A medium-sized jet has a 3.8-m-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to the
cylindrical fuselage, and aerodynamic shaping gives it a drag coefficient of 0.37. How much thrust must the jet’s engines provide to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m3?
1 answer:
Answer:
Fthrust = 110.99 kN
Explanation:
Model:
Ignore the drag on the wings and focus on the cylindrical fuselage.
Visualize:
The net force is zero as the jet cruises at 230 m/s, so the thrust must be equal in magnitude to the drag. The gravitational and lift forces are equal in magnitude and opposite in direction.
Given
D = 3.80 m ⇒ R = D / 2 = 3.80 m/ 2 = 1.90 m
C = 0.37
ρ = 1.0 kg/m³
v = 230 m/s
m = 85,000 kg
Fthrust = ?
We show the Free Body Diagram in the pic.
Then, we apply
∑Fx = 0 (→+)
Fthrust = Fdrag = (1/2)*C*ρ*A*v²
A = π*R² = π*(1.90 m)² = 11.34 m²
⇒ Fthrust = (1/2)*(0.37)*(1.0 kg/m³)*(11.34 m²)*(230 m/s)²
⇒ Fthrust = 110990.159 N = 110.99 kN
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Answer:
=> base transport factor = 0.98.
=> emitter injection efficiency = 0.99.
=> common-base current gain = 0.97.
=> common-emitter current gain = 32.34.
=> ICBO = 1 × 10^-6 A.
=> base transit time = 0.325.
=> lifetime = 1.875.
Explanation:
(Kindly check the attachment for the diagram showing the dominant current components, with proper arrows, for directions in the normal active mode).
The following parameters or data are given for a p-n-p BJT with NE 7 NB 7 NC and they are: IEp = 10 mA, IEn = 100 mA, ICp = 9.8 mA, and ICn = 1 mA.
(1). The base transport factor = ICp/IEp=9.8/10 = 0.98.
(2). emitter injection efficiency =IEp/ IEp + ICn = 10/10 + 0.1 = 0.99.
(3).common-base current gain = 0.98 × 0.99 = 0.9702.
(4).common-emitter current gain =0.97 / 1- 0.97 = 32.34.
(5). Icbo = Ico = 1 × 10^-6 A.
(6). base transit time = 1248 × 10^-2 × (1.38× 10^-23/1.603 × 10^-19). = 0.325.
(7).lifetime;
= > 2 = √0.325 + √ lifetime.
= Lifetime = 2.875.
Answer:
Force on the prototype is 5000 N
Solution:
As per the question:
Depth of water, x = 2.0 m
Flow velocity, v' = 1.5 m/s
Width of the river, w = 20 m
Force on the bridge pier model, F' = 5 N
Pressure, Ratio = Ratio of scale length
Scale = 1:10
Now,
where
P' = pressure on model
P = pressure on prototype
where
F' = Force on model
F = Force on prototype
A' = Area of model
A = Area of prototype
Now:
F = 5000 N