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AME 436 Energy and Propulsion Lecture 14 Propulsion 4: Nonideal performance of airbreathing propulsion devices

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2 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Outline Component performance - definitions AirCycles4Propulsion spreadsheet Non-ideal performance of 'conventional' propulsion systems Diffuser Compressor Burner Turbine Nozzle Fan Heat loss

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3 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Component performance Diffusers (1 2) & nozzles (7 9) subject to stagnation pressure losses due to friction, shocks, flow separation More severe for diffusers since dP/dx > 0 - unfavorable pressure gradient, aggravates tendency for flow separation If adiabatic, T t = constant, thus d T 2t /T 1t = 1 and n T 9t /T 7t = 1 If also reversible, d P 2t /P 1t = 1 and n P 9t /P 7t = 1, but d or n may may be < 1 due to losses; define performance If reversible, d = d ( -1)/ = 1, n = n ( -1)/ = 1, but if irreversible d ( -1)/ < 1 or n ( -1)/ < 1 even though d = 1, n = 1 Define diffuser efficiency d = d ( -1)/ = 1 if ideal Define nozzle efficiency n = n ( -1)/ = 1 if ideal For nozzle - also need to consider what happens if P exit ≠ P ambient, i.e. P 9 ≠ P 1

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4 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Compressors, fans & turbines Compressors (2 3) & fans (2' 3') subject to losses due to Separation of flow around compressor blades (again, unfavorable dP/dx) Gas leakage between blade tips and compressor walls How to define compression performance? (See Lecture 6) If irreversible compressor or turbine, dS > Q/T; if still adiabatic ( Q = 0) then dS > 0 For same P ratio, more work input (more T) during compression, less work output (less T) during expansion through turbine So define compressor efficiency comp as follows:

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5 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Fan efficiency fan defined similarly (note mass flow is times higher than in main flow); problems not as severe since smaller pressure ratios Turbines (4 5 and 5 6) have less fluid mechanics problems since dP/dx < 0, but need to bleed air from compressor through turbine blades for cooling (at least in first stages where T is high) Define turbine efficiency turb as follows: Same for fan turbine (if fan exists), just replace state 4 with 5 and replace 5 with 6 Compressors, fans & turbines

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6 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Component performance - combustors Combustor (4 5) and afterburner (6 7) - stagnation pressure losses due to turbulence (required for mixing of fuel and air), heat addition at M > 0 Define combustor (burner) or afterburner performance in terms of stagnation pressure ratio b or ab (= 1 if ideal) Of course for heat addition at finite M, stagnation pressure will always decrease; recall (lecture 12)

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7 AME Spring Lecture 14 - Nonideal effects in propulsion cycles AirCycles4Propulsion.xls See also lecture 9 (AirCycles4Recips) Thermodynamic model is exact, but heat loss, stagnation pressure losses etc. models are qualitative Constant is not realistic (changes from about 1.4 to 1.25 during process) but only affects results quantitatively (not qualitatively) Heat transfer model T ~ h(T wall -T gas,t ), where heat transfer coefficient h and component temperature T wall are specified - physically reasonable Unlike AirCycles4Recips.xls, wall temperature T w is specified separately for each component (diffuser, compressor, combustor, etc.) since each component feels a constant T, not the T averaged over the cycle Increments each cell not each time step Doesn't include effects of varying area, varying turbulence, varying time scale through Mach number, etc. on h Work or heat in or out going from step i to step i+1 computed according to dQ = C P (T i+1,t - T i,t ) or dW = - C P (T i+1,t - T i,t ) (heat positive if into system, work positive if out of system)

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8 AME Spring Lecture 14 - Nonideal effects in propulsion cycles AirCycles4Propulsion.xls Diffuser Mach number decrements from flight Mach number (M 1 ) to the specified value (must be zero except for hypersonic propulsion, discussed in lecture 15) after the diffuser in 25 equal steps Stagnation pressure decrements from its value at M 1 (P 1t ) to d P 1t = d /( -1) P 1t in 25 equal steps Static P and T are calculated from M and P t, T t Sound speed c is calculated from T, then u is calculated from c and M No heat input or work output in diffuser, but may have wall heat transfer Fan flow diffuser exactly same as main flow diffuser except that mass flow rate is times higher ( d,main flow = d,fan flow )

