The Carnot Gas Power Cycle Gas Power Cycles 1 2 : Reversible, isothermal expansion at T H 2 3 : Reversible, adiabatic expansion from T H to T L 3 4 : Reversible, isothermal compression at T L Dr. Md. Zahurul Haq 4 1 : Reversible, adiabatic compression from T L to T H Professor Department of Mechanical Engineering Bangladesh University of Engineering & Technology BUET) Dhaka-1000, Bangladesh T222 = q 12 = T H s 2 s 1 ) = T H s q out = q 34 = T L s 3 s 4 ) = T L s zahurul@me.buet.ac.bd http://teacher.buet.ac.bd/zahurul/ T 1 2 w net = q net = q out = T H T L ) s η th = wnet = T H T L ) s T H s = 1 T L T H TH ME 6101: Classical Thermodynamics TL T194 4 s 3 s = η th,carnot = 1 T L T H η th =: T H AND/OR T L. In Boolean Logic: η th = T H + T L c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 1 / 20 c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 2 / 20 Air-Standard Cycle Assumptions The working fluid is air, which continuously circulates in a closed loop and always behaves as an ideal gas. All the processes that make up the cycle are internally reversible. The combustion process is replaced by a heat-addition process from an external source. Overview of Reciprocating R/C) Engines TDC Top Dead Centre, BDC Bottom Dead Centre Compression Ratio r = Vmax V min = V BDC V TDC Displacement Volume V d = π 4 S B2 = V max V min Mean Effective Pressure MEP = W net V d T225 T223 T224 The exhaust process is replaced by a heat-rejection process that restores the working fluid to its initial state. Air has constant specific heats determined at room temperature. = A cycle for which the air-standard assumptions are applicable is frequently referred to as an air-standard cycle. c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 3 / 20 T226 T227 c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 4 / 20
The : Ideal Cycle for SI Engines 1 2 : Isentropic compression 2 3 : Reversible, Constant-volume heat addition 3 4 : Isentropic expansion 4 1 : Reversible, Constant-volume heat rejection T229 Isentropic processes: 1 2 & 3 4. Also, V 2 = V 3 & V 4 = V 1 T228 c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 5 / 20 r k ) k 1 ) k 1 T1 = V2 V 1 = V3 V 4 = T 4 T1 = r k 1 & T3 = T4 = u 3 u 2 = c v ) : q out = u 4 u 1 = c v T 4 ) η th,otto = wnet = 1 qout = 1 T2 [/ 1 [T 4/ 1 = 1 1 r k 1 η th,otto = 1 1 r k 1 = compression ratio, = ratio of specific heats, c p /c v c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 6 / 20 T230 Thermal efficiency of the ideal Otto cycle as a function of compression ratio k = 1.4). T231 The thermal efficiency of the Otto cycle increases with the specific heat ratio, k of the working fluid. Example: Otto cycle: P 1 = 0.1 MPa, = 300 K, r = 10, = 1800 kj/kg. P 2 /P 1 = r k P 2 = 2.511 MPa / = r k 1 = 753.6 K v = RT/P v 1 = 0.861 m 3 /kg, v 2 = 0.0861 m 3 /kg = c v ) = 3261.4 K v 2 = v 3, v 4 = v 1 T 4 = v3 v 4 ) k 1 = 1298.4 K w net = q out = 1083.4 m 3 /kg η th,otto = 1 1 r k 1 = 0.602 MEP = wnet v 1 v 2 = 1.398 MPa c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 7 / 20 c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 8 / 20
The : Ideal Cycle for CI Engines For same r: η th,otto > η th,diesel. Diesel engines operate at much higher compression ratios, and thus are usually more efficient than SI-engines. T232 T233 Isentropic processes: 1 2 & 3 4. Cut-off ratio, r c = V3 V 2, and V 4 = V 1. = h 3 h 2 = c p ) : q out = u 4 u 1 = c v T 4 ) η th,diesel = wnet = 1 qout T275 [ = 1 T2 [/ 1 k [T = 1 1 r k 4/ 1 r k 1 kr c 1) η th,diesel = 1 1 r k 1 [ r k kr c 1) c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 9 / 20 T234 T235 Dual Cycle: Two heat transfer processes, one at constant volume and one at constant pressure. c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 10 / 20 