Climb Performance Requirements (AJT)

Military trainers should climb at a much higher rate of climb than civil aircraft. The requirement of 50 m/s (10,000 ft/min) at normal training configuration (NTC) is for an unaccelerated climb for comparison with accelerated climb. Unaccelerated rate of climb varies depending on the constant speed (i. e., EAS) climb, making a comparison difficult. This section presents calculations for both rates of climb.

This section checks only the enroute climb with a clean configuration. The unac­celerated climb Equation 13.7 is used. The MTOM at the NTC is 4,800 kg (10,582 lb). The wing area SW = 17 m2 (183 ft2).

During an enroute climb, the aircraft has a clean configuration. Under maxi­mum takeoff power, it makes an accelerated climb to 800 ft (p = 0.00232 slug/ft3, a = 0.9756) from the second-segment velocity of V2 to reach a 350-KEAS speed schedule to start the enroute climb. During enroute climb, the engine throttle is retarded to the maximum climb rating. The quasi-steady-state climb schedule main­tains 350 KEAS and the aircraft accelerates with an altitude gain at a rate of dV/dh until it reaches Mach 0.8 at around 25,000 ft. From there, the Mach number is held constant until it reaches the cruise altitude. We assume that 100 kg of fuel is con­sumed to taxi and climb to an 800-ft altitude, where the aircraft mass is 4,700 kg (10,362 lb). At 350 kts (590.8 ft/s, Mach 0.49), the aircraft lift coefficient is:

Cl = MTOM/qSw = 10,362/(0.5 x 0.00232 x 590.82 x 183)

= 10,582/74,905 = 0.138

The clean aircraft drag coefficient from (see Figure 9.19) at Cl = 0.141 gives CDciean = 0.023. The clean aircraft drag, D = 0.023 x (0.5 x 0.002378 x 590.82 x 183) = 0.023 x 74,905 = 1,723 lb. The available engine-installed thrust at a maxi­mum continuous rating (95% of maximum thrust, as given in Figure 13.4) at Mach 0.49 (459.8 ft/s) is T = 0.95 x 5,000 = 4,750 lb. From Equation 13.10, the accelerated rate of climb is as follows:

c V[(T – D)/W]

‘ aacl 1 + (V/g)(dV/dh)

At a quasi-steady-state-climb, Table 13.5 gives:

– = 0.56m2 = 0.56 x 0.492 = 0.1345

g dh

From Equation 13.5, the rate of climb is:

R/Cacci = {[590.8 x (4,750 – 1,723) x 60]/10,362}/]1 + 0.1345] = 10, 355/1.1345 = 9,127 ft/min

Therefore, the unaccelerated rate of climb, R/C = 10,355 ft/min. The aircraft speci­fication is based on an unaccelerated climb of 10,000 ft/min, which is just met. (Here, the cabin area is small and the pressurization limit is high.)

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