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9 AME Spring Lecture 14 - Nonideal effects in propulsion cycles AirCycles4Propulsion.xls Compressor Mach number is assumed to be zero throughout - calculations correct only if the Mach number after the diffuser is zero, thus P = P t, T = T t Compression is done in two steps (1) Wall heat transfer at constant pressure according to the same law as for the diffuser process (step i,a to step i,b) (2) Compression according to the usual Pv relations, isentropic or with optional irreversibility comp (step i,b to step i+1,a) comp = 1 corresponds to reversible adiabatic process comp = 1 corresponds to reversible adiabatic process Compression ends at a pressure of c P 2t where c is the specified pressure ratio and P 2t is the pressure at the end of the diffuser process Stagnation pressure increments in 25 equal steps Work per step = - C P (T i+1,a,t - T i,a,t ) Fan compressor follows exactly same as rules except that mass flow rate is times higher and fan may be different from comp

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10 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Combustor and afterburner Heat generation due to combustion is added in 25 equal steps (1/25 of the total for each 1/25 of the stagnation temperature change) during the process and heat is transferred to the wall according to the same law as always Stagnation pressure drops in 25 equal increments so that the pressure after combustion P 4t = b P 3t (note b = 1 for ideal turbojet cycle), or for afterburner P 7t = ab P 6t Heat addition ends when the temperature T 4t = T 1 (i.e the turbine inlet temperature limit), or T 7t = T 1 ,ab (afterburner temperature limit) Mach number is assumed to be zero throughout AirCycles4Propulsion.xls

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11 AME Spring Lecture 14 - Nonideal effects in propulsion cycles AirCycles4Propulsion.xls Turbine Expansion to pay for compressor work is done in two steps: (1) Wall heat transfer at constant pressure according to the same law as for the diffuser process (step i,a to step i,b) (2) Expansion according to the usual Pv relations, isentropic or with optional irreversibility turb (step i,b to step i+1,a) turb = 1 corresponds to reversible adiabatic process Each of 25 expansion steps provides (after accounting for losses due to irreversibility) 1/25 of the work required to drive turbine (calculated in compressor analysis) Work per step = - C P (T i+1,a,t - T i,a,t ), Mach number is assumed to be zero throughout Expansion through turbine to pay for fan work (if fan exists, i.e. ≠ 0) - same as main turbine, except equate fan work to turbine work

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12 AME Spring Lecture 14 - Nonideal effects in propulsion cycles AirCycles4Propulsion.xls Nozzle Static (not stagnation) pressure decrements from value after afterburner to specified exhaust pressure in 25 equal steps Stagnation pressure decrements from its value after afterburner (P 7t ) to n P 7t = n /( -1) P 7t in 25 equal steps Static P and T are calculated from M and P t, T t Sound speed c is calculated from T, then u is calculated from c and M No heat input or work output in diffuser, but may have wall heat transfer Heat transfer occurs according to usual law Same rules apply to fan nozzle except We go directly from state 3' (after fan compressor) through nozzle to state 9', no combustor or turbine Mass flow rate is times higher Assume P exit = P ambient always, i.e. P 9 ' = P 1

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13 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - diffuser - effect of d Obviously thrust & overall decrease, TSFC increases as d decreases, but can have pretty bad diffuser ( d ≈ 0.55) before thrust disappears Basic turbojet No fan No afterburner M 1 = 0.8 c = 30 = 5

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14 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - diffuser - effect of d T 2 is the same, regardless of d, since air decelerated to M = 0 s 2 increases & P 2 decreases with decreasing d T 3 same (since same c ), T 4 same (same limit), T 5 same (same work required to drive compressor) If P 2 is too low (note it can be < P 1 ), after compressor/combustor/turbine, P 5 is too low, can only expand back to u 9 = u 1, so no net thrust! Basic turbojet No fan No afterburner M 1 = 0.8 c = = 5

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15 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - effect of comp Obviously thrust & overall decrease, TSFC increases as comp decreases, but (again) can have pretty bad compressor ( c ≈ 0.58) before thrust disappears Basic turbojet No fan No afterburner M 1 = 0.8 c = 30 = 5

