Solving a Cylinder Problem: Air & Water at 120°C to 180°C

[cathy88]

Homework Statement

Problem shown in attach picture

A cylinder of total internal volume 0.10 m3 has a
frictionless internal piston (of negligible mass and
thickness) which separates 0.50 kg of water below
the piston from air above it. Initially the water exists
as saturated liquid at 120°C and the air, also at
120°C, exerts a pressure such that it exactly
balances the upward force exerted by the water.
(refer to the figure right). heat is then transferred into
the entire cylinder, such that the two substances are
at the same temperature at any instant, until a final
State 2 having a uniform temperature of 180°C is
reached. The air may be assumed to behave ideally,
having R = 0.287 kJ kg-1
K-1; cp0 = 1.004 kJ kg-1
K-1;
cv0 = 0.717 kJ kg-1
K-1.
(a) What is the mass of air in the cylinder?
(b) Determine both the volume below the piston, and the dryness fraction of
the water occupying that volume when State 2 is reached.
(c) What is the change in the total internal energy of:
(i) the air; and
(ii) the water
during the entire heat addition process from State 1 to State 2.
(d) How much heat has been transferred into the complete system comprising
both water and air?

Homework Equations

I really have no idea, I was trying with pV=mRT

The Attempt at a Solution

Im totally stuned by this question, I tried doing pV=mRT but that was wrong the a) is 0.175kg. COuld someone please help me how I could start going through this problem?

 

[Astronuc]

What is the saturation pressure of water at 120°C? Think about how to use the mass of water, 0.50 kg, and the specific volume of the sat liquid at 120°C. I'm assuming that the water is completely liquid in state 1.

When heat is added the water will change phase. The steam (vapor) occupies more volume than the liquid. As liquid transforms to vapor, the piston pushes on the air, which is compressed.


The question in State 2 - does the water at 180°C remain saturated, or is it superheated or is it a compressed liquid?

 

[cathy88]

Thanks for the reply so, for saturated water at T=120 degress the Psat=0.1985MPa and v=0.001060m^3 given the mass of the saturated liquid is 0.50kg and the total volume of the piston is 0.10m3. So the air is taking up 0.09894m3 of volume. How can I link it to how much mass it is?Do I have to convert specific volume back to volume first? I could just do mv=V right?

Thanks for the help

 

[Astronuc]

Remember the specific properties are give per unit mass, so v=0.001060 m³/kg. The liquid will occupy a volume of V = m v = kg * m³/kg. Be careful in writing units.

The volume of liquid and volume of air must = 0.10 m³, so as the volume of liquid/vapor increases the volume of air must decrease.