kJ Air (Rair = 0.287 ) as an ideal gas flows through the turbine and heat exchanger arrangement kg-K) shown below. Steady-state data are given on the figure. Stray heat transfer and kinetic and potential energy effects can be ignored. Sketch a T – s diagram of processes 1-4 and determine: (a) The entropy production of turbine 1 [3.2 kW /K] (b) The isentropic efficiency of turbine 1 [75%] (c) The exit temperature of turbine 2 if its isentropic efficiency is 77, = 0.85 [960 K] (d) The entropy production within the intermediate heat exchanger [3.2 kW /K] Wn = 10,000 kW Wn =? Turbine Turbine P3= 4.5 bar T3 = ? T = 1100 K P2=5 bar P4 =1 bar Air 2 3 T, = 1400 K Pi = 20 bar www Ts = 1480 K -5 Ps 1.35 bar ms = 1200 kg/min Heat exchanger V T = 1200 K P6 = 1 bar Air in

Elements Of Electromagnetics
7th Edition
ISBN:9780190698614
Author:Sadiku, Matthew N. O.
Publisher:Sadiku, Matthew N. O.
ChapterMA: Math Assessment
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Air (Rair = 0.287 )
as an ideal gas flows through the turbine and heat exchanger arrangement
kg-K,
shown below. Steady-state data are given on the figure. Stray heat transfer and kinetic and potential
energy effects can be ignored. Sketch a T - s diagram of processes 1-4 and determine:
(a) The entropy production of turbine 1 [3.2 kW /K]
(b) The isentropic efficiency of turbine 1 [75%]
(c) The exit temperature of turbine 2 if its isentropic efficiency is nr, = 0.85 [960 K]
(d) The entropy production within the intermediate heat exchanger [3.2 kW/K]
W1 = 10,000 kW
Wn= ?
Turbine
Turbine
P3= 4.5 bar
T3 = ?
T = 1100 K
P2=5 bar
P4 = 1 bar
Air
wwww
www
2
4
T = 1400 K
Pi = 20 bar
Ts = 1480 K
5 Ps = 1.35 bar
ms = 1200 kg/min
Heat exchanger
V T = 1200 K
P6 =1 bar
%3D
Air
in
Transcribed Image Text:Air (Rair = 0.287 ) as an ideal gas flows through the turbine and heat exchanger arrangement kg-K, shown below. Steady-state data are given on the figure. Stray heat transfer and kinetic and potential energy effects can be ignored. Sketch a T - s diagram of processes 1-4 and determine: (a) The entropy production of turbine 1 [3.2 kW /K] (b) The isentropic efficiency of turbine 1 [75%] (c) The exit temperature of turbine 2 if its isentropic efficiency is nr, = 0.85 [960 K] (d) The entropy production within the intermediate heat exchanger [3.2 kW/K] W1 = 10,000 kW Wn= ? Turbine Turbine P3= 4.5 bar T3 = ? T = 1100 K P2=5 bar P4 = 1 bar Air wwww www 2 4 T = 1400 K Pi = 20 bar Ts = 1480 K 5 Ps = 1.35 bar ms = 1200 kg/min Heat exchanger V T = 1200 K P6 =1 bar %3D Air in
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