Sunday, January 18, 2015

Bernoulli’s equation P+KE+U=constant ,continuity equation for incompressible fluid, A1v1 = A2v2 .


We are going to analyses the flow of an incompressible,non-viscous and non-turbulent ie streamed lined flow of fluid.

The Continuity Equation

The mass of a moving fluid doesn’t change as it flows. This leads to an important quantitative relationship called the continuity equation. Consider a portion of a flow tube between two stationary cross sections with areas A1 and A2.The fluid speeds at these sections are v1 and v2.respectively.During a small time interval dt, the fluid at A1moves a distance v1dt,so a cylinder of fluid with height v1dt and volume dV1 = A1v1dt flows into the tube across A1.During this same interval, a cylinder of volume dV2 = A2v2dt flows out of the tube across A2.Let’s first consider the case of an incompressible fluid so that the density has the same value at all points.The mass dm1 flowing into the tube across A1in time dt is dm1 = rA1v1 dt.Similarly, the mass dm2 that flows out across A2 in the same time is dm2 = rA2v2 dt. In steady flow the total mass in the tube is constant so
So dm1=dm2.   and

                                                              rA1v1 dt = rA2v2 dt    or

                                                                     A1v1 = A2v2    (continuity equation, incompressible fluid)
The figure illustrating the above derivation



Deriving Bernoulli’s Equation

To derive Bernoulli’s equation, we apply the work–energy theorem to the fluid in a section of a flow tube.And the figure that will assist us in deriving this relation is as following:

Let’s compute the work done on this fluid element during dt. We assume that there is negligible internal friction in the fluid (i.e., no viscosity), so the only nongravitational forces that do work on the fluid element are due to the pressure of the surrounding fluid. The pressures at the two ends are p1 and p2.the force on cross section at a is p1A1 and the force at c is p2A2.The net work dW done on the element by the surrounding fluid during this displacement is therefore

The work dW is due to forces other than the conservative force of gravity, so it equals the change in the total mechanical energy (kinetic energy plus gravitational potential energy) associated with the fluid element. The mechanical energy for the fluid between sections b and c does not change.At the beginning of dt the fluid between a and b has volume A1 ds1, mass rA1 ds1 and kinetic energy 


 At the end of dt the fluid between c and d has kinetic energy
The net change in kinetic energy dK during time dt is

The net change in potential energy dU during dt is

 dW =dK + dU     (Work Energy Theorem)

This is Bernoulli’s equation. It states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the changes in kinetic and potential energies per unit volume that occur during the flow.

Summary

We can derive it using newton's second lay of motion or Bernoulli's proncipal.

 

 

 

We will come with the application of above principal in the next blog

 

 

 

https://www.youtube.com/watch?v=3zJfcafFdc4  

https://www.youtube.com/watch?v=omVTV8omGdw 

https://www.youtube.com/watch?v=Jak9_7xfMGw 

http://en.wikipedia.org/wiki/Bernoulli%27s_principle 

https://www.youtube.com/watch?v=bmczI-3qSJw&feature=youtu.be

https://www.youtube.com/watch?v=bmczI-3qSJw&feature=youtu.be

http://en.wikipedia.org/wiki/Continuity_equation

https://www.youtube.com/watch?v=bmczI-3qSJw&feature=youtu.be

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Sunday, January 11, 2015

Calculus and the concept of maxima and minima,Algebra and the concept of maxima and minima and the application of maxima and minima in mechanics

http://advancedmathematicalresearch.blogspot.in/

 Problem in mechanics

 We are going to see a problem where we have to find the minimum or maximum value of a function.we will generate the function from Newton's Second law: The vector sum of the forces F on an object is equal to the mass m of that object multiplied by the acceleration vector a of the object: F = ma.

This can be written in the limit form as 

Instantaneous acceleration is limit of the average acceleration over an infinitesimal interval of time. In the terms of calculus, instantaneous acceleration is the derivative of the velocity vector with respect to time:

\mathbf{a} = \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt} 














  This can be represented graphically in the vector form as shown above

      The second law states that the net force on an object is   equal to the rate of change (that is, the derivative) of its linear momentum p in an inertial reference frame:






\mathbf{F} = \frac{\mathrm{d}\mathbf{p}}{\mathrm{d}t} = \frac{\mathrm{d}(m\mathbf v)}{\mathrm{d}t}.




where acceleration is equal to







\mathbf{a} = \lim_{{\Delta t}\to 0} \frac{\Delta \mathbf{v}}{\Delta t} = \frac{d\mathbf{v}}{dt}

when we apply this law to any system in physics we generate a function which can be either maximized or minimized applying differential calculus or algebra.
So differential calculus come into picture.

 Derivatives, Slope, Velocity, and Rate of Change.

 Geometric Viewpoint on Derivatives

Figure 1: A function with secant and tangent lines 


Geometrically, the limit of the secant lines is the tangent line. Therefore, the limit of the difference quotient as h approaches zero, if it exists, should represent the slope of the tangent line to (a, f(a)). This limit is defined to be the derivative of the function f at a:
f'(a)=\lim_{h\to 0}\frac{f(a+h)-f(a)}{h}.
When the limit exists, f is said to be differentiable at a. Here f′ (a) is one of several common notations for the derivative

 Let f be continuous on an interval I and differentiable on the interior of I.
  • If f(x)0 for all xI, then f is increasing on I.
  • If f(x)0 for all xI, then f is decreasing on I. 


First Derivative Test
Suppose f is continuous at a critical point x0.
  • If f(x)>0 on an open interval extending left from x0 and f(x)<0 on an open interval extending right from x0, then f has a relative maximum at x0.
  • If f(x)<0 on an open interval extending left from x0 and f(x)>0 on an open interval extending right from x0, then f has a relative minimum at x0.
  • If f(x) has the same sign on both an open interval extending left from x0 and an open interval extending right from x0, then f does not have a relative extremum at x0.
In summary, relative extrema occur where f(x) changes sign.

The test states: if the function f is twice differentiable at a critical point x (i.e. f'(x) = 0), then:
  • If \ f^{\prime\prime}(x) < 0 then \ f has a local maximum at \ x.
  • If \ f^{\prime\prime}(x) > 0 then \ f has a local minimum at \ x.
  • If \ f^{\prime\prime}(x) = 0, the test is inconclusive.

 f is convex (concave-up). The slope increases from negative to positive as x increases


f is concave-down. The slope decreases from positive to negative as x increases.
Therefore, the sign of the second derivative tells us about concavity/convexity of the graph. Thus
the second derivative is good for two purposes.
1. Deciding whether a critical point is a maximum or a minimum. This is known called as the second derivative test.

 The test states: if the function f is twice differentiable at a critical point x (i.e. f'(x) = 0), then:

  • If \ f^{\prime\prime}(x) < 0 then \ f has a local maximum at \ x.
  • If \ f^{\prime\prime}(x) > 0 then \ f has a local minimum at \ x.
  • If \ f^{\prime\prime}(x) = 0, the test is inconclusive.
Or you can apply algebraic inequality to solve for maxima or minima problem.
using the quadratic condition.

                                             \Delta = b^2 - 4ac.
             The condition for maxima and minima depends upon delta.


All these information can well mapped to a physical problem to solve it.

 Q1.A small body A starts sliding down from the top of the wedge whose base is equal to l=2.10m.The coefficient of friction between the body and the wedge surface is k=.140.At what value of the angle alpha will the time of sliding be the least?what it will be equal to?
Sol.

 Consider m as the system and apply Newton's second law of motion will ease our job.

 We can apply calculus to find maxima or minima where it has been done the same way.
We can also apply algebra to find the same solution.















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