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4 years ago

7

The lift on a spinning circular cylinder in a freestream with a velocity of 30 m/s and at standard sea level conditions is 6 N/m

of span. Calculate the circulation around the cylinder

1 answer:

Evgesh-ka [11]4 years ago

5 0

Answer:

The circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

Explanation:

Given :

Velocity of spinning cylinder $v = 30$ $\frac{m}{s}$

Sea level density $\rho = 1.23$ $\frac{kg}{m^{3} }$

Sea level span $L = 6$ $\frac{N}{m}$

Lift per unit circulation is given by,

$L = \rho v c$

Where $c =$ circulation around cylinder

$c = \frac{L}{\rho v}$

$c = \frac{6}{1.23 \times 30}$

$c = 0.163$ $\frac{m^{2} }{s}$

Therefore, the circulation around the cylinder is 0.163 $\frac{m^{2} }{s}$

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hram777 [196]

Answer:

N = 38546.82 rpm

Explanation:

$D_{1}$ = 150 mm

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= 17671.45 $mm^{2}$

$D_{2}$ = 250 mm

$A_{2}= \frac{\pi }{4}\times 250^{2}$

= 49087.78 $mm^{2}$

The centrifugal force acting on the flywheel is fiven by

F = M ( $R_{2}$ - $R_{1}$ ) x $w^{2}$ ------------(1)

Here F = ( -UTS x $A_{1}$ + UCS x $A_{2}$ )

Since density, $\rho = \frac{M}{V}$

$\rho = \frac{M}{A\times t}$

$M = \rho \times A\times t$ $M = 7100 \times \frac{\pi }{4}\left ( D_{2}^{2}-D_{1}^{2} \right )\times t$

$M = 7100 \times \frac{\pi }{4}\left ( 250^{2}-150^{2} \right )\times 37$

$M = 8252963901$

∴ $R_{2}$ - $R_{1}$ = 50 mm

∴ F = $763\times \frac{\pi }{4}\times 250^{2}-217\times \frac{\pi }{4}\times 150^{2}$

F = 33618968.38 N --------(2)

Now comparing (1) and (2)

$33618968.38 = 8252963901\times 50\times \omega ^{2}$

∴ ω = 4036.61

We know

$\omega = \frac{2\pi N}{60}$

$4036.61 = \frac{2\pi N}{60}$

∴ N = 38546.82 rpm

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Answer:

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Explanation:

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solution

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