Teaching schedule Session *15 18 19 21 22 24 Topics 5. Gas power cycles Basic considerations in the analysis of power cycle; Carnot cycle; Air standard cycle; Reciprocating engines; Otto cycle; Diesel cycle; Stirling cycle; Brayton cycle; Second law analysis of gas power cycles 6. Vapor and combined power cycles Carnot vapor cycle; Rankine cycle; Deviation of actual vapor power cycles; Cogeneration; Combined gas vapor power cycles 7. Refrigeration cycles Refrigerators and heat pumps; Reversed Carnot cycle; Ideal vapor compression refrigeration cycle; Actual vapor compression refrigeration cycle
Objectives Evaluate the performance of gas power cycles for which the working fluid remains a gas throughout the entire cycle. Develop simplifying assumptions applicable to gas power cycles. Analyze both closed and open gas power cycles. Solve problems based on the Otto, Diesel, Stirling, and Ericsson cycles. Solve problems based on the Brayton cycle; the Brayton cycle with regeneration; and the Brayton cycle with intercooling, reheating, and regeneration. Perform second-law analysis of gas power cycles.
Gas power cycle Carnot cycle Otto cycle: ideal cycle for spark-ignition engines Diesel cycle: ideal cycle for compression-ignition engines Stirling and Ericsson cycles Brayton cycle: ideal cycle for gas turbine engine If the Carnot cycle is the best possible cycle, why do we not use it as the model cycle for all the heat engine? Analysis of power cycles Neglect friction Neglect the required time for establishing the equilibrium state
Carnot Cycle The Carnot cycle gives the most efficient heat engine that can operate between two fixed temperatures T H and T L ; it is independent of the type of working fluid and can be closed or steady flow. wnet Carnot Cycle η th = qin Process Description 1-2 Isothermal heat addition qin qout 2-3 Isentropic expansion η th = qin 3-4 Isothermal heat rejection 4-1 Isentropic compression in H ( ) q = T s s 2 1 T, = 1 T η th Carnot L H ( ) q = T s s out L 3 4 Area, which is enclosed by cyclic curve presents net work or heat transfer during the cycle.
Reciprocating engine TDC = Top Dead Center BDC = Bottom Dead Center Clearance volume = minimum volume formed in cylinder Compression ratio, r = Max. volume Min. volume V = V BDC TDC
Reciprocating engine Work = F s W net = MEP Piston area Stroke MEP = Mean Effective Pressure W net = MEP Displacement volume MEP net = V max W V min Larger value of MEP gives more net work per cycle, thus perform better
Gas power cycles Analysis of power cycles The working fluid remains a gas throughout the entire cycle. Neglect friction Neglect the required time for establishing the equilibrium state Air-standard 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. The exhaust process is replaced by a heat-rejection process that restores the working fluid to its initial state.
Otto cycle An ideal cycle for spark-ignition engines Nikolaus A. Otto (1876) built a 4-stoke engine Ideal Otto cycle Actual 4-stoke ignition engine
Spark plug
The air-standard Otto cycle is the ideal cycle that approximates the spark-ignition combustion engine. Process Description 1-2 Isentropic compression 2-3 Constant volume heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection The T-s and P-v diagrams are The air-standard Otto cycle
Thermal Efficiency of the Otto cycle: η = th W Q net in Q net in out out = = = 1 Q in Q Q Q Q in Q in Apply 1 st law closed system to process 2-3, constant volume heat addition E E =ΔE Q W =ΔU W = W + W in,23 out,23 23 net,23 net,23 23 net,23 other,23 b,23 Q = ΔU net, 23 23 Thus, for constant specific heats, Q = Q = mc ( T T ) net,23 in v 3 2 3 = 0+ PdV = 0 2
Apply 1 st law closed system to process 4-1, constant volume heat rejection E E =ΔE in,41 out,41 41 Q W =ΔU net,41 net,41 41 Thus, for constant specific heats, Q W = W + W = 0+ PdV = 0 net,41 other,41 b,41 = ΔU net, 41 41 Q Q mc T T net,41= out = ( ) v 1 4 Q = mc ( T T ) = mc ( T T) out v 1 4 v 4 1 4 1 η th, Otto Q = 1 Q out in mcv ( T4 T1) = 1 mc ( T T ) v 3 2 The thermal efficiency becomes ( T4 T1) η th, Otto = 1 ( T T ) 3 2 T1( T4 / T1 1) = 1 T ( T / T 1) 2 3 2
Processes 1-2 and 3-4 are isentropic, so Since V 3 = V 2 and V 4 = V 1, T T T T = or = T T T T 2 3 4 3 1 4 1 2 V T V = T 3 2 4 1 The Otto cycle efficiency becomes ( T4 T1) η th, Otto = 1 ( T T ) 3 2 T1( T4 / T1 1) = 1 T ( T / T 1) 2 3 2 η th Otto, = T T 1 1 2
Since process 1-2 is isentropic, where the compression ratio is r = V 1 /V 2 and 1, = 1 k r η th Otto 1
Thermal efficiency of the ideal Otto cycle - Increasing the compression ratio increases the thermal efficiency. - But there is a limit on r depending upon the fuel. Fuels under high temperature resulting from high compression ratios will prematurely ignite, causing knock (auto ignition).
