The received result is compared to value of take-off mass for airplanes-analogues. And after reasonable range has been obtained and designing is continued.
Then the values Mass of structure m K and its components (mass of wing m w, fuselage m F, tail-unit m TU, landing gear m LG), mass of fuel m F, mass power-plant m PP, equipment m EQ for the projected airplanes are calculated.
Mass of wing, fuselage, tail-unit, landing gear is determined with these statistical data cited in table.5. According to this table mass ratio of the airplane units is shown in share of total structural mass m K.
Values of units mass are found by the formula
m i = m i m K
The total sum of units mass ratio m i for designed airplane must be equal to 1, that is Σ m i = 1.
Mass of structure
m K = m K x m 0 = 0.25 x 292300 = 73083 kg.
Mass of power-plant
m PP = m PP x m 0 = 0.08 x 292300 = 23386 kg.
Mass of equipments
m EQ = m EQ x m 0 = 0.09 x 292300 = 26309 kg.
Mass of fuel
m F = m F x m 0 = 0.46 x 292300 = 134471 kg.
For wing, fuselage, tail unit and landing gear masses we are multiplying the ratio to the structural mass of the aircraft to the structural mass.
Chosen mass ratio of the aircraft units
mass ratio of wing m w = 0.377
mass ratio of the fuselage m FUS = 0.367
mass ratio of tail-unit m TU = 0.073
mass ratio of landing gear m LG = 0.183
Thus, the respective masses are calculated accordingly:
mass of wing m w = m w x m K = 0.377 x 53200 =20056.4kg.
mass of the fuselage m FUS = m FUS x m K = 0.367 x 53200 =19524.4kg.
mass of tail-unit m TU = m TU x m K = 0.073 x 53200 = 3883.6kg.
mass of landing gear m LG = m LG x m K = 0.183 x 53200 = 9735.6kg.
| m0 kg | mC kg | mCR kg | mF kg | mPP kg | mEQ kg | mK kg | |||
| mw kg | mFUS kg | mTU kg | mLG kg | ||||||
| 20056.4 | 19524.4 | 3883.6 | 9735.6 | ||||||
Table.1.3 Masses of Airplane parts and units.
Determination of the engine parameters
Further it is necessary to determined starting thrust of the engine P0. It is determined on the collected statistical values of starting thrust-to-weight ratio t 0.
Then it is possible to find starting total thrust of engines,
P0 = t0 m0 g,
Where g = 9.8 m/c2, t0 = 0.293 (from statiscal data)
P0 = 0.293x 190000 x 9.8
P0 = 545.57 kN.
Further starting thrust of one engine can be determined on the basis of engines number n
P01 = P0/n.
P01 = ( 545.57)/2
P01 = 272.79 kN.
The engine of necessary type is chosen on value of starting thrust from the catalogue of engines, and we got the Rolls Royce Trent 895 with a thrust of 423 kN.
Now with this real value of P01 we calculate once again P0 and t0 .
P0 = 272.79 x 2 = 545.57 kN.
to:= (564000/190000x9.8)
t0 = 0.293
A copyof the general view of the engine with a designation of overall dimension is also provided in fig.. It is placed in the place allotted for it, i.e. the wing.
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Fig1.18
Fig1.19
Royce RB-211-524 G/H
Main characteristics of Rolls Royce RB-211-524 G/H:-
Dry weight: 4387 kg
Air flow: 728kg/s
By-pass ratio: 4.3
Engine length: 3.175m
Diameter: 2.192m
The calculation of the geometrical parameters of the projected airplanes units is done after the determination of airplane mass characteristics.
