In the previous article, we discussed the types of Fluid flows, Fluid discharge (Q) and the Continuity Equations for a fluid flow. From the types of fluid flow, we can easily determine the velocity and acceleration components of the fluid flow.

## Velocity

In fluid Mechanics, the velocity is defined as the change of position with respect to time or the displacement of the fluid in unit time. velocity is a vector quantity.

So the Velocity components are a function of space coordinates (i.e., length of direction of the flow) and time.

Now let us consider u, v, and w are the components in the x, y and z directions. and t is the time component.

Mathematically the velocity components can be written as follows,

u = f_{1}(x, y, z, t)

v = f_{2}(x, y, z, t)

w = f_{3}(x, y, z, t)

Let V is the resultant velocity at any point in a fluid flow, then we can write it as follows,

V = ui + vj + wk

This is the resultant velocity at any point in a fluid flow.

## Acceleration

In fluid Mechanics, Acceleration is defined as the rate of change of velocity of the fluid flow with respect to time. Acceleration is a vector quantity.

Let us assume a_{x} a_{y} and a_{z} are the total acceleration in x y and z directions respectively.

Then by the chain rule of differentiation, we can write

For steady flow, we know,

∂V/∂t = 0

where V is the resultant Velocity.

Hence the acceleration in the x, y and z directions can be written as

Equation (b)

The acceleration vector

A = a_{x}i + a_{y}j + a_{z}k

### Local Acceleration and Convective Acceleration

**Local acceleration** is defined as the rate of increase of velocity with respect to time at a given point in a flow field.

In the equation given by (a), the following expression is known as local acceleration.

**Convective acceleration** is defined as the rate of change of velocity due to the change of position of fluid particles in a fluid flow.

The expressions Other than the following part in equation (a) are known as convective acceleration.

### Example Problems to calculate Velocity and Acceleration of a Fluid Particle

**Problem Statement:** The velocity vector in a fluid flow is given as follows

V = (4x^{3}) i – (10x^{2}y) j + (2t) k

Find the velocity and acceleration of a fluid particle at (2, 1, 3) at time t = 1.

**Answer: **

#### Velocity

The velocity components u, v and w are

u = (4x^{3})

v = -(10x^{2}y)

w = (2t)

from the point (2, 1, 3), we have x = 2, y = 1 and z=3 at time t = 1.

Hence velocity components (2, 1, 3) are

u = (4×2^{3}) = 32 units

v = -(10×2^{2}×1) = -40 units

w = (2×1) = 2 units

The velocity vector V at (2, 1, 3)

V = (32) i – (40) j + (2) k

The resultant Velocity will be

V = √[(32)^{2} + (-40)^{2} + (2)^{2}]

V = √[1024+1600+4]

V = 51.26 units

#### Acceleration

The acceleration equation we know

Now from velocity components we can write,

Substituting the values, the acceleration components at (2, 1, 3) at time t = 1 are

a_{x} = (4x^{3}) × (12x^{2}) + (-10x^{2}y) × (0) + (2t) × (0) + 0

= 48x^{5}

= 48(2)^{5}

a_{x} = 1536 units

a_{y} = (4x^{3}) × (-20xy) + (-10x^{2}y) × (-10x^{2}) + (2t) × (0) + 0

= -80x^{4}y +100x^{4}y

= -80(2)^{4}(1) +100(2)^{4}(1)

a_{y} = 320 units

a_{z} = (4x^{3}) × (0) + (-10x^{2}y) × (0) + (2t) × (0) + 2.1

a_{z} = 1536 units

The acceleration is A = a_{x}i + a_{y}j + a_{z}k

A = 1536i +320j + 2k

The resultant acceleration is

A = √[(1536)^{2} + (320)^{2} + (2)^{2}]

A = √[2359296+102400+4]

A = 1569 units

This is how you can calculate the Velocity and Acceleration of any Fluid Particle from the given velocity vector equation.

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