Notes for Fluid Mechanics chapter of class 11 physics. Dronstudy provides free comprehensive chapterwise class 11 physics notes with proper images & diagram.

** **

**FLUID STATICS**

It refers to the state when there is no relative velocity between fluid elements. In this section we will learn some of the properties of the fluid statics.

The ** pressure **exerted by a fluid is defined as the

As , the element reduces to a

When the force is constant over the surface A, the above equation reduces to

The

The other common pressure unit are the

*Example 1*

*Atmospheric pressure is about Pa. How large a force does the atmosphere exert on a 2 cm ^{2} area on the top of your head?*

*Solution*

Because p = F/A, where *F *is perpendicular to *A*, we have F = pA. Assuming that 2 cm^{2} of your head is flat (nearly correct) and that the force due to the atmosphere is perpendicular to the surface (as it is), we have

**Pressure is Isotropic **

Imagine a static fluid and consider a small cubic element of it deep within the fluid as shown in figure. Since this fluid element is in equilibrium, therefore, forces acting on each lateral face of this element must also be equal in magnitude. Because the areas of each face are equal, therefore, the pressure on each of the lateral faces must also be the same. In the limit as the cube element reduces to a point, the forces on the top and bottom surfaces also become equal. Thus, the pressure exerted by a fluid at a point is the same in all directions –– the pressure is isotropic.

Since the fluid *cannot *support a *shear stress*, the force exerted by a fluid pressure must also be *perpendicular* to the surface of the container that holds it.

**The Incompressible Fluid Model**

For an incompressible fluid, the density *r* the fluid remains constant throughout its volume. It is a good assumption for liquids. To find pressure at the point A in a fluid column as shown in the figure is obtained by integrating the above equation

or or

or

where is the density of the fluid, and

*p _{o} *is the atmospheric pressure at the free surface of the liquid.

Absolute pressure is the total pressure at a point while gauge pressure is relative to the local atmospheric pressure. Gauge pressure may be positive or negative depending upon the fact whether the pressure is more or less than the atmospheric pressure.

* **p _{gauge } = p_{absolute} – p_{atm}*

**Pascal’s Law **

According to equation

* **
*pressure at any depth

A pressure applied to a confined fluid at rest is transmitted equally undiminished to every part of the fluid and the walls of the container.

This principle is used in a hydraulic

The pressure due to a small force

or

*Consequently*, the force on the larger piston is large.

*Thus, a small force F _{1} acting on a small area A_{1} results in a larger force F_{2} acting on a larger area A_{2}.*

*Example 2*

Find the absolute pressure and gauge pressure at points A, B and C as shown in the figure.* (1 atm = 10 ^{5} Pa)
*

*Solution*

p_{atm} = 10^{5 }Pa

*Example 3*

*For the system shown in figure, the cylinder on the left, at L, has a mass of 600 kg and a cross-sectional area of 800 cm ^{2}. The piston on the right, at S, has cross-sectional area 25 cm^{2} and negligible weight. If the apparatus is filled with oil (r = 0.78 g/cm^{3}), find the force F required to hold the system in equilibrium as shown in figure*

*Solution *

The pressures at point *H _{1} *and

After solving, we get,

**The Venturi Meter**

The venturimeter is used to measure flow velocities in an incompressible fluid.

It consists of a manometer containing a liquid of density r_{m} connected to two points of a horizontal tube. The speed v of the liquid (density r) flowing through the tube at the wide neck of area A_{1} is to be measured.

From Bernoulli's equation and the equation of continuity,

also

If a body is *partially *or *wholly *immersed in a fluid, it experiences an upward force due to the fluid surrounding it.

The phenomenon of force exerted by fluid on the body is called ** buoyancy **and the force is called

A body experiences

**Archimedes Principle**

*A body immersed in a fluid experiences an upward buoyant force equivalent to the weight of the fluid displaced by it.
*

*Example 4*

*When a 2.5 kg crown is immersed in water, it has an apparent weight of 22 N. What is the density of the crown? *

*Solution*

Let *W *= actual weight of the crown

= apparent weight of the crown

= density of crown

= density of water

The *buoyant force *is given by

Since therefore

Eliminating *V *from the above two equations, we get

here,

*kg m ^{-3} *

**FLOW OF FLUID **

*In order to describe the motion of a fluid, in principle one might apply Newton’s laws to a particle (a small volume element of fluid) and follow its progress in time. This is a difficult approach. Instead, we consider the properties of the fluid, such as velocity and pressure, at fixed points in space. *

In order to simplify the discussion we make several assumptions:

**(i) **The fluid is non viscous

There is *no dissipation *of energy due to *internal friction *between adjacent layer in the fluid.

