Multi-Engine Jet Transport

Response to Request for Proposal
Small Multi-Engined Transport Jet Airplane
Georgia Institute of Technology
Atlanta, Georgia
School of Aerospace Engineering

Mission Requirements:

Total Cruise of 2800 Nm at 40,000 ft. standard day conditions.
Loiter time of 45 minutes at the end of the descent and prior to landing.
Cruise Speed of M=.80 and a dash of M=.85.
1% unusable fuel.
Total Payload is 5,000 lbs.
Two crew members each weighing 200 lbs.
Engine will be GE-CJ610-9(MOD)
SFC at cruise at 40,000 ft. is .65/lb/lb/hr
SFC at loiter is .43 lb/lb/hr
(L/D)_max = 16.17

Assumptions:

Assume standard day conditions.
Speed of sound at 40,000 ft (STP) is 968 ^ft/_sec.
Assume that the A/C will be in cruise the majority of the time rather than dash.
The A/C will be constructed of aluminum.

#	Segment	Description	Values	Fraction
1	W1/Wo	Engine Start Warm Up Taxi Take off	.990 .990 .975	.9752
2	W2/W1	Climb	.980	.980
3	W3/W2	Cruise	R=2800 Nm C=.65 lb/lb/hr V=(.8 * 968) L/D _cruise = .866(L/D)_max	.7533
4	W4/W3	Descent	.990	.990
5	W5/W4	Loiter	E=45 min C=.43 lb/lb/hr (L/D)_max = 16.17	.9803
6	W6/W5	Land Rollout Taxi	.992	.992

Initial Sizing of Jet Transport

Mission Weight Fraction (W6/Wo)	.6931
Mission Fuel Fraction (1-W6/Wo)	.3069
Total Fuel Fraction (1.01*(1-W6/Wo))	.3100

Empty Weight Fraction is given by where A is 1.02, C is -.06 and K is 1.0 as per the handout in class.

The total weight is given by where W_crew is 400 lbs, W_payload is 5000 lbs, and W_F/Wo is the total fuel fraction given on the previous page as .3100.

Wo	We/Wo	New Wo
50000	0.53292	34377.4
34377.4	0.54503	37250.2
37250.2	0.54242	36589.3
36589.3	0.543	36734.4
36734.4	0.54287	36702.2
36702.2	0.5429	36709.3
36709.3	0.54289	36707.8
36707.8	0.54289	36708.1
36708.1	0.54289	36708
36708	0.54289	36708.1
*36708.1*	*0.54289*	*36708.1*

After the iterations of Wo, the total weight of the aircraft is to be 36,708.1 lbs.

Verification:

W_E	19928.46 lbs.
W_F	11379.51 lbs.
W_crew	400 lbs.
W_payload	5000 lbs
Total Weight	36707.96 lbs

Estimation of ^T/_W and ^W/_S

The following values were used:

C_Do = .0218
e = .85
where AR = 8.5383

Dash:

M_dash = .85
q_dash = 198.361 PSF
n = 1
Lapse Factor = .18
Weight Fraction (W2/Wo) = .955696

(W/S)_dash	Term 1	Term2	(T/W)_R	(T/W)_R-w	(T/W)_R-W	(W/S)o
50	0.08847	0.01106	0.09952	0.09512	*0.52842*	*52.3179*
60	0.07372	0.01327	0.08699	0.08314	*0.46187*	*62.7815*
70	0.06319	0.01548	0.07867	0.07518	*0.41769*	*73.245*
80	0.05529	0.01769	0.07298	0.06975	*0.38749*	*83.7086*
90	0.04915	0.0199	0.06905	0.06599	*0.36661*	*94.1722*
100	0.04423	0.02211	0.06635	0.06341	*0.35226*	*104.636*
110	0.04021	0.02432	0.06454	0.06168	*0.34265*	*115.099*

Maneuver:

M_man = .30
q_man = 126.419 PSF
n = 3.2
Lapse Factor = .92
Weight Fraction (W1/Wo) = .975

