying to create perpetual motion?

Grey_Mouser67
10-03-2005, 11:26 AM

All siphoning I've done was really pulling a vacuum and the energy required that the fluid lost be replaced with the fluid in the tank providing that the output end of the system was lower that the original fluid level...gravity was working with you.

Now if you could move tons of water through a siphoning device and create the energy like a hydroelectric dam, you could create power...the issue would be how to you move water from a higher position to a lower position through a siphon hose and then replace it again using no energy....

Eureka...use the tides! Going from low tide to high tide replaces the fluid in the system and generate at periods of low tide to create the elevation difference needed to use the siphon...I wonder how much energy could be created this way?

Now for a design and patent!

danger13
10-03-2005, 11:48 AM
simple, have the lower end of the syphon tube let water flow over a wheel to utilise a simple pump to draw the water back to the top?

xTHRUDx
10-03-2005, 12:24 PM
i couldn't decide which one fit best, so i posted both.

http://img.photobucket.com/albums/v438/JodFoster/d.jpg

http://img.photobucket.com/albums/v438/JodFoster/i.jpg

Freelancer-1
10-03-2005, 12:31 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by danger13:
simple, have the lower end of the syphon tube let water flow over a wheel to utilise a simple pump to draw the water back to the top? </div></BLOCKQUOTE>

Don't start thinking like raaaid. There is no free lunch. The conservation of energy sees to that.

The energy required to lift the water will negate the energy produced.

Freelancer

raaaid
10-03-2005, 12:53 PM
iwont talk about my idea any more ill have to experiment it my self but two things

can one static kg pull down more than 3 static kg? take one siphon which right side is filled with one liter and the left with 3 at the moment you lower the 1 kg end i cm lower all the water will come out that way, how can 1 cm unbalance cause that a mass of 1 kg beats estrongly one of 3 kg? i have experimented this

the second thing is this take a 1 liter balloon filled with water at 0 mm pressure being the outside at 760 mm, if you open the balloon air will come in , but remember the uncomprensibility of water, the walls of the balloon wont suffer any stress at all since water doesnt compact as it tends to for the low pressure

now you want to increase the volume of the balloon from 1 to two liter, you can create a vacuum making a great effort or you can slowly fill the balloon with more 0 mm water slowly, it will take no effort because of the uncomprensibility of water besides in the case of my looping water happens to pass by by itself

Tully__
10-03-2005, 01:20 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by raaaid:
iwont talk about my idea any more ill have to experiment it my self but two things

can one static kg pull down more than 3 static kg? take one siphon which right side is filled with one liter and the left with 3 at the moment you lower the 1 kg end i cm lower all the water will come out that way, how can 1 cm unbalance cause that a mass of 1 kg beats estrongly one of 3 kg? i have experimented this </div></BLOCKQUOTE>Because fluid flow is controlled by pressure, not mass. As soon as the outlet is below the source end of the hose, the pressure balance will push flow to the low end.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">the second thing is this take a 1 liter balloon filled with water at 0 mm pressure being the outside at 760 mm, if you open the balloon air will come in , but remember the uncomprensibility of water, the walls of the balloon wont suffer any stress at all since water doesnt compact as it tends to for the low pressure </div></BLOCKQUOTE>If by a balloon you mean a container made of a non-rigid, elastic membrane, the pressure inside will rapidly rise to equal the pressure outside (a matter of microseconds). In fact, unless the balloon hasn't been stretched from it's original size the pressure inside will be higher than outside as it will be the sum of external atmospheric pressure and the pressure applied by the tension in the balloon skin.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">now you want to increase the volume of the balloon from 1 to two liter, you can create a vacuum making a great effort or you can slowly fill the balloon with more 0 mm water slowly, it will take no effort because of the uncomprensibility of water besides in the case of my looping water happens to pass by by itself </div></BLOCKQUOTE>In order to get fluid flow you need a pressure differential. For the fluid to flow in to the balloon you'll need greater pressure in the water source than the you have in the balloon. As I described above, the pressure in the balloon cannot remain at zero and will in fact be slightly higher than atmospheric pressure. While it's easy enough to get more water in, you cant do it with water that's currently at or near zero pressure (i.e. the water at the top of a ten metre siphon path). You need to find some way to increase the pressure above that inside the balloon or reduce the pressure in the balloon. Both will require the expenditure of energy that you might as easily have spent pumping the water by conventional means.

