Gas 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: 3 P Adiabatic reversible QH 2 4 P0 QL 5 1 V2 = V3 V1 = V4 V Figure 2A-1: Ideal Otto cycle i. ii. iii. iv. v. vi. vii. Intake stroke, gasoline vapor and air drawn into engine (5 -> 1) Compression stroke, P, T increase (1->2) 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 (4->1) Model: rejection of heat to series of reservoirs at temperatures T4 to T1 Exhaust stroke, piston pushes remaining combustion products out of chamber 1->5 2A-1 The actual cycle does not have these sharp transitions between the different processes and might be as sketched in Figure 2A-2 Not P isentropic Exhaust valve opens Spark P0 Exhaust valve closes 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= Q + QL Q work = H = 1+ L . heat input QH QH 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: QH = Q23 = ∆U23 (W23 = 0) = ∫TT23 Cv dT = Cv (T3 − T2 ) The heat rejected is given by (for a perfect gas with constant specific heats) QL = Q41 = ∆U 41 = Cv (T1 − T4 ) Substituting the expressions for the heat absorbed and rejected in the expression for thermal efficiency yields η = 1− T4 − T1 T3 − T2 2A-2 We can simplify the above expression using the fact that the processes from 1 to 2 and from 3 to 4 are isentropic: T4 V1γ −1 = T3V2γ −1 , T1V1γ −1 = T2 V2γ −1 (T4 − T1 )V1γ −1 = (T3 − T2 )V2γ −1 T4 − T1  V2  =  T3 − T2  V1  γ −1 V1 = r is called the compression ratio. In terms of compression ratio, the V2 efficiency of an ideal Otto cycle is: The quantity ηOtto = 1 −  V1   V2  γ −1 1 = 1 − γ −1 . r 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 does T2 . If T2 is too high, the mixture will ignite without a spark (at the wrong location in the cycle). Thermal efficiency, ηth 70 1 60 50 40 30 20 10 0 0 2 4 6 8 10 12 Compression ratio, rv 14 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 ˙ η W˙ Q Power = = 23 Otto ˙ pT1 ˙ pT1 mc Enthalpy flux mc 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 ˙ =m Q 23 ˙ fuel (∆h fuel ) , where • ∆h fuel is the heat of reaction, ie the chemical energy liberated per unit mass of fuel • ṁ fuel is the fuel mass flow rate . The non-dimensional power is ˙ fuel ∆h fuel  m W˙ 1 = 1 − γ −1  ˙ pT1 ˙ mc m c pT1  r  . 2A-3 16 The quantities in this equation, evaluated at stoichiometric conditions are: ˙ fuel m 1 ∆h fuel 4 × 107 ≈ , ≈ 3 ˙ m 15 c pT1 10 × 288 so, W˙ 1 ≈ 9 1 − γ −1  . ˙ pT1 mc  r  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 2 3 Adiabatic reversible 4 QL 1 V2 V3 V4 = V1 V Figure 2A-3 Ideal Diesel cycle The thermal efficiency is given by: η Diesel = 1 + η Diesel = 1 − C (T − T ) QL = 1+ v 1 4 QH C p (T3 − T2 ) (T4 T1 − 1) γT2 (T3 T2 − 1) T1 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 Muddy points When and where do we use cv and cp? Some definitions use dU=cvdT. Is it ever dU=cpdT? (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. QH Fuel Heat exchanger Combustion chamber Compressor Turbine Air wnet Compressor Products Turbine wnet Heat exchanger QL (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: Tinlet 1 = 1− η Brayton = 1 − (γ −1) / γ Tcompressor exit PR . 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 = wturbine − wcompressor where , wcompressor = − ∆h12 = − ∆hcomp wturbine = − ∆h34 = − ∆hturb 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: ∆hcomp ∆hturb , TR − 1 = = h1 h4 2A-5 ∆hturb = − ∆hcomp h4 h1 The net work is thus h  net work = ∆hcomp  4 − 1  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. h T4=Tmax 4 P3 qA wturb 3 5 wcomp qR P0 0 T0=Tinlet s 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. Station Numbers . 