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Old Dec 11th 2017, 10:37 PM   #1
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problem in thermodynamics

Hi
I am solving a problem. I have water in a piston cylinder arrangement. the water is at 150 deg C and has a volume of 24 m3. the mass of water is 25 kg. I need to find the pressure of water at this state.
I am confused... how can I find it. one more thing this water is compressed accroding to a law pV^n=c where n=0.421. I have volume after compression V2= 1.765 m3.
Kindly provide any useful guidance.
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Old Dec 12th 2017, 08:27 AM   #2
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I presume you know that, at "standard temperature and pressure" ("room temperature" and 1 atmosphere), water has a density of 1 kg/m^3. So at standard temperature and pressure, your 25 kg of water would have a volume of 25 m^3. Then your "pV^0.421= c" gives 1(25^0.421)= c. That is, c= 25^0.421. After compression to 24 m^3, you have p(24^0.421)= 25^0.421. From that, p= (25/24)^0.421.

I have no idea what you mean by "I have volume after compression V2= 1.765 m3". You have already said that the volume after compression is 24 m^3. It would take horrendous pressure to compress 25 m^3 of water to 1.765 m^3!
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Old Dec 12th 2017, 11:11 AM   #3
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Originally Posted by engrmansoor2534 View Post
Hi
I am solving a problem. I have water in a piston cylinder arrangement. the water is at 150 deg C and has a volume of 24 m3. the mass of water is 25 kg. I need to find the pressure of water at this state.
I am confused... how can I find it. one more thing this water is compressed accroding to a law pV^n=c where n=0.421. I have volume after compression V2= 1.765 m3.
Kindly provide any useful guidance.
The pressure in a volume of water depends on the depth at which you measure it. The pressure at the top equals zero while the pressure at the bottom equals the weight, W, of the water above it divided by the area on which the water is distrubuted. I.e. p = W/A.

pV = NRT is the ideal gas law and doesn't apply to liquids.
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Old Dec 12th 2017, 07:25 PM   #4
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Originally Posted by Pmb View Post
The pressure in a volume of water depends on the depth at which you measure it. The pressure at the top equals zero while the pressure at the bottom equals the weight, W, of the water above it divided by the area on which the water is distrubuted. I.e. p = W/A.

pV = NRT is the ideal gas law and doesn't apply to liquids.
Nowhere is there a suggestion of gravity being involved.....So we must assume this is in zero gravity , or more likely the extreme pressure will make such differences insignificant ....

You have the density ...25/24...= 1.041666

You have the temperature ...= 150 C

A search will get you the required chart or table....(you need a more detailed one than the one below)



Those blue lines are isotherms...the fourth down is 150C ...follow it along and you see at 1.042 ,the pressure is about 300 MPa ..

Last edited by oz93666; Dec 12th 2017 at 10:23 PM.
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Old Dec 14th 2017, 08:20 AM   #5
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Originally Posted by oz93666 View Post
Nowhere is there a suggestion of gravity being involved.....So we must assume this is in zero gravity , or more likely the extreme pressure will make such differences insignificant ....

You have the density ...25/24...= 1.041666

You have the temperature ...= 150 C

A search will get you the required chart or table....(you need a more detailed one than the one below)



Those blue lines are isotherms...the fourth down is 150C ...follow it along and you see at 1.042 ,the pressure is about 300 MPa ..
This is excellent! But I believe if there was gravity involved, would you speak of thermogravimetry?
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Old Dec 14th 2017, 04:52 PM   #6
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Originally Posted by plotter View Post
...thermogravimetry...
I don't think that comes into it ... thermogravimetry seems to deal with changes that occur in the sample ...phase changes , chemical changes ...

As for gravity , if there was a 1 meter difference in height within the apparatus this would make a pressure difference of one tenth of an atmosphere between the top and bottom of the water = 0.01MPa

Insignificant compared to the pressure required of 300 MPa (3,000 atms.)
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Old Dec 16th 2017, 02:23 PM   #7
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Originally Posted by oz93666 View Post
Nowhere is there a suggestion of gravity being involved.....
Of course there is. In all physics texts there are assumptions made in which the student is assumed to know. I.e. that the experiment is done on Earth, the rotation of the Earth is not taken into account, and in this case if there is pressure in a fluid then its assumed to be in a container in a gravitational field. In fact if there is no gravitational field there is no pressure since the pressure in a fluid is caused by the gravitational field. The only other possible cause of fluid pressure is when the container itself puts pressure on the fluid such as in hydraulics. But that requires the statement of the problem to include the force which this does not.

Originally Posted by oz93666 View Post
So we must assume this is in zero gravity , or more likely the extreme pressure will make such differences insignificant ....
You simply could not be more wrong.


Originally Posted by oz93666 View Post
You have the density ...25/24...= 1.041666

You have the temperature ...= 150 C

A search will get you the required chart or table....(you need a more detailed one than the one below)

Its irresponsible to post a URL which only contains graphs when there is no explanation of what the graph represents, which this does not.


Originally Posted by oz93666 View Post
Those blue lines are isotherms...the fourth down is 150C ...follow it along and you see at 1.042 ,the pressure is about 300 MPa ..
Wrong. E.g. if an astronaut in the ISS creates a water bubble in the cabin which will then be kept together by surface tension then the pressure in the water bubble with be zero.
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