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ID/S. Part b.) Determine the entropy production (o) of this turbine.

ID/S. Part b.) Determine the entropy production (o) of this turbine.

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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assume that the turbine is operating at a temperature of 800 degrees Rankine
PROBLEM 1 Steam at 1800 Ibf/in.? and 1100°F enters a turbine operating at steady state. As shown in the figure, 20% of the entering mass flow is extracted at 600 Ibf/in.? and 500°F. Heat transfer W. turbine The rest of the steam exits as a P1 = 1800 lbf/in.? T = 1100°F Turbine saturated vapor at 1 Ibf/in.? The turbine develops a power output of 6.8 X 106 Btu/h. Heat transfer from the turbine to the surroundings occurs at a rate of 5 X 10* Btu/h. 13 Saturated vapor P3 = 1 lbf/in.? m2 = 0.20 m, P2 = 600 lbf/in.? T2 = 500°F %3D Part a.) Neglecting kinetic and potential energy effects, determine the mass flow rate of the steam entering the turbine in Ib/s. Part b.) Determine the entropy production (0) of this turbine. 2.
Transcribed Image Text:PROBLEM 1 Steam at 1800 Ibf/in.? and 1100°F enters a turbine operating at steady state. As shown in the figure, 20% of the entering mass flow is extracted at 600 Ibf/in.? and 500°F. Heat transfer W. turbine The rest of the steam exits as a P1 = 1800 lbf/in.? T = 1100°F Turbine saturated vapor at 1 Ibf/in.? The turbine develops a power output of 6.8 X 106 Btu/h. Heat transfer from the turbine to the surroundings occurs at a rate of 5 X 10* Btu/h. 13 Saturated vapor P3 = 1 lbf/in.? m2 = 0.20 m, P2 = 600 lbf/in.? T2 = 500°F %3D Part a.) Neglecting kinetic and potential energy effects, determine the mass flow rate of the steam entering the turbine in Ib/s. Part b.) Determine the entropy production (0) of this turbine. 2.
Equation Sheets 1st Law T(K)=TCC)+273.15 COP AU +AKE +APE -Q-W dKE dPE dU 1- Properties A-u+ P Compressiblity -Q- di Closed Systems RT P P, - 2nd Law A-AC,7, -7) P. Saturated region Constant Pressure Work W- PV,-) dt v=v, +v, -v,) =", +xtu, -,) Polytropic Work Polytropic process PV" - constant 1- Open Systems Mass Subcooled region T.P)= (T)=v,() Cycles Desired Output Required Input dm T.P) = (T)=u,T) di T.P) =u, (T)-P*v, (r) VA Heat engine (Power plant) Ideal Gas P-RT 1st Law PV = mRT dE R-0.2870klkg K e-1- dt 2nd Law Air Conditioner and Refrigerator COP. W. L-1 Ideal Gas isentropic dt PP. 2. Isentropic efficiencies COP (h -h) Isentropie Turbine Work (h -h) Actual Turbine Work Isentropic Compressor Work (h, -k) Actual Compressor Work Heat Pump (h, -4) P = P Iseniropic Pump Work (P, - P) Actual Pump Work (h, -h,) Regeneration optimal flow split
Transcribed Image Text:Equation Sheets 1st Law T(K)=TCC)+273.15 COP AU +AKE +APE -Q-W dKE dPE dU 1- Properties A-u+ P Compressiblity -Q- di Closed Systems RT P P, - 2nd Law A-AC,7, -7) P. Saturated region Constant Pressure Work W- PV,-) dt v=v, +v, -v,) =", +xtu, -,) Polytropic Work Polytropic process PV" - constant 1- Open Systems Mass Subcooled region T.P)= (T)=v,() Cycles Desired Output Required Input dm T.P) = (T)=u,T) di T.P) =u, (T)-P*v, (r) VA Heat engine (Power plant) Ideal Gas P-RT 1st Law PV = mRT dE R-0.2870klkg K e-1- dt 2nd Law Air Conditioner and Refrigerator COP. W. L-1 Ideal Gas isentropic dt PP. 2. Isentropic efficiencies COP (h -h) Isentropie Turbine Work (h -h) Actual Turbine Work Isentropic Compressor Work (h, -k) Actual Compressor Work Heat Pump (h, -4) P = P Iseniropic Pump Work (P, - P) Actual Pump Work (h, -h,) Regeneration optimal flow split
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