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Draw a t-s diagram of a simple rankine diagram witht his data

Draw a t-s diagram of a simple rankine diagram witht his data

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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Draw a t-s diagram of a simple rankine diagram witht his data  

State 4s P4s PL = 30 kPa = S4s S3=6.8826 kJ/kg.K According to Table A5, when pressure 30 kPa Sf= 0.9441kJ/kg.K S.fg = 6.8234 kJ/kg.K Sg= 7.7675 kJ/kg.K hf= 191.8 kJ/kg hfg=2392.2 kJ/kg h4s=hf + x4s x hg= 28.27+0.87 ×2335.3 =2320.981 kJ/kg Since S4s is smaller than Sg and larger than Sf at the pressure of 30 kPa, the status at 4s is a saturated mixture. S4S-Sf Sfg Wturn isen = h3 h4s=3423.1-2320.981-1102.119 Wturb*act h3h4s 0.86x1102.119=947.82 kJ/kg X4s = Wout State 4a = W hsa h4a 6.826-0.9441/6.8234-0.87 P4a=PL = 30kPa turb*act=947.82g kj = h3 - Wturb act =3423.1 -947.82 = 2475.28 kJ/kg
Transcribed Image Text:State 4s P4s PL = 30 kPa = S4s S3=6.8826 kJ/kg.K According to Table A5, when pressure 30 kPa Sf= 0.9441kJ/kg.K S.fg = 6.8234 kJ/kg.K Sg= 7.7675 kJ/kg.K hf= 191.8 kJ/kg hfg=2392.2 kJ/kg h4s=hf + x4s x hg= 28.27+0.87 ×2335.3 =2320.981 kJ/kg Since S4s is smaller than Sg and larger than Sf at the pressure of 30 kPa, the status at 4s is a saturated mixture. S4S-Sf Sfg Wturn isen = h3 h4s=3423.1-2320.981-1102.119 Wturb*act h3h4s 0.86x1102.119=947.82 kJ/kg X4s = Wout State 4a = W hsa h4a 6.826-0.9441/6.8234-0.87 P4a=PL = 30kPa turb*act=947.82g kj = h3 - Wturb act =3423.1 -947.82 = 2475.28 kJ/kg
State 1 Status: Saturated Liquid P₁ = PL= 30 kPa From Table A5 when P₁ = 30 kPa v1 vf = 0.001022 m³/kg h1 hf 289.27 kJ/kg State 2 Status: Compressed liquid P2s PH=6000 kPa h2a= 296.04 kJ/kg _[h₂a = h1 + Wpact = 289.27 + 6.77 = 296.04 WP, ise V₁ (P2 - P₁) = 0.001022 (6000 - 30) = 6.101 kg Wpinsen 6.101 0.9 Wpact nisen pum Win = W State 3 pact=6.77Kg Status: Superheated Steam P3-PH=6000kpa = 6.77 h3-3423.1 kJ/Kg S3-6.8826 kJ/Kg.K kJ kg T3 Tmax= 500°C From Table 6A, When P3= 6000 kPa Heat In (Boiler) Qin=h3 - h₂a = 3423.1 - 296.04 = 3127.06 kJ/Kg Kg
Transcribed Image Text:State 1 Status: Saturated Liquid P₁ = PL= 30 kPa From Table A5 when P₁ = 30 kPa v1 vf = 0.001022 m³/kg h1 hf 289.27 kJ/kg State 2 Status: Compressed liquid P2s PH=6000 kPa h2a= 296.04 kJ/kg _[h₂a = h1 + Wpact = 289.27 + 6.77 = 296.04 WP, ise V₁ (P2 - P₁) = 0.001022 (6000 - 30) = 6.101 kg Wpinsen 6.101 0.9 Wpact nisen pum Win = W State 3 pact=6.77Kg Status: Superheated Steam P3-PH=6000kpa = 6.77 h3-3423.1 kJ/Kg S3-6.8826 kJ/Kg.K kJ kg T3 Tmax= 500°C From Table 6A, When P3= 6000 kPa Heat In (Boiler) Qin=h3 - h₂a = 3423.1 - 296.04 = 3127.06 kJ/Kg Kg
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