Example 9B - 3: Vapor Power Cycle Based on Temperature Gradients in the Ocean

9B-3 :	Vapor Power Cycle Based on Temperature Gradients in the Ocean	9 pts

A Rankine Power Cycle uses water at the surface of a tropical ocean as the heat source, T_H = 82^oF, and cool water deep beneath the surface as the heat sink, T_C = 48^oF. Ammonia is the working fluid.

The boiler produces saturated ammonia vapor at 80^oF and the condenser effluent is saturated liquid ammonia at 50^oF. The isentropic efficiencies of the pump and turbine are 75% and 85%, repectively.

Calculate the...
a.) Thermal efficiency of this Rankine Cycle
b.) Thermal efficiency of a Carnot Cycle operating between the same two thermal reservoirs

Read :

Assume the process operates at steady-state and that changes in potential and kinetic energies are negligible.

We want the boiler pressure to be as high as possible and the condenser pressure to be as low as possible in order to maximize the thermal efficiency of the cycle. But the saturation temperature of the ammonia at the boiler pressure must be less than 82^oF in order to absorb heat from the warmer seawater. This explains why the ammonia boils at 80^oF. A similar argument regarding the vapor-liquid equilibrium in the condenser leads us to the choice of 50^oF for the saturation temperature of the ammonia in the condenser.

Given:

T₂

^oF

T_C

^oF

x₂

lb_m vap/lb_m

T_H

^oF

T₃

^oF

(T_{sw, in})_boil

^oF

x₄

lb_m vap/lb_m

(T_{sw, in})_cond

^oF

h_S,pump

75%

h_S,turb

85%

Find:

a.)

h_th

???

b.)

h_max

???

Diagram:

A good flow diagram was provided in the problem statement.

TS Diagram :

Assumptions:

1 -

Each process in the cycle operates at steady-state.

2 -

The power cycle operates on the Rankine Cycle.

3 -

Changes in kinetic and potential energies are negligible.

4 -

The pump and the turbine are both adiabatic.

Equations / Data / Solve:

Let's organize the data that we need to collect into a table. This will make it easier to keep track of the values we have looked up and the values we have calculated.

Stream

State

T
(^oF)

P
(psia)

X
(lb_m vap/lb_m)

H
(Btu/lb_m)

S
(Btu/lb_m-^oR)

Sub Liq

50.18

153.13

N/A

98.132

0.21024

Sub Liq

50.27

153.13

N/A

98.233

0.21044

Sat Vap

153.13

630.36

1.1982

Sat Mix

89.205

0.9551

601.38

1.1982

Sat Mix

89.205

0.9633

605.73

1.2068

Sat Liq

89.205

97.828

0.21024

Additional data that may be useful.

State

T
(^oF)

P
(psia)

X
(lb_m vap/lb_m)

H
(Btu/lb_m)

S
(Btu/lb_m-^oR)

Sat Vap

153.13

630.36

1.1982

Sat Liquid

153.13

131.86

0.27452

Sat Vap

89.205

625.07

1.2447

Sat Liquid

89.205

97.828

0.21024

Part a.)

The thermal efficiency of a
power cycle can be determined using :

Eqn 1

We need to evaluate Q₁₂ and Q₃₄ so we can use Eqn 1 to evaluate the thermal efficiency of the cycle.

Apply the 1st Law to the boiler, assuming it operates at steady-state, changes in kinetic and potential energies are negligible and no shaft work crosses the boundary of the boiler.

Eqn 2

We can get a value for H₁ from the Steam Tables or NIST Webbook because we know that the boiler effluent in a Rankine Cycle is a saturated vapor at 80^oF.

P₂

153.13

psia

H₂

630.36

Btu/lb_m

In order to fix state 1 and evaluate H₁, we must use the isentropic efficiency of the pump. The feed to the pump is a saturated liquid at 80^oF. So, we can look-up S₄ in the Steam Tables or NIST Webbook.

P₄

89.205

psia

S₄

0.21024

Btu/lb_m-^oR

H₄

97.828

Btu/lb_m

The definition of
isentropic efficiency for a pump is :

Eqn 3

We can solve Eqn 2 for H₁ :

Eqn 4

Next, we need to determine H_1S. For an isentropic pump :

S_1S

0.21024

Btu/lb_m-^oR

Now, we know the value of two intensive properties at state 1S: S_1S and P₁ (because the boiler is isobaric in a Rankine Cycle, P₁ = P₂.

At P = 153.13 psia :

T (^oF)

H Btu/lb_m

S Btu/lb_m-^oR

97.935

0.20985

T_1S

H_1S

0.21024

T_1S

50.18

^oF

103.528

0.22077

H_1S

98.13

Btu/lb_m

Now ,we can plug values into Eqn 4 to determine H₁ :

H₁

98.23

Btu/lb_m

Next, we can plug values into Eqn 2 to evaluate Q₁₂ :

Q₁₂

532.12

Btu/lb_m

We can determine Q₃₄ by applying the 1st Law, with all the same assumptions made about the boiler.

Eqn 5

We already know H₄, so we need to evaluate H₃. To do this, we use the isentropic efficiency of the turbine.

In order to fix state 3 and evaluate H₃, we must use the isentropic efficiency of the turbine. The feed to the turbine is a saturated vapor at 80^oF. So, we can look-up S₂ in the Steam Tables or NIST Webbook.

P₄

89.205

psia

S₂

1.1982

Btu/lb_m-^oR

H₂

630.36

Btu/lb_m

The definition of
isentropic efficiency for a turnine is :

Eqn 6

We can solve Eqn 6 for H₃ :

Eqn 7

Next, we need to determine H_3S. For an isentropic turbine :

S_3S

1.1982

Btu/lb_m-^oR

Now, we know the value of two intensive properties at state 3S: S_3S and P₃ (because the condenser is isobaric in a Rankine Cycle, P₃ = P₄.

At P = 89.205 psia :

S_{sat liq}

0.21024

Btu/lb_m-^oR

Since S_{sat liq} < S_3S < S_{sat vap},
state 3S is a saturated mixture.

S_{sat vap}

1.2447

Btu/lb_m-^oR

Determine x_3S from the
specific entropy, using:

Eqn 8

x_3S

0.9551

lb_m vap/lb_m

Then, we can use the quality
to determine H_3S, using:

Eqn 9

At P = 89.205 psia :

H_{sat liq}

97.828

Btu/lb_m

H_{sat vap}

625.07

Btu/lb_m

H_3S

601.38

Btu/lb_m

Now ,we can plug values into Eqn 7 to determine H₁ :

H₃

605.73

Btu/lb_m

Next, we can plug values into Eqn 5 to evaluate Q₃₄ :

Q₃₄

-507.90

Btu/lb_m

We can now plug values into Eqn 1 to evaluate the thermal efficiency of the Rankine Cycle.

h_th

4.55%

The maximum thermal efficiency for a power cycle operating between two thermal reservoirs at T_H and T_C is the Carnot Efficiency :

Eqn 10

T_C

507.67

^oR

T_H

541.67

^oR

h_max

6.28%

Verify:

The assumptions made in the solution of this problem cannot be verified with the given information.

Answers :

a.)

h_th

4.55%

h_max

6.28%

Example Problem with Complete Solution