“unit -1] 1 Fundamentals ¢ Applications of Refrigeration and Air Conditioning Syllabus ns : Domestic Refrigerator, Domestic Air Coy eoolers, Commercial Refrigeration - Dairy, Cold sovage Contents 1.1 Introduction . .. © 42 Reverse Carnot Cycle . 4.3. Block Diagram of Heat Engine, Refrigerator and Heat Pump. 14 Modified Reversed Carnot Cycle / Bell Coleman Cycle............ 1.5 Applications of Refrigeration and Air Conditioning .........-.. 4.6 University Questions with Answers. 62... .....cceceeeeeeee ncitioners, Automotive Air Conditioners, Evaporative coolers, water Ice plant, Commercial Air Conditioning- Multiplex, Hospitals. ES ey Loe 1-13 Scanned with CamScannerScanned with CamScanner1-3 __ Fundamentals & Applications of Refrigeration and Air Conditioning expansion of air is takes place isentropically this Process pressure and temperature decreases from p; to p4 and Ty to Ty respectively. 35 specific volume increases from Vy. to Vy. Isothermal expansion process his process air is expanded isothermally (1, = 1). ocess pressure decreases from p4 to p and temperature remains constant (Ty = 4). volume ets increased from Vy to V, sorbed by air during this provess can be calculated as, Qa = Ty (8-84) oF = Ta (Sz ~S3) of T (Sz - $3). W = Heat rejected — Heat absorbed. = T, (S: - $3) ~ Ti (S2 ~ $3) = (Ty -T)) (Sp -S3) Heat absorbed COP = Work done 4 AM aT. Go. = Sa) COP TW) G - 5) cor - -2 has been made by using reversed caro cyle because of following reasons. i ‘high speed while isothermal process is extremely slow speed. ‘ is not practicable, ee Scanned with CamScannerScanned with CamScanneree buco owas te 10 ar environment at 5%. The on nos joss from the room 10 the environment is estimated to be 45000, Hiir and the power input 10 the compressor ig 4.5 kit, Determine 1) The rte Of heat absorbed from the environment in ki/h iy The COP of the heat pump sip The maximum rate of heat supply 10 the room for the given power Input and fy) The second - law efficiency of the cycle. SE ‘Ans. : Given : 1-2 = 28K, T,=S°C=278K, W=45 KW, Q = 45,000 ki/hr = 12.5 kw 1) QU_ it) COP it) (Quy iv) Ty Q = 28.800 kur swAns, Stop 3: Calculate rato of heat supply Camot COP is given as, = 258. %-T ~ 208-278 (COP)exnat = 149 But it can also be given as, Step 4 : Calculate second law efficiency Second law efficiency of cycle is given as, = (CoP) _ 27778 TH COPlama 149 ty = 0.1864 = 18.64 % sAns, Ex. 1.2 A Carnot refrigerator requires 1.2 kW per ton Of refrigeration 10 maintain a temperature of ~ 35°C, Determine. 9) COP of the refrigerator ii) The temperature at which the heat is rejected iti) COP, if the cycle is used as a hheat pump 5 IN ‘Ans. : Given data : Q, = 1 TR = 3.52 kW, W= 12 kW To find : (COP) ie pump =~ W352 (COP)retgerioe™ Go Fig. 14 (@ LESS THAN PHOTOCOPY PRICE, Scanned with CamScannerScanned with CamScanner2 reversible heat pump is used 10 maintain of OT in a refrigerator when it rejects she surroundings at 27 °C. Determine COP of the and work input required if the heat removal rate If the required input to rum the pump is ‘by reversible engine which receives heat at rejects heat 10 the atmosphere, determine of the system. data : T, = 27 +273 = 300K, #273 = 273 K on Ans, 5x 10° = 2472.80 W... Ans. “10.10 ‘absorbs heat from low temperature reservoir when work input is 250 kl. 1) COP of refiigerator ii) Q, i) (Cop)yyp in, Step 1; Calculate the (COP), ® _ 650 (COP), = 22. 