Category AIRCRAF DESIGN

An aircraft at the initial HSC is at Mach 0.74 (i. e., 716.4 ft/s) at a 41,000-ft altitude (p = 0.00055 slug/ft3). The fuel burned to climb is computed (but not shown) as 700 lb. The aircraft weight at initial cruise is 20,000 lb. At Mach 0.74, the aircraft lift coef­ficient Cl = MTOM/^SV = 20,000/(0.5 x 0.00055 x 716.42 x 323) = 20,000/45,627 = 0.438.

The clean aircraft drag coefficient from Figure 9.2 at CL = 0.438 gives CDciean = 0.0324. The clean aircraft drag, D = 0.0324 x (0.5 x 0.00055 x 716.42 x 323) = 0.0324 x 45,627 = 1,478 lb.

The available all-engines-installed thrust at maximum cruise rating at the speed and altitude from Figure 13.3 at Mach 0.74 is T = 2 x 790 = 1,580 lb (adequate). The capability satisfies the market requirement of Mach 0.74 at HSC.

13.5.1 Specific Range (Bizjet)

Specific range is a convenient way to present cruise performance. Using Equa­tion 13.27, the Sp. Rn is computed. The details of the specific range are not a direct
substantiation requirement; it is needed to compute the cruise-segment perfor­mances (i. e., fuel burned, distance covered, and time taken). Figure 13.14 shows the specific range for the Bizjet (i. e., the worked-out example). When readers redo the specific-range computations, there may be minor differences in the results. From the Sp. Rn values, the fuel burned and distance covered is worked out, which in turn gives the time taken for the distance.

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).

(a) Climb Speed Schedule 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

performances of fuel consumed, distance covered, and time taken to climb at the desired altitude.

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 first – and second – segment 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 400- to 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 first – and second-segment climb. Table 13.17 sum­marizes the first – and second-segment climbs for both 8-deg and 20-deg flap settings. At one engine failed, the aircraft must return to the base immediately.

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 (p = 0.0023 slug/ft3, a = 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 x 0.0023 x 4222 x 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 x (0.5 x 0.0023 x 4222 x 323) = 0.0242 x 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 x 2,260 = 4,520 lb. From Equa­tion 13.10, the quasi-steady-state rate of climb is given by:

R/C Vqq[(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 x 0.352 = 0.0686. Hence:

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

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:

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:

Sgtd = 6 x 168 = 1,008 ft

An aircraft in full braking with pB = 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 x 0.002378 x 110.62 = 14.54

For deceleration:

a = 32.2 x [(-0.4) – (CLq/48.92)(0.1 – 0.4)]

= 32.2 x [-0.4 + (0.15 x 14.54/48.92)]

= 32.2 x [-0.4 + 0.0445] = -11.45

The distance covered during braking, SG0Land = (158 x 79)/11.45 = 1,090 ft. The landing distance SGLand = 1,008 +1,090 = 2,098 ft.

Multiplying by 1.667, the rated LFL = 1.667 x 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.

Increasing the flap setting improves the BFL capability at the expense of a loss in climb gradient. The next section verifies the gradient requirements. With one engine inoperative, the loss of thrust percentage for a two-engine aircraft is the highest (i. e., 50%). With one engine failed, the aircraft acceleration suffers severely and the ground run from Vi to liftoff is high.

Table 13.16 summarizes the takeoff performance and associated speed sched­ules for the two flap settings and provides an example of the procedure. The ratio of speed schedules can be varied for pilot ease, as long as it satisfies FAR requirements.

At a lower flap setting of 8 deg, the decision speed Vi is close to the rotation speed Vi = 0.93VR. The situation improves with a higher flap setting of 20 deg when Vi = 0.9 Vr.

Higher flap settings provide more time between the decision speed Vi and the rotation speed VR. However, it is not problematic if V1 is close to VR. If one engine fails close to the decision speed, then the rotation speed VR is reached very quickly; that is, even if a pilot’s reaction is slow, the aircraft will still take off if there is suf­ficient runway length available (the BFL can be considerably lower than the avail­able airfield length). Also, Vmu is close to VR; hence, tail dragging is not likely. If an engine fails early enough, then a pilot has sufficient time to recognize the failure and abort the takeoff.