0 inlet 3 fuel, mf 1 exhaust 5 streamtube 4 streamtube cθ p0 T0 c3 p3 T 3 diffuser πd T4 burner τd πd c5 p5 T 5 nozzle Figure 2A-6: Ideal ramjet [adapted from J. L. Kerrebrock, Aircraft Engines and Gas Turbines] 2A-6 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, M0 → 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, is given by: F = ṁ(c 5 − c0 ) , where c5 and co are the inlet and exit flow velocities. The thrust can be put in terms of nondimensional parameters as follows: F c = 5 ṁa0 a5 a5 c0 − , where a = γRT is the speed of sound. a0 a0 F a T = M 5 5 − M0 = M 5 5 − M0 a0 T0 ṁa0 γ Using M32 , M 42 P γ − 1 2  γ −1 << 1 in the expression for stagnation pressure, T = 1 + M  2 P   , P3 ≈ PT3 = PT0 ; P4 ≈ PT4 = PT5 ; P4 ≈ P3 The ratios of stagnation pressure to static pressure at inlet and exit of the ramjet are: PT0 PT P P = 3 = 4 = e P0 P0 P0 Pe 14 24 3 1 424 3 determines M0 determines Me 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 5 = M0 . To find the thrust we need to find the ratio of the temperature at exit and the temperature at inlet. This is given by: γ −1 2 1+ M0 TT , TT5 TT T5 2 = = 5 = 4 = τb γ −1 2 T0 T0 TT0 TT3 1+ M5 2 2A-7 where τ b is the stagnation temperature ratio across the combustor (burner). The thrust is thus: F = M0 ( τ b − 1) ṁa0 Cycle efficiency in terms of aerodynamic parameters: η Brayton = 1 − T0 Tcompressor exit = 1− T0 T T = 1 − 0 , and 0 = T3 TT0 TT0 1 , so: γ −1 2 1+ M0 2 γ −1 2 M0 2 : Ramjet thermodynamic cycle efficiency in terms of flight η Brayton = γ −1 2 1+ M0 2 Mach number, M0 . For propulsion engines, the figures of merit includes more than thrust and η Brayton . The specific impulse, I sp measures how effectively fuel is used: I sp = F F = ; Specific Impulse, ˙ fg f m ˙g m where m ˙f = f m ˙ 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 T3 =Tt0? (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: 2A-8 L D F W g ṁ f = − dW F =− dt Isp ( ) ( ) W = L= D LD = F LD The rate of change of aircraft gross weight is thus dW W =− L dt D I sp . Suppose L/D and Isp remain constant along flight path: ( ) dW dt =− L I W D sp . ( ) 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 t , where Wi is the initial weight. =− L I Wi D sp ( ) If Wf is the final weight of vehicle and tinitial=0, the relation between vehicle parameters and flight time, t f , is W L I sp ln i = t f D Wf . The range is the flight time multiplied by the flight speed, or, Range = c0 t f =  L  D { × c0 × I ) ({ sp propulsion system designer aircraft designer 2A-9 × ln  Wi  W   f 123 structural designer The above equation is known as the Breguet range equation. It shows the influence of aircraft, propulsion system, and structural design parameters. Relation of overall efficiency, Isp, and thermal efficiency Suppose ∆h fuel is the heating value (‘heat of combustion’) of fuel (i.e., the energy per unit of fuel mass), in J/kg. The rate of energy release is ṁ f ∆h fuel , so F ∆h fuel Fc0 ∆h fuel c0 I sp = c0 = ˙ f g ∆h fuel m ˙ f ∆h fuel g m and Thrust power (usefulwork) Fc0 = = ηoverall (overall propulsion system efficiency) Ideal available energy ṁ f ∆h fuel g L Wi c0 I sp and Range = ∆h fuel η ln overall ∆h fuel g D Wf = ηthermalη propulsiveηcombustion ηoverall = ηoverall The Propulsion Energy Conversion Chain The above concepts can be depicted as parts of the propulsion energy conversion train mentioned in Part 0, which shows the process from chemical energy contained in the fuel to energy useful to the vehicle. Chemical 1 energy Heat ⋅ • m f ∆h fuel m ηcombustion ≅ 1 Useful work: Thrust power Mechanical work ηthermal c 2 exit 2 − c0 2 2 η propulsive = 2 1 + cexit / c0 Figure 2A-7: The propulsion energy conversion chain. The combustion efficiency is near unity unless conditions are far off design and we can regard the two main drivers as the thermal and propulsive1 efficiencies. The evolution of the overall efficiency of aircraft engines in terms of these quantities is shown below in Figure 2A-8. 1 The transmission efficiency represents the ratio between compressor and turbine power, which is less than unity due to parasitic frictional effects. As with the combustion efficiency, however, this is very close to one and the horizontal axis can thus be regarded essentially as propulsive efficiency. 