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8.11 Worked-Out Example – Civil Aircraft

247

Table 8.4. Minimum cabin-crew number for passenger load

 

 

 

 

 

 

 

 

 

 

Number

Minimum number

Number

Minimum number

of passengers

of cabin crew

of passengers

of cabin crew

 

 

 

 

 

19

1

200 to <250

7

 

19 to <30

2

250 to <300

8

 

21 to <50

3

300 to <350

9

 

50 to <100

4

350 to <400

10

 

100 to <150

5

400 to <450

11

 

150 to <200

6

450 to <500

12

 

 

 

 

 

 

8.10.13 Fuel – Civil Aircraft

The fuel load is mission-specific. For civil aircraft, required fuel is what is needed to meet the design range (i.e., market specification) plus mandatory reserve fuel. It can be determined by the proper performance estimation described in Chapter 13. At this design stage, statistical data are the only means to estimate fuel load, which is then revised in Chapter 13.

The payload and fuel mass are traded for off-design ranges; that is, a higher payload (if accommodated) for less range and vice versa.

8.11 Worked-Out Example – Civil Aircraft

The semi-empirical relations described in Section 8.10 are now applied to obtain an example of the configuration worked out in Chapters 6 and 7 in the preliminary configuration layout. This chapter more accurately estimates component and aircraft mass along with the CG locations (Figure 8.4). Therefore, the preliminary configuration needs to be refined through an iterative process with more accurate data. The iteration process may require the repositioning of aircraft components (see Section 8.13). The aircraft configuration is finalized in Chapter 11. From Chapter 6, the following specifications are obtained for the baseline-aircraft preliminary configuration. They are required to estimate aircraft component mass, as shown here:

MTOM = 9,500 kg (refined in this exercise)

Two turbofans (i.e., Honeywell TFE731), each having TSLS = 17,235 N (3,800 lbs), BPR < 4 and dry weight of 379 kg (836 lbs)

The results from this section are compared with the graphical solutions in Figure 8.3 and in Table 8.5.

8.11.1 Fuselage Group Mass

Consider a 5% weight reduction due to composite usage in nonload-bearing structures (e.g., floorboards). Use Equation 8.15:

L = 15.24 m, Dave = 1.75 m, VD = 380 KEAS = 703.76 kmph = 195.5 m/s and cfus = 0.04, ke = 1.04, kp = 1.09, kuc = 1.06, and kmat = (0.9 + 0.9 × 0.1) = 0.99

248

 

 

Aircraft Weight and Center of Gravity Estimation

Table 8.5. Bizjet mass (weight) summary

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

 

Semi-empirical

Mass

Graphical

 

 

Component

(kg)

fraction %

solution – lb (kg)

 

 

 

 

 

 

 

1.

Fuselage group

930

10

2,050 (932)

2.

Wing group

864

9.2

2,100 (946)

3.

H-tail group

124

1.32

H-Tail+V-Tail460 (209)

4.

V-tail group

63

0.67

900 (409)

5.

Undercarriage group

380

4

6.

Nacelle + pylon group

212

2.245

410 (186)

7.

Miscellaneous

 

 

 

 

 

 

Structures group total

2,591

27.56

 

 

8.

Power plant group

1,060

11.28

 

 

9.

Systems group

1,045

11.12

 

 

10.

Furnishing group

618

6.57

 

 

11.

Contingencies

143

0.7

 

 

 

 

MEM

5,457

58.05

 

 

12.

Crew

180

1.92

 

 

13.

Consumables

119

1.73

 

 

 

 

OEM

5,800

61.7

 

 

13.

Payload (as positioned)

1,100

11.7

 

 

14.

Fuel (as positioned)

2,500

26.6

 

 

 

 

MTOM

9,400

100

 

 

 

 

MRM

9,450

100.53

 

 

 

 

 

 

 

 

 

Figure 8.4 Aircraft component CG locations

8.11 Worked-Out Example – Civil Aircraft

249

Then:

MFcivil = cfus × ke × kp × kuc × kdoor ×; (MTOM × nult)x × (2 × L × Dave × VD0.5)y

For civil aircraft, (MTOM × nult)x = 1; therefore:

MFcivil = 0.04 × 1.04 × 1.09 × 1.06 × 1 × 1 × (2 × 15.24 × 1.75 × 195.50.5)1.5

=0.048 × (2 × 15.24 × 1.75 × 195.50.5)1.5 = 0.048 × (53.35 × 13.98)1.5

=0.048 × (745.95)1.5 = 0.048 × 20,373.3 = 978 kg

There is a 5% reduction of mass due to the use of composites:

