Refrigerator power vs. dissipation

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SUMMARY

The discussion clarifies the relationship between power consumption and heat dissipation in refrigerators and air conditioners. An 80-watt refrigerator does not equate to the heat it can extract from its contents; instead, it operates on the principle of moving heat, characterized by its Coefficient of Performance (CoP). The CoP for refrigerators typically ranges from 2 to 3, indicating that they can transfer two to three times the amount of heat compared to the energy consumed. For example, a refrigerator with a cooling capacity of 5600 BTU/hr consuming 510 watts achieves a CoP of 3.2.

PREREQUISITES
  • Understanding of Coefficient of Performance (CoP) in thermodynamic systems
  • Basic knowledge of heat transfer principles
  • Familiarity with BTU (British Thermal Unit) and wattage measurements
  • Awareness of refrigeration cycle mechanics
NEXT STEPS
  • Research the Coefficient of Performance (CoP) calculations for various refrigeration systems
  • Explore the principles of heat transfer in thermodynamics
  • Learn about BTU to watt conversion and its implications in appliance efficiency
  • Investigate the specifications and performance metrics of different refrigerator models
USEFUL FOR

Engineers, HVAC professionals, appliance designers, and anyone interested in understanding the efficiency and performance metrics of refrigeration systems.

Pengwuino
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OK someoone on a different forum said his refrigerator is an 80 watt refrigerator (small lil college fridge). As in it takes 80 watts from the wall. Isnt it true that this isn't the same amount of heat that can be taken out of the contents inside?

I remember someone talking about an airconditioner using 2000watts of power while dissipating like 2500 watts... any help would be ... helpful :D
 
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Just as with the air conditioner thread a week or so back, think of a refrigerator as a device which moves heat, rather than producing it.

The amount of power consumed to 'move' the heat from the refrigerator cabinet to the surroundings isn't necessarily equal to the amount of heat removed from the cabinet.

Just think of carrying a block of metal at 50000 degrees across a room. If you're wearing gloves, it doesn't matter that the block is hot. You're just moving heat.
 
Ok good.. thought so... didnt want to give these guys incorrect info.
 
Pengwuino said:
OK someoone on a different forum said his refrigerator is an 80 watt refrigerator (small lil college fridge). As in it takes 80 watts from the wall. Isnt it true that this isn't the same amount of heat that can be taken out of the contents inside?
A refrigerator performance is measured by the co-efficient of performance: Heat removed from the cold reservoir (ie inside) divided by the Work input:

CP_{ref} = \frac{Q_c}{W}

The CP depends on the temperature difference but is typically about 2 or 3 for air conditioners and a little higher for refrigerators - ie. 2 or 3 times as much heat transferred as energy input.

Refrigerators are also rated in terms of 'cooling power' or heat removed / time divided by power input (x BTU/hr cooling/y watts input) but since the units are odd you can't really tell how efficient it is in terms of heat out/input.

AM
 
AFAIK, refrigerator specs don't give cooling capacity, but often times, air conditioner specs give both cooling capacity and electrical usage. http://www.sears.com/sr/javasr/prod...vertical=APPL&pid=04273055000&tab=spe#tablink is an example. It says it has a capacity of 5600 BTU/hr and consumes 510 watts of energy to provide it. There are 3.4 BTU per watt, so that's 1647 watts of cooling. The heat rejected out the back will be the sum of the two. Using A_M's formula, that's a CoP of 3.2.
 
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