650 _ Ong 5h = wm Ans, Stop - 2: Calculate heat transfer (0, ) Q = W+0) = 650 + 250 = 900 ta o Ans. Step - 3: Calculate COP of heat engine (COP) = a =36 w- Ans. Step - 4: Calculate efficiency of engine (7,) Ww _ 250 an 0278 = 278% oA Ne = 97 9p 0278 = 278% Ans. Ex.17 The efficiency of a Carnot engine rejecting heat t0 a cooling pond at 28°C is 30 %. If the cooling Pond receives 1050 ki/min. What is the power developed by the cycle in kW. Also find the temperature of the source. Sol. : Given data : 1, = 28°C = 28 + 273 = 301 K Ncamot = 0.3, Q; = 1050 ki/min To find i) Power developed W ii) T, ‘Step - 1: Calculate the source temperature T; Refer Fig. 1.7. Scanned with CamScannerReftigeration and Air Conditioning 18 * T = 0K Ans, Stop-2: Calculate the heat Source and power ‘Camot efficieney is also given by, rejecting from heat a Q = 1500 ky/min, Power developed by the cycle is, Power = W= Qi ~Q) = 1500~ 1050 B W = 450ki/min=7.s 0/00 W = 75 kW Ans, Si) ae ee Ex.18 4 heat pump is used ‘10 maintain an auditorium hall at 25°C, when the atmospheric temperature is 10°C. The heat leakage from the hall is 1500 kimin. Calculate the power required to rim the actual heat pump, if the COP of the actual heat pump is 30% of the COP of the Carnot heat pump working. between the same temperature limits, Sol. : Given data : T, = 25 °C = 25 + 273 = 298, Ty = 10 °C = 10 + 273 = 283 K, Qy = 1500 ki/min = 25’kI/sec = 25 kW, (COP)aguat =03 (COP)camoe « To find : Power Step-1: Calculate the power required to run the actual heat pump Refer Fig. 1.8. 9.867 298-283 (COP)camot np= Sa (COP) actual hp = 03 x (COP) camot he (9 recemicat PUBLICATIONS”. Anup tnt knoe Fig, 18 ae = 03 x 19,867 = 5.96 Oo (COM peateane = Q= Gy = W 25 eo 5.96 W W = 4.195 kW w= Ans, ‘Ex. 19 A reversible heat engine operates on Cai -eyele between source and sink temperatures of 225 °C 25 °C. If the heat engine receives 40 kW from the sour find the net workdone, heat rejected to sink and of the engine, Sol. : Given data : T, = 225°C = 225 + 273 = 498 Ty =25 °C = 25 + 273 = 298 K, Q = 40 kW To.find : W, Qo and ncamot» J Stop - 1: Calculate the efficiency of heat engine Refer Fig. 19. ith CamScanner1-9. Fundamentals fe Applications of Refigeration and Air Cond 179.1 = 450-9, Q = 2709 kg Ans, Step -3: Calculate the (COP) yp and (COP) ey, ‘When the engine operates like heat pump, Qa 450 = (COP) pre ew: 2 450-2705 (COP), = 2515 Fine ‘When it operates like reftigerator, (COP): = (COP)ret, = 1515 Ex 111 An invertor claims to have developed a | refrigeration unit which maintains at ~ 5 %C in the refrigerator which is in a room of swrounding temperature 28 °C, It has COP of 9.5. Check whether his claim is right or not ? Sol. : Given data : T =~ 5 C= ~5 +273 = 268K, ‘Ty = 28 °C = 28 + 273 = 301 K, (COP)iny, = 9.5 To find : Check whether inventor's claim is right or not, Stop - 4: Calculate the (COP) ze, and check it Refer Fig. 1.10. : 2 ca Scanned with CamScannerRetrigeration and Air Conditioning It means for given temperature range inventor's claim is invalid, At Ex. 112 4 fish freezing plant of 100 tons capacity 1s to be maintained at - 40 °C for which the outside atmospheric temperature is 30 °C. The actial COP of the refrigeration system is 10 % of the theoretical Carnot Pump working between the same temperature limis. Caleulate the power required to run the plant Take, 1 1on of refrigeration = 3.5 kW Sol: Given data : T, = 30 °C = 30+ 273 = 303 K, T, =~ 40°C =~ 40 +273 = 233 K, (COP), = 0.1 (COP)p, Qi = 100 tons = 100% 3.5=350 kW To find : Power (W) Step-1: Calculate the power required to run the plant Refer Fig. al. Fig. 111 COP of Camot or theoretical heat pump is given by, (CoP); = s (CoP)y = (COP), = 0.1 COP)p = (Given) ey (COP), = 0.1% 433=0.433 er Q Now, (COP), = —2t_ - Or A a-G Ww aes : 2 0433 = : e W = 808.31 kW Ae required for a grinding mil ‘at 420 °C ts avatlable for 30°C | The power input is 30 Mdimin, A heat source Ex. 113 energy and the surrounding 1 at 1s efficient as a Carnot engin. limits. Calculer supplying the the actual engine is 25 % working between the same temperatire the energy supplied by the source per s€¢. Sol, : Given data : Power input to grinding mill 30x10? W _ 509 pw, W=30 Mi/min = =—Z>— 5, = 500 KM, 20 + 273 = 693 K, To find: Q) ‘Step - 1: Calculate the energy supplied by the source Refer Fig. 1.12. a Heat w oooh To grinding ml a, Fig. 1.12 Efficiency of Camot engine is, Q-G _T-h Nem = =@ =a Se = 693-303 9 coy Nicamot oe 0,563 = 56.3% Bat, Ta = 025 Neue = 025 x 0,563 ‘ tia = 0.141 ‘Now, = WE GT Pas a Q a 0.141 = 500 a “ Q = 3546.1 ki/sec.