The velocity potential function in a two-dimensional flow field is given by ϕ = x^{2} – y^{2}

The magnitude of velocity at point P (1, 1) is

This question was previously asked in

ESE Mechanical 2013 Official Paper - 1

- Zero
- 2
- 2√2
- 8

Option 3 : 2√2

__Concept:__

The velocity potential function, stream function, and velocity components are related as shown below

\(U = \; - \frac{{\partial ϕ }}{{\partial x}} = \; - \frac{{\partial ψ }}{{\partial y}}\;\; ...\left( 1 \right)\)

\(V = \; - \frac{{\partial ϕ }}{{\partial y}} = \;\frac{{\partial ψ }}{{\partial x}}\;\;\; ...\left( 2 \right)\)

where ϕ - velocity potential function, ψ - stream function.

__Calculation:__

__ Given__:

Velocity potential function is ϕ = x2 – y^{2};

Now the velocity components will be

\(U = \; - \frac{{\partial ϕ }}{{\partial x}} = -2x\)

\(V = \; - \frac{{\partial ϕ }}{{\partial y}} = 2y\)

At P(1,1), U = -2 m/s, V = 2 m/s;

Now the magnitude of velocity at point P (1, 1) will be

\(= \sqrt {(-2)^2 + (2)^2} = 2\sqrt2\)

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