raaaid
10-03-2005, 02:35 PM
your explanation seems quite convincing tully and i appreciate it but i still have two doubts

first why in the top of a 76 cm high mercury siphon the plastic tube doesnt stretch closing the loop being the pressure outside 760mm and inside 0 mm but no experiment says you need and extraresistent tube in order to beat the pressure gradient, as long as the mercury doesnt evaporate would it explain it its uncomprensability?

second question if i unit by a tube two mercury recipients at different height by a siphon being the end of the tube at the lowest recipient at the very bottom of the mercury so it has to stand the barometric pressure plus the column of mercury it has above and the beggining of the tube only touching slightly the surface of the higher mercury so it only has to estand the atmospheric pressure

how can be esplained that mercury flows from the first at atmospheric presure to the second at atmospheric presure plus the pressure of the mercury at the top (if i were a diver i would preffer to be in the end of the tube close to the surface not to the bottom) so it flows from less pressure to more presure

can you explain this tully and thanks in advance im not trying to challenge you you are just being very helpfull

Tully__
10-03-2005, 06:55 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by raaaid:
your explanation seems quite convincing tully and i appreciate it but i still have two doubts

first why in the top of a 76 cm high mercury siphon the plastic tube doesnt stretch closing the loop being the pressure outside 760mm and inside 0 mm but no experiment says you need and extraresistent tube in order to beat the pressure gradient, as long as the mercury doesnt evaporate would it explain it its uncomprensability? </div></BLOCKQUOTE>Atmosperic pressure is about 14lb/inch^2. As lab tubing is usually very thin, the actual force per inch of tubing is only 3 or 4 pounds and exerted at right angles to tubing with a round profile. A round profile is very resistant to forces equally distributed around the circle and at right angles to the surface and lab tubing (glass or plastic) is quite adequate to support a near vacuum. You can exert much greater local forces on these tubes by squeezing them between your fingers and still not totally collapse them (I can exert a squeeze pressure between thumb and two fingers over 11 pounds without straining). Try it.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">second question if i unit by a tube two mercury recipients at different height by a siphon being the end of the tube at the lowest recipient at the very bottom of the mercury so it has to stand the barometric pressure plus the column of mercury it has above and the beggining of the tube only touching slightly the surface of the higher mercury so it only has to estand the atmospheric pressure

how can be esplained that mercury flows from the first at atmospheric presure to the second at atmospheric presure plus the pressure of the mercury at the top (if i were a diver i would preffer to be in the end of the tube close to the surface not to the bottom) so it flows from less pressure to more presure </div></BLOCKQUOTE>Density of mercury is 13580kg/m^3. Gravity is about 9.8N/kg, so pressure at the base of a column of mercury is approximately:

(13580*9.8*h)N/m^2 = (133074.2*h)N/m^2

Atmospheric pressure is approximately 101325N/m^2.

Where h is the height of the column in metres.

A diagram of the situation you've described is here:

http://jennirivers.actewagl.net.au/Image1.jpg

h1 is 76cm
for the sake of this discussion let h2 be 100cm and h3 be 10cm.

Pressure is a vector that operates in fluids at right angles to any surface (real or arbitrary) of the fluid.

Let's first consider the arbitrary serface between the mercury in the up pipe and the surface of the source reservoir (a). The pressure in the up direction is atmospheric pressure due to air pressing on the surface of the reservoir and being transmitted through the fluid in the reservoir to the end of the pipe. The pressure in the down direction due to the 76cm column of mercury above it is:

0.76m*133074.2N/m^2

Which (not surprisingly) works out at atmospheric pressure (for those smarty pants with calculators remember that I've approximated the density of mercury by a couple of percent to reduce typing). The net vector is zero, so there is (at this point) no reason for the mercury to move.

At point (b) at the top of the arc the pressure inside the tube is about zero. I wont go into the calculations as you seem to accept and understand that.

