Category AERODYNAMICS, AERONAUTICS, AND FLIGHT MECHANICS

In flying (or flight testing) an airplane, one rarely finds a standard atmosphere. In addition, the gross weight of the airplane is generally different from that used by the manufacturer to quote performance. The following method is useful for interpreting power measurements taken at any gross weight and density altitude.

Few and Pew are defined as the velocity and required power at sea level, where “ew” signifies equivalent weight. These quantities are related to a standard weight, by,

(7.23)

(7.24)

For any other density or gross weight, the velocity and power are given

I’hus, at the same lift coefficient, V and Vcw are related by

(7.25)

Similarly,

(7.26)

A graph of Pew versus Vew will simply be the sea level power required curve at the standard gross weight. As an example, take the Cherokee Arrow at a gross weight of 2400 lb, a density altitude of 5000 ft and an airspeed of 100 kt. Using Figure 7.15 as the standard,

At 5000 ft, <T equals 0.861. Therefore,

Vew = 97.5 kt = 164.8 fps

From Figure 7.15,

Pew = 73 hp

Therefore, from Equation 6.26,

Pr = 67.8 hp

In this manner, given a sea level power required curve at a standard gross weight, one can easily determine the power required at any altitude, airspeed, and gross weight.

These relationships are particularly useful in flight testing. Power – fequired data taken at any altitude and gross weight are reduced to a plot of PewUew versus Vew – Such a plot will be a straight line having a slope proportional to / and an intercept proportional to Це. Thus all of the data collapses to a single, easily fitted line enabling one to determine accurately e and /. The equation of this straight line follows directly from Equation 7.21.

Timet

4

The time required to climb from one altitude, h, to another, h2, can be determined by evaluating the integral

Knowing the R/C as a function of h, this integral can easily be evaluated numerically. A solution in closed form can be obtained if one assumes that the rate of climb decreases linearly with altitude. A feeling of how valid this assumption is can be gained from Figures 7.13 and 7.14.

Let

R/C = (R/C)0(l – hlhabs) (7.29)

where habs = absolute ceiling.

Equation 7.28, for h = 0, then reduces to

___ ^abs j _________________ |________

~~ (R/C)0 n 1 – hlhabs

This represents the time required to climb from sea level to the altitude, h. The time required to climb from one altitude to another is obtained directly from Equation 7.30 by subtracting, one from the other, the times required to climb from sea level to each altitude.

Time to climb is presented for the Cherokee Arrow in Figure 7.16. This curve was calculated on the basis of Equation 7.30. For this particular example, it requires approximately 42.5 min to climb to the service ceiling.

The rate of climb was given previously by Equation 7.13. For a steady climb this equation becomes

dh V(T-D) dt W

It can be expanded and expressed, as noted previously, as

dh Pa – Pr dt W

where Pa and Pr are the available power and required power, respectively.

Pa = TV Pr = DV

For a gas turbine engine, T is known as a function of altitude and velocity, so that Equation 7.15 is the obvious form to use in this case to determine the rate of climb, R/C. In the case of a propeller-driven airplane,

Pa Г/F shaft

where Pshaft is the shaft power delivered to the propeller and 17 is the propeller efficiency.

As a function of V, the required power is determined by

Pr = DV

Let us apply Equation 7.15 to the 747-100 example using the rated climb thrust given in Figure 6.37. Flaps and gear during the climb are assumed to be retracted, so that / reduces to lOO. ft2 (9.29 m2). As an example, take a speed of 200 m/s at an altitude of 6000 m. The acoustic velocity of this altitude equals 318 m/s. Hence the operating Mach number equals 0.629. From Figure 6.37.

T = 311 kN (4 engines)

The mass density at 6096 m equals 0.662 kg/m3. V and p result in a dynamic pressure of

q = 13,240 Pa

With a weight of 3260 kN and a wing area of 511 m2, the lift coefficient equals 0.482. Hence Cd, =0.0152. Therefore,

D = q(f+C»S)

= 226 kN

Equation 7.15 gives, for the rate of climb,

nl„ 200(311-226) R/C 3260

= 5.2 m/s

In this way, the curves presented in Figure 7.12 were obtained. Here, the rate of climb is plotted as a function of true airspeed at four different altitudes. At the highest altitude and air speeds, the curve is probably optimistic, since the drag rise beyond the drag-divergence Mach number is not considered.

At a given altitude, there exists an optimum airspeed for maximum rate of climb. In this case, the airspeed for the best rate of climb is seen to correspond to a nearly constant indicated air-speed of 296 kt.

The speed for the best climb angle is less than the speed for the best R/C. The angle of climb, 0C, in radians, is given by

For any velocity, this angle is represented by the angle between the abscissa and a straight line from the origin to the R/C curve on Figure 7.12. As shown for the sea level curve, this angle is a maximum for the straight line that is tangent to the curve. Thus, to clear an obstacle ahead, a pilot should fly slightly slower than the speed that gives the maximum rate of climb.

The optimum airspeed for maximum rate of climb can be obtained approximately from Equation 7.15 for a turbojet or turbofan.

The maximum rates of climb obtained from Figure 7.12 are plotted as a function of altitude in Figure 7.13. The absolute ceiling’ is defined as the altitude for which the R/C equals zero. In this case, this corresponds to an altitude of 9.20 km.

Consider now the rate-of-climb calculation for a propeller-driven air­plane. We will use as an example the Piper Cherokee Arrow that is similar to the Cherokee 180 except that the former has a retractable landing gear, a 200-bhp (149-kW) piston engine, and a constant speed propeller. The engine is rated at 2700 rpm but, for a continuous climb in accordance with recom­mended practice, an rpm of only 2500 will be used. At standard sea level, the engine develops 185 bhp (138 kW) at this rpm. For higher altitudes, the engine power is estimated on the basis of Figure 6.2. The following values are known or have been estimated for this airplane.