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16 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - effect of comp State 2 is the same since air decelerated isentropically to M = 0 T 3 & s 3 increases as comp decreases (more work required to drive compressor for same comp ), T 4 same (same limit) T 5 lower as comp decreases (more turbine work to drive compressor) If comp too low after paying for compressor work in turbine, P 5 & T 5 are too low, can only expand back to u 9 = u 1, so no net thrust! Basic turbojet No fan No afterburner M 1 = 0.8 c = = 5

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17 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - burner - effect of burner Obviously thrust & overall decrease, TSFC increases as burner decreases, but can have really lousy burner ≈ 0.13 before thrust disappears Basic turbojet No fan No afterburner M 1 = 0.8 c = 30 = 5

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18 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - burner - effect of burner States 2 & 3 are the same, independent of what burner does T 4 same for all (same limit) but as burner decreases, P 4 decreases, thus s 4 increases T 5 same for all (same turbine work required to drive compressor) If burner too low after paying for compressor work in turbine, P 5 & T 5 are too low, can only expand back to u 9 = u 1, so no net thrust! Basic turbojet No fan No afterburner M 1 = 0.8 c = = 5

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19 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - turbine - effect of turb Obviously thrust & overall decrease, TSFC increases as turb decreases, but can again can have low turb ≈ 0.57 before thrust disappears Basic turbojet No fan No afterburner M 1 = 0.8 c = 30 = 5

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20 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - turbine - effect of turb States are same, independent of turb T 5 also independent of turb (same work required to drive compressor) but s 5 increases as turbine decreases If turb too low after paying for compressor work in turbine, P 5 is too low, can only expand back to u 9 = u 1, so no net thrust! Basic turbojet No fan No afterburner M 1 = 0.8 c = = 5

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21 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - nozzle - effect of n Obviously thrust & overall decrease, TSFC increases as turb decreases, but can again can have low turb ≈ 0.56 before thrust disappears Interesting that d, comp, turb, nozzle at Thrust = 0 are all about the same! Basic turbojet No fan No afterburner M 1 = 0.8 c = 30 = 5

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22 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - nozzle - effect of n States are same, independent of turb If nozzle too low, can only expand back to u 9 = u 1, so no net thrust! Note that even for zero-thrust cycle, there is still some decrease in T after turbine to create u 9 ; if u 9 = 0 then thrust is negative!) Basic turbojet No fan No afterburner M 1 = 0.8 c = = 5

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23 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - nozzle - effect of P exit ≠ P amb Thrust & overall decrease, TSFC increases if P 9 ≠ P 1, but performance very insensitive to P 9 /P 1 (note horizontal scale is logarithmic) Basic turbojet No fan No afterburner M 1 = 0.8 c = 30 = 5

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24 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - nozzle - effect of P exit ≠ P amb States are same, independent of P 9 /P 1 If P 9 too low, negative (P 9 - P 1 )A 9 term in the thrust equation balances the positive a [(1+FAR)u 9 - u 1 ], so no net thrust, even though u 9 increases as P 9 decreases If P 9 too high, u 9 = 0; analysis fails at higher P 9 although even at P 9 = P 5, there is still net thrust (although mdot a [(1+FAR)u 9 - u 1 ] term is zero, (P 9 - P 1 )A 9 term generates thrust) Basic turbojet No fan No afterburner M 1 = 0.8 c = = 5

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25 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbofan cycle - effect of fan Obviously thrust & overall decrease, TSFC increases as fan decreases, but can have low fan ≈ 0.33 before thrust disappears Actually thrust ≠ 0 at this point; at this point fan work is so large that turbine exit pressure = ambient pressure; beyond this point analysis fails since turbine work insufficient to supply greedy fan Turbofan = 2 No afterburner M 1 = 0.8 c = 30 c ' = 2 = 5

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26 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbofan cycle - effect of fan States are same, independent of turb If fan too low, fan work is so large that turbine exit pressure P 6 = ambient pressure P 9, beyond this point analysis fails since turbine work insufficient to supply greedy fan Turbofan = 2 No afterburner M 1 = 0.8 c = c ' = 2 = 5

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27 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - effect of heat loss Turbojet No fan No afterburner M 1 = 0.8 c = = 5