2A-10 0.8 0.1 Overall Efficiency 0.3 0.4 0.2 Core Thermal Efficiency 0.8 0.7 0.6 0.5 0.5 0.6 0.4 Future trend 0.7 '777' Engines 0.6 0.3 SFC Advanced UDF CF6-80C2 Low BPR 0.5 Turbojets Current High BPR 0.4 UDF Engine 0.3 Whittle 0.2 0.3 0.4 0.5 0.6 0.7 Propulsive x Transmission Efficiency 0.8 Figure 2A-8: Overall efficiency of aircraft engines as functions of thermal and propulsive efficiencies [data from Koff]. Muddy points How can we idealize fuel addition as heat addition? (MP 2A.13) 2.A.6 Performance of the Ideal Ramjet We now return to finding the ramjet fuel-air ratio, f. Using a control volume around the burner, we get: ˙ =m Heat given to the fluid: Q ˙ f ∆h fuel = m ˙ f∆h fuel . Q 3 From the steady flow energy equation: 4 ˙ 4 ht4 − m ˙ 3 ht 3 = m ˙ 3 f∆h fuel m The exit mass flow is not greatly different from the inlet mass flow, m˙ 4 = m˙ 3 (1 + f ) ≈ m˙ 3 , because the fuel-air ratio is much less than unity (generally several percent). We thus neglect the difference between the mass flows and obtain ( ) ht4 − ht3 = c p Tt4 − Tt3 = f ∆h fuel ( ) γ −1 2 Tt3 c p (τ b − 1) = f ∆h fuel , with Tt3 = Tt0 = T0 1 + 2 M0 14243 Θ0 2A-11 Fuel-air ratio, f: f= τ b −1 , ∆h fuel c p T0Θ 0 The fuel-air ratio, f, depends on the fuel properties ( ∆h fuel ), the desired flight parameters ( Θ 0 ), the ramjet performance ( τ b ), and the temperature of the atmosphere ( T0 ). Specific impulse, Isp: The specific impulse for the ramjet is given by c0 ∆h fuel  τ − ( b 1) c T F 1 p 0 I sp = = f ṁg g Θ0 (τ b − 1) The specific impulse can be written in terms of fuel properties and flight and vehicle characteristics as, a ∆h M0 . Isp = 0 fuel × gc pT0 Θ0 τ b + 1 1 424 3 14 4244 3 ( fuel properties ) flight characteristics, ramjet temp increase We wish to explore the parameter dependency of the above expression, which is a complicated formula. How can we do this? What are the important effects of the different parameters? How do we best capture the ramjet performance behavior? To make effective comparisons, we need to add some additional information concerning the operational behavior. An important case to examine is when f is such that all the fuel burns, i.e. when we have stoichiometric conditions. What happens in this situation as the flight Mach number, M0, increases? T0 is fixed so Tt3 increases, but the maximum temperature does not increase much because of dissociation: the reaction does not go to completion at high temperature. A useful approximation is therefore to take Tt4 constant for stoichiometric operation. In the stratosphere, from 10 to 30 km, T0 ≈ constant ≈ 212 K . The maximum temperature ratio is TT T τ max = max = 4 = const , T0 T0 τb = TT4 TT3 = TT4 TT3 T0 T0 = τ max Θ0 For the stoichiometric ramjet: I sp = F F a0 a0 = = M0 ( τ b − 1) ˙g m ˙ a0 f stoich g fm f stoich g 2A-12 Using the expression for τ b , the specific impulse is  τ  a0 I sp = M0  max − 1  Θ0  fstoich g Representative performance values: A plot of the performance of the stoichiometric ramjet is shown in Figure 2A-9. . m F qb = 10 τ = 10 f =max f stoic 2.5 F a0 2 . m a0 I g stoic f 1.5 a0 1 Isp gfstoich a0 0.5 f = fstoich 1 2 3 4 5 6 7 Figure 2A-9: Thrust per unit mass flow and specific impulse for ideal ramjet with stoichiometric combustion [adapted from Kerrebrock] M0 The figure shows that for the parameters used, the best operating range of a hydrocarbon-fueled ramjet is 2 ≤ M0 ≤ 4 . The parameters used are τ max = 10 , a0 ≈ 300 ms −1 in the stratosphere, a0 ≈ 450 s . fstoich = 0.067 for hydrocarbons g fstoich Recapitulation: In this section we have: Examined the Brayton Cycle for propulsion Found η Brayton as a function of M0 Found ηoverall and the relation between ηoverall and ηBrayton F Examined and I sp as a function of M0 for the ideal ramjet. ṁ a0 Muddy points What is the relation between h t4 − h t3 = f∆h f and the existence of the maximum value of Tt4? (MP 2A.14a) Why didn’t we have a 2s point for the Brayton cycle with non-ideal components ? (MP 2A.14b) What is the variable fstoich? (MP 2A.15) 2A-13 2.A.7 Effect of Departures from Ideal Behavior - Real Cycle behavior [See also charts 69-82 in 16.050: Gas Turbine Engine Cycles] What are the sources of