Answer:
[tex]F_{thrust}[/tex] ≅ 111 KN
Explanation:
Given that;
A medium-sized jet has a 3.8-mm-diameter i.e diameter (d) = 3.8
mass = 85,000 kg
drag co-efficient (C) = 0.37
(velocity (v)= 230 m/s
density (ρ) = 1.0 kg/m³
To calculate the thrust; we need to determine the relation of the drag force; which is given as:
[tex]F_{drag}[/tex] = [tex]\frac{1}{2}[/tex] × CρAv²
where;
ρ = density of air wind.
C = drag co-efficient
A = Area of the jet
v = velocity of the jet
From the question, we can deduce that the jet is in motion with a constant speed; as such: the net force acting on the jet in the air = 0
SO, [tex]F_{drag}-F_{thrust} = 0[/tex]
We can as well say:
[tex]F_{drag}= F_{thrust}[/tex]
We can now replace [tex]F_{thrust} with F_{drag}[/tex] in the above equation.
Therefore, [tex]F_{thrust}[/tex] = [tex]\frac{1}{2}[/tex] × CρAv²
The A which stands as the area of the jet is given by the formula:
[tex]A=\frac{\pi d^2}{4}[/tex]
We can now have a new equation after substituting our A into the previous equation as:
[tex]F_{thrust}[/tex] = [tex]\frac{1}{2}[/tex] × Cρ [tex](\frac{\pi d^2}{4})v^2[/tex]
Substituting our data from above; we have:
[tex]F_{thrust}[/tex] = [tex]\frac{1}{2}[/tex] × [tex](0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2[/tex]
[tex]F_{thrust}[/tex] = [tex]\frac{1}{8} (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2[/tex]
[tex]F_{thrust}[/tex] = 110,990N
[tex]F_{thrust}[/tex] in N (newton) to KN (kilo-newton) will be:
[tex]F_{thrust}[/tex] = [tex](110,990N)*\frac{1KN}{1,000N}[/tex]
[tex]F_{thrust}[/tex] = 110.990 KN
[tex]F_{thrust}[/tex] ≅ 111 KN
In conclusion, the jet engine needed to provide 111 KN thrust in order to cruise at 230 m/s at an altitude where the air density is 1.0 kg/m³.
