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442

Aircraft Performance

Table 13.15. Bizjet decision speed

Flap (deg)

8

Distance (ft) at (B + C) = (D + E)

2,100

TOFL at BFL (ft)

3,780 (3,800)

Decision speed, V1 (kts)

109

V1/ VR

0.93

 

 

13.5.2 Landing Field Length (Bizjet)

The landing weight of a Bizjet is 15,800 lb (wing-loading = 48.92 lb/ft2) and with full flap extended, it is CLmax = 2.2. Therefore:

stall@land =

(

 

0.002378 × 323 × CL

= )

 

 

=

 

 

2.2

 

 

V

 

 

2 × 15,800

 

41,141

 

 

137 ft/s

 

 

 

 

 

 

Vappr =

1.3Vstall@land = 178 ft/s

 

 

 

 

 

VTD =

1.15Vstall@land = 158 ft/s

 

 

 

 

 

The average velocity from a 50-ft height to touchdown = 168 ft/s. The distance covered before brake application after 6 s (may differ) from a 50-ft height:

SG TD = 6 × 168 = 1,008 ft

An aircraft in full braking with µB = 0.4, all engines shut down, and the average CL = 0.5, CD/CL = 0.1. Equation 13.2 for average acceleration is based on 0.7VTD = 110.6 ft/s; then:

q = 0.5 × 0.002378 × 110.62 = 14.54

For deceleration:

= 32.2 × [(0.4) (CLq/48.92)(0.1 0.4)]

=32.2 × [0.4 + (0.15 × 14.54/48.92)]

=32.2 × [0.4 + 0.0445] = −11.45

The distance covered during braking, SG 0Land = (158 × 79)/11.45 = 1,090 ft. The landing distance SG Land = 1,008 + 1,090 = 2,098 ft.

Table 13.16. Bizjet takeoff field length with 8-deg flap setting

 

 

knot

ft/s

Vstall @ 20,600 lb

106.4

179.6

V1

decision speed

109

183.8

V2

= 1.2Vstall at 35-ft height

128.2

216.5

VLO at 1.12Vstall

119.17

201.15

VR at 1.1Vstall

117

197.6

Vmu at 1.01 VR (lower than VLO)

118.2

199.5

Vmc at 0.94V1

102

172.2

BFL (ft) (TOFL requirement 4,400 ft)

 

3,800

V1 is too close to Vstall but is acceptable. If required it can be raised by a few knots.If required, V2 can be higher than 1.2Vstall.

If required, VR and VLO can be at higher speeds.

13.5 Aircraft Performance Substantiation: Worked-Out Examples (Bizjet)

443

Table 13.17. Firstand second-segment climb performance

 

 

 

 

 

 

 

 

 

 

 

 

1st Segment Climb

2nd Segment Climb

 

 

35to 400-ft altitude

400to 1,000-ft altitude

 

 

ρave = 0.00235 lb/ft3

ρave = 0.00232 lb/ft3

 

 

0.5ρSW = 0.3795

 

0.5ρSW = 0.3747

 

 

Flap (deg)

8

20

8

20

 

MTOM (kg)

20,680

20,680

20,680

20,680

 

Vstall (ft/s) (knot)

179.6 (106.4)

168.4 (99.8)

179.6 (106.4)

168.4 (99.8)

 

V2 = (ft/s) (Mach)

216.50 (0.194)

202.08 (0.181)

216.50 (0.194)

202.08 (0.181)

 

qSW

17,788

15,497.4

17,563

15,301.4

 

CL

1.16

1.33

1.177

1.35

 

CDclean (see Figure 9.2)

0.075

0.100

0.076

0.101

 

CD one eng

0.003

0.003

0.003

0.003

 

CDflap

0.013

0.032

0.013

0.032

 

CD u/c

0.022

0.022

 

CD1st seg

0.113

0.157

0.092

0.135

 

Drag (lb)

2,010

2,433

1,616

2,081

 

Thrust Available (lb)

2,740

2,750

2,740

2,750

 

Gradient (%)

+ve

+ve

5.4

3.24

 

FAR

Meets

Meets

Meets

Meets

 

 

 

 

 

 

 

Note: Aircraft flap and undercarriage extended, engine at takeoff rating (one engine inoperative). (Use drag polar Figures 9.2 and 9.12 to obtain drag.)

Multiplying by 1.667, the rated LFL = 1.667 × 1,548 = 3,597 ft, within the requirement of 4,400 ft. As expected, this is less than the BFL at an 8-deg flap. This is not always the case because at a 20-deg flap setting, LFL > BFL.

13.5.3 Climb Performance Requirements (Bizjet)

The three requirements for substantiation of the climb performance are given as follows for a two-engine aircraft (the first two are FAR requirements; Table 13.5 provides an aircraft configuration and FAR requirements for the firstand secondsegment climb):

1.Verify that the FAR first-segment climb requirement of a positive gradient is maintained.