Example An Otto cycle having a compression ratio of 9:1 uses air as the working fluid. Initially P 1 = 95 kpa, T 1 = 17 o C, and V 1 = 3.8 liters. During the heat addition process, 7.5 kj of heat are added. Determine all T's, P's, η th, the back work ratio, and the mean effective pressure. Assume constant specific heats with C v = 0.718 kj/kg K, k = 1.4. (Use the 300 K data from Table) Process 1-2 is isentropic; therefore, recalling that r = V 1 /V 2 = 9, Find T 2 Find P 2
Find T 3 Q = mc ( T T ) in v 3 2 T = T + 3 2 q C in v Let q in = Q in / m and m = V 1 /v 1 v 1 = = RT1 P1 kj 0. 287 ( 290 K) kg K 95kPa 3 mkpa kj q in Q m Q v in = = in V = = 3 m 0875. kg 75. kj 38. 10 m kj 1727 kg 1 1 3 3 3 m = 0875. kg T q = T + C 3 2 in v = 698. 4K + = 31037. K kj 1727 kg kj 0. 718 kg K
Using the combined gas law (V 3 = V 2 ) Find P 3 P P T 3 = = 915. T 3 2 2 MPa Process 3-4 is isentropic; therefore, Find T 4 k 1 k 1 1.4 1 V 3 1 1 T4 = T3 = T3 = (3103.7) K V4 r 9 = 1288.8 K Find P 4
In order to find η th, we must know net work first wnet = qnet = qin qout Process 4-1 is constant volume. So the first law for the closed system gives, on a mass basis, Find q out Qout = mcv ( T4 T1) Qout qout = = Cv ( T4 T1) m kj = 0.718 (1288.8 290) K kg K kj = 717.1 kg The first law applied to the cycle gives (Recall Δu cycle = 0) w = q = q q net net in out = (1727 717.4) = 1009.6 kj kg kj kg
The thermal efficiency is η th, Otto w net = = qin kj 1009. 6 kg kj 1727 kg = 0585. or 585%. The mean effective pressure is MEP Wnet wnet = = V V v v max min max min = v w net v 1 2 = w net v (1 v / v ) 1 2 1 = 1 w net v (1 1/ r) kj 1009.6 3 kg mkpa = = 1298 3 m 1 kj 0.875 (1 ) kg 9 kpa
The back work ratio is BWR = w comp w exp Δu12 Cv ( T2 T1) ( T2 T1 ) = = = Δu C ( T T ) ( T T ) = 34 v 3 4 3 4 0.225 or 22.5%
Diesel Cycle
Air-Standard Diesel Cycle The air-standard Diesel cycle is the ideal cycle that approximates the Diesel combustion engine Process Description 1-2 Isentropic compression 2-3 Constant pressure heat addition 3-4 Isentropic expansion 4-1 Constant volume heat rejection The P-v and T-s diagrams are
Thermal efficiency of the Diesel cycle W net η th, Diesel = = 1 Qin Q Q out in Thus, for constant specific heats Find Q in Apply the first law closed system to process 2-3, P = constant. E E =ΔE Q W =ΔU net,23 net,23 Q = Δ U + P ( V V ) net,23 23 2 3 2 in out Wnet,23 = Wother,23 + Wb,23 = 0 + PdV 3 2 ( V ) = P V 2 3 2 Q = Q = mc ( T T ) + mr( T T ) net,23 in v 3 2 3 2 Q mc T T in = ( ) p 3 2
Find Q out Apply the first law closed system to process 4-1, V = constant E E =ΔE in out Q W =ΔU net,41 net,41 41 Wnet,41 = Wother,41 + Wb,41 = 0 + PdV Thus, for constant specific heats Q = Q = mc ( T T ) net,41 out v 1 4 The thermal efficiency becomes, 1 out 1 4 Q net, 41 41 Q = mc ( T T ) = mc ( T T) η out v 1 4 v 4 1 th Diesel = ΔU Q = Qin mcv ( T4 T1) = 1 mc ( T T ) p 3 2
