Thermodynamic/Carnot efficiency

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SUMMARY

The discussion focuses on calculating the maximum power output and thermal cycle efficiency of a heat engine utilizing wastewater from a food processing plant. The wastewater temperature is 35°C, and the average ocean temperature is 10°C. Using the Carnot efficiency formula, the thermal efficiency is determined to be 8%. The conversion of 1,000 gallons per minute to mass flow rate yields 63.09 kg/s, which is essential for calculating the total heat (Q) and subsequently the work (W) and power (P) produced by the engine.

PREREQUISITES
  • Understanding of thermodynamics principles, specifically the Carnot cycle.
  • Familiarity with heat transfer calculations, including Q = mcΔT.
  • Knowledge of unit conversions, particularly between gallons and kilograms.
  • Basic proficiency in calculating power from work and time (P = W/t).
NEXT STEPS
  • Study the Carnot efficiency and its implications for real-world heat engines.
  • Learn about heat transfer methods in industrial applications, focusing on wastewater energy recovery.
  • Explore advanced thermodynamic cycles beyond the Carnot cycle for improved efficiency.
  • Investigate the impact of varying flow rates on the performance of heat engines.
USEFUL FOR

Engineers, thermodynamics students, and professionals in energy recovery systems who are interested in optimizing heat engine performance using waste thermal energy.

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Homework Statement



A novel heat engine is being considered for producing work from the thermal energy contained in the wastewater discharge from a large food processing plant. On an average day, 35 degree celsius wastewater leaves the plant at and is discharged into the ocean at 1,000 gallons PER minute. if the average annual ocean temperature is 10 degree celsius at that location, what is the maximum power that could be produced by this heat engine? What would be the thermal cycle efficiency of your maximum power engine? You can assume that the heat capacity of water at constant pressure is contant at 4,200 J/Kg K.


Homework Equations



delta T = Tf - Ti
Q = mc delta T
Carnot n=1-(Tc/Th)
W= total heat times Carnot n
P=W/t

The Attempt at a Solution



1000 Gallons per minute = 63.09 Kg per sec

T1 = 10 +273 = 283K
T2 = 35 +273 = 308K
Delta T = 25 K
Carnot η = 1- 283K/308K = 0.08 = 8%
How can I calculate the maximum power with the rate flow and without the time? I think i am missing an equation.
 
Last edited:
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You can choose a convenient time. Since you calculated the rate in kg/s, you have t and m. You can then determine Q (Total Heat), since you have C (Heat Capacity) as a given, and you calculated m/t and you have T. Once you have Q, you can calculate W and then P, using the same t that you chose earlier.
 

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