Estimate the takeoff gross weight (crude weight) of a business transport aircraf

Question

Estimate the takeoff gross weight (crude weight) of a business transport aircraft which will carry five passengers plus the pilot (1200 lb) in relative comfort in a pressurized cabin. The specific performance is to be as follows: Maximum level speed at mid cruise weight: 250 mi/h Select the appropriate aircraft from table 3.6 for this mission Select Jet or Propeller Baggage weight: 60lb/person Loiter: 20 min Range: 1200 miles Ceiling: 25000 ft Rate of climb at sea level 1000 ft/min Stalling speed: 70 mi/h Landing distance (to clear a 50ft obstacle): 2200 ft Takeoff distance (to clear a 50 ft obstacle): 2500 ft

Explanation / Answer

Take-off gross weight is the total weight of aircraft when its is begins to its motion.

Maximum take-off weight can be broken into following element

1) Crew weight 2) payload weight 3) fuel weight 4) empty weight of flight

MT0W=WC+WPY+WF+WE

the given data as

passenger weiglht = 1200 lbs,baggage weight = 60 lb/person ,max level speed = 250 mi/hr,range=1200 miles,loiter=20min,ceiling = 25000ft,range of climp 1000 ft/min,stalling speed = 70 mil/h ,landing dist=2200ft,takeup dist = 2500 ft.

1) MC+MPY=1200+60x5 = 1500 lbs.

Now calculate total travel time

travel dist for takeoff = 2200ft = 0.41 mile,travel dist for landing = 2500 ft = 0.47 mile,cruise = 1200 mile

total dist = 1200+0.41+0.47 = 1200.88

total time required for travel = 1200.88/250=4.8 hrs

as per standard jet engine 5300 lbs pe hour fuel usage for one engine

there total weightof fuel for one engine =25440 lbs

for two engine total fuel weight = 50880 lbs

fuel required for loiter = 176.66 lbs ( for 20 min)

take 10 percent extra = 1.10(50880+176.66)=56162 lbs.

total appxomate takeoff weight = 1500 + 56162+ empty weight of plane as per manufracture

= 57662+empty weight with equipment

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