E310/1 1 FLUID MECHANICS AND HYDRAULIC MACHINERY UNIT I PREPARATION OF FLUIDS Aims: This unit aims at cursory review of properties of fluids like, viscosity, Bulk Modulus, Vapour pressure, surface tension and capillarity. Objective: 1 . To distinguish between dimension and Unit. 2. To define various properties. 3. To distinguish between Ideal fluid & Real fluid Newtonian and non Newtonian fluids 4. Practical application of the above preparation in solving problem. 1.1 Introduction: Fluid Mechanics is the science of mechanics, which deals with the behaviour of fluids at rest or in motion. Both liquids and gases are called fluids. fluid is defined as a substance which is capable of flowing and deforms continuously under a shear stress however small it may be. Before review of the various properties of fluid, dimensions and units of measurement are discussed here. 1.2. Dimensions and units: Any physical quantity can be expressed in four fundamental dimensions namely, mass(M), Length(L), Time (T) and temperature (). But in incompressible fluid flows the 3 dimensions M,L, and T are adequate to define the physical quantities. The standards to measure the above dimensions of the physical quantities, are called unit of measurement. There are 4 systems of units as given below. 1.Centimeter - Gram - second (C.G.S) 2. Meter - Kilogram - Second (M.K.S) 3. Foot - Pound - Second (F.P.S) 4. International standard system (S.I) (latest system) In this present treatment of the subject , F.P.S is not used and mostly S.I. System is followed.
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E310/1 1
FLUID MECHANICS AND HYDRAULIC MACHINERY
UNIT I
PREPARATION OF FLUIDS
Aims: This unit aims at cursory review of properties of fluids like, viscosity, Bulk
Modulus, Vapour pressure, surface tension and capillarity.
Objective:
1 . To distinguish between dimension and Unit.
2. To define various properties.
3. To distinguish between Ideal fluid & Real fluid Newtonian and non Newtonian fluids
4. Practical application of the above preparation in solving problem.
1.1 Introduction: Fluid Mechanics is the science of mechanics, which deals with the
behaviour of fluids at rest or in motion. Both liquids and gases are called fluids. fluid is
defined as a substance which is capable of flowing and deforms continuously under a
shear stress however small it may be.
Before review of the various properties of fluid, dimensions and units of
measurement are discussed here.
1.2. Dimensions and units: Any physical quantity can be expressed in four fundamental
dimensions namely, mass(M), Length(L), Time (T) and temperature (θ) . But in
incompressible fluid flows the 3 dimensions M,L, and T are adequate to define the
physical quantities.
The standards to measure the above dimensions of the physical quantities, are
called unit of measurement. There are 4 systems of units as given below.
1.Centimeter - Gram - second (C.G.S)
2. Meter - Kilogram - Second (M.K.S)
3. Foot - Pound - Second (F.P.S)
4. International standard system (S.I)
(latest system)
In this present treatment of the subject , F.P.S is not used and mostly S.I. System is
followed.
E310/1 2 The dimensions and units of various quantities useful are given below.
S.No. Quantity Dimension C.C.S M.K.S S.I
1. Mass M Gram (m) m.slug Kg(m)
2. Length L C.m metre metre
3. Time T Second second Second
4. Velocity L/T = LT-1 cm/s m/s m/s
5. Acceleration LT-2 cm/s2 m/s2 m/s2
6. Force MLT-2 Dyne Kg(t) Newton
7. Pressure ML-1T-2 Dy/cm2 Kg(f)/m2 N/m2 Pascal
8. Mass density ρ= ML-3 gm(m)/cm3 m.slug/m3 kg (m)/m3
9. Specific
weight
w= ML-1T-2 Dy/cm3 kg(t)/m3
(9.81N)
N/m3
10 Viscosity ML-2 T-1
FTL-2
Dy -s/cm2
(poise) Kg( f ) − sec
m 2
(98.1 Poise)
N.Sm 2
(10 poise)
11. Kinematic
Viscosity
L2T-1 cm2 /s
(stokes)
m2/s m2/s
12 Workdone
Energy
ML2T-2 Dy-cm Kg(t) -m N-m
(Joule)
13 Power ML2T-3 Dy-cm/s kg(t) -m/s
75 kg(f) .m/s
= 1H.p
N-m/s= J/S=
Watt
1000w
=1k.w
14 Surface
Tension
MT-2 Dy/cm Kg(f)/m N/m
Self Assessment Questions
SAQ 1. Define a fluid
SAQ 2. Distinguish between Unit and Dimension
E310/1 3 SAQ 3 Derive the dimension of power
1.3 Fluid Properties:
Density or mass density or specific mass = ρ It is defined as the mass per unit volume of
fluid. It is denoted by ‘ρ’
Unit C.G.S gm(m) /Cm3
M.K.S. m slug / m3
S.I kg (m) /m3
For water at 40 c Unit of mass density
C.C.S M.K.S S.I
ρ 1g(m)/cm3 10009.81
= 101.94mslug m 3 1000kg(m) /m3
Specific weight is defined as the weight of the fluid per unit volume . It is
expressed as ‘w’ or ‘λ’
Unit is given as
C.C.S M.K.S S.I
Unit Dy| cm3 Kg(f)/m3 N/m3
For water at 40 c, the Unit is given by
C.C.S M.K.S S.I
Unit 1gm(f)/cm3
981 dy/cm3
1000 (f)/m3 9810N/m3
=9.81K.N/m3
Further the following relationship of units are also useful in the further study of Fluid
Mechanics.
Kg(f) = 1m slug ×1m/s2 = kg(m) ×9.81m/s2
N = (9.81) Kg (m) ×1m/s2 = kg (m) ×1m/s2
∴1m.slug = 9.81 kg (m) = 9810g(m)
1Kg(t) = 9.81N = 1000g(t)= 981×105Dy
1Kg(m) = 1000g(m)
Dyne = 1g(m) ×1cm/s2
N= 1000g(m) ×100cm/s2 = 105 Dynes
1g(f) = 1g(m) ×981cm/s2
=981 Dynes
E310/1 4 Thus Kg(f), g(f) are called engineering or gravitational units, where as Dyne, Newton are
called as absolute units.
Specific Volume: is defined as volume per unit mass. It is reciprocal of mass density.
It is denoted by υ = 1ρ
Unit CC.S M.K.S S.I
Cm3/g(m) m3 /m.slug m3 /kg(m)
For water at 40 c , ν = 0.001 m3 /Kg(m) in S.I System
Specific Gravity is defined as the ratio of specific weight of any fluid to the specific
weight of standard fluid which is taken as water (for liquids) at 40c it is denoted by ‘s’.
SAQ 4. What is the specific volume of a fluid whose density is 90Kg(m)/m3.
SAQ 5 10 m3 of fluid weighs 90 KN. find its specific gravity
Viscosity is defined as the property of the fluid with which it offers resistance to
shear force. It is due to both cohesion and adhesion of the fluid. It is also known as
Dynamic Viscosity,. Absolute Viscosity, Coefficient of viscosity . As force is involved in
this phenomenon, it is called as Dynamic Viscosity. It is represented by ‘µ’.
Newton’s law of Viscosity:- It states that the shear stress is proportional to the velocity
gradient normal to the direction of flow of fluid ( in rate of angular defer motion)
Thus ταdϑdy
or τ = µ dϑdy
where τ is the shear stress.
consider a fluid of viscosity in between two parallel plates at distance ‘y’ apart as
shown in figure. The lower plate is at rest and the upper plate of plane area ‘A’ is moved
by a Force “F” with a velocity ‘v’. Due to adhesion the particles at the Solid boundary
are adhered to them and so the velocity of the particles at lower plate is zero and that of
the particles at upper plate are moved by a velocity V. In between the particle move
linearly from zero velocity to v velocity, setting by a velocity gradient vy
ordϑdy
in the
direction
E310/1 5 normal to direction of plate movement. Suppose if the force is increased, then the velocity
gradient also is increased. by experiments it is proved that the shear force F αAvy
for
number of fluids.
ieFA
= ταvy
=dvdy
Where dv is change in velocity in dy as the force F is increased over the place A, the dv/dy
also is proportionately increased.
Thus τα dνdy =
dθdt
or τ = µ dudy
This equation is called Newton’s Equation of viscosity
Also = µ =τ
dϑ dy
So Viscosity is also defined as ;shear stress per unit velocity gradient in normal direction
to the movement of plate.
Dimension of Viscosity is obtained as
µ =FA
yv
=FL2
LL T
=FTL2 (force l-length- time)
or M.LT −2
L2 .LT
=MLT
C.C.S M.K.S S.I
UNIT Dy − scm 2 poise
Kg( f )− sm 2
98.1 poise
N.sm 2 10 poise
For Water 200C Viscosity = 0.01 poise (one centi poise)
E310/1 6 Based on ;the shear stress and the respective velocity gradient in normal direction,
fluids are classified as discussed below.
i).Ideal fluids and Real fluids
we know that τ = µ dϑdy
if µ= 0 τ = 0
This means that when viscosity is zero , there will be no shear stress to oppose the fluid
flow and the fluid particle ‘slip’; over the solid boundary. This leads to Uniform velocity
distribution over the cross section al flow. Such fluids are called as Ideal fluids for which
viscosity is zero and there will be no loss of energy, in the fluid flow from point to point.
On the other hand in fluids if viscosity is not zero and fluid particles satisfy ‘no-slip’
condition over solid boundary, then small fluids are called as real fluids. So for real fluids
viscosity is not zero ;and the shear stress opposes the fluids flow to make the velocity
distribution to vary from zero at solid boundary to free stream velocity to set up a velocity
gradient. In the real fluids there will be loss of energy in fluid flow from point
to point
The ideal fluid is represented in the graph shown in picture on ‘x’ axis for which shear
stress = 0.
