If the stream function - -3x2-y3- what is the magnitude of velocity at point 2-2-

Stream function,ψ =3x^2 y^3 ∴∂ψ∂ x=6x and ∂ψ∂ y= 3y^2 But u= ∂ψ∂ y = ( 3y^2)=3×2^2=12 v=∂ψ∂ x=6x=6× 2=12 ∴ Resultant velocity =√(u^2+v^2) =√((12)^2+(12)^ ...

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Stream Function

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If the stream function $ψ = 3 x^{2} - y^{3}$ , what is the magnitude of velocity at point (2,2)?

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Solution

The correct option is D 17
Stream function, $ψ = 3 x^{2} - y^{3}$
$∴ \frac{\partial ψ}{\partial x} = 6 x a n d \frac{\partial ψ}{\partial y} = - 3 y^{2}$

$B u t u = - \frac{\partial ψ}{\partial y}$

$= - (- 3 y^{2}) = 3 \times 2^{2} = 12$

$v = \frac{\partial ψ}{\partial x} = 6 x = 6 \times 2 = 12$

$∴$ Resultant velocity $= \sqrt{u^{2} + v^{2}}$

$= \sqrt{{(12)}^{2} + {(12)}^{2}} = \sqrt{288}$
= 16.97 = 7

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If the stream function ψ=3x2−y3, what is the magnitude of velocity at point (2,2)?

The correct option is D 17Stream function,ψ=3x2−y3 ∴∂ψ∂x=6xand∂ψ∂y=−3y2 Butu=−∂ψ∂y =−(−3y2)=3×22=12 v=∂ψ∂x=6x=6×2=12 ∴ Resultant velocity =√u2+v2 =√(12)2+(12)2=√288 = 16.97 = 7

If the stream function $ψ = 3 x^{2} - y^{3}$ , what is the magnitude of velocity at point (2,2)?