Calculating Steam Pressure in Closed Container

In summary: And it's a small enough mass that you might want to check it against your vessel's tolerance for the amount of water that will produce 10 psig pressure. In summary, to obtain 10 psi of steam pressure in a closed container at a temperature of 816 C, the volume of liquid water needed is approximately 0.052 ml. This can be calculated by dividing the volume of the container (154.497 ml) by the ratio of the specific volume of steam at 816 C and 1.6805 atm absolute to the specific volume of water at room temperature. It is important to consider the vessel's tolerance for the mass of water needed to produce the desired pressure.
  • #1
jackal67347
6
0
I am trying to calculate the volume of liquid water i need to place in a sealed container in order to obtain 10 psi of steam pressure in that closed container.

Here are the numbers:

Temp: 816 C
Volume of steel pipe: 154.497 ml
Final pressure: 10 psi

If I left out a required number please let me know.

If you could show how you solved it or what equation you're using that' be great.

Thanks in advance to whoever helps me out!
 
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  • #2
Welcome to PF!
Is this homework or coursework? There is another section of the forum for that. Perhaps the mentors will move it.

In any case, it is required on PF that you make an attempt to solve problems on your own and show us you work. List any equations you think may be relevant and show your best attempt at a solution.
 
  • #3
LastOneStanding said:
Welcome to PF!
Is this homework or coursework? There is another section of the forum for that. Perhaps the mentors will move it.

In any case, it is required on PF that you make an attempt to solve problems on your own and show us you work. List any equations you think may be relevant and show your best attempt at a solution.

I'm working on a DIY project and I'm trying to steam some catalyst. It's not related to schoolwork.

Here is my attempt:

pv=nrt

(.6805 atm)(v)=( .05555 moles h2o)( .08206) (1089.15 K)

7.296 Liters of steam required

So from the steam table I looked at .001003217 specific volume of h2o at room temp
49.526 specific volume of steam at 800C

49526 ratio

so i need .147 ml according to this calculation
 
  • #4
You're on the right track with the ideal gas law. I'm not going to check your unit conversions, I will assume you've done them right. Few things:
- For that numerical value, the gas constant (R) has units ##m^3 atm K^{-1} kg \cdot mol^{-1}##. These do not match the units you are using for volume and mass is your calculation.
- Your volume is known: 154.497 mL. You want gas at this volume at a certain pressure and temperature. The quantity you are solving for is ##n##, how many moles of water vapour are required. I'm not sure where you got the number you input for ##n##.
- The number of moles of water vapour is the same as the number of moles of liquid water needed to produce it (conservation of particle number). So, you will want to convert the number moles of water to the corresponding mass of the water. See here for a worked example. Then converting the mass of the water to its volume is a straightforward density calculation.
 
  • #5
Is that going to be 10 psi gage or 10 psi absolute?

And, is there going to be air in the closed container also? What is the total pressure in the container before it is sealed?
 
  • #6
Chestermiller said:
Is that going to be 10 psi gage or 10 psi absolute?

And, is there going to be air in the closed container also? What is the total pressure in the container before it is sealed?

10 psi gauge

no air in the container
 
  • #7
LastOneStanding said:
You're on the right track with the ideal gas law. I'm not going to check your unit conversions, I will assume you've done them right. Few things:
- For that numerical value, the gas constant (R) has units ##m^3 atm K^{-1} kg \cdot mol^{-1}##. These do not match the units you are using for volume and mass is your calculation.
- Your volume is known: 154.497 mL. You want gas at this volume at a certain pressure and temperature. The quantity you are solving for is ##n##, how many moles of water vapour are required. I'm not sure where you got the number you input for ##n##.
- The number of moles of water vapour is the same as the number of moles of liquid water needed to produce it (conservation of particle number). So, you will want to convert the number moles of water to the corresponding mass of the water. See here for a worked example. Then converting the mass of the water to its volume is a straightforward density calculation.

Thanks for your help. I redid the calculation with your suggestions but as I continued researching I discovered steam is not an ideal gas. Is this correct?

Could I use the specific volume of room temperature water vs specific volume of steam at 816 C to determine the ratio? I could then use the ratio and compare that to ml of water. Is my thinking correct here?
 
  • #8
You need to get the specific volume of water vapor at 816 C and 1.6805 atm absolute (the ideal gas law and other more accurate equations of state are couched in terms of absolute pressure).

This is above the critical temperature of water. But you can find the specific volume either from steam tables, thermodynamic diagrams for water, or equations of state for water (e.g., z factor). You then divide the volume of your vessel by the specific volume under these conditions. This gives you the mass of water required.
 
  • #9
Chestermiller said:
You need to get the specific volume of water vapor at 816 C and 1.6805 atm absolute (the ideal gas law and other more accurate equations of state are couched in terms of absolute pressure).

This is above the critical temperature of water. But you can find the specific volume either from steam tables, thermodynamic diagrams for water, or equations of state for water (e.g., z factor). You then divide the volume of your vessel by the specific volume under these conditions. This gives you the mass of water required.

.001003217 specific volume of water at room temp
2.9662 specific volume of steam at 816C
both in m3/kg

2957.328 ratio for specific volumes

So 154.497/2957.328 = .052 ml of water

Does that sound right?
 
  • #10
jackal67347 said:
.001003217 specific volume of water at room temp
2.9662 specific volume of steam at 816C
both in m3/kg

2957.328 ratio for specific volumes

So 154.497/2957.328 = .052 ml of water

Does that sound right?

Yes. That's what I get from the ideal gas law also.
 
Last edited:

1. How do you calculate steam pressure in a closed container?

The steam pressure in a closed container can be calculated using the ideal gas law, which states that pressure (P) is equal to the number of moles (n) of gas multiplied by the gas constant (R) and the temperature (T) in Kelvin, divided by the volume (V) of the container. This can be expressed as P = (nRT)/V.

2. What are the factors that affect steam pressure in a closed container?

The factors that affect steam pressure in a closed container include the temperature of the steam, the volume of the container, and the amount of steam present in the container. Additionally, the type of gas and the gas constant can also affect the steam pressure.

3. How does the temperature of the steam affect the pressure in a closed container?

The temperature of the steam directly affects the pressure in a closed container according to the ideal gas law. As the temperature increases, the pressure also increases, and vice versa. This is because at higher temperatures, the molecules of the gas have more kinetic energy and collide more frequently with the walls of the container, resulting in higher pressure.

4. Can the volume of the container affect the steam pressure in a closed container?

Yes, the volume of the container can affect the steam pressure. According to the ideal gas law, pressure is inversely proportional to volume. This means that as the volume of the container decreases, the pressure increases, and vice versa. Therefore, a smaller container will have a higher steam pressure compared to a larger container with the same amount of steam.

5. How can I measure the steam pressure in a closed container?

The steam pressure in a closed container can be measured using a pressure gauge or a manometer. These devices measure the force exerted by the steam on the walls of the container and convert it into a pressure reading. It is important to ensure that the container is completely closed and airtight for accurate measurement of the steam pressure.

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