PART 2 POWER AND PROPULSION CYCLES
PART 2 POWER AND PROPULSION CYCLES
PART 2- POWER AND PROPULSION CYCLE 2A -Gas Power and propulsion cycles SB&vW-118,119,11.10,11.11,11.12,1113,11.14] In this section we analyze several gas cycles used in practical applications for propulsion and power generation, using the air standard cycle. The air standard cycle is an the actual cycle behavior, and the term specifically refers to analysis using the folpproximation to assumptions Air is the working fluid(the presence of combustion products is neglected) Combustion is represented by heat transfer from an external heat source The cycle is 'completed' by heat transfer to the surroundings All processes are internally reversible Air is a perfect gas with constant specific heats 2.A. I The Internal combustion engine(Otto Cycle The different processes of an idealized otto cycle (internal combustion engine) are shown in Figure 2A-1 P Adiabatic reversible Q Po V,=V Figure 2A-1: Ideal Otto cycle Intake stroke, gasoline vapor and air drawn into engine (5->1) i1. Compression stroke, P, Tincrease(1->2) ii1. Combustion( spark), short time, essentially constant volume (2->3) Model: heat absorbed from a series of reservoir at temperatures T2 to T3 Power stroke: expansion (3->4) Valve exhaust: valve opens, gas escapes vi.(4->1)Model: rejection of heat to series of reservoirs at temperatures T4 to TI vil. Exhaust stroke, piston pushes remaining combustion products out of chamber 1->5 2A-1
2A-1 PART 2 – POWER AND PROPULSION CYCLES 2A – Gas Power and Propulsion Cycles [SB&VW - 11.8, 11.9, 11.10, 11.11, 11.12, 11.13, 11.14] In this section we analyze several gas cycles used in practical applications for propulsion and power generation, using the air standard cycle. The air standard cycle is an approximation to the actual cycle behavior, and the term specifically refers to analysis using the following assumptions: • Air is the working fluid (the presence of combustion products is neglected) • Combustion is represented by heat transfer from an external heat source • The cycle is ‘completed’ by heat transfer to the surroundings • All processes are internally reversible • Air is a perfect gas with constant specific heats 2.A.1 The Internal combustion engine (Otto Cycle) The different processes of an idealized Otto cycle (internal combustion engine) are shown in Figure 2A-1: P V P0 V2 = V3 V1 = V4 5 2 3 4 1 QH QL Adiabatic reversible Figure 2A-1: Ideal Otto cycle i. Intake stroke, gasoline vapor and air drawn into engine (5 -> 1) ii. Compression stroke, P, T increase (1->2) iii. Combustion (spark), short time, essentially constant volume (2->3) Model: heat absorbed from a series of reservoir at temperatures T2 to T3 iv. Power stroke: expansion (3 ->4) v. Valve exhaust: valve opens, gas escapes vi. (4->1) Model: rejection of heat to series of reservoirs at temperatures T4 to T1 vii. Exhaust stroke, piston pushes remaining combustion products out of chamber 1->5
The actual cycle does not have these sharp transitions between the different processes and might be as sketched in figure 2A-2 ISentropic Spark Exhaust Exhaust valve closes Figure 2A-2: Sketch of actual Otto cycle Efficiency of an ideal otto cycle The starting point is the general expression for the thermal efficiency of a cycle york L=1+L heat input engine, So QL Is negative. The heat absorbed occurs during combustion when the spark occp or The convention, as previously, is that heat exchange is positive if heat is flowing into the system or roughly at constant volume. The heat absorbed can be related to the temperature change from state 2 to state 3 as Qn=Q23=AU23(W23=0 C=C(T3-T2) The heat rejected is given by(for a perfect gas with constant specific heats) (T1-T4) Substituting the expressions for the heat absorbed and rejected in the expression for thermal efficiency yields T 2A-2
