Falling Behind in Studies? Join Our Performance Improvement Batch. Enroll For Free
One of our academic counsellors will contact you within 1 working day.
Click to Chat
1800-1023-196
+91-120-4616500
CART 0
Use Coupon: CART20 and get 20% off on all online Study Material
Welcome User
OR
LOGIN
Complete Your Registration (Step 2 of 2 )
It is proposed to build refrigeration plant for a cold storage to be maintained at -3°C. The ambient temperature is 27°C. If 5×10 6 kJ/h of energy is to be continuously removed from the cold storage, the minimum power required to run the refrigerator will be 14.3Kw 75.3 kW 154.3 kW 245.3 kW
I hope you know the relation(Q1/T1)=(Q2/T2)where Q1 =heat rejected in hot reservoir=5×106kJT1 =temperature of hot reservoir=(27+273)K=300K Q2 =heat absorbed from cold reservoirT2=temperature of cold reservoir=(-3+273)=300KUssing the relation and putting the valuesQ2 =5.55×106 kJWork input=Q1-Q2=0.55 ×106kJPower=0.55kJ/h=(0.55×106)/(60×60) kJ/s=154.3kJ/s=154.3 kW
Post Question
Dear , Preparing for entrance exams? Register yourself for the free demo class from askiitians.
points won -