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a. A crude oil pipe’s radius is reduced by 5%. What is the corresponding percentage change in the pressure drop per unit length? b. A pipe’s radius can be thought of as a characteristic dimension, D. What is the surface-to-volume ratio (S/V) of the inside of the pipe as a function of D?

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Answer: a) 10% b) 1/D

Explanation:

a)

The change in pressure is inversely proportional to the square of the radius.

Now if radius of pipe originally= 1

So, radius of pipe after reduction= 0.95

Now percentage change in pressure drop per unit length= [tex](\frac{1^{2} }{0.95^{2} }- 1) * 100[/tex]

Percentage change in pressure= 0.1 or 10%

b)

Surface area in terms of D= D²

Volume in terms of D= D³

Surface area to volume ratio in terms of D= [tex]\frac{D^{2} }{D^{3} }[/tex]

S/V ratio= 1/D

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