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BRAYTON CYCLE. THE IDEAL CYCLE FOR GAS-TURBINE ENGINES We /S=const Qin COMPRESSOR 1. Military aviations. 3. Electric Generation 5. Industrial B. IS ENTROPIC PV = constant IV"! = constant TV EQUATIONS: A. HEAT ENTERING & Q OUT Qin = H₂-H₂ = Cp (Ig-1₁) Qout = H₁ - Hy = Cp (T₁ - 14) PROCESS 1-2 ISENTROPIC COMPRESSION PROCESS 2-3: CONSTANT PRESSURE HEAT ADDITION.... PROCESS 3-4: ISENTROPIC EXPANSION PROCESS 4-1 · CONSTANT PRESSURE HEAT REJECTION APPLICATIONS DA PROCESS TURBINE M² S=const. IP (+7)=k y = SP./ CV. ឬ processes - power, heat WI 2. Commercial aviation 4. Transportation (ships, tanks) C. FROM COMPRESSION & EXPANSION Wc - Cp(T2-TD) WT - Cp (14-13) D- PRESSURE RATIO ** [p = P₂ 1 E EFFICIENCY A=1-11 T₂ REQ'D.. ADDITIONAL PROBLEM 1.4 The ideal air-standard Brayton Cycle operates with air entering the compressor at as kPa, 22°C. The pressure ratio is 6:1, and the air leaves the beat addition process at 1100K. Determine the compressor work and the turbine work per unit mass flow, The cycle efficiency, and the back work ratio. Assume constant properties. GIVEN: p Gin fp = 6/1 -1 10분 P₁= askra... 1₁ = 22°C 3 Bog. 295.15 K:: F. BACK WORK RATIO, row row= We -W₁. ...T3 = 1100. K a. Wc & WT in kJ/kg b. n C. row V (p = (71₂) R Cv = (³/₂) R Y=1.4 j DATE: SOLN 9 Wc = (p (T₂-T₁) W₁ = Cp (T4-13) CONDITION 2: TP²²+ = K₂ P₂ P₁ 95 kPa P₂ = 570 kPa LY L-V 1₁P₁ Y = 1₂ P₂ Y (295-15) (95) + 4 = 1₂ (516) +4 1₂ = 492.4609 R 6 rp = 1/² 4 - CONDITION 3. P₁ = P₂ = 570kPa T3 = 1100 K CONDITION 4: Py = P₁ =...
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Alternative transcript:
95 kPa TP ²= = K p T₂P₂²= = T₁ P. ² Борз Y = T4 (95) 41-44 659. 2707 K (1100) (570) +++ = Ty ¡" ht = *(૪.૭૭પ) (૫૧૨.૫.૧ – ૨૧૩-૭) ( kmoll29kg) Wc = 197. 9845 kJ/kg W₁ = (²/2) (8-314) ( kmol / Ja kg) (659.2707 - 1100 k) W₁ = -442. 2339 kJ/xg| T₁ 0 = 1 - 17/12/2 n = = 0.4007 = 40.07% = Low: Low = ·|- Wo -W₁ ૨૧૩/5 493.4609 1.4-1. = 0.4007 = 40.07% 197-9845 KU/k9 --(-442, 2389 kJ/kg) 0.4477
Important formulas and concepts in thermodynamic brayton cycle.
Important formulas and concepts in thermodynamic brayton cycle.
