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

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Discussion Overview

The discussion revolves around calculating the thermal efficiency of a Rankine cycle using superheated steam at a pressure of 40 bar and a temperature of 500°C, with a condenser pressure of 0.03 bar. Participants explore the application of steam tables and various equations related to thermodynamics to derive the efficiency of the cycle.

Discussion Character

  • Homework-related
  • Mathematical reasoning
  • Debate/contested

Main Points Raised

  • One participant presents a calculation for thermal efficiency, arriving at approximately 40.13% using specific enthalpy values from steam tables.
  • Another participant claims to have calculated an efficiency of 31%, questioning the differences in results based on the steam tables used.
  • Some participants note the similarity in steam table values and express concern over a 9% difference in efficiency results.
  • A later reply suggests that the method used by one participant was correct, attributing earlier discrepancies to a misreading of the steam tables.
  • Additional calculations are presented, including a method involving specific volume and pump work, leading to a consistent efficiency of 40%.
  • Participants discuss the validity of different methods for calculating efficiency and express a preference for the one they are more comfortable with.

Areas of Agreement / Disagreement

There is no consensus on the thermal efficiency, with participants reporting different values (31% and 40.13%). The discussion reflects multiple competing views and methods for calculation, with some participants affirming their results while others question the accuracy of their approaches.

Contextual Notes

Participants rely on steam tables, which may have slight variations in values, leading to different efficiency calculations. The discussion highlights the importance of careful reading and interpretation of data from these tables.

Who May Find This Useful

Students and professionals interested in thermodynamics, particularly those studying Rankine cycles and efficiency calculations in steam systems.

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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