Download The Bernoulli equation

The Bernoulli equation
Work and energy
A general approach to obtaining the parameters of motion of solids and fluids
is to apply the principle of conservation of energy. When friction is negligible
the sum of kinetic energy and gravitational potential energy is constant.
Kinetic energy =0.5mv2
Gravitational potential energy =mgh
(m is the mass, v is the velocity and h is the height above the datum).
Now apply this to a droplet, initial velocity of zero, falling through a height h.
Initial KE = 0, Final KE = 0.5mv2 , Initial PE =mgh, Final PE = 0
Initial kinetic energy + Initial potential energy = Final kinetic energy + Final
potential energy
ie
mgh = 0.5mv2 so v = (2gh)0.5
Although this is applied to a drop of liquid, a similar method can be applied to
a continuous jet of liquid.
Trajectory of a jet of water
Consider the situation as in the figure above - a continuous jet of water
flowing from a pipe with velocity v1. A particle of the liquid, mass m, travels
with the jet and falls from height h1 to h2. The velocity of the particle changes
from v1 to v2. The jet of water travels in air where the atmospheric pressure is
the same everywhere so there is no additional force due to atmospheric
pressure acting on the fluid. The only force acting on the fluid is due to gravity.
The sum of the kinetic and potential energies remains constant (neglecting
energy loss due to friction) so
.
mgh1 + 0.5mv12 = mgh2 + 0.5mv22
Since m is constant gh1 + 0.5v12 = gh2 + 0.5v22
This formula is satisfactory as long as the weight of the jet of water is large
compared to the frictional forces. Also it is only applicable while the jet is
whole - before it breaks up into droplets.
Flow from a reservoir
Consider the 'idealised reservoir' below.
An idealised reservoir
The level of the water in the reservoir is h1. There is no movement of water so
KE is zero (v1= 0) and PE is mgh1. If a pipe is attached at the bottom of the
reservoir water flows out, through the pipe, at a level h2. A mass m flows from
the top of the reservoir to the nozzle and gains a velocity u2.
KE is 0.5mv22 and PE is mgh2.
Now mgh1 = 0.5mv22 + mgh2 so v2 = ( 2g(h1- h2))0.5
This gives an expression for the velocity of the water as it flows from of a pipe
nozzle at a height (h1- h2) below the surface of the reservoir.(Ignoring friction
losses in the pipe and the nozzle).
Volume of flow.
The volume of flow/sec = CSA x length = Area x velocity = A.v m3/sec
Mass of flow: The mass of flow = Volume x density = V.ρ = A.v.ρ
Example A reservoir of water has the surface at 310m above the outlet
nozzle of a pipe with diameter 15mm.
Calculate (a) the velocity of the water at the nozzle.
(b) the volume discharged from the nozzle and
(c) mass flow rate. (Ignore all friction in the nozzle and the pipe).
Ans (a) v2 = ( 2g(h1- h2))0.5 = (2x10x310)0.5 =78m/s
(b) Volume flow rate = area of the nozzle x velocity of flow
Volume flow rate Q = A.v = π x 0.00752 x 78 = 0.01378m3/s
(c) The density of water is 100kg/m3 so the mass flow rate = Q.ρ
where Q.ρ = 1000 x 0.01378 = 13.78kg/s
Pressure Difference
In the examples above the pressure force on the fluid is zero as the pressure
surrounding the fluid is the same everywhere. Ie atmospheric.
If the pressures in the fluid are different then an extra force is acting which
has to be accounted for when calculating the final velocity of flow.
Below is an example of pressure difference.( fluids at different heights).The
fluid is stationary so the velocities are zero.
A pipe filled with stationary fluid of density ρ has pressures p1and p2 at levels
h1 and h2 respectively. The pressure difference (p1 - p2) is
.
(p2 – p1) = ( ρgh1 - ρgh2) = ρg.(h1 - h2) or (p1 + ρgh1) = ( p2 + ρgh2)
.
so (p1 + h1) = (p2 + h2) ….. (1)
.
ρg
ρg
Refer to the equation above derived for a moving fluid but constant pressure:
.
gh1 + 0.5v12 = gh2 + 0.5v22
.
or h1 + v12 = h2 + v22 ….(2)
.
2g
2g
2. Bernoulli's Equation If both pressure and velocity vary then Bernoulli’s
equation is used. Bernoulli's equation is one of the most important/useful
equations in fluid mechanics. It may be written,
.
h1 + v12 + p1
2g ρg
= h2 + v22 + p2 = Constant H ….(3)
2g ρg
ie Potential Energy + Kinetic Energy + Pressure Energy = Total Energy H
Equations (1) and (2) are special cases of Bernoulli's equation. If p1 = p2,
equation (2) applies. If v1 =v2 (zero velocity difference) equation (1) applies.
Bernoulli's equation has some restrictions in its applicability, they are:
(a) Flow of fluid is steady.
(b) Fluid density ρ is constant (ie fluid is incompressible)
(c) Friction losses are negligible.
(d) The equation relates the states at two points along the same streamline,
All these conditions cannot be satisfied at the same time but the equation gives very
good results in real situations where the conditions are approximately satisfied,
Equation (3) can be rewritten (by multiplying by ρg)
. ρgh1 + ρ v12 + p1 = ρgh2 + ρv22 + p2 = Constant ….(4)
.
.
2
2
When the Bernoulli equation is combined with the continuity equation
A1v1 = A2v2 the two can be used to find velocities and pressures at points in
the streamline flow . Here is an example of using the Bernoulli equation to
determine pressure and velocity at within a contracting and expanding pipe.
A contracting expanding pipe
Q A fluid of density ρ = 960kg/m3 flows steadily through the tube. The radii at
the sections are r1 =0.05m and r2 = 0.04m. The pressure p1 = 200kpa and v1 =
5m/s.What is the pressure p2 ?
Ans The Bernoulli equation is applied along the section1 to 2.
The tube is horizontal, so h1 =h2 .
Also, the flow is steady so the volume/sec is the same ie continuity equation
holds A1v1 = A2v2. v2 = (0.05)2/(0.04)25 = 7.813m/s
From Bernoulli’s equation (4)
ρ v12 + p1 = ρv22 + p2 ie p2 = p1 + ρ (v12-v22) = 200000 + 480(25 -61) =187kpa
. 2
2
2
Note that the velocity increased in the contraction while the pressure
decreased. This is how a carburettor in many car engines work - pressure
reduces in a contraction allowing a small amount of fuel to enter.