250 kts below 10000?

[jimmyg]

The Climb is far and away the most fuel expensive phase of the flight and every effort should be made to optimize fuel use per mile in this phase.

Power increases linearly with speed. Ram recovery; the degree of compression is exponential to speed. At low speed numbers this compression is not very significant, but does partially compensate for 'Thrust Equation' losses. They are about equal at approximately M0.5 (250 Kt in ball-park figures). With further speed increase, the exponentially increasing ram effect exceeds the linearly decreasing 'thrust equation' effect, and at about M0.75, thrust is back to static thrust. Much more of the engine's internal power is now used to propel the exhaust gases, much less is used to drive the compressor. Above this mach number, net thrust can and does exceed static thrust. Ram recovery is an important factor and directly affects thrust. At a fixed engine setting, due to the effects of the 'thrust equation', thrust declines with increasing speed up to about M0.5, where the decline bottoms out and then, due to ram recovery, begins to rise again reaching parity with static thrust at about M0.75 and may exceed static thrust thereafter. If there were no ram recovery, jet aircraft would do no better than propeller aircraft.

The best rate of climb occurs at a speed where there is maximum excess power. As jet engines directly produce thrust, not power, it is necessary to consider thrust multiplied by speed (Power = Force X Velocity). Thus, for a given thrust setting, power increases as TAS increases. Thrust actually ‘dips’ as speed increases, but then there is significant ram recovery at higher mach numbers, thus further increasing Power at higher speeds.

I frequently accelerate 'slingshot' above the low altitude speed limit of 250 KIAS from 5000 feet instead of the ‘standard’ 10000 feet. For the A320, this repeatedly saves 100 Kg of fuel per SECTOR and that’s only over 5000 feet. The risk is nominal due to bird strike versus performance rewards.