Rankine Cycle Efficiency Calculation - Superheated Steam at 40 bar and 500°C

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SUMMARY

The discussion focuses on calculating the thermal efficiency of a Rankine cycle using superheated steam at 40 bar and 500°C, with a condenser pressure of 0.03 bar. The calculated thermal efficiency is approximately 40.13%, derived from specific enthalpy values obtained from steam tables. Participants confirmed the accuracy of their calculations and discussed variations in efficiency results due to different steam table references. The final consensus is that both methods of calculation yield similar efficiency results, affirming the reliability of the approach.

PREREQUISITES
  • Understanding of Rankine cycle thermodynamics
  • Familiarity with steam tables and their application
  • Knowledge of specific enthalpy and entropy calculations
  • Proficiency in thermal efficiency equations
NEXT STEPS
  • Study the use of steam tables for various pressures and temperatures
  • Learn about the impact of condenser pressure on Rankine cycle efficiency
  • Explore the differences between superheated and saturated steam in thermodynamic cycles
  • Investigate alternative methods for calculating thermal efficiency in power cycles
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Mechanical engineers, thermodynamics students, and professionals involved in power generation and thermal system design will benefit from this discussion.

MCTachyon
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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?
 
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MCTachyon said:

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:
williamcarter said:
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?
 
MCTachyon said:
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
 
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Phew, I've struggled with this so glad to hear I've sussed it out.

Thanks.
 
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MCTachyon said:
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:
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williamcarter said:
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?
 
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MCTachyon said:
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)William
 
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Thanks.

I will add this method to my notes as well.

Cheers.
 

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