STAT CH2 Confidential Interval
STAT CH2 Confidential Interval
STAT CH2 Confidential Interval
Phnom Penh
Falls, 2017 (Third Edition)
OUTLINE
- General Introduction
- Hydrostatics
- Kinematics of Fluids
Fluid Definition
In contrast to a solid, a fluid does not have a well defined form as it
takes the form of the container that it stays inside. Thus, the fluid is
deformable.
Fluids have a flowing property due to the fact that its particles are
weakly linked to each others and thus can easily move around each
others.
Fluid Definition
Liquids and gas are fluids. A liquid does not have a proper form but
possess a proper volume. A gas neither has a proper form nor a proper
volume, i.e., it always occupies all the available space.
A gas may be defined as a highly superheated vapor; that is, its state is
far removed from the liquid phase. Dr. Keang Sè POUV 5
General Introduction
Fluid Definition
The molecules of a gas are much farther apart than those of a liquid.
Liquid Gas
There are many materials that are not easily categorized into solid,
liquid, or gas. Gels (cross-linked networks of polymer molecules) and
colloids (suspensions of macromolecules or microscopic particles) are
materials, many of which are important biomaterials, that can exhibit
both liquid-like and solid-like properties depending on the conditions.
Fluid Definition
Ideal Fluid
Real Fluid
Physical Properties
Density
Physical Properties
Density
Physical Properties
Density
Physical Properties
Exercise 1
1. The specific weight of water is 9.81 kN/m3. The specific gravity of mercury is
13.56. Compute the density of water and the specific weight of mercury.
2. A fluid that occupies a volume of 24L weighs 225N. Determine its density if g
= 9.80 m/s2.
Exercise 2
A plastic cylinder tank has a radius of 25.23 cm, a height of 1 m, and a mass of
3 kg. Determine the weight of the combined system for the two cases below:
1. The tank is filled with water.
2. The tank is filled with water until the 2/3 of the tank’s height and with olive
oil (density = 800 kg/m3) for the remaining 1/3 height. Dr. Keang Sè POUV 11
General Introduction
Physical Properties
Exercise 3
Given:
Physical Properties
Viscosity
Physical Properties
Viscosity
Physical Properties
Viscosity
The relationship between the shear stress and the shear strain rate is:
d du
(1.5)
dt dy
Physical Properties
Viscosity
Dynamic viscosity:
(1.6)
d / dt
Kinematic viscosity: (1.7)
Physical Properties
Viscosity
(shear thickening)
(shear thinning)
Physical Properties
Viscosity
Rotational viscosimeter:
Physical Properties
Exercise 4
Note: Before calculation, convert the units of all the given parameters
into SI (International Standard) units. (1in = 2.54cm, (°F-32)x5/9 = °C, 1ft =
30.48 cm, 1lb = 0.4535kg = 4.448N)
Physical Properties
Compressibility
Physical Properties
Compressibility
dp
Ev v (1.8)
dv
Where v = specific volume, and p = pressure.
Where Ev is the mean value of the modulus for the pressure range.
21
General Introduction
Physical Properties
Exercise 5
(a) What will be the change in specific volume between that at the
surface and at that depth?
(b) What will be the specific volume at that depth?
(c) What will be the specific weight at that depth?
General Conditions
Hydrostatics is concerned with the behaviour of fluids at rest.
Notions of Pressure
The magnitude of the force per unit area in a static fluid is called the
pressure:
F
p (2.1)
S
Notions of Pressure
Illustration of absolute, gage and vacuum pressure readings:
Notions of Pressure
Considering a small wedge of fluid at rest as shown below.
F x
0 p x bz pn bs sin (2.4)
F
1
z
0 p z bx pn bs cos bxz (2.5)
2 Dr. Keang Sè POUV 27
Hydrostatics
Notions of Pressure
Since: s sin z and s cos x
Then: p x pn (2.6)
1
p z pn z (2.7)
2
These relationships indicate that:
Pascal’s Principle
If a fluid is confined in a closed container and an external force is
applied to a region of the surface bounding the fluid, there is an
external pressure being applied. The external pressure applied does not
remain localized near the surface where the pressure is applied, but the
external pressure on a confined fluid increases the pressure uniformly
throughout the fluid by the same amount. This is known as Pascal’s
principle.