MFcivil = 0.95 MFcivil all metal 930 kg

This is checked using Torenbeek’s method (Equation 8.13); refer to Chapters 6 and 9 for dimensions:

WFcivil = 0.021Kf VDLHT / (W + D) 0.5 (Sfus gross area)1.2

Kf = 1.08,L = 50 ft, W = 5.68 ft,D = 5.83 ft,LHT = 25 ft,VD = 380 KEAS,

Sfus gross area = 687 ft2

Therefore,

WFcivil = 0.021 × 1.08 × 1.07 × {380 × 25/(5.68 + 5.83)}0.5 × (687)1.2

= 0.0243 × (825.4)0.5 × 2,537.2 = 0.0243 × 28.73 × 2,537.2 = 1,770 lb (805 kg)

The higher value of the two (i.e., 930 kg) is retained, which gives a safer approach initially.

8.11.2 Wing Group Mass

Consider a 10% composite secondary structure; that is:

kmat = (0.9 + 0.9 × 0.1) = 0.99

It has no slat, making ksl. = 1, and without a winglet, kwl = 1. For the spoiler, ksl = 1.001, and for a wing-mounted undercarriage, kuc = 1.002.

SW = 30 m2, nult = 4.125, b = 15 m, AR = 6.75, λ = 0.375, fuel in wing, MWR

=1,140 kg, = 14, t/c = 0.105, and VD = 380 knots = 703.76 kmph

=195.5m/s. The load factor n = 3.8.

Equation 8.21 becomes:

MW = 0.0215 × 0.99 × 1.002 × 1.001 × (9,500 × 4.125)0.48 × 290.78

× 6.75 × (1 + 0.375)0.4 × (1 1,140/9,500)0.4/(Cos14 × 0.1050.4)

MW = 0.0213 × 1.003 × 160.2 × (13.8 × 6.75 × 1.136) × 0.880.4/(0.97 × 0.406) MW = 3.42 × 105.8 × 0.95/(0.3977) = 351.8/0.3977 = 864 kg

250

Aircraft Weight and Center of Gravity Estimation

8.11.3 Empennage Group Mass

For a H-tail, a conventional split tail has a kconf = 1.0. Consider a 20% composite secondary structure; that is:

kmat = (0.8 = 0.9 × 0.2) = 0.98

MTOM = 9,500 kg, nult = 4.125, SHT = 5.5 m2 (exposed), AR = 3.5, λ = 0.3,= 16, t/c = 0.105, and VD = 380 knots = 703.76 kmph = 195.5 m/s.

MHT = 0.02 × 0.98 × (9,500 × 4.125)0.484 × 5.50.78 × 3.5

× (1 + 0.3)0.4/(Cos16 × 0.1050.4)

=0.0196 × 167.1 × 3.8 × 3.5 × 1.11/(0.961 × 0.406) = 47.7/0.39 = 124 kg

For a V-tail (T-tail), kconf = 1.1. Consider a 20% composite in a secondary structure; that is:

kmat = (0.8 + 0.9 × 0.2) = 0.98

MTOM = 9,200 kg, nult = 4.125, SVT = 3.5 m2(exposed), AR = 2.0, λ = 0.5,= 20, t/c = 0.105, and VD = 380 knots = 703.76 kmph = 195.5 m/s

MVT = 0.0215 × 0.98 × 1.1 × (9,500 × 4.125)0.484 × 3.50.78 × 2 × (1 + 0.5)0.4/

(Cos20 × 0.1050.4)

=0.02318 × 167.1 × 2.66 × 2 × 1.176/(0.94 × 0.406)

=23.877/0.382 = 63 kg

Figure 8.5 provides the graphical solution. It reads µ/Wf = 0.02 (top line corresponding to span, b = 49.2 ft). Therefore, wing weight µ = 22,000 × 3.8 × 0.021 = 1,756 lb = 800 kg, a difference of about 6%.

8.11.4 Nacelle Group Mass

Engine thrust = 17,230 N (3,800 lb) per engine with a BPR < 4.

Mnac+ pylon = 6.2 × thrust (kN) per nacelle = 6.2 × 17.23 = 106 kg per nacelle Two nacelles = 212 kg.

8.11.5 Undercarriage Group Mass

MTOM = 9,500 kg low-wing mount; MU/C wing = 0.04 × 9,500 = 380 kg

8.11.6 Miscellaneous Group Mass

Fortunately, there are no miscellaneous structures in the examples considered herein.

8.11.7 Power Plant Group Mass

This is determined from statistics until it is sized in Chapter 10. A typical engine is of the class Allison TFE731–20 turbofan with thrust per engine = 15,570 N to 17,230 N (3,500 to 3,800 lbs).

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