(Thermodynamics) A tank with water & air heated

In summary, the problem involves a rigid, impervious, closed tank filled with 70% water and the rest with air at an initial temperature of 20 °C and absolute pressure of 1 bar. The final pressure needs to be determined when the temperature is increased to 200 °C. The specific volume of liquid water, vapor pressure of water, and specific volume of saturated water vapor at 20 °C need to be calculated in order to find the initial mass of liquid water, water vapor, and total water in the tank. Then, the partial pressure of air and number of moles of air initially in the tank need to be determined. Finally, the total volume of liquid water and water vapor needs to be equal to
  • #1
Baybora
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0

Homework Statement


For a rigid, impervious, closed tank - The volume of tank is filled with 70% water and the rest is with air- The initial temperature of the tank is 20 °C, and the absolute pressure is 1 bar. - What would be the final pressure if we raise the temperature to 200 °C?

Homework Equations

The Attempt at a Solution


I tried finding some values at the water-vapour table but I just don't know what to do about it because I can't be sure if the water will become vapour or will stay in its liquid form. ANY help is appreciated
 
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  • #2
Do not be afraid!
It is not dangerous to start calculating with an assumption that turns out to be wrong.
(The vessel will not explode, you won't get burns, the only risk is that you would have to try something else.)
 
  • #3
The problem is that I can't even be sure of my assumption. How am i supposed to know if the water will stay liquid or will turn into gas?
 
  • #4
Baybora said:
The problem is that I can't even be sure of my assumption. How am i supposed to know if the water will stay liquid or will turn into gas?
Water can't stay liquid at 200° C unless the internal pressure in the tank is greater than the vapor pressure of water at that temp.
 
  • #5
Baybora said:
The problem is that I can't even be sure of my assumption. How am i supposed to know if the water will stay liquid or will turn into gas?
Are you currently learning how to use the Steam Tables in your course, or are you learning about the ideal gas law and vapor pressure?

Chet
 
  • #6
Yes and any way to find out if it will or not?
 
  • #7
Chestermiller said:
Are you currently learning how to use the Steam Tables in your course, or are you learning about the ideal gas law and vapor pressure?

Chet

Its a off course question, asked by mentor at a company. He said it has the same principle for calculations for extruders which contain screw and barrel. He has given me time for me to solve it until the end of 1st semester and said I could ask anybody if I can't make it myself. I asked it to my teacher but she didnt help at all not sure if she wants me to think, study and learn by myself or she also doesn't know it either. So its probably higher than thermodynamics 1 course
 
  • #8
Take as a basis a 1 cubic meter tank (the actual volume you take doesn't matter). First focus on the initial conditions. What is the specific volume of liquid water at 20 C and 1 atm? What is the equilibrium vapor pressure of water at 20 C? What is the specific volume of saturated water vapor at 20 C? Based on these answers, what is the initial mass of liquid water in the tank? What is the initial mass of water vapor in the tank? What is the initial total mass of water in the tank. What is the partial pressure of air initially? How many moles of air are in the tank initially?

When you have the answers to these questions, let's see what you get.

Chet
 
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  • #9
Baybora said:
The problem is that I can't even be sure of my assumption. How am i supposed to know if the water will stay liquid or will turn into gas?
You know that by the outcome of the calculation. When you numbers are consistent, that is the answer.

The water does not "know" either.
 
  • #10
Baybora said:
The problem is that I can't even be sure of my assumption. How am i supposed to know if the water will stay liquid or will turn into gas?
If you know the total mass M of water in the tank, you let the final amount of water vapor be x and the final amount of liquid water be M-x. You then look up the specific volume of saturated liquid water at 200 C and the specific volume of saturated water vapor at 200 C. Then you solve for x under the constraint that the total volume of liquid water and water vapor must be equal to the volume of the tank.

Chet
 
  • #11
There is no need to solve for x. The only question is what the pressure will be.

If there is any water in the liquid phase, you just look at the vapor pressure curve.
 
  • #12
PietKuip said:
There is no need to solve for x. The only question is what the pressure will be.

