Where is Point 1 in the Rankine and Carnot Cycle?

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SUMMARY

The discussion centers on the Rankine and Carnot cycles, specifically the implications of setting the dryness fraction, x1, to 0 instead of 0.364. The participants analyze how this change affects the efficiency and heat load of the condenser, emphasizing that x1 represents the dryness fraction before compression in a fully saturated system. The entropy equations are critical, with the relationship between entropies s1, s2, sg1, and sf1 being established to determine the correct dryness fraction. The entropy of the saturated vapor leaving the boiler is defined as s2 = sg2 = 5.614 kJ/(kg-K).

PREREQUISITES
  • Understanding of thermodynamic cycles, specifically the Rankine and Carnot cycles.
  • Familiarity with the concept of dryness fraction in thermodynamics.
  • Knowledge of entropy and its role in phase changes of fluids.
  • Ability to manipulate and solve thermodynamic equations involving entropies and heat loads.
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  • Learn about the Carnot cycle and its theoretical implications for maximum efficiency.
  • Explore the effects of varying the dryness fraction on system performance in thermodynamic cycles.
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ShaunG123
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Homework Statement
By using data from the plant (given below) and thermodynamic properties tables and/or charts, you need to calculate the overall efficiencies for turbine ‘A’ which is operating on a Rankine cycle

Turbine ‘A’ uses steam as the working fluid and operates on the ideal Rankine cycle (without superheat). The boiler pressure is 10 MPa and the condenser pressure is 5.5 kPa. Assume that the operating cycle is reversible.

My attempt is below however when I change x1 to 0 the numbers don't work.
Relevant Equations
Rankine/Carnot Efficiencies
thermo (2).png
 
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Hi,

What does ##x_1=0\ ## mean according to you ?
 
it would change h1 to 145(2564-145) = 350755 however when worked through the numbers don't come out
 
Are you able to post a schematic of the setup with a list of symbols or do we have to reverse-engineer it ? I like the subject, but doing that would take some time !
 
I am just trying to figure out if i change x1 to 0 rather than the 0.364 how the efficiency would change however whenever i change this to zero the numbers change drastically.
 
x1 is the dryness fraction before compression however the system is fully saturated so this should =0
 
The fluid leaving the boiler is vapor, not liquid.
 
how would this change the math?
 
It would change the dryness fraction entering the condenser, and the heat load of the condenser.
 
  • #10
could you inform me what the equation would become as I am really struggling to understand this.
 
  • #11
Well, the entropy of the saturated vapor leaving the boiler and entering the turbine is ##s_2=s_{g2}##. Since the turbine is operating adiabatically and reversibly, this is also the entropy of the stream leaving the turbine and entering the condenser ##(s_1=s_2)##. This stream is comprised of a mixture of saturated vapor and saturated liquid at entropies ##s_{g1}## and ##s_{f1}##, respectively. In order for the overall stream to have entropy ##s_2##, the dry fraction ##x_1## of the stream must satisfy: $$s_1=s_2=xs_{g1}+(1-x)s_{f1}$$Solving for this dry fraction yields: $$x_1=\frac{s_2-s_{f1}}{s_{g1}-s_{f1}}=\frac{s_{g2}-s_{f1}}{s_{g1}-s_{f1}}$$where ##s_{g2}=5.614\ kJ/(kg-K)##.
 
Last edited:
  • #12
Where is point 1 supposed to be?
 

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