With more than two engines, the decision speed Vi is farther from the rotation speed VR. A pilot must remain alert as the aircraft speed approaches the decision speed Vi and must react quickly if an engine fails.

Figure 13.11. Balanced field length

Three decision speeds are worked out to establish the V1 for the BFL computation. Equation 13.2 gives average acceleration as:

a = g[(T/ W – ,i) – (CLSq/ W)(Cd/Cl – n)]

* Takeoff at 8- and 20-deg flaps. Landing at 35- to 40-deg flap, engines at idle, and Vstaii at aircraft landing weight of 15,800 lb.

The average acceleration a is at 0.7V of the segment of operation. For each segment of the BFL, a is computed.

The lift coefficient during the ground run is changing with speed gain and is not easy to determine. During the ground run, the angle of attack is low, even when the aircraft reaches the stall speed, Vstall. Up to the decision speed V;, only a fraction of the aircraft weight is taken up by the wing as a result of lift generation. Liftoff is not achieved until a pilot rotates the aircraft just above the Vstall. There is a rapid gain in lift generation because the angle of attack increases rapidly with rotation. Table 13.9 lists typical CL and CD/CL values.

Segment A: All Engines Operating up to the Decision Speed V1

Using Equation 13.2 and data from Table 13.6, the average acceleration becomes:

a = 32.2 X [(0.34 – 0.03) – (Cl?/64)(0.1 – 0.03)] = 32.2 x (0.31 – Ctq/914.3)

At a representative speed of 0.7V;, the average q = 0.5 x 0.002378 x 0.49V;2 = 0.0006 x V;2. At this segment, the average CL = 0.5 (yet to reach the full value). Then:

a = 32.2 X (0.31 – Cq/914.3)

= 32.2 x (0.31 – 0.0003 x V2/914.3) ft/s2

Equation 13.3 gives the ground distance covered as:

Sg = Vave x (V/ – Vi)/a

Table 13.10 computes the ground distance covered for all engines operating up to V1 .

Segment B: One-Engine Inoperative Acceleration from V1 to Liftoff Speed, VLO

Because one engine is inoperative, there is a loss of power by half (T/W = 0.17) plus an asymmetric drag rise (CD/CL = 0.102). As the speed increases, the aver­age CL increases to 0.8, making the weight on the wheels lighter; therefore, the ground friction, /л, is reduced to 0.025. The acceleration Equation 13.2 is rewritten as follows:

a = 32.2 x [(0.17 – 0.025) – (Cl^/64)(0.102 – 0.025)] = 32.2 x (0.145 – 0.8 x q/831.2)

The velocity that would give the average acceleration is:

V0.7 = 0.7 x (Vlo – V1) + V1

a = 32.2 x (0.145 – 0.000951 x Vi.72/831.2 = 32.2 x (0.145 – 0.00000114 x V027) Equation 13.3 gives the ground distance covered as:

Sg = Vave x (Vf – Vi)/a

Table 13.11 computes the ground distance covered from Vi to VLO for the two flap settings.

Table 13.11. Segment B: Bizjet one-engine ground distance Vi to VLO (8-degflap)

Guess V1 (kt) (1.688 ft/s) 90 (151.92)

Vstaii at 20,600 lb (ft/s) 177.6

VLo at 1.12 Vstaii 199

V0.7 = 0.7 x (VLO – V1) + V1 (ft/s) 185 (0.166M) T/W (from Figure 13.1) 0.138

q (dynamic head at 0.7V1) 41.09

a (ft/s2) 3.21

Ground distance, SG_VLO (ft) 2,664

Segment C: Flaring Distance with One Engine Inoperative from VLO to V2

The flaring distance reaches V2 from VLO; from statistics, the time to flaring is 3 s. Table 13.12 computes the ground distance covered from VLO to V2 with one engine inoperative for the two flap settings. In this segment, an aircraft is airborne; hence, there is no ground friction. Taking the average velocity between V2 and VLO gives the distance covered during flare.

