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IRINA_888 [86]
3 years ago
8

Explain the underlying physical reason why when we conduct various heat treatments on 1018 steel we expect the modulus of elasti

city to stay about the same for every heat treatment.
Engineering
1 answer:
pashok25 [27]3 years ago
6 0

Answer:

Modulus of  elasticity is dependent on the elemental constitution of  1080  steel and various heat treatment don't affect this elemental composition  but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel

Explanation:

Generally the underlying physical reason why when we conduct various heat treatments on 1018 steel we expect the modulus of elasticity to stay about the same for every heat treatment is  

 

Modulus of  elasticity is dependent on the elemental constitution of  1080  steel and various heat treatment don't affect this elemental composition  but only affects the mechanical properties(like strength , hardness, ductility, malleability) of the 1080 steel that are affected by plastic deformation of the 1080 steel

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A medium-sized jet has a 3.8-mm-diameter fuselage and a loaded mass of 85,000 kg. The drag on an airplane is primarily due to th
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Answer:

F_{thrust} ≅ 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:

F_{drag} = \frac{1}{2} × 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, F_{drag}-F_{thrust} = 0

We can as well say:

F_{drag}= F_{thrust}

We can now replace F_{thrust} with F_{drag} in the above equation.

Therefore, F_{thrust} = \frac{1}{2} × CρAv²

The A which stands as the area of the jet is given by the formula:

A=\frac{\pi d^2}{4}

We can now have a new equation after substituting our A into the previous equation as:

F_{thrust} = \frac{1}{2} × Cρ (\frac{\pi d^2}{4})v^2

Substituting our data from above; we have:

F_{thrust} = \frac{1}{2} × (0.37)(1.0kg/m^3)(\frac{\pi(3.8m)^2 }{4})(230m/s)^2

F_{thrust} = \frac{1}{8}   (0.37)(1.0kg/m^3)({\pi(3.8m)^2 })(230m/s)^2

F_{thrust} = 110,990N

F_{thrust}  in N (newton) to KN (kilo-newton) will be:

F_{thrust} = (110,990N)*\frac{1KN}{1,000N}

F_{thrust} = 110.990 KN

F_{thrust} ≅ 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.

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