Example: Diesel cycle: P 1 = 0.1 MPa, = 300 K, r = 20, = 1800 kj/kg. P 2 /P 1 = r k P 2 = 6.629 MPa / = r k 1 = 994.3 K v = RT/P v 1 = 0.861 m 3 /kg, v 2 = 0.0430 m 3 /kg = c p ) = 2785.6 K P 2 = P 3 v 3 = v 2, v 4 = v 1 T 4 = v3 v 4 ) k 1 = 1270.0 K w net = q out = 1104.5 m 3 /kg η th,diesel = wnet = 0.613 r c = v 3 /v 2 = 2.80 [ η th,diesel = 1 1 r k r k 1 kr c 1) = 0.613 MEP = wnet v 1 v 2 = 1.355 MPa c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 11 / 20 Stirling & Ericsson Cycles Regenerative Cycles Stirling and Ericsson cycles involve an isothermal heat-addition at T H and an isothermal heat-rejection at T L. They differ from the Carnot cycle in that the two isentropic processes are replaced by two constant-volume regeneration processes in the Stirling cycle and by two constant-pressure regeneration processes in the Ericsson cycle. Both cycles utilize regeneration, a process during which heat is transferred to a thermal energy storage device called a regenerator) during one part of the T276 cycle and is transferred back to the working fluid during another part of the cycle. Stirling cycle is made up of four totally reversible processes: 1 2: Isothermal expansion heat addition from the external source) 2 3: Isochoric regeneration internal HT from working fluid to regenerator) 3 4: Isothermal compression heat rejection to the external sink) 4 1: Isochoric regeneration internal HT from regenerator back to fluid) c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 12 / 20
Regenerative Cycles Brayton Cycle: Ideal Cycle for Gas Turbines T237 An open-cycle gas-turbine T236 c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 13 / 20 T238 A closed-cycle gas turbine c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 14 / 20 T239 = h 3 h 2 = c p ) q out = h 4 h 1 = c p T 4 ) r p P2 P 1 T2 = r p k 1)/k = T3 T 4 η th,brayton = 1 qout = 1 T1T4/T1 1) / 1) = η th,brayton = 1 1 r k 1)/k p w net = c p [ T 4 )+ ) [ [ wnet 1 c p = 1 +T r p k 1)/k 1 For max. work: w net r p = 0: 1 r k 1)/k p = r p = T3 ) k/2k 1) : for max. wnet If = 1000 K, = 300 K r p = 8.2. c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 15 / 20 T240 T241 Brayton Cycle with Regeneration T244 T245 q regen,act = h 5 h 2 q regen,max = h 4 h 2 ǫ qregen,act q regen,max = h5 h2 h 4 h 2 T5 T2 T 4 ) η th,regen = 1 T1 r p k 1)/k ǫ = effectiveness. c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 16 / 20
T243 T247 Example: Gas Turbine with r p = 8: η c = wc,s w c,a T2s T1 a : η t = wt,a w t,s T3 T2a T 4s a) η isen = 1.0 & ǫ = 1.0: s = 543.4 K, T 4s = 717.7 K w c,s = 244.6 kj/kg, w t,s = 585.2 kj/kg = 760.2 kj/kg, η th = 0.448 b) η isen = 0.8 & ǫ = 1.0: a = 604.3 K, T 4a = 834.1 K w c,a = 305.7 kj/kg, w t,a = 468.1 kj/kg = 699.0 kj/kg, η th = 0.232 c) η isen = 0.8 & ǫ = 0.8: ǫ = qregen,act q regen,max T5 T2a T 4a a T 5 = 788.2 K = 514.3 kj/kg, η th = 0.317 c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 17 / 20 Turbines with Reheat, Compressors with Intercooling c T246 T252 T250 Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 18 / 20 Brayton Cycle with Intercooling, Reheating & Regeneration Example: GT with reheating & intercooling: r p = 8, η isen = 1.0, ǫ = 1.0. In case of staging, for min. compressor work or for max. turbine work: P i = P min P max w turb = h 6 h 7 )+h 8 h 9 ) w comp = h 2 h 1 )+h 4 h 3 ) T249 c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 19 / 20 T251 w net = w turb w comp, bwr = wcomp w turb Without regen: w turb = 685.28 kj/kg w comp = 208.29 kj/kg = h 6 h 4 )+h 8 h 7 ) = 1334.30 kj/kg = η th = 0.358 bwr = 0.304 With regen: turbine and compressor works remains unchanged. = h 6 h 5 )+h 8 h 7 ) = 685.28 kj/kg = η th = 0.696 bwr = 0.304 c Dr. Md. Zahurul Haq BUET) Gas Power Cycles ME 6101 2017) 20 / 20