Rankine Cycle Efficiency

[MCTachyon]

Homework Statement

Superheated steam at a pressure of 40 bar and a temperature of 500°C is supplied to the turbine of a Rankine cycle. If the condenser pressure is 0.03 bar.

Find the thermal efficiency of the cycle. (Neglect feed pump work).

I used steam tables found https://www.slideshare.net/SGhallab/steam-tables-fifth-edition-by-rogers-and-mayhew

Homework Equations

S₁ = S_g

S_g = (1 - x_g)S_f2 + x_gS_g2

h₂ = (1 - x_g)h_f2 + x_gh_g2

Specific work (W) = h₁ - h₂

Specific Heat (Q) = h₁ - h_f2

Eff (η) = W / Q

The Attempt at a Solution

From steam tables:

At 40 Bar and 500°C (Before turbines):

h_g = 3445 kJ Kg^-1
S_g = 7.089 kJ Kg^-1 K^-1

At 0.03 Bar (After turbines):

h_f2 = 101 kJ Kg^-1
h_g2 = 2545 kJ Kg^-1
h_fg2 = 2444 kJ Kg^-1

S_f2 = 0.354 kJ Kg^-1 K^-1
S_g2 = 8.576 kJ Kg^-1 K^-1
S_fg = 8.222 kJ Kg^-1 K^-1
-------------------------------------------------------------------------------------------------------------------------
Attempt at working out the thermal efficiency of the cycle:

S_g = (1 - x_g)S_f2 + x_gS_g2

7.089 = (1 - x_g) * 0.354 + x_g 8.576
∴
x_g = (7.089 - 0.345) / (8.576 - 0.345)

x_g = 0.819

Now

h₂ = (1 - x_g)h_f2 + x_gh_g2

h₂ = (1 - 0.819) * 101 + 0.819 * 2545

h₂ = 2103 kJ kg^-1
∴
Specific work (W) = h₁ - h₂

W = 3445 - 2103

W = 1342 kJ kg^-1

And

Q = h₁ - h_f2

Q = 3445 - 101

Q = 3344 kJ kg^-1
∴
Eff (η) = W / Q

η = 1342 / 3344

η = 0.4013 or 40.13% efficient


Am I looking at the right area to solve this?

 

[williamcarter]

Homework Statement

Superheated steam at a pressure of 40 bar and a temperature of 500°C is supplied to the turbine of a Rankine cycle. If the condenser pressure is 0.03 bar.

Find the thermal efficiency of the cycle. (Neglect feed pump work).

I used steam tables found https://www.slideshare.net/SGhallab/steam-tables-fifth-edition-by-rogers-and-mayhew

Homework Equations

S₁ = S_g

S_g = (1 - x_g)S_f2 + x_gS_g2

h₂ = (1 - x_g)h_f2 + x_gh_g2

Specific work (W) = h₁ - h₂

Specific Heat (Q) = h₁ - h_f2

Eff (η) = W / Q

The Attempt at a Solution

From steam tables:

At 40 Bar and 500°C (Before turbines):

h_g = 3445 kJ Kg^-1
S_g = 7.089 kJ Kg^-1 K^-1

At 0.03 Bar (After turbines):

h_f2 = 101 kJ Kg^-1
h_g2 = 2545 kJ Kg^-1
h_fg2 = 2444 kJ Kg^-1

S_f2 = 0.354 kJ Kg^-1 K^-1
S_g2 = 8.576 kJ Kg^-1 K^-1
S_fg = 8.222 kJ Kg^-1 K^-1
-------------------------------------------------------------------------------------------------------------------------
Attempt at working out the thermal efficiency of the cycle:

S_g = (1 - x_g)S_f2 + x_gS_g2

7.089 = (1 - x_g) * 0.354 + x_g 8.576
∴
x_g = (7.089 - 0.345) / (8.576 - 0.345)

x_g = 0.819

Now

h₂ = (1 - x_g)h_f2 + x_gh_g2

h₂ = (1 - 0.819) * 101 + 0.819 * 2545

h₂ = 2103 kJ kg^-1
∴
Specific work (W) = h₁ - h₂

W = 3445 - 2103

W = 1342 kJ kg^-1

And

Q = h₁ - h_f2

Q = 3445 - 101

Q = 3344 kJ kg^-1
∴
Eff (η) = W / Q

η = 1342 / 3344

η = 0.4013 or 40.13% efficient


Am I looking at the right area to solve this?

