Aircraft Design Projects - part 5 doc

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128 Aircraft Design Projects Climb rate (m/s)

sea level

25 000 ft

36 000 ft

30

40

50

60

0

–20

–10

10

20

Aircraft speed (m/s)

0 50 100 150 200 250 300

Fig. 5.19 Aircraft rate of climb

0 5000 10 000 15 000 20 000 25 000 30 000 35 000 40 000

Aircraft altitude (ft)

Max. climb rate (m/s) 20

30

40

50

60

0

10

Fig. 5.20 Aircraft climb and ceiling evaluation

• The maximum speed even at 85 per cent thrust is 505 kt. This easily exceeds the

specified requirement of 450 kt.

• All the turn performance criteria are easily met.

• Take-off ground run at 1856 ft is below the specified 2000 ft but with a derated engine

of 85 per cent thrust, this increases to 2184 ft.

• The approach speed requirement of 100 kts cannot be met except by a lighter aircraft

(no weapon load). In this case an approach speed of 95 kt is achieved.

• Landing ground run at 2215 ft also exceeds the specification of 2000 ft. Only with

aircraft at lighter landing mass can the specification be met.

• The ferry mission of 1000 nm cannot be met with internal fuel but can be achieved

if 833 kg of fuel is carried externally. The maximum range that could be flown is

estimated at 1464 nm. This is substantially less than the 2000 nm specified. It is

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suggested that this requirement be reviewed as in its present form it would seriously

compromise the overall aircraft design.

• Climb and ceiling requirements are easily achieved.

5.8 Constraint analysis

From the project brief there are six separate constraints to be considered in this

analysis:

1. Take-off distance less than 2000 ft.

2. Approach speed no greater than 100 kt.

3. Landing distance less than 2000 ft.

4. Combat turn, at least 4g at sea level.

5. Combat turn, at least 2g at 25 000 ft.

6. Climb rate to provide for 7 min climb to 25 000 ft.

5.8.1 Take-off distance

The equations to be used to determine the effect of the take-off criterion can be found

in most textbooks (e.g. reference 4) as shown below:

(T/W) = (constant) (W/S)/(Stake-off .CLtake-off)

Obviously this represents a straight line on the (T/W) versus (W/S) graph. For our

aircraft the lift coefficient in the take-off configuration (CLtake-off) is assumed to be

1.7. The value Stake-off represents the total take-off distance (i.e. ground roll plus climb

distance to 50 ft). Assuming a climb gradient from zero to 50 ft of 5◦ gives a ground

distance covered of 571 ft. Adding this to the specified ground roll of 2000 ft gives

Stake-off = 2571 ft (784 m).

The constant in the above equation is assessed from Nicholi’s book4 as 1.27 (in SI

units with wing loading in kg/m2), so

(T/W) = 1.27/(784 × 1.7)(W/S) = 0.00095(W/S)

5.8.2 Approach speed

Assuming the approach speed VA = 1.2VSTALL then:

(W/S)landing = β(W/S) = 0.5 × ρ(VA/1.2)

2 × CLlanding/g

VA is specified at 100 kts (52 m/s). β is the ratio of landing mass to take-off mass. At

a maximum landing weight β = 0.9. At minimum landing weight (i.e. empty aircraft

plus pilot plus 10 per cent fuel = 3311 + 136 + 90 = 3537 kg) β = 0.62.

Assuming the lift coefficient in the landing configuration (CLlanding) = 2.1

(W/S) = (0.5 × 1.225 × 52 × 52 × 2.1)/(1.2 × 1.2 × 9.81) = 273.6 @ β = 0.9

= 397.1 @ β = 0.62

Note: these constraints are constant (vertical) lines on the (T/W) versus(W/S) graphs.

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130 Aircraft Design Projects

5.8.3 Landing distance

The approximate equation to determine ground run in landing can be rewritten as

shown below:

(W/S) = (Slanding run × CLlanding)/(constant × β)

The landing ground run Slanding run is specified as 2000 ft (610 m). The lift coefficient in

the landing configuration (CLlanding) is assumed to be 2.1 (as above). The expression

will be evaluated for the two landing mass fractions used above (i.e. β = 0.9 and 0.62).

