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zloy xaker [14]

3 years ago

11

A 1,040 N force is recorded on a hemispherical vane as it redirects a 2.5 cm- blade diameter water jet through a 180 angle. Dete

rmine the velocity of the flowing water jet if the blade diameter water jer through a 180' angle. Determinethe velcity of the flowing water jet if is assumed to be frictionless.

1 answer:

Alex777 [14]3 years ago

8 0

This question is incomplete, the complete question is;

A 1,040 N force is recorded on a hemispherical vane as it redirects a 2.5 cm- blade diameter water jet through a 180 angle.

Determine the velocity of the flowing water jet if the blade is assumed to be frictionless.

Answer: the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless

Explanation:

Given that;

Force Ft = 1040 N

diameter d = 2.5 cm = 0.025 m

we know that; force acting on Hemispherical plate is;

Ft = 2δav²

where

a is area = π/4(0.025)²

δ is density of water = 1000 kg/m³

v is velocity = ?

now we substitute

1040 = 2 × 1000 × (π/4(0.025)²) × v²

1040 = 0.9817v²

v² = 1040 / 0.9817

v² = 1059.3867

v = √1059.3867

v = 32.5482 ≈ 32.55 m/s

Therefore the velocity of the flowing water jet is 32.55 m/s assuming the blade is frictionless

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Compute the volume percent of graphite, VGr, in a 2.5 wt% C cast iron, assuming that all the carbon exists as the graphite phase

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Answer:

The volume percent of graphite is 91.906 per cent.

Explanation:

The volume percent of graphite ( $\% V_{Gr}$ ) is determined by the following expression:

$\%V_{Gr} = \frac{V_{Gr}}{V_{Gr}+V_{Fe}} \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\frac{V_{Gr}}{V_{Fe}} }\times 100\,\%$

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$V_{Gr}$ - Volume occupied by the graphite phase, measured in cubic centimeters.

$V_{Fe}$ - Volume occupied by the ferrite phase, measured in cubic centimeters.

The volume of each phase can be calculated in terms of its density and mass. That is:

$V_{Gr} = \frac{m_{Gr}}{\rho_{Gr}}$

$V_{Fe} = \frac{m_{Fe}}{\rho_{Fe}}$

Where:

$m_{Gr}$ , $m_{Fe}$ - Masses of the graphite and ferrite phases, measured in grams.

$\rho_{Gr}$ , $\rho_{Fe}$ - Densities of the graphite and ferrite phases, measured in grams per cubic centimeter.

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$\%V_{Gr} = \frac{1}{1 +\frac{\frac{m_{Gr}}{\rho_{Gr}} }{\frac{m_{Fe}}{\rho_{Fe}} } } \times 100\,\%$

$\%V_{Gr} = \frac{1}{1+\left(\frac{m_{Gr}}{m_{Fe}} \right)\cdot \left(\frac{\rho_{Fe}}{\rho_{Gr}} \right)}\times 100\,\%$

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$m_{Gr} = \frac{2.5}{100}\times (100\,g)$

$m_{Gr} = 2.5\,g$

$m_{Fe} = 100\,g - 2.5\,g$

$m_{Fe} = 97.5\,g$

If $m_{Gr} = 2.5\,g$ , $m_{Fe} = 97.5\,g$ , $\rho_{Fe} = 7.9\,\frac{g}{cm^{3}}$ and $\rho_{Gr} = 2.3\,\frac{g}{cm^{3}}$ , the volume percent of graphite is:

$\%V_{Gr} = \frac{1}{1+\left(\frac{2.5\,gr}{97.5\,gr} \right)\cdot \left(\frac{7.9\,\frac{g}{cm^{3}} }{2.3\,\frac{g}{cm^{3}} } \right)} \times 100\,\%$

$\% V_{Gr} = 91.906\,\%$

The volume percent of graphite is 91.906 per cent.

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