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EightsOnPylons
12th Feb 2004, 02:39
Hello,

Is there a way to calculate the horsepower output per engine on the CFM56-7B6? I know it has 27100 lbs of thrust (121 kn), but how can I convert this into Horsepower? Does anybody have a conversion factor or formula??

thanks,
EightsonPylons

Smoketoomuch
12th Feb 2004, 03:00
It depends how fast you are going.

EightsOnPylons
12th Feb 2004, 05:47
I am sorry, do you mean it is related to the phase of flight in relation to the surrounding atmospheric pressure? If possible I would like to find out what it is like just during the first application of thrust when taking off from a runway.

thanks,
eightsonpylons

DDG
12th Feb 2004, 07:26
Because power is determined by using the product of a force and a distance,it is not possibleto make a direct comparison of thrust and horsepower of a jet engine.When the engine is driving the aeroplanethrough the air,however,we can compute the EQUIVALENT horsepower being developed.When we convert the foot pound per minute of 1 hp to mile-pounds per hour we obtain the figure of 375.That is , 1 hp is equal to 33,00ft.lb/min or 375mi.lb/h.Thrust Horsepower (Thp) is then obtained by using the following formula;


Thp= (thrust(lbs) x airspeed(mph)) divided by 375



ie using the B737NG with 26k (cfm56-7b6) motors producing 27100 lbs of thrust(don`t know how you got the 27.1k figure) at stand still would be

(27100 x 0) divide by 375= 0

at say 50mph;

(27100 x 50) divide by 375 = 3,613.33 Thp

at say 100mph it would be;

(27100 x 100) divde by 375 = 7,226.66 Thp

at say 150 mph;

(27100 x 150) divide by 375 = 10,840 Thp


at 200mph;

(27100 x 200) divide by 375 = 14,453.33 Thp