(W/S)_F	Term 1	Term 2	(T/W)_R	(T/W)_R-w	(T/W)_R-wl	(W/S)_o-w
50	0.05512	0.17763	0.23275	0.22693	*0.24667*	*51.2821*
60	0.04593	0.21316	0.25909	0.25261	*0.27458*	*61.5385*
70	0.03937	0.24869	0.28806	0.28085	*0.30528*	*71.7949*
80	0.03445	0.28421	0.31866	0.31069	*0.33771*	*82.0513*
90	0.03062	0.31974	0.35036	0.3416	*0.37131*	*92.3077*
100	0.02756	0.35527	0.38282	0.37325	*0.40571*	*102.564*
110	0.02505	0.39079	0.41585	0.40545	*0.44071*	*112.821*

The design point was found to have the following values:

(T/W)o = .37

(W/S)o = 92.1 PSF

Taking the base line engine thrust of 4250 lbs. and scaling it up by 10% for an engine thrust of 4675 lbs. each, it will take 3 engines to power the aircraft.

3 engines * (4675 lbs/engine) = 14025 lbs. This value is slightly higher than the actual need thrust, but engines do not always perform at maximum specification, so this leaves a safety margin.

Landing Data

S_FL = .3*(V_A)² where S_FL is given in the RFP as 5500 ft, thus (V_A)² = 18333.33

V_A = 135.401 knots

V_SL = V_A/1.3 = 104.15 knots or 175.793 ft/sec.

but (W_L/Wo) = .88 so

(C_lmax)_landing	(W/S)o ~PSF
2.0	83.51
2.2	91.86
2.4	100.21

Interpolating, this gives a maximum coefficient of lift when landing of 2.21.

Take-off Data

S_TOFL = 5800 ft.

TOP₂₅ = 5800/37.5 = 154.67lbs/ft²

(W/S)TO			(T/W)TO
	C_Lmax=1.4	C_Lmax=1.6	C_Lmax=1.8	C_Lmax=2
60	0.27709	0.24245	0.21551	0.19396
70	0.32327	0.28286	0.25143	0.22629
80	0.36945	0.32327	0.28735	0.25862
90	0.41563	0.36368	0.32327	0.29094
100	0.46181	0.40409	0.35919	0.32327

From the design plot, C_Lmax in Take off = 1.6.

Fuselage
	Length	61.5 ft.
	Diameter	9 ft.
	Fineness Ratio	6.8333
Main Wing
	Area	398.57 ft²
	Aspect Ratio	8.5382
	Wing Span	58.34 ft.
	Leading Edge Sweep	28
	Taper Ratio	.35
	Root Chord	10.12 ft.
	Tip Chord	3.54 ft.
	Quarter Chord Sweep	25.42
	Mean Aerodynamic Chord	7.36 ft.
	Spanwise Location MAC	12.24 ft.
	Mean Geometric Chord	6.83 ft.
	Thickness Ratio	.14
	Wing Fuel Volume Avail.	166.32 ft³
	Fuel Volume Required	232.23 ft³
	Dihedral	5
Weights
	Design Take-off Weight	36708.1 lbs.
	Design Empty Weight	19928.46 lbs.
	Design Fuel Weight	11379.51 lbs.
	Crew Weight	400 lbs.
	Total Payload	5000 lbs.
Aerodynamic
	Clean C_Do	.0218
	Cruise "e"	.85
Engine
	Total S.L. Thrust	13582 lbs.
	Thrust/Engine	4527.3 lbs.
	Number of Engines	3
	Core Length	49.2 in.
	Core Diameter	19.82 in.
	Core Weight	536.00 lbs.
Empennage
	V_H	1.0
	S_H	117.34 ft²
	L_H	25 ft.
	V_V	.09
	S_V	77.5 ft²
	L_V	27 ft.

Validation of Assumed C_Do

The equation log₁₀ S_wet = c + dlog₁₀ Wo can be used to find the wetted area. The constants c and d are found in Roskam in Table 3.5 in Part 1, page 122.

c = .0199

d = .7531

The wetted area was found to be 2868.28 ft².

The equation log₁₀ f= a + blog₁₀ S_wet can be used to find a value for the equivalent parasite area where the constants a and b are found in Roskam part 1, page 122, Table 3.4.

C_f = .0030

a = -2.5229

b = 1.0000

The equivalent parasite area was found to be 8.60441 ft².

Finally, the C_Do can be found by dividing f by the wing area. This gives:

C_Do = .02158 ~ .0218