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?

Logic May 29, 2019

#1**+2 **

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 !

ElectricPavlov May 29, 2019

#1**+2 **

Best Answer

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 !

ElectricPavlov May 29, 2019