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 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_im_K

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

Mass of structure

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

Mass of power-plant

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

Mass of equipments

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

Mass of fuel

m_F = m_F 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 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 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.