The (constant) in the expression above (in SI units with W/S in kg/m2) is 5.0.

(W/S) = (610 × 2.1)/(5.0 × β) = 284.7 @ β = 0.9 and 413.2 @ β = 0.62

Note: these are also constant vertical lines on the constraint diagram.

5.8.4 Fundamental flight analysis

The fundamental equation used in the flight cases can be found in most textbooks. In

terms of sea level, take-off thrust loading the equation is:

(T/W)TO = (β/α)[(q/β){CDO/(W/S)TO + k1(nβ/q)

2

(W/S)TO}

+ (1/V)(dh/dt) + (1/g)(dV/dt)]

where (T/W)TO is the take-off thrust loading

α1 = T/TSLS

TSLS = sea level static thrust (all engines)

β = W/WTO

CDO and k1 are coefficients in the aircraft drag equation, see below

D = qS(CDO + k1C2

L)

(W/S)TO is the take-off wing loading (N/m2)

n is the normal acceleration factor = L/W

g = gravitational acceleration

V is the aircraft forward speed

q is the dynamic pressure = 0.5ρV2

(dh/dt) = rate of climb

(dV/dt) = longitudinal acceleration

5.8.5 Combat turns at SL

In this flight condition the aircraft is in ‘sustained’ flight with no change in height

and no increase in speed therefore the last two terms in the fundamental equation are

both zero.

At sea level α = 1

Assume that the turn requirement is appropriate to the mean combat mass (i.e. air￾craft empty + pilot + half fuel + half weapon load = 3311 + 136 + 450 + 680 =

4577 kg/10 092 lb)

Hence β = 4577/5707 = 0.8.

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From previous analysis (in SI units) the best speed for turning at SL is about 150 m/s.

∴ q = 0.5 × 1.225 × 1502 = 13 781

From the drag analysis done earlier (at 4577 kg with an increase in drag coefficient to

represent the stores on the wing) at a speed of 150 m/s, CD = 0.03 + 0.017C2

L.

As specified, the aircraft is subjected to a normal acceleration n = 4 in the turn.

T/W = 13 781{(0.03/(W/S) + 0.017 × [4/13 781]

2 × (W/S)}

5.8.6 Combat turn at 25 000 ft

This is similar to the analysis above but with α = 0.557/1.225 = 0.455.

At 25 000 ft the best speed for excess power is 200 m/s (in SI units)

∴ q = 0.5 × 0.557 × 2002 = 11 140

With β and CD values the same but with load factor n = 2 gives:

T/W = (0.8/0.445)[(11 140/0.8){(0.03/(W/S)+0.017×[(2×0.8)/11 140]

2×(W/S)}

5.8.7 Climb rate

This criterion assumes a non-accelerating climb, so the last term in the fundamental

equation is zero but the penultimate term assumes the value relating to the specified

rate of climb.

We will use an average value of climb rate of 18.15 m/s (i.e. 25 000 ft in 7 min) and

make the calculation at the average altitude of 12 500 ft, at a best aircraft speed of

150 m/s.

At 12 500 ft α = 0.841/1.225 = 0.686

At 150 m/s q = 0.5 × 0.841 × 1502 = 9461

Using the standard values for β at mean combat mass, and the drag coefficients (CDO

and K) previously specified, we get:

T/W = (0.8/0.686)[(9461/0.8){(0.03/(W/S) + 0.017 × [(1 × 0.8)/9461]

2 × (W/S)}

+ 18.15(1/150)

5.8.8 Constraint diagram

The above equations have been evaluated for a range of wing loading values (150 to

550 kg/m2). The resulting curves are shown in Figure 5.21.

The constraint diagram shows that the landing constraints (approach speed and

ground run) present severe limits on wing loading.

To identify the validity of the constraints relative to other aircraft, values appropriate

to specimen (competitor) aircraft that were identified earlier in the study have been

plotted on the same constraint diagram Figure 5.21. Some interesting conclusions can

be drawn from this diagram:

• The S212, T45, MiG, L159 and, to a lesser extent, the Hawk aircraft appear to fit

closely to the climb constraint line. This validates this requirement.