Example Problem with Complete Solution

2B-2 : Quality of a Two-Phase Ammonia Mixture in a Rigid Tank 4 pts
Ammonia exists as a saturated mixture at 240.21 kPa and -14.6oC in a rigid vessel with a volume of 1.0 m3. The specific volume of the saturated liquid and saturated vapor are 1.5195 L/kg and 0.50063 m3/kg, respectively.
The quality of the ammonia is 0.275 kg vap/kg. What is the total mass of Ammonia inside the vessel in kg?

Read : There are two keys to this problem. The first is the relationship between the total mass, the total volume and the overall, average or mixture specific volume. The other key is how to use quality and specific properties of saturated liquid and saturated vapor to determine a specific property of a saturated mixture.
Given: P 240.21 kPa Vsat liq 1.5195 L/kg
T -14.6 oC Vsat vap 0.50063 m3/kg
Vtotal 1 m3 x 0.275 kg vap/kg
Find: Mtotal ??? kg
Assumptions: None.
Equations / Data / Solve:
Let's begin with the relationship between mass, volume and specific volume for the entire system.
Eqn 1
We want to determine Mtotal and we know Vtotal, so all we need to do is determine Vmix and we will be able to use Eqn 1 to solve this problem.
The specific volume of the two-phase mixture is related to the quality and the specific volumes of the saturated liquid and saturated vapor by the following equation.
Eqn 2
We know the values of all of the variables on the right-hand side of Eqn 2, so we can plug-in values to determine Vmix.
Vmix 0.1388 m3/kg
Be careful with the units in Eqn 2. You must convert L to m3 in Vsat liq to make all of the units work out properly.
Now, we can plug values into Eqn 1 to complete this problem.
Mtotal 7.2059 kg
Verify: There are no assumptions to verify in this problem.