η th, Diesel = 1 Cv ( T4 T1) C ( T T ) p 3 2 = 1 1 T1( T4 / T1 1) kt( T / T 1) 2 3 2 PV T T T 3 3 3 2 3 PV 2 2 = where P = T V3 = = V 2 2 r c P 3 2 where r c is called the cutoff ratio, defined as V 3 /V 2, and is a measure of the duration of the heat addition at constant pressure. Since the fuel is injected directly into the cylinder, the cutoff ratio can be related to the number of degrees that the crank rotated during the fuel injection into the cylinder.
η th, Diesel T( T / T ) = 1 1 1 1 4 1 k T ( T / T 1) 2 3 2 k T rc = 1 1 1 1 k T ( r 1) 2 k 1 rc 1 = 1 k 1 r kr ( 1) c c T T V = = V 3 3 2 2 r c When r c > 1 for a fixed r, η th,diesel < η th,otto But, since r Diesel > r Otto, η th,diesel > η th,otto.
Example An ideal Diesel cycle with air as the working fluid has a compression ratio of 18 and a cutoff of 2. At the beginning of the compression process, the working fluid is at 14.7 psia, 80 O F,and 117 in 3. Utilizing the cold-airstandard assumptions, determine (a) the temperature and pressure of air at the end of each process, (b) the net work output and the thermal efficiency, (c) the mean effective pressure Properties R = 0.3704 psia. ft 3 /lbm.r At room temp., c p = 0.204 Btu/lbm.R c v = 0.171 Btu/lbm.R k = 1.4 From compression ratio, r = 18 V V r r V = = V V V max 1 min 2 117 in 18 3 1 2 = = = 6.5 in 3
From cutoff ratio, r c = 2 r c = V V 3 2 ( )( ) V 3 = 2 6.5 in = 13 in 3 3 V4 = V1 = 117 in 3 T 2 and P 2 Process 1-2 (isentropic compression of an ideal gas, constant specific heat) k 1 V 1 T2 = T1 V 2 P V 1 2 = P1 V 2 k ( )( ) 1.4 1 = 540 R 18 = 1716 R ( psia)( ) 1.4 = 14.7 18 = 841 psia
T 3 and P 3 Process 2-3 (constant pressure heat addition to an ideal gas) P3 = P2 = 841 psia PV 2 2 PV 3 3 T T 3 = 3 2 ( )( ) 2 3 V T = T = 1716 R 2 = 3432 R V 2 T 4 and P 4 Process 3-4 (isentropic expansion of an ideal gas, constant specific heats) k 1 V 3 T4 = T3 V 4 V P P V 3 4 = 3 4 k 1.4 1 13 = ( 3432 R) = 1425 R 117 1.4 13 = ( 841 psia) = 38.8 117 psia
(b) Net work Wnet = Qin Qout in ( ) ( ) Q = m h h = mc T T 3 2 p 3 2 ( )( ) 3 ( 0.3704 psia.ft /lbm.r )( 540 R) 14.7 psia 117 1ft = = = 3 PV 1 1 m RT 3 1 1728 in 0.00498 lbm ( )( )( ) Q = 0.00498 0.24 3432 1716 = 2.051 Btu in out ( ) ( ) Q = m u u = mc T T 4 1 v 4 1 ( )( )( ) Q = 0.00498 0.171 1425 540 = 0.754 Btu out W = 2.051 0.754 = 1.297 Btu net
Thermal efficiency (c) Mean effective pressure Wnet 1.297 η th, Diesel = = = 0.632 Q 2.051 in W W MEP net net = = V max V V V min 1 2 1297 Btu 778.17 lbf.ft 12 in = = 110 psia 117-6.5 in 1 Btu 1 ft ( ) 3
Summary of Diesel cycle Cutoff ratio, r c Defined as V 3 /V 2, and is a measure of the duration of the heat addition at constant pressure. Thermal efficiency η = 1 1 r th, Diesel k 1 k rc 1 k( r 1) c
Stirling engine Robert Stirling patented his Heat Economiser in 1816 The power piston compresses the enclosed air in the cold end of the displacer cylinder. The displacer then shifts the air from cold to hot chambers. The piston is driven back, the power stroke, by the air expanding in the hot end.