Newtonian and non-Newtonian fluids: If fluids like air. water. kerosene. glycerine etc.
obey the law of viscosity and the shear stress in these fluid is proportional to velocity
gradient. Then such fluids are called as Newtonian fluids. If the shear stress is not
proportional to velocity gradient, then these fluids are called as Non-Newtonian fluids as
shown in figure.
E310/1 7 Plastic and Thixotropic fluids
Some fluids flow after certain yield stress. Thus plastic will flow with linear
relationship shear stress and velocity gradient, after certain yield stress. Thixotropic fluids
after yield stress, flow with a non linear relationship between shear stress and velocity
gradient. printers ink is called as Thixotropic fluid. These fluids are represented in the
figure.
Kinematic Viscosity is defined as the ratio of dynamic viscosity for mass density of the
fluids. It is represented by. ‘ν’.
∴υµρ
=
In this unit only Kinematic of fluid, like length, Time are involved and so it is called as
Kinematic Viscosity.
Dimension of Kinematic Viscosity
υµρ
= = × = −MLT
LM
L T3
2 1
Unit of Kinematic Viscosity
Unit C.G.S M.K.S S.I
Cm2/s stokes m2/s m2/s
Kinematic Viscosity of water at 200 c is 0.01 stokes.
SAQ 6. Define ideal fluid
SAQ 7 Why shear stress is zero for a fluid at rest
SAQ 8 Why Viscosity is called as Dynamic Viscosity
SAQ 9 Distinguish between ideal and real fluids
SAQ 10 The Viscosity of a fluids 0.09 poise Determine its Kinematic Viscosity if
its specific weight is 9Kn/m3
SAQ 11 What are the unit and dimension of velocity gradient
SAQ 12 Why the ratio of viscosity to mass density is called as ‘Kinematic
Viscosity’
SAQ 13 (a). of velocity gradient in normal direction is 50m/s/m . Determine the
rate of angular deformation
E310/1 8 (b) If the rate of deformation is 1 rad/s find its velocity gradient.
SAQ 14 If the Viscosity is o/.s/m2 compute the angular deformation for a shear
stress of 0.1N .S/m2.
Worked Out Examples
Worked Example (1) Two parallel plates are 2 cm part. This space is filled with glycerine
Find the force required to drag this plate of area 0.5 sqm at a speed of 0.6m/s when it is
placed at 0.75cm from the lower plate Viscosity of glycerine is 8.0 poise.
F = µAϑ1y1
+1y2
810
× 0.5 × 0.61001.25
+1000.75
=51.2 N
Worked Example (2) A piston of 496 mm dia and 150 cm long slides vertically down a
cylinder of 500 mm dia. If the clearance is filled with lubricating oil of Viscosity 5×10-2
poise find the terminal speed of the piston The weight of piston is 5N.
Sol:
p= µ vy
× A
E310/1 9 A = ΠDL =Π×0⋅496×0⋅15=0⋅233m2
∴5 =5 × 10−2
10
×
v0.002
× 0.233
∴v= 8.5 m/s
Worked Example (3) A 15 cm dia vertical cylinder rotates at 100 r.p.m concentrically
inside another cylinder of 15.10 cm dia. Both the cylinders are 25 cm high. If the
viscosity 8.0 poise is filled in between the space in between the cylinder, find the power
absorbed in overcoming the shear resistance.
sol
Ω =2ΠN60 −
=2 × Π ×100
60= 10.47
velocity = ϑ =ΠDN
60=
Π ×15100
×10060
= 0.785m / s
Shear force =τ×area
µ vtΠDL
=8
10×
0.7850.05100
× Π × 0.15 × 0.25 = 148N
Torque T = F×r = 148 ×0.15
2= 11.1Nm
Power absorbed = T ×Ω = or F×U =116 Watts
Worked Example (4) If the velocity distribution of a fluid over a plate is given by
u =3/4 y-y2, where u = velocity in m/s at a distance of ym above the plate, determine
shear resistance at y = 0.15 m from the plate . viscosity = 0.00833 poise
E310/1 10 sol
Given u =34
y − y2
∴dudy
=34
− 2y ;dudy
=34
− 2 × 0.15 = 0.45m / s / m
y = 0.15
So shear resistance = τ = µ dudy
=0.00833
10× 0.45
= 0.375 ×10-3 N/m2
Surface Tension: The surface of contact between a gas and a liquid or between two
immiscible liquids acts as a film capable of resisting small tensile forces. It is due to
cohesion between the liquid particles at surface. This phenomenon is known as surface
tension which is due to the imbalance of the intermolecular attractive forces between
liquid particles at the interface. This imbalance of forces causes the film or interface to
behave like an elastic film under tension.
So surface tension is defined as the force required so keep unit length of the film in
equilibrium condition. It is denoted by ‘σ’.
C.G.S. M.K.S S.I
Unit Dy/cm kg(f)/m N/m
For water with air at 18 0c surface tension σ = 0.074 N/m (S.I)
= 75 Dy/cm (C.G.S)
Effect of surface tension : The internal pressure inside a fluid drop is increased due to
surface tension.
Determination of excessive pressure inside a drop
Consider a drop of a diameter ‘d’ is divided into two halves. Now consider the force that
are acting on it to keep it in equilibrium condition as shown in figure
E310/1 11 The external force and internal force acting on the drop in horizontal direction. are given
by pressure intensity and projected area of the drop in vertical plane
ie ΠD 2
4Pa and
ΠD 2
4pi respectively
Now consider the forces in equilibrium condition i.e. the ∑F in horizontal direction = 0
∴ΠD2
4pi − ΠD2
4pa − ΠDσ = 0
(Pi-Pa)/ΠD /2
4= Π /Dσ
Pe = 4σD
is the expression for the inside excessive pressure.
SAQ 15 Find the excessive pressure inside a cylindrical jet of water of diameter ‘d’
SAQ 16 Find the expression for excessive pressure inside a soap bubble of dia d.
(hint there will be two surfaces) in contact with air.
SAQ 17 If the dia of drop is increased so 4d from d, find the excessive pressures
ratio inside the drops.
Worked Example (5) Air is introduced through a nozzle into a tank of water to produce
a stream of bubbles of 2 mm dia. calculate the excessive pressure inside the bubble over
surrounding water pressure. Surface tension of water 74 dy/cm
Pe=4σd
=4 × 74 ×10−5
10−2 ×1000
2
=2×74=148N/m2
Capillarity We know that cohesion is intermolecular attractive force between
molecules of the same liquids, where as Adhesion is the attractive force between solid and
liquid or between two different liquids.
When a liquid wets a surface, adhesion is greater than Cohesion and so the liquid rises in a
capillary tube immersed in the liquid. Similarly when liquid does not wet a solid,
Cohesion is greater than adhesion and so the liquid drops in side the capillary tube. This
rise or fall of liquid in capillary tube is known as capillarity which is due to with adhesion
and cohesion of the liquid.
Derivation of an expression for capillary rise or fall :Consider the rise or fall of liquid
in the capillary tube in water and mercury as shown in figure. Water wets the tube
surface where mercury does not.
E310/1 12
In case of water which wets the surface of the tube, the adhesion is more than the
cohesion and so the density of water decreases in the tube at the liquid surface. So to
make the pressure over the liquid surface equal, the water rises in the tube till the pressure
is equal in side and outside over the surface.
In case of mercury , cohesion is more than adhesion, as it does not wet the tube
There fore the density of liquid increases inside the tube near the liquid surface and so to
make the pressure in side and outside of the tube to be equal over the surface, the liquid
falls in the tube.
So consider the equilibrium condition between the forces in vertical direction on
the liquid column in the tube.
upward force = downward force
due for surface tension due to weight of liquid
ie //Π /dσ cosθ =/Π /d 2
4h × w
h =4σ cosθ
wdWhere ‘θ’ is the angle of the surface tension with vertical
For pure water ‘θ’ with glass = 00
For mercury ‘θ’ with glass = 1400
SAQ 18 What is the inclination of surface tension with glass tube with vertical for
pure water.
SAQ 19 Determine the capillary rise in two parallel plates ‘d’ apart immersed in a
liquid.
SAQ 20 Capillary rise in 5 mm dia tube is 0.6 c.m. Find the same in 10 m.m tube.
E310/1 13 Worked Example (6) What should be the minimum dia of a glass tube used to measure
water level if the capillary rise is to be limited to 1mm. Surface tension of water = 75
dy/cm. θ = 0
h =4σ cosθ
wd= 0.1 =
4 × 75 ×1981 × d
∴d= 3cm
Vapour Pressure When a liquid with a free surface is combined in a closed vessel, the
vapour molecules from the free surface will escape into the space above free surface in the
vessel. The escape of vapour molecules continue till the saturation of the space within
vessel with vapour molecules. This vapour exerts a pressure on the liquid surface which is
called as partial vapour pressure.
If, in any fluid flow, the fluid pressure in dropped to its vapour pressure at that
room temperature, then the fluid starts to vapourise. This vapour and the liberated
dissolved gases form as vapour bubble which grow in size. After some time these
bubbles move to higher pressure zone, where they collapse (decrease in size due to high
pressure) releasing very high pressures which are nearly equal to 20 atmospheric pressure.
This high pressure erode the solid boundaries by repeated blows. This phenomenon is
called as Cavitation. So care should be taken in design of Hydraulic machines to avoid
cavitation which causes damage of solid surface of machines.
Mercury vapour pressure is very low and so it is an excellent fluid for manometers.