Pascal’s Principle
A smaller force Fin acting over a smaller area Ain determines the applied
pressure P=Fin/Ain. The output end of the lift has a much larger area,
Aout, and because the pressure within the fluid is essentially constant
(exactly so if the heights are the same), the output force Fout is
determined from P=Fin/Ain=Fout/Aout, so that the output force is
amplified to be:
Aout
Fout Fin
Ain
p1 p2 g z1 z 2 (2.8)
p2 p1
z1 z 2 ( 2.9 )
g g
p
z const (2.10)
g
- z: geometric height
- p/ρg: pressure head (hauteur piézométrique) of the fluid
- z + p/ρg: total height or head
dp / dx f
x
grad p dp / dy ; f fy force density
dp / dz f
z Dr. Keang Sè POUV 33
Hydrostatics
p pa gz
Dr. Keang Sè POUV 34
Hydrostatics
Exercise 7
Pressure Measurement
Mercury Barometer
Pressure Measurement
Mercury Barometer
0 pa Hg g h 0
pa Hg gh
pa
h
Hg g
101.3 103
h 0.762m 762mm
13546 9.81
Pressure Measurement
Manometer
Pressure Measurement
Manometer
pdown p up g z (2.12)
Without worrying too much about which point is “z1” and which is “z2”,
the formula simply increases or decreases the pressure according to
whether one is moving down or up.
p5 p1 0 g z1 z 2 w g z 2 z3 G g z3 z 4 M g z 4 z5
That is, keep adding on pressure increments as you move down through
the layered fluid.
Pressure Measurement
Manometer
Pressure Measurement
Exercise 8
Pressure Measurement
Exercise 9
Pressure Measurement
Exercise 10
Pressure Measurement
Exercise 11
Pressure Measurement
Exercise 12
Two non miscible liquids are filled in an opened tank equipped with two
piezometric tubes. Given: Patm = 1 bar. ρoil = 850kg/m3, ρwater =
1000kg/m3, h1 = 6m and h2 = 5m. Determine ZE, and ZD.
Pressure Measurement
Exercise 13
Water Manometer
Mercury
Dr. Keang Sè POUV 46
Hydrostatics
Pressure Measurement
Exercise 14
F pdA p a h dA pa A hdA
(2.13)
Since, h sin
1
CG
AdA
F pa A sin dA
pa A sin CG A
( pa hCG ) A
F pCG A (2.14)
FyCP sin y ( CG y )dA sin y 2 dA (2.16)
1 I
yCP
F
sin y 2 dA sin xx
pCG A
(2.17)
The term ydA 0 by definition of centroidal axes.
The negative sign in eq. (2.17) shows that ycp is below the centroid at a
deeper level.
Dr. Keang Sè POUV 49
Hydrostatics
For positive Ixy, xcp is negative because the dominant pressure force
acts in the third or lower left, quadrant of the panel.
Area:
Centroid:
Moment of inertia
Exercise 16
2.4m
water
The figure below presents a gravity dam with the length in y-direction
taken as a unit. The dam is made of a material of a density of ρ0d, (d is
the density of the material in relation to the density of water ρ0).
The vertical component of the force exerted by the soil on the dam is
defined by:
dFz
ax b
dx
By expressing the equilibrium conditions
of the dam, calculate the values of a and
b in function of ρ0, d, e, h and g
water
(atmospheric air penetrate between the
soil and the dam, but there is no water
seepage).
soil
Dr. Keang Sè POUV 56
Hydrostatics
The resultant force acting on the curved solid surface is equal and
opposite to the force acting on the curved liquid surface (Newton’s third
law).
Noting that the fluid block is in static equilibrium, the force balances in
the horizontal and vertical directions give:
FH Fx (2.21a )
FV Fy W (2.21b)
W is the weight of the enclosed liquid block. The summation (Fy + W) is
a vector addition (i.e. add magnitudes if both act in the same direction
and subtract if they act in opposite directions).