At point (d) the force in the up direction as a result of pressure is atmospheric plus the pressure resulting from the 10cm mercury column in the destination reservoir:

(101325+(133074.2*0.10))N/m^2=
114632.42N/m^2

The force in the down direction as a result of pressure at point (d) is:

133074.2*(h2 + h3)N/m^2=
133074.2*(1.0m + 0.10m)N/m^2=
146381.62N/m^2

The net force down is:

(146381.62 - 114632.42)N/m^2=
31755.2N/m^2

Which is about 25% of atmospheric. The mercury below the tube is under considerable force pushing it out of the way to make room for mercury in the tube. Gravity will do the rest and flow from the higher reservoir to the lower will ensue.

Further analysis would show that the height at the top of the tube is irrelevant. The pressure differential between the inlet and the outlet of the siphon pipe is determined wholly by the height diference between the surface of the source reservoir and the surface of the destination reservoir.

Disclaimer: It's 10:55am and I've been at work all night so please excuse any typos http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

raaaid
10-04-2005, 06:12 AM
thanks tully with your view of pressure that pushes up and pulls down i see it and understand the pressure gradient will be the same than the two levels difference

although i think that is vacuum who makes the work:

i have a mercury siphon that loops up 35 cm and then down 70
being a level difference of 35cm so a pressure gradient of 350 mm

the 70cm end of the loop makes a pressure down of 700mm which is balanced with the 700 hundred of the atmosphere

but the beggining of the loop only pushes down 350 mm of mercury while the atmosphere pushes up 700 so the water goes up in the beggining of the loop not because of going down in the end but because the atmosferic pressure is greater on one side while the other is balanced , is not necesary for the water to come out to keep the atmosferic pressure, there is balance of forces in one side and suction in the other, is a constant suction to avoid vacuum

if i stop the water in the descend of the loop it will still be balanced the mercury and the atmosphere weight but water will still flow up because in the up part of the loop atmospheric pressure pushes the 35cm of mercury up and what if i gather this mercury thats flowing up, i really have to experiment it

what better place to search from energy from the vacuum that vacuum itself

Tully__
10-04-2005, 06:34 AM
The vacuum does not pull, the pressure in the liquid pushes towards the lower pressure. Please avoid thinking of the vacuum as pulling, it causes unecessary confusion. You will only get movement if there is pressure to push.

Woof603
10-04-2005, 02:39 PM
You're a good man, Tully.

10-04-2005, 05:08 PM
He never ceases to amaze me! And he's a moderator http://forums.ubi.com/images/smilies/354.gif

raaaid
10-05-2005, 07:08 AM
i take one liter of ice up to space i heat it so it first melts and then evaporates which will happen very soon because the presure in space is close to 0, i take the vapour into a balloon which lets say its now 1000 liters

now i steal all the heat i gave to the ice by putting inside the balloon very cold liquid nitrogen till the vapor rises a temperature of -1âÂºC so it becomes ice besides i covered the balloon with an steel armour

what will happen? my bet is the steel ballon will colapse from a volume of 1000 liters to a volume of 1 liter i cant accept vapour to be at -1âÂºC nor the existence 0f 999 liters of perfect vacuum

Tully__
10-05-2005, 05:24 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">what will happen? my bet is the steel ballon will colapse from a volume of 1000 liters to a volume of 1 liter </div></BLOCKQUOTE>

Not if it's still in space, there's no pressure on the outside to collapse it!! Vacuum does not suck, you need a positive pressure difference on the other side to push. Despite the freezing, the pressure inside the balloon, though very near total vacuum, will still likely be higher than that outside the balloon.

If you bring it back to earth sea level it may be different but then there's about 14lb/inch^2 pressure on the outside and close to zero pressure. What happens then will depend on the thickness and strength of the steel. As a balloon is close to spherical (a shape fairly resistant to external pressure) it wouldn't have to be more than about 1/4 inch thick steel to retain shape under that pressure. Modern submarine hulls easily stand many times that pressure when diving deep.

Klarkash
10-05-2005, 05:41 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Tully__:

atmospheric pressure (for those smarty pants with calculators remember that I've approximated the density of mercury by a couple of percent to reduce typing). The net vector is zero, so there </div></BLOCKQUOTE>

I've got a calculator and I've still no idea what you're on about

major_setback
10-05-2005, 06:03 PM
Cal-cu-lat-or.