S.I M.K.S
Unit N/m2 Kg(f)/m2
For water at 200c , vapour pressure = 1.23 K Pa(abs)
For pure water v.p = 18 to 24 Kpa(cabs)
SAQ 21 Which fluid is good for manometers
Bulk Modulus:
When ever a pressure is applied over a fluid in a container it will be compressed .
Compressibility of a fluid is defined as change in volume under a pressure It is the inverse
of Bulk Modulus of Elasticity ‘K’
which is defined as
K∆p−∆v
v
=stressstrain
E310/1 14 C -ve sign indicates the decrease in volume due to increase in pressure.
Unit N/m2 (S.I) or kg (f) /Cm2
K for water = 2.06×109 N/m2 at N.T.P
K for air = 1.03 ×105 N/m2
ie K of water = 20,000 × k of air
So water is considered to be incompressible as its K is very high. That means
the density of incompressible fluid remains constant whatever may be the pressure over it.
Air is compressible fluid.
Summary
1. Fluid is defined as a substance which is capable of flowing and deforms
continuously under a shear stress however small it may be
2. In fluid mechanics dealing with in compressible fluid, 3 fundamental
dimensions namely Mass (M), Length (L) & Time (T) are used to express
any physical quantity.
3 Units are the standards of measurement of the above dimensions.
4 3 systems of units namely C.G.S, M.K.S, and S.I. are used in present
treatment of the subject. But mostly S.I system which is the latest system is
followed
5 Kg(f) = 9.81 N ; 1M slug = 9.81 kg(m)
6. Viscosity is defined as the property of fluid with which it offers resistance to
shear force. Its unit in C.G.S is poise
7 Kinematic Viscosity is the ratio of viscosity to the fluid mass density
8 Ideal fluid is that for which viscosity and shear stress are zero.
9 Surface tension is defined as the force per unit length of the film to keep it in
equilibrium condition.
10 Vapour pressure is the partial pressure exerted by the vapour on the fluid.
11 Compressibility of fluid is the change in volume of fluid under a pressure and is
the inverse of bulk modulus of a fluid.
Answers to self assessment questions:
3. ML2T-3 4.0.0011m3/kg(m) 5. s= 0.91
7. dv/dy = 0 10. v =0.098 stokes 11. m/s/m/ -7-1
13. tan θ = 250, θ=890 .7, dθ/dt = 0.498 rad/s
E310/1 15 Vel gradient = 1.55 m/s/m
14. tan θ = 1, θ= 450 , 15. 2σd
16.pe = 8σd
17. p1
p 2
= 4 18. θ=0 19. h = 2σ cosθ
wd
20. h2 = 0.3cm
EXERCISE 1.(1) Of 5.27 m3 of a certain oil weights 44 KN, calculate the specific weight,
mass density and specific gravity of oil (8349N/m3 , 85109 Kg/m3 , 0.85)
1.(2) A certain liquid has a dynamic viscosity of 0.073 poise and specific gravity
of 0.87 compute the kinematic viscosity of the liquid in stakes of SI.I
system (0.08 stakes, 0.083x10-4m2)
1.(3) of the equation of a velocity distributing over a plate in given by v=2y-y2,
where v is velocity in m/s at a distance ‘y’ in from the solid surface,
determine the shear stress normality at 7.5 cm and 15cm from it.
Given m =8.6 poise.[0.175 Kg(f)/m2 , 0.162 Kg(f)/m2 , 0.149Kg(f)/m2 ]
1.4 Two large place surfaces are 20mm apart and the gap contained oil of
dyamic viscosity 0.6 poise. A this plate of 0.5m2 surface area is to be
parallel although the gap at a constant velocity of 0.6 m/s. The plate is
placed at 8mm from one of the surfaces. Find the face required to parallel
the plate. [F=3.75 N].
1.5 A flat plate of area 0.15m2 is to be tossed up an inclined plane of slope 1
vertical to 3 horizontal on a thin layer of oil m=0.75 poise of thickness
0.2mm. Of the weight of the plate in 250N , estimate the force required to
pull the place at 1.6m/s velocity [F=979N].
1.6 A piston of 7.95 cm diameter and 30 cm long works in a cylinder of 8.0 cm
diameter. the annular space of the piston is filled with an oil of viscosity 2
poise. Of an axial load of 10N is applied to the piston, calculate the speed
of the piston(v=16.68 cm/s)
1.7 A 90mm diameter shaft states at 1200rpm in a 100mm long cylinder
90.5mm internal dia mater. The annular space in the shaft and cylinder in
filled with oil of viscosity 0.12poise. Find the power absorbed to overcome
shear friction. (P=434 w)
E310/1 16 1.8 A tube of internal diamter 2mm is dipped vertically into a vessel containing
mercury. The lower and of the tube is 2cm blow the mercury surface.
Estimate the pressure of air inside the tube to blow a semi spherical bubble
at the lower end. Surface tension of the mercuty is 0.4N/m. [∆P=3.468
Kg/cm].
1.9 A soap bubble 51mm in diameter has an internal pressure in excess of the
external pressure of 0.00021Kg(f)/m2 . Calculate the tension in the soap
film [0.0134kg(f)/m].
1.10 Calculate the capillary rise ’h’ a glass tube of 3mm diameter when
immersed in water at 20oC. The surface tension at 20oC is 0.075kg(f)/m.
What will be the percentage increase in the value of ‘h’ if the diameter of
the tube in 2mm [10mm,50%].
1.11 By how much does the pressure in a cylindrical set of water 4mm in
diameter exceed the pressure of the sorrounding atmosphere of
8.4 Steady Laminar flow through horizontal pipe Hagen- Porsenille law:
Consider a concentric cylinder of ‘r’ radius and ‘dx’ long as shown in figure in a
laminar flow through a horizontal pipe of diameter D.
E310/1 156
Fig.
By applying Bernoulli’s equation between (1) and (2) we have,
p1w
+ z1 + v12
2g= p1
w+ z2 + v1
2
2g+ hf
i.e hf =p1 − p2
w
Qv1=v2 , z1=z2
The forces acting on the elemental volume are shown in figure. Then for steady flow, ∑F
= 0 , Qacceleration is zero, we have
pΠr2 − p +∂p∂x
dx
Πr2 = τ2Πrdx
τ = −∂p∂x
r2
But for laminar flow we know that
τ = µ dvdy
where y = R-r is the distance form boundary.
So dy = - dv
i.e., τ = −µ dvdr
= −∂p∂u
r2
i.e., dv =12u
∂p∂x
r dr
Integrating and substituting the B.C v = 0 at r = R, we have
E310/1 157
v =
14µ
∂p∂x
r2( )+ C
and C = −1
4µ−∂p∂x
R2( )
∴ v =1
4µ−∂p∂x
R2 − r2( ) is the velocity distribution over the cross
section of flow, which is parabolic. Velocity is maximum at r = 0
Fig.
i.e., vmax =1
4µ−∂p∂x
R2
=1
16µ−∂p∂x
D2
Then V = vmax 1−rR
2
Mean Velocity:
To get the mean velocity, discharge is to be calculated based on velocity
distribution. So consider at radius ‘r’ a ring of thickness ‘dr’ as shown in Figure.
Fig.
The discharge passing through this ring area is
dQ = dav = 2Πrdr ×1
4µ−∂p∂x
R2 − r2( )
So by integrating we have
E310/1 158
Q = dQ = 2Πrdr 14µ
− ∂p∂x
R2 − r2( )
0
R
∫∫
Q = 2Π1
4µ−∂p∂x
R2 r2
2−
r4
4
0
R
=Π
8µ−∂p∂x
R4
=Π
128µ−∂p∂x
D4
So the mean velocity is v
v = Qa
= Π8µ
− ∂p∂x
R4 × 1ΠR2
=1
8µ−∂p∂x
R2
=1
32µ−∂p∂x
D2
So v =12
vmax
Fig.
Further v = v at a radius r given by
vmax2
= vmax 1 − rR
2
i.e.,rR
=12
∴ r = 0.707 R
E310/1 159 Loss of head:
We know that the mean velocity is given by
v =1
32µ−∂p∂x
D2
− ∂p∂x
= 32µvD2
−∂p( )p1
p2
∫ = 32µvD2
0
L
∫ dx
p1 − p2( )=32µvL
D2
orp1 − p2( )
w= hf =
32µvLwD2 is known as Hagen Poisemille equation.
But hf =fLv2
2gD=
32µvLwD2
Then f =64µρvD
=64Rn
where Rn =Raynolds no = ρvD
µ
So for laminar flow, the friction factor is 64Rn
Shear stress:
The shear stress at any radius ‘r’ is given by
τ = −∂p∂x
r2
So at boundary τ0 = −∂p∂x
R2
( )=
−
=
p pL
D
whL
Df
1 2
4
4
=wL
fLvgD
D2
2 4
E310/1 160
τρ
0
2
8=
f v
Then
τ0ρ
= vA = shear velocity
vA = vf8
Power required to maintain the laminar flow is
P = F × v
= τ × A × v
=whf
2D4
× πDL × v
= w πD2
4v
hf
= wQhf N-m / sec or Kg-m / sec
Also P = Force ×v
=(p1-p2)A ×v
=(p1-p2) Q N-m / sec or Kg-m / sec
SAQ(11). Determine the shear stress at wall in case of a pipe of 10 cm diameter in which
laminar flow is taking place, given the -ve pressure gradient in the direction of motion as
20 KN / m2 / m.
SAQ(12). Determine the maximum velocity in a pipe of 10 cm diameter in which laminar
flow is taking place . Given -ve pressure gradient in the direction of motion as 2 KN / m2 /
m and µ=2.5 NS / m2.
