Given A gas turbine power plant operating with air-standard Brayton cycle
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1 ME-200 Fall 2017 HW-38 1/4 Given A ga turbine power plant operating with air-tandard Brayton cycle Find For ientropic compreion and expanion: (a) Net power (kw) produced by the power plant (b) Thermal efficiency (%) of the cycle (c) Back work ratio of the power plant For non-ientropic compreion and expanion: (d) Net power (kw), thermal efficiency (%), and back work ratio Sytem/EFD III Q in p 2 = 1000 kpa I m 2 Heat Exchanger p 3 = p 2 T 3 = 1400 K m air 6 m W turbine 3 W comp Compreor m 1 p 1 = 100 kpa T 1 = 300 K comp 80% Turbine turbine 80% Aumption - Steady tate, teady flow - One-dimenional, uniform flow - Air behave a an ideal ga - Ignore KE and PE change - Ignore heat tranfer for the compreor and turbine: Q 0 - Neglect preure drop for the heat exchanger (combutor) - No work for the heat exchanger (combutor): W 0 W II m 4 p 4 = 100 kpa Baic Equation dm m m m1 m2 m3 m4 m dt de dt i e air i e 2 2 i 2 e Q W mi hi gzi me he gze i e 2
2 ME-200 Fall 2017 HW-38 2/4 Solution State 1: p 1 = 100 kpa, T 1 = 300 K Ideal ga table for air: h 1 = /, 1 0 = /-K, p r1 = State 2: p 2 = p 2 = 1000 kpa, 2 = 1 (ientropic) 0 0 p2 0 0 p Rair 2 1 Rair p1 p kpa K -K 100 kpa -K pr2 p2 p kpa Alternatively: pr2 pr pr1 p1 p1 100 kpa Ideal ga table for air: T2 574 K, h2 580 State 3: p 3 = p 2 = 1000 kpa, T 3 = 1400 K Ideal ga table for air: h 3 = 1515 /, 0 3 = /-K, p r3 = State 4: p 4 = p 4 = 1000 kpa, 4 = 3 (ientropic) 0 0 p4 0 0 p Rair 4 3 Rair p3 p kpa K -K 1000 kpa -K pr4 p4 p4 100 kpa Alternatively: pr4 pr pr3 p3 p kpa Ideal ga table for air: T4 788 K, h (a) Conidering energy balance for the compreor ( I), the ientropic power for the compreor: W comp, m air h1h kw ; negative ign indicate power input Conidering energy balance for the turbine ( II), the ientropic power for the turbine: W turbine, m air h3h kw ; poitive ign indicate power output Net power produced by the power plant: W W turbine W comp W 2558 kw
3 ME-200 Fall 2017 HW-38 3/4 (b) Conidering energy balance for the heat exchanger (combutor) ( III), the rate of energy tranfer into the power plant: Q in m air h3h kw 2558 kw Thermal efficiency of the cycle: W thermal Q 5610 kw thermal 45.6% (c) Back work ratio of the power plant: W comp, kw bwr W kw bwr turbine, (d) Conidering irreverible compreion through the compreor. State 2: p 2 = 1000 kpa, comp = 80% wientropic h1 h comp. 0.8 h h1h h2 Ideal ga table for air: T2 640 K Conidering irreverible expanion through the turbine. State 4: p 4 = 100 kpa, turbine = 80% h3 h h4 turbine 0.8 h4 950 wientropic h3h Ideal ga table for air: T4 915 K in Conidering energy balance for the compreor ( I), the actual power for the compreor: W comp, actual m air h1h kw ; negative ign indicate power input Conidering energy balance for the turbine ( II), the actual power for the turbine: W turbine, actual m air h3h kw ; poitive ign indicate power output Net power produced by the power plant: W W turbine W comp W 1290 kw Conidering energy balance for the heat exchanger (combutor) ( III), the rate of energy tranfer into the power plant: Q in m air h3h kw
4 ME-200 Fall 2017 HW-38 4/ kw Thermal efficiency of the cycle: W thermal Q kw thermal Back work ratio of the power plant: W comp, actual 2100 kw bwr W 3390 kw bwr 0.62 turbine, actual in 24.8% Note: Irreveribilitie in the compreor and turbine affect cycle performance ( power and thermal efficiency) ignificantly; compreor power input increae and turbine power output decreae cauing power output and thermal efficiency to decreae
