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H.W Example 3 Steam enters an adiabatic turbine at 450°C and 20 MPa with the mass flow rate of 0.5 kg/s and leaves at 10 kPa and 85% quality. Assume changes in kinetic and potential are negligible, determine the work produces by the turbine. Steam P₁ = 20 MPa 450 T₁ = 350°C m= 0.5 kg/s Turbine 9=0 P₂ = 10 kPa X2= 0.85 Control Volume: Turbine Assumptions: SSSF Process 9cv = 0, AKE =0,APE = 0, CV Solution: apply 1st law SSSF Proc. q+h+V+zg=w+ h; + ½ V² +zeg i h₁ =w+h₂ → w = h₁- h₂ and steam table 2 state 2 → mixture, h = h₂+x₂hfg

H.W Example 3 Steam enters an adiabatic turbine at 450°C and 20 MPa with the mass flow rate of 0.5 kg/s and leaves at 10 kPa and 85% quality. Assume changes in kinetic and potential are negligible, determine the work produces by the turbine. Steam P₁ = 20 MPa 450 T₁ = 350°C m= 0.5 kg/s Turbine 9=0 P₂ = 10 kPa X2= 0.85 Control Volume: Turbine Assumptions: SSSF Process 9cv = 0, AKE =0,APE = 0, CV Solution: apply 1st law SSSF Proc. q+h+V+zg=w+ h; + ½ V² +zeg i h₁ =w+h₂ → w = h₁- h₂ and steam table 2 state 2 → mixture, h = h₂+x₂hfg

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
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H.W Example 3 Steam enters an adiabatic turbine at 450°C and 20 MPa with the mass flow rate of 0.5 kg/s and leaves at 10 kPa and 85% quality. Assume changes in kinetic and potential are negligible, determine the work produces by the turbine. Steam P₁ = 20 MPa 450 T₁ = 350°C m= 0.5 kg/s Turbine 9=0 P₂ = 10 kPa X2= 0.85 Control Volume: Turbine Assumptions: SSSF Process 9cv = 0, AKE =0,APE = 0, CV Solution: apply 1st law SSSF Proc. q+h+V+zg=w+ h; + ½ V² +zeg i h₁ =w+h₂ → w = h₁- h₂ and steam table 2 state 2 → mixture, h = h₂+x₂hfg
Transcribed Image Text:H.W Example 3 Steam enters an adiabatic turbine at 450°C and 20 MPa with the mass flow rate of 0.5 kg/s and leaves at 10 kPa and 85% quality. Assume changes in kinetic and potential are negligible, determine the work produces by the turbine. Steam P₁ = 20 MPa 450 T₁ = 350°C m= 0.5 kg/s Turbine 9=0 P₂ = 10 kPa X2= 0.85 Control Volume: Turbine Assumptions: SSSF Process 9cv = 0, AKE =0,APE = 0, CV Solution: apply 1st law SSSF Proc. q+h+V+zg=w+ h; + ½ V² +zeg i h₁ =w+h₂ → w = h₁- h₂ and steam table 2 state 2 → mixture, h = h₂+x₂hfg
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