=3546,1 kW .. Ans. Scanned with CamScannerfo have invented B optratig! Betweer=23°°C and 1 RW electrical. power and gives ng effect in one how. Comment on Fundamentals & Applications of Refrigeration and AirCenlitoning 4-2) 035 = ara (i) Dividing equation (i) and (ii), L167) -1.167T) = Ty Ty +25 0.167 1, = 0.167% + 25 T= +497 (ii) Substituting this value in equation (i), noel (+197) T, = 493K on Ans, and Ty = Ty + 149.7= 349.34149.7 1 = 49K Ex. 116 A reversible heat engine operates between two reservoirs at temperature of 600 °C and 40 °C, The engine drives a reversible refrigerator, which operates between at temperature 40 and -20°C. The heat transfer 10 the heat engine is 2000 kJ and the net work ‘output of the combined engine refrigerator plant is 360 ki. Sol, : Given data : T, = 600 °C = 600 + 273 = 873 K, = 404279313 K, Scanned with CamScannerRetngeraton ac Air Cooning W = Ww. +W2 @ T; But, =1-2 "Tr 7 * ar Now, 11 0.6415 ‘Substituting in equation (i), 1283 = 360+ W2 S Wr = 923 kt = Ans. But | We know that, (COP)et = fe. (COP) et. = signe But, (COP ep = Wr 1:12 Ait Conditionin, Qs = 3895.06 kd and Wa = Q4-95 923 = Q4 - 3895.06 Qy = 4818.06 ks +. Total heat transfer to the reservoir =Q, +Q4 = 717+ 4818.06 = 5535.06 kd os Ans, Step -2: Calculate heat transfer to the refrigerato, and reservoir at 40 % of their maximum values ma 2566 Now, Ta = : ae a 0.2566 = 1-53, Q, = 1486.8 kd w=» Ans. w d =i ani Mano, : 02566 = 2 W= 5132 But, we Similarly(COP), = 4X 4,22 = 1.688 Qs Q& N so 2 1.688 = 3 Br as We He 1532 Q = 258645 But, Wa = Q4-Q3 153.2 = Qy-258.6 oe Q, = 4118 +, Total heat transfer to the reservoir = Qz +Q, = 148684 411.8= 1898.6 kJ. Ans. 3.2 | Tonne of Refrigeration + Rating or refrigeration machine is given by standard unit of refrigeration called as tonne of refrigeration (TR) ap reorca. “PUBLICATIONS”. Aa up trust fer krowedgs (@ LESS THAN PHOTOCOPY PRICE Scanned with CamScannerYTR = 211 ka/min = 3.517 kp/see, d Carnot Cycle / Boll Coleman Cycle ined with CamScannerScanned with CamScannerGT) = Ty) eyele, ‘Heat rejected ~ heat absorbed Gl = T) «or, ~ 14) nance of cycle, Heat absorbed ? £y My - T) 1) = T4) oh) 7 =T,) we (Ld) ADM Seed Saatendin rd sepa roms a takes place as por the law PY" = Constant, *# Work done by compressor during process 1-2 We Pi (p2 Ys - 1 %) qn (RT - Bh) ‘+ Work done by expander during process 3-4, W, = 2 (p Va ~ Pa Ya)- Th ~ RT) + Not work done can be find out as, wet wal -11)-( te] +» (5) Heat absorbed. ‘Work done pom > (UE Ma)iar |S ag atk -T)- (ts -T.)] | Weaow, Cp -Cy=R cee ) COP = Scanned with CamScanner1 | Advantages of Bell - Coleman Cycle + It uses air as @ working medium which is freely available, + As the air is non flammable this system is safe, « The weight of air refrigeration system per ton of refrigeration is low compared with other refrigeration system 1.4.2 | Disadvantages of Bell - Coleman Cycle + COP of system is low as compare to VCC. ‘+ For UTR large volume of air is required which results in large size of compressor and expander. 