Stirling engine
Ericsson Engine
Stirling and Ericsson cycles -Involve an isothermal heat addition process at T H and isothermal heat rejection process at T L - 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 cycle and is transferred back to the working back to the working fluid during the other part of the cycle.
Ideal stirling cycle 1-2 T = constant expansion (heat addition from the external source) 2-3 v = constant regeneration (internal heat transfer from the working fluid to the regenerator 3-4 T = constant compression (heat rejection to the external sink) 4-1 v = constant regeneration (internal heat transfer from the regenerator back to the working fluid)
Stirling and Ericsson cycles differ from the Carnot cycle in that two isentropic processes are replaced by two constant volume regeneration processes in Stirling cycle and by two constant pressure regeneration processes in Ericsson cycle.
Thermal efficiency of Stirling and Ericsson cycles Process 1-2 and 3-4 q = TΔs Stirling cycle Ericsson cycle Entropy change of an ideal gas during an isothermal process Te Pe Pe Δ s = cp ln Rln = Rln Ti Pi Pi P P qin = TH ( s2 s 2 1 1) = TH Rln = RTH ln P1 P2 P P qout = TL s4 s3 = TL Rln = RTL ln P P q RTL ln ( P4 / P out 3) T η th = 1 = 1 = 1 q RT ln P / P T ( ) 4 4 η =η =η = L th, Stirling th, Ericsson th, Carnot 1 TH T 3 3 ( ) in H 1 2 H L
Merit and demerit of Stirling and Ericsson engines Demerit Difficult to achieve in practice They involve heat transfer through a differential temperature difference in all components including regenerator. Need large surface for allowing heat transfer or need long time for the process. Pressure losses in the generator are considerable. Merit Due to external combustion in both cycles - A variety of fuel can be used. - There are more time for combustion, resulting in more complete combustion (less pollution) Because these cycles operate on closed cycles, - A working fluid (e.g.,h and He) that has most desirable characteristics (stable, chemically inert, high thermal conductivity) can be used.
Brayton Cycle The Brayton cycle was first proposed by George Brayton around 1870. The Brayton cycle is the air-standard ideal cycle approximation for the gas turbine engine. This cycle differs from the Otto and Diesel cycles - Processes making the cycle occur in open systems or control volumes. A open cycle gas turbine engine A closed cycle gas turbine engine with air standard assumptions
Brayton Cycle 1-2 Isentropic compression (in a compressor) 2-3 Constant pressure heat addition 3-4 Isentropic expansion (in a turbine) 4-1 Constant pressure heat rejection
Thermal efficiency of the Brayton cycle W net η th, Brayton = = 1 Qin Q Q out in Process 2-3 for P = constant (no work), steady-flow, and neglect changes in kinetic and potential energies. The conservation of mass gives m in = m out m = m = m 2 3 For constant specific heats, the heat added per unit mass flow is E in = E out mh + Q = mh 2 2 in 3 3 Q in = m ( h3 h2) Q = mc ( T T ) in p 3 2 Qin q = in Cp ( T3 T2) m =
Process 4-1; constant specific heats Q out = m ( h4 h1) Q = mc ( T T) q out out p 4 1 Q out = = Cp( T4 T1) m The thermal efficiency becomes η th, Brayton Q 1 1 Q out = = in p q q Cp ( T4 T1) = 1 C ( T T ) 3 2 ( T4 T1) η th, Brayton = 1 ( T T ) 3 2 out in T1( T4 / T1 1) = 1 T ( T / T 1) 2 3 2
Processes 1-2 and 3-4 are isentropic Since P 3 = P 2 and P 4 = P 1, T T T T = or = T T T T 2 3 4 3 1 4 1 2 The Brayton cycle efficiency becomes T1, = 1 T η th Brayton 2 Process 1-2 is isentropic, 1 η th, Brayton = 1 ( k 1)/ k rp where the pressure ratio is r p = P 2 /P 1