SAQ(13).A laminar flow is taking place at 2 lps in pipe of 0.01 m2 C.S. Determine the
maximum velocity.
SAQ(14). Find the velocity in a pipe of 10 cm diameter at a radius of 2 cm from centre, in
which diameter flow is taking place. Given max velocity as 0.5 m / s.
SAQ(15). Find the radius at which local velocity is equal to mean velocity.
SAQ(16).If Reynolds no =1280 , find the friction for laminar flow through pipe.
SAQ(17).Pressure drop over a length in a pipe in which the flow is laminar, is
2 KN / m2 .Find the power lost, if the ratio of flow is 2 lps.
E310/1 161 SAQ(18). Given the pressure drop over 20m of pipe is 200 KN / m2 in a pipe of 10 cm
diameter in which the flow is laminar. Find the shear stress at wall.
WE(3). Under oil of dynamic viscosity 1.5 poise and specific gravity 0.9 flaws through a
20 mm diameter vertical pipe .Two pressure gauges have been fixed at 20m apart. The
pressure gauges fixes at higher level reads 200 KN / m2 and that at lower level reads 600
KN / m2 .Find the direction of flow and rate of flow. Verify whether the flow is laminar.
Then find the power lost in overcoming the friction.
Solution Taking datum through A
HA = Total energy at A
= pAw
+ zA + vA2
2g
=600 ×10000.9 × 9810
+ 0 +vA
2
2g
= 67.95 +vA
2
2g
Fig.
HB = Total energy at B
= pBw
+ zB + vB2
2g
=200 ×10000.9 × 9810
+ 20 +vB
2
2g
= 42.65 +vB
2
2g
E310/1 162 Q VA = VB ,HA > HB
∴ flow is from A to B (upward).
Loss of head is hf = HA - HB
= 67.95 +VA
2
2g
− 42.65 +
VB2
2g
= 25.30 m.
Assuming laminar flow
hf = 25.30 = 32µLvwD2 =
32 × 1.510
× 20 × v
0.9 × 9810 × 0.02( )2
= 27.18 v.
∴ v =25.3027.18
= 0.93 m / s.
Q =π4
0.02( )2 × 0.93 = 2.92 × 10−4m3 / s
RN =ρVD
µ=
0.9 × 98109.810
×0.93 × 0.02
0.15
= 111.6 < 2000
So the flow is laminar .
Power lost = WQ Hf =0.9 × 9810 × 2.92 ×10−4 × 25.3
= 65.3 Watts.
8.5. Turbulent flow- Smooth and rough pipes
We know that for flow through pipes when Reynolds no is > 4,000 the flow is
turbulent. In this case analytical treatment is very different due to fluctuations in the
velocity in the turbulent flow and so in the deviation of velocity distribution over the C.S
of flow, experimental values are to be considered. Here due to mixing the velocity
distribution is almost uniform over the C.S of flow as shown in Figure.
E310/1 163
Fig.
Due to mixing there will be some change in momentum which offers shear
resistance to flow. In turbulent flow there will be fluctuations in velocity. Let in 2.D flow
the fluctuations in velocity be Vx and Vy in x ans y direction respectively. Now if a fluid
mass ρAvy while moving over C.S of flow ‘A’ in ‘y’ direction fluctuates with velocity Vx
in x direction. Therefore the momentum change is ρAVyVx which is equal to shear stress
over C.S area A.
∴Shear stress τ =Forcearea
=ρAvx vy
A= ρvxvy
The transverse distance in which the fluctuating velocity of a lamp is equal
to mean velocity is called as mixing length ‘l’ by Prandtle in 1925. This fluctuating
velocities are related to mixing length by Prandtle as follows.
vx = ldvdy
= vy
Where v is the mean velocity.
Here vy is of the same order as vx.
∴Shear stress in turbulent flow is ρvxvy
τ = ρl2 dvdy
2
Further Prandtle assumes the mixing length proportional to ‘y’ transverse distance
from boundary i.e.,
l ∝ y
or l = k y
Where k is called Karmen’s constant ‘Roppa’
So τ = ρk2y2 dvdy
2and for small values of y , τ = τ 0
E310/1 164
∴τ 0 = ρk2y2 dvdy
2
τ0p
= k2y2 dvdy
2
dvdy
= 1ky
τ oρ
= v0ky
where vo =Shear velocity = τ0ρ
Now by integrating we have
v =v0k
loge y + C -------------------A
which is a logarithmic velocity distribution in turbulent flow.
Smooth and rough pipes
In fluid flows over a boundary, Prandtle recognises that there would be a small
layer adjacent to the boundary in which velocity varies gradually from zero at boundary to
free stream velocity (undisturbed velocity), as shown in Figure.
Fig.
This layer is known as boundary layer. The thickness of this layer normal to the flow , at
which the velocity variation is with 1% of free stream velocity, is called as boundary layer
thickness ‘S’. When the flow in the boundary layer is of laminar nature , then it is called as
laminar boundary layer and if turbulent nature , it is called as turbulent boundary layer .
But very near to the boundary even in turbulent flow , in a very thin layer fluid can not
maintain normal velocity component and so the flow will be always laminar. Then this
thin layer is turbulent boundary layer, adjacent to the boundary is called as laminar
sublayer whose thickness is ‘s’.
E310/1 165
Fig.
But as shown in fig any boundary surface will have certain irregularities rendering the
surface a roughness with average height ‘k’ where these roughness projections ‘k’ are
complete by submerged by laminar sublayer then the flow is not affected by the surface
roughness. Such pipe is called a hydraulically smooth pipe . If the roughness projections
project into the flow by penetrating the laminar sublayer , then such pipe is called as
hydraulically roughness pipe as the flow is disturbed by the roughness projections.
Velocity distribution in smooth and rough pipes:
We have already obtained an expression for the velocity distribution over a cross
section of flow over a boundary , in turbulent flow is given by v =v∗k
loge y + c
But at y = 0 , v is equal to ‘-α‘ indirectly that the velocity must be zero at certain distance
y’ normal to the boundary as shown in fig.
Fig
i.e., at y = y’ v = 0
∴ = −
C
vk
yelog '
So the velocity distribution is given by
E310/1 166
v =v∗k
logeyy'
Of y’ is known for both smooth and rough pipes, then the velocity distribution is
known for turbulent flow through pipes.
Before arriving at this distance y’ , we have to remember that there is a laminar
sublayer in the turbulent flow , adjacent to the boundary , in which the flow is laminar
type. In this layer then the velocity distribution is parabolic . But as the layer is very thin it
can be considered as linear with y.
Fig.
As shown in fig there will be a transition layer normal to the surface in between
laminar flow , in which the velocity distribution varies gradually from logarithmic to
parabolic. But in the absence of clear cut demarcation between each zone, the
intersection point of these two velocity distributions can be assumed to be the laminar
sublayer itself.
Velocity distribution laminar sublayer:
We know that the shear stress at boundary in laminar flow is given by
τ0 = µ vy
E310/1 167 ie., v = τ0
yµ
= ρτ0ρ
yµ
where τ 0ρ
= v∗
vv∗
= ρv∗yµ
=v∗y2
Velocity distribution in smooth pipes
By Nikaradse’s(student of Parandtle) experiments on flow through sand coated
pipes , the parameter v yvs = 116. at y =δ’
i.e.,v∗δ '
v= 11.6
or δ ' =11.6v
v∗− − − − − − − (1)
and at y = y' ,v∗y'
v= 0.108
i.e., y' =0.108v
v∗− − − − − −(2)
But we have from equation (1)
vv∗
=δ'
11.6
substituting y = g this in equ (2) We have y’ =0.108 ×δ’ / 11.6=δ’ / 107
Substituting this value of y’ in the velocity distribution
v =v∗k
logeyy1
, we have using k = 0.4
v =2.30.4
v* log10v∗y
0.108v
vv∗
= 5.75 log10v∗yv
+ 5.75log10(9.254)
= 5.75log10v∗y
v
+ 5.5
This equation is known as Karman -Prandtl equation for velocity distribution for
hydrautically smooth pipes. The velocity distribution for smooth pipe may also be given
E310/1 168 by an exponential equation empirically for v*y / v between 70 and 700 as
vv∗
= 8.74v∗yv
17
Velocity distribution in rough pipes
Nikuradse and others found by experiments that the roughness height k ∝y| and
y|= k / 30.
Now substituting k = 0.4 and y| = k / 30 in v =v∗k
logey′y
we have
vv∗
=2.30.4
log1030y
k
= 5.75log10yk
+ 5.75log10(30)
vv∗
= 5.75log10yk
+ 8.5
which is known as Karman-Prandtl equation for velocity distribution in rough pipes.
Criteria for smooth and rough pipes
This criteria depends on relative magnitude of laminar sub layer and roughness
hight, i.e., k′δ
=k
11.6vv∗
=v∗kv
×1
11.6
i.e.,k′δα
v∗kv
By Nikuradses experiments it was found that when v∗kv
≤ 3 or k′δ≤ 0.25 the pipe is
hydrautically smooth, when 3 <v∗kv
< 70 or 0.25 ≤k′δ
< 60 then the flow is transition
and when v∗kv
≥ 70 or k′δ≥ 6.0 the pipe is called as hydraulically rough pipe.
Mean velocity for smooth and rough pipes:
To get the mean velocity discharge is to be calculated base on the velocity
distribution as explained below.
E310/1 169
Fig.
Consider an elemental ring of ‘dr’ thickness at radius ‘r’ in a pipe of radius R.
The discharging following in this elemental area is dQ = 2Πr dr v.
