Propane phase change in a container

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SUMMARY

The discussion centers on the behavior of liquid propane in a container under varying pressures, specifically at 10 bara and 5 bara. At 10 bara, the boiling point of propane is approximately 25°C, which corresponds to its vapor pressure at that temperature. The conversation highlights that the phase change of propane is influenced by pressure, and once the temperature drops below the boiling point, evaporation occurs. Additionally, increasing the pressure with nitrogen does not lead to condensation of propane unless the temperature is lowered.

PREREQUISITES
  • Understanding of phase change principles
  • Knowledge of propane properties, including boiling point and vapor pressure
  • Familiarity with concepts of adiabatic and isothermal processes
  • Basic grasp of pressure units, specifically bar and bara
NEXT STEPS
  • Research the phase diagram of propane to understand its behavior under different pressures and temperatures
  • Learn about the thermodynamic principles governing adiabatic processes
  • Explore the effects of pressure on boiling points for various substances
  • Investigate the properties of nitrogen as an inert gas in pressure applications
USEFUL FOR

Engineers, chemists, and professionals involved in thermodynamics, particularly those working with refrigerants and phase change materials.

Kirmandarren
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If I had a container containing 10 bara of nitrogen, and pumped in liquid propane at room temperature would the propane boil?
If I carried out the same experiment but this time at 5 bara, what would be the temperature of the material once the propane had completed its phase change i.e would it drop to around 0C and stop boiling, would the liquid then just evaporate into the vapour space?
 
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Room temperature is what? "bara" is what?
 
Bystander said:
Room temperature is what? "bara" is what?
Take room temperature as 25C, I'm using bar absolute pressure rather than gauge
 
What is the boiling point of propane at 10 bar? What is the vapor pressure at 25?
 
The boiling point at 10bar is approximately 25C. The vapour pressure would therefore also be 10 bar at 25C.
 
Is this an isothermal process? Adiabatic?
 
adiabatic process, I'm aware that there isn't enough info to answer the second question. What I was getting at is: the propane liquids boiling point changes due the the pressure above it no matter what it vapour/gas consists of. Once the temperature drops below its boiling point, evaporation takes over. Also if all the propane turns to vapour (not saturated with vapour), by increasing the pressure by addition of more nitrogen (volume remains constant) the propane will not condense, lowering the temperature is the only way to condense the propane. Is the above logic correct?
 
Kirmandarren said:
adiabatic process, I'm aware that there isn't enough info to answer the second question. What I was getting at is: the propane liquids boiling point changes due the the pressure above it no matter what it vapour/gas consists of. Once the temperature drops below its boiling point, evaporation takes over. Also if all the propane turns to vapour (not saturated with vapour), by increasing the pressure by addition of more nitrogen (volume remains constant) the propane will not condense, lowering the temperature is the only way to condense the propane. Is the above logic correct?
Yes
 

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