The propeller performance curves for this airplane were presented in Figures 6.16 and 6.17. These curves, together with the engine power, are used to estimate the available power. As an example, consider an altitude of

10,0 ft (305 m). The engine power at this altitude equals 130 bhp. At a speed of, say, 140 fps, the advance ratio will be

J = VlnD

= 140(0.1467)/(2500/60)/6.17 = 0.389

The power coefficient equals

Lp pn3D5

= 130(550)/(0.00176X41.7)3(6.17) = 0.063

This power coefficient and advance ratio lead to a blade angle of 2 Г from Figure 6.16. Using this blade angle, together with J, results in a propeller efficiency, 17, of 0.70 from Figure 6.17. Thus the available power at this speed
and altitude equals 91 thp where thp stands for “thrust horsepower.” The equivalent term in the SI system would be tkW, for “thrust kilowatts.”

The power required for the Arrow is calculated from

Pr = DV

which can be expressed in the form

p _ Pfvз і 2(Wlb)2 1 r 2 пре V

The first term on the right side is the parasite power and the second term the induced power. The shape of this relationship will be discussed in more detail later. For now let us simply evaluate Equation 7.21 at the altitude of 10,000 ft and a speed of 140 fps. The obvious substitutions result in a required power of 65.3 hp. The rate of climb can now be calculated from Equation 7.16. However, in so doing, the excess power must be expressed in foot-pounds per second.

(91 – 65.3)550
2650

= 5.34 fps

It is current practice in the American aviation industry to express the R/C in feet per minute, so that the above becomes 320 fpm (1.63 m/s).

In this manner, the rate of climb can be calculated over a range of speeds for several altitudes for the Arrow. The maximum rates of climb thus determined are presented in Figure 7.14, where the R/C is seen to decrease almost linearly with altitude. The altitude for which the R/C equals 100 fpm (0.5 m/s) is shown on Figure 7.14 and is called the service ceiling. In this example, the calculated service ceiling and sea level rate of climb are close to the corresponding values quoted by the manufacturer.

The power-required and power-available curves calculated for this example at sea level are presented in Figure 7.15. Similar to the drag curve in Figure 4.52, the power-required curve has a minimum value at some speed. Below this speed, it actually requires more power to fly slower. This part of the curve is referred to as the backside of the power curve.

‘ The speed for minimum required power can be found by setting to zero

the derivative of equation 7.21 with respect to V. This leads to

(7.22)

The preceding speed is, of course, not necessarily the speed for maxi­mum rate of climb, since the available power varies with airspeed. In the case of the Arrow, Figure 7.15 shows the optimum speed to be greater than the value given by Equation 7.22. Nevertheless, one might expect the two speeds to be related. Equation 7.22 shows the minimum required power to occur at a constant indicated airspeed. This suggests that the maximum rate of climb will also occur at a constant indicated airspeed. This was found to be true for the 747-100 and also holds for piston engine airplanes.

During the takeoff run the pilot of a jet transport has the option of aborting the takeoff up to the speed Vj. Above this speed, in the event of an engine failure, the takeoff should be continued. Figure 7.11 (from Ref. 7.1) clearly illustrates these options. A balanced field length is defined as one where the distance to continue the takeoff following the recognition of an engine failure at Vi is equal to the distance required to stop if the takeoff – should be aborted. On Figure 7.11, the field length is balanced if the sum of seginents В and C equals the sum of D and E. FAR Part 25 stipulates the field—length to be the greatest of the accelerate-and-go distance, the ac- celerate-and-stop distance, or 115% of the all-engine-operating distance to a 35-ft height. The stop-and-go portions need not be balanced.

The distance to stop can be found by numerically integrating Equation 7.1. In this case, T will be negative and equal to the reverse thrust. ^ is the braking coefficient and, with antiskid systems, can be as high as 0.6 on a dry, hard surface. Detailed considerations relating to the deceleration of an air­plane will be deferred until the later section on landing distance.

Becoming airborne at the speed FLof, an airplane continues to accelerate to the speed V2 over the obstacle height. An airplane that is accelerating both normal to and along its flight path is pictured in Figure 7.9. As shown, V(deldt) is the acceleration normal to the flight path and dVIdt is the acceleration along the flight path. The equations of motion normal to and along the flight path can be written as

The rate of climb, dh/dt, and denoted by Vc in Chapter One, is found * from

dhldt =V sine (7.12)

Rate of climb is also denoted in the literature by R/C. Solving for sin в from Equation 7.10, Equation 7.12 can be written as

dh_ Г(Г-£>) dV/dtl

dt V[ W t g J

This can also be expressed as

(T-d>V-W£+£(Wv!) <714)

n

TV is the available power and DV is the required power. Thus (T-D)V is the excess power that, as Equation 7.14 shows, can be used either to climb or to accelerate. Actually, Equation 7.14 is an energy relationship that states that the excess power equals the sum of the time rates of change of the potential energy and the kinetic energy.