1.4.3 | Numericals on Bell - Coleman Cycle Ex. 1.17 In refrigeration plant working on Bell - coleman cycle, air is compressed to 1 bar from 7 bar. Its init temperature is 10 °C. After compression air is cooled upto 20 °C in cooler before expanding back to pressure of 1 b Determine theoretical COP of the plant and net refrigerating effect, Take Cp = 1.005 ki/kg K and Cy = 0.718 ki/kg K. ‘Sol: Given data : Py = 1 bar,p2 = 7 bar Ty = 10°C = 10+ 273 = 283K Ty = 20°C = 204 273 = 293K Cp = 1.005 ki/kg K Cp = 0.718 Ki/kg K Fig. 1.16 precio PuLICATIONS™ An tiara @ LESS THAN PHOTOCOPY PRICE Scanned with CamScanner= 168.03 Scanned with CamScannerScanned with CamScanner1-19 __ Fundamentals & Applications of Refrigeration and Air Conditioning, = icsWe = gay (P2¥2 ~ pavn)- 4 (Pa¥s - ava) . poi ln- T1) - (1 -T.)] R= C, ~Cy =1-070= 03 kike K Woet = = «03 [ (477-11 ~ 283) - (303 - 179-72)) = 85.63 kykg K Heat extracted _ Cp (Ti - Ta) _ 1 (283 ~ 179-72) i COR Work done oS 2). ese | = 120 a or works between pressure linit of 1 bar and 5 bar. The temperature of air entering the Scanned with CamScannerS 2 & 5 3 € S & Ff = 3 B 2 — 5 3anand Alr Conditioning To find out net work done, (43712 258) ~ £25 (303 - 219.)] 4745 tujhe cop = RE. _ 68:39 aa NNT Win, ides 2 : 68:39 x 25 Capac of refrigerating system = REX ms (sss) Dw §.14 tons. Ex 12 5 TR refrigerating machine based on Bell coleman cycle operates between 5 bar and 20 bar The alr temperature afier heat rejection 10 the surrounding is 37 °C and air temperature at compressor inlét is 7 °C. The no and Ty = 48 % respectively determine i) COP, it) Power per TR, il) Mass of circulation of refrigerant in kg/hr, ty) carnot COP Jor sane entry temperature to compressor and expander. (Take C, =\ kl/kg K , Cy = 0.718 ki/hg K). Sol: Given data: Reftigerating capacity = 5 TR = 5 x 3-52 = 17.6 kW py=Sber pp = 20bar Ty = 37°C = 37+ 273 = 310K T= 7°C = 7+ 273 = 280 KScanned with CamScannerScanned with CamScannerScanned with CamScannerRefgeraton and Ait Conditioning + Reffgerior ## a common appliance used in. domestic housing, Yendor shop, ofc, laboratories, chemists etc «The capacity of refrigerator is expressed in lites, The feffigerators are available in standard capacities of 65, 90, 165, 210, 240, 300, 380, 420, 540 titres eto, + Reffigerant used in refrigerator is R-12 or R-22, + Domestic Tefrigerator works on vapour compression system. «The refrigerators may be single door or double door. Domestic reftigerator consists of following major components. Compressor : + In domestic reftigerator hermatically sealed compressor is used. {s Jocated at the bottom andbehind the compartment of refrigerator. + Vapour reftigerstors are sucked into the compressor for ‘compression purpose, “2 ~ + The power of the compressor vary according to the size of the compressor (ic. 75 W, 92 W, 125 W, 180 W). Evaporator : Evaporator is mounted on top protion of the refrigerator. cr jeved in evaporator. + Evaporator generally made in box type construction + Jn evaporator, evaporation will takes place ie. liquid reffigerants can be convered into vapour reftigerants ‘+ In evaporaior negative temperature is maintained to. produce the ice, ‘Condenser : Condenser which is used in domestic refrigerator is air ‘cooled condenser. 25 _ Fundamental & Applications of Refrigeration and Air Condoning_ + Capillary tube reduce the pressure of reftigerant by keeping enthalpy constant rll also provide constant flow of refrigerant to the ‘evaporator. 41.5.2 | Domestic Air Conditioner Domestic air conditioner is normally installed in a window or wall opening of the room and these are known as window type air conditioner. + It works on vapour compression cycle. ( «The domestic. air conditioner consists of reffigration system. Control system, electrical protection system, ait circulation system, ventilation and exhaust system, *Reffigeration system consists of hermatically sealed compressor, forced air cooled -conderser, cooling, col, capillary tube and refrigerant drier. +The refrigerant used for domestic air conditioner is R-12_ or R22, | [fs The condenser is a coil which is made up_of copper tubing along with aluminium fins. These fins are used to increase the heat transfer rate +A propellar fan pravides the air to cool the reftigerants + The room_air conditioner is installed