Chapter7, Lesson B - Reversible Adiabatic Compression of R-134a

7B-1 :

Reversible Adiabatic Compression of R-134a

5 pts

A piston-and-cylinder device contains saturated R-134a vapor at -5^oC. This vapor is compressed in an internally reversible, adiabatic process until the pressure is 1.0 MPa. Determine the work per kg of R-134a for this process.

Read :

The key to solving this problem is to recognize that any process that is both adiabatic and reversible is ISENTROPIC. This means that S₂ = S₁ and this allos you to fix state 2 and evaluate U₂. Use U₂ in the 1st Law to evaluate W.

Given:

T₁

-5

^oC

Find:

P₂

1000

kPa

???

kJ/kg

Diagram :

Assumptions:

1 -

Process is internally reversible.

2 -

Changes in kinetic and potential energies are negligible.

3 -

Boundary work is the only form of
work that crosses the system boundary.

Equations / Data / Solve :

Begin by applying the 1st Law to the process, assuming changes in kinetic and potential energies are negligible:

Eqn 1

The process is adiabatic so Eqn 1 can be simplified to :

Eqn 2

Use the NIST Webbook to obtain properties for state 1, saturated vapor at -5^oC :

P₁

243.34

kPa

U₁

375.51

kJ/kg

S₁

1.7300

kJ/kg-K

Because the process is both reversible and adiabatic, it is isentropic.

Therefore :

S₂

1.7300

kJ/kg-K

At this point we know values of two intensive variable for state 2, so we can use the NIST Webbook to determine the value of any other property. In this case, we need U₂. First we need to determine the phases that exist at state 2.

At P₂ :

T_sat

39.388

^oC

S_{sat vap}

1.7113

kJ/kg-K

S_{sat liq}

1.1876

kJ/kg-K

Because S₂ > S_{sat vap} at P₂, we can conclude that state 2 is a superheated vapor. We could have reached the same conclusion after careful consideration of a TS Diagram.

We can get the following data from the superheated R-134a Tables or from the NIST Webbook :

T (^oC)

S (kJ/kg-K)

U (kJ/kg)

1.7113

399.45

1.7312

404.32

Interpolation yields :

T₂

44.70

^oC

U₂

404.03

kJ/kg

Now, we can plug values back into Eqn 2 :

-28.52

kJ/kg

Verify:

None of the assumptions made in this problem solution can be verified.

Answers:

-28.5

kJ/kg

Example Problem with Complete Solution