Regeneration Brayton Cycle - Irreversible

Regeneration Brayton cycle - irreversible An actual gas turbine differs from the ideal due to inefficiencies in the turbines and compressors and pressure losses in the flow passages (heat exchangers in closed cycle). The T - s diagram may be as shown: static data for plot T-s diagram 1200 temperature 1000 800 600 400 1 1.2 1.4 1.6 1.8 2 2.2 entropy reversible cycle irreversible cycle regeneration inlet temperature reversible irreversible, heat exchanger maximum regeneration inlet temperature irreversible 3 6 T 7ss 5 5s 2 2s 7s 7 4 4 - outlet of turbine 5s - inlet to regenerator T5s = T4 8 irreversible 1 - start 2 - irreversible compressor outlet 6 - outlet of heat addition T6 = Tmax 4 - outlet of turbine 5 - inlet to regenerator T5 = T7 1 s 11/21/2005 state - reversible process 1 - start 2s - reversible compressor outlet 3 - outlet of heat addition T3 = Tmax 1 irreversible processes can be described by some efficiencies and heat transfer effectiveness: N.B. the efficiencies are defined wrt irreversible overall cycle h6 − h7 ηt = turbine efficiency compressor efficiency h 6 − h 7s ηc = h 2s − h 1 h2 − h1 T5 − T2 = = T6 − T7 η t := 0.8 T6 − T7s T2s − T1 η c := 0.78 T2 − T1 heat exchanger effectiveness ε = pressure loss in heater δp H   p 6 = p 3 − δp H = p 3 ⋅  1 − p3   delta_p_over_p_H := 5% δp L   p 7 = p 1 + δp L = p 1 ⋅  1 + p1   delta_p_over_p_L := 3% pressure loss (increase) in cooler, relative to p 1 ε := 94% T7ss − T2 we will combine these as follows as for efficiency only ∆p across turbine matters:  p6 p7 p3⋅  1 − δp H  p3   = p 2 ⋅ ( 1 − δp%) p1 δp L   p1⋅  1 + p1   = this combines losses into effect on turbine for these calculations γ := 1.4 power := γ−1 γ one compressor no intercooling Nc = 1  1 − delta_p_over_p_H    1 + delta_p_over_p_L  delta_p_over_p := 1 − delta_p_over_p = 7.767 % taking advantage of constant c po T1 := 300 Tmax := 1200 maximum T3 := Tmax T6 := Tmax start with 1+ as η = 1 mathematically pr := 1.3 , 1.4 .. 5 reversible relationships are developed in brayton_cycle_summary.mcd (may be 2005) reversible .... T2s ( pr) := pr T4 ( pr) := irreversible ..... power 1   pr  ⋅ T1 T2s ( 2 ) = 365.704 T2 ( pr) := T1 + T2s ( pr) − T1 T2 ( 2 ) = 384.236 ηc p6_over_p7 ( pr) := pr⋅(1 − delta_p_over_p) power ⋅ T3 T4 ( 2 ) = 984.402 p6_over_p7 ( 2 ) = 1.845 1  reversible turbine calc in irreversible cycle ... T7s ( pr) := T6 ⋅  p6_over_p7 ( pr)  power  ( ) T7 ( pr) := T6 − T6 − T7s ( pr) ⋅ η t 11/21/2005 2 T7s ( 2 ) = 1007 T7 ( 2 ) = 1046 at this point we can compute the thermal efficiency without regeneration reversible η th = wnet qH = wt + wc qH irreversible ( T3 − T4 − T2s − T1 = T3 − T2s rev ) =  T6 − T7 − ( T2 − T1 )    ( QH = T3 − T2s = T6 − T2   T6 − T2 irrev ) so thermal efficiency becomes η th_basic_rev ( pr) := ( T3 − T4 ( pr) − T2s ( pr) − T1 ) η th_basic_irr( pr) := T3 − T2s ( pr) ( T6 − T7 ( pr) − T2 ( pr) − T1 ) T6 − T2 ( pr) efficiency Brayton cycles 0.4 η t = 0.8 thermal efficiency 0.3 η c = 0.78 ε = 0.94 0.2 0.1 delta_p_over_p = 7.8 % basic cycle - irreversible basic cycle - reversible 0 2 2.5 3 3.5 4 4.5 5 pressure ratio with regeneration, all