2A-2 The actual cycle does not have these sharp transitions between the different processes and might be as sketched in Figure 2A-2 Spark Exhaust valve opens Not isentropic Exhaust valve closes P P0 V Figure 2A-2: Sketch of actual Otto cycle Efficiency of an ideal Otto cycle The starting point is the general expression for the thermal efficiency of a cycle: η = = + = + work heat input Q Q Q Q Q H L H L H 1 . The convention, as previously, is that heat exchange is positive if heat is flowing into the system or engine, so QL is negative. The heat absorbed occurs during combustion when the spark occurs, roughly at constant volume. The heat absorbed can be related to the temperature change from state 2 to state 3 as: QQ U W C dT C T T H T v T v == = ( ) = ∫ = − ( ) 23 23 23 2 3 3 2 ∆ 0 The heat rejected is given by (for a perfect gas with constant specific heats) Q Q U CT T L v == = − 41 41 1 4 ∆ ( ) Substituting the expressions for the heat absorbed and rejected in the expression for thermal efficiency yields η = − − − 1 4 1 3 2 T T T T
We can simplify the above expression using the fact that the processes from l to 2 and from 3 to 4 are isentropic Tw-=T3w-,T;w-=T2V2-1 (T4-7)W-1=(T3-72) 74-7_(V T-72(V The quantity ==r is called the compression ratio. In terms of compression ratio, the efficiency of an ideal Otto cycle is ne The ideal otto cycle efficiency is shown at the right, as a function of the compression ratio As the compression ratio, r, increases .E月 noo increases, but so does T. If T is too high, the mixture will ignite without a park(at the wrong location in the cycle) Ideal Otto cycle thermal efficiency Engine work, rate of work per unit enthalpy flu he non-dimensional ratio of work done(the power )to the enthalpy flux through the engine is given by Powe 22 Enthalpy flux mcpl cpTl There is often a desire to increase this quantity, because it means a smaller engine for the same power. The heat input is given by fuel(an fu Ah fuel is the heat of reaction, ie the chemical energy liberated per unit mass of fuel I fuel is the fuel mass flow rate The non-dimensional power is 2A-3
2A-3 We can simplify the above expression using the fact that the processes from 1 to 2 and from 3 to 4 are isentropic: TV TV TV TV T TV T TV T T T T V V 4 1 1 3 2 1 1 1 1 2 2 1 4 11 1 3 22 1 4 1 3 2 2 1 1 γγ γγ γ γ γ −− −− − − − = = ( ) − = − ( ) − − = , The quantity V V r 1 2 = is called the compression ratio. In terms of compression ratio, the efficiency of an ideal Otto cycle is: ηOtto γ γ V V r =− =− − − 1 1 1 1 1 2 1 1 . The ideal Otto cycle efficiency is shown at the right, as a function of the compression ratio. As the compression ratio, r, increases, ηOtto increases, but so doesT2 . If T2 is too high, the mixture will ignite without a spark (at the wrong location in the cycle). Ideal Otto cycle thermal efficiency Engine work, rate of work per unit enthalpy flux: The non-dimensional ratio of work done (the power) to the enthalpy flux through the engine is given by Power Enthalpy flux = = ˙ ˙ ˙ ˙ W mc T Q p mc T Otto 1 p 23 1 η There is often a desire to increase this quantity, because it means a smaller engine for the same power. The heat input is given by ˙ Qm h ˙ 23 = fuel fuel ( ) ∆ , where • ∆hfuel is the heat of reaction, ie the chemical energy liberated per unit mass of fuel • m˙ fuel is the fuel mass flow rate. The non-dimensional power is ˙ ˙ ˙ ˙ W mc T m m h p c T r fuel fuel 1 1 p 1 1 1 = − − ∆ γ . 0 0 10 20 30 40 50 60 70 2468 Compression ratio, rv Thermal efficiency, ηth 10 12 14 16
The quantities in this equation, evaluated at stoichiometric conditions are fuel 0. pTT Muddy points Ho △ h., calculated?(MP2A1) What are"stoichiometric conditions"?(MP 2A.2 2. 4.2. Diesel cycle The Diesel cycle is a compression ignition(rather than spark ignition)engine. Fuel is sprayed into the cylinder at P,(high pressure)when the compression is complete, and there is ignition without a spark. An idealized Diesel engine cycle is shown in Figure 2A-3 Adiabatic Figure 2A-3 Ideal Diesel cycle The thermal efficiency is given by eL C,(T-T) (T This cycle can operate with a higher compression ratio than otto cycle because only air is compressed and there is no risk of auto-ignition of the fuel. Although for a given compression ratio the Otto cycle has higher efficiency, because the Diesel engine can be operated to higher compression ratio, the engine can actually have higher efficiency than an Otto cycle when both are operated at compression ratios that might be achieved in practice 2A-4
