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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?

 May 29, 2019

Best Answer 

 #1
avatar+18754 
+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  !

 May 29, 2019
 #1
avatar+18754 
+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

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