Dr. Keang Sè POUV 58
Hydrostatics
The exact location of the line of action of the resultant force (e.g., its
distance from one of the end points of the curved surface) can be
determined by taking a moment about an appropriate point. These
discussions are valid for all curved surfaces regardless of whether they
are above or below the liquid.
FH
FV
F F p
i CGi Ai (2.22)
i g sin i I xxi
yCP i (2.23)
pCGi Ai
i g sin i I xyi
xCP i (2.24)
pCGi Ai
2m
1.25 m
3.4 m
4.9 m
Buoyancy
Archimedes’ laws of buoyancy:
(b) Summation
of elemental
(a) Forces on vertical-
upper and pressure forces
lower curved
surfaces Dr. Keang Sè POUV 67
Hydrostatics
Buoyancy
The body in the previous figure experiences a net upward force:
FB p
body
2 p1 dAH z 2 z1 dAH V (2.26)
V body volume
Buoyancy
Floating bodies are a special case, only a portion of the body is
submerged, with the remainder poking up out of the free surface. The
eq. (2.26) is modified to apply to this smaller volume:
Buoyancy
Exercise 25
Exercise 26 W = 45 kg
Buoyancy
Exercise 27
Exercise 28
Buoyancy
Exercise 29
Buoyancy
Exercise 30
A ship that the sides are vertical to the height of water surface line
weighs 4000 tons and has a draft (tirant d’eau) of 6.7 m in sale water
(density 1025 kg/m3). When it is unloaded by 200 tons from the ballast,
the draft is reduced to 6.4 m. What should be the draft of the ship in
fresh water?
There exist two ways to describe the motion of a fluid: Lagrangian and
Eulerian descriptions.
Dr. Keang Sè POUV 75
Kinematics of Fluid
This description is suited for multiphase flows, where bubble and solid
particles can be easily tracked using the Lagrangian formulation.
This can be
visualized as
sitting in a plane
and watching
vehicles flow.
x xt , x0
dxt , x0
vt , x0
dt
dvt , x0 d 2 xt , x0
a (t , x0 )
dt dt 2
In the case of a uniform flow (2D) parallel to the x axis, the velocity
is:
V
v
0
The position of the particle at instant t is:
x1 t , x0 x01 Vt
xt , x0
2
x t , x
0 02
x
x01
where x0 is the position of the particle at time t 0
x02
dvt , x0 0
The acceleration is: at , x0
dt 0
Dr. Keang Sè POUV 78
Kinematics of Fluid
Looking at a
fixed location Dr. Keang Sè POUV 79
Kinematics of Fluid
v v x, t
Dvx, t
a x, t
Dt
Dr. Keang Sè POUV 80
Kinematics of Fluid
Material Derivative
In the Eulerian description, the fluid field is given as fluid properties at
fixed or arbitrary points. Therefore, if evolution of the particle properties
is desired, we need specific mathematical transformations to recover
the derivative following the fluid particle.
rx t
r t ry t
r t
z
drx t
dt
x
v t
dr t dry t
vt v y t
v t dt dt
z dr
z t
dt Dr. Keang Sè POUV 81
Kinematics of Fluid
Material Derivative
Let a scalar Eulerian field, such as a velocity component, be given by
c(x,y,z,t) where x, y, z are spatial coordinates.
c c rx t , ry t , rz t , t Fluid field
In this case,
Material Derivative
The derivative can also be expressed in tensor notation as:
c
v x x
Dc c c
v y
Dt t y
v z c
c
z
xc
c
v c
Dc
where c
Dt t y
c
z
The derivative is made from the temporal term and the convective
term. The latter represents the transport of a property in the fluid due
to its macroscopic motion.
Dr. Keang Sè POUV 83
Kinematics of Fluid
Material Derivative
If the velocity is expressed in components vi and the Cartesian
coordinates as xi where i = 1, 2, 3.