non-ideal performance and departures from reversibility? - Losses (entropy production) in the compressor and the turbine - Stagnation pressure decrease in the combustor - Heat transfer We take into account here only irreversibility in the compressor and in the turbine. Because of these irreversibilities, we need more work, ∆hcomp (the changes in kinetic energy from inlet to exit of the compressor are neglected), to drive the compressor than in the ideal situation. We also get less work, ∆hturb , back from the turbine. The consequence, as can be inferred from Figure 2A-10 below, is that the net work from the engine is less than in the cycle with ideal components. P3 h T4 = Tmax 4 wturb qA P0 3 5 3s 5s wcomp qR 0 T0 = Tinlet s Figure 2A-10: Gas turbine engine (Brayton) cycle showing effect of departure from ideal behavior in compressor and turbine To develop a quantitative description of the effect of these departures from reversible behavior, consider a perfect gas with constant specific heats and neglect kinetic energy at the inlet and exit of the turbine and compressor. We define the turbine adiabatic efficiency as: ηturb = actual h − h5 wturb = 4 ideal h4 − h5S wturb actual ideal where wturb is specified to be at the same pressure ratio as wturb . (See charts 69-76 in 16.050 Gas Turbine Cycles.) There is a similar metric for the compressor, the compressor adiabatic efficiency: 2A-14 ηcomp = ideal wcomp actual wcomp = h3S − h0 h3 − h0 again for the same pressure ratio. Note that the ratio is the actual work delivered divided by the ideal work for the turbine, whereas the ratio is the ideal work needed divided by the actual work required for the compressor. These are not thermal efficiencies, but rather measures of the degree to which the compression and expansion approach the ideal processes. We now wish to find the net work done in the cycle and the efficiency. The net work is given either by the difference between the heat received and rejected or the work of the compressor and turbine, where the convention is that heat received is positive and heat rejected is negative and work done is positive and work absorbed is negative.  qH + q L = (h4 − h3 ) − (h5 − h0 ) { { Net work = heat in heat out wturb + wcomp = (h4 − h5 ) − (h3 − h0 )  The thermal efficiency is: Net work ηthermal = Heat input We need to calculate T3 , T5 From the definition of ηcomp , T3 − T0 With ( T3S T0 ) (T = 3S − T0 ηc )=T ( T3S 0 γ −1  γ ) T0 −1 ηc γ −1 P γ = isentropic temperature ratio =  exit  = Π comp  Pinlet  comp  γ −1  γ  Π comp −1   + T0 T3 = T0 ηcomp Similarly, by the definition of η = turb ( T4 − T5 = ηturb T4 − T5S actual work received , we can find T5 : P ideal work for same ext Pinlet ) γ −1    T5S  γ = ηturb T4 1 − = η 1 − Π T   turb 4  Turb   T4     γ −1   γ T5 = T4 − ηturb T4 1 − Π turb     2A-15 The thermal efficiency can now be found: ηthermal = 1 + T −T QL = 1− 5 0 QH T4 − T3 γ −1 1 with Π c = = Π , and τ S = Π γ the isentropic cycle temperature ratio, Πt   1  T4 1 − ηturb 1 −   − T0  τS  ηthermal = 1 −   1  τ S − 1) + 1 T4 − T0  (  ηcomp  or, ηthermal   T4 1  −τS 1 −  ηcompηturb τ S  T0  =  T4  1 + ηc  − 1 − τ S  T0  There are several non-dimensional parameters that appear in this expression for thermal efficiency. We list these just below and show their effects in subsequent figures: Parameters reflecting design choices γ −1 γ τ S = Π : cycle pressure ratio T4 : maximum turbine inlet temperature T0 Parameters reflecting the ability to design and execute efficient components ηcomp : compressor efficiency ηturb : turbine efficiency In addition to efficiency, net rate of work is a quantity we need to examine, W˙ net = W˙ turbine − W˙ compressor Putting this in a non-dimensional form: W˙ net 1 1 T  = − ηturb 4 1 −  (τ S − 1) + ˙ c p T0 m ηcomp T  τS 14402443 144 2443 work to drive compressor work extracted from flow by turbine η T4  T W˙ net 1 T 0  = (τ S − 1) − ˙ c p T0 τ η m S comp     2A-16 Trends in net power and efficiency are shown in Figure 2A-11 for parameters typical of