If for a two-dimensional potential flow, the velocity potential is given by ϕ = 4x(3y - 4), determine the velocity at the point (2, 3) . Determine also the value of stream function ψ at the poins (2, 3).
Velocity potential Function
u=−∂ϕ∂x; v=−∂ϕ∂yu = - \frac{{\partial \phi }}{{\partial x}};\;v = - \frac{{\partial \phi }}{{\partial y}}u=−∂x∂ϕ;v=−∂y∂ϕ
Stream Function
u=∂ψ∂y; v=−∂ψ∂xu = \frac{{\partial ψ }}{{\partial y}};\;v = -\frac{{\partial ψ }}{{\partial x}}u=∂y∂ψ;v=−∂x∂ψ
ϕ=4x(3y−4)ϕ = 4x(3y - 4)ϕ=4x(3y−4)
u=−δϕδxu=-\frac{\delta \phi}{\delta x}u=−δxδϕ and v=−δϕδy=−δψδxv=-\frac{\delta \phi}{\delta y}=-\frac{\delta ψ}{\delta x}v=−δyδϕ=−δxδψ
u=−4(3y−4)=δψδyu = -4 (3y - 4) =\frac{\delta ψ}{\delta y}u=−4(3y−4)=δyδψ
ψy=16y−6y2+C1ψy = 16y - 6y2 + C1ψy=16y−6y2+C1
and
v=−12x=−δψδxv = -12x =-\frac{\delta ψ}{\delta x}v=−12x=−δxδψ
ψx=6x2+C2ψx = 6x2 + C2ψx=6x2+C2
Therefore, ψ=ψx+ψyψ = ψx + ψyψ=ψx+ψy
ψ=6(x2–y2)+16yψ = 6(x2 – y2) + 16yψ=6(x2–y2)+16y
ψ∣(2,3)=6(4−9)+16×3{\left. ψ \right|_{\left( {2,3} \right)}} = 6\left( {4 - 9} \right) + 16 \times 3ψ∣(2,3)=6(4−9)+16×3
=18= 18=18 units
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