A pipe with inside diameter 10'' is to carry water from a reservoir to a small town in an arid land. Neglecting the friction and turbulence of the water against the inside of the pipes, what is the minimum number of 2''-inside-diameter pipes of the same length needed to carry the same volume of water to the arid town?
The sum of the cross sectional area of the 2" pipes must equal the cross sectional area of the larger 10" pipe
Cross sectional area = pi r^2
x pi r^2 = pi 5^2 where x = # of 2" pipes pi 5^2 = area of 10" pipe
x pi (1)^2 = pi 5^2 divide both sides by pi
x = 5^2
x = 25 Yep, you'll need 25 of the 2" pipes to match the 10" pipe !
The sum of the cross sectional area of the 2" pipes must equal the cross sectional area of the larger 10" pipe
Cross sectional area = pi r^2
x pi r^2 = pi 5^2 where x = # of 2" pipes pi 5^2 = area of 10" pipe
x pi (1)^2 = pi 5^2 divide both sides by pi
x = 5^2
x = 25 Yep, you'll need 25 of the 2" pipes to match the 10" pipe !