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28 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - effect of heat loss With moderate heat loss, T 2 & T 3 decrease (given my choice of component wall temperatures, 300K for diffuser & compressor, 900K thereafter), and since P 3 /P 2 is set by choice of c, s 3 decreases Heat addition is greater, because need to add more heat to get to limit limit (starting from lower T, plus heat loss during heat addition) Since heat addition at constant pressure and same maximum temperature, T 4 is the same with heat loss Heat loss during turbine & nozzle expansion also result in decreased T; since P 9 is fixed, lower T 9 means lower s 9 Note that s decreases during nozzle expansion, even though T 9 < T wall - heat transfer is based on stagnation temperature of gas, not static temperature, since heat loss occurs at the walls, and gas must decelerate to M = 0 at wall, T t, not T, is a better temperature to use in heat tranfer estimates (Actually, how nearly this happens depends on the wall “accommodation coefficient" and other gasdynamic effects that we won't discuss in this course)

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29 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - effect of heat loss For ridiculously high wall heat loss coefficient, gas temperature follows wall temperature during compression (constant-T compression, like Stirling cycle), then heat is added at constant P, then during turbine & nozzle expansion, gas must follow constant- P path again until T t = 900K (turbine & nozzle wall temperature) is reached, then remainder of turbine work can be extracted isothermally (again like Stirling cycle), then gas can expand adiabatically through nozzle (no change in T t during nozzle expansion if adiabatic, and it will be adiabatic if gas T t = wall T) Effect of heat losses on performance (plot next page) Heat losses barely affect specific thrust, can get about the same thrust per unit mass flow independent of heat losses since the pressures don't change much, thus velocities don't change much Similarly, since pressures don't change much, velocities don't change much, thus propulsive efficiencies don't change much HOWEVER, much more heat must be added when heat losses are present, thus TSFC goes WAY up, similarly thermal efficiency and overall efficiency go way down

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30 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbojet cycle - effect of heat loss Turbojet No fan No afterburner M 1 = 0.8 c = = 5

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31 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Turbofan, no afterburner M 1 = 0.8, c = 10.97, = 5 c ' = 2, = 2 diffuser = 1 or 0.9 comp = 1 or 0.9 π burner = 1 or 0.9 turb = 1 or 0.9 nozzle = 1 or 0.9 fan = 1 or 0.9 h = 0 or 0.02 (all wall temperatures = average gas temp for that component) Drag coefficient = 0 or 0.2 Combination of all non-ideal effects Spec. Thrust TSFCThermal Eff. Prop. Eff. Overall Eff. Specific impulse (sec) Ideal Non-ideal

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32 How accurate is aircycles4recips.xls? Real engine: GE90-85B turbofan, no afterburner Static sea level (M 1 = 0, T 1 = 300K, P 1 = 1 atm) or cruise (M 1 = 0.83, T 1 = 219K, P 1 = atm) Reported properties: c = 39.3; c ' = 1.65; = 8.4; Inlet area = 7.66 m 2 ; C D = 0 (drag not included in reported performance) Guessed properties Turbine inlet temp. = 1700K = 5.67 (static) or 7.76 (cruise) Turbine inlet temp. = 1700K = 5.67 (static) or 7.76 (cruise) diffuser = 1 (static) or 0.97 (cruise); comp = 0.9; burner = 0.98; turb = 0.9; nozzle = 0.98; fan = 0.9 h = 0.01 (all wall temperatures = average gas temp for that component) Results (very sensitive to diffuser because of huge fan air flow!) AME Spring Lecture 14 - Nonideal effects in propulsion cycles Spec. Thrust (static) TSFC (static) Spec. Thrust (cruise) TSFC (cruise) Spreadshee t Actual ??? (air flow not reported) 2.213

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33 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Example - numerical For a nonideal turbofan with bypass ratio ( ) = 8, = 1.35, compressor pressure ratio ( c ) = 30, fan pressure ratio ( c ') = 1.8, flight Mach number (M 1 ) = 0.8, turbine inlet temperature = 1800K, ambient pressure (P 1 ) = 0.25 atm and ambient temperature (T 1 ) = 225 K, with P 9 = P 1 and FAR << 1, and component efficiencies as follows: diffuser ( d ) = 0.97, compressor ( c ) = 0.90, combustor ( b ) = 0.98, turbine ( t ) = 0.90, nozzle ( d ) = 0.98, fan ( f ) = 0.90, determine the following (this is the same as the example in lecture 13 except non-ideal components are presumed here): (a) (a)T, P and M after the diffuser (station 2) (b) (b)T, P and M after the compressor (station 3) (c) T, P and M after the combustor (station 4) M 4 = 0; P 4t = P 4 = ( b )P 3t = 0.98(11.0 atm) = 10.7 atm ; T 4t = T 4 = 1800K

