Sunday, October 21, 2012

Proof of Bernoulli Equation


The Bernoulli Equation for an incompressible, steady Fluid Flow:

The Bernoulli Equation is a statement derived from conservation of energy and work-energy ideas that come from Newton's Laws of Motion.
Statement:It states that the total energy (pressure energy, potential energy and kinetic energy) of an incompressible and non–viscous fluid in steady flow through a pipe remains constant throughout the flow, provided there is no source or sink of the fluid along the length of the pipe.
This statement is based on the assumption that there is no loss of energy due to friction.
To prove Bernoulli’s theorem, we make the following assumptions:
1. The liquid is incompressible.   
2. The liquid is non–viscous. 
3. The flow is steady and the velocity of the liquid is less than the critical velocity for the liquid.

Proof of Bernoulli’s Theorem:
Imagine an incompressible and non–viscous liquid to be flowing through a pipe of varying cross–sectional area as shown in Fig. The liquid enters the pipe with a normal velocity v1 and at a height h1 above the reference level (earth’s surface). It leaves the pipe with a normal velocity v2 at the narrow end B of cross–sectional area a2 and at a height h2 above the earth’s surface.
Fluid flow to the right.

We examine a fluid section of mass m traveling to the right as shown in the schematic above. The net work done in moving the fluid is
Net work done = work 2 - work 1.Eq.(1)
where F denotes a force and an x a displacement. The second term picked up its negative sign because the force and displacement are in opposite directions.
Pressure is the force exerted over the cross-sectional area, or P = F/A. Rewriting this as F = PA and substituting into Eq.(1) we find that
Net work done = (PFA)_2 - (PFA)_1.Eq.(2)
The displaced fluid volume V is the cross-sectional area A times the thickness x. This volume remains constant for an incompressible fluid, so
Volume is constant for an incompressible fluid.Eq.(3)
Using Eq.(3) in Eq.(2) we have
dW = (P_1 - P_2) V.Eq.(4)
Since work has been done, there has been a change in the mechanical energy of the fluid segment. This energy change is found with the help of the next diagram.
Diagram for derivation of Bernoulli's equation for an imcompressible unrestricted steady fluid flow.


The energy change between the initial and final positions is given by
Energy change = E_2 - E_1.Eq.(5)
Here, the the kinetic energy K = mv²/2 where m is the fluid mass and v is the speed of the fluid. The potential energy U = mgh where g is the acceleration of gravity, andh is average fluid height.
The work-energy theorem says that the net work done is equal to the change in the system energy. This can be written as
Net work = energy change.Eq.(6)
Substitution of Eq.(4) and Eq.(5) into Eq.(6) yields
Expansion of dW = dE.Eq.(7)
Dividing Eq.(7) by the fluid volume, V gives us
Expansion of dW = dE.Eq.(8)
    where
Density = mass / volume.Eq.(9)
is the fluid mass density. To complete our derivation, we reorganize Eq.(8).
Expansion of dW = dE.Eq.(10)
Finally, note that Eq.(10) is true for any two positions. Therefore,
P + mgh + mv^2/2 = Constant.Eq.(11)
Equation (11) is commonly referred to as Bernoulli's equation. Keep in mind that this expression was restricted to incompressible fluids and smooth fluid flows.

Sunday, October 7, 2012

Work, Power & Energy

Work

Work is said to be done when a force applied on the body displaces the body through a certain distance in the direction of force.
Let a constant force  be applied on the body such that it makes an angle θ with the horizontal and body is displaced through a distance s

By resolving force  into two components :
(i) F cos θ   in the direction of displacement of the body.
(ii) F sin θ  in the perpendicular direction of displacement of the body.
Since body is being displaced in the direction of , therefore work done by the force in displacing the body through a distance is given by
                
or             
Thus work done by a force is equal to the scalar or dot product of the force and the displacement of the body.
If a number of force  are acting on a body and it shifts from position vector  to position vector  then .

Unit of Work

Since work is the product of force and distance, unit of work is unit of force and unit of distance. If force is in N and distance is in metres, unit of work will be N-m. In SI system of unit, 1N-m = 1 J. The term Joule s used for one N-m work done, and letter J is used as a symbol of Joule. Hence, one Joule may be defined as the amount of work done by one Newton force when it moves one metre distance in its direction.

Energy

The energy of a body is defined as its capacity for doing work. Since energy of a body is the total quantity of work done therefore it is a scalar quantity.

Power

Power of a body is defined as the rate at which the body can do the work.
Average power 
Instantaneous power          [As ]
                                                           [As ]
i.e. power is equal to the scalar product of force with velocity.
If work done by the two bodies is same then power 
i.e. the body which perform the given work in lesser time possess more power and vice-versa. As power = work/time, any unit of power multiplied by a unit of time gives unit of work (or energy) and not power.

Work Done by Variable Force

Let a varying force P move a body by distance s as shown by force-displacement curve in Fig.. At an instance, when force P is acting and the body moves by an elemental distance ' δs', then work done at that instance
= P δ s
Hence work done by the varying force in moving the body by distance ‘s’
= Σ P δ s
and it is equal to the area under the force-displacement curve. The variation of force is in a regular fashion so that its value at any instance can be represented by an expression. Then the work done can be evaluated by suitable integration. In such a case,
 work done= ΣPs = ∫ Pds
WORK DONE BY A VARYING FORCE.JPG