Let us now apply these relationships to the calculation of the horizontal

distance required during the takeoff flare to attain a specified height. The actual flight path that is followed during the flare, or transition, segment of the takeoff depends on pilot technique. Referring to Equations 7.10 and 7.11, V and в are the independent variables, while g, W and T are known, the latter as a function of V. L and D are functions of V and the airplane’s angle of attack a. By controlling a, and hence CL, the pilot can fly a desired trajectory (i. e., the pilot can accelerate or climb or do some of each). During the takeoff, however, in attempting to clear an obstacle, FAR Part 25 limits the operating CL to approximately CimJ.2 at FLof and to C^J 1.44 at V2. Therefore, in calculating the flare distance, it will be assumed that CL varies linearly with V between these limits and is constant for speeds above V2. Thus, if V2 is £ ’ і before the specified obstacle height is reached, the stall margin on CL

і tained.

turning to the example of the 747 at 733,000-lb takeoff weight, let us assume a with partially deflected flaps of 1.8. In addition, let FLOf be 10% above the stall speed. Thus, at sea level,

Flop — 83.7 m/s CLlof= 1.49 V2 = 91.2 m/s = 1.25

Assuming an / of 18.6 m2 (200 ft2) and calculating the induced drag obtained using CL, the effective aspect ratio, and Figure 7.4, the drag can be determined at V’lqf to be

where q = 4289 Pa and CD; = 0.0203. Thus,

D = 225.8 kN

The rated takeoff thrust at this speed for the four JT9D-7A engines equals 654.7 kN. From Equation 7.10 for an initial в of zero,

Initially at Vlof, F and W are exactly equal, so that dd/dt is zero from Equation 7.11. For illustrative purposes, however, let us take a time increment of 1 sec and assume the preceding acceleration to be constant over that time. At the end of 1 sec, V then becomes 85.0 m/s and CL is reduced slightly to 1.45. Thus, at this time,

L = 3280 kN (737,400 lb)

dd/dt from Equation 7.11 then becomes 0.000693 rad/sec. Averaged over the
1 sec, this value of d8/dt gives a climb angle of approximately 0.02° 1 sec after lift-off.

The rate of climb, dhldt is obtained from Equation 7.12. Over the 1 sec, the average rate of climb will equal

dhldt (average) = 1(85) sin (0.02)

= 0.0148 m/s

Using this average rate of climb results in an altitude gain of 0.0148 m over the 1 sec.

The complete numerical solution of this example is shown in Figure 7.10. A time increment of 0.1 sec was used in the numerical integration. It can be see that a distance of 3140 ft (957 m) and 10.3 sec are required to attain an altitude of 35 ft (10.7 m). For the first half of the flare following lift-off, the airplane is primarily accelerating with little of the excess power going into climb. After approximately 7 sec, the acceleration decreases and the rate of climb increases rapidly.

The flare distance to clear 35 ft together with the ground roll distance of 6350 ft shown in Figure 7.6 to reach Vlof gives a total calculated takeoff distance of 9490 ft. Reference 5.11 quotes a value of 9450 ft at the same gross weight. This agreement may be fortuituous in view of the uncertainty on / with the gear and flaps down.

A headwind will always reduce the ground roll distance required for an airplane to attain a desired airspeed. Although the headwind increases the drag and decreases the thrust for a given ground speed, it increases the lift and adds directly to the ground speed to increase the airspeed so that the net effect on takeoff distance is favorable.

The effect of jthe wind is most easily determined by using the ap­proximation of Equation 7.4, keeping in mind that Г, Д and L depend on the airspeed and not the ground speed.

If Va denotes the ground speed and Vw the headwind then, for this case, Equation 7.4 becomes

The factor к is presented graphically in Figure 7.8.

Suppose, in the previous example, that the 747 was taking off into a 30-kt (50.7-fps) headwind. Using the same lift-off velocity of 274.5 fps gives a ratio s of headwind to airspeed of 0.185. From Figure 7.8, к equals 0.0753. Thus the average acceleration should be evaluated at an airspeed of 207 fps. At this speed,

T = 153,7001b D= 12,736 lb L = 280,200 lb

Therefore, a = 5.79ft/sec2.

The ground roll distance is thus determined to be

(274.5 – 50.7)2
2(5.79)

= 4325 ft (7.9)

Compared to the no-headwind case, the 30-kt headwind decreases the ground roll distance by more than 30%.

It should be noted that FAR Part 25 requires a conservative estimate of

headwind effects. Takeoff and landing distances must be calculated on the basis of 50% of reported headwinds and 150% of reported tailwinds. Since the wind can vary from one instant to another, the intent of FAR Part 25 is to use only half of any wind that improves performance but 150% of the opposite.

The forces acting on an airplane during the ground roll portion of the takeoff are shown in Figure 7.2. From Newton’s second law of motion,

T – D – p(W – L) = MV

is the coefficient of rolling friction, and a dot above the V denotes differentiation with respect to time, ц values can range from approximately 0.02 to 0.1, depending on the surface. The lower value corresponds to a hard, dry surface; the higher value might correspond to moderately tall grass.

Before rotation, the attitude of the airplane on the ground is constant and hence CL and CD are constant. After rotation, CL and CD increase, but still remain constant until lift off. The most direct means of solving Equation 7.1 is numerically, using a digital computer. This will require a table lookup or a curve fit for T as a function of D. After V is obtained as a function of time, it can then be numerically integrated to obtain s so that V will be known as a function of s.

A word of caution regarding CD is in order. Ground effect may reduce the induced drag significantly. Hence CD as a function of CL is less during the ground roll than it is in the air. In view of this, a fairly good approximation is to neglect the induced drag for calculating the total airplane drag during the ground roll, particularly for tricycle landing gears where the wing is nearly level during the ground roll.

