ES-7A Thermodynamics HW 8: 9-22, 29, 67, 84; 10-11, 19, 38, 86 Spring 2003 Page 1 of 9

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1 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page of 9 9- Ideal Ranke Cycle Given: Steam power plant wit ideal Ranke cycle. Net power output i MW. Steam enter te at 7 MPa and 00 C. e team i cooled te at 0 kpa ug coolg water at a rate of 000 kg/. Fd: a) termal efficiency of te cycle, b) ma flow rate of te team, and c) temperature rie of te coolg water. Draw te - diagram. At tate, P 7 MPa and 00 C. We ave upereated vapor: 0. kj/kg and.797 kj/kgk. boiler At tate, and P 0 kpa. At 0 kpa, f 0.9 kj/kgk and fg kj/kgk x f 9. kj/kg and fg 9. kj/kgk. kj/kg. At tate, we ave aturated liquid at 0 kpa v v f m /kg, f 9. kj/kg. e work conumed by te i: w v (P P ) (0 7000) kj/kg. State i found by: w w 9. (-7.099) 9.9 kj.kg. e work produced by te i: w kj/kg. a) e termal efficiency i: w w + w net η t 0.9, or.9 percent. q b) e ma flow rate i: MW/w net,000/0..99 kg/. Q & m& MW. c) e eat given off te i: ( ) ( ) i amount of eat i beg aborbed by te coolg water: Q & m& c Q& m& c 7000 ( 000. ). C. cw cw p cw cw p e - diagram i own on te rigt. Some of te feature are: o tate i aturated liquid o tate i aturated mixture o tate i upereated o and are ientropic o P P ; P P.

2 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page of Reeat Ranke Cycle Given: Steam power plant tat operate on a reeat Rake cycle wit 0 MW of net output. Steam enter te ig preure at 0 MPa and 00 C, and te low preure at MPa and 00 C. Steam leave te a a aturated liquid at 0 kpa. Ientropic efficiencie of te and compreor are 0 percent and 9 percent, repectively. Fd: a) quality or temperature of team at exit, b) termal efficiency of te cycle, and c) ma flow rate of te team. Draw te - diagram. State i aturated liquid at 0 kpa: f 9. kj/kg, v v f m /kg. State i at 0 MPa. e ideal work of te i: w v (P P ) v (P P ) (0 0,000) 0.9 kj/kg. State a i found from te ientropic efficiency of te compreor: η c ( )/( a ) a ( )/η c + (0.9 9.)/ kj/kg. State i at 0 MPa and 00 C: 7.7 kj/kg,.9 kj/kgk. State i at MPa and : terpolate between at and 00 C to get 7.7 kj/kg. State a i found from te ientropic efficiency of : η t ( a )/( ) a η t ( ) + 0.( ) kj/kg. State i at MPa and 00 C: 7. kj/kg, 7.7 kj/kgk. State i at 0 kpa and : i i aturated mixture. f 0.9 kj/kgk, fg kj/kgk. x ( f )/ fg ( )/ f 9. kj/kg, fg 9. kj/kg 0. kj/kg. State a i found from te ientropic efficiency of : a η t ( ) + 0.(0. 7.) kj/kg. a) i i greater tan g, o we ave a upereated vapor. Interpolate between 0 C and 00 C to fd 7. C. b) e termal efficiency of te reeat cycle i given by: η t w q net , or.0 percent. 79. c) e ma flow rate i given by: 0 MW/w net 0,000 /7.. kg/. e important feature of te - diagram are: o tate i aturated liquid o tate i aturated mixture o tate,, a,, a are upereated o ; P P ; P P ; P P. boiler a a a

3 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page of Combed ga-team power plant Given: Combed ga-team power plant wit net power output of 0 MW. Ga cycle a a preure ratio of. Air enter te compreor at 00 K and te at 00 K. e combution gae at exaut i ued to eat te team at MPa to 00 C a eat excanger. e combution gae leave te eat excanger at 0 K. An open feedwater eater te team cycle operate at 0. MPa. e preure i 0 kpa. Fd: a) ma flow rate ratio of air to team; b) required eat put to te combution camber; c) termal efficiency of te combed cycle. e air cycle i an ideal Brayton cycle. Ug variable pecific eat: State : 00 K 00.9 kj/kg, P r.0 State : P P r Pr P Interpolate between 0 K and 0 K. kj/kg. State : 00 K. kj/kg, P r 0. State : P P r Pr 0./.79 P Interpolate between 70 K and 70 K 7. kj/kg. comp. State *: At te exaut of te eat excanger, * 0 K *.0 kj/kg. * combution camber eat excanger open 7 FWH 0 9 e team cycle i a regenerative Ranke cycle wit open feedwater eater. State i a aturated liquid at 0 kpa:.0 kj/kg, v m /kg. State i at te feedwater preure of 0. MPa. w, v (P P ) (0 00) kj/kg. w,.0 (-0.99).99 kj/kg. State 7 i aturated liquid at 0. MPa: kj/kg, v m /kg. State i at MPa. w, v 7 (P 7 P ) 0.000(00 000) -.7 kj/kg. 7 w, 70. (-.7) 7.7 kj/kg. State 9 i at MPa and 00 C: 9. kj/kg, 9. kj/kgk. State 0 i at 0. MPa and 0 9 : f.9 kj/kgk and fg. kj/kgk x f 70. kj/kg, fg 0. kj/kg 0. kj/kg State i at 0 kpa and 9 : f 0.0 kj/kgk and fg 7.07 kj/kgk x 0.7. f.0 kj/kg, fg. kj/kg 09.7 kj/kg Heat balance around te open FWH to fd y: y 0 + ( y) y

