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Y_Kistochka [10]

3 years ago

13

a sprue is 12 in long and has a diameter of 5 in at the top. The molten metal level in the pouring basing is taken to be 3 in fr

om the top of the sprue for design purposes. If a flow rate of 40 in3/s is to be achieved, what should be the diameter at the bottom of the sprue

2 answers:

vampirchik [111]3 years ago

8 0

Answer:

See explaination

Explanation:

We can describe Aspiration Effect as a phenomenon of providing an allowance for the release of air from the mold cavity during the metal pouring.

See the attached file for detailed solution of the given problem.

prohojiy [21]3 years ago

3 0

Answer:

The diameter at the bottom of the sprue is =0.725 in, and the sprue will not occur when 0.021<0.447.

Explanation:

Solution

The first step to take is to define the Bernoulli's eqaution

h effective = v²top/2g + ptop /ρg = hbottom + v² bottom/2g + p bottom/ ρg

h effective + 0 +0= 0 +v² bottom/2g + 0

Thus,

v bottom = √ 2gh total

=√ 2 (32. 6 ft/ s²) + (12/12 ft)

Which is = 8.074 ft/s

We now, express the relation for flow rate.

Q =π/4 D² bottom v bottom

= 40 in 3/s = π/4 D²₃ ( 8.074 ft/s) (12 in/ ft)

so,

D bottom = 0.725 in.

Then,

We express the relation to avoid aspiration

A₃/A₂ < √ h top /h total

= π/4 D²₃/ π/4 D²₂ < √3/15

= 0.725²/5² < √3/15

=0.021<0.447

Therefore, the aspiration will not happen or occur

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1. Consider a city of 10 square kilometers. A macro cellular system design divides the city up into square cells of 1 square kil

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Answer:

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$\Delta t_{max} = \frac{\sqrt{2} }{30}\,h$

The approximate time is giving by the average of minimum and maximum times:

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$\Delta t_{min} = \frac{1}{3000}\,h$

The maximum time for traversing a cell is:

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$\Delta t_{max} = \frac{\sqrt{2} }{3000}\,h$

The approximate time is giving by the average of minimum and maximum times:

$\Delta t_{mean} = \frac{1+\sqrt{2} }{2}\cdot\frac{l}{v}$

$\Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h$

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