?
http://forums.ubi.com/images/smilies/16x16_smiley-indifferent.gif

Tully__
10-05-2005, 07:13 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by BBushe:
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by Tully__:

atmospheric pressure (for those smarty pants with calculators remember that I've approximated the density of mercury by a couple of percent to reduce typing). The net vector is zero, so there </div></BLOCKQUOTE>

I've got a calculator and I've still no idea what you're on about </div></BLOCKQUOTE>
The calculator comment is because I've got a couple of approximate values in the calculations and the actual numbers don't quite balance as a result.

The theory is a quick & dirty explanation so I don't blame you if you're not following. To explain it properly would require a whole bunch of diagrams that I'm not going to bother making/finding http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

AA_Double_Tap
10-06-2005, 04:06 AM
Vacuum?

When I recently taught my eight-year old son about how a siphon works, I harkened back to my physics teacher who explained the physical attraction of water molecules.

I got a 10 metre heavy chain and put it in the back tray of my utility truck and pulled the chain over the rim of the tray (about 30 cm high). Letting go, the chain will feed over the rim and keep going till all off the chain is on the ground.

Water acts the same way. Why does it pull? Simple . The weight of the chain hanging over the edge is greater than the weight of the 30 cm chain in the tray on the back of the truck.
Same as the weight of the water molecules which are attracted together seeks a balance.Of course, you need to give it a start to overcome the friction that will hold the chain at rest, or the attraction of water to the sides of the vessel.

To hell with thirty centimetres. Why not get a longer chain and pull that sucker sky high over a really high rail? What is the limit to how high you can go? The weight of the chain versus the strength of each individual link. There is a limit. Eventually the weight of the chain will snap your weakest link.

Don't think about vacuum, all that does is yank your chain. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

And giraffe's neck have one-way valves and the height of a giraffe is limited by strength of the heart pump.They have nothing to do with this argument.

raaaid
10-06-2005, 07:31 AM
i almost understand it is the high presure that pushes not the low that pulls

but theres something i cant explain take a inyection without a needle and cover the hole of the inyection with your finger, you pull the piston and create vacumm at 0 mm so the 760 mm of the atmosphere pushes the piston

but why the force of atmospheric presure pushes 3 times harder the piston if there are 15ml of vacuum than if there are 5 ml

the logical would be the same push because the surface of the piston is the same and the presure gradient is the same as well

teoretically 3 times more vacuum means 3 times more work because the piston has to run triple distance with the same force from the atmospheric pressure but in reality the force of the atmospheric pressure is much bigger when there is more vacuum, i remember trying this when i was a kid

this is something i cant explain at all

Tully__
10-07-2005, 05:42 AM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by raaaid:
...but why the force of atmospheric presure pushes 3 times harder the piston if there are 15ml of vacuum than if there are 5 ml... </div></BLOCKQUOTE>Because it's not a perfect vacuum, there's still a small amount of air in the syringe. When you expand that air to 3 times the volume, you will reduce the pressure inside the syringe to 1/3 of the pressure with the smaller volume. This dramatically increases the pressure difference between outside the syringe and inside and hence the force due to external pressure on the plunger trying to push it back in.

raaaid
10-07-2005, 06:31 PM
you start with 1 ml of vacuum in the inyection and make it 2 then 4 then 8 and so on and the pressure will be 1mm then 0.5 ,0.125 and so on,
then perfect vacuum or 0mm is imposible

the presure gradient goes from 759 to 758 to 757.5 to 757. 125 and so on

how can a pressure gradient of 759 be double force to one of 758 and so on

Tully__
10-07-2005, 07:00 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by raaaid:
you start with 1 ml of vacuum in the inyection and make it 2 then 4 then 8 and so on and the pressure will be 1mm then 0.5 ,0.125 and so on,
then perfect vacuum or 0mm is imposible </div></BLOCKQUOTE>Where does 1ml come from? Is this the graduation on the syringe corresponding to how far the plunger has been withdrawn? Because 1ml is a volume not a pressure and doesn't tell us anything about vacuum or pressure unless we know the initial residual volume and pressure before the plunger is withdrawn.

