Soldato
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ok boring thread but im just wondering what would happen if there was water trapped in a metal ball and it was heated? would it explode because of the pressure? thanks
Water in ball + heat?
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Eventually. Depends on the strength of the metal ball.
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what would happen if it didnt explode, would it boil?
It depends on the size/stength of the ball, the volume of water, and the amount it's heated by.
If it didn't explode, the water would either boil, or not boil, depending on the pressure inside the ball. The pressure would be dependant again on the size of the ball, the volume of water, and the amount it's heated by.
If it didn't explode, the water would either boil, or not boil, depending on the pressure inside the ball. The pressure would be dependant again on the size of the ball, the volume of water, and the amount it's heated by.
Boiling is a form of evaporation. If the pressure was high enough, the water would just not boil.
Soldato
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The ball essentially in this scenerio is a pressure vessel.
The ideal gas equation states:
Pv = RT
Where
P - pressure
v - Volume
R - Gas constant
T - Temperature
Since this is a pressure vessel, the volume is not allowed to increase. In this case as temperature rises, pressure will also rise. This will lead to an eventual rupture in the ball.
The ideal gas equation states:
Pv = RT
Where
P - pressure
v - Volume
R - Gas constant
T - Temperature
Since this is a pressure vessel, the volume is not allowed to increase. In this case as temperature rises, pressure will also rise. This will lead to an eventual rupture in the ball.
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Sorry, I meant dissapate, as in the water would evaporate cool and return to the bottom to be heated again.
Ricochet J,
does it all depend on the temp applied then?
Gas constant.
Soldato
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Soldato
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It does. There are other factors which effect the process. Material properties, fluid properties, specific heat capacity, specific latent heat, T(atmos) etc.
If you're interested in this stuff, read up on the topics: Thermodynamics and Heat Transfer. Do not start at Wiki. You'll get confused.
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Soldato
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You never stated what metal, the thickness of the metal ball - or what type of water. All have a massive effect on your rather vague question.
Imagine I had a Bunsen burner, & heated salt laden water, in a tin ball. The ball itselff would begin to melt at around 230C (much less hotter than most flames required to heat the water inside) - Then imagine I make the same ball out of Tungsten Carbide, or something as malleable as gold (that would probably expand many times before actually exploding)
Imagine I had a Bunsen burner, & heated salt laden water, in a tin ball. The ball itselff would begin to melt at around 230C (much less hotter than most flames required to heat the water inside) - Then imagine I make the same ball out of Tungsten Carbide, or something as malleable as gold (that would probably expand many times before actually exploding)
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would it still explode if a bear of weight equal to the force exerted radially by the water stood on it?
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ok, if i had a metal which attracted heat but was very strong, would the water just boil at a certain point and stay boiling or will italways explode after a certain time#?
Simple question needs a simple answer! Of course there are complex ways to answer the question, but for the sake of clarity I'll stick to simple terminology and principles. Obviously it's a bit vague, but all that means is we need to make a few assumptions! My main assumption here is that the steel ball has an infinite yield strength. ie. it will never explode!
With the above assumption, from the point at which heat is applied, the fluid within the ball will go through the processes below:
1) Water and air inside the ball will increase in temperature and pressure.
2) Water will reach it's boiling point (this will likely be in excess of 100C)
3) Water will boil away until the contents of the ball is a steam mixture.
4) Steam mixture will continue to rise in temperature and pressure forever.
Now, lets take away our assumption of infinite yield strength. The ball can now potentially rupture at any one of the 4 stages above. The point at which it does this will depend on our now defined strength. ie. The stronger the ball, the later in the above process it will rupture.
Hope this is clear enough.
EDIT: By the way, the plane takes off
With the above assumption, from the point at which heat is applied, the fluid within the ball will go through the processes below:
1) Water and air inside the ball will increase in temperature and pressure.
2) Water will reach it's boiling point (this will likely be in excess of 100C)
3) Water will boil away until the contents of the ball is a steam mixture.
4) Steam mixture will continue to rise in temperature and pressure forever.
Now, lets take away our assumption of infinite yield strength. The ball can now potentially rupture at any one of the 4 stages above. The point at which it does this will depend on our now defined strength. ie. The stronger the ball, the later in the above process it will rupture.
Hope this is clear enough.
EDIT: By the way, the plane takes off
I'm pretty sure you forgot the n which is the number of moles of gas present too.
Unless thats just a chemistry thermo term and it's somehow hidden in the above equation
What about if you put the whole apparatus on a treadmill?