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BRAYTON CYCLE. THE IDEAL CYCLE FOR GAS-TURBINE ENGINES We /S=const Qin COMPRESSOR 1. Military aviations. 3. Electric Generation 5. Industrial B. IS ENTROPIC PV = constant IV"! = constant TV EQUATIONS: A. HEAT ENTERING & Q OUT Qin = H₂-H₂ = Cp (Ig-1₁) Qout = H₁ - Hy = Cp (T₁ - 14) PROCESS 1-2 ISENTROPIC COMPRESSION PROCESS 2-3: CONSTANT PRESSURE HEAT ADDITION.... PROCESS 3-4: ISENTROPIC EXPANSION PROCESS 4-1 · CONSTANT PRESSURE HEAT REJECTION APPLICATIONS DA PROCESS TURBINE M² S=const. IP (+7)=k y = SP./ CV. ឬ processes - power, heat WI 2. Commercial aviation 4. Transportation (ships, tanks) C. FROM COMPRESSION & EXPANSION Wc - Cp(T2-TD) WT - Cp (14-13) D- PRESSURE RATIO ** [p = P₂ 1 E EFFICIENCY A=1-11 T₂ REQ'D.. ADDITIONAL PROBLEM 1.4 The ideal air-standard Brayton Cycle operates with air entering the compressor at as kPa, 22°C. The pressure ratio is 6:1, and the air leaves the beat addition process at 1100K. Determine the compressor work and the turbine work per unit mass flow, The cycle efficiency, and the back work ratio. Assume constant properties. GIVEN: p Gin fp = 6/1 -1 10분 P₁= askra... 1₁ = 22°C 3 Bog. 295.15 K:: F. BACK WORK RATIO, row row= We -W₁. ...T3 = 1100. K a. Wc & WT in kJ/kg b. n C. row V (p = (71₂) R Cv = (³/₂) R Y=1.4 j DATE: SOLN 9 Wc = (p (T₂-T₁) W₁ = Cp (T4-13) CONDITION 2: TP²²+ = K₂ P₂ P₁ 95 kPa P₂ = 570 kPa LY L-V 1₁P₁ Y = 1₂ P₂ Y (295-15) (95) + 4 = 1₂ (516) +4 1₂ = 492.4609 R 6 rp = 1/² 4 - CONDITION 3. P₁ = P₂ = 570kPa T3 = 1100 K CONDITION 4: Py = P₁ =...
BRAYTON CYCLE. THE IDEAL CYCLE FOR GAS-TURBINE ENGINES We /S=const Qin COMPRESSOR 1. Military aviations. 3. Electric Generation 5. Industrial B. IS ENTROPIC PV = constant IV"! = constant TV EQUATIONS: A. HEAT ENTERING & Q OUT Qin = H₂-H₂ = Cp (Ig-1₁) Qout = H₁ - Hy = Cp (T₁ - 14) PROCESS 1-2 ISENTROPIC COMPRESSION PROCESS 2-3: CONSTANT PRESSURE HEAT ADDITION.... PROCESS 3-4: ISENTROPIC EXPANSION PROCESS 4-1 · CONSTANT PRESSURE HEAT REJECTION APPLICATIONS DA PROCESS TURBINE M² S=const. IP (+7)=k y = SP./ CV. ឬ processes - power, heat WI 2. Commercial aviation 4. Transportation (ships, tanks) C. FROM COMPRESSION & EXPANSION Wc - Cp(T2-TD) WT - Cp (14-13) D- PRESSURE RATIO ** [p = P₂ 1 E EFFICIENCY A=1-11 T₂ REQ'D.. ADDITIONAL PROBLEM 1.4 The ideal air-standard Brayton Cycle operates with air entering the compressor at as kPa, 22°C. The pressure ratio is 6:1, and the air leaves the beat addition process at 1100K. Determine the compressor work and the turbine work per unit mass flow, The cycle efficiency, and the back work ratio. Assume constant properties. GIVEN: p Gin fp = 6/1 -1 10분 P₁= askra... 1₁ = 22°C 3 Bog. 295.15 K:: F. BACK WORK RATIO, row row= We -W₁. ...T3 = 1100. K a. Wc & WT in kJ/kg b. n C. row V (p = (71₂) R Cv = (³/₂) R Y=1.4 j DATE: SOLN 9 Wc = (p (T₂-T₁) W₁ = Cp (T4-13) CONDITION 2: TP²²+ = K₂ P₂ P₁ 95 kPa P₂ = 570 kPa LY L-V 1₁P₁ Y = 1₂ P₂ Y (295-15) (95) + 4 = 1₂ (516) +4 1₂ = 492.4609 R 6 rp = 1/² 4 - CONDITION 3. P₁ = P₂ = 570kPa T3 = 1100 K CONDITION 4: Py = P₁ =...
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SuSSan, iOS User
Love this App ❤️, I use it basically all the time whenever I'm studying
Alternative transcript:
95 kPa TP ²= = K p T₂P₂²= = T₁ P. ² Борз Y = T4 (95) 41-44 659. 2707 K (1100) (570) +++ = Ty ¡" ht = *(૪.૭૭પ) (૫૧૨.૫.૧ – ૨૧૩-૭) ( kmoll29kg) Wc = 197. 9845 kJ/kg W₁ = (²/2) (8-314) ( kmol / Ja kg) (659.2707 - 1100 k) W₁ = -442. 2339 kJ/xg| T₁ 0 = 1 - 17/12/2 n = = 0.4007 = 40.07% = Low: Low = ·|- Wo -W₁ ૨૧૩/5 493.4609 1.4-1. = 0.4007 = 40.07% 197-9845 KU/k9 --(-442, 2389 kJ/kg) 0.4477