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victus00 [196]
3 years ago
14

Velocity profile of a moving fluid over a flat (fixed) plate is given as: ???????? = from the fixed plate (yy = 0) in mm, u is t

he velocity in xx direction in mm/????????. Calculate the shear stress on the fluid at yy = 0.15 m if the viscosity of the fluid 8.6 poise (1 Poise= 1 ????????yy????????????????.????????) cccc2

Physics
1 answer:
schepotkina [342]3 years ago
8 0

Answer: The question has some details missing which is the velocity profile .The velocity profile is given as u = (8y - 0.3y^2) mm/s

the shear stress on the fluid at y = 0.15 m = 6.80 x 10^-3N/m^2

Explanation:

The detailed steps is the use of the relationship between the shear stress, viscosity and the velocity profile. appropriate differentiation and substitution were made to calculate the shear stress as is shown in the attachment.

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  • The heat absorbed is 7475.69 joules

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<h3>Work</h3>

We know that the differential work made by the gas  its defined as:

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As n, R and T are constants

\int\limits_{v_1}^{v_2} P \ dv = \ n \ R \ T \int\limits_{v_1}^{v_2} \frac{1}{V} \ dv

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\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ( ln (v_2) - ln (v_1 )

\Delta W = \ n \ R \ T  ln (\frac{v_2}{v_1})

But the volume is:

V = \frac{\ n \ R \ T}{P}

\Delta W = \ n \ R \ T  ln(\frac{\frac{\ n \ R \ T}{P_2}}{\frac{\ n \ R \ T}{P_1}} )

\Delta W = \ n \ R \ T  ln(\frac{P_1}{P_2})

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The temperature its:

T = 27 \° C = 300.15 \ K

The ideal gas constant:

R = 8.314 \frac{m^3 \ Pa}{K \ mol}

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