Hi,

*edit* I've got efficiency=0.40 or 40% so what u did is fine

but I've used those tables:
https://www.physicsforums.com/attachments/thermodynamic_tables_si-pdf.88587/

 

[MCTachyon]

Hi,

I've got efficiency=0.31 or 31%

but I've used those tables:
https://www.physicsforums.com/attachments/thermodynamic_tables_si-pdf.88587/

Looking at the values on the tables you've used they are very similar. Probably shouldn't give a difference in efficiency of 9%

Is there anything in my method you can comment on? Have I used the correct method? Correct equations?

 

[williamcarter]

Looking at the values on the tables you've used they are very similar. Probably shouldn't give a difference in efficiency of 9%

Is there anything in my method you can comment on? Have I used the correct method? Correct equations?

Yeah I did it again and got correct 0.400 or 40% efficiency, My mistake was on 3rd point on superheated I looked by mistake upwards a bit on other bit of table.Everything seems fine

 

[MCTachyon]

Phew, I've struggled with this so glad to hear I've sussed it out.

Thanks.

 

[williamcarter]

Looking at the values on the tables you've used they are very similar. Probably shouldn't give a difference in efficiency of 9%

Is there anything in my method you can comment on? Have I used the correct method? Correct equations?

Let 1= point after turbines when enters condenser

Data for1: Pressure P1=3 KPa(KiloPascal)
h1(entalphy1)=100.98 Kj/Kg
v1(specific volume)=0.001003 m³/kg

So we can use that Wpump(work that goes IN the pump)=v1(P2-P1)=h2-h1 (I)
v1(P2-P1)=0.001003*(4000 KPA-3KPA) =4.008 m^3/kg
from (I) 4.008=h2-h1 =>h2=h1(from above)+4.008 =>h2=104.988991 kj/kg

For state 3

We can say it is superheated because the saturation temp at 4 Mpa is 250.35 degrees celsius and we have 500 Celsius

so h3=3446.0 kj/kg(table)
s3=7.0922 kj/kg*K

s3=s4(isentropic) => 7.0922=0.3543 +8.2222*x4 (use the fact that is 2 phase region)

x4=0.8193=>h4=100.98+0.8193*(2443.9) =>h4=2103.26 kj/kg

efficiency eta= wnet,in/qin=1-qout/qin=((h3-h4)-(h2-h1))/(h3-h2)
=>efficiency = ((3446-2103)-(104.98-100.98))/(3446-104.98)= 0.40

=>efficiency=40%

 

[MCTachyon]

Let 1= point after turbines when enters condenser

Data for1: Pressure P1=3 KPa(KiloPascal)
h1(entalphy1)=100.98 Kj/Kg
v1(specific volume)=0.001003 m³/kg

So we can use that Wpump(work that goes IN the pump)=v1(P2-P1)=h2-h1 (I)
v1(P2-P1)=0.001003*(4000 KPA-3KPA) =4.008 m^3/kg
from (I) 4.008=h2-h1 =>h2=h1(from above)+4.008 =>h2=104.988991 kj/kg

For state 3

We can say it is superheated because the saturation temp at 4 Mpa is 250.35 degrees celsius and we have 500 Celsius

so h3=3446.0 kj/kg(table)
s3=7.0922 kj/kg*K

s3=s4(isentropic) => 7.0922=0.3543 +8.2222*x4 (use the fact that is 2 phase region)

x4=0.8193=>h4=100.98+0.8193*(2443.9) =>h4=2103.26 kj/kg

efficiency eta= wnet,in/qin=1-qout/qin=((h3-h4)-(h2-h1))/(h3-h2)
=>efficiency = ((3446-2103)-(104.98-100.98))/(3446-104.98)= 0.40

=>efficiency=40%

Makes perfect sense that. Seem to get similar numbers at the various stages as well. Can both methods be used in this example?

Or is one preferred?

 

[williamcarter]

Makes perfect sense that. Seem to get similar numbers at the various stages as well. Can both methods be used in this example?

Or is one preferred?

Hi,

Both work well, choose and use the one you are more comfortable with, and don't forget to pay attention in state 3 (you might have superheated or saturated)

Best,
William

 

[MCTachyon]

Thanks.

I will add this method to my notes as well.

Cheers.