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">the presure gradient goes from 759 to 758 to 757.5 to 757. 125 and so on </div></BLOCKQUOTE> What units are those numbers in? How did you arrive at them? Are you sure you're talking about pressure gradient, or do you mean pressure in the syringe. Pressure gradient is messured in units of pressure per unit of distance... perhaps you mean pressure difference between inside & outside the syringe?

<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">how can a pressure gradient of 759 be double force to one of 758 and so on </div></BLOCKQUOTE> Sort out the previous question and I might be able to work on this one as well http://forums.ubi.com/groupee_common/emoticons/icon_smile.gif

raaaid
10-08-2005, 04:14 AM
i have the syringe initally with 0.01 ml of air at 760 mm of Hg,lets consider that when i increase the volume to 1 ml the presures lows to 1mm, i double size presure reduces half

when theres a presure inside of 1mm and outside of 760mm theres a pressure difference of 759mm which will be traduced to a force that depends also with the surface of the piston of the syringe
if i pull the piston and expand the air to 2ml the pressure inside will be 0.5 mm so the difference between the inside and the outside will be 758.5mm, the surface of the piston the same but the force will be double than before

if the force exerted on the piston depends on pressure diference between the outside and inside and surface in contact how can this doubling of force can be explained if the pressure difference is not double nor the surface in contact?

Tully__
10-08-2005, 04:38 AM
If the initial volume is 0.01ml and atmospheric is 760mm Hg,

@ 1ml, internal pressure = 0.01/1 *760mm Hg = 7.6mm Hg, pressure differential is 752.4mm Hg

@ 2ml, internal pressure = 0.01/2 *760mm Hg = 3.8mm Hg, pressure differential is 756.2mm Hg

provided:
<UL TYPE=SQUARE><LI>temperature of the air inside the syringe is constant (it isn't, lowering the pressure will also considerably lower the temperature resulting in a greater pressure loss than the linear calculation I've used will show); and:-
<LI>the seals on the syringe are all perfect.[/list]

Were those two conditions satisfied, you could expect an increase in required force of approximately 1/2% plus the force required to overcome friction in the syringe to extend the plunger from 1ml to 2ml. Have you measured the relevant forces or are you just going by feel? In my experience, the friction in medical syringes is significantly greater than 1/2% of atmospheric would account for, so the increase in force would mostly be to overcome the friction, not the pressure.....

tigertalon
10-08-2005, 05:40 AM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by raaaid:
how would you make 1000 tons of water go all the way up 10 km and then down again with no friction?

have you heard about a tiphon? your toilet has one, you would make use of that principle if you wanted to take gas from a car deposit with a plastic tube

with an ideal fluid you could make the tube go 10 km up and move all the fluid you wanted up always that the end of the tube is lower than the level of water where the other end of the tube is

now imagine that in the top of the 10 km tiphon you trick the system and divide the tube in two, one which goes down again ending one meter below the level of water but the other that lets half of the water up not letting air come through the tube so the tiphon keeps alive

for what i have studied as long as density*gravity*height is higher in one end of the tube the fluid will move but it doesnt talk about mass so the fluid would not be noticing the splitting in two of the tubes at the top keepeng constant the variables that count but not the mass that doesnt count in the equation

the key would be not letting air entering in the stealing top tube what may not be posible

by the way if you havent noticed this is not posible because then you could move 500ton 10 km up by moving 500 tons 1 meter down but why not </div></BLOCKQUOTE>

Hey Raaaid,

sorry but it cannot be done. You cannot pull water up the hose more than 9.81 meters (because of g), at that point you would have to apply vacuum at the top of the hose. And you cannot apply less than a vacuum to lift it more.

Plus it has a lot of friction, but if it moves slowly it can be done (of course not higher than 9.81 meters ~ 30 feet).

Plus: Let's go totally hypotetical and assume G is high enough to lift water to 10000 m. Thing is, that water which is falling back down in a down-leading hose, produces negative pressure, which lifts the water in the up-leading hose. If you take half of that down-leading water away, also the pressure will reduce, so you will be unable to lift the water in first hose.