Dc c 3
c
Dt t
j 1
vj
x j
Dv x
Dt
Dv Dv y
a
Dt Dt
Dv z
Dt
Dr. Keang Sè POUV 84
Kinematics of Fluid
2c 2c 2c
c 2 2 2
x1 x2 x3 Dr. Keang Sè POUV 85
Kinematics of Fluid
0
t
For the uniform flow, as sketched in the figure below; the velocity
profile and other properties such as pressure, is uniform across the
section of pipe.
If along their path the fluid particles translate and rotate about the
particle center, the flow is rotational.
V 0
t
A special case which affords great simplification is incompressible flow,
where the density changes are negligible (i.e., / t 0 regardless of
whether the flow is steady or unsteady).
v x v y v z
0
x y z
Dr. Keang Sè POUV 92
Kinematics of Fluid
If the particles simply deform and do not rotate, we refer to the flow, or
a region of the flow, as an irrotational flow. If the particles do rotate,
they possess vorticity.
1 v z v y
x
2 y z
1 v v
y x z
2 z x
1 v y v x
z
2 x y
The streamline is the line tangent at every point to the velocity vector.
It indicates the velocity direction.
dx v x
dl v dy v y 0
dz v
z
vx v y vz
dx dy dz Dr. Keang Sè POUV 95
Kinematics of Fluid
Since the particle velocity is known at each spatial point, the trajectory
coordinates can be obtained by integrating the equation of motion:
dx dy dz
vx vy vz
dt dt dt
(x,y,z) is the position of the particle as a function of time.
2. Calculate the integration constants, such that at time ξ<t the fluid particle
was at (x0,y0). Here ξ is the parameter that designates the particle, by the
time it passed through the injection point. What we have done is to obtain all
the trajectories of the particles that were injected in the flow field before the
present time t.
3. Eliminate ξ.
Exercises
Exercise 31
Given the Eulerian fluid field: v x, y , z , t 3ti xzj ty k
2
Exercise 32
Exercise 33
A velocity field in a plane flow is given by: vx, y, t 2 yti xj
Find the equation of the streamline passing through (4,2) at t = 2.
Exercises
Exercise 34
Exercise 35
Find the acceleration (Eulerian description), the angular velocity and the
vorticity at the point (2,-1,1) at t = 2.
Dr. Keang Sè POUV 100
Kinematics of Fluid
Exercises
Exercise 36
The velocity field of a fluid is given by: V (α, β, k constants)
kt
1. What is the nature of the flow (i.e., steady, rotational, and/or
compressible flow)? Justify it.
2. Give the equation of streamline and the equation of particle
trajectory. Are they similar? Why?
Exercise 37
Exercises
Exercise 38
We study the compressible flow of an ideal fluid. The velocity field is:
V k xi yj
1. Determine the equation and the nature of the streamlines.
2. Calculate the acceleration vector of the flow using Eulerian
description.
3. Determine the equation of particle trajectories. Show that they are
the same as the streamline equation. Justify it.
4. Recalculate the acceleration vector using Lagrangian description.
Motivation
In analysing fluid motion, we might take one of two paths: (1) seeking
to describe the detailed flow pattern at every point (x, y, z) in the field
or (2) working with a finite region, making a balance of flow in versus
flow out, and determining gross flow effects such as the force or torque
on a body or the total energy exchange. The second is the control-
volume method and is the subject of this chapter. This analysis is the
most valuable tool to the engineer for flow analysis.
It has constant density: the density is the same for all fluid elements
and for all time (a consequence of incompressibility).
It is inviscid (frictionless).
Dr. Keang Sè POUV 104
Dynamics of Ideal Fluids
dV v dt dA cos v n dA dt
dV
Qv
s dt
(v n)dA Vn dA
s s
Qm (v n)dA vn dA
s s
Conservation of Mass
For an ideal fluid, the velocity is constant over the cross-sectional area.
For a real viscous fluid, the velocity profile varies over the cross-
sectional area because of the drag forces slowing the fluid flow.
If the fluid of density ρ has a velocity v1 in the portion of the tube with a
constant cross-sectional area A1, then in a time Δt, the mass of fluid
that passes a given point in this section of the tube is given by:
m A1V1t
A1V1 A2V2
or Qv AV cte
This is known as the continuity equation.
