64% n.th + COP4 = more heat out of fuel than its calorific value?

  • Context: Undergrad 
  • Thread starter Thread starter cmb
  • Start date Start date
  • Tags Tags
    Fuel Heat Value
Click For Summary
SUMMARY

The discussion centers on the efficiency of Combined Cycle Gas Turbines (CCGT) operating at 64% efficiency and the implications of using a COP4 heat pump to generate thermal power. The calculation suggests that this combination could yield 256% heat output relative to the fuel's calorific value. However, this figure does not violate thermodynamic laws, as it reflects the movement of energy rather than its creation. The key takeaway is that the heat pump's Coefficient of Performance (COP) exceeds 100% due to its ability to extract environmental heat, emphasizing the importance of temperature differentials in energy transfer.

PREREQUISITES
  • Understanding of Combined Cycle Gas Turbine (CCGT) systems
  • Familiarity with Coefficient of Performance (COP) in heat pumps
  • Basic principles of thermodynamics and energy transfer
  • Knowledge of heat pump technology and efficiency metrics
NEXT STEPS
  • Research the thermodynamic principles governing heat pumps and their COP
  • Explore the efficiency metrics of CCGT systems and their operational parameters
  • Investigate alternative methods for direct heat extraction from combustion processes
  • Examine case studies on the integration of heat pumps with renewable energy sources
USEFUL FOR

Energy engineers, HVAC professionals, and anyone involved in optimizing energy systems and heat recovery technologies.

cmb
Messages
1,128
Reaction score
128
TL;DR
If a CCGT generates electrical power at 64% efficiency, and a heat pump generates 4 times the thermal power to electrical power, is that more heat energy out than the fuel has?
Something that has been on my mind a while.

If a CCGT generates electrical power at 64% efficiency, and a COP4 heat pump, powered by that generator, generates 4 times the thermal power to electrical power, wouldn't that mean we can get 256% heat out of the fuel's calorific value?

Maybe that is OK, but seems non-intuitive.

Where does the extra heat come from, or what am I not understanding?

I presume the answer is that because we're going from 'very hot' heat to colder heat, we can suck up some of the environmental heat? If so, is there not a more direct way to use the combustion temperature from a burning fuel to directly suck in extra heat from the environment, missing out the electrical generation bit?
 
Science news on Phys.org
Short summary of @Llewelyn 's point: COP is like efficiency, but it is not efficiency. It is always way over 100%. That's not a violation of the laws of thermo. The energy isn't being created, just moved, and there are hard restrictions on the available temperature deltas.
 
  • Like
Likes   Reactions: Lnewqban

Similar threads

  • · Replies 6 ·
Replies
6
Views
4K
Replies
7
Views
3K
  • · Replies 4 ·
Replies
4
Views
3K
Replies
9
Views
8K
Replies
1
Views
2K
  • · Replies 4 ·
Replies
4
Views
2K
  • · Replies 5 ·
Replies
5
Views
4K
  • · Replies 2 ·
Replies
2
Views
4K
  • · Replies 7 ·
Replies
7
Views
2K
  • · Replies 4 ·
Replies
4
Views
5K