FLIGHT VEHICLE DESIGN PROJECT 2

Professor: Dr. Steven Lu
Written By: Joey Haripersaud

Design Specifications for a Particular Jet Transport

Payload: 304 Passengers
Crew: Two pilots and three cabin attendants
Range: 4200 nm following by ¾ hour loiter
Altitude: 35,000 ft
Cruise speed: M = 0.84 at 35,000 ft
Climb: Direct climb to 35,000 ft at maximum take-off weight WTO
Take-off and landing: FAR 25 fieldlength 9,800 ft at an altitude of 5,300 ft and 98°F day.
Landing performance at WL = 0.8WTO
Engines: Four turbofans
Certification base: FAR 25

Specification Project 1

WTO= 357,100
WF used= 106,722
WOE TENT=188,008
WE TENT=185,197
WE= 185,240

Procedure
Step 1:
The Temperature ratio (φ) has to be …show more content…
CLTOmax is assumed to be 2.0, therefore CL = 2.0 / 1.44 = 1.389.
The drag polar for this case is:
CD = 0.0310+ CL^2 / 25.1
CD = 0.1051
This yields: (L / D) = 13.2125
(T / W)TO = 0.1754 (T / W)TO @ 98F = 0.2192

FAR 25.121 (Gear down T.O Flaps) (T / W)TO = 2 [1 / (L / D) + 0] between VLOF and V2. | It will be assumed, that VLOF = 1.1VSTO. | Because CLTOmax = 2.0, CLLOF = 2.0 / 1.1^2 = 1.6529 | The drag polar is: | CD = 0.0310 + CL^2 / 25.4 | CD = | 0.1521 FAR 25.121 (gear down, t.o flaps) | | | AT V2 | (CL)LOF | 1.6529 | 1.3889 | CD | 0.1521 | 0.1201 | L/D | 10.8692 | 11.5626 | (T/W)TO | 0.1840 | 0.1730 | (T/W)TO @ 98°F | 0.2300 | 0.2162 | |

FAR 25.121 (Gear Up Take Off Flaps)
(T / W)TO = 2 [1 / (L / D) + 0.024] at 1.2VSTO.
CLTOmax is assumed to be 2.0, therefore CL = 2.0 / 1.4 = 1.3889
The drag polar is:
CD = 0.0310 + CL^2 / 25.1
CD = 0.1051 FAR 25.121 (gear up, t.o. flaps) | | | CL | 1.3889 | CD | 0.1051 | L/D | 13.2125 | (T/W)TO | 0.1994 | (T/W)TO @ 98°F | 0.2492 |

FAR 25.121 (Gears Up, Flaps Up)
(T / W)TO = 2 [1 / (L / D) + 0.012] at 1.25VS
CL is assumed to be 1.4, therefore CL = 1.4 / 1.25^2 = 0.8960
The drag polar is:
CD = 0.016 + CL^2 / 26.7
CD = 0.0434 FAR 25.121 (gear up, flaps up) | | | CL | 0.8960 | CD | 0.0434 | L/D | 20.6288 |