ENGINEERING.com has updated it's forum.  To post a question please visit the new Ask@ Forum.   

With a database of over 10000 questions the library will remain available for an extended period.
Q&A


SHUVRA		 1 kW/ton = 3.516/COP . How to derive it? View All
In Refrigeration Course, I have found this equation. But I can't derive this. You see, how did I approach :
COP=(Refrigeration Effect) / (Work Input)
COP is dimensionless. So, I can write:
COP=(1 kW of Refrigeration Effect) / (1 kW of Work Input)
So, (1 kW of Work Input) / (1 kW of Refrigeration Effect) =1/COP
or, (3.516 kW of Work Input) / (3.516 kW of Refrigeration Effect) =1/COP
or, (3.516 * 1 kW of Work Input) / ( 1 Ton of Refrigeration Effect) = 1/COP
[3.516 kW = 1 Ton of Refrigeration Effect]
or, (3.516 * 1 kW) / ( 1 Ton) = 1/COP
which implies: 1 kW/ton = 1/(COP * 3.516)
What is the error in my calculation? If anybody knows, please help me. It is driving me crazy. Thanks.
14 years ago - 9 months left to answer. - 1 response - Report Abuse
Respond to question
    0      [lnkReport]        0       0       
Share |
  Responses

Niel
 Actually based on your equations

COP = 3.516 kW/ton of Cooling / 1.0 kW/ton of work

This is based on the definition of the cooling CoP
COP =|deltaQ| / |deltaW|

deltaQ = heat change in the reservoir of interest
deltaW = energy consumed by the heat pump.

For more information about how to derive CoP go to [1] Wikipedia.

Niel

14 years ago

Source: [1] http://en.wikipedia.org/wiki/Coefficient_of_performance


  0     0         

ENGINEERING.com does not provide engineering advice. The Ask@ service is a forum for members to exchange ideas relating to the world of engineering. We caution users not to accept any responses that they receive without further validation, and not to rely on any engineering advice that they may get from other members of the Ask@ forum. ENGINEERING.com specifically disclaims any obligation to validate or verify any information posted within the Ask@ service. ENGINEERING.com encourages users to seek the services of a professional engineer for any engineering advice they may require.