Jet Propulsion - Lecture Thrust from a jet engine is essentially achieved by mass flow through five stages. 1. First, the air enters an intake, which provides uniform subsonic air to the compressor blades. 2. Second, the air in the compressor is then compressed to a high pressure for the combustion chamber. 3. Third, when the air arrives in the combustion chamber, it is mixed with fuel and ignited to produce very high energy air to drive the turbine. 4. Fourth, the turbine has two functions: drive the compressor and accelerate the high energy air. 5. Fifth, the exhaust nozzle discharges the air at the highest possible velocity. Recalling Newton’s 2nd law, F = m a , we can express T = Q (V2 V1) (Equation 6.1, Dole et al., 2017) where Q is the mass Flow, V2 is the exit velocity, and V1 is the intake velocity. Q can also be expressed, using Equation 2.6 (Dole et al., 2017): Q = ρ A V. Therefore, T = ρ A V1 (V2 - V1) Thrust is directly proportional to density. As the density increases so does available thrust. See Figure 6.9 (Dole et al., 2017). Thrust is essentially constant with airspeed; as airspeed increases, the change in velocity decreases, so thrust stays the same. This is an approximation and each engine will be different. The higher by-pass engines (turbofans) will lose available thrust as airspeed increases. Thrust is affected by density, so temperature indirectly affects engine thrust because temperature affects density. Turbo-jet engine RPM also affects thrust. The higher the RPM, the higher the thrust. But this is not linear. Looking at Figure 6.11 (Dole et al., 2017), it shows thrust available at 100% RPM and thrust available at 95%. In this example, the 5% RPM reduction yields a 50% reduction in thrust. Fuel consumption is also affected by temperature and engine RPM. We usually express fuel consumption as specific fuel consumption, (Ct), the fuel flow per pound of thrust. This is an indication of the efficiency of the engine. Examining Figure 6.10 (Dole et al., 2017), you see it varies with engine RPM and with airspeed. The higher the RPM, the more efficient the engine is. This doesn’t mean you use less fuel. It’s just less fuel flow per pound of thrust. Examining Figure 6.11 (Dole et al., 2017), you can see that specific fuel consumption is less at higher altitudes. This is due mainly to temperature. You can see that above the tropopause, where the temperature actually increases, specific fuel consumption actually increases again. Prop Propulsion-Lecture Jet aircraft performance is related to mass airflow and differential velocity; or thrust. And thrust required is equal to drag. So, determining jet performance (fuel consumption) is based on interpreting drag (thrust required) curves. Propeller aircraft do not produce thrust directly. Propeller engines produce power by driving a shaft that turns a propeller. So, determining propeller performance (fuel consumption) is based on interpreting power required (thrust * velocity) curves. First, let’s relate jet performance to power. 1. What is the equivalent horsepower of a jet aircraft engine producing 32,000 lb of thrust at 450 KTAS? Using Equation 1.13 (Dole et al., 2017): (325) = 44,307 HP P[HP] = T V[kts]/325 = (32,000) (450) / 2. What is the power (ft-lb/s) of a jet engine aircraft engine producing 32,000 lb of thrust at 450 KTAS? P = T V = (32,000 lb) (450 nm/hr) (1.69 ft/s *hr/nm) = 24336000 ft-lb/s You recall from the jet performance lecture and examples that induced drag increases inversely to velocity squared while parasite drag increases directly with velocity squared. Propeller performance is proportional to power required. Comparing power to drag, the power required due to induced drag is inversely proportional to velocity, and power required due to parasite drag is directly proportionally to velocity cubed. A comparison of drag or thrust required to power required (due to drag) is shown in Figure 8.7 (Dole et al., 