significant difference to the computed propulsive efficiency. By computing the exhaust velocity from the equation for thrust (while still assuming ), we can also obtain the propulsive efficiency as a function of specific thrust (): A corollary of this is that, particularly in air breathing engines, it is more energy efficient to accelerate a large amount of air by a small amount, than it is to accelerate a small amount of air by a large amount, even though the thrust is the same. This is why turbofan engines are more efficient than simple jet engines at subsonic speeds. Rocket engines A rocket engine's is usually high due to the high combustion temperatures and pressures, and the long converging-diverging nozzle used. It varies slightly with altitude due to changing atmospheric pressure, but can be up to 70%. Most of the remainder is lost as heat in the exhaust. Rocket engines have a slightly different propulsive efficiency () than air-breathing jet engines, as the lack of intake air changes the form of the equation. This also allows rockets to exceed their exhaust's velocity. Similarly to jet engines, matching the exhaust speed and the vehicle speed gives optimum efficiency, in theory. However, in practice, this results in a very low specific impulse, causing much greater losses due to the need for exponentially larger masses of propellant. Unlike ducted engines, rockets give thrust even when the two speeds are equal. In 1903, Konstantin Tsiolkovsky discussed the average propulsive efficiency of a rocket, which he called the utilization (utilizatsiya), the "portion of the total work of the explosive material transferred to the rocket" as opposed to the exhaust gas. Propeller engines The calculation is somewhat different for reciprocating and turboprop engines which rely on a propeller for propulsion since their output is typically expressed in terms of power rather than thrust. The equation for heat added per unit time, Q, can be adopted as follows: where H = calorific value of the fuel in BTU/lb, h = fuel consumption rate in lb/hr and J = mechanical equivalent of heat = 778.24 ft.lb/BTU, where is engine output in horsepower, converted to foot-pounds/second by multiplication by 550. Given that specific fuel consumption is Cp = h/Pe and H = 20 052 BTU/lb for gasoline, the equation is simplified to: expressed as a percentage. Assuming a typical propeller efficiency of 86% (for the optimal airspeed and air density