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
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 = 300.1 /, 1 0 = 1.703 /-K, p r1 = 1.386 State 2: p 2 = p 2 = 1000 kpa, 2 = 1 (ientropic) 0 0 p2 0 0 p2 0 2 1 Rair 2 1 Rair p1 p1 0 1000 kpa 2 1.703 0.287 2.364 -K -K 100 kpa -K pr2 p2 p2 1000 kpa Alternatively: pr2 pr1 1.386 13.86 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 = 3.362 /-K, p r3 = 450.5 State 4: p 4 = p 4 = 1000 kpa, 4 = 3 (ientropic) 0 0 p4 0 0 p4 04 3 Rair 4 3 Rair p3 p3 0 100 kpa 4 3.362 0.287 2.701 -K -K 1000 kpa -K pr4 p4 p4 100 kpa Alternatively: pr4 pr3 450.5 45.05 pr3 p3 p3 1000 kpa Ideal ga table for air: T4 788 K, h4 808.7 (a) Conidering energy balance for the compreor ( I), the ientropic power for the compreor: W comp, m air h1h2 6 300.1580 1679.4 kw ; negative ign indicate power input Conidering energy balance for the turbine ( II), the ientropic power for the turbine: W turbine, m air h3h4 6 1515 808.7 4237.8 kw ; poitive ign indicate power output Net power produced by the power plant: W W turbine W comp W 2558 kw
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 h3h2 6 1515 580 5610 kw 2558 kw Thermal efficiency of the cycle: W thermal 0.456 Q 5610 kw thermal 45.6% (c) Back work ratio of the power plant: W comp, 1679.4 kw bwr W 4237.8 kw bwr 0.396 turbine, (d) Conidering irreverible compreion through the compreor. State 2: p 2 = 1000 kpa, comp = 80% wientropic h1 h2 300.1 580 comp. 0.8 h2 650.1 h1h2 300.1h2 Ideal ga table for air: T2 640 K Conidering irreverible expanion through the turbine. State 4: p 4 = 100 kpa, turbine = 80% h3 h4 1515 h4 turbine 0.8 h4 950 wientropic h3h4 1515 808.7 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 h1h26 300.1650.1 2100 kw ; negative ign indicate power input Conidering energy balance for the turbine ( II), the actual power for the turbine: W turbine, actual m air h3h46 1515 950 3390 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 h3h26 1515 650.1 5189.4 kw
ME-200 Fall 2017 HW-38 4/4 1290 kw Thermal efficiency of the cycle: W thermal 0.248 Q 5189.4 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
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 = 300.1 /, 1 0 = 1.703 /-K, p r1 = 1.386 m 9
ME-200 Fall 2017 HW-39 2/4 State 2: p 2 = p 2 = 316 kpa, 2 = 1 (ientropic) 0 0 p2 0 0 p2 0 2 1 Rair 2 1 Rair p1 p1 0 316 kpa 2 1.703 0.287 2.033 -K -K 100 kpa -K pr2 p2 p2 316 kpa Alternatively: pr2 pr1 1.386 4.38 pr1 p1 p1 100 kpa Ideal ga table for air: T2 416 K, h2 417.3 State 2: p 2 = 316 kpa, comp,1 = 80% wientropic h1 h2 300.1 417.3 comp,1 0.8 h2 446.6 h1h2 300.1h2 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 = 300.1 /, 3 0 = 1.703 /-K, p r3 = 1.386 State 4: p 4 = p 4 = 1000 kpa, 4 = 3 (ientropic) 0 0 p4 0 0 p4 04 3 Rair 4 3 Rair p3 p3 0 1000 kpa 4 1.703 0.287 2.033 -K -K 316 kpa -K pr4 p4 p4 1000 kpa Alternatively: pr4 pr3 1.386 4.38 pr3 p3 p3 316 kpa Ideal ga table for air: T4 416 K, h4 417.3 State 4: p 4 = 1000 kpa, comp,2 = 80% wientropic h3 h4 300.1 417.3 comp,2 0.8 h4 446.6 h3h4 300.1h2 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 = 3.362 /-K, p r6 = 450.5
ME-200 Fall 2017 HW-39 2/4 State 7: p 7 = p 7 = 316 kpa, 7 = 6 (ientropic) 0 0 p7 0 0 p7 07 6 Rair 7 6 Rair p6 p6 0 316 kpa 2 1.703 0.287 3.031 -K -K 100 kpa -K pr7 p7 p7 316 kpa Alternatively: pr7 pr6 450.5 142.6 pr 6 p6 p6 1000 kpa Ideal ga table for air: T7 1058 K, h7 1112.7 State 7: p 7 = 316 kpa, turbine,1 = 80% h6 h7 1515 h7 turbine,1 0.8 h7 1193 w h h 1515 1112.7 ientropic 6 7 Ideal ga table for air: T7 1128 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 = 3.362 /-K, p r8 = 450.5 State 9: p 9 = p 9 = 100 kpa, 9 = 8 (ientropic) 0 0 p9 0 0 p9 09 8 Rair 9 8 Rair p8 p8 0 316 kpa 9 1.703 0.287 3.031 -K -K 100 kpa -K pr9 p9 p9 100 kpa Alternatively: pr9 pr8 450.5 142.6 pr8 p8 p8 316 kpa Ideal ga table for air: T9 1058 K, h9 1112.7 State 9: p 9 = 316 kpa, turbine,2 = 80% h8 h9 1515 h9 turbine,2 0.8 h9 1193 wientropic h8 h9 1515 1112.7 Ideal ga table for air: T9 1128 K State 10: p 10 = p 9 = 316 kpa (no preure drop) Conidering energy balance for the regenerator: h5 h4 h9 h10 h10 593.4 Ideal ga table for air: T10 587 K
ME-200 Fall 2017 HW-39 2/4 State Abolute Preure (kpa) Temperature (K) Specific Enthalpy (/) 1 100 300 300.1 2 316 445 446.6 3 316 300 300.1 4 1000 445 446.6 5 1000 1000 1046 6 1000 1400 1515 7 316 1128 1193 8 316 1400 1515 9 100 1128 1193 10 100 587 593.4 (b) Conidering energy balance for the two compreor: W comp m air h1h2h3h4 6 2300.1446.6 1758 kw ; negative ign indicate power input Conidering energy balance for the two turbine: W turbine m air h6 h7h8 h9 6 21515 1193 3864 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 h7 6 1515 10461515 1193 4746 kw 2106 kw Thermal efficiency of the cycle: W thermal 0.443 Q 4746 kw thermal 44.3% in Back work ratio of the power plant: W comp 1758 kw bwr W 3864 kw bwr 0.455 turbine
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 0.396 (~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 = 1193 7 9 T 5 = 1000 5 7 9 T 10 = 587 T 2 = T 4 = 445 4 4 2 2 10 p 1 = p 9 = p 10 = 100 kpa T 1 = T 3 = 300 3 1 (/-K) Note: Preure ratio acro both tage of compreor and turbine are equal