question in aeronautical engineering

Question

For Gulfstream IV, The following data on this aircraft is given.

A) Calculate the required available power per engine (in MW, or MegaWatt) for the cruise at the given cruise speed and atmospheric conditions.

Number of engines: 2
Wing surface: S=88.3 m2
Diameter of engine inlet: d=1.1176 m
Oswald efficiency: e=0.67
Span: b=23.7 m
Mass: m=30000 kg
(ISA) Air density at cruise altitude: ρ=0.42 kg/m3
Cruise speed: VTAS=460 kts
Zero-lift drag coefficient: CD0=0.015
For the International Standard Atmosphere (ISA) please use:

Gravity acceleration: g=9.80665 m/s2
Gas constant for air: R=287.00 J/ kgK
Sea-level pressure: p0=101325 Pa
Sea-level temperature: T0=15∘C
Sea-level density: ρ0=1.225 kg/m3

Answer 1

Assumptions:

SLUF: Steady, Level, Unaccelerated Flight

Asking for Power Required (P_R), not Power Available (P_A) which cannot be calculated with givens

Givens:

W = 30,000 kg = 66,139 lbs

S = 88.3 m^2 = 950.5 ft^2

C_D0 = 0.015

V = 460 KTAS = 776.393 ft/s

ρ = .42 kg/m^3 = 0.0008149345 slug/ft^3 (at altitude)

e=0.67

b = 23.7 m = 77.76 ft

AR = b^2 / S = (77.76 ft)^2 / 950.5 ft^2 = 6.362

Power Required: Drag * Velocity

Equations:

D = .5 * ρ * V^2 * S * C_D

C_D = C_D0 + C_L²/(π * e * AR)

C_L = 2 * W / (ρ * S * V^2)

Solving:

C_L = 2 * 66,139 lbs / (0.0008149345 slug/ft^3 * 950.5 ft^2 * (776.393 ft/s)^2) = 0.2833

C_D = 0.015 + 0.2833^2 / (π * .67 * 6.632) = 0.0207

D = .5 * 0.0008149345 slug/ft^3 * (776.393 ft/s)^2 * 950.5 ft^2 * 0.0207 = 4832.57 lbs

P_R = D * V = 4832.57 lbs * 776.393 ft/s = 3751973.52 ft*lbs/s = 6821.77 HP (TOTAL)

P_R (one engine) = P_R / 2 = 6821.77 HP / 2 = 3410.88 HP = 2.544 MW