let us discuss about 1 of the most important
concept in fluid dynamics that is bernawlli’s
theorem. we can state about it that, this
is a theorem, about, conservation of energy,
for a flowing fluid, from 1 end to another,
in stream line flow. when a fluid is flowing
in a stream line manner then from 1 end to
another, we apply energy conservation, the
mathematical analysis is deduced, by the concept
of bernawlli’s theorem. let’s analyze
the practical situation, based on which we’ll
also derive an relation for, energy conservation,
which we term as, bernawlli’s equation.
say we are given with a flowing fluid line,
and say for 1 end, if we talk with respect
to ground, say 1 end is located at a height
h 1, other end is located at a height h 2
above the ground. and say, the cross sectional
area at these ends are a-1 and a-2. say, this
is end 1, this is end 2. and fluid which is
getting into the fluid line, is at a speed
v 1, due to the pressure p 1 acting on it.
and at the other end if the fluid pressure
is p 2, and fluid velocity which is coming
out is v 2. as we know well that if cross
sectional area reduces flow velocity increases,
but if pressures are p 1 p 2 v1 v2 are the
velocities and a-1 a-2 are the cross section
areas, then, we apply a basic work energy
theorem for the flowing fluid, in flow from
end 1 to end 2. let’s apply this. we use
work energy theorem, for a volume, delta v
of fluid, from end 1 to end 2, let’s see
what we’ll get. if we talk about end 1,
the mass of fluid, having a volume delta v
can be written as ro delta v, ro is the fluid
density, then the initial ki-netic energy
at end 1 can be written as, then the initial
ki-netic energy we can write, half, mass is
ro delta v, multiplied by v 1 square. this
is the initial ki-netic energy. and if the
pressure p 1 is pushing the volume delta v
into the fluid, the work done onto the fluid
can be written as, plus p 1 delta v. and,
the mass ro delta v, is pushed from end 1
to 2. so its height level with respect to
ground is increased by h 2 minus h 1. so in
this situation as it is going up, gravity
will be doing a negative work on the fluid,
so we can write, minus, the mass is ro delta
v, multiplied by g into, h 2 minus h 1, we
used m g h. and this must be, the ki-netic
energy of fluid at the other end, but here
a pressure p 2 is acting, so when the fluid
will come out this p 2 will be opposing the
fluid which is coming out so another negative
work is done, in ejection of liquid against
the pressure p 2. so we’ll write it minus
p 2 delta v, this must be the final ki-netic
energy with which, the fluid will eject out
from the other end, so it can be written as,
half, ro delta v, mass of the fluid as we
take delta v as volume, multiplied by v 2
square, which is the final ki-netic energy.
now in this expression, the term delta v can
be cancelled from all terms. and now if we
simplify, let’s continue on the next sheet.
here if we further simplify the expression,
you can see we’ll get, half ro v 1 square
plus p 1, in this expression we’ll get minus,
ro g h 2 plus, ro g h 1 minus p 2 is equal
to, half ro v 2 square. now if we slightly
rearrange these terms, for 1st end, we take
all the terms on left hand side, and for 2nd
end we take all terms on right hand side of
equality we can see we’ll get, half ro v
1 square, plus p1, plus ro g h 1, is equal
to half ro v 2 square, plus p 2, plus ro g
h 2. this is the expression that we get. so
here if we just carefully analyze we can write,
at all sections, of the flowing fluid body,
we can use, in this situation we can write,
half ro v square, plus pressure, plus ro g
h. this term will remain constant, as here
we have written, the work energy theorem for
only 2 ends. so we can see, summation of these
terms is, coming out to be a constant. so,
in this situation we can state, the 1st term
we can directly see, half ro v square where
ro is the mass per unit volume, so half ro
v square can be simply written as, ki-netic
energy, per unit volume of the fluid. and,
now if we talk about the last term, it is
m g h, this can be written as, gravitational
potential energy, per unit volume, with respect
to given reference. here reference of potential
energy we have taken the ground level. so
obviously as these 2 terms are giving us,
the energies per unit volume for the flowing
fluid, this p is also given a term, pressure
energy, per unit volume, of the flowing fluid.
so whenever a fluid flows, in a given manner,
we can simply state, at 1 point we can define
3 different kind of energies per unit volume
for the flowing fluid, that is ki-netic energy,
pressure energy and gravitational potential
energy. and for a flowing fluid, sum of these
3 always remains a constant. and this equation
we term as, bernawlli’s equation. and, in
various numerical applications we are going
to use this equation, as, it is 1 of the most
useful mathematical expression, for describing
the analysis of, fluid flow.