If there is any water in the liquid phase, you just look at the vapor pressure curve.
Well, if there is a change in the volume of liquid, the volume of the air also changes, and this can affect the partial pressure of the air and total pressure.
 
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  • #13
Bump
 
  • #14
Still looking for answers. Need help with the calculation part mostly
 
  • #15
Baybora said:
Still looking for answers. Need help with the calculation part mostly
You still haven't answered my questions in posts #5 and #8. How are we supposed to help you if you don't answer our leading questions? I know you can answer the question in post #5, because it doesn't require any analysis.
 
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  • #16
for your fifth post my reply was
Baybora said:
Its a off course question, asked by mentor at a company. He said it has the same principle for calculations for extruders which contain screw and barrel. He has given me time for me to solve it until the end of 1st semester and said I could ask anybody if I can't make it myself. I asked it to my teacher but she didnt help at all not sure if she wants me to think, study and learn by myself or she also doesn't know it either. So its probably higher than thermodynamics 1 course

Chestermiller said:
Well, if there is a change in the volume of liquid, the volume of the air also changes, and this can affect the partial pressure of the air and total pressure.
and for your 8th post you already pointed out the fact that makes it very complicated and that doesn't give me a clue what are the constants in here. Everything in this system are effected by the parameters
 
  • #17
OK. Let's do it without using the steam tables. Let's take as the basis of the calculations a tank with a volume of 1 cubic meter. That means that there is 0.7 cubic meters of liquid water plus 0.3 cubic meters of a gaseous mixture of air an water vapor at 20 C and 1 bar. The first step is to figure out the mass of water and the mass of air in the tank to start with (this won't change when the contents is heated)? Do you know how to figure this part out?

Chet
 
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  • #18
Baybora said:
and for your 8th post you already pointed out the fact that makes it very complicated and that doesn't give me a clue what are the constants in here. Everything in this system are effected by the parameters
I expect you to look up the data asked for in post #8 on your own. This is what is required to start solving this problem. In real life, people don't spoon feed you data like they do in school. Welcome to the real world.

If you follow the steps I am leading you through, starting with the answers to the questions in post #8, I can guarantee you will get your answer. Otherwise, good luck.

Chet
 
  • #19
Chestermiller said:
OK. Let's do it without using the steam tables. Let's take as the basis of the calculations a tank with a volume of 1 cubic meter. That means that there is 0.7 cubic meters of liquid water plus 0.3 cubic meters of a gaseous mixture of air an water vapor at 20 C and 1 bar. The first step is to figure out the mass of water and the mass of air in the tank to start with (this won't change when the contents is heated)? Do you know how to figure this part out?

Chet

Ok so the volume density of saturated water at 20 C is 0.001002 m^3/kg ( neglecting the small difference between compressed water at 20 C and 1 bar) and the volume density of air at 20 C and 1 bar is around 0,833 m^3 / kg

Assuming we have a 1 m^3 tank we have

0,7 / 0,001002 = 698,6 kg of water and
0,3 / 0,8333 = 0,36 kg of air in the tank
 
  • #20
Chestermiller already told you that the 0.3 m^3 also contains water. Yet you refuse to calculate how much.
 
  • #21
Baybora said:
Ok so the volume density of saturated water at 20 C is 0.001002 m^3/kg ( neglecting the small difference between compressed water at 20 C and 1 bar) and the volume density of air at 20 C and 1 bar is around 0,833 m^3 / kg

Assuming we have a 1 m^3 tank we have

0,7 / 0,001002 = 698,6 kg of water and
0,3 / 0,8333 = 0,36 kg of air in the tank
Nice job. Your answer is close, but not quite. There is a small amount of water in the gas phase. It isn't very important at 20 C, but the amount of water in the gas phase will be important at 200 C. Would you prefer to neglect it and move on to 200 C, or would you rather get the practice accounting for the water vapor even at 20 C? It's your call.

Chet
 
  • #22
Chestermiller said:
Nice job. Your answer is close, but not quite. There is a small amount of water in the gas phase. It isn't very important at 20 C, but the amount of water in the gas phase will be important at 200 C. Would you prefer to neglect it and move on to 200 C, or would you rather get the practice accounting for the water vapor even at 20 C? It's your call.