the states are the same with reversible - regen inlet temperature irreversible ... T5s := T4 ( T5 ( pr) := T2 ( pr) + ε ⋅ T7 ( pr) − T2 ( pr) with regeneration reversible η th_ic = wnet qH = wt + wc qH η th_reg_rev ( pr) := 1 − 11/21/2005 = ( T3 − T4 − T2s − T1 T3 − T5s = T4 irreversible ) =  T6 − T7 − ( T2 − T1 )    T6 − T5   T2s ( pr) − T1 1 T3 − T3 ⋅    pr  power η th_reg_irr( pr) := 3 ) T5 ( 2 ) = 1006 rev irrev ( QH = T3 − T5s = T6 − T5 ( T6 − T7 ( pr) − T2 ( pr) − T1 T6 − T5 ( pr) ) ) efficiency regeneration irreversible 0.8 thermal efficiency (ideal) 0.7 0.6 0.5 0.4 0.3 0.2 0.1 with regeneration - irreversible with regeneration - reversible 1 1.5 2 2.5 3 3.5 4 4.5 5 pressure ratio also look at magnitude of compressor work compared to turbine, say for pr = 2 (since these states are the same for w & w/o regeneration, the work is also the same ratiorev = workcomp = workturb  T2s − T1    T3 − T4  ratioirr = T2s ( pr) − T1 ratiorev( pr) := T3 − T4 ( pr) T2s := 0 T2 := 0  T2 − T1    T6 − T7  T2 ( pr) − T1 T6 − T7 ( pr) T7 := 0 T7s := 0 parameters from above ... power = 0.286 T1 = 300 3 T6 = 1.2 × 10 T4 := 0 maximum reset to insure calculation for these calculations one stage intercooling two compressors η t = 0.8 pr := 1.1 , 1.2 .. 5 η c = 0.78 T2s (pr , N) := rc(pr , N) power efficiencies from above ... delta_p_over_p = 7.767 % range for pressure ratio reversible ..... 11/21/2005 = ratioirr( 2 ) = 54.7 % Intercooled Irreversible (and reversible) N := 1 workturb ratioirr( pr) := ratiorev( 2 ) = 30.5 % γ = 1.4 workcomp assuming equal pressure ratios across multiple compressors, the ratio for each is ... 1 rc(pr , N) := pr N+ 1 temperature out of all compressors (isentropic) intercooling occurs along p = constant to same T 1 . Subsequent compressions ⋅ T1 are at the same ratio so temperatures after each compression are the same. 4 T2s (2 , 1 ) = 331.227 irreversible ... all compressors T2 (pr , N) := T1 + T2s (pr , N) − T1 ηc T2 (2 , 1 ) = 340.034 p6_over_p7(pr) := pr⋅(1 − delta_p_over_p) 1 T4 ( pr) := T3 ⋅   pr power 1  T7s ( pr) := T6 ⋅   p6_over_p7 ( pr)  T4 ( 2 ) = 984.402   ( power 3 T7s ( 2 ) = 1.007 × 10 ) T7 ( pr) := T6 − T6 − T7s ( pr) ⋅ η t reversile .... η th_ic = wnet qH = wt + wc qH = 3 irreversible ( T3 − T4 − (N + 1)⋅ T2s − T1 T3 − T2s ) = T6 − T7 − (N + 1)⋅ (T2 − T1) T6 − T2 η th_ic_irr( pr , N) := η th_ic_rev( pr , N) := T7 ( 2 ) = 1.046 × 10 (T6 − T7(pr)) − (N + 1)⋅ (T2(pr , N) − T1) T6 − T2 (pr , N) (T3 − T4(pr)) − (N + 1)⋅ (T2s( pr , N) − T1) T3 − T2s (pr , N) efficiency Brayton cycles thermal efficiency 0.4 0.3 η t = 0.8 0.2 η c = 0.78 0.1 delta_p_over_p = 7.767 % 0 intercooled - irreversible intercooled - reversible 0.1 1 1.5 2 2.5 3 3.5 4 4.5 5 pressure ratio ratiorev = workcomp workturb ratiorev( pr) := = ( (N + 1)⋅ T2s − T1 ) ratioirr = T3 − T4 ( (N + 1)⋅ T2s (pr , N) − T1 ) workcomp workturb ratioirr( pr) := T3 − T4 ( pr) ratiorev( 2 ) = 29 % ( (N + 1)⋅ T2 − T1 = T6 − T7 ( (N + 1)⋅ T2 ( pr , N) − T1 ) T6 − T7 ( pr) ratioirr( 2 ) = 52 % calculations with reheat and multiple turbines are similar and will not be done here. see brayton_plot.mcd for general calculations and plotting 11/21/2005 5 )