www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS Vpour ompression refrigertion system Introdution In vpour ompression system, the refrigernts used re mmoni, ron dioxide, freons et. the refrigernts lterntely undergoes ondenstion nd evportion during the yle. When refrigernt enters the evportor it will e in liquid stte nd y soring ltent het it eome vpours. Thus the C.O.P of this system is lwys muh higher tht ir refrigertion systems. Shemti Digrm Anlysis of the yle The vrious proesses re Proess. The vpour refrigernt entering the ompressor is ompressed to high pressure nd temperture in isentropi mnner. Proess. This high pressure nd high temperture vpour then enters ondenser where the temperture of the vpour first drops to sturtion temperture nd susequently the vpour refrigernts ondenses to liquid stte. Proess d. This liquid refrigernt is olleted in the liquid storge tnk nd lter on it is expnded to low pressure nd temperture y pssing it through the throttle vlve. At point d we hve low temperture liquid refrigernt wuitn smll mount of vpour. Proess d. This low temperture liquid then enters the evportor where it sors het from the spe to e ooled nmely the refrigertor nd eome vpour. Re frigertion effet = d But proess - d is throttling proess = d, R. E = work done = RE C. O. P = = work
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS Effet of under ooling the liquid Effet of super heting the vpour Advntges of Vpour ompression refrigertion system over ir refrigertion system Sine the working yle pprohes loser to rnot yle, the C.O.P is quite high. Opertionl ost of vpour ompression system is just ove 1/4th of ir refrigertion system. Sine the het removed onsists of the ltent het of vpour, the mount of liquid irulted is less nd s result the size of the evportor is smller. Any desired temperture of the evportor n e hieved just y djusting the throttle vlve. Disdvntges of Vpour ompression refrigertion system over ir refrigertion system Initil investment is high Prevention of lekge of refrigernt is mjor prolem Refrigernt A refrigernt is fluid in refrigerting system tht y its evporting tkes the het of the ooling oils nd gives up het y ondensing the ondenser.
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS Identifying refrigernts y numers The present prtie in the refrigertion industry is to identify refrigernts y numers. The identifition system of numering hs een stndrdized y the Amerin soiety of heting, refrigerting nd ir onditioning engineers (ASRAE), some refrigernts in ommon use re Refrigertion R-11 R-12 R-22 R-717 R114(R40) R-500 R502 R-764 Nme nd Chemil Formul Trihloromonofluoromethne CCl 3 F Dihlorodifluoromethne CCl 2 F 2 Monohlorodifluoromethne CClF 2 Ammoni N 3 Azeotropi mixture of 73.8% (R-22) nd 26.2% R-152 Azeotropi mixture of 48.8% (R-22) nd 51.2% R-115 Sulphur Dioxide SO2 Properties of Refrigernts Toxiity: It oviously desirle tht the refrigernt hve little effet on people Inflmmility: Although refrigernts re entirely seled from the tmosphere, leks re ound to develop. If the refrigernt is inflmmle nd the system is loted where ignition of the refrigernt my our, gret hzrd is involved. Boiling Point. An idel refrigernt must hve low oiling temperture t tmospheri pressure Freezing Point An idel refrigernt must hve very low freezing point euse the refrigernt should not freeze t low evportor tempertures. Evportor nd ondenser pressure. In order to void the lekge of the tmosphere ir nd lso to enle the detetion of the lekge of the refrigernt, oth the Evportor nd ondenser pressure should e slightly ove the tmosphere pressure. Chemil Stility An idel refrigernt must not deompose under operting onditions.. Ltent het of Evportion.
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS The Ltent het of Evportion must e very high so tht minimum mount of refrigernt will omplish the desired result; in other words, it inreses the refrigertion effet Speifi Volume The Speifi Volume of the refrigernt must e low. The lower speifi volume of the refrigernt t the ompressor redues the size of the ompressor. Speifi het of liquid vpour. A good refrigernt must hve low speifi het when it is in liquid stte nd high speifi het when it is vporized Visosity The visosity of the refrigernt t oth the liquid nd vpour stte must e very low s improved the het trnsfer nd redues the pumping pressure.. Corrosiveness. A good refrigernt should e non-orrosive to prevent the orrosion of the metlli prts of the refrigertor. Coeffiient of performne The oeffiient of performne of refrigernt must e high so tht the energy spent in refrigertion will e less. Odour. A good refrigernt must e odourless, otherwise some foodstuff suh s met, utter, et. loses their tste Lekge A good refrigernt must e suh tht ny lekge n e deteted y simple test. Oil solvent properties. A good refrigernt must e not ret with the luriting oil used in the refrigertor for luriting the prts of the ompressor. Cost The ost of the refrigernt is the mjor importnt, it will esily ville nd low ost.
