The rectangular channel if 2 feet tall and 4 feet wide. Water enters the channel

Question

The rectangular channel if 2 feet tall and 4 feet wide.  Water enters the channel with a linear velocity profile as shown and exits in a steady flow with a parabolic profile given by the equation u = umax (1-(y/h)^2). Find the value of Umax.

Please explain in detail.

The rectangular channel if 2 feet tall and 4 feet wide. Water enters the channel with a linear velocity profile as shown and exits in a steady flow with a parabolic profile given by the equation u = umax (1-(y/h)^2). Find the value of Umax.

Explanation / Answer

at Entry velocity profile

(Vi - 2)/(4-2) = (y+1)/(1-(-1)

Vi = 2 + y+1 = 3 + y (-1<y<1)

Mass flow rate is constant

A*V = constant

Integral(Vi*W*dy) = integral(Ve*W*dy)

W is width = 4 ft constant

Integral(Vi*dy) = integral(Ve*dy)

Integral of(Vi*dy) = 3*y + y^2/2 (limits -1 to 1)

Integral of(Vi*dy) = 3 + 1/2 - (-3 + 1/2) = 6--(1)

Integral of (Ve*dy) =Umax (y - (1/h)^2*y^3/3) (limits -1 to 1 and h = 1)

Integral of (Ve*dy) = Umax{1 - 1/3 - (-1 + 1/3)} = Umax*(2+2/3) = Umax*8/3----(2)

But Both equations are equal

Umax*8/3 = 6

Umax = 9/4 m/s

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