advanced civil engines. Some points to note in the figure: • For any ηcomp , ηturb ≠ 1 , the optimum pressure ratio ( Π ) for maximum ηth is not the highest that can be achieved, as it is for the ideal Brayton cycle. The ideal analysis is too idealized in this regard. The highest efficiency also occurs closer to the pressure ratio for maximum power than in the case of an ideal cycle. Choosing this as a design criterion will therefore not lead to the efficiency penalty inferred from ideal cycle analysis. • There is a strong sensitivity to the component efficiencies. For example, for ηturb = ηcomp = 0.85 , the cycle efficiency is roughly two-thirds of the ideal value. • The maximum power occurs at a value of τ S or pressure ratio ( Π ) less than that for max η . (this trend is captured by ideal analysis). • The maximum power and maximum ηthermal are strongly dependent on the maximum temperature, T4 . T0 Muddy points T How can 4 be the maximum turbine inlet temperature? (MP 2A.16) T0 When there are losses in the turbine that shift the expansion in T-s diagram to the right, does this mean there is more work than ideal since the area is greater? (MP 2A.17) 2A-17 Figure 2A-11 available from: Cumpsty, Jet Propulsion. Cambridge University Press, February 1998. Figure 4.3 p.33. Figure 2A-11: Non-dimensional power and efficiency for a non-ideal gas turbine engine - (a) Non-dimensional work as a function of cycle pressure ratio for different values of turbine entry temperature divided by compressor entry temperature, (b) Overall cycle efficiency as a function of pressure ratio for different values of turbine entry temperature divided by compressor entry temperature, (c) Overall cycle efficiency as a function of cycle pressure ratio for different component efficiencies. [from Cumpsty, Jet Propulsion] 2A-18 A scale diagram of a Brayton cycle with non-ideal compressor and turbine behaviors, in terms of temperature-entropy (h-s) and pressure-volume (P-v) coordinates is given below as Figure 2A12. Figure 2A-12: Scale diagram of non-ideal gas turbine cycle. Nomenclature is shown in the figure. Pressure ratio 40, T0 = 288, T4 =1700, compressor and turbine efficiencies = 0.9 [adapted from Cumpsty, Jet Propulsion] Muddy points For an afterburning engine, why must the nozzle throat area increase if the temperature of the fluid is increased? (MP 2A.18a) Why doesn’t the pressure in the afterburner go up if heat is added? (MP 2A.18b) Why is the flow in the nozzle choked? (MP 2A.18c) What’s the point of having a throat if it creates a retarding force? (MP 2A.18d) Why isn’t the stagnation temperature conserved in this steady flow? (MP 2A.18e) 2A-19 Muddiest points on part 2A 2A.1 How is ∆hfuel calculated? For now, we rely on tabulated values. In the lectures accompanying Section 2.C of the notes, we will see how one can calculate the heat , ∆h fuel , liberated in a given reaction. 2A.2 What are "stoichiometric conditions"? Stoechiometric conditions are those in which the proportions of fuel and air are such that there is not an excess of each one--all the fuel is burned, and all the air (oxidizer) is used up doing it. See Notes Sections 2.C. 2A.3 When and where do we use cv and cp? Some definitions use dU=cvdT. Is it ever dU=cpdT? The answer is no. The definitions of cp and cv are derived in the notes on page 0-6. cp is the specific heat at constant pressure and for an ideal gas dh = cp dT always holds. Similarly cv is the specific heat at constant volume and for an ideal gas du = cv dT always holds. A discussion on this is also given in the notes on pages 0-6 and 0-7. If you think about how you would measure the specific heat c = q/(Tfinal – Tinitial) for a certain known change of state you could do the following experiments. For a process during which heat ∆q is transferred (reversibly) and the volume stays constant (e.g. a rigid, closed container filled with a substance, or the heat transfer in an Otto engine during combustion – the piston is near the top-dead-center and the volume is approximately