2A-4 The quantities in this equation, evaluated at stoichiometric conditions are: ˙ ˙ , m m 1 15 h c T 4 10 10 288 fuel fuel p 1 7 3 ≈ ≈ × × ∆ so, ˙ ˙ W mc Tp 1 r 1 9 1 1 ≈ − γ − . Muddy points How is ∆ h fuel calculated? (MP 2A.1) What are "stoichiometric conditions"? (MP 2A.2) 2.A.2. Diesel Cycle The Diesel cycle is a compression ignition (rather than spark ignition) engine. Fuel is sprayed into the cylinder at P2 (high pressure) when the compression is complete, and there is ignition without a spark. An idealized Diesel engine cycle is shown in Figure 2A-3. QH P V2 V V3 V4 = V1 QL Adiabatic reversible 2 3 4 1 Figure 2A-3 Ideal Diesel cycle The thermal efficiency is given by: η η γ Diesel L H v p Diesel T T T T Q Q CT T CT T T T =+ =+ ( ) − ( ) − = − ( ) − ( ) − 1 1 1 1 1 1 4 3 2 1 4 1 2 3 2 This cycle can operate with a higher compression ratio than Otto cycle because only air is compressed and there is no risk of auto-ignition of the fuel. Although for a given compression ratio the Otto cycle has higher efficiency, because the Diesel engine can be operated to higher compression ratio, the engine can actually have higher efficiency than an Otto cycle when both are operated at compression ratios that might be achieved in practice
Muddy points When and where do we use c, and c? Some definitions use du=c dT. Is it ever Explanation of the above comparison between Diesel and Otto. ( MP 2A. 4) 2.A.3 Brayton Cycle The Brayton cycle is the cycle that represents the operation of a gas turbine engine. The simple gas turbine "can be operated in open cycle or closed cycle(recirculating working fluid) modes. as shown below H Fuel Heat Turbine Co Turbine Air Products Q Figure 2A-4: Gas turbine engine operating on the Brayton cycle-(a)open cycle operation, (b) closed cycle operatio Efficiency of the Brayton cycle We derived the ideal Brayton cycle efficiency in Section 1.A Net work per unit mass flow in a Brayton cycle: The net mechanical work of the cycle is given by Net mechanical work/unit mass= wrurbine-Wcompressor compressor=-4h12 burbine =-Ah34=-Ahrurb If kinetic energy changes across the compressor and turbine are neglected, the temperature ratio TR, across the compressor and turbine is related to the enthalpy changes R-1= 2A-5
2A-5 Muddy points When and where do we use c v and c p ? Some definitions use dU=c v dT. Is it ever dU=c p dT? (MP 2A.3) Explanation of the above comparison between Diesel and Otto. (MP 2A.4) 2.A.3 Brayton Cycle The Brayton cycle is the cycle that represents the operation of a gas turbine engine. The “simple gas turbine” can be operated in open cycle or closed cycle (recirculating working fluid) modes, as shown below. Fuel QH w wnet net Air Products QL Combustion chamber Compressor Turbine Heat exchanger Heat exchanger Compressor Turbine (a) (b) Figure 2A-4: Gas turbine engine operating on the Brayton cycle – (a) open cycle operation, (b) closed cycle operation Efficiency of the Brayton cycle: We derived the ideal Brayton cycle efficiency in Section 1.A: η Brayton γ γ inlet compressorexit T T PR =− =− − 1 1 1 ( )/ 1 . Net work per unit mass flow in a Brayton cycle: The net mechanical work of the cycle is given by: Net mechanical work/unit mass = − w w turbine compressor, where w hh w hh compressor comp turbine turb =− =− =− =− ∆ ∆ ∆ ∆ 12 34 If kinetic energy changes across the compressor and turbine are neglected, the temperature ratio, TR, across the compressor and turbine is related to the enthalpy changes: TR h h h h comp turb −= = 1 1 4 ∆ ∆
