Shendal
24th Mar 2006, 09:05
Hello,
My questions are:
If a climb is done in the thrust position of Max Climb, and with C.I.=0 the airplane may climb at V_x and with a C.I. = 200 it may climb at V_y. Then if we apply the calculation of Fuel Consumption:
FuelConsumption = SpecificFuelConsumption x Thrust x Time.
Why when we are climbing at V_x (in theory less fuel consumption) we are consuming more as long as SFC is the same for V_y and V_x, thrust is the same as well, and only time changes because we climb more quickly at V_y.
There is something wrong but I don't know what :bored:
Another question is:
Which is the average angle of climb of the civil aviation airplanes ? 5 degrees ? 15 degrees ? 30 degrees ? ...
Again...
How much ground distance will fly an airplane taking-off at 0 ft and climbing to 36000 ft at a climb gradient of 4% and 250 Kt of EAS (FAA minimum all engines working) ? I calculated it and I obtained it will fly for 274Km with a climb rate of 1000 ft/min more or less but I am not sure at all :O
And finally...
If we suppose our airplane may flight with a C.I. of 200 (Minimum trip time), cannot we flight at mach maxi on the altitude where sqrt(CL)/CD (Max range lift_to_drag coefficient) is maximum, or it will not be flying at C.I. = 200 ? In this conditions, the airplane will flight at maxi range (requirement of C.I.=0) and at mach maxi (requirement of C.I.=200).
I suppose that climbing to the optimal altitude for sqrt(CL)/CD maxi and mach maxi consume extra time that implies we will not be flying at C.I.=200, may be it will be C.I.=100. I am not sure at all; I hope some of you can help me :)
Thank you very much in advance.:)
Edu
My questions are:
If a climb is done in the thrust position of Max Climb, and with C.I.=0 the airplane may climb at V_x and with a C.I. = 200 it may climb at V_y. Then if we apply the calculation of Fuel Consumption:
FuelConsumption = SpecificFuelConsumption x Thrust x Time.
Why when we are climbing at V_x (in theory less fuel consumption) we are consuming more as long as SFC is the same for V_y and V_x, thrust is the same as well, and only time changes because we climb more quickly at V_y.
There is something wrong but I don't know what :bored:
Another question is:
Which is the average angle of climb of the civil aviation airplanes ? 5 degrees ? 15 degrees ? 30 degrees ? ...
Again...
How much ground distance will fly an airplane taking-off at 0 ft and climbing to 36000 ft at a climb gradient of 4% and 250 Kt of EAS (FAA minimum all engines working) ? I calculated it and I obtained it will fly for 274Km with a climb rate of 1000 ft/min more or less but I am not sure at all :O
And finally...
If we suppose our airplane may flight with a C.I. of 200 (Minimum trip time), cannot we flight at mach maxi on the altitude where sqrt(CL)/CD (Max range lift_to_drag coefficient) is maximum, or it will not be flying at C.I. = 200 ? In this conditions, the airplane will flight at maxi range (requirement of C.I.=0) and at mach maxi (requirement of C.I.=200).
I suppose that climbing to the optimal altitude for sqrt(CL)/CD maxi and mach maxi consume extra time that implies we will not be flying at C.I.=200, may be it will be C.I.=100. I am not sure at all; I hope some of you can help me :)
Thank you very much in advance.:)
Edu