4E-1 : |
Isobaric Expansion of Steam in a Closed System |
6 pts |
|
|
|
|
|
|
|
|
|
|
A piston and cylinder device with a free-floating piston has an initial volume of 0.1 m3 and contains 0.5 kg of steam at 400 kPa. Heat is transferred into the steam until the temperature reaches 300oC while the pressure remains constant. Determine the heat transfer and work for this process. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Read : |
We know the values of two intensive variables for state 1: P and specific volume, so we can determine the values of all other properties in this state. |
|
|
|
|
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 ΔU directly. |
|
|
|
|
|
We can also use the definition of work for an isobaric process to evaluate W12. |
|
Once we know W12 and ΔU, we can use the 1st Law to evaluate Q12. |
|
|
|
|
|
|
|
|
|
|
Given : |
V1 |
0.1 |
m3 |
Find : |
Q12 |
??? |
kJ |
|
|
m |
0.5 |
kg |
|
W12 |
??? |
kJ |
|
|
P1 |
400 |
kPa |
|
|
|
|
|
|
P2 |
400 |
kPa |
|
|
|
|
|
|
T2 |
300 |
oC |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Diagram : |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
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 |
|
|
|
|
|
|
|
|
|
|
|
We still need to lookup the same amount of data in the Steam Tables, but the calculations are just a little bit simpler and faster. Now, let's lookup the specific volumes and specific enthalpies. First, we need to determine the state of the water in the cylinder. |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Let's combine Eqns 2 and 3: |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Eqn 4 |
|
|
|
|
|
|
|
|
|
|
|
We still need to lookup the same amount of data in the Steam Tables, but the calculations are just a little bit simpler and faster. Now, let's lookup the specific volumes and specific enthalpies. First, we need to determine the state of the water in the |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
V1 |
0.2 |
m3/kg |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
At 400 kPa : |
|
|
Vsat liq |
0.001084 |
m3/kg |
|
|
|
|
|
|
Vsat vap |
0.4625 |
m3/kg |
|
|
|
|
|
|
|
|
|
|
|
Since Vsat liq < V1 < Vsat vap, we conclude that a saturated mixture exists in the cylinder at state 1. |
|
|
|
|
|
|
So, we must begin by evaluating the quality of the steam. |
|
|
|
|
|
|
|
|
|
|
|
|
|
Eqn 5 |
|
x |
0.4311 |
kg vap/kg |
|
|
|
|
|
|
|
|
|
|
|
Then, we can use the quality to evaluate the specific enthalpy : |
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
|
Eqn 6 |
|
|
|
|
|
|
|
|
|
|
|
Hsat liq |
604.7 |
kJ/kg |
|
|
|
|
|
|
Hsat vap |
2738.5 |
kJ/kg |
|
H1 |
1524.6 |
kJ/kg |
|
|
|
|
|
|
|
|
|
|
|
Next, we need to determine the phases present in State 2. We can do this by comparing T2 to Tsat(P2). |
|
|
|
|
|
|
Tsat(P2) |
143.6 |
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 |
0.6548 |
m3/kg |
|
H2 |
3066.7 |
kJ/kg |
|
|
|
|
|
|
|
|
|
|
|
Now, we can plug values back into Eqns 3 and 4 to evaluate Q12 and W12: |
|
|
|
|
|
|
|
|
|
|
|
W12 |
90.96 |
kJ |
|
Q12 |
771.1 |
kJ |
|
|
|
|
|
|
|
|
|
|
Verify: |
The assumptions made in this problem solution cannot be verified. |
|
|
|
|
|
|
|
|
|
|
Answers : |
W12 |
91.0 |
kJ |
|
Q12 |
771 |
kJ |
|