In actual air-conditioning applications for R-12 and R-22, and operating at a condenser temperature of 40°C and an evaporator temperature of 5°C, the heat rejection factor is about
A. 1
B. 1.12
C. 2.15
D. 5.12
Answer: Option B
Solution(By Examveda Team)
T1 = 5°C or 273 + 5 = 278KT2 = 40°C or 273 + 40 = 313K
$$\eqalign{ & {\text{C}}{\text{.O}}{\text{.P}}{\text{.}} = \frac{{{\text{T1}}}}{{{\text{T2}} - {\text{T1}}}} \cr & = \frac{{278}}{{313 - 278}} \cr & = \frac{{278}}{{35}} \cr & {\text{C}}{\text{.O}}{\text{.P}}{\text{.}} = 7.942 \cr & {\text{Heat rejection factor (HRF)}} \cr & = 1 + \frac{1}{{{\text{C}}{\text{.O}}{\text{.P}}{\text{.}}}} \cr & = 1 + \frac{1}{{7.942}} \cr & = \frac{{8.942}}{{7.942}} \cr & = 1.12 \cr} $$
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Comments ( 3 )
Nusselt number (NN) is given by
A. $${{\text{N}}_{\text{N}}} = \frac{{{\text{h}}l}}{{\text{k}}}$$
B. $${{\text{N}}_{\text{N}}} = \frac{{\mu {{\text{c}}_{\text{p}}}}}{{\text{k}}}$$
C. $${{\text{N}}_{\text{N}}} = \frac{{\rho {\text{V}}l}}{\mu }$$
D. $${{\text{N}}_{\text{N}}} = \frac{{{{\text{V}}^2}}}{{{\text{t}}{{\text{c}}_{\text{p}}}}}$$
In case of sensible heating of air, the coil efficiency is given by (where B.P.F. = Bypass factor)
A. B.P.F. - 1
B. 1 - B.P.F.
C. $$\frac{1}{{{\text{B}}{\text{.P}}{\text{.F}}{\text{.}}}}$$
D. 1 + B.P.F.
The undesirable property of a refrigerant is
A. Non-toxic
B. Non-flammable
C. Non-explosive
D. High boiling point
The desirable property of a refrigerant is
A. Low boiling point
B. High critical temperature
C. High latent heat of vaporisation
D. All of these
Brife
T1= 5 °C or 273+5=278 K
T2= 40 °C or 273+40= 313K
COP = T1/T2-T1
= 278/313-278
= 278/35
COP = 7.942
Heat rejection factor (HRF) = 1+1/COP
=1+1/7.942
= 8.942/7.942
= 1.12 ( Ans.)
i cant understand ?