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Juliette [100K]
4 years ago
12

for high-volume production runs, machining parts from solid material might not be the best choice of manufacturing operations be

cause
Engineering
1 answer:
Naddik [55]4 years ago
8 0

Answer:

There are actually multiple types of processes a manufacturer uses, and those can be grouped into four main categories: casting and molding, machining, joining, and shearing and forming.

Explanation:

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Air flows steadily and isentropically from standard atmospheric conditions to a receiver pipe through a converging duct. The cro
Liula [17]

Answer:

The answer is "0.0728"

Explanation:

Given value:

P_0= 14.696\ ps\\\\\ p _{0}= 0.00238 \frac{slue}{ft^{3}}\\\\\ A= 0.05 ft^2\\\\\ T_0= 59^{\circ}f = 518.67R\\\\\ air \ k= 1\\\\ \ cirtical \  pressure ( P^*)=P_0\times \frac{2}{k+1}^{\frac{k}{k-1}}\\

                                     = 14.696\times (\frac{2}{1.4+1})^{\frac{1.4}{1.4-1}}\\\\=7.763 Psia\\\\

if P flow is chocked

if P>P^{*} \to flow is not chocked

When  P= 10 psia < P^{*} \to not chocked

match number:

\ for \ P= \ 10\ G= \sqrt{\frac{2}{k-1}[(\frac{\ p_{0}}{p})^{\frac{k-1}{k}}-1]}

                       = \sqrt{\frac{2}{1.4-1}[(\frac{14.696}{10})^{\frac{1.4-1}{1.4}}-1]}

M_0=7.625

p=p_0(1+\frac{k-1}{2} M_0 r)^\frac{1}{1-k}

  =0.00238(1+\frac{1.4-1}{2}0.7625`)^{\frac{1}{1-1.4}}\\\\\ p=0.001808 \frac{slug}{ft^3}

\ T= T_0(1+\frac{k-1}{2} Ma^r)^{-1}\\\\\ T=518.67(1+\frac{1.4-1}{2} 0.7625^2)^{-1}\\\\\ T=464.6R\\\\

\ velocity \ of \ sound \ (C)=\sqrt{KRT}\\\\

                                    =\sqrt{1.4\times1716\times464.6}\\\\=1057 ft^3\\\\

R= gas constant=1716

m=PAV\\\\

    =0.001808\times0.05\times(Ma.C)\\\\=0.001808\times0.05\times0.7625\times 1057\\\\=0.0728\frac{slug}{s}

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stimate the force required in extruding 70-30 brass at 700C, if the billet diameter is 125 mm and the extrusion ratio is 20
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If safeguarding is not possible what must be used instead
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