A simple model that can be used to predict the relative effect of the ground on CD, can be obtained by replacing the wing by a simple horseshoe vortex. The span of the vortex is taken to equal 7t/4 times the span of the wing, b, since the vortex sheet shed from an elliptic wing can be shown to roll up into a pair of vortices spaced this distance apart. This horseshoe vortex system, together with the image system required to cancel vertical velocities at the ground, is shown in Figure 7.3. The ratio of the downwash induced at the midspan of the bound vortex, including the image system to the down – wash without the image, equals approximately the ratio of the induced drag in ground effect (IGE) to the induced drag, out-of-ground effect (OGE). Using the Biot-Savart law, this ratio becomes

Cn(IGE)’ = (16 hlb)2

Cd,(OGE) 1+(16hlb)1 4 ]

Equation 7.2 is plotted in Figure 7.4 where, for a typical value of h/b of 0.1, the induced drag IGE is seen to equal only 20% of the induced drag OGE for the same CL-

During the ground roll an increment to the parasite drag is required for airplanes with retractable landing gear. This increment can be estimated using the material presented in Chapter Four or on the basis of Figure 7.5. (Ref. 7.4). This figure presents the equivalent flatyjlate area, /, as a function of gross weight for three different types of landing gear.

Returning to Equation 7.1, let us apply a simple Euler method to the numerical integration of this equation. For illustrative purposes, a Boeing 747-100 will be used. This airplane is powered by four JT9D-7A turbofans having the rated takeoff thrust shown in Figure 6.36. This thrust at sea level for one engine is approximated closely by

Г = 46,100-46.7 V + 0.0467 V2

where T is in pounds and V is in feet per second.

Standard sea level conditions are chosen together with a gross weight of

733,0 lb (3260 kN). hlb for this airplane is approximately 0.08, giving an 86% reduction in the induced drag due to ground effect. Since takeoff is made with partially deflected flaps, a Cl of 1.0 will be assumed. Clean, the equivalent flat-plate area for this airplane is estimated to be 100 ft2 (9.3 m2). With gear

0 0.1 0.2 0.3

h

j

Figure 7.4 Reduction in CD, resulting from ground effect.

and flaps down, / is assumed to double, having a value of 200 ft2 (18.6 m2). (See Equation 3.46 and Figure 7.5.) These numbers are strictly estimates and may not agree with those used by the ntanufacturer. These numbers, together with other constants to be used for this example, are listed below.

W = 733,000 lb S = 5500 ft2 A = 6.69 e = 0.7 b = 196 ft H = 0.02

CD.(IGE)/CD,(OGE) = 0.14 / = 100 ft2 clean = 200 ft2 gear and flaps down

The numerical integration uses the following approximations.

" V(t + At) = V(t)+ V(t)At

s(/ + AO = s(0 + [V(f)+V(/ + A/)]y

Initially the airplane is at rest, so that V is calculated from

= 7.46 ft/s2

Choosing a time increment, At, of 0.1 sec for the numerical integration, at the next time step,

/=0.1 sec V = 0 + (7.46)(0.1)

= 0.746 fps

s =0 +(0 + 0.746X0. l)/2 = 0.0378 ft

With V now finite, the drag must be calculated. For this time step, D is negligible, being equal to only 0.171b. However, if this numerical integration is continued to, say, 32 sec, a distance of 3509 ft and a velocity of 210.3 fps are calculated. At this instant of time and for this velocity, the total thrust equals 153,377 lb and the lift and drag are calculated to be 289,200 lb and 13,1461b respectively. At this speed the drag is indeed significant. The in­stantaneous acceleration is found from Equation 7.1.

= 5.77 ft/sec2 (7.3)

Figure 7.6 presents the numerical solution of Equation 7.1 up to a velocity of 274.5 fps for the 747-100. This particular velocity is 10% above the stalling speed, assuming a of 1.8. In the next section, this particular value of V is chosen as the lift-off speed for calculating the flare distance over an obstacle.

Since CL and / are uncertain, Figure 7.7 was prepared to shovv the sensitivity of the ground roll distance to these parameters for a constant lift-off speed for the 747-100. s is seen to increase almost linearly with / while being rather insensitive to CL.

For preliminary design studies, an approximate method is frequently used tQ calculate the ground roll distance. The method is based on assuming that

I___ I___ I___ I I I *’ I I’ I

0 0.2 0.4 0.6 0.8 1.0 1.2 1.4 1.6

Cl

Figure 7.7 Sensitivity of predicted ground roll distance to flat-plate area and CL for Boeing 747-100.

•i

the inverse of the acceleration is a linear function of the square of the velocity. There appears to be no real rational basis for this assumption other than the fact that the results to which it leads are reasonable. To begin, let

dV=Wdi= ydV _ 1 dV2 dt ds dt ds 2 ds

The term in parentheses can be identified as the reciprocal of the acceleration evaluated at V divided by V2. Thus, Equation 7.5 becomes

(7.6)

^ where a is an average acceleration evaluated at V/V2. For example, consider ‘_phe foregoing 747 example for a CL of 1.0, an / of 2000 ft2, and a V of 274.5 fps.

-7== 194.1 fps

V2

At this speed,

T = 155,1801b D = 11,2001b L = 247,100 lb

so that, from Equation 7.3,

d = 5.90 ft/sec2

The ground roll distance is then calculated from Equation 7.6 to be

= 247.52 S 2(5.9)

= 6386 ft

This result, by comparison to Figure 7.6, is seen to be within 1% of the more exact value obtained by the numerical methods.

When a customer buys an airplane, whether it be a private individual, a corporation, an airline, or the military, the buyer wants to know what the airplane will do. How fast will it fly, how high, and how far? How long a runway is required from which the airplane will operate? How expensive will it be to operate, and what are the operating limitations? How fast will it climb? Will it take a half hour to get up to cruising altitude or only 5 min? This chapter provides methods for answering these questions and others related to the general subject of airplane performance. The groundwork for doing so has been presented in the preceding chapters. With the use of this material, one can calculate lift, drag, and thrust. Aside from weight, these are the principle forces acting on an airplane that determine its performance. We will begin, as every flight begins, by first considering the problem of calculating the distance required to take off safely over a prescribed obstacle height.