Qv
cs
(V n)dA
Vav
Qv 1
V n dA
A A
If the density varies across the section, we define an average density in
the same manner:
1
av
AdA
But the mass flow would contain the product of density and velocity,
and the average product (ρV)av would in general have a different value
from the product of the averages:
1
Em E p Ec dm1 gz1 Mgz dm1V12
S2 1
2 S1 ' 2
V 2 dm
At the instant t’ = t + dt, the fluid of mass (M + dm2) is taken between
S1’ and S2’. The mechanical energy is:
E 'm E ' p E 'c Mgz dm2 gz 2
S2 1 1
S1 ' 2
V 2 dm
2
dm2V22
The potential energy equals the work required to move the system of mass m from the
origin to a position vector (r=xi+yj+zk) against a gravity field. The kinetic energy equals the
work required to change the speed of the mass from zero to velocity v.
1 1
We have: Em dm2 gz 2 dm2V22 dm1 gz1 dm1V12
2 2
since dm1 dm2 dm (mass conservation), we obtain :
V22 V12
Em dm g z 2 z1
2
Work done by the pressure forces in the fluid can be found from the
following argument. First because the walls of the tube only exert
normal forces on our ideal fluid and the fluid flows along the walls,
there is no work done by the pressure supplied by the walls. The fluid
column to the left exerts a pressure on our system by supplying a force
F1 = P1S1 toward the right in order for the fluid to flow toward the right.
Similarly, the fluid to the right of our system exerts a pressure to the
left resulting in a force F2 = P2S2 that must be less than that acting
toward the right in order for the fluid to flow to the right. Each of these
forces does work on the system fluid.
At the left end, positive work is done on the fluid in the amount:
dm1
W1 F1dx1 P1S1dx1 P1dV1 P1
1
Dr. Keang Sè POUV 113
Dynamics of Ideal Fluids
P1 P2
dm1 dm2 dm
Wnet P1 P2
1 2
To write the change in mechanical energy as equal to the net work done
by the external pressure forces, we find:
P1 P2 V22 V12
dm dm g z 2 z1
2 Dr. Keang Sè POUV 114
Dynamics of Ideal Fluids
v2
2
p1 p2
Pnet
Pa
In the case of a turbine, the efficiency is given by:
Pa
Pnet
Between the instant t and t’=(t+dt), the fluid exchanged a net work
Wnet = Pnetdt with the hydraulic machine. Wnet is supposed to be positive
for a pump and negative for a turbine.
The unit of the work is Joul (J) and the unit of the power is Watt (W) or
J/s. Dr. Keang Sè POUV 119
Dynamics of Ideal Fluids
Using the work-energy theorem by considering this time the work done
by the hydraulic machine (Wnet), we can write:
V22 V12 P2 P1
g z 2 z1 Pnet
dt
2 dm
V22 V12 P2 P1
g z 2 z1 net
P
2 qm
The unit of each term in this equation is J/kg. Dr. Keang Sè POUV 120
Dynamics of Ideal Fluids
Fext qm V2 V1 Dr. Keang Sè POUV 121
Dynamics of Ideal Fluids
Exercises
Exercise 39
1. Kerosen (0OC) flows under the action of gravity in the pipe shown.
Determine the rate of flow in the pipe in l/s.
Exercises
Exercise 40
Exercises
Exercise 41
Let consider that we wish to measure the water flow rate in a horizontal
pipe of diameter D = 9 cm. We insert a Venturi tube (D = 9 cm, d = 3
cm). The difference in level h of the mercury in a U-tube can be
measured precisely. (ρwater = 1000 kg/m3, ρmercury = 13600 kg/m3)
Exercises
Exercise 42
Exercises
Exercise 42
2. A cylindrical pipe carries the water from a dam via a turbine. The level ZA is
kept constant. The level ZB is supposed to be constant. The mass flow rate
passing through the turbine is Qm = 175 kg/s. Given H = ZA-ZB = 35 m.