constant for the heat transfer) the first law is du = dq since v = const. Using the definition du = cv dT we obtain for the specific heat at constant volume cv = ∆q/ ∆T , where both the heat transferred ∆q and the temperature difference ∆T can be measured. Similarly we can do an experiment involving a process where the pressure is kept constant during the reversible heat transfer ∆q (e.g. a rigid container filled with a substance that is closed by a lid with a certain weight, or the heat transfer in a jet engine combustor where the pressure is approximately constant during heat addition). The first law can be written in terms of enthalpy as dh – vdp = dq, and since p = const we obtain dh = dq. Using the definition dh = cp dT we obtain for the specific heat at constant pressure cp = ∆q/ ∆T . 2A.4 Explanation of the above comparison between Diesel and Otto. Basically we can operate the diesel cycle at much higher compression ratio than the Otto cycle because only air is compressed and we don't run into the auto-ignition problem (knocking problem). Because of the higher compression ratios in the diesel engine we get higher efficiencies. 2A.5 What is shaft work? I am not sure how best to answer, but it appears that the difficulty people are having might be associated with being able to know when one can say that shaft work occurs. There are several features of a process that produces (or absorbs) shaft work. First of all the view taken of the process is one of control volume, rather than control mass (see the discussion of control volumes in Section 0 or in IAW). Second, there needs to be a shaft or equivalent device (a moving belt, a row of blades) that can be identified as the work carrier. Third, the shaft work is work over and above the “flow work” that is done by (or received by) the streams that exit and enter the control volume. 2A.6 What exactly is the specific impulse, Isp, a measure of? The specific impulse is a measure of how well the fuel is used in creating thrust. For a rocket engine, the specific impulse is the effective exit velocity divided by the acceleration of gravity, g. In terms of relating the specific impulse to some characteristic time, we can write the definition of I sp as I sp m& g = F . From this, one can regard the specific impulse as the time that it would take to flow a quantity of fuel that has a weight equal to the thrust force 2A.7 How is Isp found for rockets in space where g ~ 0? The impulse I given to a rocket is the thrust force integrated over the burn time. Traditionally, for the case of constant exhaust velocity cex, the specific impulse has been used Isp = I /(mp g) = cex / g0, where mp is the propellant mass and g0 is the Earth's surface gravity. Thus Isp is measured in seconds and is a force per weight flow. Often today, however, specific impulse is measured in units meters/second [m/s], recognizing that force per mass flow is more logical. The specific impulse is then simply equal to the exhaust velocity Isp = cex. 2A.8 Why does industry use TSCP rather than Isp? Is there an advantage to this? I am not sure why Thrust*Specific Fuel Consumption was originally used. The gas turbine industry uses TSFC; the rocket propulsion industry uses essentially its inverse, Specific Impulse. Perhaps an advantage is that TSFC is a number of magnitude unity, whereas specific impulse is not. 2A.9 Why isn't mechanical efficiency an issue with ramjets? As defined, the mechanical efficiency represents bearing friction, and other parasitic torques on the rotating shaft in a gas turbine. The work associated with this needs to be provided by the turbine, but does not go into driving the compressor. The ramjet has no shaft, and hence does not encounter this. 2A.10 How is thrust created in a ramjet? You can look at thrust in several ways. One is through the integral form of the momentum equation, which relates thrust to the difference between exit and inlet velocities, multiplied by the mass flow. Another way, however, is to look at the forces on the ramjet structure, basically the summation of pressure forces on all the surfaces. I attempted to do this in class using the turbojet with an afterburner. For the ramjet, from the same considerations, we would have an exit nozzle that was larger in diameter than the inlet so that the structural area on which there is a force in the retarding direction is smaller than the area on which there is a force in the thrust direction. 