Transcript Euler equation of motion

Unit: IV-Fluid Dynamic
Fluid Dynamics
Types of fluid forces
- Euler equation of motion
- Bernoulli’s equations
Application
-Bernoulli’e equation
-Venturimeter
- Orifice meter
- Pitot-tube
- Momentum equation
- Moment of Momentum eq.
- Application of momentum
Section I
Fluid Dynamic:
It is the branch of applied science of fluid which deals with fluid in
motion condition under the force.
Types of fluid forces:
1) Fg, gravity force.
2) Fp, the pressure force
3) Fv, force due to viscosity
4) Ft, force due to turbulence.
5) Fc, force due to compressibility
Reynol’d Equation of
Motion
Navier-Stokes Equation
Euler’s Equation of
motion
Euler’s equation of motion:
The equation of motion in which the forces due to gravity and
pressure are taken into consideration.
The forces acting on the cylindrical element are:
1. Pressure force pdA in the direction of flow
2. Pressure force
opposite to the direction
3. Weight of element
The resultant force on the fluid elements
as is the acceleration in the direction of s and t
ds
Ө
dz
Bernoulli’s equation from Euler’s equation:
= Pressure energy per unit weight of fluid or pressure Head
= Kinetic energy per unit weight or kinetic head
= Potential energy per unit weight or potential head
Assumptions:
1)
2)
3)
4)
The fluid is ideal, i.e. viscosity is zero
The flow is steady
The flow is incompressible
The flow is irrotational
Bernoulli’s equation for real fluid:
Ex Water is flowing through a pipe of 5 cm diameter under a
pressure of 29.43 N/cm2 (gauge) and with mean velocity of 2.0 m/s.
find the total head or total energy per unit weight of the water at a
cross-section, which is 5 m above the datum line.
Total Head = Pressure Head + Velocity Head + Datum Head
Ans Total head = 35.204 m
Ex A Pipe, through which water is flowing, is having diameter, 20 cm
and 10 cm at the cross-section 1 and 2 respectively. The velocity of
water at section 1 is given 4.0 m/s. Find the velocity head at section 1
and 2 and also rate of discharge.
D1 = 20 cm
V1 =4.0 m/s
D2 = 10 cm
A1 V1 = A2 V2
Ans 1) Velocity head1= 0.815 m, velocity head at 2 = 83.047 m, 3)
rate of discharge = 0.1256 m3/s
Ex The water is flowing through a pipe having diameter 20 cm and
10 cm at section 1 and 2 respectively. The rate of flow through pipe is
35 lit/s. the section 1 in 6 m above datum and section 2 is 4 m above
datum. If the pressure at section 1 is 39.24 N/cm2, find the intensity
of pressure at section 2.
A1 V1 = A2 V2
Ans v1=1.14 m/s, v2= 4.45 m/s, p2 = 40.27 N/cm2
Ex Water is flowing through a pipe having diameter 300 mm and 200
mm at the bottom and upper end respectively. The intensity of
pressure at the bottom end is 24.525 N/cm2 and the pressure at the
upper end is 9.81 N/cm2. Determine the difference in datum head if
the rate of flow through pipe is 40 lit/s..
2
Z2
1
Z1
Ans Difference in datum head= z2-z1= 13.70m
Practical application of Bernoulli’s equation:
1) Venturimeter: It is device used for measuring the rate of flow of a
fluid flowing through a pipe
Expression:
Let d1 = diameter at inlet or at section
p1 = Pressure at section
v1 = velocity of fluid at section
a = area at section
Applying Bernoulli’s equation at section
Value of ‘h’ given by differential U-tube manometer:
Case I: Diff. Manometer contains liquid which is heavier than the
liquid flowing through the pipe
Case II: Diff. Manometer contains liquid which is lighter than the
liquid flowing through the pipe
Case III: Inclined Venturimeter with Differential U-tube manometer
Case IV: Inclined Venturimeter with Differential manometer
contains a liquid which is lighter than the liquid flowing
through the pipe
Ex A horizontal venturimeter with inlet and throat diameters 30 cm
and 15 cm respectively is used to measure the flow of water. The
reading of differential manometer connected to the inlet and the
throat is 20 cm of mercury. Determine the rate of flow.
