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Adiabatic Efficiency of a Turbine
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BASIC CONSIDERATIONS IN THE ANALYSIS OF POWER CYCLES ONORTVEL OZJA IDELIZATIONS & SIMPLICICATIONS 1. Thy cycle does not involve fiction any 2. All expansion and compression of processes take place in a quasi- Comalar shastrapce) 41 equilibrium manner... 3. The pipe connecting the various components of a system are well insulated, and heat transfer through them is negligible Roll WHY IS THE CARNOT CYCLE PRATICAL Ubecause it is ideal very MO BOILER CONDENSER 1 I. REVERSIBLE ISOTHERMA GAS EXPANSION I. REVERSIBLE ADIABATIO GAS EXPANSION III. REVERSIBLE SOTHERMAL GAS COMPRESSION IV. REVERSIBLE ADIABATIC GAS COMPRESSION STEADY STATE FLOW PROCESS NO: DATE: turbine A. STEADY STATE: ACCUMULATION IS ZERO B. IDEAL PROCESS WILL SERVE AS M MODEL EVEN THOUGH IT IS UNAVOIDABLE, OPERATIONS ARE ASSUMED TO BE ADIABATIO. C. IDEAL PROCESSES DO NOT HAVE IRREVERSIBILITIES. NO: DATE: D. SINCE IT IS REVERSIBLE AND ADIABATIC, IDEAL PROCESS IS ALSO ISENTROPIO. AS=0; S₁ S₂ ´ADIABATIO EFFICIENCY OF TURBINES (EXPANDERS) VP (expandable volume) #H H₂ act Heisent P₁ H₁ actual shaft Work Teentropio turbine w Wact * Wact *Wisen W AH + APE + OKE SH n TURBINE = Wact Wisen Wisen 25 = она otts = + Ws Ws actual process 2à Isentropic process IDEAW 7 Wact Wisent = P₂ H₂a-7 actual H₂5isentropic SOUS CONS H₂a - H₁ H₂5 - H₁ P2 75 :!! • ADDITIONAL PROBLEM # 1 STEAM ENTERS AT AN ADIA BATIC TURBINE STEADILY AT 3 MPa AND 400°C AND LEAVES AT 50 kPa AND 100°C. IF THE POWER OUTPUT AND THE TURBINE IS 2 MW AND THE KINETIO E CHANGE OF THE SYSTEM IS NEGLIGIBLE DETERMINE (A) THE ADIABATIC EFFIELENCY OF THE TURBINE, AND (B) THE MASS FLOWRATE OF THE STEAM FLOWING THROUGH THE...
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Alternative transcript:
TURBINE. GIVEN. P₁₂₁ = 3 MPa = 3000 kPa 1₁ = 400 °C REQ'D: a.) nt b. m SOLUTION a.) n₁ = ni Wact Wisen TURBINE actual process #₂9₁-H₁ Нга = INITIAL CONDITION: P₁ = 3000kPa @3000 kPg . Wact. = 2 MW #₂5 - HIT ACTUAL FINAL CONDITION: P2=50 кра @50 kPa, T₁ = 400 °C 233-84 ° C P2 = Teat T T₁7 Teat superheated H₁ = 3232 5 kJ/ kg. S₁ = 6.9246. KJ/kg-1 OK -.. +10=100 °C 5 T₂ = 100°C Isat = $1.35°C T27 Tssat .. superheated steam = 2682.6 kJ/kg 50kPa • DATE: steam. SU₁ = ? ISENTROPIC FINAL CONDITION @ 50 kPq b. W saturated mixture H₂S = (1-xv) (H₂) + (xv) (Hu) #1 = 340 564 kJ 1kg Hv nT S₂ =S₁ = 6-9246 KJ | xg. K P₂ - 50kPa Se = 1.0912 kJ/kg. K SV = 7-6977 KJ / kg. k Se L S₂ L SV S₂ = (1-xv) (Se) + (xv) (Sv) 6.9246 - (1-Xv) (1-6912) + (XU) (7.59 417) Xv = 0.8970 H₂S = (1 - 0.8970) (340- 564) + (0-8970) (26 46.0) H₂S 2408.54 kJ/kg H₂a - Hi #25-H₁ = -2000 K³|s = = 2646.0 = H₂a-8₁ W = m (H₂a-H₁) = 26 826-3232.5 2408-54-3232-5 m (2682.6 m = 3.6370 kg || 3230.5) KJ kg = 0.6674 = 66.74 %
Key notes and important details for an adiabatic efficiency of a turbine.