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS Prolem1: 20 tons of ie is produed from wter t 20ºC to ie t -6ºC in dy of 24 hours, when the temperture rnge in the ompressor is from -15ºC to 25ºC. The ondition of the vpour is dry t the end of ompression. Assuming reltive C.O.P s 80%, lulte the power required to drive the ompressor. Tke Cpie=2.1kJ/kg, Ltent het of ie=335k/kg Vpour Liquid Temp Enthlpy hf Entropy Sf Enthlpy hg Entropy Sg ºC 25 100.04 0.347 1319.2 4.4852-15 -54.55-2.1338 1304.99 5.0585 To find the ondition of vpour t point ''. Entropy t = Entropy t s f - 2.1338 + x x + x s = 0.92 f g = s + x work = g 5.0585 ( 2.1338) = 4.4852 h g = 1196.22kJ f g Re frigertion effet R.E = = 54.55 + 0.92[1304.99 ( 54.55)] = 100.04kJ = 1319.2kJ = 1196.22-100.04 = 1096.18kJ/kg = 1319.2 1196.22 = 122.98kJ
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS RE 1096.18 C. O. P = = = 8.913 work 122.98 Reltive C.O.P = 0.8 Atul C.O.P = 0.8x8.913 = 7.13 et extrted/kg of ie = C pw + Ltent het + C (20 0) pie [0 ( 6)] = 4.187 x 20 + 335 + 2.1x 6 = 431.34 20x1000 Mss of ie produed/se = = 0.231kg / s 24x3600 Atul het extrted/se = 431.34x0.231 = 99.84kJ/s Atul het extrted/se Atul C.O.P = Atul work/se 99.84 Atul work/se = 7.13 Power = 14kW Prolem2: A vpour ompression refrigertor working with Freon-12 hs its temperture rnge -10ºC nd 30ºC. The Vpour enters the ompressor dry nd under ooled y 5ºC in the ondenser. For pity of 15 TOR, find: ()C.O.P () mss of freon () Power required. Cp for vpour = 0.56kJ/kgK Cp for liquid = 1.003kJ/kgK Solution: Refrigertion pity=15 TOR From tles the properties of Freon 12 re Entropy Enthlpy hf fg hg Sf Sg Temp ºC 30 64.59 135.03 199.62 0.24 0.6853-10 26.87 156.31 183.19 0.108 0.7019
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS To find the ondition of vpour t point ' '. Entropy t = Entropy t T sg' + C pυ Ln = sg T ' T 0.6853+ 0.56Ln = 0.7019 303 T = 312.15K h' = 100.62 + 0.56(312.15-303) = 204.74kK/kg = 183.19kJ g' f' + C = g pυ C PL ( T ( T T ) = 64.59-1.003(30-25) = 59.575kJ/kg R.E = = 183.19 59.575 = 123.61kJ work = = 204.74 183.19 = 21.55kJ ' T ) ' RE 123.61 C. O. P = = = 5.73 work 21.55 Re frigertion pity = 15tons = 15x3.5 = 52.5kJ/kg Re f. pity Mss of feron = RE 52.5 = = 0.424kJ / s 123.61 Power required = work/kg xmss of freon/s = 21.55x0.424 = 9.152kW Prolem3: A food storge loker requires refrigertion system of 12 tons pity t n evportor temperture of -8ºC nd ondenser temperture of 30ºC. The refrigernt freon-12 is su ooled to 25ºC efore entering the expnsion vlve nd the vpour is superheted to -2ºC efore entering the ompressor. The ompression of the refrigernt is reversile diti. A doule tion ompressor with stroke equl to 1.5 times the ore is to e used operting t 900 rpm. Determine COP Theoretil piston displement/min Mss of refrigernt to e irulted/min Theoretil ore nd stroke of the ompressor.