Brayton cycle 1 η th, Brayton = 1 ( k 1)/ k rp
Example The ideal air-standard Brayton cycle operates with air entering the compressor at 95 kpa, 22 o C. The pressure ratio r p is 6:1 and the air leaves the heat addition process at 1100 K. Determine the compressor work and the turbine work per unit mass flow, the cycle efficiency, the back work ratio, and compare the compressor exit temperature to the turbine exit temperature. Assume constant properties. For steady-flow and neglect changes in kinetic and potential energies to process 1-2 for the compressor. Note that the compressor is isentropic. E in = E out mh + W = mh 1 1 comp 2 2 The conservation of mass gives m in = m out m = m = m 1 2
For constant specific heats, the compressor work per unit mass flow is W = m ( h h ) W = mc ( T T) w comp comp comp 2 1 p 2 1 W comp = = Cp ( T2 T1) m Since the compressor is isentropic w = C ( T T) comp p 2 1 kj = 1005. ( 492. 5 295) K kg K kj = 19815. kg
Process 3-4; constant specific heats W turb = m ( h3 h4) W = mc ( T T ) w turb turb p 3 4 W turb = = Cp ( T3 T4) m Since process 3-4 is isentropic Since P 3 = P 2 and P 4 = P 1, T T 4 3 1 = r p ( k 1)/ k ( k 1)/ k (1.4 1) /1.4 1 1 T4 = T3 = 1100K = 659.1K r p 6
kj wturb = Cp ( T3 T4) = 1005. ( 1100 659. 1) K kg K kj = 442. 5 kg Process 2-3 m 2 = m 3 = m mh + Q = mh Qin q = in h3 h2 m = kj = Cp ( T3 T2) = 1.005 (1100 492.5) K = 609.6 kg K The net work done by the cycle is w = w w net turb comp = ( 442. 5 19815. ) = 2 2 in 3 3 244. 3 kj kg kj kg kj kg
The cycle efficiency becomes kj w 244.3 net kg η th, Brayton = = = 0.40 q kj in 609.6 kg or 40% The back work ratio is defined as w w in comp BWR = = w out w turb kj 198.15 kg = = 0.448 kj 442.5 kg Note that T 4 = 659.1 K > T 2 = 492.5 K, or the turbine outlet temperature is greater than the compressor exit temperature.
What happens to η th, w in /w out, and w net as the pressure ratio r p is increased? (combustor) w η = = 1 net th, Brayton ( k 1)/ k qin rp 1 Efficiency of Brayton cycle depends on Specific heat ratio, k Pressure ratio, P 2 /P 1 or r p Limitation in an actual gas turbine cycle Max. temp. is at the outlet of combustor (T 3 ) - Limitation temp. in materials Same T max and plot w net in various r p
What happens to η th, w in /w out, and w net as the pressure ratio r p is increased? The effect of the pressure ratio on the net work done. Net work is zero when wnet = wturb wcomp p = C ( T T ) C ( T T) p 3 4 p 2 1 = CT(1 T/ T) CT( T/ T 1) k/( k 1) T 3 = 1 p = T1 r and r p 3 4 3 p 1 2 1 1 = CT ( k 1)/ k p 3(1 ) CT ( 1)/ p 1( rp 1) k k rp
Find Max. compression ratio 1 w C T C T r ( k 1)/ k net = p 3(1 ) ( 1)/ p 1( p 1) k k rp For fixed T 3 and T 1, the pressure ratio that makes the work a maximum is obtained from: dw dr net p = 0 Let X = r p (k-1)/k Solving for X w dw dx net 1 = C T3( 1 ) CT1( X 1) X net p p 2 = CT[ 0 ( 1) X ] CT[ 1 0] = 0 p 3 p 1
Then, the r p that makes the work a maximum for the constant property case and fixed T 3 and T 1 is For the ideal Brayton cycle T T T T = or = T T T T 2 3 4 3 1 4 1 2 And show that the following results are true. 2 4 When r p = r p, max work, T 4 = T 2 When r p < r p, max work, T 4 > T 2 When r p > r p, max work, T 4 < T 2 2 4
Example: ideal gas turbine Air standard assumptions The cycles does not involve any friction. Therefore, the working fluid does not experience any pressure drop as it flow in pipe or devices such as heat exchanger. All expansion and compression processes take place in a quasiequilibrium manner. The pipe connecting the various components of a system are well insulated, and heat transfer through them is negligible.
