17brayton_cycle - Brayton Cycle Summary define some units 3...

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Unformatted text preview: Brayton Cycle Summary define some units 3 kJ := 10 J compressor turbine ma_dot air QH_dot QL_dot 1 2 3 4 Gas Turbine represented by air standard Brayton cycle W_dotnet= Wt_dot+Wc_dot Brayton cycle consists of: 1-2 adiabatic compression 2-3 heat addition ~ constant pressure 3-4 adiabatic expansion in turbine 4-1 heat rejection ~ constant pressure p-v and T - s plots for Brayton cycle shown below for reversible cycle. in irreversible cycle, p 2 > p 3 and p 4 > p 1 , s 2 > s 1 , s 4 > s 3 starting conditions p 1_plot := 1 T 1_plot := 25 + 273.15 s 1_plot := 1 after compression p 2_plot := 10 max temperature after heat addition T 3_plot := 1000 + 273.15 calculations 11/14/2005 1 p-v plot of Brayton cycle 0.2 0.4 0.6 0.8 1 1.2 1.4 1.6 1.8 volume 5 10 adiabatic compression heat addition adiabatic expansion in turbine heat rejection pressure {3} {2} {1} {4} T-s plot of Brayton cycle (reversible) 0.9 1 1.1 1.2 1.3 1.4 1.5 1.6 1.7 1.8 1.9 entropy 200 400 600 800 1000 1200 1400 adiabatic compression heat addition adiabatic expansion in turbine heat rejection temperature {3} {4} {2} {1} 11/14/2005 2 Ideal (reversible) basic Brayton cycle compressor work w c = ( h 2 h 1 ) heat addition q H = h 3 h 2 turbine work w t = h 3 h 4 heat rejection q L = ( h 4 h 1 ) q H + q L q L w t + w c th = = 1 + = q H q H q H h 4s := C p C p ( T 4s T 1 ) + h 1 assuming perfect gas, constant specific heat. (2) h only a function of temperature; (5.23) VW &S, Joule's h 3 := C p C p ( T 3 T 2s ) + h 2s experiment shows u is f(T) only, pv = RT => h=f(T). factor out T1 / T2s T 4s (1) th := 1 h 4s h 4s h 1 1 T 4s T 1 := 1 T 1 T 1 T 1 1 h 3 h 2s th T 3 T 2s th T 2s T 3 1 T 2s 1 this is reversible adiabatic isentropic compression (and expansion) ... T 2s p 2s process with ideal gas and T 1 = p 1 constant specific heat 1 1 p 2s p 3 T 2s p 2s p 3 T 3 T 4s T 3 since = = = = => = p 1 p 4s T 1 p 1 p 4s T 4s T 1 T 2s 1 T 1 p 1 1 1 th = 1 T 2s = 1 p 2s = 1 1 = 1 1 r = pressure_ratio p 2s r p 1 example; for 50 % efficiency, and some typical gas constants ... 1.29 CO2 th = 1 1 1 1 1 = 1 th 1 := 1.4 air r r r = ( 1 th ) 1.67 monotonic gasses, He, Ar, Ne, He th := 0.5 i := .. 0 2 i 21.83 so for air as the working fluid, i 1 a pressure ratio of 11.3 will r := 1 r = 11.31 i ( th ) provide 0.5 isentropic 5.63 efficiency 11/14/2005 3 effect of pressure ratio on isentropic efficiency th ( r , ) := 1 1 1 r := 0 25 1.29 .....
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17brayton_cycle - Brayton Cycle Summary define some units 3...

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