2A.11 Why don't we like the numbers 1 and 2 for the stations? Why do we go 0-3? A common convention in the industry is that station 0 is far upstream, station 1 is after the shock in the inlet (if there is one), station 2 is at inlet to the compressor (after the inlet/diffuser) and station 3 is after the compressor. In class, when we examined the ramjet we considered no changes in stagnation pressure between 0 and 2, so I have used 0 as the initial state for the compression process. It would be more precise to differentiate between stations 0 and 2, and I will do this where appropriate. 2A.12 For the Brayton cycle efficiency, why does T3=Tt0? The ramjet is operating as a Brayton cycle where ηb= 1 – Tinlet / T compressor exit. For the ramjet discussed in class the inlet temperature is T0 and since there is no compressor (no moving parts) the only compression we get is from diffusion. We assumed isentropic diffusion in the diffuser and found for very low Mach numbers that the diffuser exit or combustor inlet temperature T3 is Tt3. From first law we know that for a steady, adiabatic flow where no work is done the stagnation enthalpy stays constant. Assuming perfect gas we thus get Tt0 = Tt3 = T3. So we can write for the ramjet thermal efficiency ηb= 1 – T0 / Tt0 . 2A.13 How can we idealize fuel addition as heat addition? The validity of an approximation rests on what the answer is going to be used for. We are seeking basically only one item concerning combustor exit conditions, namely the exit temperature or the exit enthalpy. The final state is independent of how we add the heat, and depends only on whether we add the heat. If it is done from an electrical heater or from combustion, and if we neglect the change in the constitution of the gas due to the combustion products (most of the gas is nitrogen) the enthalpy rise is the same no matter how the temperature rise is achieved. 2A.14a What is the relation between h t4 − h t3 = f∆h f and the existence of the maximum value of Tt4? The two are very different physical statements. The first is the SFEE (steady flow energy equation) plus the approximation that inlet and exit mass flows to the control volume are the same. The heat received within the volume is represented by the quantity f∆h f , where ∆h f is the heat liberated per kilogram of fuel. The second statement is a representation of the fact that the degree of completion of the reaction in the combustor depends on temperature, so that even though the inlet temperature increases strongly as the Mach number increases, the combustor exit temperature does not change greatly. This is an attempt to represent a complex physical process (or set of processes) in an approximate manner, not a law of nature. 2A.14b Why didn’t we have a 2s point for the Brayton cycle with non-ideal components? If we didn’t, we should have, or I should at least have marked the point at which the compressor exit would be if the compression process was isentropic. 2A.15 What is the variable fstoich? fstoich is the fuel-to-air ratio for stoichiometric combustion, or in other words the fuel-toair ratio for a chemically correct combustion process during which all fuel is burnt. 2A.16 How can T4 be the maximum turbine inlet temperature? T0 I agree that the T4/T0 is a temperature ratio. If we assume constant ambient temperature then this ratio reflects the maximum cycle temperature. The main point was to emphasize that the higher your turbine inlet temperature the higher your power and efficiency levels. 