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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