Key notes and important details for an adiabatic efficiency of a turbine.
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BASIC CONSIDERATIONS IN THE ANALYSIS OF POWER CYCLES ONORTVEL OZJA IDELIZATIONS & SIMPLICICATIONS 1. Thy cycle does not involve fiction any 2. All expansion and compression of processes take place in a quasi- Comalar shastrapce) 41 equilibrium manner... 3. The pipe connecting the various components of a system are well insulated, and heat transfer through them is negligible Roll WHY IS THE CARNOT CYCLE PRATICAL Ubecause it is ideal very MO BOILER CONDENSER 1 I. REVERSIBLE ISOTHERMA GAS EXPANSION I. REVERSIBLE ADIABATIO GAS EXPANSION III. REVERSIBLE SOTHERMAL GAS COMPRESSION IV. REVERSIBLE ADIABATIC GAS COMPRESSION STEADY STATE FLOW PROCESS NO: DATE: turbine A. STEADY STATE: ACCUMULATION IS ZERO B. IDEAL PROCESS WILL SERVE AS M MODEL EVEN THOUGH IT IS UNAVOIDABLE, OPERATIONS ARE ASSUMED TO BE ADIABATIO. C. IDEAL PROCESSES DO NOT HAVE IRREVERSIBILITIES. NO: DATE: D. SINCE IT IS REVERSIBLE AND ADIABATIC, IDEAL PROCESS IS ALSO ISENTROPIO. AS=0; S₁ S₂ ´ADIABATIO EFFICIENCY OF TURBINES (EXPANDERS) VP (expandable volume) #H H₂ act Heisent P₁ H₁ actual shaft Work Teentropio turbine w Wact * Wact *Wisen W AH + APE + OKE SH n TURBINE = Wact Wisen Wisen 25 = она otts = + Ws Ws actual process 2à Isentropic process IDEAW 7 Wact Wisent = P₂ H₂a-7 actual H₂5isentropic SOUS CONS H₂a - H₁ H₂5 - H₁ P2 75 :!! • ADDITIONAL PROBLEM # 1 STEAM ENTERS AT AN ADIA BATIC TURBINE STEADILY AT 3 MPa AND 400°C AND LEAVES AT 50 kPa AND 100°C. IF THE POWER OUTPUT AND THE TURBINE IS 2 MW AND THE KINETIO E CHANGE OF THE SYSTEM IS NEGLIGIBLE DETERMINE (A) THE ADIABATIC EFFIELENCY OF THE TURBINE, AND (B) THE MASS FLOWRATE OF THE STEAM FLOWING THROUGH THE...
