The discussion focuses on calculating the total pressure and partial pressure of dry air in a closed system containing water, specifically when the temperature is raised from 90°C to 120°C. The correct approach involves using the polytropic process to determine the partial pressure of dry air at 1 atm, resulting in an increase of 0.3375 bar. The initial conditions include a water vapor pressure of 0.7 bar and a total pressure of 1 atm, leading to a final water vapor pressure of 1.99 bar. It is established that the change in the volume of liquid water is negligible for these calculations.
PREREQUISITESThis discussion is beneficial for thermodynamics students, engineers working with closed systems, and professionals involved in pressure calculations in fluid dynamics.
Why do you need to use a polytropic process? This is an equilibrium problem.dbag123 said:what about using the polytropic process to figure out the partial pressure of the dry air at 1 atm?, the increase would be 0.3375 bar and that is the right answer, initial temperature is 90C
You need to assume that the change in the volume of liquid water is negligible.dbag123 said:in a jar that has an airtight lid there is water. Temperature is raised to 120C, from 90C, where the pressure of water vapor is 0,7 bar and the air 1 atm and as a result the pressure of the water vapor is 1,99bar. Calculate the partial pressure of dry air and the total pressure.
One could account for the density of the liquid water. However, since liquid water is approximately 1000 times as dense as steam at 2.0 bars, one might expect such an accounting to affect the result only in the third significant digit. The inputs here look to be good to only two digits at best.Chestermiller said:You need to assume that the change in the volume of liquid water is negligible.