2.Verify that the FAR second-segment climb gradient requirement exceeds 2.4%.

3.Verify that the market requirement of the initial enroute rate of climb equals or exceeds 2,600 ft/min. The cabin-pressurization system should handle the rate of climb. (This is a customer requirement, not a FAR requirement.)

The second segment starts at a 400-ft altitude with flaps extended and the undercarriage retracted (i.e., one engine inoperative). From a 400to 1,000-ft altitude, the undercarriage is retracted in the second-segment climb. An aircraft is maintained at the V2 speed for the best gradient – a 50% loss of thrust does not favor an accelerated climb, which will be low in this case. The engine is at the takeoff rating. The available one-engine-installed thrust is from Figure 13.1. The thrust is kept invariant at the takeoff rating through the firstand second-segment climb. Table 13.17 summarizes the firstand second-segment climbs for both 8-deg and 20-deg flap settings. At one engine failed, the aircraft must return to the base immediately.

444 Aircraft Performance

When verifying initial enroute rate of climb, the specification requirement is 2,600 ft/min. When the initial enroute climb starts at a 1,000-ft altitude (ρ = 0.0023 slug/ft3, σ = 0.9672) and all engines are throttled back to maximum climb rating, an aircraft has a clean configuration. An aircraft makes an accelerated climb from V2 to reach 250 KEAS, which is kept constant in a quasi-steady-state climb until it reaches Mach 0.7 at about a 32,000-ft altitude. From there, the Mach number is held constant in the continued quasi-steady-state climb until it reaches the cruise altitude. Fuel consumed during the second-segment climb is small and assumed empirically (from statistics) to be 120 lb (see Table 13.17). Therefore, the aircraft weight at the beginning of the enroute climb is M = 20,600 lb. At 250 kts (422 ft/s, Mach 0.35), the aircraft lift coefficient CL = M/qSW = 20,600/(0.5 × 0.0023 × 4222 × 33) = 20,600/66,150 = 0.311.

The clean aircraft drag coefficient from Figure 9.2 at CL = 0.311 gives CDclean = 0.0242. The clean aircraft drag, D = 0.0242 × (0.5 × 0.0023 × 4222 × 323) = 0.0242 × 66,150 = 1,600 lb. The available all-engine-installed thrust at the maximum climb rating from Figure 13.2 at Mach 0.378 is T = 2 × 2,260 = 4,520 lb. From Equation 13.10, the quasi-steady-state rate of climb is given by:

R/C

V[(T D)/ W]

 

 

 

 

 

 

 

accl =

1

+

(V/g)(dV/dh)

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

At the quasi-steady-state climb, Table 13.5 gives

V

dV

=

0.56 m2

=

0.56

×

0.352 = 0.0686. Hence:

 

 

 

g

dh

 

 

R/Caccl = {[422 × (4,520 1,600)]/20,680}/[1 + 0.0686] = 55.8 ft/s

= 3,345 ft/min(17 m/s)

This capability satisfies the market requirement of 2,600 ft/min (13.2 m/s). (The civil aircraft rate of climb is limited by the cabin-pressurization schedule. An aircraft is limited to 2,600 ft/min at an altitude where the cabin pressurization rate reaches its maximum capability. Naturally, at the low altitude of 1,000 ft, this limit is not applicable.)

13.5.4 Integrated Climb Performance (Bizjet)

Integrated climb performance is not a requirement for substantiation – it is used to obtain the aircraft payload-range capability, which is a requirement for substantiation. Section 13.1 explains why only results of the integrated climb performances in graphical form are given as shown in Figures 13.12 and 13.13. Section 13.4.3 describes the theory for deriving the climb equations. Instructors may assist with the computational work to obtain similar performance graphs for the coursework projects. Readers redoing the graphs as given here may have minor differences in their results, which is understandable.

It is convenient to establish first the climb velocity schedule (Figure 13.12a) and the point performances of the rate of climb (Figure 13.12b) up to the ceiling altitudes and for at least three weights for interpolation. A Bizjet carries out the quasi-steady- state climb at a constant VEAS = 250 knots until it reaches Mach 0.7; thereafter, it continues at a constant Mach number until it reaches the ceiling (i.e., rate of climb = 100 ft/min).

13.5 Aircraft Performance Substantiation: Worked-Out Examples (Bizjet)

445

(a) Climb Speed Schedule

(b) Rate of Climb

Figure 13.12. Climb point performances

The next step is to perform the computations for the integrated climb performance in increments of approximately 5,000-ft altitudes (as convenient; at higher altitudes, smaller steps are appropriate) in which the variables are kept invariant using their mean values (see Figure 13.10). Equations 13.15 through 13.18 are used to compute the integrated performances. Figure 13.13 shows the integrated

Figure 13.13. Integrated climb performances for a Bizjet

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