in such a way that evaporator faces the room, Evaporator Condenser 'e" fan, Condenser Evaporator 9 rer Capillary tube Scanned with CamScannerScanned with CamScanner1-27 _ Fundamentals & Applications of Refrigeration and Air Conditioning, From tank ‘through pads Motor support beams Water pump Pump motor Water supply with ‘ Fig, 1.23 2 ey (BEE ttective ‘cooling. depends upon the relative humidity-of it, spesd of the air and the time of contact between air and water. ‘These evaporative coolers are used in summer, season. ter Coolers +The purpose of water cooling is to make the water available at constant temperature to quench the thirst, “+ There are two types of water cooler. 4) Storage type water cooler. }) Instantaneous water cooler. 8) Storage type water cooler “in storage type water cooler the machine has to run for long time to bring down the temperature of water which ‘AS present in storage tank, 7 Chee he temperature i ine cycle wil be switched of automatically. tein femperature will rise slowly even when water is drawn from cooler because equal amount of fresh water enters into the tank. ‘way there will not be frequent @ LESS THAN PHOTOCOPY PRICE Scanned with CamScannerFundamental pplestionso Regen er 128 : ernie ‘ y ‘Strainer, Capillary Fig. 1.24 ») Instantaneous type water cooler ‘kt In this type of water cooler. evaporator consists of two separate cylindrical wound coils made up of c tube. (See Fig. 125 on next page) ag Hr ses the to tubes, in ist el evaporating reigerant sallow to low and in second coil water whichis be cooled is allow to flow. + These 1wo tbe are bonded together by soldering. ¥ ‘+ In case of instantaneous water cooler it is very important to-adjust the. rate of flow of water to the cooler to ‘match the capacity with higher rate of flow. 1.5.6 | Commercial Refrigeration Dairy (Milk Processing) Milk is a most perishable food. + Iegets spoiled, if itis not maintained at sufficient temperature because of growth of bacteria, | +o decrease the growth of bacteria itis required to cool the milk at 4 °C to 5 °C but still all the bacterias can not bé killed. «+ To eliminate bacteria processing on milk is required that process is called as milk pasteurization, FE reonencns Pusucarions™. An up trust for krone @ LESS THAN PHOTOCOPY PRICEae e A_A Evaporator # Wary tube i Reciever 1 Wtar tap | Capillary | Cone iy Fig, 1.25 sHnithis process, the milk is heated upto, the temperature of 62.°C to 65 °C and holding it at that temperature for about half an_hour. ‘sThere afer fo liminate bacterial grows and preservation of milk, the milk is cooled up to 4 °C to 5 °C He THis process of heating the milk and immediate cooling to control the bacterial growth is called as pasteurization. _*TThis process is done in batch type process. *To control the fat content of the milk itis desired to chum the milk +The removed fat is processed as buiter and stored it at 4°C to 5 °C. 7 The chease is also another product of milk which is stored at about 4 °C. 457] pes Storage "+ Cold storage is an application of air conditioning, 4 13m the cold storage certain goods like food stuf, fut, dairy products can be stored with well defined temperature range and relative humi i cold stor sir is cooled by passing it over a cooling col. "por the storagé of different products different requirements are required. |) Plexee, a. cold storage temperature end humidity conditions are maintained depends upon the type of the ‘ _ Product stored in it : @LESS THAN PHOTOCOPY PRICE 4 Scanned with CamScannerScanned with CamScannerAit Conditioning » Mutiplex on ait Conditioning eystem is in theaters, mE Fequires higher capacity dewing 10 Wight shor, ee Jon outdoor nit Yequired ie 0.15 cx m, on veiling nt 10.5 cu.m, 308 design conditions are 26 °C DBT and 55% presence of large number of people in the increases the cooling load to comparatively short time oveupancy % Subamantate te Aypienta ch Weliaration wed hit Coming, Eaplain reverse carmen cycle Leplain difference between heat engine, heat pump and reftiqerion 3, Explain Bell-coleman cycle 4. Explain mile processing - commercial application. 