Example Problem with Complete Solution

4E-1 : Isobaric Expansion of Steam in a Closed System 6 pts
Steam is contained in a piston and cylinder device with a free-floating piston. Initially, the steam occupies a volume of 0.18 m3 at a pressure of 500 kPa. 
The steam is slowly heated until the temperature is 300oC, while the pressure remains constant. If the cylinder contains 0.65 kg of steam, determine the heat transfer and the work in kJ for this process.
Read : We know the values of two intensive variables for state 2: T and P, so we can determine the values of all other properties in this state.
Therefore we can calculate DU directly.
We can also use the definition of work for an isobaric process to evaluate W12.
Once we know W12 and DU, we can use the 1st Law to evaluate Q12.
Given: V1 0.18 m3 Find: Q12 ??? kJ
m 0.65 kg W12 ??? kJ
P1 500 kPa
P2 500 kPa
T2 300 oC
Assumptions: 1 - Changes in kinetic and potential energies are negligible.
2 - The process is a quasi-equilibrium process.
Equations / Data / Solve:
Choose the water inside the cylinder as the system.
Apply the integral form of the 1st Law to the process:
Eqn 1
If we assume that changes in kinetic and potential energies are negligible, then Eqn 1 simplifies to :
Eqn 2
We can evaluate W12 from the definiton of work applied to an isobaric process.
Eqn 3
Let's combine Eqns 2 and 3:
Eqn 4
We still need to lookup the same amount of data in the Steam Tables, V and H, but the calculations are just a little bit simpler and faster using H than using U.
Before we can look up H, we need to determine the state of the water in the cylinder.
Calculate V1 from :
V1 0.2769 m3/kg
At 500 kPa : Vsat liq 0.0010925 m3/kg
Vsat vap 0.37481 m3/kg
Since Vsat liq < V1 < Vsat vap, we conclude that a saturated mixture exists in the cylinder at state 1.
So, we must next evaluate the quality of the steam.
Eqn 5 x 0.7381 kg vap/kg
Then, we can use the quality to evaluate the specific enthalpy :
Eqn 6
Hsat liq 640.09 kJ/kg
Hsat vap 2748.1 kJ/kg H1 2196.0 kJ/kg
Next, we need to determine the phases present in State 2.  We can do this by comparing T2 to Tsat(P2).
In the saturation pressure table of the Steam Tables we find: Tsat(P2) 151.8 oC
Because T2 > Tsat(P2), state 2 is a superheated vapor.
From the NIST Webbook or the Superheated Tables of the Steam Tables we obtain the following data:
V2 0.52261 m3/kg H2 3064.6 kJ/kg
Now, we can plug values back into Eqns 3 and 4 to evaluate Q12 and W12:
W12 79.8 kJ Q12 564.6 kJ
Verify: The assumptions made in this problem solution cannot be verified.
Answers : W12 80 kJ Q12 565 kJ