Water flows on a pipe whose diameters are different ends with a 1:3 ratio. If the water velocity that flows on the large pipe portion is 20 ms^(-1), the velocity of the water in the small pipe is ?

1 Answer
G_Ozdilek
Oct 26, 2017

Discharge#=vtimesA#

Explanation:

Since it is the same discharge (within the small diameter pipe and within the large dimater pipe), velocities are different. A stands for cross sectional area of pipes and v stands for velocities. Just equalize:

#v_1timesA_1=v_2timesA_2#

#20times(pitimes3^2/4) = v_2times(pitimes1^2/4)#

#20times(pitimes9/4) = v_2times(pi/4)#

when you arrange the above equation:

#20times9=v_2#

#180=v_2#

Velocity in the second (less dimater pipe) is 180 #m s^-1#

Related questions
Impact of this question
301 views around the world
You can reuse this answer
Creative Commons License