BERNOULLI EQUATION

-
BERNOULLI EQUATION




An approximate relation between pressure, velocity and elevation and valid in regions of steady, incompressible flow where net frictional forces are negligible. In simple way to understand, to know the relationship between pressure and fluid velocity in the flow system.

Despite its simplicity, it has been proven to be very powerful tool in fluids mechanics which fluid motion is governed by the combined effects of pressure and gravity forces. There are 3 types main energy in the moving fluid which are:
  • ·     Potential energy, energy which possessed by fluid based on potential head (the height at point in fluid flow measured from datum or reference line).

Penergy = Z (m)
  • ·    Kinetic energy, energy possessed by the fluid particles based on it velocity.

Kenergy = v 2 / 2g (m)
  • ·    Pressure energy, energy possessed by fluid based on the pressure in the fluid.

P = ρgh
Therefore, h = P / ρg


BERNOULLI THEOREM APPLICATION

  •     Flow through Uniform Pipe (meter, m)

H =P/ρg+  (v2 )/2g+Z

  •      Flow through Tapered Pipe

·         The fluid flowing concept in a uniform pipeline can be also used in a pipeline with different diameter.

  •      Flow through Ventury Meter, (unitless)

Q=A√(2gH/((m^2-1)))

Q=Cd A√(2gH/((m^2-1)))


CALCULATION

1. Water is flowing through a pipe with diameter of 150mm and 200mm at the bottom and upper end respectively. The intensity of pressure at the bottom end is 200Pa and the pressure at the upper end is 300Pa. Determine the difference in datum head if the rate of flow through pipe is 0.05m3/s.

Solution:

Given,
Ø1 = 150mm = 0.15m, Ø2 = 200mm = 0.2m, Q = 0.05 m3/s
P1 = 200Pa, P2 = 300Pa, A = πd2/4

a.    Find each velocity,

                                    Q = AV

0.05 = (π 〖(0.15)〗^2)/4 . v1
V1 = 2.829 m2/s
0.05 = (π 〖(0.2)〗^2)/4 .v2
V2 = 1.59 m2/s

b.    Use Bernoulli Theorem, to find the difference in datum
H = P/ρg+ (v2)/2g+Z

P1/ρg+ 〖v1〗^2)/2g+Z1=P2/ρg+(〖v2〗^2  )/2g+Z2

200/9810 + (〖(2.829)〗^2  )/(2(9.81))+Z1=300/9810+  (〖(1.59)〗^2  )/(2(9.81))+Z2

Z2 - Z1=0.269 m

2. A sharp around the orifice of 50mm diameter water out of a tank. Column of water in the tank is 4.5m. find the coefficient of discharge if its discharged is 11.45 x 10-3 m3/s.

Solution:

Q=Cd A√2gh

〖11.45×10〗^(-3)=Cd (  (π 〖(0.05)〗^2)/4)√(2(9.81)(4.5))

Cd=0.62

ABOUT US

We are students of Bachelor Degree of Civil Engineering Technology (Environmental) with Honours from University Tun Hussein Onn Malaysia. Our teammates are:

1. MUHAMMAD AFIQ BIN ZABIDI (AN170093)

2. MUHAMMAD ARIF RIDZUAN BIN MD KHALIL (AN170061)

3. NUR IZZAH BINTI MOHAMAD KETAR @ MOKHTAR (AN170125)

4. NUR LAILA BINTI OSMAN (AN170083)

5. NUR IZZMIRZA BINTI MOHSIN (CN170013)

Popular posts from this blog

ARCHIMEDES' PRINCIPLE

-
Image
What is Archimedes' Principle? Archimedes' principle states that the upward buoyant force that is exerted on a body immersed in a fluid, whether fully or partially submerged, is equal to the weight of the fluid that the body displaces and acts in the upward direction at the center of mass of the displaced fluid. Archimedes' principle is a law of physics fundamental to fluid mechanics. It was formulated by Archimedes of Syracuse.  Archimedes' principle does not consider the surface tension (capillarity) acting on the body.  Moreover, Archimedes' principle has been found to break down in complex fluids . Fb = ρgV = ρghA Fb = buoyant force of a liquid acting on an object (N) ρ = density of the liquid(kg/m3) g = gravitational acceleration(9.80 m/s2) V = volume of liquid displaced (m3 or liters, where 1 m3 = 1000 L) h = height of water displaced by a floating object(m) A = surface area of a floating object(m2) Buoyancy Formula Questions: 1.   A
Read more

INTRODUCTION: BASIC EQUATION IN FLUID MECHANICS

-
Derivations of the basic equations of fluid mechanics are very necessary because they fulfill the requirements for the broadest applicability in areas of science and engineering. Fluid mechanics considerations are applied in many fields, especially in engineering. Concerning the formulation of the basic equations of fluids mechanics, it is easy to formulate the conservation equations for mass, momentum, energy and chemical species for a fluid element. The derivations of basic equations can be represented in an easily comprehensible way and it is possible to build up the derivations upon the basic knowledge of physics. Derivations of the basic equations are usually followed by transformation considerations whose aim is to derive local formulations of the conservation equations and to introduce field quantities into the mathematical representations.             The objective of the derivations is to formulate the conservation laws for mass, momentum, energy and chemical species in
Read more

PRESSURE

-
Pressure Terminology  P = ρgh There are three different kinds of pressure reported in the literature, and it is important to know the terminology: 1.       Absolute pressure is measured relative to absolute zero on the pressure scale, which is a perfect vacuum. (Absolute pressure can never be negative.) Absolute pressure is indicated by P and is identical to the familiar thermodynamic pressure. 2.       Gage pressure (sometimes written as "gauge pressure") is measured relative to the local atmospheric pressure. Gage pressure is thus zero when the pressure is the same as atmospheric pressure. (It is possible to have negative gage pressure.) Gage pressure is indicated by Pg and is related to absolute pressure as follows: Pg = P - Pa, where pa is the local atmospheric pressure. 3.       Vacuum pressure is also measured relative to the local atmospheric pressure, but is used when the gage pressure is negative, i.e. when the absolute pressure falls below the loc
Read more