Exercises
Exercise 43
Exercises
Exercise 44
Exercises
Exercise 45
Exercises
Exercise 46
Introduction
For real fluids, the flow is more complex than that of ideal fluids. In
fact, there exists friction forces due to fluid viscosity acting between the
fluid particles as well as between the particles and the wall of the
container.
A fluid is called real if during its motion, the contact forces are not
perpendicular to the surface elements to which they are acting; i.e.,
they possess tangential components opposing to the sliding between
fluid layers. Such resistance is characterized by the viscosity.
Head Losses
Let us consider a flow of a real fluid between the points (1) and (2) in a
duct as shown in the figure.
Head Losses
Like in the previous chapter of dynamics of ideal fluids, we use the
theorem of mechanical energy for fluid between the instants t and t’,
but by considering the work of viscous forces dτ. So we can write:
W
1 1 P1 P2
dm2 gz 2 dm2V2 dm1 gz1 dm1V1 dm1
2 2
dm2
1 2
dt
2 2
V22 V12 P2 P1
g z 2 z1
W dt
2 dm
Head Losses
We define the head loss between the points (1) and (2) by:
J12
W dt
dm
V22 V12 P2 P1
g z 2 z1 J12
2
The unit of each term in the equation above is J/kg. By dividing this
expression by g, we obtain:
V22 P2 V12 P1 J
z2 z1 12
2 g g 2 g g g
Head Losses
The equation above can be graphically interpreted as below:
J12 = Js + JL
Head Losses
For example, in the pipe system presented below, the section BC, DE,
FG, HI and JK are the bents with different angles, so they present
singular head losses. The sections AB, CD, EF, GH, IJ and KL are
straight pipes, so they present linear head losses.
Head Losses
Singular or local head loss
V2
J s K s
2
s is the index of the type of the duct singularity. Ks is the coefficient of
head loss that depends on the nature and the geometry of the duct
singularity. Generally, the values of Ks are given by the constructors in
their products catalogues.
Head Losses
Singular or local head loss
2
A
For brutal section enlargement, we have: K s 1 1
A2
Head Losses
Singular or local head loss
Head Losses
Singular or local head loss
Head Losses
Singular or local head loss
Head Losses
Singular or local head loss
Head Losses
Singular or local head loss
Head Losses
Linear head loss
The linear head losses are the losses which are distributed regularly
along the pipes. At each point of permanent flow, the characteristics of
flow are well defined and independent on time. The graphical
representation of flow can be seen below.
Head Losses
Linear head loss
If the velocity is constant, the piezometer head and the total head are
parallel. The variation of piezometer head, evaluated in liquid level is
equal to the linear head loss between the two measuring points.
V2 L
J L
2 D
Head Losses
Linear head loss
Head Losses
Moody diagram
Head Losses
Linear head loss
Recommended roughness values for commercial ducts
Head Losses
Linear head loss
Additive
At the same flow rate, the pressure loss can be reduced by more than 2/3.
At the same pumping pressure, the flow rate can be increased by 30 to 40%.
Dr. Keang Sè POUV 152
Dynamics of Real Fluids
Head Losses
Linear head loss
Instantaneous velocity (CTAC 75 ppm, Re21000, T=20°C et D=17mm) [F. Hadri 2009]
Dr. Keang Sè POUV 153
Dynamics of Real Fluids
V22 V12 P2 P1
g z 2 z1 J12
Pn
2 qm
J12 is the sum of all the head losses between the points (1) and (2).
Pn is the mechanical power exchanged between the fluid and the
machines existing between the points (1) and (2).
Pipes in series
Q1 Q2 Q3 constant
The total head loss through the system equals the sum of the head loss
in each pipe:
H A B H1 H 2 H 3
Dr. Keang Sè POUV 155
Dynamics of Real Fluids
Q Q1 Q2 Q3
H A B H1 H 2 H 3
Q1 Q2 Q3 0 (a )
This implies that one or two of the flows must be away from the
junction. The pressure must change through each pipe so as to give the
same static pressure PJ at the junction.