Q = 2πr v∗ 5.75log10v∗yv
+ 5.5
0
R
∫
is the discharge through smooth pipe.
But y = R - r.
∴Q = 2πv∗ 5.75log10v∗(R − r)
vrdr + 5.5rdr
0
R
∫
So mean v =Q
πR2 = v∗ 5.75log10v∗Rv
+1.75
i.e., v
v∗= 5.75log10
v∗Rv
+1.75
is mean velocity for smooth pipe.
Similarly using equation for velocity distribution for rough pipes , the mean velocity in
rough pipes is given by
v = 1πR2 2πrv∗ 5.75log10
(R− r )k
+ 8.5
0
R
∫
dv
vv∗
= 5.75log10Rk
+ 4.75
Subtracting these mean velocity distribution in the respective local velocity distribution we
have
v − vv∗
= 5.75log10yR
+ 3.75 is identical to both smooth and rough pipes.
At y = R , v = vmax
Then for both smooth and rough pipes we have
E310/1 170
vmax − vv∗
= 3.75
v − vmaxv∗
= 5.75log10yR
Friction factor for smooth and rough pipes:
By dimensional analysis we have fraction factor in turbulent flow is function of RN
and relative magnitude roughness height to diameter of pipe , i.e.,
f = φvDv
k
D
(a) For laminar flow f = 64 / RN for RN ≤2000
(b) For turbulent flow:
(i)Blasius developed
f =0.316
RN( )1/ 4 for RN =4000-105
(ii) For RN > 105 ‘f’ is developed as below. The velocity distribution for smooth pipes is
vv∗
= 5.75log10v∗R
v
+1.75
Substituting v∗ = v f8
in above
v
v f8
= 5.75log10v f
8 R
V
+1.75
1f
= 2.03log10 RN f( )− 0.91
But by Nikuradse experiments , it is corrected as
1f
= 2 log10 RN f( )− 0.8 for RN = 5 ×104 − 4 ×107
This is known as Kurmon-Prandtl resistance equation for smooth pipes, which is to be solved
by Trial and error .
Further Nikuradse gave another equation for the same as
f = 0.0032 +0.221
RN( )0.237
(c) For rough pipes friction factor is obtained as below.
E310/1 171 The velocity distribution for rough pipes in turbulent flow is
vv∗
= 5.75log10Rk
+ 4.75
Substituting v∗ = v f8
, in above we have
v
v fRk
8
5 75 4 7510=
+. log .
( )12 03 0 9110f
R fN= −. log .
which is corrected by Nikuradse’s data as
1f
= 2 log10Rk
+1.74
which is known as Kurman-Prandtl resistance equation .
Thus in smooth pipes ‘f’ depends on Reynold’s no. only where as in rough pipes it is
independent of RN but depends on (R / k) only. So criterion for smooth and rough pipes is
RN f
Rk
< 17 smooth pipes
> 400 Rough pipes
=17-400 Transition pipes.
For any commercial pipe , the friction factor ‘f’ can be obtained by L.F Moody, diagram
shown below on log-log scale , knowing RN and relative magnitude of radius of pipe ‘R’
or roughness height ‘k’ i.e., (R / k). The roughness height of commercial pipes can be
obtained by equating the loss of energy of both sand coated pipes and commercial pipes.
E310/1 172
Fig.
Identify the following as true or false:
SAQ(19) In turbulent flow the velocity distribution is logarithmic distribution.
SAQ(20). If laminar sublayer submerges the roughness , then it is called as smooth pipe, if it
projects out of the sublayer, it is called as rough pipe.
SAQ(21). In turbulent flow, the velocity is zero at certain depth from boundary for
logarithmic velocity distribution.
SAQ(22). The velocity distribution is zero at y| = S| / 107 where S| is the laminar sublayer
thickness, in smooth pipes.
SAQ(23). For rough pipes roughness value k = 3. y ‘ where y| is the depth of flow from
boundary at which velocity is zero.
SAQ(24). The criterion to distinguish between smooth and rough pipes, depends on relative
magnitude of laminar sublayers and roughness height.
SAQ(25).The difference of local velocity and mean velocity in turbulent flow is identical for
both smooth and rough pipes.
SAQ(26). Difference of Maximum and mean velocities in turbulent flow through pipes is
3.75×v* .
SAQ(27). Friction factor in turbulent flow is function of RN and R / k, relative magnitude of
radius and roughness height.
E310/1 173 SAQ(28). For rough pipes ‘f’ is independent of RN.
SAQ(29). For smooth pipes ‘f’ is dependent on RN but not on roughness height of pipe.
SAQ(30). Same pipe will behave smooth and rough if v is varied.
SAQ(31). Given v* =0.5 m / s in a turbulent flow v = 0.01 stokes, roughness height k=1.0 mm
determine whether the flow is smooth or rough.
WE(4) For turbulent flow in pipes , show that vmax
v= 1.33 f +1
Sol:
The velocity distribution for turbulent flow in pipes is
v − vv∗
= 5.75log10yR
+ 3.75 for both smooth and rough pipes.
Now at y = R , v = v max
sovmax − v
v∗= 3.75
substituting v∗ = v f8
vmax − v
v f8
= 3.75
vmaxv
= 3.75 f8 +1
vmaxv
= 1.33 f +1
But by Nikuradse experiments
vmaxv
= 1.43 f +1
WE(5) A turbulent flow of water is flowing in a pipe of 10 cm diameter with roughness height
0.5 mm with a mean velocity of 10 m / s. Given v = 0.01 stokes. Find whether the flow is
smooth or rough. If it is rough for what velocity it will behave like smooth pipe. f = 0.02 .
Find the decmeter of pipe.
Solution:
Criterion for smooth and rough pipe is v∗kv
E310/1 174
v∗ = vf8
= 10 ×0.02
8= 0.5m / s
So v∗kv
=0.5 ×100 × 0.5
0.01= 250 > 70
∴The pipe is rough pipes.
So for smooth pipe v∗kv
≤ 3
or v∗ =3 × v
k=
3 × 0.010.05
= 0.6 = 0.05 × v cm / s
v =0.6
0.05= 12cm /s.
Further for smooth pipes
RN f
Rk= 17
i.e., RN f = 17 × Rk
But for smooth pipe
1f
= 2.0 log RN f − 0.8
= 2.0 log17 × R
k
− 0.8
1
0.02= 2.0 log
17 × R0.05
− 0.8
Where R is in cm.
7.07 = 5.06 - 0.8 + 20 log (R)
= 4.26 + 2 log ( R )
log R = 2.8
R = 25 cm.
WE(6)
The velocities in a 30 cm pipe carrying oil are 4.5 m / s and 4.2 m / s on the central line
and at a radius of 5 cm from the axis. Calculate the discharge and shear stress at boundary.
Solution:
For smooth and rough pipes
v − vv∗
= 5.75logRy
E310/1 175
4.5 − 4.2( )
v∗= 5.75log
0.150.15 − 0.5
=1.01252.
v* =0.296 m / s = v f8
But mean velocity is related to vmax as vmax − v
v∗= 3.75 for both the smooth and rough pipes.
Now at y = R , v = vmax
So vmax − v
v∗= 3.75
Substituting v∗ = vf8
vmax − v
v f8
= 3.75
4.5 − v0.2963
= 3.75
∴v = 3.389 m / s.
Q =πD2v
4
=π4
0.3( )2 × 3.389
= 0.2395 m3 / s.
Further v* =v f8
0.2963 = 3.389 f8
f = 0.0612
Shear stress at boundary τ0:
v∗ = τ0ρ = 0.2963
τ0 = (0.2963)2 ×ρ
= (0.2963)2×1000 N / m2
= 87.79 N / m2
E310/1 176 WE (7). A 1000m long pipe line of diameter 0.3 m carries oil at the rate of 540 lps. of the
specific gravity of oil is 0.8 and Kinematic viscosity is 0.023 stokes, determine the loss of
head, max velocity and at 10cm from axis and the maximum roughness height uph which
it behaves like smooth pipe.
Solution:
v =QA
=540 ×10154
0.3( )2= 7.64 m / s.
RN =vDv
=7.64 × 0.3
0.023 ×10−4 = 9.97 ×105
For smooth pipe
1f
= 2.0 log10 RN f( )− 0.8
=2.0 log10 9.97 ×105 f( )− 0.8
=2.0 log10 f +11.2
Solving by Trial and error
f = 0.0116
For smooth pipe
RN f
Rk≤ 17
Rk
>RN f
17=
9.97 ×105 × 0.011617
= 6317.65
k =R
6317.65=
0.15 ×10006317.65
= 0.0237mm
Loss of head hf =fLv2
2gd=
0.0116 ×1000 × 7.64( )2
2 × 9.81 × 0.3
= 115.0 m
Shear velocity v* = v f8
= 7.64 0.0116
8= 0.29 m / s.
Maximum velocity for smooth pipe:
E310/1 177
vv∗
= 5.75log10v∗y2
+ 5.5
at y = 0.15 v = vmax
vmaxv∗
= 5.75log100.29 × 0.15
0.023 ×10−4
+ 5.5
= 5.75×4.278+5.5 = 30.0
vmax = v* ×30 = 8.7 m / s.
Velocity at r = 10 cm ,
y = R - r
=15 - 10 = 5
vv∗
= 5.75log100.29 × 0.05
0.023 ×10−4
+5.5
= 21.84 + 5.5 = 27.348
v = 7.9 m / s.
WE(8):
In problem (7) what should be roughness height . If the pipe behaves like rough
pipe. Then determine the maximum velocity and loss of head.
