Course: UNDERSTANDING THE BASICS OF REFRIGERATION Lesson 1 INTRoDuCTION To REFRIGERATION Lesson 1 : INTRODUCTION TO REFRIGERATION Presented By: Harshit Aggarwal
ABOUT ME Graduated from NIT Nagpur in 2008 . Cleared Indian Engineering Services (IES-UPSC) Exam . Got the offer letter from most of the Maharatna and Navratna Companies Cleared GATE Exanmm Course fee: CONTRIBUTE . Follow me on Unacademy at: https://unacademy.in/user/harshit_aggarwal
REFRIGERATION . It is a process of maintaining lower temperature compare to surroundings. In order to maintain lower temperature continuously, refrigeration system must run on a cycle. . Refrigerant: Refrigerant is a substance used for producing lower temperature. E.g NH3 , Water, Air, R-11,R-12, R-134.
REFRIGERATION EFFECT (RE) . It is the amount of heat which is to be extracted from storage space in order to maintain lower temperature. It is the desired effect of refrigerator. SURROUNDIN Win COB= RE Nin NW In %-RE 2. STORAGE SPACE
SIGNIFICANCE OF COP . COP (Coefficient of Performance): It represents running cost of the system for a given refrigeration capacity . Greater the COP lesser is the work input and hence lower is the running cost.
COP OF REFRIGERATOR For ReFrigerator: Atm COP= Desired Effect-9 Work Input of] BODY A COP= Desired Effect / Work Input = Q2 /w [COP]ref =Q2 / (Q,-Q)
COP OF HEAT PUMP For Hest ump umo. BODY AT 8, G1-81 Aim at COP= Desired Effect / Work Input =Q, /w
UNITS OF REFRIGERATION Refrigeration is expressed in terms of" tonne of refrigeration" . A tonne of refrigeration is defined as amount of refrigeration effect produced by the uniformmelting of one tonne (1000 Kg) ofice from and at 0 C in 24 hours. . I TR = 1000 x 335 kJ in 24 hours 1 000 x 335/ 24 x 60 x 60 = 232.6 KJ/min In actual practice it is equivalent to 210 KJ/ min or 3.5 Kw
Cleared UPSC ESE twice with Rank 63 and 90 in mechanical engg. Got 99 percentile in GATE. Cracked ONGC, BHEL,ISRO, SAIL, GAIL successfully