4 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page of (contued) a) e air-to-team ratio i found from eat balance around te eat excanger: m& m& air ( * ) m& team ( 9 ) m& b) We firt need to fd te ma flow rate. air team * e net work of te Brayton cycle i: w net,b kj/kg of air. e work from te Ranke cycle i: w,r y( 9 0 ) + ( y)( 9 ) 0.79(..) + ( 0.79)(. 09.7) 9.70 kj/kg of team. e net work from te Ranke cycle i: w net,r w, + w, + w,r (-0.99) + (-.7) kj/kg. e net work i: & & & & + & W net mair w net, B + m teamw net, R. 9m teamw net, B m teamw net, R Rearrange to olve for ma flow rate: W& net 0,000 m& team 9.9 kg/ of team.9w net, B + w net, R.9(. ) m &. 9 &. kg/ of air air m team e required eat put i: Q m& ( ) & air.(..) 70. MW. c) e termal efficiency i: η W & Q& , or. percent. t net 9- Reeat-Regenerative Ranke Cycle Given: Ideal reeat-regenerative Ranke cycle wit one open feedwater eater. e boiler preure i 0 MPa, preure i kpa, reeater preure i MPa, and feedwater preure i 0. MPa. Steam enter te ig and low preure at 00 C. Fd: a) fraction of team extracted for regeneration (y), and b) termal efficiency of te cycle. Draw te - diagram. State i aturated liquid at te preure of kpa:.9 kj/kg, v m /kg. State i at te feedwater preure of 0. MPa: w, v (P P ) 0.000( 00) -0.9 kj/kg. w,.9 (-0.9). kj/kg. State i aturated liquid at 0. MPa: 70. kj/kg, v m /kg. State i at te boiler preure of 0 MPa: w, v (P P ) 0.000(00 0,000) -0.9 kj/kg. w, 70. (-0.9) 0.9 kj/kg. boiler open FWH 7 9

5 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page of 9 9- (contued) State i at 0 MPa and 00 C: 7.7 kj/kg,.9 kj/kgk. State i at te reeater preure of MPa and : Interpolate between at. temperature and 00 C 7.7 kj/kg. State 7 i at MPa and 00 C: 7 7. kj/kg, kj/kgk. State i at 0. MPa and 7 : terpolate between 00 C and 00 C 09. kj/kg. State 9 i at kpa and 9 7 : x 9 0.9, 9. kj/kg. a) e fraction of team extracted for regeneration i found from a eat balance around te open feedwater eater: y + ( y) 70.. y b) e termal efficiency of te cycle i found from w net /q. e work for te ig-preure i: w, kj/kg. e work for te low-preure i: w, y( 7 ) + ( y)( 7 9 ) 0.(7. 09.) + ( 0.)(7..). kj/kg. e net work i: w, + w, + w, + w, kj/kg. e eat put i: q kj/kg. e termal efficiency i: w net /q./. 0.09, or. percent. e - diagram ould ave te followg feature: o tate and are aturated liquid 7 o tate,, 7, are upereated vapor o tate 9 i aturated mixture o P P 9 ; P P P ; P P ; P P 7 o,,, and 7 9 are ientropic o tate and 7 ave te ame temperature 9

6 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page of 9 0- Ideal Refrigeration Cycle Given: An ideal vapor-compreion refrigeration cycle ue R-a between 0. and 0.7 MPa. e ma flow rate i 0.0 kg/. Fd: a) Rate of eat removal from refrigerated pace and te power put to te compreor; b) rate of eat rejection to te environment; and c) coefficient of performance. Draw te - diagram. State i aturated vapor at 0. MPa g. kj/kg, g 0.9 kj/kgk. State i at 0.7 MPa, and. Interpolate beween 0 C and 0 C to get 70. kj/kg. State i aturated liquid at 0.7 MPa. f.7 kj/kg. State i at 0. MPa, and. a) e eat removal from refrigerated pace i between tate and : ( ) 0.0. (.7) & m& 7. kw. Q e power put to te compreor i between tate and : ( ) 0.0. ( 70.) & m& -. kw, or. kw of power put. W b) Heat i rejected between tate and : & m& kw, or 9.7 kw of eat rejected. Q out ( ) ( ) c) COP Q & & W eat excanger comp. e - diagram i caracterized by: o tate i aturated vapor o tate i upereated vapor o tate i aturated liquid o tate i aturated mixture o P P ; P P. o i ientropic

