Unit 12 Refrigeration and air standard cycles November 30, 2010
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1 Unit Rerigeration and air tandard cycle Noveber 0, 00 Unit Twelve Rerigeration and Air Standard Cycle Mechanical Engineering 70 Therodynaic Larry Caretto Noveber 0, 00 Outline Two oewhat related topic thi week Ue auption or cycle analyi developed in the unit on Rankine cycle Rerigeration cycle luid that exit in both a liuid and a ga during the cycle Air-tandard cycle odel cobution engine gae a air (an ideal ga a the working luid Internal v. external cobution Review Cycle Analyi Baic Baic auption No line loe (output tate o one device i input to the next device Work device are ientropic (w = h Heat traner ha no work and = 0 Exit ro two-phae device i aturated Ue actual data i available Account or dierent a low rate Unit Twelve Goal A a reult o tudying thi unit you hould be able to calculate heat, work and coeicient o perorance or rerigeration cycle undertand the auption or airtandard cycle copute heat, work, and eiciency or otto, brayton, dieel and iilar cycle Rerigeration Cycle Only reuired input i preure (or teperature o evaporator and condener outlet i aturated vapor Ientropic copreor No work and = 0 in condener and evaporator h = 0 or throttling valve 5 Rerigeration Calculation evaporator = = = at (T = T condener = = = at (T < T State i aturated vapor at evaporator State, ga, = and = condener State i aturated liuid at condener State, ixed, h = h and = evaporator ME 70 Therodynaic
2 Unit Rerigeration and air tandard cycle Noveber 0, 00 Rerigeration Cycle or R-a 0 R-a Rerigeration Cycle T > 0, reure (ka, condener Saturation Throttle Evaporaor Teperature (K Saturation 0 evaporator Enthalpy (kj/ 7 50 T evaporator Entropy (kj/-k Rerigeration Calculation Rerigeration Calculation II Given evaporator and condener h ( evaporator and ( evaporator h = h( condener, = h = h = h ( condener L = evaporator = h h w = w copreor = h h CO = L / w 9 Given T evaporator and h = = condener = at ( h = h( condener, = h = h = h ( L = evaporator = h h w = w copreor = h h CO = L / w Do not ue T condneer to ind h. T > 0 R-a Exaple Calculation evaporator = 00 ka; condener = Ma Find: Coeicient o erorance, CO h (00 ka =. kj/ (00 ka =.958 kj/k h = h( = cond = Ma, = = 8.5 kj/ h = h ( Ma = 07. kj/ h = h = 07. kj/ R-a Exaple Anwer w = w cop = h h =.5 kj/ 79. kj/ = 8.09 kj/ L = evap = h h =.5 kj/ 05.9 kj/ = 9.5 kj/ evap h - h CO wcop h - h 9.5 kj CO kj ME 70 Therodynaic
3 Unit Rerigeration and air tandard cycle Noveber 0, 00 evap = 0 ka; cond = 0.9 Ma h ( evaporator ( evaporator I T evaporator i given, evaporator = at at T = T evaporator o that h and h = h( condener, = Note i i 9. given, kj you 0. ut kj ind condener = at (T. h h h CO = h = h ( evap condener h.5 - h h 80. CO9. h kj kj wcop h - h CO =? Air-Standard Cycle Otto cycle i odel or park-ignition (gaoline powered engine Dieel (copreion-ignition cycle Brayton cycle or ga turbine Stirling cycle eicient but proble in ipleentation Suggeted ue with olar collector Other: Erickon, Atkinon, Dual All have iilar analyi Air-Standard Cycle Analyi Ue air propertie a ideal ga with variable or contant heat capacity Model cheical energy releae a heat addition (~,00 Btu/lb or,800 kj/ or Otto cycle engine Heat addition at contant preure, volue or teperature Ientropic work Cloed yte except Brayton Cycle 5 Otto Cycle Deinition Start at initial point with piton at botto o cylinder illed with air at v = v Ientropic copreion to v ratio, CR = v /v Contant volue heat addition at v = v Ientropic expanion to v = v Model exhaut a contant volue heat rejection Given: CR,, T, H v = RT / v = v /CR Ientropic copreion to = (CR k T = v /R Otto Cycle reure (Ma Ratio = 0: Eiciency = 0.