2017). Now’s lets discuss propeller performance. Here are a few definitions: • • • • Brake horsepower (BHP) - measured at crankshaft Shaft horsepower (SHP) - measured at propeller shaft Thrust horsepower (THP) - usable horsepower due to propeller loses Propeller efficiency, η o η = THP / SHP (Equation 8.2, Dole et al., 2017) 3. What is the useable horsepower if the Shaft horsepower is 450 HP and the propeller efficiency of .85? THP = η SHP = (.85) (450 HP) = 382.5 HP Factors affecting propeller efficiency are: • • • Diameter of the propeller increases efficiency Compressibility effects and high tip speeds decrease efficiency Blade angle Jet Aircraft Applied Performance – Example If the T-38 produces 4200 lb of thrust at sea level at 100% RPM, what is the amount of thrust available at 100% RPM at 33,000 ft. See Figure 6.9 (Dole et al., 2017). Use Figure 6.9 (Dole et al., 2017), at 33,000 ft; the thrust available is 40% of sea level thrust. Ta = (.40) (4200) = 1680 lb Compare max level airspeed at 100% RPM to 95% RPM for aircraft in Figure 6.13 (Dole et al., 2017). Where thrust available = thrust required at a given thrust setting determines max level airspeed. Where there is no excess thrust, there is no climb. @100% RPM Max level speed is ~610 KTAS @95% RPM Max level speed is ~510 KTAS Climb Performance Find angle of climb for 100 % RPM at (L/D)max for an aircraft at W = 12,000 lb. Best angle of climb is also called Vx. This is where the thrust available, compared to the thrust required, is greatest. Given a constant thrust available curve vs. airspeed, an assumption for turbo-jet aircraft but not so for turbofan engines, Vx occurs at (L/D)max. Looking at Figure 6.14 (Dole et al., 2017) and rearranging Equation 6.5b (Dole et al., 2017) produces: Angle of climb ( γ) = sin-1 (Ta−TrW) Now using Figure 7.2 (Dole et al., 2017) and using the thrust available from Figure 6.13 (Dole et al., 2017), @ 100% RPM Ta = 4200 lb. At W = 12,000 lb, the min thrust occurs at ~260 KTAS and is ~1000 lb. γ = sin-1 So, (4200−100012,000)= sin-1 (.267) = 15.5 deg At any other airspeed, the aircraft cannot sustain an angle higher than this for this weight. If you decrease the weight to 8,000 lb, then the climb angle will increase, and the best angle airspeed will decrease because, with the decrease in induced drag, the thrust required curve goes down and to the left on the airspeed vs. thrust required graph. At 8,000 lb the Min Drag or (L/D)max occurs at ~210 KTAS. Assuming the same thrust available = 4200 lb and using in Figure 7.2 (Dole et al., 2017) the 8,000 lb curve where min thrust = ~700 lb: γ = sin-1 (4200−70012,000) = sin-1 (.438) = 25.9 deg Find rate of climb using 95% power at best climb angle airspeed from Figure 7.2 (Dole et al., 2017) for the W = 12,000 lb T-38. The thrust available at 95% RPM is 2100 lb (from Figure 6.13, Dole et al., 2017). The best angle of climb airspeed for the W = 12,000 lb curve is at the minimum drag point of that curve in Figure 7.2 (Dole et al., 2017), at ~260 KTAS, for which thrust required is 1,000 lb. Using Equation 6.6 (Dole et al., 2017): ROC = Vk 23.9 kts (Ta−TrW) = 260 kts (2100−100012,000)= 260 (0.092) = We traditionally think of rate of climb in the units of ft/min. So, (23.9 nm/hr) x (6076 ft/nm) / (60 min/h) = 2420 ft/min If the weight is decreased the rate of climb will increase. Using Figure 7.2 (Dole et al., 2017) for the 8,000 lb curve, and still 95% RPM with Ta = 2100 lb. ROC = 220 (2100-700)/8,000 = 38.5 nm/hr = 38.5(101.3) ft/min = 3900 ft/min Cruise Performance Find max endurance airspeed (VBE) and fuel flow of the jet aircraft depicted in Figure 6.13 (Dole et al., 2017). Specific fuel consumption (Ct) is assumed constant at 1.4 lb/hr/lb. Max endurance airspeed occurs at (L/D)max for a jet aircraft. Flying at max endurance airspeed, it is assumed that the aircraft thrust is set for level flight and therefore thrust required = thrust available. This is the