In a parallel one-dimensional flow in the positive x direction, the velocity varies linearly from zero at y = 0 to 32 m/s at y = 1.6 m. (a) Determine an expression for the stream function, Ψ. (b) Also determine the y coordinate above which the volume flow rate is half the total between y = 0 and y = 1.6 m.

Ψ = 10(y^2) + c

y = 1.067m

Explanation:

since the flow is one dimensional in positive X direction, the only velocity component is in X, which is denoted by u

while u is a function of y

we find the u in terms of y; u varies linearly wih y

we use similiraty to find the relation

32/1.6 =u/y

u = 20y

Ψ = ∫20ydy

Ψ = 10(y^2) + c

(b)

the flow is half below y = 1.6*(2/3)=1.067 m

this is because at two third of the height of a triangle lies the centroid of triangle. since the velocity profile forms a right angled triangle , its height is 1.6 m . the flow is halved at y = 1.067m

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