The next step is to compute the stopping distance with the maximum application of brakes.

Segment D: Distance Covered in 1 s as Pilot-Recognition Time and 2 s for Brakes to Act from V to VB (Flap Settings Are of Minor Consequence)

Table 13.13 computes the ground distance covered from Vi to VB.

Segment E: Braking Distance from VB to Zero Velocity (Flap Settings Are of Minor Consequence)

The reaction time to apply the brakes, after the decision speed, V1t is 3 s. The aircraft continues to accelerate during the 3 s.

For an aircraft in full braking with pB = 0.4, all engines shut down, and the average CL = 0.5, Equation 13.2 for average acceleration, based on 0.7VB (« 0.7V1), reduces to:

a = 32.2 X [(-0.4) – (Ciq/64)/(0.1 – 04)] = 32.2 x [-0.4 + (0.15q/64)]

= 32.2 x [-0.4 + q/426.7)] = 0.075q – 12.88

Table 13.14 computes the ground distance covered from VB to stopping.

The TOFL (see Table 13.12, Segments A + B + C) and the stopping distance (see Table 13.14, Segments A + D + E) are plotted in Figure 13.11 to obtain the BFL for a flap setting of 8 deg and summarized in Table 13.16. It satisfies the specified TOFL requirement of 4,400 ft.

This section computes aircraft performance to substantiate capabilities as required by the FAR and the operators. Table 13.4 gives the speed schedules appropriate to an aircraft takeoff; aircraft drag polar is given in Figure 9.1. The wing area, Sw = 30 m2 (323 ft2) and the MTOM = 9,400 kg (20,723 lb). Assuming that 20 kg of fuel is consumed during the taxi, the MTOM for the takeoff estimation = 9,380 kg (20,680 lb), and wing-loading, W/Sw = 64 lb/ft2. The known stalling speed is com­puted by using the following:

V 2 x MTOM I 2 x 20,680 /53,847.6

stal1 = pSwCl = 0.002378 x 323 x Cl = CL

Table 13.8 provides the Bizjet aircraft data generated thus far. To make the best use of the available data, all computations are in the FPS system. The results subse­quently can be converted to the SI system.

Civil aircraft maximum speed is executed in HSC in a steady, level flight when the available thrust equals the aircraft drag. The first task is to compute drag at the maximum cruise speed and then check whether the available thrust (at the maxi­mum cruise rating) is sufficient to achieve the required speed. Sometimes, the avail­able maximum cruise thrust is more than what is required; in that case, the engine is adjusted to a slightly lower level. The LRC schedule is meant to maximize range and is operated at a lower speed to avoid the compressibility drag rise. Section

13.5.6 explains the worked-out example.

13.4.3 Payload Range Capability

Finally, a civil aircraft must be able to meet the payload range capability as specified by the market (i. e., customer) requirements. The mission range and fuel consumed during the mission are given by the following two equations:

mission range — Rclimb + Rcruise + Rdescent (13.22)

mission fuel — Fuelclimb + Fuelcruise + Fueldescent (13.23)

mission time — Timedimb + Timewise + Timedescent

The method to compute fuel consumption, distance covered, and time taken during a climb and a descent is discussed in Section 13.4.3. In this section, the governing equations for cruise range (Rcruise), cruise fuel (Fuelcmise), and time taken during cruise are derived.

Let Wi — aircraft initial cruise weight (at the end of a climb) and Wf — aircraft final cruise weight (at the end of a cruise). Then:

fuel burned during cruise — Fuelcruise — Wi – Wf (13.24)

At any instant, rate of aircraft weight change, dW — rate of fuel burned (consumed). In an infinitesimal time dt, the infinitesimal weight change, dW — sfc x thrust (T) x dt, or:

dt — dW/(sfc x T)

Integrating Equation 13.25 gives the time taken for the Rcruise. At cruise, T — D and L — W.

