# 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 = Sg

Sg = (1 - xg)Sf2 + xgSg2

h2 = (1 - xg)hf2 + xghg2

Specific work (W) = h1 - h2

Specific Heat (Q) = h1 - hf2

Eff (η) = W / Q

## The Attempt at a Solution

From steam tables:

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

hg = 3445 kJ Kg-1
Sg = 7.089 kJ Kg-1 K-1

At 0.03 Bar (After turbines):

hf2 = 101 kJ Kg-1
hg2 = 2545 kJ Kg-1
hfg2 = 2444 kJ Kg-1

Sf2 = 0.354 kJ Kg-1 K-1
Sg2 = 8.576 kJ Kg-1 K-1
Sfg = 8.222 kJ Kg-1 K-1
-------------------------------------------------------------------------------------------------------------------------
Attempt at working out the thermal efficiency of the cycle:

Sg = (1 - xg)Sf2 + xgSg2

7.089 = (1 - xg) * 0.354 + xg 8.576

xg = (7.089 - 0.345) / (8.576 - 0.345)

xg = 0.819

Now

h2 = (1 - xg)hf2 + xghg2

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 - hf2

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 = Sg

Sg = (1 - xg)Sf2 + xgSg2

h2 = (1 - xg)hf2 + xghg2

Specific work (W) = h1 - h2

Specific Heat (Q) = h1 - hf2

Eff (η) = W / Q

## The Attempt at a Solution

From steam tables:

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

hg = 3445 kJ Kg-1
Sg = 7.089 kJ Kg-1 K-1

At 0.03 Bar (After turbines):

hf2 = 101 kJ Kg-1
hg2 = 2545 kJ Kg-1
hfg2 = 2444 kJ Kg-1

Sf2 = 0.354 kJ Kg-1 K-1
Sg2 = 8.576 kJ Kg-1 K-1
Sfg = 8.222 kJ Kg-1 K-1
-------------------------------------------------------------------------------------------------------------------------
Attempt at working out the thermal efficiency of the cycle:

Sg = (1 - xg)Sf2 + xgSg2

7.089 = (1 - xg) * 0.354 + xg 8.576

xg = (7.089 - 0.345) / (8.576 - 0.345)

xg = 0.819

Now

h2 = (1 - xg)hf2 + xghg2

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 - hf2

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/

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

Thanks.

williamcarter
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%

Last edited:
MCTachyon
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
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
MCTachyon
Thanks.

I will add this method to my notes as well.

Cheers.