Information Ideal Brayton cycle Pressure ratio,r p = 8 Temperature at the inlet compressor = 300 K Temperature at the inlet turbine = 1300 K From Table A-17 From table and s 1 = s 2
From Table A-17 From table and s 3 = s 4 (b) Back work ratio
(c) Thermal efficiency η = th w q net in = w w turb, out comp, in h h 3 2 362.4 kj/kg = = 0.426 851.62 kj/kg which is sufficiently close to the value obtained by accounting for the variation of specific heat with temperature
Deviation of actual gas turbine cycles from idealized ones Pressure drop during the heat-addition and heat rejection Due to irreversiblility, actual work input to the compressor is more and the actual work output from the turbine is less Isentropic efficiency Compressor Turbine η = c η = T ws h h w h h a s 2s 1 2a 1 wa h h w h h 3 4a 3 4s Pressure & Entropy generation in actual processes
Example: actual gas turbine Assuming a compressor efficiency of 80 percent and a turbine efficiency of 85 percent, determine (a) the back work ratio, (b) the thermal efficiency, and (c) the turbine exit temperature of the gas turbine cycle discussed in the previous example (ideal gas turbine) From previous example (ideal gas turbine) w scomp, = 244.16 kj/kg w sturb, = 606.6 kj/kg r sbw, = 0.403 η = 0.426 th For actual gas turbine turbine w =η w turb, out T s, turb Compressor w scomp, w comp, in = η c 244.16 kj/kg = = 0.8 ( 0.85)( 606.6 kj/kg) 515.61 kj/kg = = 305.2 kj/kg
Back work ratio r bw = w w comp, in turb, out 305.2 = = 515.61 0.592 back work ratio in actual case is larger than that in ideal case, where r s,bw = 0.403 (b) Thermal efficiency η = th w q net in = w w turb, out comp, in h h 3 2a 515.61 605.39 = = 1395.97 605.36 0.266 wcomp, in = h2a h1 h2a = h1 + w, comp in = 300.19 + 305.2 = 605.36 kj / kg Thermal efficiency in actual case is lower than that in ideal case, where η th = 0.426
(c) Air temperature at the turbine exit Apply energy balance on turbine w = h h h = h w turb, out 3 4a 4a 3 turb, out h4 a = 1395.97 515.61 = 880.36 kj / kg Find T 4a from Table T 4a = 853 K The temperature at turbine exit is considerably higher than that at the compressor exit (T2a = 598 K), which suggests the use of regeneration to reduce fuel cost
The Brayton cycle with regeneration Temperature of the exhaust gas from turbine is higher than the temperature of the air leaving the compressor. Use regenerator or recuperator as an heat exchanger
Actual heat transfer from exhaust gas to the air q = h h regen, act 5 2 Max. heat transfer from exhaust gas to the air q = h h = h h regen,max 5 2 4 2 T 4 is temp from exhaust turbine gas Extent to which a regenerator approaches an ideal regenerator is called the effectiveness ε, q h h ε= = q h h regen, act 5 2 regen,max 4 2
For the cold-air-standard assumptions are utilized, it reduces to ε T T T 5 2 T 4 2 Higher effectiveness requires the use of larger regenerator Due to higher price of components and larger pressure drop, the use of larger regenerator with very high effectiveness does not justify saving money Thermal efficiency of an ideal Brayton cycle with regenerator T 1 η th, regen = 1 T3 ( r ) p k k 1 T 3 is T max, and T 1 is T min
Example Use the information from previous example (actual gas turbine), and determine the thermal efficiency if a regenerator having an effectiveness of 80 percent is used. The effectiveness q h h ε= = q h h regen, act 5 2a regen,max 4a 2a 0.8 = h5 605.39 880.36 605.39 h5 = 825.37 kj / kg qin = h3 h5 = 1395.97 825.37 = 570.6 kj / kg Thus a saving of 220 kj/kg from the heat input requirement, i.e., 970.61-570.6 kj/kg
With regenerator wnet 210.41 η th, regen = = = 0.369 q 570.6 Without regenerator w net η th = = qin in 0.266
