Chapter4, Lesson F - Performance of an Ideal Gas Cycle

4F-5 :

Performance of an Ideal Gas Cycle

10 pts

An ideal gas, initially at 30°C and 100 kPa, undergoes an internally reversible, cyclic process in a closed system. The gas is first compressed adiabatically to 500 kPa, then cooled at a constant pressure of 500 kPa to 30°C, and finally expanded isothermally to its original state.

a.)

Carefully draw this process in a traditional piston-and-cylinder schematic.

b.)

Sketch the process path for this cycle on a PV Diagram. Put a point on the diagram for each state. Be sure to include and label all the important features for a complete PV diagram for this system.

c.)

Calculate Q, W, ΔU and ΔH, in J/mole, for each step in the process and for the entire cycle. Assume that CP = (7/2) R.

d.)

Is this cycle a power cycle or a refrigeration cycle ? Explain. Calculate the thermal efficiency or COP of the cycle, whichever is appropriate.

Read :

Sketch carefully. Understanding what is going on in the problem is half the battle. Apply the 1st Law, the definitions of boundary work, CP and CV to a cycle on an ideal gas with constant heat capacities. Take advantage of the fact that step 1-2 is both adiabatic and reversible, so it is isentropic. Power cycles produce a net amount of work and proceed in a clock-wise direction on a PV Diagram.

Given :

303.15

100

kPa

???

500

kPa

303.15

500

kPa

Q12

J/mole

8.314

J/mole-K

29.099

J/mole-K

Find :

For each of the three steps and for the entire cycle:

ΔU

???

J/mole

ΔH

???

J/mole

???

J/mole

???

J/mole

Diagrams :

Part a.)

Part b.)

Assumptions:

Step 1-2 is adaibatic, Step 2-3 is isobaric, Step 3-1 is isothermal.
The entire cycle and all of the steps in the cycle are internally reversible.
Changes in kinetic and potential energies are negligible.Changes in kinetic and potential energies are negligible.
Boundary work is the only form of work interaction during the cycle.
The PVT behavior of the system is accurately described by the ideal gas EOS.

Equations / Data / Solve :

Part c.)

Let's begin by analyzing step 1-2, the adiabatic compression.

Begin by applying the 1st law for closed systems to each step in the Carnot Cycle. Assume that changes in kinetic and potential energies are negligible.

Eqn 1

Because internal energy is not a function of pressure for an ideal gas, we can determine DU by integrating the equation which defines the constant volume heat capacity. The integration is simplified by the fact that the heat capacity for the gas in this problem has a constant value.

Eqn 2

Eqn 3

Combining Eqns 1 & 3 yields:

Eqn 4

The problem is that we do not know T2. So, our next task is to determine T2.

Since the entire cycle is reversible and this step is also adiabatic, this step is isentropic.

The fastest way to determine T2 is to use one of the PVT relationships for isentropic processes.

Eqn 5

Solve Eqn 5 for T2:

Eqn 6

Now, we need to evaluate γ :

Eqn 7

But for ideal gases :

Eqn 8

Solving Eqn 8 for CV yields :

Eqn 9

Plugging values into Eqn 9 and then Eqn 7 yields :

20.785

J/mole-K

1.4

Now, plug values into Eqn 5 to get T2
and plug that into Eqn 4 to get W12 :

480.13

W12

-3678.6

J/mole

Plugging values into Eqn 1 yields :

ΔU12

3678.6

J/mole

Now, we can get ΔH
from its definition :

Eqn 10

But, the gas is an ideal gas:

Eqn 11

Combining Eqns 10 & 11 gives us :

Eqn 12

Now, we can plug values into Eqn 12 :

ΔH12

5150.1

J/mole

Next, let's analyze step 2-3, isobaric cooling.

480.1349

500

kPa

303.15

500

kPa

The appropriate form of the 1st Law is:

Eqn 13

Because we assumed that boundary work is the only form of work that crosses the system boundary, we can determine work from its definition.

Eqn 14

Isobaric process:

Eqn 15

Because the system contains and ideal gas:

Eqn 16

W23

-1471.5

J/mole

Next we can calculate ΔU by applying Eqn 3 to step 2-3:

Eqn 17

ΔU23

-3678.6

J/mole

Now, solve Eqn 13 to
determine Q23 :

Eqn 18

Q23

-5150.1

J/mole

Now, we apply Eqn 12 to
step 2-3 to determine ΔH :

Eqn 19

ΔH23

-5150.1

J/mole

Next, we analyze step 3-1, isothermal expansion.

For ideal gases, U and H are
functions of T only. Therefore :

ΔU31

0.0

J/mole

ΔH31

0.0

J/mole

The appropriate form
of the 1st Law is:

Eqn 20

Again, because we assumed that boundary work is the only form of work that crosses the system boundary, we can determine work from its definition.

Eqn 21

Ideal Gas EOS :

Eqn 22

Solve Eqn 22 for P and
substitute the result into
Eqn 21 to get :

Eqn 23

Eqn 24

Integrating Eqn 24 yields :

Eqn 25

We can use the Ideal Gas EOS to avoid calculating V1 and V3 as follows:

Apply Eqn 22 to both states 3 and 1 :

Eqn 26

Cancelling terms and rearranging leaves :

Eqn 27

Use Eqn 27 to eliminate the V's from Eqn 25 :

Eqn 28

Now, plug values into Eqn 28 and then Eqn 20 :

W31

4056.4

J/mole

Q31

4056.4

J/mole

Finally, we can calculate Qcycle and Wcycle from :

Eqn 29

Eqn 30

Plugging values into Eqns 29 & 30 yields :

Wcycle

-1093.7

J/mole

Qcycle

-1093.7

J/mole

This result confirms what an application of the 1st Law to the entire cycle tells us:
Qcycle = Wcycle.

Part d.)

The cycle is a refrigeration cycle because both Wcycle and Qcycle are negative.

The coefficient of performance of a refrigeration cycle is defined as :

Eqn 31

QC is the heat absorbed by the system during the cycle. In this case, QC =Q31.

W is the work input to the system during the cycle. In this case, W = - Wcycle.

Therefore :

4056.4

J/mole

1093.7

J/mole

Plug values into Eqn 31 to get :

COPR

3.709

Verify :

The ideal gas assumption needs to be verified.

We need to determine the specific
volume at each state and check if :

Eqn 32

25.20

L/mol

Eqn 33

7.98

L/mol

5.04

L/mol

The specific volume at each state is greater than 5 L/mol for all states and Air can be considered to be a diatomic gas, so the ideal gas assumption is valid.

Answers:

a.)

See diagram above.

b.)

See diagram above.

c.)

Step

ΔU

ΔH

1 - 2

3679

5150

-3679

All values in this table are in J/mole.

2 - 3

-3679

-5150

-1471

3 - 1

4056

Cycle

-1094

d.)

Refrigeration or Heat Pump Cycle.

COPR

3.7

COPHP

4.7

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