In Equation 13.25, multiply both the numerator and the denominator by weight, W, and then equate T — D and W — L.

Equation 13.25 reduces to:

_ / W (dW_ / М /dW

t — sfc T W — sfc D W

The elemental range:

(13.27)

Therefore, the range covered during cruise (Remise) is the integration of Equation 13.27 from the initial to the final cruise weight. At cruise, V and sfe remain nearly constant. Using the midcruise L/D, the change in L/D can be ignored and taken out of the integral sign:

The value of ln(Wi/Wf) = ki_range varies from 0.2 to 0.5; the longer the range, the higher is the value.

In Equation 13.28, the terms Wi and Wf are concerned with fuel consumed dur­ing cruise and the term sfe stems from the matched-engine characteristics. The other terms (VL/D) are concerned with aircraft aerodynamics. Aircraft designers aim to increase the VL/D as best as it is possible to maximize the range capability. The aim is not just to maximize the L/D but also to maximize the VL/D. Expressing this in terms of the Mach number, it becomes ML/D. To obtain the best of engine-aircraft gain, it is to maximize (ML)/(sfcD).

Specific range (Sp. Rn) is defined as range covered per unit weight (or mass) of fuel burned. Using Equation 13.28:

Sp. Rn = Reruise/cruise fuel = [kj_range x (VL)/(sfeD)]/(Wi – Wf) (13.29)

The cruise fuel weight (Wi – Wf) can be expressed in terms of the MTOW and varies from 15 to 40% of the MTOW; the longer the range, the higher is the value. Let k2_range = ki^ange/(0.15 to 0.4). Then, Equation 13.29 reduces to:

Equation 13.30 provides insight to what can maximize the range; that is, a good design to stay ahead of the competition:

1. Design an aircraft to be as light as possible without sacrificing safety. Mate­rial selection and structural efficiency are key; integrate with lighter bought-out equipment.

2. Use superior aerodynamics to lower drag.

3. Choose a better aerofoil for good lift, keeping the moment low.

4. Design an aircraft to cruise as fast as possible within the Mcrit.

5. Match the best available engine with the lowest sfe.

However, these points do not address cost implications. In the end, the DOC dictates the market appeal and designers must compromise performance with cost. These points comprise the essence of good civil aircraft design, which is easily said but not so easy to achieve, as must be experienced by readers.

Equation 13.25 can be further developed. From the definition of the lift coeffi­cient, CL, the aircraft velocity, V, can be expressed as follows:

Substituting in Equation 13.25, the cruise range, Rcruise, can be written as:

R fWi 1 I 2W / L dW ПГ rW 1 / УCl dW

Rmise Wf sfc P SwCl D W p Sw Wf sfc CD JW

(13.31)

As mentioned previously, over the cruise range, changes in the sfc and L/D typically are minor. If the midcruise values are taken as an average, then they may be treated as constant and are taken outside the integral sign. Then, Equation 13.31 becomes:

This equation is known as the Breguet range formula, originally derived for propeller-driven aircraft that had embedded propeller parameters (jet propulsion was not yet invented).

The LRC is carried out at the best sfc and at the maximum value of VCL/CD (i. e., L/D) to maximize range. Typically, the best L/D occurs at the midcruise con­dition. For a very high LRC (i. e., 2,500 nm or more), the aircraft weight difference from initial to final cruise is significant. It is beneficial if cruise is carried out at a higher altitude when the aircraft becomes lighter, which can be done either in a stepped altitude or by making a gradual, shallow climb that matches the gradual lightening of the aircraft. Sometimes a mission may demand HSC to save time, in which case Equation 13.29 is still valid but not operating for the best range.