The net work is thus h1 The turbine work is greater than the work needed to drive the compressor. The thermodynamic states in an enthalpy-entropy (h, s)diagram, and the work of the compressor and turbine, are shown below for an ideal Brayton cycle 4 T4 q urb q P To=t Figure 2A-5: Brayton cycle in enthalpy-entropy (h-s)representation showing compressor and turbine work Muddy points What is shaft work?(MP 2 2. A 4 Brayton Cycle for et Propulsion: the Ideal ramjet A schematic of a ramjet is given in Figure 2A-6 below Station numbers exhaust fuel 5 streamtube streamtube diffuser burner nozzle Figure 2A-6: Ideal ramjet [adapted from J. L. Kerrebrock, Aircraft Engines and Gas Turbines] 2A-6
2A-6 ∆ ∆ h h h h turb comp = − 4 1 The net work is thus net work = − ∆h h h comp 4 1 1 The turbine work is greater than the work needed to drive the compressor. The thermodynamic states in an enthalpy-entropy (h,s) diagram, and the work of the compressor and turbine, are shown below for an ideal Brayton cycle. h 4 T4=Tmax wturb qR s 0 T0=Tinlet 5 P0 wcomp qA P3 3 Figure 2A-5: Brayton cycle in enthalpy-entropy (h-s) representation showing compressor and turbine work Muddy points What is shaft work? (MP 2A.5) 2.A.4 Brayton Cycle for Jet Propulsion: the Ideal Ramjet A schematic of a ramjet is given in Figure 2A-6 below. 0 inlet 1 3 4 5 streamtube cθ p0 T0 c3 p3 T3 diffuser πd burner τd πd nozzle Station Numbers fuel, mf . exhaust streamtube T4 c5 p5 T5 Figure 2A-6: Ideal ramjet [adapted from J. L. Kerrebrock, Aircraft Engines and Gas Turbines]
In the ramjet there are"no moving parts". The processes that occur in this propulsion device are 0->3 isentropic diffusion(slowing down) and compression, with a decrease in Mach number,M→M34 Constant pressure combustion, 4->5 Isentropic expansion through the nozzle Thrust of an ideal engine ramjet The coordinate system and control volume are chosen to be fixed to the ramjet. The thrust. F where cs and co are the inlet and exit flow velocities. The thrust can be put in terms of non- dimensional parameters as follows F c yRT is the speed mao as ao a0 F Mo=M Using M3, M2 <<lin the expression for stagnation pressure, P=1+Y- P=Pr =Pr: P=Pt=Ps: P=P3 The ratios of stagnation pressure to static pressure at inlet and exit of the ramjet are Pio PP Pr PP The ratios of stagnation to static pressure at exit and at inlet are the same, with the consequence that the inlet and exit mach numbers are also the same To find the thrust we need to find the ratio of the temperature at exit and the temperature at inlet. This is given by TTs TT T T 2A-7
2A-7 In the ramjet there are “no moving parts”. The processes that occur in this propulsion device are: 0->3 isentropic diffusion (slowing down) and compression, with a decrease in Mach number, M M 0 3 → 4 Constant pressure combustion, 4->5 Isentropic expansion through the nozzle. Thrust of an ideal engine ramjet The coordinate system and control volume are chosen to be fixed to the ramjet. The thrust, F, is given by: F cc = − m˙ ( ) 5 0 , where c5 and co are the inlet and exit flow velocities. The thrust can be put in terms of nondimensional parameters as follows: F a c a a a c m˙ a 0 5 5 5 0 0 0 = − , where a RT = γ is the speed of sound. F a M a a M M T T M m˙ 0 5 5 0 0 5 5 0 = −= − 0 Using M M 3 2 4 2 , <<1in the expression for stagnation pressure, P P T = + M − − 1 1 2 2 γ 1 γ γ , P P PP P PP P 3 ≈= ≈= ≈ TT TT 3 0 4 4 5 4 3 ; ; The ratios of stagnation pressure to static pressure at inlet and exit of the ramjet are: P P P P P P P P T M Te e Me 0 0 3 0 0 4 0 === determines determines 1 24 34 124 34 The ratios of stagnation to static pressure at exit and at inlet are the same, with the consequence that the inlet and exit Mach numbers are also the same. M M 5 0 = . To find the thrust we need to find the ratio of the temperature at exit and the temperature at inlet. This is given by: T T T M M T T T T T T T T T T b 5 0 5 5 2 0 2 0 5 0 4 1 3 1 2 1 1 2 = + − + − === γ γ τ