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS Tke liquid speifi het of refrigernt s 1.23 kj/kg K nd the speifi het of vpour refrigernt is 0.732 kj/kg K. Solution: From tles the properties of Freon 12 re Entropy Enthlpy Temp ºC hf hg Sf Sg 30 64.59 199.62 0.24 0.6853-8 25.75 184.2 0.1142 0.7002 C pυ = 0.732kJ K CPL = 1.235kJ K Entropy t = entropy t T T Sg' + C pυ Ln = Sg' + C pυ Ln T ' T ' T 0.6853+ 0.732 Ln = 0.7002 + 0.733 Ln 303 T = 317. 22K g' g' + C pυ pυ ( T T ' = 184.2 + 0.732(271 265) = 188.59kJ/kg C ( T T ) g' ' = 199.62 + 0.732(318.22 303) = 210.02kJ/kg C PL ( T T ) ) 271 265 = 64.59-1.235(303-298) = 58.41kJ/kg R.E = = 188.59 58.41 = 130.181kJ work = = 210.02 188.59 = 21.43kJ ' RE 130.18 C. O. P = = = 6.07 work 21.43
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS Re f. pity Mss of refrigernt = RE 12x3.5 = = 0.322kg / s 130.18 = 0.322x60 = 19.35kg/min From tles t -8 C, Vg' = 0.0441995m PV T g' ' PV = T 3 T xvg' 271 V = = x0.0441995 = 0.0452 T ' 265 Theoretil piston displement V = mss xv = 19.35x0.0452 = 0.87462m / min 2π 2 V = d LN ( L = 1.5d) 4 2 2πxd x1.5d 0.87462 = x900 4 = 0.0203m 3 d = 0.0738m = 7.38m L = 1.5d = 1.5x7.38 = 11.08m
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS Prolem4: A vpour ompression refrigertion system of 5kW ooling pity opertes etween -10ºC nd 30ºC. The enthlpy of refrigernt vpour fter ompression is 370kJ/kg. Find the COP, refrigerting effet, mss flow rte of the refrigernt nd the ompressor power. The extrt of the refrigernt property tle is given elow Temp Pressure Vf Vg hf hg Sf Sg C r m3/kg kj/kg kj/kgk -10 226 0.7x10-3 0.08 190 345 0.95 1.5 30 7.5 0.77x10-3 0.02 220 220 1.10 1.45 Solution: Assume the ondition efore ompression s dry sturted vpour f g' = 220kJ = 345kJ = 370kJ ( given) R.E = = 345 220 = 125kJ work = = 370 345 = 25kJ RE 125 C. O. P = = = 5 work 25 Refrigertion pity = 5kW or kj/s Re f. pity Mss of refrigernt = RE 5 = = 0.04kg / s 125 Compressor work = work. kg x mss of - 25x0.04 = 1kW refrigernt
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS Prolem5: A vpour ompression refrigertor uses methyl hloride nd works in the pressure rng of 1.19 r nd 5.67 r. At the eginning of ompression, the refrigernt is 0.96 dry nd t the end of isentropi ompression, its temperture is 55ºC. The refrigernt liquid leving the ondenser is sturted. If the mss flow of refrigernt is 1.8kg/min, Determine COP The rise in temperture of ooling wter if the wter flow rte is16 kg/min. the properties of methyl hloride is given elow Entropy Enthlpy hf fg hg Sf Sg Temp ºC Pressure r 30 1.19 64.59 135.03 199.62 0.24 0.6853-10 5.67 26.87 156.31 183.19 0.108 0.7019 Tke speifi het of super het methyl hloride s 0.75kJ/kg K Solution x T = 55 C = 0.96 f + x ( h fg ) f + x ( h h = 430.1+ 0.96(455.2-30.1) = 438.196kJ g' f R.E = work = C p = 100.5kJ ( T = 438.196 100.5 = 337.669kJ T ) ' = 476.5 + 0.75(55-25) = 499kJ/kg g = 499 438.196 = 60.8kJ RE 337.669 C. O. P = = = 5.55 work 60.8 et lost y the vpour in the ondenser et gin y ooling wter f )
www.ookspr.om VTU NEWS VTU NOTES QUESTION PAPERS FORUMS RESULTS m C r p ( T T ') + mrh fg' = m w C pυ 1.8 x 0.75(55-25) + 1.8(476.5-100.5) = 16 x4.187 x temperture rise Temperture rise = 10.7 C x temperture rise Vpour sorption refrigertion system Generl The sorption refrigertion system is het-operted unit whih used refrigernt tht is lterntely sored nd lierted y the sorent. Simple Asorption system The minimum numer of primry units essentil in n sorption system inlude n evportion, sorer, genertor nd ondenser. An expnsion vlve, pressure reduing vlve, nd pump re used in onventionl two-fluid yle, ut the pump n e eliminted y dding gseous third fluid. A simple sorption yle is shown in figure This yle differes from vpour ompression yle y the sustitution of n sorer, genertor, pumps nd reduing vlve for the ompressor. Vrious omintios of fluids my e used, ut tht of mmoni, strong solution tht ontins out s muh mmoni s possile; wek solution ontins onsiderly less mmoni. The wek solution ontining very little mmoni is spryed or otherwise exposed in the sorer nd sors mmoni oming from the evportion. Asorption of the mmoni lowers the pressure in the sorer, whih in turn drws more mmoni vpour from the evportor. Usully some forms of ooling is employed in the sorer to remove the het of ondenstion nd the het of solution evolve there. The strong solution is then pumped into genertor, whih is t higher pressure nd is where het is pplied the het vpourises the mmoni driving it out of solution nd into the ondenser, where it is liquefied. The liquid mmoni psses on to the reeiver if seprte one is used, or through the expnsion vlve nd into the evportor. The wek solution left in the genertor fter the mmoni hs een drive off flows through the reduing vlve k to the sorer
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