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can you draw and schematic and t-s fiagram for this actual reheat rankine cycle

can you draw and schematic and t-s fiagram for this actual reheat rankine cycle

Elements Of Electromagnetics
Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
Section: Chapter Questions
Problem 1.1MA
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can you draw and schematic and t-s fiagram for this actual reheat rankine cycle 

 

 

State 1:

 

 Status: Saturated Liquid

30 kPa

From Table A5 when

ʋ1=ʋf = 0.001022 m3/kg

h1=hf = 289.27 kJ/kg

 

State 2a:

Status: Compressed liquid

P2s=PH= 6000 kPa

h2a= 296.04 kJ/kg    [                                                                

     WP, ise=                                                

Wpact =                                                                

 

     

                                                    

State 3

 

Status: Superheated Steam

P3=PH=6000kpa

T3=Tmax= 550°C

     From Table 6A, When = 6000 kPa and =550

    

     =3423.1 kJ/Kg

=6.8826 kJ/Kg⸼K

 

 

State 4s P4s = P₁= 2000 kPa S4s S3=6.8826 kJ/kg.K According to Table A5, when pressure 2000 kPa Sf= 2.4467 kJ/kg.K Sfg=3.8923 kJ/kg.K Sg= 6.3390 kJ/kg.K hf=908.47 kJ/kg hfg= 1889.8 kJ/kg hg=2798.3 kJ/kg Since S4s is larger than Sg and larger than Sf at the pressure of 2000 kPa, the status at 4s is superheated steam. By interpolation at S4s 6.8826 kJ/kg from table A6 6.8826-6.7684 6.9583-6.784 Wturn*isen = h3 h4s=3423.1-3098.56 =324.54 kJ/kg Wturb*act = WHPTX - nisenxturb= 0.86x324.54=279.1044 kJ/kg h4s = State 4a P4a = PI h4a h4a = = (3137.7 3024.2) (3024.2) 3098.56 kj/kg 2000kPa h3 -Wturb act =3423.1 -279.1044 = 3143.99 kPa Since ha is larger than hg = 2798.3 kJ/ kg at the pressure of 2000 kpa, the status at 4a is superheated steam. State 5 Status: Superheated Steam P5= P₁= 2000 kPa Tr=Tr = 400 From Table A6 when P5 = 2000 kPa and T5 = 400 °C h5 = 3248.4 kJ/kg S5 7.1292 kJ/kg x k = Qin2 = h5 -h4a = 3248.4-3143.99 = 104.41 kJ/Kg State 6s Status: Superheated Steam Pos PL= 30 kPa S6S = S5 = 7.1292 kJ/kg According to Table A5, when pressure of 30 kPa, the status at 6, is saturated mixture. Sf= 0.9441 kj/kg k Sfg= 6.8234 kj/kg k Sg- 7.7675 kj/kg k hf 289.27 kj/kg hfg=2335.3 kj/kg hg=2624.6 kj/kg X6s = Since S6s is smaller than Sf at the pressure of 30 kPa, the status at 6s is saturated mixture. - S6s-sf 7.4337-0.9441 Sfg 6.8234 hos = 2405.0518 kj/kg WLpt isen= = 0.906 n=h5-h6s =3248.4-2405.0518=843.3482 kJ/kg
Transcribed Image Text:State 4s P4s = P₁= 2000 kPa S4s S3=6.8826 kJ/kg.K According to Table A5, when pressure 2000 kPa Sf= 2.4467 kJ/kg.K Sfg=3.8923 kJ/kg.K Sg= 6.3390 kJ/kg.K hf=908.47 kJ/kg hfg= 1889.8 kJ/kg hg=2798.3 kJ/kg Since S4s is larger than Sg and larger than Sf at the pressure of 2000 kPa, the status at 4s is superheated steam. By interpolation at S4s 6.8826 kJ/kg from table A6 6.8826-6.7684 6.9583-6.784 Wturn*isen = h3 h4s=3423.1-3098.56 =324.54 kJ/kg Wturb*act = WHPTX - nisenxturb= 0.86x324.54=279.1044 kJ/kg h4s = State 4a P4a = PI h4a h4a = = (3137.7 3024.2) (3024.2) 3098.56 kj/kg 2000kPa h3 -Wturb act =3423.1 -279.1044 = 3143.99 kPa Since ha is larger than hg = 2798.3 kJ/ kg at the pressure of 2000 kpa, the status at 4a is superheated steam. State 5 Status: Superheated Steam P5= P₁= 2000 kPa Tr=Tr = 400 From Table A6 when P5 = 2000 kPa and T5 = 400 °C h5 = 3248.4 kJ/kg S5 7.1292 kJ/kg x k = Qin2 = h5 -h4a = 3248.4-3143.99 = 104.41 kJ/Kg State 6s Status: Superheated Steam Pos PL= 30 kPa S6S = S5 = 7.1292 kJ/kg According to Table A5, when pressure of 30 kPa, the status at 6, is saturated mixture. Sf= 0.9441 kj/kg k Sfg= 6.8234 kj/kg k Sg- 7.7675 kj/kg k hf 289.27 kj/kg hfg=2335.3 kj/kg hg=2624.6 kj/kg X6s = Since S6s is smaller than Sf at the pressure of 30 kPa, the status at 6s is saturated mixture. - S6s-sf 7.4337-0.9441 Sfg 6.8234 hos = 2405.0518 kj/kg WLpt isen= = 0.906 n=h5-h6s =3248.4-2405.0518=843.3482 kJ/kg
WLpt act=WLpt isen xnisenxturb =843.3482 x 0.86 = 725.279 kJ/kg State 6a h6a-h5-WLpt act= 3248.4 - 725.279= 2523.121 kj/kg Since ha is larger than hg at the pressure of 30 kPa, the status at 6a is superheated steam Qout-hoa-h₁-2523.121 — 289.27 = 2233.851 kJ/kg Checking if W, Qin-lin1+Qin2 = 3127.06+109.659= 3236.711 kJ/kg Onet-lin-Qout = 1002.860 kJ/kg Wout WHPT act + WLPT act = 725.279+284.359= 1009.6384 kJ/kg Wnet Wout + Win =13009.63+6.77= 1002.860 kJ/kg Since Wnet-Qnet, the Rankine cycle obey the 1st law of thermodynamics and the calculation is correct. net=Qout
Transcribed Image Text:WLpt act=WLpt isen xnisenxturb =843.3482 x 0.86 = 725.279 kJ/kg State 6a h6a-h5-WLpt act= 3248.4 - 725.279= 2523.121 kj/kg Since ha is larger than hg at the pressure of 30 kPa, the status at 6a is superheated steam Qout-hoa-h₁-2523.121 — 289.27 = 2233.851 kJ/kg Checking if W, Qin-lin1+Qin2 = 3127.06+109.659= 3236.711 kJ/kg Onet-lin-Qout = 1002.860 kJ/kg Wout WHPT act + WLPT act = 725.279+284.359= 1009.6384 kJ/kg Wnet Wout + Win =13009.63+6.77= 1002.860 kJ/kg Since Wnet-Qnet, the Rankine cycle obey the 1st law of thermodynamics and the calculation is correct. net=Qout
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