Climb is possible when the available engine thrust is more than the aircraft drag; the excess thrust (i. e., thrust minus aircraft drag, (T – D)) is converted into the potential energy of height gain. The total energy of an aircraft is the sum of its PE and KE, expressed as follows:

Total energy:

E = mgh + (mgV2/2g) = mg(h + V2/2g)

(13.5)

(13.6)

The term for the rate of change of specific energy is specific excess power (SEP):

SEP = dh/dt + V/g(dV/dt) = V(T – D)/(mg) = dhe/dt (13.7)

Equation 13.7 shows that he > h by the term V2/2g in Equation 13.6. In other words, an aircraft can continue to climb by converting KE to PE until the speed is decreased to the point where the aricraft is unable to sustain the climb.

An enroute climb is performed in an accelerated climb. The equation for an accelerated climb is derived as follows (Figure 13.10). For simplicity, the subscript to to represent aircraft velocity is omitted. From Figure 13.10, the force equilibrium gives:

(T – D) = mg sin y + (m)dV/dt This gives the gradient:

sin y = [T – D – (W/g)dV/dt]/W = [(T – D)/W] – [(1/g) x dV/dt] (13.8)

Write:

dV/dt = (dV/dh) x (dh/dt)

Then, rate of climb:

R/Cacd = dh/dt = Vsin y = V(T – D)/ W – (V/g) x (dV/dh) x (dh/dt) (13.9) By transposing and collecting dh/dt:

V[(T – D)/ W]

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

Combining Equations 13.7 and 13.9, the rate of climb is written as: dhe /dt = V(T – D)/ W – (V/g)(dV/dt) + V/g(dV/dt) = V(T – D)/ W (13.11)

The rate of climb is a point performance and is valid at any altitude. The term V (dV) is dimensionless. It penalizes the unaccelerated rate (i. e., the numerator in Equation 13.10) of climb depending on how fast an aircraft is accelerating during the climb. Part of the propulsive energy is consumed for speed gain rather than altitude gain. Military aircraft make an accelerated climb in the operational arena when the g (ddh) term reduces the rate of climb depending on how fast the aircraft is acceler­ating. Conversely, civil aircraft has no demand for a high-accelerated climb; rather, it makes an enroute climb to cruise altitude at a quasi-steady-state climb by holding

Table 13.7. Y (dY) value (dimensionless quantity)

Below tropopause Above tropopause

At constant EAS 0.566 m2 0.7 m2

At constant Mach number -0.133 m2 0 (Mach held constant) the climb speed at a constant EAS or Mach number. A constant-EAS climb causes the TAS to increase with altitude gain. A constant speed indication eases a pilot’s workload. During a quasi-steady-state climb at a constant EAS, the contribution by the V (dY) term is minimal. The magnitude of the acceleration term decreases with altitude gain and becomes close to zero at the ceiling (i. e., defined as when R/Caccl = 100 ft/min). (Remember that V = VeasU/° and Yeas = Ma^/ст.)

Constant EAS Climb Below Tropopause (y = 1.4, R = 287 J/kgK, g = 9.81 m/s2)

The term V (dY) can be worked out in terms of a constant EAS as follows:

V /dV VeasVeas (d(1/a) = _ V^ /da = _M2a2_ /da

g dh g^a dh 2ga2 dh 2ga dh

In SI, Equation 3.1 gives a troposphere T = (288.16 – 0.0065h), and p = 1.225 x (T/288.16)(g/00065R)-1 = 1.225 x (T/288.16)(9 81/0 0065 x 287)-1 = 1.225 x (T/288.16)4 255 derives the density ratio (up to the tropopause) by replacing T in terms of its lapse rate and h:

a = p/p0 = (288.16 – 0.0065h/288.16)4’255 = (1 – 2.2558 x 10-5 x h)4 255

This gives (da/dh) = -9.6 x 10-5 x (1 – 2.2558 x 10-5 x h)3 255 Therefore:

M2a2

x [9.6 x 10-5 x (1.2558 x 10-5 x h)3 255]

2ga

M2 x 1.4 x 287 x (288.16 – 0.0065h)

2 x 9.81

x [9.6 x 10-5)/(1 – 2.2558 x 10-5 x h)] M2 1.4 287 288.16

2x 9.81

These equations are summarized in Table 13.7.