5. Keplatn with heat sketch We plant 6 Explate alr conditioning system usad tn multiple. University Questions with Answers Dec. - 2011 Derive expression for COP of refrigerator and | heat pump working on carot cycle. | (Sections 1.2 and 1.3) a May - 2013 State different applications of refrigerction and explain any one. (Section 1.5) Bl) Dec. - 2013 | Q3 Prove COP of Bell-coleman cycle by cop = =. Section 1.4) 3 - Te Q4 Define 1 TR, (Section 1.3.2) Explain reverse Bryaton cycle. (Section 1.4) [8] ise nk Scanned with CamScannerRefrigerants Syllabus Classification of refrigerants, Desi depletion and global warming, ODP. GP & ignation of refrigerants, Desirable properties of refrigerants, environmental issues, Ozone LCCP, selection of environment friendly refrigerants, seconc refrigerants, MARIN Zt pon decraps eigenen sen oe Contents 24. Introduction. . 22 Classification of Refrigerants. . 23 Designation of Refrigerants . . 24 Desirable Properties of Refrigerants........ EEE 25 Environmental Issues. 21 Green House Effect 28 Global Warming Potential (GWP) 2.9 Life Cycle Climate Performance (LCCP) ......2-10 L May-16 J 2.10 TEWL. Mao) eC 2.11 Selection of Environment Friendly Refrigerants 2-11 22 Anti-Freeze Solution . 2.13 Zeotropie and Azeotropic Refrigerants 2-12 2.14 Montreal and Kyoto Protocol... . 2.15 Recovery, Recycling and Reclaiming of Refrigerant (RRR) .. 2.16 Recharging of Refrigerant .. 2.17 University Questions with Answers... « 2-14 Scanned with CamScannerScanned with CamScannery 23 Refrigerants Hydrocarbons-Refrigerants ‘+ Most of the hydrocarbon reffigerants are successfully used in industrial and commereial applications. ‘= They passes satisfactory thermodynamic properties but are highly flamable and explosive. Table 2.3 : Hydrocarbon Refrigerant Chemical Name Ethane Propane = Butane oe 1s0-buane ‘Trichloro ethylene a Dichloro ethylene Ethylene Propylene Tnorganic - Compounds + The in-organic refrigerants were exclusively used before the introduction of halo-carbon refrigerants. ‘* These refrigerants are still in use due to their inherent thermodynamic and physical properties. Table 2.4 : Inorganic Rettig Unsaturated Organic Compounds : These are the hydro-carbons with ethylene and propylene Table 2.5 : Unsaturated organic compounds =‘Refrigeration and Air Conditioning at 2.22 | Secondary Refrigerants EUAN # The_reffigerants which are_first_gooled_by_primary refrigerants and then used for cooling purpose are known as E ‘Secondary reftigerants are used in ice plant and many big installations to have advantages like- 4) Different temperature in different zone can be ‘maintained by varying circulation rate of secondary reftigerans. ii) Use of secondary refrigerant is advisable in order to keep toxic reftigerant away from load places. “Commonly used secondary refrigerants are i) Water ii) Brine + When required temperature is above the freezing temperature of water then water is used as }air conditioning plants, chilled water is used for g and dehumidifying air, + When required temperature to be maintained freezing point of water then brine solution : ised as secondary reftigerant, Brine is a solution containing salt in dissolved - condition in water. “a Salt used is sodium chloride (NaCI) or_calium pSieaal of brine is ~ 20 °C. —— Designation of Refrigerants ‘+The reftigerants are intemationally designated as R followed by certain number such as R-11, R-2 etc. The ‘numbers assigned 19 hydro-carbon and halocarbon reffigerant have special meaning * Example - R-12 should be designed