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igor_vitrenko [27]
3 years ago
11

A copper wire of original diameter .80 m exhibits a maximum tensile load/ strength at an engineering stress= 248.2 mpa. its duct

ility is measured as 75 reduction of area. Determine the true strain at failure.
Engineering
1 answer:
ddd [48]3 years ago
4 0

Answer:

True Strain at failure = 1.386

Explanation:

For the question, ductility is given in reduction In the area to be 0.75

Let the initial Cross sectional Area of wire be Ao

And the final cross sectional Area of wire of cross sectional Area of wire at fracture be A

(Ao - A)/Ao = ductility = 0.75

Ao - A = 0.75Ao

A = Ao - 0.75Ao

A = 0.25 Ao

True Strain = In (Lf/Lo)

To obtain the ratio of the lengths of wire,

The volume of the wire stays constant, that is, Vo = Vf; Vo = Ao × Lo and Vf = A × Lf

AoLo = ALf

Lf/Lo = Ao/A

In (Lf/Lo) = In (Ao/A) = In (Ao/0.25Ao) = In 4 = 1.386

True Strain = 1.386

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lana [24]

Answer:

the overall heat transfer coefficient of this heat exchanger is 1855.8923 W/m²°C

Explanation:

Given:

d₁ = diameter of the tube = 1 cm = 0.01 m

d₂ = diameter of the shell = 2.5 cm = 0.025 m

Refrigerant-134a

20°C is the temperature of water

h₁ = convection heat transfer coefficient = 4100 W/m² K

Water flows at a rate of 0.3 kg/s

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First at all, you need to get the properties of water at 20°C in tables:

k = 0.598 W/m°C

v = 1.004x10⁻⁶m²/s

Pr = 7.01

ρ = 998 kg/m³

Now, you need to calculate the velocity of the water that flows through the shell:

v_{w} =\frac{m}{\rho \pi (\frac{d_{2}^{2}-d_{1}^{2}  }{4} )} =\frac{0.3}{998*\pi (\frac{0.025^{2}-0.01^{2}  }{4}) } =0.729m/s

It is necessary to get the Reynold's number:

Re=\frac{v_{w}(d_{2}-d_{1}) }{v} =\frac{0.729*(0.025-0.01)}{1.004x10^{-6} } =10891.4343

Like the Reynold's number is greater than 10000, the regime is turbulent. Now, the Nusselt's number:

Nu=0.023Re^{0.8} Pr^{0.4} =0.023*(10891.4343)^{0.8} *(7.01)^{0.4} =85.0517

The overall heat transfer coefficient:

Q=\frac{1}{\frac{1}{h_{1} }+\frac{1}{h_{2} }  }

Here

h_{2} =\frac{kNu}{d_{2}-d_{1}} =\frac{0.598*85.0517}{0.025-0.01} =3390.7278W/m^{2}C

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5 0
3 years ago
Investigating how slime molds reproduce is an example of applied research.<br> True<br> False
azamat

Answer:

false

Explanation:

3 0
3 years ago
Two steel plates are to be held together by means of 16-mm-diameter high-strength steel bolts fitting snugly inside cylindrical
dusya [7]

Answer:

The outer diameter of the spacers that yields the most economical and safe design is 25.03 mm

Explanation:

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Pressure = Stress* Area

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Pbolt = 210 N/mm2 * 200.96 mm^2 = 42201.6  N

For Brass spacer

Pressure = 42201.6  N

Area of Brass spacer = Pressure/Stress

Area of Brass spacer = 42201.6  N/145 N/mm^2 = 291.044 mm^2

Area of Brass spacer = (pi) (d^2 - 16^2)/4 =  291.044 mm^2

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5 0
3 years ago
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alekssr [168]

Answer:

d. All of the above would require an EIS.

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Answer:

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8 0
3 years ago
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