Cajun76
10-08-2005, 07:17 PM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by AA_Double_Tap:
Vacuum?

When I recently taught my eight-year old son about how a siphon works, I harkened back to my physics teacher who explained the physical attraction of water molecules.

I got a 10 metre heavy chain and put it in the back tray of my utility truck and pulled the chain over the rim of the tray (about 30 cm high). Letting go, the chain will feed over the rim and keep going till all off the chain is on the ground.

Water acts the same way. Why does it pull? Simple . The weight of the chain hanging over the edge is greater than the weight of the 30 cm chain in the tray on the back of the truck.
Same as the weight of the water molecules which are attracted together seeks a balance.Of course, you need to give it a start to overcome the friction that will hold the chain at rest, or the attraction of water to the sides of the vessel.

To hell with thirty centimetres. Why not get a longer chain and pull that sucker sky high over a really high rail? What is the limit to how high you can go? The weight of the chain versus the strength of each individual link. There is a limit. Eventually the weight of the chain will snap your weakest link.

Don't think about vacuum, all that does is yank your chain. http://forums.ubi.com/groupee_common/emoticons/icon_wink.gif

</div></BLOCKQUOTE>

Either you misunderstood your teacher, or he/she had it wrong. It happens: As an Airman Basic straight out of Basic Training, I corrected the tech school instructor when he got Bernoulli's Principle backwards. The rest of the class was giving me these shocked "Who do you think YOU are" looks while I was trying to explain his error.

Tully explains it in his posts. Kudos to him for his patience. http://forums.ubi.com/images/smilies/11.gif

raaaid
10-09-2005, 08:01 AM
i know im being kind of obnoxious so this will be my last argument

lets make vacuum in a tube and sink it in mercury so it will rise 76 cm

is it vacuum that pulls or high pressure that pushes?

if it pushes you wouldnt have to hold the weight of the mercury inside the tube just the weight of the tube itself because its the atmospheric pressure which pushes the weight of the mercury up

but if it pulls you would have to hold both the weight of the mercury and tube as is the case in reality

how can this be explained if mainstream is correct?

Cajun76
10-09-2005, 09:05 AM
In laymans terms: Science never sucks. Vaccums are relative states that have a higher pressure acting on it, not creating a force of its own in that sense.

Take a space capsule with an astronaut (or cosmonaut for our Russian friends) inside. For whatever reason, the emergency hatch explosive bolts go off and the hatch departs. Most people might say he is sucked out into open space. The unfortunate astronaut is actually blown out with the higher pressure inside the capsule.

The other common example is the boiled egg going through the neck off a jar that it normally wouldn't fit through. You can actually try this without causing a national tragedy. http://forums.ubi.com/images/smilies/16x16_smiley-wink.gif

Drop three lit matches into a glass bottle that has a narrow neck (an old-style 600 mL milk bottle works well). Quickly put a peeled, hard-boiled egg on the mouth of the bottle.

The flames heat the air in the bottle. As the heated air expands, some of it escapes out the bottle. When the matches go out, the air inside the bottle cools and contracts, thus creating a lower pressure inside the bottle than outside. The greater pressure outside the bottle forces the egg into the bottle.

It's absolutely vital in physics to look at the forces acting on something from the proper perspective.

Cajun76
10-09-2005, 09:11 AM
Found another cool one:

Fill a plastic (PET) bottle with hot water (NOT boiling water -- see if you can work out why I said that!) and fill a bowl with cold water. Let them sit for one minute, then empty the bottle quickly. Stretch a balloon over the open end of the bottle and push the bottle down into the cold water.

Results:

The warm water heats the bottle which, in turn, heats the air inside the bottle after the water is poured out. When the bottle is placed in the cold water, the air inside cools and contracts, causing outside air to be drawn in, pulling the balloon in and inflating it inside the bottle.

Try sitting the bottle back in the hot water again.

