The x component of velocity in a steady, incompressible flow...

The $x$ component of velocity in a steady, incompressible flow field in the $x y$ plane is $u=A\left(x^{5}-10 x^{3} y^{2}+5 x y^{4}\right),$ where $A=2 \mathrm{m}^{-4} \cdot \mathrm{s}^{-1}$ and $x$ is measured in meters. Find the simplest $y$ component of velocity for this flow field. Evaluate the acceleration of a fluid particle at point $(x, y)=(1,3)$.

Key Concepts

Incompressible Flow

In incompressible flows, the fluid density is constant, and as a result, the velocity field must satisfy a divergence?free condition. This means that the sum of the spatial derivatives of the velocity components is zero (i.e., ?u/?x + ?v/?y = 0 in two dimensions). This condition constrains how one component of velocity can vary with respect to the other, ensuring mass conservation throughout the flow.

Deriving Complementary Velocity Components

When one velocity component is given in an incompressible flow, the divergence-free condition allows you to determine the complementary component by integrating the relationship ?v/?y = -?u/?x. This process typically involves integrating with respect to y and choosing integration constants (or functions of the other variable) judiciously to achieve the simplest possible form, while still ensuring consistency with boundary conditions or other physical constraints.

Material Derivative and Fluid Particle Acceleration

The acceleration of a fluid particle in a flow field is determined by the material derivative of the velocity, which, for a steady flow, primarily consists of the convective terms. The material derivative combines local changes in velocity with changes due to the movement of the particle in the spatially varying velocity field, and is given by (u?/?x + v?/?y) when time derivatives vanish. This concept is critical for relating the velocity field to the forces experienced by fluid particles.

Transcript

00:01 Hi everyone here it is given x component of velocity u is equal to a into x key power 5 minus 10 x cube y square plus 5 x by key power 4 here value of a is given to meter to the power minus 4 second to the power minus 1 x is measured in meter find by component of the velocity and acceleration at 1 .3 x component of velocity is defined as dale of zi upon del y and y is defined as del jai upon dale x so using it be no u is called to a x to the power 5 10 x cube y square plus 5 x y to the power 4 is called to del of zai upon del of y so streamline function you will get zy to v integration of a x to the power 5 10 x cube b square 5 x k power by 4 del of 5 so here you can write zi having the value a x to the power 5 into y 10 by 3 x cube y cube plus x into y to the power 5 upon 5 plus function of x say equation number 1 so y component of velocity is defined as d -zai upon dale of x so a into x x x x x x x x x x x x x x x x x x x x x x x x x x x x x minus 3 x q pi cube plus x y to the power 5 upon 5 plus function of x so it can be written as minus a 5 x k power 4 y 10 x square y cube y to the power 5 plus capital f x for simplest form function of x must be 0 v is equal to you will write minus a 5 x to the power 4y 10 x square y cube plus y to the power 5 this is the answer of part a now acceleration is defined as x component of acceleration is u into dale u upon dale x plus v dare u upon day y substitute the value a x to the power 5 minus 10 x cube y square plus 5 x y to the power 4 del of del of x x to the power 5 a into 10 x cube y square plus 5 x y to the power 4 minus a 5 x to the power 4 minus a 5 x to the power 4 by minus 10 x square y cube y to the power 5 deal over deal y a x to the power 5 minus 10 x q 5 5 5 plus 5 x y to the power 4 so on solving this x component of acceleration you will get 5 a square to x x square plus y square to the whole to the power 4.

06:51 Now finding the y component of acceleration which is defined as u tail v upon dale x plus v, del v upon day y substitute the value...

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