Structural mass of main aircraft assemblies, power plant mass, fuel mass, mass of equipment and control

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 mK and its components (mass of wing mw, fuselage mF, tail-unit mTU, landing gear mLG), mass of fuel mF, mass power-plant mPP, equipment mEQ 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 mK.

Values of units mass are found by the formula

mi = mimK

The total sum of units mass ratio mi for designed airplane must be equal to 1, that is Σmi = 1.

Mass of structure

mK = mK x m0 = 0.25 x 292300 = 73083 kg.

Mass of power-plant

mPP = mPP x m0 = 0.08 x 292300 = 23386 kg.

Mass of equipments

mEQ = mEQ x m0 = 0.09 x 292300 = 26309 kg.

Mass of fuel

mF = mF x m0 = 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 mw = 0.377

· mass ratio of the fuselage mFUS = 0.367

· mass ratio of tail-unit mTU = 0.073

· mass ratio of landing gear mLG = 0.183

 

Thus, the respective masses are calculated accordingly:

· mass of wing mw = mw x mK = 0.377 x 53200 =20056.4kg.

· mass of the fuselage mFUS = mFUS x mK = 0.367 x 53200 =19524.4kg.

· mass of tail-unit mTU = mTU x mK = 0.073 x 53200 = 3883.6kg.

· mass of landing gear mLG = mLG x mK = 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 t0.

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 P01we 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.

 

 

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.