Solution:
RN fR
k
≥ 400
f ≥400 R k( )
9.97 ×105 ≥ 40.12 ×10−5 Rk
For rough pipes
1f
= 2.0 log10R
k( )+1.74
E310/1 178
140.12 ×10−5 R
k( )= 2.0log10R
k( )+1.74
Solving by trial, R / k = 363.3
k =0.15363.3
= 4.13 ×10−4 = 0.413mm
∴ f = 40.12 ×10−5 0.154.13× 10−4
= 0.1457
f = 0.0212
hL =fLv2
2gd=
0.0212 ×1000 × 7.642
2 × 0.3 × 9.81= 210.4m
Max velocity :
We know velocity distribution in rough pipe is
vv∗
=v
v f8
=v
9.64 0.02128
=v
0.393= 5.75log10
yk
+ 8.5
v = vmax at y = 0.15.
vmax
v∗= 5.75log10
0.154.13× 10−4
+ 8.5
= 14.72 + 8.5 = 23.2
∴vmax= 23.2 ×0.393 = 9.12 m / s
Summary:
(1). For laminar flow the velocity distribution is v =1
2µ−∂p∂x
By − y2( )for parallel plates at
rest.
The mean velocity =B2
12µ−∂p∂x
Max velocity = B2
8µ−∂p∂x
∴ v =23
vmax
Loss of head hf =12µvLωB2
E310/1 179 Shear stress at boundary
τ0 =12
−∂p∂x
B
(2) The velocity distribution in couette flow is
v =vyB
−1
2µ∂p∂x
(By − y2 )
Which is a super position of simple couette flow on laminar flow through stationary plates.
The non dimensional form of velocity distribution for general couette flow is
vV
=yB
+ P 1 −yB
yB
where P =B2
2µ2 −∂p∂x
where V is the velocity of top plate.
(3). For laminar flow through pipes the velocity distribution is
v =1
4µ−∂p∂x
R2 − r2( ) which is a parabolic distribution.
vmax =1
4µ−∂p∂x
R2
v =vmax 1 −rR
2
vav = 1
32µ−∂p∂x
D2
So vav = 1 / 2 vmax
Loss of head hL =32µvLωD2
and f = 64 / RN
Shear stress τ = −∂p∂x
r2
(4) Laminar sub layer thickness S’’ =11.6v
v∗
where v* = shear velocity = vf8
(5) Criterion for smooth and rough pipe is
(1) k / S| ≤0.25 smooth pipe
E310/1 180 k / S| ≥ 6.0 rough pipe
(2) v∗kv
≤ 3 smooth pipe
≥ 70 rough pipe
(6)
(a) Velocity distribution for smooth pipe v / v* =5.75 log10 (v∗yv
)+5.5
rough pipe = v / v* = 5.75 log10(y / k)+8.5 known as Kurman -
Prandtl equations.
(b) Mean velocity for
Smooth pipe v
v∗= 5.7log10
v∗Rv
+1.75
Rough pipe v
v∗= 5.75log10
Rk
+ 4.75
v − vmaxv∗
= 5.75log10yR
for both Smooth and rough pipes
and v − vmax
v∗= 5.75log10
yR
+ 3.75 for both smooth and rough pipe.
(7) Friction factor ‘f’
(a) smooth pipe
1f
= 2 log10(RN f ) − 0.8
(b) For rough pipe
1f
= 2 log10Rk
+1.74
(8)RN f
Rk
≤ 17 smooth pipe
≥ 400 rough pipe
Answers for S.A.Q.
1) - 4) True 5) 2.0 K N / m2 6) - 9) True
10) y = 0.051 11)500 n / m 2 12) 0.5 m / s
13) 0.4 m / s 14)0.42 m / s 15)r = 7.07 cm
E310/1 181 16) f = 0.05 17) 4 N m / s 18) τ0 =250 N/m2
19)- 30) True 31) Rough.
Exercise
8.1) An oil of specific gravity 0.92 and dynamic viscosity of 0.082 pose flows in an 80 mm
diameter pipe. In a distance of 20m the flow has a heat loss of 2m. Calculate (1) The mean
velocity , discharge velocity and shear stress at a radial distance of 38 mm from the pipe
axis and boundary shear stress.
[v =2.197 m / s , Q = 11.04 l/ s
v = 0.4284 m / s , τ = 17.114 Pa τ0 = 18.02 Pa]
8.2) What power will be required for kilometre length of a pipeline to overcome viscous
resistance to the flow of an oil of viscosity 2.0 poises through a horizontal 10 cm diameter
pipe at the rate of 200 l/ min? Find the Reynolds number of the flow if the relative density
of the oil is 0.92.
[ RN = 194.8 , P = 0.905 KW]
8.3) A flow of 60 L / s per meter width of glycerine of specific gravity 1.25 and dynamic
viscosity 1.5 poises takes place between two parallel plates having a gap of 25 mm
between them. Calculate the (1)maximum velocity (2) Boundary shear stress and (3)
Energy gradient
[ Vmax = 3.6 m / s , τ0 =864 Pa hf / L = 5.648]
8.4) Two parallel plates are placed horizontally 10 mm apart. The bottom plate is fixed and
the top plate is moved at a uniform speed of 0.25 m / s. The fluid between them has a
dynamic viscosity 1.472 N S / m2 . Determine the pressure gradient which corresponds to
the condition of zero discharge between the plates and the shearing stress at each plate.
[22.08 K N / m2 / m , -73.6 N / m2 , 147.2 N / m2]
8.5) A smooth pipe line 0.1 m in diameter and 1000m long carries water at the rate of 7.5 lps
of the kinematic viscosity of water is 0.02 stokes. Calculate the head loss , wall shear
stress , centreline velocity ,shear stress and velocity at 40 mm from the centreline and the
thickness of laminar sublayer.
[9.95m ,2.44 N/m2 1.15 m /s 1.95 N /m2
0.95 m /s ,0.47 mm]
8.6). A pipe of diameter 0.3m is to convey water at 40 oC at the rate of 200 lps of the power
required to maintain the flow in 100m length is 61.8 KW, Calculate the value of k, vmax .τ 0
and v*. Take v = 0.0075 stokes (at 40 oC for water)
E310/1 182 [0.57 mm ,3.4m / s ,23.18 N / m2 , 0.152 m /s]
8.7). A 300 mm diameter pipeline carries water at 20 oC with a mean velocity of 7.5 m / s .
The pipeline is new with no surface irregularities at the beginning , but it was found that
the surface irregularities grow at the rate of 0.075 mm per year. Find the number of years
after which the surface irregularities will affect the flow. Take v = 0.01 stokes.
[2.67 years]
8.8). Field tests on a 30 cm cast pipe carrying 0.25 m3/s of water [v = 1×10-6 m2/s] indicate
that the height of roughness projections has incurred to 1.5 mm after many years of
service. What increase in flow can be expected if the pipe is replaced by a new pipe with k
= 0.26 mm of the same diameter .
[40 lps].
*****
Unit 9
Open Channel flow Aims: The aims of this unit are to define open channel flow, types of flow, to explain velocity distribution over the cross section of flow, energy and momentum correction factors, to review velocity equations, in uniform flow and to analyse economical sections. Objectives :
1. To define open channel flow and compare it with pipe flow. 2. To explain types of flow in open channel. 3. To discuss about velocity distribution over cross section of open channel flow. 4. To calculate energy and momentum of fluid flowing in open channels and then to obtain their correction factors. 5. To review uniform flow and to obtain expressions for velocity of flow by chezy, manning and Bazio. 6. To define economical section and to obtain conditions for economical sections of rectangular, trapezoidal, triangular and circular open channels. 7. To apply the above to solve practical problems. 9.1 Introduction:
Like pipe flow, open channel flow is another important branch of Hydraulics, which is very useful to civil engineers. The flow of rain water in streams, rivers comes under open channel flow. In Irrigation the application of theory of open channel flow is very much necessary. So study of open channel flow is very important to civil Engineer without which he can not be a successful civil Engineer.
E310/1 183 9.2. Definition of Open Channel flow Open channel flow is defined as that flow which takes place in a passage with a free surface subjected to atmospheric pressure. Here the water surface is exposed to atmosphere.
Figure
When this flow is compared to the flow through a pipe it is seen as shown in figure. Here the bottom of open channel refers to centre line of the pipe, the free surface is similar to hydraulic gradient, and the total energy is same in both the cases over the datum. In open channels the bed will be slopped down towards the direction of flow, to make the gravity force (component of weight of water) to overcome resistance and to cause the flow of water. Channels are broadly classified into natural and artificial channels. Rivers streams etc are examples of natural channels. Artificial channels are made artificially to carry water with different cross sections like rectangular, trapezoidal, triangular, parabolic and circular sections. Closed conduits flowing partially, with free surface exposed to atmospheric pressure are also called as open channels. Under ground drains are the examples of closed conduit open channels. A channel which has the same shape of cross sections along its length is called as prismatic channel, otherwise it is called as non prismatic channel.