Brayton cycle with intercooling, reheating, and regeneration w = w w net turb, out comp, in Work required to compress a gas between two specified pressure can be decreased by carrying out the compression process in stages and cooling gas, i.e., multi-stage compression with intercooling. Comparison of work inputs to a single-stage compressor (1AC) and a two-stage compressor with intercooling (1ABD)
Intercooling When using multistage compression, cooling the working fluid between the stages will reduce the amount of compressor work required. The compressor work is reduced because cooling the working fluid reduces the average specific volume of the fluid and thus reduces the amount of work on the fluid to achieve the given pressure rise. To determine the intermediate pressure at which intercooling should take place to minimize the compressor work For the adiabatic, steadyflow compression process, the work input to the compressor per unit mass is vdp 4 3 2 0 4 w = v dp = v dp + v dp + v dp comp 1 3 1 2
For the isentropic compression process k k wcomp = ( Pv 2 2 Pv 1 1) + ( Pv 4 4 Pv 3 3) k-1 k-1 k kr = RT ( 2 T1) + ( T4 T3) k-1 k-1 k = RTT 1 2 T 1 + T 3 T 4 T 3 k-1 [ ( / 1) ( / 1) ] ( k 1)/ k ( k 1)/ k k P 2 P 4 = R T 1 1 + T 3 1 k-1 P 1 P 3 Notice that the fraction kr/(k-1) = C p.
For two-stage compression, let s assume that intercooling takes place at constant pressure and the gases can be cooled to the inlet temperature for the compressor, such that P 3 = P 2 and T 3 = T 1. The total work supplied to the compressor becomes To find the unknown pressure P 2 that gives the minimum work input for fixed compressor inlet conditions T 1, P 1, and exit pressure P 4, we set dw ( comp P ) 2 dp 2 = 0
This yields P = PP 2 1 4 or, the pressure ratios across the two compressors are equal. P2 P 1 P4 = = P 2 P P 4 3 Intercooling is almost always used with regeneration. During intercooling the compressor final exit temperature is reduced; therefore, more heat must be supplied in the heat addition process to achieve the maximum temperature of the cycle. Regeneration can make up part of the required heat transfer.
A gas-turbine engine with two-stage compression with intercooling, two-stage expansion with reheating, and regeneration Work output from turbine can be increased by multi-stage expansion with reheating.
Intercooling P = PP 2 1 4 Pressure ratios across the two compressors P2 P 1 P4 = = P 2 P P 4 3 Reheating P = PP 7 6 9 or the pressure ratios across the two turbines are equal. P P 6 7 P7 = = P 9 P P 8 9
Example An ideal gas-turbine cycle with two stages of compression and two stages of expansion has an overall pressure ratio of 8. Air enters each stages of the compressor at 300 K and each stage of the turbine at 1300 K. Determine the back work ratio and the thermal efficiency of this gas-turbine cycle, assuming (a) no regenerators and (b) an ideal regenerator with 100 percent effectiveness. Assumptions - Assume steady operating conditions - Use air-standard assumptions - Neglect KE & PE Information Overall pressure ratio of 8 - each stage of compressor and each stage of turbine have same pressure ratio
At turbine Under these conditions, the work input to each stage of the compressor will be same
(a) No regeneration case Determine the back work ratio & thermal efficiency
From table A-17
(b) Add an ideal regenerator (no pressure drop, 100 percent effectiveness) Compressed air is heated to the turbine exit temperature T 9, before it enters the combustion chamber. Under the air-standard assumptions, h 5 = h 7 = h 9
Two stage compression and expansion with intercooling, reheating, and regeneration Two stage compression and expansion with intercooling, reheating (without regeneration)