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Tatiana [17]
2 years ago
9

(SI units) Molten metal is poured into the pouring cup of a sand mold at a steady rate of 400 cm3/s. The molten metal overflows

the pouring cup and flows into the downsprue. The cross section of the sprue is round, with a diameter at the top = 3.4 cm. If the sprue is 20 cm long, determine the proper diameter at its base so as to main- tain the same volume flow rate.
Engineering
1 answer:
maxonik [38]2 years ago
6 0

Answer:

diameter of the sprue at the bottom is 1.603 cm

Explanation:

Given data;

Flow rate, Q = 400 cm³/s

cross section of sprue: Round

Diameter of sprue at the top d_{top} = 3.4 cm

Height of sprue, h = 20 cm = 0.2 m

acceleration due to gravity g = 9.81 m/s²

Calculate the velocity at the sprue base

V_{base} = √2gh

we substitute

V_{base} = √(2 × 9.81 m/s² × 0.2 m )

V_{base} = 1.98091 m/s

V_{base} = 198.091 cm/s

diameter of the sprue at the bottom will be;

Q = AV = (πd_{bottom}^2/4) × V_{base}

d_{bottom} = √(4Q/πV_{base})

we substitute our values into the equation;

d_{bottom} = √(4(400 cm³/s) / (π×198.091 cm/s))

d_{bottom}  = 1.603 cm

Therefore, diameter of the sprue at the bottom is 1.603 cm

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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
kakasveta [241]

Answer:

a) n = 1000\,users, b)\Delta t_{min} = \frac{1}{30}\,h, \Delta t_{max} = \frac{\sqrt{2} }{30}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{60}\,h, c) n = 10000000\,users, \Delta t_{min} = \frac{1}{3000}\,h, \Delta t_{max} = \frac{\sqrt{2} }{3000}\,h, \Delta t_{mean} = \frac{1 + \sqrt{2} }{6000}\,h

Explanation:

a) The total number of users that can be accomodated in the system is:

n = \frac{10\,km^{2}}{1\,\frac{km^{2}}{cell} }\cdot (100\,\frac{users}{cell} )

n = 1000\,users

b) The length of the side of each cell is:

l = \sqrt{1\,km^{2}}

l = 1\,km

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{1\,km}{30\,\frac{km}{h} }

\Delta t_{min} = \frac{1}{30}\,h

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\Delta t_{max} = \frac{\sqrt{2} }{30}\,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} }{60}\,h

c) The total number of users that can be accomodated in the system is:

n = \frac{10\times 10^{6}\,m^{2}}{100\,m^{2}}\cdot (100\,\frac{users}{cell} )

n = 10000000\,users

The length of each side of the cell is:

l = \sqrt{100\,m^{2}}

l = 10\,m

Minimum time for traversing a cell is:

\Delta t_{min} = \frac{l}{v}

\Delta t_{min} = \frac{0.01\,km}{30\,\frac{km}{h} }

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

The maximum time for traversing a cell is:

\Delta t_{max} = \frac{\sqrt{2}\cdot l }{v}

\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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