Aircraft design projects - part 8 pps

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250 Aircraft Design Projects

(E) Although point D above gives the max. sustained turn rate, the tangent of a radial

from the origin to the zero SEP curve gives the smallest sustained turn radius. In

the dry thrust case, the values are 7.5◦/s at 200 kts with a radius of 2582 ft. The

wet thrust intersection coincides with the minimum speed boundary.

∗ The minimum speed boundary in this calculation assumes that the high angle of attack

required to achieve the max. CL value is controllable. It is likely that in our design, this

may only be possible with a contribution from vectored thrust. The component of force

from the thrust vectoring has not been included in the calculations because this would

require more aircraft and propulsion details than are available at this stage. As we are

relying on assistance from thrust vectoring for landing control, it may be possible to

design the system to provide an integrated aerodynamic and propulsion control system

in the turn manoeuvre without much additional complexity. We have easily met the

instantaneous turn requirement so we will assume that the minimum speed boundary

is achievable and not critical.

Recommendations

All of the specified manoeuvre and turn requirements have been easily met with the

current design but a word of caution is appropriate. As the value of SEP at a particular

flight condition is dependent on the difference between two relatively large numbers

(thrust and drag), small percentage changes in either will result in large variations in

SEP. At this stage in the design process, when only crude estimates have been made

about aerodynamic and propulsion characteristics, this must concern us. For example,

when considering flight at high manoeuvre load factors, the lift-induced drag becomes

a significant component of drag. As this is dependent on the estimation of the induced

drag factor, which is difficult to predict accurately for our planform, there could be

uncertainty in the ‘high g’ performance. Also, the engine performance is affected by the

detail layout and control mechanisms in the intake. A poor estimate of intake efficiency

will significantly affect the net thrust available. For these reasons it is important to

obtain better (higher confidence) estimates of these parameters in the next phase of the

design process.

8.9.2 Mission analysis

There are four cruise stages to be assessed: outbound supercruise, outbound dash,

return dash and return supercruise. Each of these stages is to be flown at M1.6. It is

necessary to determine the optimum (based on minimum fuel burn) cruise height for

each stage. From the engine data it is possible to extract the performance (net thrust

and sfc) versus altitude for the cruise speed of M1.6, see Figure 8.29.

This provides the variation of sfc to be used to predict fuel burn by assuming that

the thrust required equals the aircraft drag. From section 8.8.4, the aircraft drag polar

at M1.6 is:

CD = 0.0205 + 0.514C2

L

The dynamic pressure q (=0.5ρV2) in the stratosphere (above 36 089 ft) where the

speed of sound is constant at 986 kts, can be determined for each height using the ISA

formula for relative density multiplied by the sea-level value (0.002378 slugs/cu. ft) as

shown below:

q = 0.5(0.2971e−x)0.002378(1.6 × 986)

2

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Project study: advanced deep interdiction aircraft 251

TSFC (1/hr) and net thrust (lb) for M1.6

0

5000

10 000

15 000

20 000

25 000

30 000

35 000 1.40

1.30

1.20

1.10

20 000 25 000 30 000 35 000 40 000 45 000 50 000 55 000 60 000 65 000 70 000

Altitude (ft)

Net thrust (lb) TSFC/hr

SFC/hr

(a)

(b) (c) (d)

Net thrust

per engine (lb)

Fig. 8.29 Engine performance at M1.6

where x = (H − 36089)/20806.7

H = altitude (ft)

For our aircraft, the reference area is 1430 sq. ft (133 sq. m).

As we are unaware of the fuel burnt in each segment at this stage in the design process,

it will be necessary to make some assumptions regarding the weight of the aircraft at

the start of each stage, as shown below:

(a) Outbound supercruise = 0.9 MTOW

(b) Outbound dash = 0.8 MTOW

(c) Return dash = 0.7 MTOM

(d) Return supercruise = 0.6 MTOW

And if the return stages follow the release of munitions:

(e) Return dash = 0.7 MTOM – 8000 lb

(f) Return supercruise = 0.6 MTOW – 8000 lb

where, from section 8.8.2, MTOM = 114 082 lb (51 739 kg).