Identify the following as True or False SAQ1 When water surface in a passage exposed to atmospheric pressure it is called as open channel flow. SAQ2 In open channels the flow is gravity flow SAQ3 If the closed conduit running helpful it is called as pipe flow 9.3 Classification of Open Channel flow i. Steady and Unsteady flow If the flow characteristics like velocity depth do not vary with time at any given cross section of the channel, then it is called as steady flow
E310/1 184 ie
dvdt
= 0,dydt
= 0
In prismatic channels since cross section is constant along the length, then the flow is
steady if dydt
= 0
If velocity and depth vary with time at any cross section of channel then it is called as unsteady flow is
ie dvdt
≠ 0 and dydt
≠ 0
Flow in a river during rainy season is unsteady. ii. Uniform and non uniform flow (varied flow) When depth slope, cross section and velocity do not change along a given length, then the flow is called as uniform flow.
ie dyds
= 0, dϑds
= 0, dAds
= 0, dSds
= 0
So uniform flow occurs only in prismatic channels. If the above flow characteristics vary along the length of channel then it is called as non uniform (varied) flow.
ie dy/ds ≠ 0 etc Further varied flow is classified as gradually varied flow and rapidly varied flow. If the depth of flow varies abruptly over a short length of channel, then it is called as rapidly varied flow (R.V.F). Hydraulic jump is the example of R.V.F. iii. Laminar flow and Turbulent flow Flow in open channel can also be classified as laminar, transition and turbulent flows, like in pipe flow, based on Reynolds number which is given by
RN =ρVR
µWhere R = Hydraulic Mean radius
V = Velocity of flow By experiments it is shown that when RN = 500-600 , the flow is laminar in open channels and RN > 2000, the flow is turbulent flow and between 500-2000 the flow in transitional flow. iv. Subcritical, critical and super critical flow
Based on relative magnitudes of gravity force and inertia forces, the flow in channel can be classified as subcritical, critical and super critical flow. The relative magnitude of gravity force and inertia force is given by Froude Number which is square root of ratio of Inertia force to gravity force ie
E310/1 185
ie FN =Inertiaforcegravityforce
(ρL3 VT ) (ρL3 g)[ ]
12
= (ρL2 LT v
(ρL3g)
12
= ρL2 v2
ρL3g
12
=vLg
where L is taken is depth of flow = y
ie FN =vgy
If v< gy , FN <1, then the flow is called as subcritical or tranquil or streaming flow. If v= gy , FN =1, then the flow is called as critical flow.
and of V> gy FN >1, then the flow is called as supercritical, rapid or shooting or torrential flow. Hydraulic jump occurs when the flow changes from super critical to sub critical flow. Identify the following as True or False. SAQ4 During floods, the flow in a river is unsteady. SAQ5 In a prismatic channel ifdischarge is constant, then the flow is uniform flow SAQ6 When mean velocity is equal to gy then the flow is called as critical flow. Geometrical Properties of Channel section
y = Vertical Depth of flow T = Top width of flow A = Cross sectional area of flow P = Wetted Perimeter which is in contact with water R = Hydraulic mean radius or depth = A/P D = Hydraulic depth = A/T
Z = Section factor = A D = A AT = A3
T( )12 for critical flow
Z` = AR2/3 for uniform flow 9.4 Velocity distribution in open channel
Due to free surface and frictional resistance of the boundary surface, the velocity distribution over the cross section of flow is non-uniform. The velocity distributions as measured by pitot tube on various cross sections are shown in figure which are function of shape of section, roughness of channel and bends in the channel.
E310/1 186
The maximum velocity will occur at a depth equal to 0.05 to 0.15 y from free surface. The mean velocity can be computed from the velocity distribution and it is equal to local velocity at a depth o.6y from the free surface. A better approximation for the mean velocity is equal to average of velocities at 0.2 depth and 0.8 of depth from free surface.
ie Vmean = 12
(vel ab 0.2y+vel at 0.8y)
Due to non-uniform velocity distribution over the cross section, the computation of kinetic energy and momentum of flow based on mean velocity are to be corrected by multiplying those by ‘α’ kinetic energy correction factor and ‘β’ momentum correction factor respectively as shown below. α - Kinetic energy correction factor (coriolis coefficient) Kinetic energy based on velocity distribution =
------ do----- based on mean velocity = α12ρAv
v 2 = α
12ρAv3
∴α =ϑ 3
A∫ dA
Av 3 > 1
for turbulent flow α = 1.03 - 1.36β - Momentum Correction factor (Boussining co efficient)
β =ρϑdAϑ
A∫ρVAV
=ϑ 2 dA
A∫
AV 2
β = 1.01 - 1.12 for turbulent flow But generally for turbulent flow , these are taken as unity.
Identify the following as True or False SAQ7 Maximum velocity occur at a depth of 0.05 - 0.15 of depth from free surface. SAQ8 Mean velocity is equal to local velocity at a depth of 0.6 depth from free surface SAQ9 Kinetic and momentum correction factors are taken as unity in turbulent flow.
E310/1 187 9.5 Uniform flow in open channel
Review : We know already that the depth of flow wetted area, velocity and discharge are constant along the prismatic channel in case of uniform flow. So the water surface , the bed of channel are parallel to total energy line. Consider a steady and uniforms flow in a prismatic channel of length L is shown in figure
Figure
The forces acting are 1. Gravity force which ;is equal to the weight component in the direction of flow = w sinθ2. Frictional resistance given by τox PL where τo - shear resistance per unit surface area. 3. Hydrostatic forces p1 & p2 which are equal and opposite as depths y1 = y2 in uniform
flow Now as the flow is steady ∑F = m × a = 0 --- acceleration = 0 in steady flow ∴w sinθ - τ .PL + p1-p2 = 0w sinθ = τ.PL This shows that the gravity force is equal and opposite to frictional resistance in uniform flow w=wΑL
∴τ o =WALsinθ
PL= W
Ap
sinθ = ωRsinθ
But sinθ ≈(z1 − z2 )
l= s. bed slope = tan θ
for small angle ‘θ’∴ τ o = w R So
But τ0 = f8 ρv 2 by pipe flow analysis
ie τ 0 = wRSo = f 8ρv2
E310/1 188
v = 8wpf
Rs0 . = C Rs0
where C = 8gf
is called as chezy’s coefficient
and V= C Rs o is called chezy’s equation. for mean velocity in uniform flow in open channels. Here C is inversely proportional to ‘f’ Darcy weisback coefficient of friction.
The dimensions of C = L12 T −1 and so C is not constant and varies for each system of unit.
It is simple but its determination is difficult. Further by applying Bernoulli’s equation between (1) and (2) , we have Z1+ y1 + V1
2 /2g = Z2 + +y2 + V22 / 2g + hf
But y1 = y2 V1=V2 in uniform flow ∴ hf = Z2 -Z1
slope of TEL = z2 − z1
L=
hf
L= s f = so = sw
Thus all the slopes are parallel to each other in uniform flow. The depth of flow in uniform flow is called as normal depth ‘yn’
To determine C Empirical formulae have been developed to find C as given below. a. The Ganguillet - Kutter formula
Based on flow measurement, in open channels above two Swiss engineers proposed in 1869 an empirical formula to find C in M.K.S as
C = 23 +
0.00155s
+1n
1 + (23 + 0.00155s
) nR
where n- Kutter’s roughness coefficient = Manning roughness Coefficient n - depends on channel surface and its condition. The typical values of n for different surfaces are given below channel surface value of n 1. Very smooth concrete, planed wood 0.012 2. Ordinary concrete lining 0.013 3. B.W lined with CM
E310/1 189 7. Unlined earth channels in good condition 0.02 8. Rubble masonry 0.02 9. Rivers and earth Channels in fair condition 0.025 10. Earth channels with gravel bottom 0.025 11. Earth channels with dense weed 0.035 12. Mountain stream with rock bed 0.045 b. Bazin formula He proposed in 1897 the following formula to find C in M.K.S as
C =157.6
1.81 + mR
where ‘m’ is proposed by Bazin as below
channel surface value of mVery smooth cement , planed wood 0.11 Concrete, brick or unplan ed wood 0.21 Ashlar, rubble masonry or poor B.W 0.83 Earth channels in very good condition 1.54 ” ordinary condition 2.36 ” rough condition 3.17
c. Manning’s formula In 1889 he proposed the following formula to find mean velocity in M.K.S as
v= 1n
R2/ 3 S1/2 which is very simple and given satisfactory results and is widely
used in practice
Here C = 12
R1 6
Identify the following as True or False
SAQ10 Water weight component in the direction of flow is equal to frictional resistance in uniform flow. SAQ11 Chezy’s Coefficient is in verbally proportional to square root of Darcy weisback Coefficient of friction ‘f’ SAQ12 In uniform flow bed slope, water surface slope and energy line slope are all parallel to each other. SAQ13. Chezy’s coefficient C is dependent on surface of channel and its condition SAQ14 Chezy’s coefficient is inversely proportional to Manning’s roughness coefficient
Worked Example (1) A rectangular channel conveys a discharge of 10 m3/s . If the width of channel is
6m, find the depth of flow if C = 54.62 and bed slope = 1
5000Sol:
E310/1 190 Q = AC RS
1.0 = b × yn × 54.626yn
6 + 2yn
1
50006yn
3
6 + 2yn
=10
6 × 54.62
2
× 5000
y3n = 4.655 + 1.552yn
yn = 1.95
Worked Example (2) A discharge of 100 l.p.s is flowing in a rectangular channel of 60 cm wide with a normal depth of 30 cm. Find the necessary slope if C = 56
A = 0.6×0.3 = 0.18 m2
velocity = Q A =0.1
0.18= 0.555 m s
p = b+2d = 0.6+2×0.3 = 1.2 m
Hydraulic mean depth R = A p =0.181.2
= 0.15m
But ϑ = c Rs0.555 = 56 × 0.15 × 3
s =1
1500
9.6 Economical Sections Economical section is that when maximum discharge can pass through it for a given cross section, roughness coefficient and bed slope. By this definition it is clear that discharge is maximum for a given cross sectional area when velocity is maximum since by continuity equation Q = AV. Velocity is maximum
when R is maximum because V = C RS or 1n
R2 3S 1 2 ie R = A/P , R is maximum
whenP is minimum So for given slope and roughness value V is maximum when P is minimum. Therefore section is economical when P is minimum for a given cross section, roughness coefficient and bed slope . By making use of this condition expressions can be obtained for different channels whose sections are to be economical. a. Rectangular channel.