TAKEOFF

The takeoff of an airplane certified in the transport category is illustrated in Figure 7.1. Starting from a resting position at the far left, the airplane accelerates under takeoff power. At some point the velocity exceeds the stalling speed, Vs. Beyond this point, the airplane is capable of flying. However, the airplane continues to accelerate on the ground until the mini­mum control speed, Vmc is reached. At this speed, if a critical engine fails, the manufacturer has demonstrated that the airplane is able to maintain straight flight at that speed with zero yaw or with a bank angle of less than 5°. Under these conditions at this speed, the required rudder force may not exceed ‘ 1801b. Continuing to accelerate on the ground, the airplane reaches a cali­brated airspeed of Vlt the critical engine failure speed. This speed may not be less than Umc and represents the speed at which the average pilot could safely continue with the takeoff in the event of a critical engine failure. At a speed that can equal Vj but that must be 5% higher than Vmc, the pilot rotates the airplane. This speed, VR, is called takeoff rotation speed.

Because of tail interference with the ground, the angle of attack at VR may not be sufficient to lift the airplane. The pilot therefore continues to

0 К vm v-, Kmu vL0F v2

Figure 7.1 FAR Part 25 takeoff.

accelerate up to a speed of Vmu, the minimum unstick speed. At this speed, the pilot could lift the airplane off of the runway and continue the takeoff, even with one engine inoperative, without any hazardous characteristics. However, to provide an additional margin of safety, the airplane continues to accelerate to the lift-off speed, Vlof, at which point the airplane becomes airborne. Vlof must be at least 10% higher than Vmu, with all engines operating or 5% higher than Vmu with one engine inoperative. After lift-off, the airplane continues to accelerate up to the takeoff climb speed, V2. V2 is the speed attained at a height of 35 ft (10.7 m) above the ground. V2 must be greater than 1.2 V",, in the takeoff configuration and 1.1 Vmc.

This description, applicable to the takeoff of a turbojet or turbofan transport, is in accordance with the definition of these various speeds as presented in the Federal Air Regulations (FAR) Part 25. These regulations govern the airworthiness standards for airplanes in the transport category. Similar regulations for other categories of nonmilitary airplanes can be found in FAR Part 23. The Cherokee 180, which has been used as an example in the preceding chapters, is certified under FAR Part 23. The total horizontal distance, ground roll and airborne, which is required to reach the altitude of 35 ft, starting from rest, is referred to as the FAR takeoff distance.

FAR Part 23 is simpler in specifying the takeoff procedure. For airplanes over 6000 lb (26,700 N), maximum weight in the normal, utility, and acrobatic categories, it is stated simply that the airplane must attain a speed at least 30% greater than the stalling speed with one engine out, Vs,. For an airplane weighing less than 6000 lb, the regulations state simply that the takeoff should not require any exceptional piloting skill. In addition, the elevator power must be sufficient to lift the tail (fctr a “tail dragger”) at 0.8 Vs, or to raise the nose for a nose-wheel configuration at 0.85 VS|.

The performance curves that have been presented for the JT4A-3 tur­bojet, the JT9D-7A turbofan, and the PT6A-27 turboprop are all optimistic, since they do not include installation losses. These losses result from:

• Total pressure loss in the inlet ducting.

• Total pressure loss in the exhaust nozzle.

• Bleed air requirements.

• Power extraction for accessories.

• Deicing requirements.

Methods for calculating these losses are not included here because of the extensive information that is required. In practice, an engine manufacturer supplies a computer deck to the airframe manufacturer in order to estimate corrections to the engine performance resulting from installation losses. Typically, these losses equal approximately 0.4% for inlet, 5% for antiicing, and 8 to 22 hp/engine for accessories.

TRENDS IN AIRCRAFT PROPULSION

The title of this section was borrowed directly from an interesting paper (Ref. 6.11) by Rosen. From the preceding material and examples of specific engines, it is hoped that you now have a pretty good feeling of engine performance capabilities in the 1978 time frame. Let us now consider what developments we might expect in the near future. Our considerations will be limited to subsonic airspeeds. To do otherwise is beyond the limitations of this text. It may also be beyond the price that society is willing to pay for speed with the emphasis on fuel economy and noise.

Regarding fuel consumption, Figure 6.47 presents the static TSFC as a function of net thrust for turbojets with afterburners, turbojets, and tur­bofans. The points represent engines that are currently operational. Generally,

there is a tendency, as with piston engines, for the TSFC to improve with size for any given engine type, particularly at the lowest thrust values. However, the important point is the obvious gain to be realized by going to higher bypass ratios. This point is emphasized by Rosen in a slightly different manner, as shown in Figure 6.48 (taken from Ref. 6.11).

* It is interesting that, in a sense, the application of gas turbines to commercial aircraft propulsion has nearly completed a cycle. The sudden transition to the turbojet for commercial transportation in the 1950s intro­duced the air traveler to above-the-weather flying at significantly higher speeds with a power plant that was almost vibrationless. In doing so, the bypass ratio went from a high value, where most of the air goes through the propulsor as compared to the air that goes through the power plant, to a value of zero, where all of the aif goes through the power plant. Over the years, the

BPR has gradually increased, but at no sacrifice in comfort or convenience to the passenger. Indeed, today’s high bypass ratio turbofan is quieter, consumes less fuel, and is relatively much lighter (Figure 6.49) than the turbojet.