The aircraft weight defines the CL which in turn defines the CD from which the aircraft

drag is calculated. This is multiplied by the engine sfc to obtain the fuel used per hour.

This procedure is easily performed using a spreadsheet method.

The results are shown in Figure 8.30.

This clearly shows an optimum altitude for each stage. The optimum heights are

cross-plotted against aircraft weight in Figure 8.31. The associated fuel consumption

is also plotted on this graph.

At this point it is possible to use the fuel consumption results to determine the

overall fuel burnt on the mission (assuming that the fuel consumption in each stage

is the average between the start and end values). The time spent on each stage is the

stage distance divided by the aircraft speed. As the speed is constant (933.5 kt), the

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252 Aircraft Design Projects

10 000

15 000

20 000

25 000

30 000

35 000

40 000

35 000 40 000 45 000 50 000 55 000 60 000 65 000 70 000

Cruise altitude (ft)

(a) Outward supercruise
= 0.9

(b) Outward dash
= 0.8

(c) Return dash
= 0.7

(c) Less weapons

(d) Return dash
= 0.7

(d) Less weapons

Fuel burn (lb/hr)

Fig. 8.30 Aircraft fuel burn versus altitude

Best alt

50 000

55 000

60 000

65 000 25 000

20 000

15 000

70 000

65 000 70 000 75 000 80 000 85 000

Aircraft weight (lb)

90 000 95 000 100 000 105 000

Optimum (min

fuel burn) altitude

Fuel burn

optimum altitude

Altitute (ft) Fuel burn (lb/hr)

Fig. 8.31 Optimum cruise versus aircraft weight

supercruise stages of 1000 nm take 1.07 hr and the dash stages of 750 nm take 0.80 hr.

The analysis is shown in Table 8.8.

This is much larger than originally estimated due to the lower lift to drag ratio

(4.87 compared to 5.56 assumed earlier). This would mean that the aircraft MTOW

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Project study: advanced deep interdiction aircraft 253

Table 8.8

Fuel burn per hour

Stage Start End Average Time Fuel used (lb)

(a) 25 521 22 742 24 131 1.07 25 820

(b) 22 742 19 968 21 355 0.80 17 148

(c) 19 968 17 178 18 573 0.80 (14 914)

(d) 17 178 14 000∗ 15 589 1.07 (16 680)

(e) 18 012 15 219 16 620 0.80 13 346

(f) 15 219 13 000∗ 14 109 1.07 15 097

71 411 lb

∗ Guessed values.

0

5000

10 000

15 000

20 000

25 000

15 000 20 000 25 000 30 000 35 000 40 000 45 000 50 000 55 000 60 000 65 000

Altitude (ft)

Thrust and drag (lb)

Thrust (single eng.) @ M0.9

Thrust (single eng.) @ M1.6

Aircraft drag (50%):

(a) 1st supercruise

(b) 1st dash

(c) 2nd dash

(d) 2nd supercruise

a

b

c

d

Fig. 8.31

optimum

Fig. 8.32 Engine thrust at M0.9 versus altitude

must be increased. As the above mission assumed that only 8000 lb of weapon load

would be dropped and about 13 000 lb was used in the mass statement to account for

different missions, we could substitute 5000 lb into the fuel load. This would mean

that the aircraft weight would need to be raised by 11 400 lb. However, some of this

penalty could be set against potential improvements in engine design as mentioned in

section 8.8.5. (For example, if the sfc could be reduced from 1.2 to 1.1 the fuel load

would reduce by 6000 lb.)

The cruise analysis predicted the drag. This can be compared to the available thrust

which has been extracted from the engine data and shown on Figure 8.32.

The analysis shows that the engine needs to be slightly more powerful to fly at

optimum (minimum fuel burn) altitudes. However, as the aircraft L/D ratio is lower

than expected no change should be made to the engine until a more accurate estimation

of aircraft drag is available.