A refrigerating system operating on reversed Brayton refrigeration cycle is used for maintaining 250 K. If the temperature at the end of constant pressure cooling is 300 K and rise in the temperature of air in the refrigerator is 50 K, then the net work of compression will be (assume air as working substance with Cp = 1 kJ/kg)
A. 25 kJ/kg
B. 50 kJ/kg
C. 100 kJ/kg
D. 125 kJ/kg
Answer: Option B
Solution(By Examveda Team)
A refrigerating system operating on reversed Brayton refrigeration cycle is used for maintaining 250 K. If the temperature at the end of constant pressure cooling is 300 K and rise in the temperature of air in the refrigerator is 50 K, then the net work of compression will be 50 kJ/kgThis Question Belongs to Mechanical Engineering >> Refrigeration And Air Conditioning
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Comments ( 1 )
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
Here ambient air will be at 350K.
Air after cooling effect = 300K (air temp inc as it act as refrigerant).
Air before cooling effect = 250K.
Cooling effect = (300-250) = 50 KJ/KG.
Total heat rejected to atm= (350-250) = 100KJ/KG.
Hence work done by compressor= thr - ce = 100-50= 50KJ/KG.