Take Cd = 0.98
Ans h = 252 cm, Q = 125.756 lit/s
Ex A oil of sp.gr. 0.8 is flowing through a venturimeter having inlet
diameter 20 cm and throat diameter 10 cm. The oil-mercury
differential manometer shows a reading 0f 25 cm. calculate the
discharge of oil through the horizontal venturimeter. Take Cd =0.98
Ans h = 400 cm, Q = 70.465 lit/s
Ex A horizontal venturimeter with inlet diameter 20 cm and throat
diameter 10 cm is used to measure the flow of oil of sp.gr. 0.8. the
discharge of oil through venturimeter is 60 lit/s. Find the reading of
the oil-mercury differential manometer.
Take Cd = 0.98
Ans h = 289.98 cm, x = 18.12 cm
Case : Inclined Venturimeter with Differential U-tube manometer
Ex A 30 cm x 15 cm venturimeter is inserted in a vertical pipe
carrying water, flowing in the upward direction. A differential
mercury manometer connected to the inlet and throat gives a
reading of 20 cm. Find the discharge. Take Cd = 0.98
Ans h = 252 cm, Q = 125.75 lit/s
Ex A 20 cm x 10 cm venturimeter is inserted in a vertical pipe
carrying oil of sp.gr. 0.8, the flow of oil is in upward direction. The
difference of levels between the throat and inlet section is 50 cm. the
oil mercury differential manometer gives a reading of 30 cm of
mercury. Find the discharge of oil. Neglect losses.
Ans Q = 78.725 lit/s
2) Orifice Meter or Orifice Plate: It is device used for measuring
the rate of flow of a fluid flowing through a pipe
Expression:
Let p1 = Pressure at section (1)
v1 = velocity of fluid at section (1)
a1 = area of pipe at section (1)
P2, v2, a2 are corresponding values.
Applying Bernoulli’s equation to Sec (1) and (2)
Ex An orifice meter with orifice diameter 10 cm is inserted in apipe
of 20 cm dia. The pressure gauge fitted u/s and d/s of the orifice
meter gives readings of 19.62 N/cm2 and 9.81 N/cm2 res. Co-effiecnt
of discharge for the orifice meter is given as 0.6 find the discharge of
water through pipe..
3) Pitot-tube: It is device used for measuring the velocity of flow at
any point in a pipe or a channel
Expression:
Consider two points (1) and (2) at the same level in such a way that
point (2) is just at the inlet of the pitot-tube and point (1) is far
away from the tube
Let p1 = Pressure at section (1)
v1 = velocity of fluid at section (1)
p2 = pressure at point (2)
v2 = Velocity at point (2)
H = depth of tube in the liquid
h = rise of liquid in the tube
Applying Bernoulli’s equation at points (1) and (2)
The momentum equation:
It is based on the law of conservation of momentum or on the
Momentum principle, which states that the net force acting on a
Fluid mass is equal to the change in momentum of flow per unit time
In that direction.
Force acting on the fluid mass ‘m’ is given by Newton’s second law
Force exerted by a flowing fluid on a Pipe-bend:
Consider two section (1) and (2).
v1 = velocity of fluid at section (1)
p1 = pressure intensity at point (1)
A1 = Area of cross-section of pipe at section (1)
v2, p2, A2 = Corresponding valuse of velocity, pressure and area (2)
Applying Bernoulli’s equation at points (1) and (2)
Net force acting on fluid in the direction of x = Rate of change of
momentum in x-dir.
Resultant force (FR) acting on bend:
Angle made by the resultant force:
Moment of Momentum Equation:
Moment of momentum equation is derived fom moment of
momentum principle which states that the resulting torque acting on
rotating fluid is equal to the rate of change of moment of
momentum.
V1 = velocity of fluid at section (1)
r1 = Radius of curvature at section 1
Q = rate of flow of fluid
P = density of fluid
V2 and r2 = Velocity and radius at section 2
Moment of momentum per second at section 1
Moment of momentum per second at section 2
Rate of change of moment of momentum
This equation is known as moment of momentum.
This equation is applied:
1. For Analysis flow problem in turbines and centrifugal pumps
2. For finding torque exerted by water on sprinkler
Free liquid Jets: Free liquid jet is defined as the jet of water coming
out from the nozzle in atmosphere.
The path travelled by the free jet is parabolic.
x = Velocity component in x-direction x t
= U cosθ x t
1) Maximum Height attained by the jet:
2) Time of flight:
3) Time reach highest point:
4) Horizontal range of the jet:
5) Value of θ for maximum range:
Prepared by,
Dr Dhruvesh Patel