In other words, let the Hydraulic Grade Line (HGL) at the junction have
the elevation: p
hJ z J J
g
where pJ is in gage pressure for simplicity. Then the head loss through
each, assuming p1 = p2 = p3 = 0 (gage) at each reservoir surface, must
be such that:
V12 L1 We guess the position hJ and solve Eq. (b)
h1 1 z1 hJ
2 g d1 for V1, V2 and V3 and hence Q1, Q2 and Q3,
iterating until the flow rates balance at the
V22 L2 junction according to Eq. (a). If we guess
h2 2 z 2 hJ (b) hJ too high, the sum Q1 + Q2 + Q3 will be
2g d2
negative and the remedy is to reduce hJ,
V32 L3 and vice versa.
h3 3 z 3 hJ
2g d3 Dr. Keang Sè POUV 158
Dynamics of Real Fluids
Exercises
Exercise 47
1. Determine the critical velocity for fuel oil (νfuel = 4.47 10-6 m2/s) and
for water (νwater = 10-6 m2/s) flowing each in a pipe of 150 mm of
diameter.
h?
Exercises
Exercise 48
Exercises
Exercise 49
Determine:
1. The volume flow rate of the petroleum.
2. The flow regime of the petroleum using Reynolds number.
3. The linear head loss coefficient.
4. The minimal length L after which the pressure drop between the stations A
and B exceeds 3 bar.
Exercises
Exercise 50
Exercises
Exercise 51
Exercises
Exercise 51 (continue)
Exercises
Exercise 52
A pump with a volume flow rate of Qv = 2.8 l/s is used to carry the water from a
pool to a tank through a pipe of diameter d = 135 mm. Hypotheses: Z1 = 0, Z2 =
35 m, P1 = P2 = 1013 mbar, dynamic viscosity of water μ = 10-3 Pa.s, length of
the pipe L = 65 m. All the singular head losses are neglected.
Exercises
Exercise 53
In a water supply system, we use an electric pump of hydraulic power Ph (or Pnet)
to be determined. The pump sucks the water from the point G up to the air at
the point O. The diameter of the duct of the system is d = 120 mm and the flow
velocity is V = 0.5 m/s. The absolute pressure of the water at G is: pG = 1.5 105
Pa.
For connecting
different ducts, we
use 4 bends at 90o
and of curved radius
of R0 = 100 mm.
Exercises
Exercise 53 (continue)
Given:
Lt=68.6m, total length of the linear ducts between the points O and G.
Kv=0.24, coefficient of head loss at the butterfly valve V.
KG=0.15, coefficient of head loss at the point of water sucking up G.
Kc=Kc’=0.45, coefficients of head loss at the connections at the entry and the
exit of the pump.
1. Calculate the volume flow rate and mass flow rate of the pump.
2. Calculate the Reynolds number Re and determine the nature of the flow.
3. Calculate the total linear head loss.
4. Calculate the total singular head loss.
5. Calculate the total head loss between the points O and G, ∆pGO.
6. Calculate the mechanical power Pm (or Pa) supplied to the pump by the
electrical motor knowing that its efficiency is 85%.
Exercises
Exercise 54
Let consider a hydraulic installation connected to a very big free surface tank
under a pressure of 1.4patm. The installation consists of circular pipes, a pump
and a convergent tube opened to free air. The water in the tank is pumped in
order to supply a jet. Given: h1 = 2 m, h2 + h3 = 1.5 m, h4 = 8 m, D = 0.15 m,
d = 0.08 m, patm = 1.013 105 N/m2, ρw = 1000 kg/m3, g = 9.81 m/s2. All the
head losses are neglected.
1. Calculate the velocity of the jet U4
and deduce the volume flow rate Q.
2. Determine the hydraulic power P of
the pump necessary for maintaining
the jet.
3. Calculate the dynamic heads
between the section 3 and 4.
4. Represent graphically the piezometer
head and the total head as well as
indicate their values on the drawing.
Dr. Keang Sè POUV 168
Dynamics of Real Fluids
Exercises
Exercise 55
Exercises
Exercise 56
Derive the expression of the total volume flow rate QV in the pipe below.