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2 years ago

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A viscous liquid is sheared between two parallel disks of radius �, one of which rotates with angular speed Ω, while the other i

s fixed. The velocity field is purely tangential, and the velocity varies linearly with z from: �; = 0 at � = 0 (the fixed disk) to the velocity of the rotating disk at its surface (� = ℎ). Derive an expression for the velocity field between the disks.

1 answer:

Alexus [3.1K]2 years ago

5 0

Answer:

Upper disk rotates at a constant angular velocity. The velocity at any height from stationery disk, say at x metres

$U_o=v(\frac {x}{h})$ where v is tangential velocity at radius r from the centre of disk

$U_o=r\omega (\frac {x}{h})$

The radial component of velocity is given as

$U_r=0$

The z component of velocity is also given as

W=0

Total velocity, $v= r\omega (\frac {x}{h})\hat e_{o}$

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Answer:

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The electron diffusion coefficient, $D_{n} = \frac{(\triangle x)^{2} }{16 t_{d} }$

$\triangle x = (\triangle t )V_{d}$ , where Δt = separation of pulse seen in an oscilloscope in time( it should be in micro second range)

$\triangle x = \frac{(\triangle t) L}{t_{d} } \\\triangle x = \frac{180*10^{-6} * 1.8}{0.608*10^{-3} }\\\triangle x =0.533 cm$

$D_{n} = \frac{0.533^{2} }{16 * 0.608 * 10^{-3} }\\D_{n} = 29.20 cm^2 /s$

For the Einstein equation to be satisfied, $\frac{D_{n} }{\mu_{n} } = \frac{KT}{q} = 0.025 V$

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