7 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page 7 of Non-Ideal Refrigeration Cycle Given: R-a enter te compreor of a refrigeration cycle at 0 kpa and -0 C at a rate of 0. m /m, and leave at MPa. e ientropic efficiency of te compreor i 7 percent. e refrigerant enter te trottlg valve at 0.9 MPa and 0 C, and leave te evaporator a aturated vapor at -. C. Fd: a) Power put to te compreor; b) rate of eat removal from te refrigerated pace; and c) preure drop and te rate of eat gaed te le between evaporator and compreor. Draw te - diagram. State i upereated at 0 kpa and -0 C:.0 kj/kg, 0.90 kj/kgk, v 0.9 m /kg. e ma flow rate i: m m V& 0. m 0ec m& 0.07 kg/ v 0.9 State i at MPa, and. Interpolate between 0 C and 0 C to get.0 kj/kg. State can be found from te ientropic efficiency of te compreor. η c ( )/( a ) a ( )/η c + (.0.0)/ kj/kg. State i at 0.9 MPa and 0 C, wic i a compreed liquid. We will ue te propertie of aturated liquid at 0 C: 9.9 kj/kg. State i a aturated mixture at -. C, and. State * i a aturated vapor at -. C. We can terpolate between -0 C and - C to fd *. kj/kg and P * P at.7 kpa. a) e power put to te compreor i between tate and : ( ) ( 9.07) W & & -.79 kw, or.79 kw of power put. m a b) e eat removal from refrigerated pace i between tate and *: & m& kw. Q ( ) ( ) * c) e preure drop between tate * and i: kpa. & 0. kw e rate of eat gaed i: Q m& ( * ) (.) evaporator * comp. e - diagram i caracterized by: a o tate * i aturated vapor o tate,, and a are upereated vapor o tate i compreed liquid o tate i aturated mixture * o P P ; P P * o i ientropic

8 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page of 9 0- Cacade Refrigeration Cycle Given: wo-tage cacade refrigeration ytem operatg between 0. MPa and 0. MPa. Eac tage i an ideal vapor-compreion cycle wit R-a. Heat rejection from te lower cycle to te upper cycle take place an adiabatic eat excanger were bot tream enter at 0. MPa. e ma flow rate of te refrigerant troug te upper cycle i 0. kg/. Fd: a) ma flow rate of te refrigerant troug lower cycle; b) rate of eat removal from te refrigerated pace and te power put to te compreor; c) coefficient of performance. Lower Cycle: State i a aturated vapor at 0. MPa.0 kj/kg, 0.9 kj/kgk. State i at 0. MPa and. Interpolate between 0 C and 0 C to get 7.9 kj/kg. State i a aturated liquid at 0. MPa.00 kj/kg. State i a aturated mixture at 0. MPa, and. Upper Cycle: State i a aturated vapor at 0. MPa. kj/kg, 0.9 kj/kgk. State i at 0. MPa and. Interpolate between at and 0 C to get.9 kj/kg. State 7 i a aturated liquid at 0. MPa 9. kj/kg. State i a aturated mixture at 0. MPa, and. 7 eat excanger evaporator comp. comp. a) e ma flow rate of te lower cycle i found from a eat balance around te eat excanger: m& U ( ) m& ( ) L. 9. m& m& L U kg/ b) e eat removal from refrigerated pace i between tate and : ( ) (.00) Q& m&.97 kw. L e total power put to te compreor i: W& m& + m& L ( ) U ( ) ( 7.9) + 0.(..9) kw, or 7. kw power put. c) e coefficient of performance i: COP Q & & W

9 ES-7A ermodynamic HW : 9-, 9, 7, ; 0-, 9,, Sprg 00 Page 9 of 9 0- Heat Pump Given: A eat operate on an ideal vapor-compreion cycle wit R-a. e ma flow rate i 0. kg/. e condenor and evaporator preure are 900 kpa and 0 kpa, repectively. Fd: a) rate of eat upplied to te oue; b) volume flow of te refrigerant at te compreor let; and c) coefficient of performance of te eat. Sow te - diagram. State i a aturated vapor at 0 kpa.09 kj/kg, 0.9 kj/kgk, v 0.0 m /kg. State i at 900 kpa and. Interpolate between 0 C and 0 C to get 7. kj/kg. State i a aturated liquid at 900 kpa 99. kj/kg. State i a aturated mixture at 0 kpa, and. eat excanger comp. a) e eat upplied to te oue come from te, : ( ) ( 7.) & m& -. kw, or. kw upplied to te oue. Q out b) e volume flow rate of te refrigerant at te compreor let (tate ) i: V & m & v m /. ( ) c) e coefficient of performance i: Q& COP W& & W out ( ) 0.(.09 7.) m& COP./ kw, or.7 kw power put. e - diagram a te followg feature: o tate i aturated vapor o tate i upereated vapor o tate i aturated liquid o i ientropic o P P ; P P.