Thermodynamics -- Calculate the pressure and temperature

AI Thread Summary
A rigid vessel with a volume of 5m^3 contains 1.0 kg of water and steam at 19°C, requiring calculations for pressure and mass of liquid and vapor. The vessel is heated until all water evaporates, maintaining constant volume throughout the process. The specific volume of the vapor is identified as 50 m^3/kg, indicating a superheated vapor state. To find the corresponding temperature and pressure, interpolation from steam tables is necessary. The discussion emphasizes the importance of understanding phase changes and specific volume in thermodynamic calculations.
ps3stephen
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Homework Statement


A rigid vessel of volume 5m^3 contains 1.0 kg of water and steam at 19 degree C.Find the pressure and the mass of liquid and vapour in the vessel.

The vessel is now heated unit the water is just evaporated. Calculate the the pressure and temperature of this process. The volume should be constant since it is a rigid vessel. Please help with this part. Thank you.

T= 19 degree C
v=50m^3/kg
Psat=0.02196 bar (from steam tables)
vf=0.100164*10^-2 m^3/kg
vg=61.34 m^3/kg
mvapour=0.0815 kg
mliquis=0.0185 kg

Homework Equations


v = (1 – x)vf + x vg

The Attempt at a Solution

 
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ps3stephen said:
Please help with this part
You're going to have to move your feet at least one step --- what is one thing you know?
ps3stephen said:
the water is just evaporated
 
Bystander said:
You're going to have to move your feet at least one step --- what is one thing you know?

Well, the volume is constant.
 
Anything else?
 
Bystander said:
Anything else?
everything i know is stated up there.
 
How many phases?
 
It should be at the superheated vapour state since it state the water is just evaporated.
 
ps3stephen said:
superheated vapour
Mass, volume, number of phases --- need anything else?
 
Mass and volume are given. I just have no clue about what its about when it just evaporated. Volume should be constant since it is a rigid tank.
 
  • #10
Your answer for liquid water in part 1 should be 0.9185 kg. You must have a typo in post #1.

For part 2, if there is no liquid water remaining in the container, what does the specific volume of the water vapor have to be?

Chet
 
  • #11
specific volume should be 50m^3/kg. so i should be find the temperature and pressure from the steam table?
 
  • #12
ps3stephen said:
specific volume should be 50m^3/kg. so i should be find the temperature and pressure from the steam table?
Sure
 
  • #13
so, i have specific volume of 50 m^3/kg from the steam table and do interpolation to find the temperature and pressure correspond to the specific volume of 50 m^3/kg?
 
  • #14
ps3stephen said:
so, i have specific volume of 50 m^3/kg from the steam table and do interpolation to find the temperature and pressure correspond to the specific volume of 50 m^3/kg?
Sure
 
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