BASIC CONSIDERATIONS IN THE ANALYSIS OF POWER CYCLES ONORTVEL OZJA IDELIZATIONS & SIMPLICICATIONS 1. Thy cycle does not involve fiction any 2. All expansion and compression of processes take place in a quasi- Comalar shastrapce) 41 equilibrium manner... 3. The pipe connecting the various components of a system are well insulated, and heat transfer through them is negligible Roll WHY IS THE CARNOT CYCLE PRATICAL Ubecause it is ideal very MO BOILER CONDENSER 1 I. REVERSIBLE ISOTHERMA GAS EXPANSION I. REVERSIBLE ADIABATIO GAS EXPANSION III. REVERSIBLE SOTHERMAL GAS COMPRESSION IV. REVERSIBLE ADIABATIC GAS COMPRESSION STEADY STATE FLOW PROCESS NO: DATE: turbine A. STEADY STATE: ACCUMULATION IS ZERO B. IDEAL PROCESS WILL SERVE AS M MODEL EVEN THOUGH IT IS UNAVOIDABLE, OPERATIONS ARE ASSUMED TO BE ADIABATIO. C. IDEAL PROCESSES DO NOT HAVE IRREVERSIBILITIES. NO: DATE: D. SINCE IT IS REVERSIBLE AND ADIABATIC, IDEAL PROCESS IS ALSO ISENTROPIO. AS=0; S₁ S₂ ´ADIABATIO EFFICIENCY OF TURBINES (EXPANDERS) VP (expandable volume) #H H₂ act Heisent P₁ H₁ actual shaft Work Teentropio turbine w Wact * Wact *Wisen W AH + APE + OKE SH n TURBINE = Wact Wisen Wisen 25 = она otts = + Ws Ws actual process 2à Isentropic process IDEAW 7 Wact Wisent = P₂ H₂a-7 actual H₂5isentropic SOUS CONS H₂a - H₁ H₂5 - H₁ P2 75 :!! • ADDITIONAL PROBLEM # 1 STEAM ENTERS AT AN ADIA BATIC TURBINE STEADILY AT 3 MPa AND 400°C AND LEAVES AT 50 kPa AND 100°C. IF THE POWER OUTPUT AND THE TURBINE IS 2 MW AND THE KINETIO E CHANGE OF THE SYSTEM IS NEGLIGIBLE DETERMINE (A) THE ADIABATIC EFFIELENCY OF THE TURBINE, AND (B) THE MASS FLOWRATE OF THE STEAM FLOWING THROUGH THE...
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Knowunity was a featured story by Apple and has consistently topped the app store charts within the education category in Germany, Italy, Poland, Switzerland and United Kingdom. Join Knowunity today and help millions of students around the world.
Still not sure? Look at what your fellow peers are saying...
iOS User
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SuSSan, iOS User
Love this App ❤️, I use it basically all the time whenever I'm studying
Alternative transcript:
TURBINE. GIVEN. P₁₂₁ = 3 MPa = 3000 kPa 1₁ = 400 °C REQ'D: a.) nt b. m SOLUTION a.) n₁ = ni Wact Wisen TURBINE actual process #₂9₁-H₁ Нга = INITIAL CONDITION: P₁ = 3000kPa @3000 kPg . Wact. = 2 MW #₂5 - HIT ACTUAL FINAL CONDITION: P2=50 кра @50 kPa, T₁ = 400 °C 233-84 ° C P2 = Teat T T₁7 Teat superheated H₁ = 3232 5 kJ/ kg. S₁ = 6.9246. KJ/kg-1 OK -.. +10=100 °C 5 T₂ = 100°C Isat = $1.35°C T27 Tssat .. superheated steam = 2682.6 kJ/kg 50kPa • DATE: steam. SU₁ = ? ISENTROPIC FINAL CONDITION @ 50 kPq b. W saturated mixture H₂S = (1-xv) (H₂) + (xv) (Hu) #1 = 340 564 kJ 1kg Hv nT S₂ =S₁ = 6-9246 KJ | xg. K P₂ - 50kPa Se = 1.0912 kJ/kg. K SV = 7-6977 KJ / kg. k Se L S₂ L SV S₂ = (1-xv) (Se) + (xv) (Sv) 6.9246 - (1-Xv) (1-6912) + (XU) (7.59 417) Xv = 0.8970 H₂S = (1 - 0.8970) (340- 564) + (0-8970) (26 46.0) H₂S 2408.54 kJ/kg H₂a - Hi #25-H₁ = -2000 K³|s = = 2646.0 = H₂a-8₁ W = m (H₂a-H₁) = 26 826-3232.5 2408-54-3232-5 m (2682.6 m = 3.6370 kg || 3230.5) KJ kg = 0.6674 = 66.74 %