As Rosen notes in Figures 6.48 and 6.49, for a given BPR, there is a gradual improvement in engine performance with time. This improvement is the result of better materials and cooling techniques, which allow operation at higher pressure ratios and turbine inlet temperatures. It would. therefore appear that improved propulsion efficiency in the future will depend on further increases in the pressure ratio, turbine inlet temperatures, and bypass ratios.

In 1971, Rosen was a fairly accurate soothsayer when he targeted turbine inlet temperatures of 2700 °F (1480 °С) and pressure ratios of 30:1 in the 1980s. We are not there yet, but the numbers are getting close. He also recommended a further increase in the bypass ratio by advocating a so-called prop-fan configuration. This configuration is a controllable pitch, ducted fan (or one might call it a propeller) with 8 to 12 blades, a 1.1 to 1.2 pressure ratio across the fan, and a tip speed of around 750 fps (230 m/s).

More recently, NASA has been taking another look at propellers for application to high subsonic speeds. Reference 6.12 discusses the design philosophy of these propellers and reports on some early test results. These propellers are multibladed and incorporate thin, transonic airfoil sections. One such propeller that has been tested by NASA’s Lewis Research Center is shown in Figure 6.50. Preliminary test data obtained in their supersonic wind tunnel and presented here as Figure 6.51 appear promising.

The design characteristics for this propeller are:

O. g cruise Mach number.

10.7 km (35,000 ft) cruise altitude.

8 blades.

203 activity factor per blade.

301 kW/m2 (37.5 hp/ft2) power loading. 243.8 m/s (800 fps) tip speed.

0.08 integrated design lift coefficient.

Mach number, M

Figure 6.51 Preliminary test data on advanced turboprop NASA Lewis Research Center 8- by 6-ft wind tunnel. A1.

If this propeller meets its design goal of 80% efficiency at 0.8 Mach

number, this would represent a reduction of approximately 30% in fuel

consumption compared to existing turbofan engines.

The Pratt & Whitney JT9D-7A is representative of a modern high bypass turbofan engine. This engine has a dry weight of 8850 lb (39,365 N) and delivers a maximum continuous static thrust at SSL of 39,6501b (176,363 N). The static dry takeoff rating of 45,500 lb (202,384 N) is flat rated up to 27 °С (80 °F). The diameter of the engine is 2.43 m (95.6 in.) with a length of 3.92 m (154.2 in.). The compressor incorporates one fan stage, three low-pressure stages, and eleven high-pressure stages. The turbine has two high-pressure stages and four low-pressure stages. The bypass ratio equals 5.1 at the dry takeoff rating, with a total airflow of 1545 lb/sec (6872 N/s). The dash 7 A model of the JT9D is installed on several versions of the Boeing 747, including the 747-100, -200B, C, F, 747 SR, and 747 SP.

Figure 6.36 presents the rated takeoff thrust for this engine as a function of ambient temperature for altitudes up to 6000 ft. The net thrust is seen to drop approximately 14% in going from sea level to 6000 ft and to decrease rapidly with increasing Mach number.

Figures 6.37 and 6.38 give the rated maximum climb thrust and cruise thrust, respectively, as a function of Mach number for constant values of altitude up to 45,000 ft (13,700 m). Both figures also include lines of constant TSFC values. At the lower altitudes, the net thrust is seen to decrease rapidly as фе Mach number increases. However, at the higher altitudes, T is nearly constant and even increases slightly with Mach number above 30,000 ft and M values greater than 0.7.

The range of operating Mach number decreases in the preceding figures at the higher altitudes. This is a reflection of the limitations of the operating envelope presented in Figure 6.39. Such an envelope can result from several limitations, including temperature restrictions, stress limits, surge, and com­pressor stall.

9D-7A maximum cruise thrust. One hundred percent ram reco – , no power extraction, Pratt & Whitney Aircraft reference exhaust ss for ICAO standard day +10°C and below. (Courtesy, Pratt &

Temperature restrictions are normally associated with the turbine inlet temperature (TIT). High-pressure turbines in the latest high bypass turbofan engines operate with gas temperatures in the 2000 to 2300 °F (1094 to 1260 °С) range. Various techniques have been developed that keep blade metal tem­peratures equal to those of uncooled blades used in earlier turbine designs. Normally, blade cooling is only required in the first or first and second turbine stages. After these stages, sufficient energy has been extracted from the burner exhaust to cool the hot gases to a tolerable level.

Three forms of air cooling are described in Reference 6.7; these are used singly or in combination, depending on the local temperatures. The air for this cooling is bleed air taken from the compressor section. Even though this air is warmer than ambient air, it is still considerably cooler than the burner exhaust.

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Convection Cooling Cooling air flows inside the turbine vane or blade through serpentine paths and exits through the blade tip or through holes in the trailing edge. This form of cooling is limited to blades and vanes in the area of the lower gas temperatures.

Impingement Cooling Impingement cooling is a form of convection cooling, accomplished by directing cooling air against the inside surface of the airfoil through small, internal, high-velocity air jets. Impingement cooling

is concentrated mostly at critical sections, such as the leading edges of the vanes and blades.

Film Cooling Film cooling is a process whereby a layer of cooling air is maintained betwediklhe high-temperature gases and the external surfaces of the turbine blades and vanes.

Of the three forms of air cooling, film cooling is the most effective and the least demanding as far as airflow is concerned.

These types of cooling are illustrated in Figure 6.40, which shows their application to both stationary and rotating turbine stages.

Surge and compressor stall are related but are not the same thing. Surge refers to oscillations in the rotational speed of the entire engine. This surge is usually related to compressor stall, where the local angles of attack of the rotor blades, for various reasons, achieve sufficiently high values to cause local stalling. Some of these reasons include inlet airflow distortion from gusts, inlet design or uncoordinated maneuvering, rapid power changes, water ingestion, and Reynolds number effects.

