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Evgen [1.6K]
2 years ago
7

A cylindrical 1040 steel rod having a minimum tensile strength of 865 MPa (125,000 psi), a ductility of at least 10%EL, and a fi

nal diameter of 6.0 mm (0.25 in.) is desired. Some 7.94 mm (0.313 in.) diameter 1040 steel stock, which has been cold worked 20% is available. Describe the procedure you would follow to obtain this material. Assume that 1040 steel experiences cracking at 40%CW
Engineering
1 answer:
gtnhenbr [62]2 years ago
4 0

Answer:

procedure attached below

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

Explanation:

<em>Given data:</em>

Minimum tensile strength = 865 MPa

Ductility = 10%EL

Desired Final diameter = 6.0 mm

20% cold worked 7.94 mm diameter 1040 steel stock

<u> Describe the procedure you would follow to obtain this material.</u>

assuming 1040 steel experiences cracking at 40%CW

<em>attached below is a detailed procedure of obtaining the material</em>

The material to be used will have a %Cw of 34.5%Cw which is < 40%Cw

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Answer:

<u><em>To answer this question we assumed that the area units and the thickness units are given in inches.</em></u>

The number of atoms of lead required is 1.73x10²³.    

Explanation:

To find the number of atoms of lead we need to find first the volume of the plate:

V = A*t

<u>Where</u>:

A: is the surface area = 160

t: is the thickness = 0.002

<u><em>Assuming that the units given above are in inches we proceed to calculate the volume: </em></u>

V = A*t = 160 in^{2}*0.002 in = 0.32 in^{3}*(\frac{2.54 cm}{1 in})^{3} = 5.24 cm^{3}    

Now, using the density we can find the mass:

m = d*V = 11.36 g/cm^{3}*5.24 cm^{3} = 59.5 g

Finally, with the Avogadros number (N_{A}) and with the atomic mass (A) we can find the number of atoms (N):

N = \frac{m*N_{A}}{A} = \frac{59.5 g*6.022 \cdot 10^{23} atoms/mol}{207.19 g/mol} = 1.73 \cdot 10^{23} atoms    

Hence, the number of atoms of lead required is 1.73x10²³.

I hope it helps you!

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A viscous fluid flows in a 0.10-m-diameter pipe such that its velocity measured 0.012 m away from the pipe wall is 0.8 m/s. If t
maksim [4K]

Answer:

A) centerline velocity = 1.894 m/s

B) flow rate = 7.44 x 10^(-3) m³/s

Explanation:

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U(r) = (Δp•D²/16μL) [1 - (2r/D)²]

Velocity at the centre of the tube can be expressed as;

V_c = (Δp•D²/16μL)

Thus,

U(r) = (V_c)[1 - (2r/D)²]

From question, diameter = 0.1m,thus radius (r) = 0.1/2 = 0.05m

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Thus, let's put 0.038 for r in the velocity intensity equation and put other relevant values to get the velocity at the centre.

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0.8 = (V_c)[1 - (0.76)²]

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B) flow rate is given by;

ΔV = Average Velocity x Area

Now, average velocity = V_c/2

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So, flow rate = 0.947 x 0.007854 = 7.44 x 10^(-3) m³/s

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