Chet

I think we will have to assume that they are separated as just liquid water and air with no water vapor because i couldn't find a way to find out the amount of water in air without a given humidity ratio or other info. That's why I think we should neglect it.

Here's one of the sites I have checked if I can calculate the water vapor in air.

http://home.howstuffworks.com/humidifier1.htm

There it says "At 20 C air can hold up to 18 grams of water" so we can just know the limits of contained vapor unless we are given a humidity ratio which in this case we aren't
 
  • #23
Baybora said:
I think we will have to assume that they are separated as just liquid water and air with no water vapor because i couldn't find a way to find out the amount of water in air without a given humidity ratio or other info. That's why I think we should neglect it.

Here's one of the sites I have checked if I can calculate the water vapor in air.

http://home.howstuffworks.com/humidifier1.htm

There it says "At 20 C air can hold up to 18 grams of water" so we can just know the limits of contained vapor unless we are given a humidity ratio which in this case we aren't
The partial pressure of the water vapor in the head space is equal to the equilibrium vapor pressure of water at 20 C. In other words, the air above the liquid water in the closed container equilibrates with the liquid water and becomes saturated. So the relative humidity in the air in the head space is 100%. What is the equilibrium vapor pressure of water at 20 C?
 
  • #24
Chestermiller said:
The partial pressure of the water vapor in the head space is equal to the equilibrium vapor pressure of water at 20 C. In other words, the air above the liquid water in the closed container equilibrates with the liquid water and becomes saturated. So the relative humidity in the air in the head space is 100%. What is the equilibrium vapor pressure of water at 20 C?

17,535 mm Hg = 0,0233 bar

Source :

http://genchem.rutgers.edu/vpwater.html
 
  • #25
Baybora said:
17,535 mm Hg = 0,0233 bar

Source :

http://genchem.rutgers.edu/vpwater.html
Excellent. So the partial pressure of water vapor in the head space is 0.0233 bars, and the partial pressure of air in the head space is 0.9767 bars. That gives a total pressure of 1 bar. Using the ideal gas law, how many gram moles of water are there in 0.3 cubic meters of head space at a partial pressure of 0.0233 bars and a temperature of 293 K (20 C)? Using the ideal gas law, how many gram moles of air are there in 0.3 cubic meters of head space at a partial pressure of 0.9767 bars and a temperature of 293 K (20 C)? Based on this, what is the mass of the water in the head space? What is the mass of air in the head space? What is the total mass of water (liquid and vapor) in the container?
 

1. How does heating a tank with water and air affect the pressure inside?

As the temperature of the water and air inside the tank increases, the molecules gain more kinetic energy and begin to move faster, causing an increase in pressure. This is due to the ideal gas law which states that pressure is directly proportional to temperature, assuming constant volume and number of particles.

2. Is there a limit to how much the temperature of the water and air can increase in the tank?

Yes, there is a limit to how much the temperature can increase in the tank. This is known as the boiling point for water and the limit for the air is determined by its specific heat capacity. Once the boiling point or maximum temperature is reached, any additional heat added will cause the water to evaporate and the air to expand without a significant increase in temperature.

3. How does the heat transfer between the water and air in the tank?

The heat transfer between the water and air in the tank is through convection. As the water is heated, it becomes less dense and rises to the top of the tank, causing the cooler air to sink to the bottom. This creates a convection current, allowing the heat to be evenly distributed throughout the tank.

4. What happens to the density of the water and air as the temperature increases?

As the temperature of the water and air increases, their density decreases. This is because the molecules gain more energy and begin to spread out, resulting in a decrease in the number of molecules per unit volume. This is why hot air rises and cold air sinks.

5. Can the tank with heated water and air reach equilibrium?

Yes, the tank can reach equilibrium, but only if the heating source is turned off and the tank is left to cool. At equilibrium, the temperature, pressure, and density throughout the tank will be uniform, and there will be no further heat transfer between the water and air.

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