Efficiency of a steam power plant

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SUMMARY

The efficiency of a steam power plant is determined by the product of the efficiencies of its components rather than their sum. In this discussion, the maximum efficiency was calculated using the formula η = (1 - (T_res/T_Sun)^4)(1 - (T_sink/T_res)), resulting in an overall efficiency of 59%. This value is less than the individual efficiencies of the processes, which were noted as η_1 = 99% and η_2 = 60%. The importance of understanding the relationship between component efficiencies is emphasized, particularly in tandem systems.

PREREQUISITES
  • Understanding of thermodynamic principles, specifically the Carnot efficiency.
  • Familiarity with temperature scales and conversions (e.g., Kelvin).
  • Knowledge of efficiency calculations in thermodynamic systems.
  • Basic mathematical skills for manipulating equations and understanding products of efficiencies.
NEXT STEPS
  • Study the Carnot efficiency and its implications for thermal systems.
  • Learn about the impact of temperature reservoirs on efficiency calculations.
  • Research the concept of combined efficiencies in multi-stage processes.
  • Explore advanced thermodynamic cycles and their efficiencies.
USEFUL FOR

Engineers, thermodynamics students, and professionals involved in energy systems and power plant design will benefit from this discussion, particularly those focusing on efficiency optimization in steam power plants.

TheBigDig
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Homework Statement
Assess the thermodynamic maximum efficiency for a steam solar thermal power plant operating at 750K steam temperature. Assume ideal concentration and negligible optical and insulation losses. Assume that the low temperature reservoir is at 300K and the sun is at 5600K. Calculate the thermodynamic maximum efficiency of the total system for a steam engine of 1/2 and 1/3 of the Carnot efficiency
Relevant Equations
[tex]\eta = \frac{T_1-T_2}{T_1}[/tex]
My inital assumption was to just take T1 = 5600 and T2= 300K, find the maximum efficiency and then divide it by two and three but I don't believe this question to be that simple. I'm confused as to where the 750K fits in as I thought no matter what occurred in between the heat reservoir and heat sink, the efficiency would only depend on T1 and T2.
 
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I think you want to consider that the Sun makes the steam and the steam runs the machinery of the plant. So you have two efficiencies in tandem to consider.
 
Ah okay, so I would just take the sum of each efficiency for the total and then divide by 2 and 3?
 
TheBigDig said:
Ah okay, so I would just take the sum of each efficiency for the total ...
Nope. If process A is 60% efficient and process B is 50 % efficient the sum would be ... 110%? How do you think the overall efficiency works for two efficiencies in tandem? Should it be more than either one, in between the two or less than either one? Give the matter some thought.
 
kuruman said:
How do you think the overall efficiency works for two efficiencies in tandem? Should it be more than either one, in between the two or less than either one? Give the matter some thought.
I'm thinking it should be somewhere in between the two since one process is more efficient than the other. I've done a bit of further reading and found that total efficiencies are given by the product of component efficiencies. This gives me
\frac{5600-750}{5600}\times\frac{750-300}{750} \approx 52\%
So then by this calculation, \frac{\eta}{2} \approx 26\% \quad \frac{\eta}{3} \approx 17\%
I hope my reasoning is correct.

Source:https://www.e-education.psu.edu/egee102/node/1944
 
TheBigDig said:
I'm thinking it should be somewhere in between the two since one process is more efficient than the other.
Think again. Both efficiencies are less than unity. The combined efficiency is the product ##\eta_{comb.}=\eta_1 \times \eta_2.## Say ##\eta_2## is the smaller of the two efficiencies. Can ##\eta_{comb.}## be greater than ##\eta_2##?
 
Hey,
I was able to find some further information on the problem and I believe this to be the correct solution
\eta = \bigg(1-\big(\frac{T_{res}}{T_{Sun}}\big)^4\bigg)\bigg(1-\big(\frac{T_{sink}}{T_{res}}\big)\bigg) = 59\%.
This result is less than the individual efficiencies of each system
\eta_1 = 99\% \quad \eta_2 = 60\%
 

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