**(ii) **The flow is steady

The *velocity *and *pressure *at each point are *constant *in *time
*

A tiny paddle wheel placed in the liquid will

In rotational flow, for example, in eddies, the fluid has net angular momentum about a given point.

**The Principle of Continuity**

In general, the velocity of a particle will not be constant along a *streamline*. The density and the cross-sectional area of a tube of flow will also change. Consider two sections of a tube of flow, as shown in figure. The mass of fluid contained in a small cylinder of length * *and area *A _{1} *is

Since fluid does *not *leave the tube of flow, this mass will later pass through a cylinder of length and area *A _{2}*. The mass in this cylinder is . The lengths

, and

This is called the ** equation of continuity**. It is a statement of the conservation of mass.

If the fluid is *incompressible*, its density remains unchanged. This is a good approximation for liquid, but not for gases. If , the equation becomes,

*A _{1}v_{1} = A_{1}v_{2}*

The product *Av *is the *volume rate of flow *(m^{3}/s). Figure shows a pipe whose cross section narrows. From equation we conclude that the *speed of a fluid is greatest where the cross-sectional area is the least*.** Notice **that the

**Bernoulli’s Equation**

Let us focus our attention on the motion of the shaded region. This is our “system”. the lower cylindrical element of fluid of length *D**l _{1} *and area

A pressure force *F _{1} *acts on the lower cylinder due to fluid to its left , and a pressure force

where we have used the relations

*DV = A _{1}Dl_{1} = A_{2}Dl_{2}*. The net effect of the motion of the system is to raise the height of the lower cylinder of mass

These changes are brought about by the

Since the *density *is , we have

Since the points *1 *and *2 *can be chosen arbitrarily, we can express this result as *Bernoulli’s Equation*

*Daniel Bernoulli* derived this equation in 1738. It is applied to all points along a streamline in a nonviscous, incompressible fluid.

*Example 5*

*A siphon tube is discharging a liquid of specific gravity 0.9 from a reservoir as shown in the figure
*

*Solution *

Assume datum at the free surface of the liquid.

**(a) **Applying Bernoulli’s equation on point 1 and 2 , as shown in the figure.

Here *p _{1} = p_{2} = p_{0} = 10^{5} *N/m

Since area of the tube is very small as compared to that of the reservoir, therefore,

thus

(

Here,

= 41.5 kN/m

Applying

=

and

For working of siphon , H ¹ 0, and H should not be high enough so that p

*Example 6*

*A garden hose has an inside cross-sectional area of 3.60 cm ^{2}, and the opening in the nozzle is 0.250 cm^{2}. If the water velocity is 50 m/s in a segment of the hose that lies on the ground*

*Solution*

(a) The problem is illustrated in figure. We first apply the *Equation of Continuity*, to find the velocity of the fluid at the nozzle.

(b) We next apply *Bernoulli’s Equation *to find the pressure *p _{1}*. We know that

The pressure at the nozzle is atmospheric pressure

p

Solving for p

r = 1 ´ 10

= (1.01 ´ 10

**Reynold's number**

Osbrone Reynolds observed that viscous fluids flowing at low rates are less likely to turbulent flow. He defined a dimensionless number whose value gives an approximate idea whether the flow would be turbulent. The number called as Reynold's number is defined as

where symbols have their usual meaning and d is the dimension of the boundary.

For > 2000, the flow is often turbulent.

* *

* *

Contact Us

awesome notes.. but where are the notes of fluid dynamics....I like to read that too..!!

Spot i’ll carry on with this write-up, I really believe this amazing site needs a great deal more consideration. I’ll more likely once again to learn to read much more, many thanks for that info.

http://idmfullversionfreedownload.bravesites.com

Brilliantly explained