where tb is the stagnation temperature ratio across the combustor(burner). The thrust is thus Cycle efficiency in terms of aerodynamic parameters nBravton=l 0-1-=1-, d lo compressor exit y Ramjet thermodynamic cycle efficiency in terms of flight Mach number, Mo For propulsion engines, the figures of merit includes more than thrust and nBravt The specific impulse, I s, measures how effectively fuel is used ri/gfms, Specific Impulse, where ni =f ri is the fuel mass flow rate To find the fuel-air ratio, f, we employ a control volume around the combustor and carry out an energy balance. Before doing this, however, it is useful to examine the way in which Isp appears In expressions for rang Muddy points What exactly is the specific impulse, Isp a measure of?(MP 2A.6) How is Isp found for rockets in space where g-0?(MP 2A.7) Why does industry use TSCP rather than Isp? Is there an advantage to this?(MP Why isnt mechanical efficiency an issue with ramets?(MP 2A. 9) How is thrust created in a ramjet?(MP 2A. 10) Why don' t we like the numbers 1 and 2 for the stations? why do we go 0-3?(MP 2A.11) For the Brayton cycle efficiency. why does T:=L2(MP 2A. 12) 2. 4.5 The Breguet Range equation See Waitz Unified Propulsion Notes, No IV(see the 16.050 Web site) Consider an aircraft in level flight, with weight W. The rate of change of the gross weight of the vehicle is equal to the fuel weight flow 2A-8
2A-8 where τ b is the stagnation temperature ratio across the combustor (burner). The thrust is thus: F a M b m˙ 0 = − 0( ) τ 1 Cycle efficiency in terms of aerodynamic parameters: η Brayton compressor exit T T T T T T T =− =− =− 1 11 0 0 3 0 0 , and T TT M 0 0 0 2 1 1 1 2 = + γ − , so: η γ γ Brayton M M = − + − 1 2 1 1 2 0 2 0 2 : Ramjet thermodynamic cycle efficiency in terms of flight Mach number, M0 . For propulsion engines, the figures of merit includes more than thrust and η Brayton . The specific impulse, Isp measures how effectively fuel is used: I F g F f g sp f = = m m ˙ ˙ ; Specific Impulse, where m m ˙ ˙ f = f is the fuel mass flow rate. To find the fuel-air ratio, f, we employ a control volume around the combustor and carry out an energy balance. Before doing this, however, it is useful to examine the way in which Isp appears in expressions for range. Muddy points What exactly is the specific impulse, Isp, a measure of? (MP 2A.6) How is Isp found for rockets in space where g ~ 0? (MP 2A.7) Why does industry use TSCP rather than Isp? Is there an advantage to this? (MP 2A.8) Why isn't mechanical efficiency an issue with ramjets? (MP 2A.9) How is thrust created in a ramjet? (MP 2A.10) Why don't we like the numbers 1 and 2 for the stations? Why do we go 0-3? (MP 2A.11) For the Brayton cycle efficiency, why does T 3 =T t0 ? (MP 2A.12) 2.A.5 The Breguet Range Equation [See Waitz Unified Propulsion Notes, No. IV (see the 16.050 Web site)] Consider an aircraft in level flight, with weight W. The rate of change of the gross weight of the vehicle is equal to the fuel weight flow:
L W dw F W=L-D(LD=F(D) The rate of change of aircraft gross weight is thus dw Suppose L/D and Isp remain constant along flight path dw We can integrate this equation for the change in aircraft weight to yield a relation between the weight change and the time of flight W W Dl, where w is the initial weight If w is the final weight of vehicle and initia 0, the relation between vehicle parameters and flight The range is the flight time multiplied by the flight speed, or, W Range= cot aircraft designer propulsion system structural designer 2A-9
2A-9 L D F W g dW dt F Isp f m˙ =− =− WLD L D F L D = = ( ) = ( ) The rate of change of aircraft gross weight is thus dW dt W L D Isp = − ( ) . Suppose L/D and Isp remain constant along flight path: dW W dt L D Isp = − ( ) . We can integrate this equation for the change in aircraft weight to yield a relation between the weight change and the time of flight: ln W W t L D i Isp = − ( ) , where Wi is the initial weight. If Wf is the final weight of vehicle and tinitial=0, the relation between vehicle parameters and flight time, t f , is L D I W W t sp i f f ln = . The range is the flight time multiplied by the flight speed, or, Range c t L D c I W W f sp i f = = × × ( ) × 0 0 aircraft designer propulsion system designer structural designer ln { { 123