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Steam at specific volume/pressure 
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#1
Nov911, 03:24 PM

P: 5

hi all!
i'm trying to figure out how much water i need to heat and to what temperature, to end up with 1 cubic meter of steam at 70atm. i used the calculator found here: http://www.efunda.com/materials/wate...mtable_sat.cfm i punched in 70atm, and got the following data: Temperature (T) 286.72 C Saturated Vapor (g) 37.067 kg/m3 is it correct that i need to get 37.067 liters of water to 286.72 celsius to get 1 cubic meter of steam at 70atm? eventually, i'm trying to find how much energy is needed for this process, but i'm guessing that's more of an engineering question. i just want to get the quantities right first. thanks in advance to anyone taking the time to help me with this! :) 


#2
Nov911, 04:50 PM

P: 412

How did you come up with those particular values of temperature and vapor pressure? Are they postulates? 


#3
Nov911, 05:37 PM

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P: 6,187

Welcome to PF, tikipu!
If you approximate this with an ideal gas (assuming *only* steam), you have: nRT = PV So given P and V, you can choose any temperature and calculate n, the amount of matter. 


#4
Nov911, 06:06 PM

Sci Advisor
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P: 2,905

Steam at specific volume/pressure
Hi tikipu, welcome to the board.



#5
Nov911, 10:36 PM

P: 412




#6
Nov1011, 04:23 AM

P: 5

first of all, klimatos, I like Serena, Q_Goest and klimatos again,
thank you all for your reply and the warm welcome! i knew physicists party hard, now i know they also party well. :) let me zoom out and give you a wider picture. i need to pressurize 1 cubic meter of liquid from STP to 70atm. if a vertical cylinder of 1 cubic meter volume with a piston is completely filled with the liquid at the bottom and then i start pushing steam at 70atm into the other side of the piston at the top, then the liquid should be pushed out at 70atm. right? if so, then as the liquid gets pushed out, more steam needs to be added to the cylinder to occupy the enlarged volume. right? (or just to heat the existing steam more to increase it's volume) if so, then by the end of the process all the liquid was pushed out at 70atm, and the cylinder is now filled with 1 cubic meter of steam at 70atm. right? so: 


#7
Nov1011, 07:44 AM

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P: 6,187

I know of 3 flavors: PV=nRT where n is the number of moles and R is the universal gas constant PV=mRT where m is the mass and R is the specific gas constant (for steam in this case). PV=NkT where N is the number of molecules and k is the Boltzmann constant. I intended the first, but it is arbitrary. Just out of curiosity, when would the molar volume become awkward? 


#8
Nov1011, 07:59 AM

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P: 6,187

Is the steam pure water, or is it saturated in air? For reference, with steam of pure water, the formula PV=nRT combined with a molar mass of water of 18 g/mol, yields that: At P=70 atm, V=1 m3, T=286.72 °C, you need 27 kg water which is approximately 27 liters of liquid water. If the steam is saturated in air, you need to consult the saturated steam table. 


#9
Nov1011, 03:07 PM

P: 412

The formula P = nkT is the kinetic gas theory equivalent of your equations. This is sometimes written P = nmσ^{2} in statistical mechanics and statistical thermodynamics. In both equations, P is the pressure in Pascals per square meter, n is the molecular number density in number of molecules per cubic meter, k is Boltzmann's Constant in joules per molecule per Kelvin, T is the gas temperature in Kelvins, m is the mean molecular impulse mass (see my subsequent posting), and σ is the standard deviation of the axial distribution of molecular velocities (rootmeansquare axial velocity). This is easily derived from the equivalence n = N_{A} / V where N_{A} is Avogadro's Number and V is the molar volume and the equivalence R = N_{A} * k where R is the Universal Gas Constant. 


#10
Nov1011, 03:12 PM

P: 412

1) n = P/kT = (70 x 1013.25)/(1.3806504E23 x 559.87). This is the number of water molecules per cubic meter at the specified temperature and pressure. k is Boltzmann’s Constant. 2) mwater = 2.99144976E26 kg. Note that this is the mean impulse mass of the water molecule, not the mean mass which is 2.99150512E26 kg (VSMOW standard). Technically speaking, because water is a mixture of molecular masses, the impulse mass should be used in any calculation involving pressure. The difference is negligible in most calculations. 3) n mwater = the mass of one cubic meter of steam at the specified temperature and pressure. 4) Go to Engineering Toolbox and calculate how many joules it will take to heat that amount of water to that temperature and pressure. Note A: You mention vaporizing a cubic meter of water to a cubic meter of steam. It can’t be done. It is physically impossible. You will end up with slightly more than a cubic meter of superheated water, not steam. Since water is not very compressible, you will likely end up with a ruptured container. Note B: Having worked with water vapor for many decades, my gut tells me your steam mass of 37.067 kg is way, way, out of whack. Using the calculations shown above, I get only 2.17 grams of steam. 


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