as R012. * Three numerals of this code follow the formula + First number of carbon atoms minus 1, © Second number of hydrogen atoms plus 1 and © Finally the number of fluorine atoms in that order. © Any unfilled positions are normally occupied by chlorine atoms. nearer Ref «Thus chemical formula is CyHyCly F, in nt p+o=2m+2 «The chemical formula for any compound derive hydrocarbon is denoted by R - (m ~ 1) (1+ 1) (» “aelalabae he aa Where, ‘m = Number of carbon atoms sefei n = Number of hydrogen atoms p = Number of fluorine atoms ‘© = Number of chlorine atoms Za | Examples to Study Designation ot Refrigerants Example 4: R-22 or RUZ) which means number carbon atoms is 'I! 2-.n = 1 which means numbe; ‘hydrogen atoms is '!’ 2 it means number of fluorine 4, We know, n+P+0 2m +2 1+2+0=(@«1+2) 34024 fs] O-l Number of chlorine atoms = | We can write chemical formula for R-22 as, Cy HyCloFy > CHCIR, Example 2: R-12 or R-012 m-1=0 m= 1 which means ‘number of carbon atoms is n+1=1 2 n=0 which means number of hydrogen atom \ P= 2 It means number of fluorine atoms '2! We know, n+P+0 = Qm+2) 0+240=@x1+2) @ LESS THAN PHOTOCOPY RIC: Scanned with CamScannerts am Refrigerant m=1=1 sm =2 it means number of carbon atoms '2' n+1=1 — sn=0 it means number of hydrogen atoms ‘Te means number of fluorine atoms "4" pa ae Scanned with CamScannerRefrigeration and Air Conditionin [ean] ‘Thermodynamic si ° Properties: + It the boiling temperature of refiizerant is high’at atmospheric pressure, compressor should be operated st ‘vacuum, - Table 2.6 ; Bolling temperature of refrigerant Freezing Temperature : ‘The freezing temperature 0 ‘« Freezing temperature of most of the refrigerants are below ~ refrigerant should be well below the operating evaporator emperatur. a aC. Table 2.7 : Freezing temperature of refrigerants Evaporator and Condenser Pressure : «Both the evaporator (low side) and condenser pressure (high side) should be positive (ie. above atmospheric) and it should be as near to the atmospheric pressure as possible. = The positive i event of air and moisture into the refrigera system bs ‘SA small difference between suction pressure and discharge pressure OF low pressure ratio (Pe {i} (Ratio “condenser to ressure)is desirable for iiency.. : Table 2.8 : Evaporator and condenser pressure Critical Temperature and Pressure : ‘eIt is temperature above which there will not be any phase change. “+The rica! temperature of the reftigerent should be viell above the condensing temperature for C10+02 ‘Im this way ozone (3 )is depleted to O» gas. 2.6.1 | Effect of Ozone Depletion + UV radiation causes skin cancer. ‘« Due to UV-radiation eyes may get damaged. 4 Tt increases global warming, « Lower yield of the crop. 4 It leads to increase in UY-radiations at ground level. ‘Ozone Depletion Potential (ODP) 26.2 |_ «The ozone depletion potential (ODP) is the relative amount of degradation 10 it_ean_cause with trichlorofluromethane (R-11_ or CFC-11) which is being fixed at an ODP of 1.0. + ODP of given substance is defined as ratio of global loss of ozone due to given substance to global toss of ozme ‘due fo CFC-L1_of the same-mass.. + ODP can be estimated from molecular structure of given Scanned with CamScannerSeen A Conitoning < pee ra Refriye faz] Green House Effect * Global warming increases the temperature which ¢ + The term green house effect indicates physical property the melting of ice which rises the sea level of earth atmosphere. ‘* Global warming has increased ocean level by more ; > Sun can emit the energy in the form of short wavelength. * The earth can absorb the heat from the sun, + Some of the heat which is absorbed by earth from the ‘sun is reflected back to the space, + Because of this carth’s temperature does not rise above the certain level, * The film of gas envelopes the earth's