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SCORPION-xisa [38]
3 years ago
10

What will happen in a wire drawing operation when the cross-sectional area has a reduction of 60% in a single pass?

Engineering
1 answer:
Fofino [41]3 years ago
4 0

Answer:

DRAWING LOAD IS  3.67 A_{O}\sigma

Explanation:

wire drawing is a method of obtaining wire of bigger large diameter from iron rod . it is cold process which need die to obtain wire

drawing load for wire drawing is given as P = A_{F}*\sigma*ln(\frac{A_{O}}{A_{F}})

Where A f is initial area, Ao is original area, σ is yield stress

as given in question sectional area reduce 60%, therefore

A_{f} = A_{O}- 0.6A_{O}

    = 0.4 A_{O}

Due to change in area ,drawing load p is

p = 0.4A_{O}*\sigma*ln(\frac{A_{O}}{0.4A_{O}})

p = 3.67 A_{O}\sigma

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The power that must be supplied to the motor is 136 hp

<u>Explanation:</u>

Given-

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According to the equation of motion:

F = ma

3(500) - 1000 = \frac{1000}{32.2} * a

a = 16.1 ft/s²

We know,

v^2 - u^2 = 2a (S - So)\\\\v^2 - (0)^2 = 2 * 16.1 (27-0)\\\\v = 29.48m/s

To calculate the output power:

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As efficiency is given and output power is known, we can calculate the input power.

ε = Pout / Pin

0.65 = 44220 / Pin

Pin = 68030.8 lb.ft/s

Pin = 68030.8 / 500 hp

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A motorcycle starts from rest with an initial acceleration of 3 m/s^2, and the acceleration then changes with the distance s as
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Follows are the solution to this question:

Explanation:

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A = as

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Calculating the kinematics equation:

\to v^2 = v^2_{o} + 2as\\\\

        =0+ \sqrt{2as}\\\\ = \sqrt{2(A)}\\\\= \sqrt{2(950 \frac{m^2}{s^2})}\\\\= 43.59 \frac{m}{s}

Calculating the value of acceleration:  

\to a= \frac{dv}{dt}

=\frac{dv}{ds}(\frac{ds}{dt}) \\\\=v\frac{dv}{ds}\\\\\to \frac{dv}{ds}=\frac{a}{v}

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2 years ago
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jok3333 [9.3K]

Answer:

Explanation:

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\dfrac{\partial^2T}{\partial x^2}=  \ 0  \  ;  \ if \  T = f(x)  \\ \\ \dfrac{\partial^2T}{\partial y^2}=  \ 0  \  ;  \ if \  T = f(y)  \\ \\ \dfrac{\partial^2T}{\partial z^2}=  \ 0  \  ;  \ if \  T = f(z)

b) For a transient, 1-D, constant with energy generation

suppose T = f(x)

Then; the equation can be expressed as:

\dfrac{\partial^2T}{\partial x^2} + \dfrac{Q_g}{k} = \dfrac{1}{\alpha} \dfrac{dT}{dC}

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c) The heat equation for a cylinder steady-state with 2-D constant and no compressible energy generation is:

\dfrac{1}{r}\times \dfrac{\partial}{\partial r }( r* \dfrac{\partial \ T }{\partial \ r}) + \dfrac{\partial^2 T}{\partial z^2 }= 0

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\dfrac{\partial ^2 T}{\partial z^2} = \dfrac{1}{\alpha}\Big [\dfrac{\partial ^2 T}{\partial ^2 t}+ V_z \dfrac{\partial ^2T}{\partial ^2z} \Big ]

since;

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e) The heat equation for a sphere that is transient, 1-D, and incompressible with energy generation is:

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3 years ago
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