An incompressible fluid flows horizontally in the x-y plane with a velocity give

Question

An incompressible fluid flows horizontally in the x-y plane with a velocity given by v = 20 (z/h)^(1/3),
where z and h are in meters and h is a constant.
Determine the average velocity for the portion of the flow between z= 0 and z=h. Answer is between 10 and 20 m/s

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this problem deals with the chapter dealing with velocity fields and the reynolds transport theorem
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Explanation / Answer

Assume unit width in z-directon.

Take strip of infinitesimal thickness dz along the flow direction.

Infitesimal area dA = 1*dz = dz

Flowrate Q = Integral v*dA

Flowrate Q = Integral (z varies from z = 0 to z = h) [v dz]

Q = Integral (z varies from z = 0 to z = h) [20*(z/h)^(1/3) dz]

Q = [20 / h^(1/3)]*Integral (z varies from z = 0 to z = h) [z^(1/3) dz]

Q = [20*(3/4) / h^(1/3)]* (z varies from z = 0 to z = h) [z^(4/3)]

Q = [15 / h^(1/3)]* [h^(4/3) - 0]

Q = 15*h

Area A = 1*h = h

Average velocity = Q / Area

= 15*h /h

= 15 m/s

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