atmosphere, * But there are certaim gases known as green house gases traps some reflected heat, preventing it to reilect back to space which results in increase in average temperature of earth, which leads global warming. ‘Thus higher the concentration of green house gases in the atmosphere, it leads to increase the temperature of ‘earth surface. Carbon dioxide (CO, } methane (CH, } nitrous oxide are ‘the main green house gases. * All the hologenated refrigerants (CFC, HFC) are green house gases. [28] ctobat warming Potential (GWP) *+ The reffigeration and air conditioning system contributes in global warming. 1) While servicing of the equipment the reftigerant ‘may get leak and spread into atmosphere. 2) Indirectly by the way of consumption of energy for its operation as carbon dioxide is relead to generate the energy. +The ODP and GWP. are the common parameters that describes the potential of refrigerant on global warming. + To minimize the global warming following care should be taken - i) Refrigeration system should be leak proof. ii) While handling the refrigerant proper care should be taken, so that there will not be a leakage of refrigerant, 1 mm a year in past century, But since 1960 ocean level is increasing 1.3 annually Life Cycle Climate Performance (LCCP) + Life cycle climate performance is an evaluation me by which heating air-condition reftigeration (HVAC and R) systems can be evaluated ventilation, * Life cycle climate performance is calculated as the of direct and indirect emissions generated over lifetime of the system. ‘+ There are two types of emissions 1) direct emission and 2) indirect emissions Direct emissions include all effects from the releas reffigerant into the atmosphere during the lifetime 0! system. * Annual leakage and leakage during the disposal oi ‘unit comes under direct emission. +The indirect emissions include emissions from manufacturing process, energy consumption and dis; ‘of the system. + To reduce the environmental impacts of air condito ‘and refrigeration systems the heating, ventilation ‘conditioning and refrigeration (HVAC and R) industr spending a large amount of effort. * Several indices are developed to measure this impact It includes Total Equivalent Warming Impact (TI Life Cycle Analysis (LCA) and Life Cycle Clin Performance (LCCP). + LCCP is an important tool in analyzing reftigers systems and needs a globally harmonized approach Scanned with CamScannerRefrigerants | | ‘Selection of Environment Friendly Refrigerants Selection of reigerant for a particular application is based ‘on following requirements : i) Thermodynamic and thermophysical properties ii) Environmental and safety properties ii) Beonomics Table 214 Scanned with CamScannerScanned with CamScannerFirst machanical tefrigeration R744 carbon dioxide 1950 ‘17 ammonia RT44 carbon dioxide | 4980 2000 2008 } in 2001, and are referred to as the "Marrakesh Accords". aa started in 2008 and ended in cae involves the removal of Bie peng fil eign io Fig. 2.4 + RECYCLING of teftigerant involves processing of reffigerants ro reduce contaminants. + Then these reftigerants can be reused. ‘+ Result of contamination is failure of system. * Contaminants includes oil, moisture, acid, particulars etc + RECLAMATION of reftigerant involves purifying used feftigerant to meet industry product specification ‘The reclamation is used to identify bad or mixed reftigerant which could results in damage of the “equipment, The reclamation process includes filtering, separation, distillation, dilution of recovered refrigerant. ‘* The reclamation facilities are designed to minimize the emission, Recharging of Refrigerant “caf is a process in which fresh refiigerant is added in the ‘existing refrigeration or air conditioning system. je acta sk fein he eg in eng Scanned with CamScannerScanned with CamScanner