Thermodynamics: Carnot efficiency

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SUMMARY

The discussion focuses on calculating the power consumption of a heat pump operating at 25% of Carnot efficiency for heating and cooling applications. In winter, the Carnot efficiency is calculated as n = 0.25(1 - 0/30) = 0.25, while in summer, it is n = 0.25(1 - 15/35) = 0.1429. The participants explore how to derive the heat transfer (Q) for both air source and ground source scenarios, emphasizing the relationship between work (W) and heat transfer (Q) rates.

PREREQUISITES
  • Understanding of Carnot efficiency and its formula
  • Basic knowledge of heat pumps and their operational principles
  • Familiarity with thermodynamic cycles and heat transfer concepts
  • Ability to perform calculations involving sinusoidal temperature variations
NEXT STEPS
  • Research the principles of heat pump efficiency and performance metrics
  • Learn about the mathematical modeling of sinusoidal temperature variations
  • Explore the implications of ground source vs. air source heat pumps
  • Study the thermodynamic equations governing heat transfer in HVAC systems
USEFUL FOR

Students studying thermodynamics, HVAC engineers, and professionals involved in energy efficiency optimization for heating and cooling systems.

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Homework Statement



In one location, the average daily temperature varies sinusoidally between 35C in summer and 0C in winter. The ground temperature below 2 m underground is the annual average temperature. A heat pump / air conditioner unit must deliver air to a house at 30C in winter and 15 C in summer. If the unit operates at 25% of Carnot efficiency, how much power is consumed per each kW of heating or cooling provided if the unit operates air source and ground source in winter and summer?

Homework Equations



How do i get Q for each case?

The Attempt at a Solution



Winter: efficiency n = 0.25(1 - 0/30) = 0.25
Summer: n = 0.25(1 - 15/35) = 0.1429

n = Q_hot/W_cycle, so we need Q_hot, where both W and Q are rates of work and heat transfer
 
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