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34 AME Spring Lecture 14 - Nonideal effects in propulsion cycles Example - numerical (continued) (d) T, P and M after the compressor turbine (station 5). Balance turbine & compressor work: (e) T, P and M after the fan turbine (station 6) For the fan stream, T 2 ' = T 2 = 250.2K and P 2 ' = P 2 = atm (same as main stream) Then we need to compute the fan stream work and equate it to the turbine work:

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35 Example - numerical (continued) (e) T, P and M after the nozzle (station 9, and 9' for the fan stream) (note that for the fan stream, nothing happens between stations 3 and 6) (f) (f)Specific thrust (ST) (recall FAR << 1 and P 9 = P 1 by assumption) AME Spring Lecture 14 - Nonideal effects in propulsion cycles

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36 Example - numerical (continued) (g) TSFC and overall efficiency Recalling that only 1/(1+ ) of the total air flow is burned, note that compared to the ideal cycle, TSFC suffers less (2.52 vs = 8.6% increase) than ST (0.647 vs = 12.5% decrease) (h) Propulsive efficiency AME Spring Lecture 14 - Nonideal effects in propulsion cycles

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37 Example - numerical (continued) (i) Thermal efficiency Note that th p = (0.516)(0.615) = = o as advertised. Note that compared to the ideal cycle, thermal and overall efficiencies decreased (as expected) but propulsive efficiency actually increased slightly due to lower exit velocities. (j) Specific impulse AME Spring Lecture 14 - Nonideal effects in propulsion cycles

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38 In an ideal -limited afterburning turbojet, how would the T-s diagrams be affected if (a) The turbine is irreversible, but all other components are still ideal The cycles are the same up to the end of the first heat addition process. The turbine work (thus T) does not change, but the irreversibility during expansion increases the entropy, so that afterburning heat addition occurs along a lower constant-pressure curve up to the same temperature limit. Although not shown, since the afterburner T is the same for the modified cycle, the heat addition, thus area under the T-s curve, is the same for the modified afterburner. (b) A new compressor is used with a higher pressure ratio, but this compressor is irreversible (all other components are still ideal) The compression process ends at a higher pressure and (since it is irreversible) a higher entropy than the base cycle. Heat addition then occurs along this constant-pressure curve up to the turbine temperature limit. More work, thus more T, is required from the turbine, so the afterburning heat addition occurs along a lower constant pressure curve. Example - graphical AME Spring Lecture 14 - Nonideal effects in propulsion cycles Same T (a) (b)

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39 In an ideal turbofan, how would the T-s diagrams be affected if (a ) The fan is removed and the redesigned turbine that drives the compressor is irreversible The cycle is the same up to the end of heat addition. The T and thus the T after the compressor turbine is the same but the entropy is higher. There is no fan turbine. Expansion through the nozzle continues to ambient pressure. (b) There are substantial pressure losses in the combustor The cycle is the same up to the end of the compression process. Heat addition occurs at decreasing pressure but up to the same temperature limit as the base cycle. The compressor and fan work requirements are unchanged, so the temperatures at states 5 and 6 are unchanged. Example - graphical AME Spring Lecture 14 - Nonideal effects in propulsion cycles (a) (b) (a) '5'5'5' 6'6'6'6' 4'4'4'4' 4

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40 Summary - non-ideal effects Irreversible diffusion, compression, expansion all reduce thrust & o Stagnation pressure loss in diffuser - less expansion possible at end of cycle More T (thus more work) to drive compressor More P needed to extract turbine work required to drive compressor Surprisingly, component efficiency at thrust = 0 nearly same for diffuser, compressor, turbine, nozzle! ( ) Effect of combustor pressure loss relatively small P exit ≠ P ambient (P 9 ≠ P 1 ) reduces thrust & o but effect is small Heat transfer to/from gas during cycle Modeling of heat losses different for steady flow device »Each component has its own T (unlike unsteady flow engines - use 1 wall temperature) »Stagnation vs. static temperature As with unsteady flow engines, it takes more work to compress a hot gas than a cold gas, lowers o - best performance at h = 0

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