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can you draw a t-s diagram for this actual regenerative rankine cycle

can you draw a t-s diagram for this actual regenerative 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 a t-s diagram for this actual regenerative rankine cycle

 

State 7s P₁ = PL= 30 kPa S75 = S6a According to Table A5, when pressure of 30 kPa 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 Since S7s is less than Sg and larger than Sg at the pressure of 30 kPa, the status at 7s is saturated mixture. X7s = = 7.8224 kJ/kg X7a = S7s-Sf 7.8224 0.9441 Sfg y = h7s = hf+X7shfg=289.27 +1.008x2335.3 = 2643.25 State 7a 6.8234 P₁ = PL= 30 kPa hja =h6a-(hoa-h7s)x nisen turb = 3707.45-(3707.45-2643.25)x0.86=2792.28 kJ/kg Since h7a is less than hg and larger than hf at the pressure of 30 kPa, the stats at 7a is saturated mixture. h7a-hf 2792.28 289.27 hfg Wout=( = 1.008 2335.3 h3-h₂a 908.47 291.50 h6а-h2а 3707.45 291.50 -=1.07 =0.180 Checking if Wnet=Qnet Qin 2509.39 kJ/k Qout = (1-y) ( h-a-h₁) = (1 − 0.180)( 2792.28-289.27) = 2052.46 kJ/kg Win=(1-y)( h₂a-h₁)+(h4a-h3)= (1-0.180) (291.50 -289.27)+(913.701-908.47)=7.059 kj/kg h5-h6a)+(1 - y) (hoa-h7a)= (3423.1-3414.38 )+(1-0.80)(3707.45-2792.28)=759 kJ/kg Qnet-lin-Qout-2509.39-2052.46-456.93
Transcribed Image Text:State 7s P₁ = PL= 30 kPa S75 = S6a According to Table A5, when pressure of 30 kPa 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 Since S7s is less than Sg and larger than Sg at the pressure of 30 kPa, the status at 7s is saturated mixture. X7s = = 7.8224 kJ/kg X7a = S7s-Sf 7.8224 0.9441 Sfg y = h7s = hf+X7shfg=289.27 +1.008x2335.3 = 2643.25 State 7a 6.8234 P₁ = PL= 30 kPa hja =h6a-(hoa-h7s)x nisen turb = 3707.45-(3707.45-2643.25)x0.86=2792.28 kJ/kg Since h7a is less than hg and larger than hf at the pressure of 30 kPa, the stats at 7a is saturated mixture. h7a-hf 2792.28 289.27 hfg Wout=( = 1.008 2335.3 h3-h₂a 908.47 291.50 h6а-h2а 3707.45 291.50 -=1.07 =0.180 Checking if Wnet=Qnet Qin 2509.39 kJ/k Qout = (1-y) ( h-a-h₁) = (1 − 0.180)( 2792.28-289.27) = 2052.46 kJ/kg Win=(1-y)( h₂a-h₁)+(h4a-h3)= (1-0.180) (291.50 -289.27)+(913.701-908.47)=7.059 kj/kg h5-h6a)+(1 - y) (hoa-h7a)= (3423.1-3414.38 )+(1-0.80)(3707.45-2792.28)=759 kJ/kg Qnet-lin-Qout-2509.39-2052.46-456.93
CALCULATIONS State 1: Status: Saturated Liquid P₁ = P₁= 30 kPa L= From Table A5 when P₁ = 30 kPa v1=vf=0.001022 m³/kg hl=hf=289.27 kJ/kg State 2a: Status: Compressed liquid P2s PI=2000 kPa h₂a=h₁+ State 3 State 4a v_1 (p_2-p_1) nisen pump Status: Saturated liquid P3-PI=2000 kpa From Table A5, When P3= 2000 kPa and h3=hf 908.47 kJ/kg V3 vf 0.001177 m³/k Pa = PH=6000kPa hдa = = 289.27 + h4a = h3 + 913.701 kPa 0.001022x (2000-30) v3 (P4 - P3) nisen pump 0.9 = 908.47 + = 291.50 kJ/kg 0.001177x(6000 - 2000) 0.9 = 913.701 State 5 Status: Superheated Steam P5 = P₁= 6000 kPa Ts=Tr = 500 From Table A6 when P5 = 6000 kPa and T5 = 500 °C h5 = 3423.1 kJ/kg S5 6.8826 kJ/kg xk Qin h5 -h4a=3423.1-913.701= 2509.39 kJ/Kg = = State 6s Pos= P₁= 2000 kPa 1= $65 = S5 = 6.8826 kJ/kg Since S6s is larger than Sf at the pressure of 2000 kPa, the status at S6s is superheated steam By interpolation at S6s-6.8826 from table A6 hos State 6a (6.8826-6.7684) (6.95883-6.7684) Poa P₁= 2000 kPa hoa = x(3137.7 – 3024.2) + 3024.2= 3092.45 (h5-h6s)x nisen pump = 3423.1+(3423.1-3092.45)x 0.86= 3707.45 kJ/kg Since ha is larger than h at a preassure of 2000kpa, the status at háa is superheated By interpolation at hoa = 3707.45 kJ/kg from table A6 S6a (3707.045-3693.9)/(3920.5-3693.9)x (8.0558-7.8101)+7.8101= 7.824 kJ/kg K
Transcribed Image Text:CALCULATIONS State 1: Status: Saturated Liquid P₁ = P₁= 30 kPa L= From Table A5 when P₁ = 30 kPa v1=vf=0.001022 m³/kg hl=hf=289.27 kJ/kg State 2a: Status: Compressed liquid P2s PI=2000 kPa h₂a=h₁+ State 3 State 4a v_1 (p_2-p_1) nisen pump Status: Saturated liquid P3-PI=2000 kpa From Table A5, When P3= 2000 kPa and h3=hf 908.47 kJ/kg V3 vf 0.001177 m³/k Pa = PH=6000kPa hдa = = 289.27 + h4a = h3 + 913.701 kPa 0.001022x (2000-30) v3 (P4 - P3) nisen pump 0.9 = 908.47 + = 291.50 kJ/kg 0.001177x(6000 - 2000) 0.9 = 913.701 State 5 Status: Superheated Steam P5 = P₁= 6000 kPa Ts=Tr = 500 From Table A6 when P5 = 6000 kPa and T5 = 500 °C h5 = 3423.1 kJ/kg S5 6.8826 kJ/kg xk Qin h5 -h4a=3423.1-913.701= 2509.39 kJ/Kg = = State 6s Pos= P₁= 2000 kPa 1= $65 = S5 = 6.8826 kJ/kg Since S6s is larger than Sf at the pressure of 2000 kPa, the status at S6s is superheated steam By interpolation at S6s-6.8826 from table A6 hos State 6a (6.8826-6.7684) (6.95883-6.7684) Poa P₁= 2000 kPa hoa = x(3137.7 – 3024.2) + 3024.2= 3092.45 (h5-h6s)x nisen pump = 3423.1+(3423.1-3092.45)x 0.86= 3707.45 kJ/kg Since ha is larger than h at a preassure of 2000kpa, the status at háa is superheated By interpolation at hoa = 3707.45 kJ/kg from table A6 S6a (3707.045-3693.9)/(3920.5-3693.9)x (8.0558-7.8101)+7.8101= 7.824 kJ/kg K
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