**Fluid Dynamics**

Some fluids are moving and studying this movement of the fluid is complicated. Thus, in studying the dynamics of the fluid, the topic is limited on the study of the *ideal fluid. *The following are the four assumptions on the properties of an ideal fluid. These properties concern on how the fluid flows.

**1. Steady Flow**

Fluid is traveling at the same path and it does not change with time. The path taken by every particle in a fluid is called **flow line**. In this type of flow, the velocity of every particle passing through a point is the same. Steady flow is also called as **laminar flow.**

**2. Incompressible Flow**

An ideal fluid is incompressible, meaning, it’s density is constant and has uniform value.

**3. Nonviscous Flow**

Viscosity is referred to as the internal friction in a fluid. Since ideal fluid is nonviscous, this means that there is no internal fiction in it. It follows that an object flowing in a nonviscous fluid experiences no internal friction or viscous drag force, it will move at constant speed though the fluid.

**4. Irrotational Flow**

The fluid has no angular momentum at any point. This implies that an object placed in an irrotational fluid will not rotate about its center of mass.

Some terms we may also encounter in studying fluid dynamics are:

**Streamline or flow line **is the path followed by an individual fluid particle.

**Tube of flow*** *is a bundle of streamlines.

**Turbulent flow** is irregular flow characterized by small whirlpool-like regions.

**The Continuity Equation**

The speed of flow of fluids may vary with varying cross-sectional area. The aim here is to derive an equation to show the relationship between the speed and area for the steady flow of an ideal fluid passing through a tube of different cross-sectional area. This idea and the fact that the mass of a fluid does not change as it flows leads to the discovery of an important relationship called as the **continuity equation**.

In getting this equation, a portion of a flow tube between two stationary cross sections with areas ** A_{1} **and

**is being considered in which the speed of the fluid flowing is indicated as**

*A*_{2}**and**

*v*_{1}**. It is determined that the speed of flow of the fluid varies inversely with the cross-sectional area, thus, decreasing the cross-sectional area means the faster the speed flow. Mathematically, the continuity equation is written as**

*v*_{2}\( A_1v_1 = A_2v_2\)

The volume past a given point per unit time is said to be the same for all cross sections of the tube of flow. This is called as the volume flow rate, denoted as ** R_{v} **and is expressed mathematically as

\(R_v = Av = constant\)

The unit for volume flow rate is the cubic meter per second, m^{3}/s. If the density of the flowing fluid is given with constant value, the mass flow rate can be determined by multiplying the volume flow rate with the density. Denoted as ** R_{m}**, the mass flow rate is represented by the equation

\(R_m = \rho R_v = \rho Av = constant\)

The SI unit for the mass flow rate is kilogram per second, kg/s. The equation implies that the mass of the fluid flowing in and out the tube segment per second must be equal.

**Example**

The speed of the water flowing through a garden hose of diameter is 1.2 cm is 2.5 m/s. If the opening of the hose is reduced to 0.8 cm, what is the speed flow of the fluid as it escapes out of the nozzle?

Given:

diameter, d_{hose} = 1.2 cm

diameter, d_{nozzle} = 0.8 cm

speed, v_{1} = 2.5 m/s

Find: v_{2}

Solution:

We first solve for the cross-sectional area of the hose and the nozzle. Since the cross section of the hose and the nozzle is in circular shape, we use the formula for the area of circle which is \(A=\pi r^2\).

For the cross-sectional area of the hose, * A_{1}*:

Since the given in the problem is the diameter, we solve for the radius \(r = {d \over 2} = {1.2\;cm \over 2}=0.6\; cm\).

\(A_1=\pi r^2 = \pi (0.6\;cm)^2 = 1.13\;cm^2\)

For the cross-sectional area of the nozzle, * A_{2}*:

The diameter of the nozzle is 0.8 cm. Its radius is \(r = {0.8\;cm \over 2} = 0.4\;cm\). The cross-sectional area is

\(A_2 = \pi r^2 = \pi (0.4\;cm)^2 = 0.50\;cm^2\)

We can now solve for the speed flow of the fluid using the continuity equation.

\( A_1v_1 = A_2v_2 \implies v_2 = {A_1v_1 \over A_2}\)

Substitute all the known values to get the value for * v_{2}*,

\(v_2 = {(1.13\;cm^2)(2.5\;m/s) \over 0.50\;cm^2} = 5.65\;m/s\)

Thus, the speed flow increases when the cross-sectional area is decreased.