Simple coefficient of performance question

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Homework Help Overview

The discussion revolves around evaluating the coefficient of performance (CoP) for cold room installations, specifically comparing two manufacturers with differing CoP values. The original poster seeks to understand the implications of these values in terms of efficiency and cost-effectiveness.

Discussion Character

  • Exploratory, Assumption checking, Conceptual clarification

Approaches and Questions Raised

  • Participants explore the relationship between CoP and efficiency, questioning whether higher CoP indicates better performance. There is discussion about the calculations based on the Carnot cycle and the implications of the CoP values provided by the manufacturers.

Discussion Status

Participants are actively questioning the definitions and calculations of CoP, with some suggesting that the provided values may not align with theoretical expectations. There is a recognition of potential discrepancies in how CoP is defined by the companies, and some guidance is offered on seeking clarification from the manufacturers.

Contextual Notes

There are concerns regarding the historical definitions of efficiency and how they may affect the CoP values presented. Participants note that the measurement of input power and the recovery of latent heat in modern systems could influence perceived efficiency.

Joon
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Homework Statement


Your company needs to install a cold room which will maintain a temperature of -15 degrees operating in a room where the temperature is 20 degrees. A new member of staff has considered the specifications of two manufacturers and asked you to check their figures. Explain what your decision would be.

Company A : coefficient of performance 8.5
Company B: coefficient of performance 9.7

Homework Equations


CoP = desired output / required output


The Attempt at a Solution


Does higher CoP simply mean that the efficiency of the cold room is higher?
I tried to use the temperature values given:
T(cold) / (T(hot)-T(cold)) = 7.37
 
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Yes, higher COP means higher efficiency (and therefore less costs). However you calculated the COP based on the Carnot Cycle (which should perform at theoretically minimal cost):

$$COP = \frac{Q_0}{W} = \frac{Q_0}{Q_A-Q_0}=\frac{T_0\Delta S_0}{T_A\Delta S_A-T_0\Delta S_0}$$

Since in the Carnot process ##\Delta S_A =\Delta S_0## you obtain your formula

$$\frac{T_0}{T_A-T_0}=7.37$$

Since the companies offer cold rooms with higher COPs, maybe this is kind of a "trick question"...
 
The Carnot cycle based COP means the most efficient, right?
Since the companies offer cold rooms with higher COPs than Carnot COP, should I tell my staff not to work with these companies? (as it doesn't make sense or doesn't fit my company's demand)
 
General answer: higher COPs are better
Specific answer: something seems to be wrong with the COPs "offered" by the companies
 
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I'd ask the companies how they define the COP.

Here in the UK we can buy gas heaters for houses that are 103% efficient because of the strange/hostoric way that 100% was defined.
 
CWatters said:
I'd ask the companies how they define the COP.

Here in the UK we can buy gas heaters for houses that are 103% efficient because of the strange/hostoric way that 100% was defined.

There is also the possibility that water or rooms are heated with the discharged heat to increase the benefit (and subsequently the COP) ...
 
stockzahn said:
There is also the possibility that water or rooms are heated with the discharged heat to increase the benefit (and subsequently the COP) ...

I believe its to do with the way the input power is measured. The efficiency is output power / input power. The output power into the circulating water can be measured easily but the input power depends on the energy density of the fuel and how that's measured. When they originally measured the calorific value of the fuel they ignored the latent heat of steam lost up the exhaust. These days boilers are condensing and recover that making some appear more than 100% efficient.
 

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