Constant Mach Climb Below Tropopause (Y = 1.4, R = 287 J/kgK, g = 9.81 m/s2)

The term Y (dV’j can be worked out in terms of a constant Mach-number climb as follows:

Y dV MaM da aM2Y7R dVT

g dh = g dh = g dh

From Equation 3.1, the atmospheric temperature, T, can be expressed in terms of altitude, h, as follows:

T = (288 – 0.0065h)

where h is in meters. Substituting the values of y , R, and g, the following is obtained:

V /dVMVyR(dVT_ 0.00664aM2 (i3i3)

g dh = g dh = – (288 – 0.0065h) ‘

When evaluated for altitudes, the equation gives the value as shown in Table 13.7.

In a similar manner, the relationships above tropopause can be obtained. Up to 25 km above tropopause, the atmospheric temperature remains constant at 216.65 K; therefore, the speed of sound remains invariant.

With the loss of one engine at the second-segment climb, an accelerated climb penalizes the rate of climb. Therefore, a second-segment climb with one engine inoperative is achieved at an unaccelerated climb speed, at a speed a little above V2 due to the undercarriage retraction. The unaccelerated climb equation is obtained by omitting the acceleration term in Equation 13.10, yielding the following equa­tions:

T – D = WsinY becomes sinY = (T = D)/ W

The unaccelerated rate of climb:

R/C = dh/dt = VsinY = V x (T – D)/ W (13.14)

The climb performance parameters vary with altitude. An enroute climb perfor­mance up to cruise altitude is typically computed in discreet steps of altitude (i. e., 5,000 ft; see Figure 13.10), within which all parameters are considered invariant and taken as an average value within the altitude steps. The engineering approach is to compute the integrated distance covered, the time taken, and the fuel consumed to reach the cruise altitude in small increments and then totaled. The procedure is explained herein. The infinitesimal time to climb is expressed as dt = dh/( R/Cacd). The integrated performance within the small altitude steps is written as:

At = final tinitial = (hfinal hinitial) / (R/ Caccl) ave (13.15)

and

AH = (hfinal – hinitial) (13.16)

Using Equation 13.8, the distance covered during a climb is expressed as:

Ax = At x Vave = At x Vcosy (13.17)

where V = the average aircraft speed within the altitude step.

Fuel consumed during a climb can be expressed as:

Afuel = average fuel flow rate x At (13.18)

Summary

The time used to climb, timeclimb = J2 At, is obtained by summing the values obtained in the small steps of altitude gain. The distance covered during a climb,

Rciimb = J2 As, is obtained by summing the values obtained in the small steps of altitude gain. The fuel consumed for a climb, Fuelclimb = J2 A fuel, is obtained by summing the values obtained in the small steps of altitude gain.

Descent

A descent uses the same equations as for a climb except that the thrust is less than the drag; that is, the rate of descent (R/Daccl) is the opposite of the rate of climb. The rate of descent is expressed as follows:

Unlike in a climb, gravity assists a descent; therefore, it can be performed without any thrust (i. e., the engine is kept at an idle rating, producing zero thrust). However, passenger comfort and structural considerations require a controlled descent with the maximum rate limited to a certain value depending on the aircraft design. A controlled descent is carried out at a partal-throttle setting. To obtain the maximum range, an aircraft should ideally make its descent at the desired minimum rate. These adjustments entail varying the speed at each altitude. To ease the pilot’s workload, a descent is made at a constant Mach number; when the Veas limit is reached, the aircraft adapts to a constant Veas descent, similar to a climb. Special situations may occur, as follows:

1. For an unaccelerated descent, Equation 13.19 becomes:

R/Dunacci = dh/dt = V[(DW T)] (13.20)

At a higher altitude, the prescribed speed schedule for a descent is at a constant Mach number; therefore, the previous tropopause Vtas is constant and the descent is maintained in an unaccelerated flight.

This indicates that at a constant V(L/D), the R/Cdescent is the same for all weights.

