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→M3<<1 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 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 → <<1 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: