Example 7D - 3: Work, Efficiency and the T-S Diagram for an Ideal Gas Power Cycle

7D-3 :	Work, Efficiency and the T-S Diagram for an Ideal Gas Power Cycle	8 pts

Air contained in a piston-and-cylinder device undergoes a power cycle made up of three internally reversible processes.
Step 1-2: Adiabatic compression from 20 psia and 570^oR to 125 psia

Step 2-3: Isothermal expansion to 20 psia
Step 3-1: Isobaric compression
a.) Sketch the process path for this power cycle on both PV and TS diagrams
b.) Calculate T₃ in ^oF

c.) Calculate the boundary work in Btu/lb_m
d.) Calculate the thermal efficiency of the power cycle

Read :

For part (a) sketch the cycle first to get a better understanding of the processes.

For part (b) recall that the Process 2-3 is isothermal and therefore T₃ = T₂. Determine S^o(T₂) and look it up in the Ideal Gas Entropy Table for air to determine T₂.

For part (c) determine the net work by determining the work for each process and then adding them together.

For part (d) determine the thermal efficiency as the ratio of the net work to the heat going into the system.

Given:

T₁

570

^oR

P₂

125

psia

P₁

psia

P₃

psia

Find:

Part (a)

Sketch PV and TS diagrams

Part (c)

W_cycle

Btu/lb_m

Part (b)

T₃

^oR

Part (d)

Diagram:

Assumptions:

1 -

The system is the air inside the cylinder.

2 -

The air is modeled as an ideal gas.

3 -

Each process is internally reversible.

4 -

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

5 -

There is no change in kinetic or potential energy for either of the two processes.

Equations / Data / Solve:

Part b.)

Since Process 2-3 is isothermal:

Eqn 1

So we should work on determining T₂.

Let's apply the 2nd Gibbs Equation
for ideal gases to Process 1-2 :

Eqn 2

We can determine T₂ and thus T₃ from:

Eqn 3

Lookup S^o(T₁) in the Ideal Gas Entropy Tables:

S^o(T₁)

0.014381

Btu/lb_m-^oR

Now, plug values into Eqn 3 to determine S^o(T₂):

1.987

Btu/lbmol-^oR

S^o(T₂)

0.14007

Btu/lb_m-^oR

28.97

lb_m/lbmol

Now that we know the value of S^o at T₂, we can interpolate on the air Ideal Gas Property Table to determine T₂.

T (^oR)

S^o (Btu/lb_m-^oR)

950

0.13939

T₂

0.14007

Interpolation yields :

T₂ = T₃ =

952.60

^oR

960

0.14202

492.93

^oF

Part c.)

The net work is the sum of the
work done during each process:

Eqn 4

We need to determine the work involved in each process. Begin with Process 1-2.

Since Process 1-2 is adiabatic, and changes in internal and kinetic energies are negligible, the appropriate form of the 1st Law is :

Eqn 5

Eqn 6

Because we know both T₁ and T₂, we can look up the U's in the Ideal Gas Property Table:

At T₁ = 570 ^oR, no interpolation is required:

U₁

5.6705

Btu/lb_m

T (^oR)

U^o(Btu/lb_m)

950

72.806

952.60

U₂

Interpolation yields :

U₂

73.281

Btu/lb_m

960

74.631

Now, we can plug values back into Eqn 6 to determine W₁₂ :

W₁₂

-67.610

Btu / lb_m

Boundary work done by the system during Process 2-3 can be calculated from the definition of boundary work :

Eqn 7

For an ideal gas substitute P = nRT/V into Eqn 7 :

Eqn 8

Integrate Eqn 2
(the process is isothermal, T₂ = T₃ ) :

Eqn 9

Since P₂V₂ = nRT₂ and P₃V₃ = nRT₃ and
T₂ = T₃, we conclude that P₂ V₂ = P₃ V₃ , or :

Eqn 10

Combining Eqn 10 and Eqn 9 yields :

Eqn 11

We can plug numbers into Eqn 11 to evaluate W₂₃ :

W₂₃

119.74

Btu / lb_m

Boundary work done by the system during Process 3-1 can be calculated from the definition of boundary work :

Eqn 12

Since Process 3-1 is isobaric
(P=constant), Eqn 12 simplifies to :

Eqn 13

Since :

Eqn 14

Eqn 15

And since P₁ = P₃, Eqns 13, 14 & 15 can be combined to obtain :

Eqn 16

Now, we can plug values into Eqn 16 to evaluate W₃₁ :

W₃₁

-26.24

Btu / lb_m

Now, we can calculate W_cycle from the sum of the work terms for each step, using Eqn 4 :

W_cycle

25.88

Btu / lb_m

Part d.)

The thermal efficiency of the cycle is defined by :

Eqn 17

We know W_cycle, so we need to determine Q_in. We also know that Q₁₂ = 0 (adiabatic process) and from the TS Diagram it can be concluded that Q₂₃ > 0 and Q₃₁ < 0. Therefore, Q_in = Q₂₃.