As in a climb, the other parameters of interest during a descent are range cov­ered (Rdescent), fuel consumed (Fueldescent), and time taken (timedescent). There are no FAR requirements for the descent schedule. The descent rate is limited by the cabin-pressurization schedule for passenger comfort. FAR requirements are enforced during an approach and a landing. At high altitude, the inside cabin pres­sure is maintained as an approximate 8,000-ft altitude. Depending on the structure design, the differential pressure between the inside and the outside is maintained at approximately 8.9 lb/in[26].

Integrated performances for a climb to cruise altitude and a descent to sea level are computed, and the values for distance covered, time taken, and fuel consumed are estimated to obtain the aircraft payload range. Textbooks may be consulted for details of climb and descent performances.

The rated TOFL at the MTOM is determined by the BFL in the event of an engine failure. The BFL must comply with FAR requirements. A normal takeoff with all engines operating needs a considerably shorter field length than the rated TOFL. Designers must provide the decision speed V1 for pilots that below which the takeoff must be aborted for safety reasons if an engine fails. Figure 13.9 shows the segments involved in computing the BFL.

Figure 13.9. Balanced field length consideration

The figure shows that taking off with one engine inoperative (i. e., failed) has three segments to clear a 35-ft height, as follows:

Segment A: Distance covered by all-engine operating ground run until one engine fails at the decision speed V1.

Segment B: Distance covered by one-engine inoperative acceleration from V1

to Vlo.

Segment C: Continue with the flare distance from liftoff speed VLO to clear a

35-ft obstacle height reaching aircraft speed V2.

For stopping at the decision speed Vi, there are two segments (which replace seg­ments B and C), as follows:

Segment D: Distance covered during the reaction time for a pilot to take brak­ing action. (Typically, 3 s is used as the pilot recognition time and braking to act, spoiler deployment, and so on. At engine failure, the thrust decay is gradual; within this reaction time before brake application, there is a minor speed gain, shown in Figure 13.9.)

Segment E: Distance to stop from VB to Vo (maximum brake effort).

The BFL is established when Segments (B + C) = Segments (D + E).

Takeoff Equations

During takeoff, the aircraft accelerates. At the conceptual design phase, the average values of speed, acceleration, and thrust are taken of 0.7 of the velocity of the ground run segments. In later stages of a project, the computation is figured more accurately in smaller steps of speed increments within which average values of the variables are considered constant. CL also varies with speed changes; typical values of CL and Cd/Cl are given in Section 13.5.1.

Section 11.3.1 derives the associated governing equations to compute the TOFL. Equations 11.2 and 11.4 give:

Table 13.6. FAA second-segment climb gradient at missed approach

The landing configuration is with full flaps extended and the aircraft at land­ing weight. The approach segment at landing is from a 50-ft altitude to touch down. At approach, the FAR requires that an aircraft must have a minimum speed Vapp = 1.3Vstaii@iand. At touchdown, aircraft speed is Vtd = 1-15Vstaii@iand. Brakes are applied 2 s after all wheels touch down. A typical civil aircraft descent rate at touch­down is between 12 and 22 ft/s. The landing runway length should be 1.667 times the computed landing distance. Generally, this works out to be slightly less than the BFL at the MTOM (but not necessarily).

For a balked landing or missed approach at landing weight, the FAR require­ments are given in Table 13.6. An aircraft is configured with full flaps, undercar­riage extended, and engine in full takeoff rating. In general, this is not a problem because all engines are operational and the aircraft is lighter at the end of the mis­sion. Military aircraft requirements are slightly different: Vapp = 1.2Vstall@land and Vtd = 1.1Vstall@land.

The approach has two segments, as follows:

• a steady, straight glide path from a 50-ft height

• flaring in a nearly circular arc to level out for touchdown, which incurs a higher g

The distances covered in these two segments depend on how steep is the glide path and how rapid is the flaring action. This book does not address these details of anal­ysis; instead, a simplified approach is taken by computing the distance covered dur­ing the time from a 50-ft height to touchdown before the brakes are applied; it is assumed to be 6 s herein.