% Speciic volue (/ Otto Cycle Continued T = T + H /c v v = v = RT /v 0 v = v = (CRv = /(CR k 8 T = v /R L = c v T T = - L / H = (CR -k 0 reure (Ma Ratio = 0: Eiciency = 0.% Speciic volue (/ Expanion Heat addition7 Heat rejection Expanion Heat addition8 Heat rejection ME 70 Therodynaic
4 Unit Rerigeration and air tandard cycle Noveber 0, 00 Otto Cycle Calculation Given: = 00 ka; T = 00 K; CR = 0; H =,800 kj/ For air, R = 0.87 ka /K v = RT / = (0.87 ka /K (00 K/(00 ka = 0.8 / v = v / CR = 0.08 / Continue with parallel coputation or contant and variable heat capacity 9 Otto Cycle Calculation II = (CR k = (00 (0. =,50 ka T = v /R = (50 (0.08 / (0.87 = 75. K v r (00 K =. v r (T =./0 Find: T = 70.0 K u(t = 5.0 kj/ = RT /v = (0.87 (70.0 / (0.08 =, ka 0 Otto Cycle Calculation III T = T + H /c v = 75. K + (800 kj/ / (0.78 kj/k = 5 K v = v (both calculation = RT /v = (0.87 (5 / (0.08 = 5 50 ka u(t = u(t + H = 5.0 kj/ kj/ =.0 kj/ Find: T = 5 K = RT /v = (0.87 (5 / (0.08 = 707 ka Otto Cycle Calculation IV = /(CR k = (00 /(0. = 7. ka v = v = 0.8 / (both calculation T = v /R = (7. (0.8/(0.87 = 85 K L = c v (T T v r (5 K =.0998 v r (T =.0998*0 Find: T = 85 K u(t = 57. kj/ u(t = u(00 K =. kj/ Otto Cycle Calculation V L = c v (T T = (0.78 kj/k (85 K 00 K =. kj/ = - L / H = (. kj/ / (800 kj/ = 0.% L = u(t u(t = 57. kj/. kj/ =.0 kj/ = = - L / H = (.0 kj/ / (800 kj/ = 5.7% Given: Rankine cycle with data hown Find: Eiciency or (a ideal cycle, (b cycle with,turbine = 85% and,pup = 80% oint i aturated liuid; h = 9. Btu/lb,v = 0.0 t /lb h = h + w = 9. Btu/ +.00 Btu/lb = w 97. Btu/lb Quiz Solution = 000 pia T = 800 o F Stea Generator Turbine = pia up 0.0 t v 000 pia Btu.00 Btu pia pia t lb ME 70 Therodynaic
5 Unit Rerigeration and air tandard cycle Noveber 0, 00 Quiz Solution II h = h(t, = h(800 o F, 000 pia = 88. Btu/lb and = (T, =.5 Btu/lb R h = h( cond, = i in ixed region h h xh h h g g g.5 Btu 0.75 Btu 9. Btu R R 0.8 Btu Btu lb.75 Btu lb R wturb w SG h h w h h 8. Btu Btu.00 Btu 8. Btu 97. Btu lb lb.% 5 Quiz Solution III With 00% h and h do not change Turbine work i 85% o work ound above up work =.00 Btu/lb /80% h = h +.00 Btu/lb /80% wturb w SG w,, T h h, w, h h, 8. Btu Btu.00 Btu Btu 9. Btu.00 Btu 0.80 =.0% Quiz Eleven Solution Given: Rankine cycle with data hown Find: (a Eiciency or ideal cycle oint i aturated liuid; h = 9.78 kj/; v = / h = h + w = 9.78 kj/ kj/ kj/ = 8 Ma T = 500 o C Stea Generator up 5 MW Net Output Turbine = 0 ka w v 8000 ka kj kj 0 ka 7 ka Quiz Eleven Solution II h = h(t, = h(500 o C, 8 Ma = 97.8 kj/; = (8 Ma, 500 o C =.7 kj/ K h = h( cond, = i in ixed region.7 kj ( kj ka K K x (0 ka kj g K h = h + x h g = 9.78 kj/ + (0.8099(9. kj/ = 9. kj/ H = h h = 97.8 kj/ kj/ = 98.0 kj/ L = h h = 9.8 kj/ 9. kj/ = 97.5 kj/ Net work, w = H - L =.5 kj/ kj/ = 0.5 kj/ The eiciency = w / H = (0.5 kj/ / (98.0 kj/ = 9.% 8 Q Quiz Eleven Solution III Find: (b Ma low rate o tea in cycle W 5 MW 000 kj 5.70 w 0.5 kj MW Find: (c Cooling water teperature rie i it a low rate i 000 / h out h in c p T out T in,,,,, c p T QL, Q L T c tea c kj kj 8. K L o 7 p, p, C 9 ME 70 Therodynaic 5
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