minimum drag; therefore, minimum thrust required. At a given altitude and atmospheric conditions, the thrust specific fuel consumption (Ct) can be assumed fairly constant at different airspeeds. So, fuel flow = thrust required * (Ct) From Figure 6.13 (Dole et al., 2017), the min drag point or (L/D)max is about 240 KTAS and the thrust required is 830 lb. FF = (830 lb) (1.4 lb/hr/lb) = 1162 lb/hr Find specific endurance at VBE. SE = 1/FF = 1/ 1162 lb /hr = .00086 hr/lb = .86 hr/ 1000 lb fuel Find stall speed (VS), thrust required, and fuel flow for aircraft shown in Figure 6.13 (Dole et al., 2017). Ct is assumed constant at 1.4 lb/hr/lb. According to the graph estimate, the %RPM to maintain flight at VS. CL(max) occurs at slowest point on thrust required curve on Figure 6.13 (Dole et al., 2017), estimated to be at 160 KTAS. Thrust required there is 2000 lb. FF = (2000lb) (1.4 lb/hr/lb) = 2800 lb/hr At 95% RPM, the engines produce 2100 lb of thrust. At 100% RPM the engines produce 4200 lb of thrust. It is not linear, but 5% RPM yields 2100 lb in this range. So, 1% is 2100 / 5 = 420 lb reduction. Therefore, a 100 lb reduction is equivalent to 100 / 420 = 0.24 % RPM reduction. Likewise, at VS, 2000 lb of thrust is required which is approximately 95% - 0.24% = 94.76% PRM. Compare max range speed (VBR) for aircraft shown in Figure 7.6 (Dole et al., 2017). W = 12,000 lb and W=8,000 lb @ 8,000 lb VBR = ~290 KTAS @ 12,000 lb VBR = ~320 KTAS Therefore, as weight decreases best range speed decreases. Compare max range speed (VBR) for aircraft shown in Figure 7.5 (Dole et al., 2017). W = 10,000 lb Altitude Sea Level Altitude 20,000 ft @sea level VBR = ~290 KTAS @ 20,000 VBR = ~375 KTAS Therefore, as altitude increases KTAS for best range increases. Find maximum range airspeed during a 100 KT headwind in Figure 6.23 (Dole et al., 2017). Max range airspeed is ~ 300 KTAS with no wind. To account for the headwind, you draw a tangent line from 100 KTS on the x axis. It will touch the thrust required curve at a greater airspeed. In this example the new best range airspeed is 320 KTAS. Find maximum specific range of aircraft in Figure 6.13 (Dole et al., 2017). Thrust specific fuel consumption (Ct) is 1.4 lb/hr/lb. No wind. Max range airspeed (VBR) is 300 KTAS and thrust required is 1000 lb. Fuel flow = (Thrust) (Ct) = (1000 lb) (1.4 lb/hr/lb) = 1400 lb/hr Specific range = V / FF = (300 nm/h) / (1400 lb/h) = 0.21 nm/lb Cruise Climb If air traffic control (ATC) allows achieving increased maximum range, a cruise climb can be used. As fuel burns down, as shown in Figure 7.6 (Dole et al., 2017), the best range airspeed is actually less. And as shown in Figure 7.5 (Dole et al., 2017), an increased altitude yields a higher true airspeed for close to the same fuel flow. So, it is best to fly at the higher altitude, but on the initial climb out the aircraft weight might be too high to climb to a high altitude, so if you do a slow climb remaining at the best range AOA or (CLCD)max after your initial climbout, as fuel burns you allow the aircraft to climb. This can extend your range ~10%. Glide Performance If you lost all engine thrust at 10,000 ft what angle of attack would be the best to fly and what would the max glide range be? Use Figure 5.20 (Dole et al., 2017). Max range occurs at best (L/D)max. In Fig 5.20, (L/D)max occurs at ~6 deg AOA at a L/D of ~12.5/1. The L/D represents the glide path ratio. If you fly at best L/D then for every ft of altitude lost you will glide 12.5 ft horizontally forward. Thus, at 10,000 ft of altitude, you will glide ~12.5 * (10,000 ft) = 125,000 ft or 125,000 ft / 6076 ft/nm = 20.6 nm. The glide path is independent of weight. If the L/D is 12.5/1 the glide range is 20.6 nm. If the weight is greater than the speed at which (L/D)max is achieved is greater, but the L/D is still the same. So, heavier, faster, but same range, for gliding.
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