2A.17 When there are losses in the turbine that shift the expansion in T-s diagram to the right, does this mean there is more work than ideal since the area is greater? We have to be careful when looking at the area enclosed by a cycle or underneath a path in the T-s diagram. Only for a reversible cycle, the area enclosed is the work done by the cycle (see notes page 1C-5). Looking at the Brayton cycle with losses in compressor and turbine the net work is the difference between the heat absorbed and the heat rejected (from 1st law). The heat absorbed can be found by integrating TdS = dQ along the heat addition process. The heat rejected during the cycle with losses in compressor and turbine is larger than in the ideal cycle (look at the area underneath the path where heat is rejected, this area is larger than when there are no losses dsirrev = 0 – see also muddy point 1C.1). So we get less net work if irreversibilities are present. It is sometimes easier to look at work and heat (especially shaft work for turbines and compressors) in the h-s diagram because the enthalpy difference between two states directly reflects the shaft work (remember, enthalpy includes the flow work!) and / or heat transfer. 2A.18a For an afterburning engine, why must the nozzle throat area increase if the temperature of the fluid is increased? The Mach number of the flow is unity at the throat with and without the afterburner lit. The ratio of static pressure to stagnation pressure at the throat is thus the same with and without the afterburner lit. The ratio of static temperature to stagnation temperature at the throat is thus the same with and without the afterburner lit. Tt P γ −1 2 γ − 1 2  γ /( = 1+ M ; t = 1 + M  2 2 T P  The flow through the throat is P γRT Athroat . ṁ = ρcAthroat = ρaAthroat = RT The flow through the throat thus scales as  P  Athroat   ˙ mA / B  T  A/ B =  P  m˙ noA / B A    T throat  noA / B γ −1 ) From what we have said, however, the pressure at the throat is the same in both cases. Also, we wish to have the mass flow the same in both cases in order to have the engine operate at near design conditions. Putting these all together, plus use of the idea that the ratio of stagnation to static temperature at the throat is the same for both cases gives the relation Athroat A / B Tt A / B = AthroatnoA / B TtnoA / B The necessary area to pass the flow is proportional to the square root of the stagnation temperature. If too much fuel is put into the afterburner, the increase in area cannot be met and the flow will decrease. This can stall the engine, a serious consequence for a single engine fighter. 2A.18b Why doesn’t the pressure in the afterburner go up if heat is added? From discussions after lecture, the main point here seems to be that the process of heat addition in the afterburner, or the combustor, is not the same as heat addition to a gas in a box. In that case the density (mass/volume) would be constant and, from P = ρRT , increasing the temperature would increase the pressure. In a combustor, the geometry is such that the pressure is approximately constant; this happens because the fluid has the freedom to expand so the density decreases. From the equation P = ρRT if the temperature goes up, the density must go down. 2A.18c Why is the flow in the nozzle choked? As seen in Unified, choking occurs when the stagnation to static pressure ratio ( Pt / P) gets to a certain value, 1.89 for gas with γ of 1.4. Almost all jet aircraft operate at flight conditions such that this is achieved. If you are not comfortable with the way in which the concepts of choking are laid out in the Unified notes, please see me and I can give some references. 2A.18d What’s the point of having a throat if it creates a retarding force? As shown in Unified, to accelerate the flow from subsonic to supersonic, i.e., to create the high velocities associated with high thrust, one must have a convergingdiverging nozzle, and hence a throat. 2A.18e Why isn’t the stagnation temperature conserved in this steady flow? Heat is added in the afterburner, so the stagnation temperature increases.