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