A typical compressor map is given in Figure 6.41. This map shows qualitatively the relationship among the corrected rpm, corrected airflow, and total pressure ratio across the compressor. A small insert in the figure illustrates an airfoil on the compressor rotor under the influence of two velocities, one proportional to the, airflow and the other proportional to the rotational speed. At a fixed blade angle, the angle of attack of this section obviously increases as N increases or Wa decreases. This is reflected in the map, which shows one approaching the surge zone as N increases for a constant Wa or as Wa decreases for a constant N.

As the altitude increases, the surge zone drops down, mainly because of Reynolds number effects. At the same time, the steady-state operating line moves up. Thus compressor stall and surge are more likely to be encountered at the higher altitudes.

Accelerating the engine can also lead to compressor stall. Suppose, in attempting to get from the steady operating point A to point B, the rpm is suddenly increased. The airplane may be unable to accelerate rapidly enough to follow the rpm, so the airflow is less than the steady-state value. Surge can be alleviated by unloading the compressor during certain operating conditions. This is accomplished by bleeding air near the middle or end of the com­pressor. Stators having variable blade angles are also used to delay com­pressor rotor blade stall. The JT9D, for example, has variable stators automa­tically positioned by hydraulic actuators on the first four stages of the high-pressure compressor. Their function is to provide an adequate surge margin during engine starting, acceleration, and partial thrust operation.

A small, modern turbofan engine typical of others of this size is shown in Figure 6.42a and 6.42b. Only the rotating assemblies of the JT15D-1 engine

are shown in order to emphasize its relative simplicity and general configura­tion. The JT15D-9 is a growth version of the -1 model and was achieved by adding one additional axial stage to the low compressor. Although this decreases the bypass ratio, the pressure ratio across the compressor is increased so that the rated thrust of the -4 engine is approximately 14%

greater than that of the -1 engine. Notice that the high-pressure section of the compressor for the JT15D employs a centrifugal compressor similar to that of the Garrett TPE331/T76 engine. Specifics on the JT15D-1 and -4 turbofan engines are given in Table 6.2.

Turboprop

The PT6A-27 turboprop engine manufactured by Pratt & Whitney Air­craft of Canada, Ltd. is representative of the latest technology for turboprop engines of this size. Airplanes in which this engine is installed include the Beech 99A, the de Havilland Twin Otter, and the Pilatus Turbo Porter. The arrangement of this engine is shown in Figure 6.43. Compared to a turbojet, the core engine is reversed with the air being taken toward the rear of the engine and flowing forward through the compressor, burner, turbines, and then exhausted through stacks at the front of the engine.

A typical nacelle installation for this engine is illustrated in Figure 6.44.

An inlet duct channels the air to the rear of the engine, where it must turn in order to enter the engine. In good weather, the vane is raised from the position shown to lie flush with the surface above it, and the bypass door is closed. When ice or other types of particles are encountered, the bypass door is opened and the vane is lowered, as pictured. The air-particle mixture is then deflected downward. Because of their inertia, the heavier particles continue

on out through the bypass door, while the air is turned and drawn into the engine.

The specifications for this engine are presented in Table 6.3, and its performance is given in Figures 6.45 and 6.46a and 6.46b. These curves, taken from the installation handbook, assume no losses. Figures 6.45 and 6.46a are

V_y

Table 6.3 PT6A-27 Turboprop Engine Specifications

6.46a PT6A-27 maximum cruise performance. •2200 rpm. (Courtesy, Pratt & Whitney of Canada.)

for the rated propeller rpm of 2200, while Figure 6.46b presents a correction to the power for operation at lower rpm values. Only the maximum cruise and maximum takeoff ratings are given, since these are the same as the maximum climb and maximum continuous ratings, respectively. The power curves are for actual shaft power and do not include an effective power increment for the residual thrust. Statically, from Table 6.3, this thrust appears to equal approximately ЩЖ This represents about a 5% correction to the shaft power.

For illustrative purposes, this section will consider the characteristics and performance of specific engines.

Turbojet

The Pratt & Whitney JT4A-3 engine will be used as an example of a turbojet. This engine, installed on the Boeing 707-320 and McDonnell-Douglas DC-8-20, is a two-spool engine. The low-pressure compressor section has eight stages, with seven stages in the high-pressure section. The turbine has two low-pressure stages and one high-pressure stage. Other characteristics of this engine are presented in Table 6.1. This particular model of the JT4 engine has a takeoff thrust-to-dry weight ratio of 3.15. Later versions of this engine, such as the JT4A-12, develop T/W ratios of 3.58.

Table 6.1 Characteristics of the JT4A-3 Engine

Type—turbojet SSL static thrust

The net thrust and fuel consumption curves for this engine are reproduced from the manufacturer’s installation handbook in Figure 6.32a, 6.32b, 6.32c, and 6.32<f. At this point, a definition of net thrust is needed. To do this we first define gross thrust, Fg, as the product of the mass flow rate in the jet exhaust and the velocity attained by the jet after expanding to ambient static pressure.

Fg = rrijVj

The net thrust, F„, is then defined by

Fn = Fg – niiVж

where m, is the inlet mass flow and V^ is the velocity of the ambient air. For static operation, Fg and Fn are equal.

Net thrust for the takeoff rating of this engine is presented in Figure 6.33a, 6.33b, and 6.33c for speeds of 0, 100, and 200 kt and altitudes from sea level to 14,000 ft.

The curves of Figure 6.32 are for standard atmospheric conditions. One rarely finds a standard day, so it is usually necessary to correct engine performance for deviations from the standard. Without delving into the details of compressor design, one can argue that, for the same flow geometry (ratio of rotor speed to axial velocity and M„), the pressure increase across the compressor can be written as

Дp oc pJV2

where N is the rotor angular velocity.

