Calculating Vapor Quality in Tank A: Detailed Solution & Step-by-Step Guide

[fishinbs]

I'm having trouble beginning this problem. It doesn't seem that I have enough information to begin the problem, but I know that can't be the case. Can someone please provide a little direction?

Consider two tanks, A and B, connected by a valve. Each has a volume of 200L, and tank A has R-12 at 25 degrees C, 10% liquid and 90% vapor by volume, while tank B is evacuated. The valve is now opened and saturated vapor flows from A to B until the pressure in B has reached that in A, at which point the valve is closed. This process occurs slowly such that all temperatures stay at 25 degrees C throughout the process. How much has the quality changed in tank A during the process?


EDIT---> Ok, here's what I've got so far:

Before the valve is opened:

specific volume of R-12 liquid is .000763 m3/kg
mass of the R-12 liquid= 20L/.763L/kg= 26.21 kg

specific volume of R-12 vapor is .02685 m3/kg
mass of the R-12 vapor= 180L/26.85L/kg= 6.70 kg

After the valve is opened:

mass of the R-12 vapor= 380L/26.85L/kg= 14.15 kg

What I don't understand is how I determine how much vapor is still in tank A. To determine the quality, I need to know the specific volume of the vapor still in the tank. The formula I have for quality is:

x(quality)= (υ – υf)/ υfg where υf is the specific volume of the liquid and υfg is the specific volume of the vapor-liquid mixture.
Thanks for your help so far. If you could provide any more assistance I would be grateful.

 

[andrevdh]

The vapour pressure for a pure liquid is only temperature dependent. If this R-12 can be considered such then the amount of vapour will increase as it expands from 180 L to 380 L in order to keep its pressure the same. Since the expansion is adiabatically the vapour will stay in a state of saturation during all stages of the expansion. My physics intuition tells me that the density of the vapour (and the liquid) should stay the same under such conditions.