5 ME-200 Fall 2017 HW-39 1/4 Given A ga turbine power plant operating with intercooling, reheating, and regeneration Find (a) Temperature (K) and pecific enthalpy (/) at variou tate in the cycle (b) Net power (kw), thermal efficiency (%), and back work ratio of the cycle (c) T- diagram of the cycle Sytem/EFD m 10 m 1 Compreor 1 W comp1 m 2 p 1 = 100 kpa T 1 = 300 K m 3 Intercooler Compreor 2 W comp2 Regenerator m 4 m 5 Q in Combutor 1 m 6 Turbine 1 W turbine1 m 7 m air 6 comp,1 comp,2 turbine,1 turbine,2 80% Q in Combutor 2 m 8 Turbine 2 W turbine2 Aumption - Steady tate, teady flow - One-dimenional, uniform flow - Air behave a an ideal ga - Ignore KE and PE change - Ignore heat tranfer for compreor, turbine, and regenerator (both tream): Q 0 - Neglect preure drop for the intercooler, combutor, and regenerator - No work for the intercooler, combutor, and regenerator: W 0 Baic Equation dm m m dt i e m1 m2 m3 m4 m5 m6 m7 m8 m9 m10 m 2 2 de i 2 e Q W mi hi gzi me he gze dt i e 2 i e air Solution (a) State 1: p 1 = 100 kpa, T 1 = 300 K Ideal ga table for air: h 1 = /, 1 0 = /-K, p r1 = m 9
6 ME-200 Fall 2017 HW-39 2/4 State 2: p 2 = p 2 = 316 kpa, 2 = 1 (ientropic) 0 0 p2 0 0 p Rair 2 1 Rair p1 p kpa K -K 100 kpa -K pr2 p2 p2 316 kpa Alternatively: pr2 pr pr1 p1 p1 100 kpa Ideal ga table for air: T2 416 K, h State 2: p 2 = 316 kpa, comp,1 = 80% wientropic h1 h comp,1 0.8 h h1h h2 Ideal ga table for air: T2 445 K State 3: p 3 = p 2 = 316 kpa (no preure drop), T 3 = 300 K Ideal ga table for air: h 3 = /, 3 0 = /-K, p r3 = State 4: p 4 = p 4 = 1000 kpa, 4 = 3 (ientropic) 0 0 p4 0 0 p Rair 4 3 Rair p3 p kpa K -K 316 kpa -K pr4 p4 p kpa Alternatively: pr4 pr pr3 p3 p3 316 kpa Ideal ga table for air: T4 416 K, h State 4: p 4 = 1000 kpa, comp,2 = 80% wientropic h3 h comp,2 0.8 h h3h h2 Ideal ga table for air: T4 445 K State 5: p 5 = p 4 = 1000 kpa (no preure drop), T 5 = 1000 K Ideal ga table for air: h 5 = 1046 / State 6: p 6 = p 5 = 1000 kpa (no preure drop), T 6 = 1400 K Ideal ga table for air: h 6 = 1515 /, 6 0 = /-K, p r6 = 450.5
7 ME-200 Fall 2017 HW-39 2/4 State 7: p 7 = p 7 = 316 kpa, 7 = 6 (ientropic) 0 0 p7 0 0 p Rair 7 6 Rair p6 p kpa K -K 100 kpa -K pr7 p7 p7 316 kpa Alternatively: pr7 pr pr 6 p6 p kpa Ideal ga table for air: T K, h State 7: p 7 = 316 kpa, turbine,1 = 80% h6 h h7 turbine,1 0.8 h w h h ientropic 6 7 Ideal ga table for air: T K State 8: p 8 = p 7 = 316 kpa (no preure drop), T 8 = 1400 K Ideal ga table for air: h 8 = 1515 /, 8 0 = /-K, p r8 = State 9: p 9 = p 9 = 100 kpa, 9 = 8 (ientropic) 0 0 p9 0 0 p Rair 9 8 Rair p8 p kpa K -K 100 kpa -K pr9 p9 p9 100 kpa Alternatively: pr9 pr pr8 p8 p8 316 kpa Ideal ga table for air: T K, h State 9: p 9 = 316 kpa, turbine,2 = 80% h8 h h9 turbine,2 0.8 h wientropic h8 h Ideal ga table for air: T K State 10: p 10 = p 9 = 316 kpa (no preure drop) Conidering energy balance for the regenerator: h5 h4 h9 h10 h Ideal ga table for air: T K
8 ME-200 Fall 2017 HW-39 2/4 State Abolute Preure (kpa) Temperature (K) Specific Enthalpy (/) (b) Conidering energy balance for the two compreor: W comp m air h1h2h3h kw ; negative ign indicate power input Conidering energy balance for the two turbine: W turbine m air h6 h7h8 h kw ; poitive ign indicate power output Net power produced by the power plant: W W turbine W comp W 2106 kw Conidering energy balance for the two combutor, the rate of energy tranfer into the power plant: Q in m air h6 h5h8 h kw 2106 kw Thermal efficiency of the cycle: W thermal Q 4746 kw thermal 44.3% in Back work ratio of the power plant: W comp 1758 kw bwr W 3864 kw bwr turbine
9 ME-200 Fall 2017 HW-39 2/4 For the ame overall preure ratio and ma flow rate of air, uing HW-38(a), (b), and (c), ientropic compreion and expanion reult in power output of 2558 kw, thermal efficiency of 45.6%, and back work ratio of (~40% of turbine work conumed by the compreor). Irreveribilitie in compreor and turbine decreae power output and thermal efficiency and increae back work ratio ignificantly a een from HW-38(d). However, uing two-tage compreion with intercooling (to decreae work input per unit ma), two-tage expanion with reheating (to increae work output per unit ma), and regeneration (to decreae external energy tranfer into the cycle), power output, thermal efficiency, and back work ratio become nearly comparable to the ientropic imple cycle performance. (c) T- diagram T (K) T 6 = T 8 = 1400 p 4 = p 5 = p 6 = 1000 kpa 6 p 2 = p 3 = p 7 = p 8 = 316 kpa 8 T 7 = T 9 = T 5 = T 10 = 587 T 2 = T 4 = p 1 = p 9 = p 10 = 100 kpa T 1 = T 3 = (/-K) Note: Preure ratio acro both tage of compreor and turbine are equal
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