If Дp is expressed as a ratio to the ambient pressure, then

Д p N2

— a —- 7—

Poo Poo/Poo

or

Др N2

—— oc ——–

Poo Too

Thus, if N denotes the rpm of a compressor operating at an ambient temperature of T„, the rpm required to deliver the same pressure ratio at standard sea level conditions is known as the corrected rpm, Nc, given by

(6.79)

9 being the ratio of the absolute temperature to the standard absolute temperature at sea level.

Similarly, one can say that the thrust, F, must be proportional to Др, or

P<X>9 for a constant pressure ratio. Thus, the corrected thrust, Fc, corresponding to the corrected rpm, is defined by

Fc = j (6.80)

where S is the ratio of the ambient pressure to standard sea level pressure. Similarly, corrected values for fuel flow, airflow, and exhaust gas tern-

3500

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True airspeed, V, knots

Figure 6.326 Pratt & Whitney Aircraft JT4A-3 turbojet engine. Estimated thrust, and airflow at 45,000 ft. Standard atmospheric conditions, 100% ram recovery. ^Courtesy, Pratt & Whjtney.)

perature (EGT) are defined by

A more elegant derivation of these corrected parameters, based on Buckingham’s ж theorem of dimensional analysis, can be found in Reference 6.9.

Excluding scale effects, the important point is made that the corrected thrust of a gas turbine engine is a unique function of the corrected values of N, Wa, and Wf. These, in turn, assure a constant value of the pressure ratio. In practice, the pressure ratio used to monitor the corrected thrust is

referred to as the engine pressure ratio (EPR), defined by

EPR = S£L (6.84)

Pt2

The subscript t refers to the total stagnation pressure, the 7 and 2 refer to the engine stations shown in Figure 6.23. Thus EPR is the ratio of the total pressure at the turbine nozzle to the total pressure at the compressor inlet.

в and 5, used to correct the operating parameters, are also based on the total#temperature and pressure, respectively, at the compressor inlet.

(6.85)

ST2 = — (6.86)

P о

where T0 and p0 are the standard sea level values of temperature and pressure. ‘Assuming 100% ram pressure recovery, (i. e., that M = 0 at station

2), вт2 and 5T2 can be calculated from

0T2 = в [1 + (-у – 1) МІІ2] (6.87)

ST2 = 8 [1 + (-у – 1) Mil2yh~’ (6.88)

The operating curves for the JT4A-3 turbojet are presented in Figure 6.34a and 6.34b. Turbine discharge temperature, compressor speeds, and fuel flow are presented in Figure 6.34a as a function of EPR. Figure 6.34b shows the net thrust as a function of Mach number for constant values of EPR. All of the curves presented thus far for the JT4A-3 engine assume 100% ram recovery (no inlet duct loss) and a standard nozzle installation prescribed by the manufacturer. They also assume zero power extraction or compressor air bleed. In. an actual airplane installation, corrections must be made for these factors. The details of these corrections are too lengthy to be presented here.

As an example of the use of the performance curves presented thus far for the JT4A-3 engine, consider its operation at an airspeed of 400 kt at an altitude of 30,000 ft. For the maximum continuous thrust rating, a net thrust of

53001b is read from Figure 6.32c. Thus, for this altitude and airspeed,

p – = 17,785

^am

Moo = 0.679

From Figure 6.34b,

EPR = 2.56

It follows from Figure 6.34a that

-7= = 6420 rpm

Vel2

-7== = 8850 rpm

Vet2

= 1550°R

W

—Г = 12,700 lb/hr

where Nt = rpm of low-pressure compressor and turbine, N2 = rpm of high – pressure compressor and turbine, and Kc = correction factor yet to be read from Figure 6.34a. At 30,000 ft, 0 = 0.794 and 5 = 0.298. Equations 6.87 and 6.88 give values of

0(2 = 0.867 5,2 = 0.406

From the preceding 0,2, Tt2 = -23 °С, so that Kc = 0.915. The actual values for the operating parameters can now be determined as

AT, = 5978 rpm

N2 = 8240 rpm *

T,7 = 473 °С

W, = 4718 lb/hr

Now consider operation at standard sea level conditions at this same MaAi number and thrust rating. For this case, V equals 448 kt which gives a net thrust of 10,5001b. From Figure 6.32a. Using the same procedure as that followed at 30,000 ft gives, in order,

EPR = 1.92

Ж

КсЬа

Note that the engine rotational speeds and exhaust gas temperature are approximately the same in both cases. Indeed, if other speeds and altitudes at the maximum thrust rating are examined, Nj, N2, and T,7 values ap­proximately equal to those just calculated are found. Thus the thrust available from a turbojet engine at a given speed and altitude depends on the maximum stress and temperature levels that can be tolerated by the engine materials. As

a result, the net thrust of a turbojet will not decrease with altitude in proportion to the density ratio, as with piston engines. As a rough ap­proximation, one can assume Fn to be proportional to a but, in practice, F„ will not decrease with altitude as rapidly as this approximation predicts. For the example just presented, one might predict a net thrust at 30,000 ft at 400 kt based on er and the SSL value of Fn of 3941 lb. This value is 25.6% lower than the rated value previously noted. To illustrate further the accuracy of the approximation, Figure 6.35 presents the rated maximum continuous thrust at 200,400 and 600 kt as a function of altitude and compares this thrust with that obtained by multiplying the sea level values by a. The approximation to Fn is seen to improve for the lower airspeeds and certainly predicts the proper trend. However, at the higher altitudes, the differences between the thrust curves are significant at all airspeeds.