Another variation: Try shaking a small amount of very hot water in a bottle to heat the air inside, then quickly fit a balloon to the neck. Then wait a few minutes for things to cool down before you sit the bottle in an ice bath. Some of the pressure inside the bottle will have been a result of water vapour, which now condenses.

sammy4657
10-09-2005, 09:33 AM
Wow! Science lectures on a flight sim site. You get so much for your money these days... http://forums.ubi.com/images/smilies/16x16_smiley-tongue.gif

Tully__
10-10-2005, 04:12 AM
<BLOCKQUOTE class="ip-ubbcode-quote"><div class="ip-ubbcode-quote-title">quote:</div><div class="ip-ubbcode-quote-content">Originally posted by raaaid:
i know im being kind of obnoxious so this will be my last argument

lets make vacuum in a tube and sink it in mercury so it will rise 76 cm

is it vacuum that pulls or high pressure that pushes?

if it pushes you wouldnt have to hold the weight of the mercury inside the tube just the weight of the tube itself because its the atmospheric pressure which pushes the weight of the mercury up

but if it pulls you would have to hold both the weight of the mercury and tube as is the case in reality

how can this be explained if mainstream is correct? </div></BLOCKQUOTE>

What you're feeling is not the weight of the mercury column, it is the atmospheric pressure pushing downwards on the top of the tube. If there were atmospheric pressure inside the tube instead of a vacuum, this pressure would be balanced by an equal force upwards on the inside of the top of the tube, but as there is no pressure inside the tube you feel only the downwards pressure on the outside of the tube.

The mercury in the tube is supported by atmospheric pressure on the top of the reservoir. You can check this by reducing the pressure on the top of the reservoir to match the pressure inside the tube. The mercury in the tube will descend to the level of the mercury in the reservoir, but the downward force on the tube will not change (apart from a very minor component related to surface tension perhaps).

raaaid
10-10-2005, 11:59 AM
you are totally right the weight of the mercury can be explained by the atmospheric pressure that only pushes down

but what if i take a conical tube which apex is in contact with the surface of the mercury

the same pressure that pushes the cone down would push the cone up so the column of mercury should be weightless, you would only have to hold the light plastic cone, but i think that in practice you would still have to hold all the weight of the mercury though this would have to be experimented

i think i could prove with this inverted cone barometer that is vacuum that pulls

by the way i had never discussed this kind of things with someone so patient and intelligent http://forums.ubi.com/images/smilies/11.gif

raaaid
10-10-2005, 12:22 PM
" You can check this by reducing the pressure on the top of the reservoir to match the pressure inside the tube. The mercury in the tube will descend to the level of the mercury in the reservoir, but the downward force on the tube will not change"

i dont see this if i have a pressure of 1mm outside and 1 mm inside why should i make any effort to keep the tube up except for the weight of the tube?

thanks again for your patient tully i stopped asking questions to teachers long ago since they ended hating me

Tully__
10-10-2005, 10:18 PM
This is going to need some more diagrams to properly explain. I'm on night shifts for the next two nights so it may be two or three days before I get to the diagrams.

In general raaaid, your confusion always seems to stem from neglecting one or more of the forces involved, not just in this thread but in all of your threads about physics & energy.

DHC2Pilot
10-10-2005, 10:45 PM
Raaaid, as stated before by numerous individuals, you have to take into account ALL forces and influences acting on each body in your "machines". This includes gravity, mass, velocity, air density, friction, air resistance, bouyancy, viscosity, heat, energy transfer, volume, etc. etc. etc. I have to hand it to you - you are a thinker. If you keep it up you may one day invent something fantastic. But you need to think deeper into the equation. You seem to always neglect some major forces that effectively negate the whole purpose of your minds creations. Don't let this discourage you....just fill in the missing pieces and then once you've made your time machine go back to 1944 and retrieve some mint examples of rare WW2 fighters and bombers of which there are ony a handful of today and put them in long term storage somewhere. You'll be filthy rich when you come back to 2005. http://forums.ubi.com/images/smilies/88.gif

raaaid
10-11-2005, 05:27 AM
thanks tully but dont bother, since now i think my pump problably wont work so im busy now with my antigravity engine which i really think works

its funny that the only similar thing i found is the time bell developed by the nazis which was two counterotatory cylinders just like my engine

to understand how much appreciate you see this discussion which is taking my time now, so dont bother with those diagrams tully:http://physicsforums.com/showthread.php?t=93319

any way if you want to throw some light into this angular momentum disscusion i would be glad