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Refrigeration 001

The document contains formulas and examples related to refrigeration cycles. It provides formulas for heat added and rejected, compressor work, coefficient of performance for refrigeration and heat pumps, and the relationship between the two COP values. It then shows worked examples applying the formulas to calculate heat rejection, COP and compressor power requirements.

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Elisif DeFair
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15 views15 pages

Refrigeration 001

The document contains formulas and examples related to refrigeration cycles. It provides formulas for heat added and rejected, compressor work, coefficient of performance for refrigeration and heat pumps, and the relationship between the two COP values. It then shows worked examples applying the formulas to calculate heat rejection, COP and compressor power requirements.

Uploaded by

Elisif DeFair
Copyright
© © All Rights Reserved
We take content rights seriously. If you suspect this is your content, claim it here.
Available Formats
Download as PPTX, PDF, TXT or read online on Scribd
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Formulas:

  

a.) Heat Added in the Evaporator


QA=T1 (S1-S4) ; T1=T4=TL

b.)Heat Rejected in the Condenser

QR=T2 (S2-S3) ; T2=T3=TH


c.) Compressor Work

From:QR=QA+W

W=QR-QA=T2(S2-S1)-T1(S1-S4)

Since; S2-S3=S1-S4

W=(T2-T1)(S1-S4)
 
d.) COP of Refrigeration

COPR= ====
Where;
  
e.)COP of Heat Pump or Performance Factor

COPH====

COPH==
  
f.) Relation of COPC and COPH

From;
COPH=
Since;
=

COPH===COPR+1
Examples:

A carnot refrigeration cycle has a coefficient


of performance of 5…The power input to the
compressor is 5 Kw..Compute the rate of heat
rejection from the refrigerant to the
environment?
 
Solution:

Given: Required:
COPR=5 QR=?
W=5Kw
Formula:
COPR= QR=QA+W
Substitute; Substitute;
5= QR=25+5,KW
QA=25KW QR=30KW
  
Example:

A reversed Carnot refrigerator is


operating under the temperature limits -10
and 20..Find the COP?
 
Solution:

Given:
TL=-10 +273=263K
TH=20 +273=293K

Formula:
COPR===8.77
 
Example:

A building has to be maintained at 18 at


all times…When the temperature outside the
building drops to -6 ,the building losses heat
at a rate of 120,000KJ/hr…Compute the least
power necessary to drive the heat pumps?
 
Solution:
Given:
QR=120,000kJ/Hr=120000kJ/3600sec
TH=18+273=291K
TL=-6+273=267K
Formula:
COPH====97/8
QA=30.5842KW
W=QR-QA=2.7491KW,

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