Consider a rectangular channel of bottom width ‘B’ and depth of flow as y
Then P = B+ 2y, A = By
So p = Ay
+ 2y which is function of y only
E310/1 191 If p is to be miximum
dpdy
= 0
ie dpdy
= −Ay2 + 2 = 0
A = 2y2
By = 2y2
B = 2y or y = B 2is the condition for rectangular channel to be economical Further Hydraulic mean radius R = A/P
R =By
B + 2y=
2y × y4y
=y2
ie R =y2
is another condition for rectangular channel to be economical
b. Trapezoidal channel
Consider a Trapezoidal channel with bottom width B and side slope Z : 1 as shown in figure with depth of flow as y Here the cross sectional area is constant The parameters involved are B,y, Z . So there will be three cases as explained below to get conditions for economical sections case (1) for given cross section are A, Z is constant and y is variable.
figure
A = y(B+YZ) and P = B+ 2Y 1 + Z 2
So B = Ay
− yZ
and then p2 =Ay
− yZ + 2y 1 + Z2
which is function of ‘y’ only
for P to be minimum dρ dy = 0
ie dpdy
= −Ay2 − Z + 2 1+ Z
2= 0
Ay2 + Z = 2 1 + Z
2
But substituting for A = y(B+yZ), we have
y(B + yZ)
y 2 + Z = 2 1 + Z2
B+2yZ = 2y 1 + z 2
ie 12
(top width ) = side slope is the condition for Trapezoidal section to be economical
E310/1 192 Further R = A/p =
y(B + yZ)B + 2y 1+ Z
2
But 2y 1 + Z2
= B + 2yZ So substituting this in above equation , we have
R =y(B + yZ)2(B + yZ)
=y2
ie Hydraulic mean radius = 12
depth of flow is another condition for trapezoidal section to
be economical. Further draw a perpendicular line OA from centre of Top width t side of the section as shown in figure.
figure.
OC = 12
top width = (B + 2yZ)
2sin θ =
yy 1+ Z
2 =1
1 + Z2
OA = OC sin θ =(B + 2yZ)
2×
11 + Z
2
But (B + 2yz)
2= y 1 + Z 2
ie OA = y 1+ Z1 + Z
2
2
= y
This means that Trapezoidal channel will be economical when a semi circle of radius ‘y’ with centre as mid point of top width will be tangential to both sides and bottom. So trapezoidal section will be economical for a given cross sectional area keeping B and Z as constant, when
1. 12
top width = side slope
2. Hydraulic mean radius R = y/2 3. Semicircle of radius y and mid point of top width as centre will be tangential to both sides and bottom. case 2. For given cross section B is constant here also we have
A = (B +zy)y p= B+ 2y 1 + Z 2
From (1) Z = A/y2 - B/y
E310/1 193 substituting this for p, we have
p = B+2y 1 +Ay2 − B y
2
dpdy
= 2 1 +Ay 2 −
By
2
+ 2y ×12
1
1 +Ay2 −
By
2 × 2Ay 2 −
By
−2Ay 3 +
By2
= 0
2 1 + (Ay2 −
By
)2
+ 2y
Ay2 −
By
−
2Ay3 +
By2
= 0
1 = ZAy 2
Z =y2
A=
y 2
(B + Zy)Y;
(B+Zy) = yZ
By
+ Z =1Z
,
By =
1Z
− Z =1 − Z 2
Z
ie By
=1 − Z 2
Z
are the conditions required for Trapezoidal section to be
&` A =y 2
Zeconomical when B is constant
case 3 Depth of flow y = constant here also we have A = (B+Zy)y A/Y-Zy=B p = B+2y 1 + Z 2
p =Ay
− Zy + 2y 1 + z 2
For p to be min dp/dz = 0
ie dpdz
= −y +2y
2 1 + z 22z = 0
2z = 1 + z2
1+z2 = 4z2
z= 13
or tanθ = 3
or θ = 600
So when depth is constant for a given cross sectional area, Trapezoidal section is economical when the inclination of slides with horizontal is 600.
c. Triangular channel
E310/1 194 Consider a triangular section with depth of flow as y and side slope Z=1. Let apex angle be 2θ.
figure
Then A =
12
2y tan θ y = y2 tan θ
p = 2y sec θ
but y = A
tanθ(fromA)
s0 p = 2A
tansecθ
for p to be minimum dp/dθ = 0
ie dpdθ
= 2 Asecθ tanθ
tanθ−
secθ2 tan 32 θ
sec2 θ
= 0
2 tan2 θ - sec 2 θ =0
sin θ =12
or θ = 450 ie Z = 1 So triangular section is economical when included angle is 900 i.e. side slope is 1:1 , i.e. half square on diagonal.
figure
Further R =Ap
=y2 tanθ2y secθ
=y sinθ
2
But sin θ=12
∴R =y
2 2
NOTE: Q = ACAp
s
E310/1 195 Q is maximum when A is ;maximum for a given p. Then the conditions will be same for triangular section to be economical as derived in the previous section , for a given p. d. Circular channel
In this case both cross sectional area and wetted Peri meter vary with depth of flow y. Hence in circular channels two conditions will be derived (1) for maximum dischargeand (2) for maximum mean velocity as arrived below. i. Condition for maximum discharge
Let in a circular channel of radius r, the depth of flow be ;y as shown in figure and ‘θ’ bethe angle subtended at centre by wetted perimeter. Then we have
A =Πr 2
2θΠ
− 2 ×12
2sinθ
22 cosθ
2
=r 2θ
2−
12
r 2 sinθ
=r 2
2(θ − sinθ )
p = r θ
Then Q = AC RS = CA Ap
s = c A3
ps
Then for given C & S, Q is maximum when A3/p is maximum
ie d A3 / p( )
dθ= 0
=p3A2 dA dθ − A3 dp dθ
p 2 = 0
but dA dθ =r 2
2(1− cosθ )
and dp dθ = rby substituting dA dθ , dp dθ in above equation we have
3p r 2
2(1− cosθ ) − Ar = 0
2θ-3θ cosθ + sinθ= 0ie = 3080
for maximum discharge depth of flow is y = r+r cos(180-θ/2) = r(1+cos 260)
= 1.8988 r ≈ 0.95D so the depth of flow is 0.95D for maximum Q and hydraulic mean radius is
R = A p =r 2
2(θ − sinθ) ×
12θ
E310/1 196 =
r(θ − sinθ )2θ
=r2
×1
308
=r2
×1
308180
× Π(Π
180308 − sin 308)
= 0.5733r 0.29D So for maximum discharge R = 0.29D The above conditions slightly vary if Manning’s equation is used for velocity i.e y = 0.938D Condition for maximum velocity
We have velocity = V = C RS= C A p s
So V is maximum when A/p is max
i.e d(A / p)
dθ=
pdA dθ − A dp dθp2 = 0
and substituting for dA/dθ as dp/dθ in above equation we have
rθ r 2
2(1 − cosθ ) −
r 2
2(θ − sinθ )r = 0
θ = tan θie θ = 257.50
So the depth of flow for maximum velocity is y = r+ r cos (180-θ/2) r[ 1+cos 51025’]
=1.626r = 0.81D So the depth of flow y = 0.81D for maximum velocity Further R = A/p
= r
2θ(θ − cosθ )
=r
2Π 257.5180
Π180
257.5 − sin 257.5
≈ 0.6086r = 0.3D is maximum velocity the hydraulic mean radius in 0.3D State whether the following are true or false.
SAQ15. When discharge passing through a given cross section is maximum, then it is called as economical. SAQ16. For a given cross section Q is maximum when p is maximum SAQ17 Width of a rectangular channel is 10 cm Find the depth of flow when it is to be economical. SAQ18. When for rectangular section hydraulic mean radius is equal to B/4, then it is economical.
E310/1 197 SAQ19. Half of Top width of Trapezoidal section is equal to side slope, then it is called as economical section.
SAQ20. When hydraulic mean depth is equal to 12
nthe depth of flow , then both
Trapezoidal and rectangular sections are economical SAQ21. The sides and bottom of a Trapezoidal section are tangential to a semi circle of radius equal to depth of flow and its centre being mid point of top width, then the section is economical . SAQ22. When depth of is kept constant, the trapezoidal section is economical if side slope is 600 with horizontal. SAQ23 When the slope of side with vertical is 45 0 then the triangular section is economical. SAQ24 In case of circular section analysis of economical section with cross section being constant is not possible. SAQ25 When depth of flow = 0.95 diameter, then circular channel is economical SAQ26. For maximum discharge to pass through circular channel the hydraulic mean depth is 0.29 diameter SAQ27 For circular channel Q is maximum when (A3/p) is maximum SAQ28 For circular channel velocity is maximum when (A/p) is maximum SAQ29 The mean velocity through a circular section is maximum when depth of flow = 0.81 diameter and hydraulic mean radius = 0.3 dia SAQ30 For a given cross section the economical section circular channel is semi circle. Worked Example (3) A lined rectangular channel with Manning’s n = 0.02, is 5m wide and the depth of flow is 2m with a bed slope of 1 in 1500 keeping the same rectangular shape of section wetted perimeter and slope, find the maximum extent increase in discharge. sol : A1 = 5×2 = 10 m2
p1 = 5+2×2 = 9 m
R1 = A1 /p1 =109
−1.11m
Q1 =12
A1 R1
23 so
12
=1
0.02×10 × (1.11)2 3 (
11500
)1 2
=13.8558 m3 /s For maximum Q, y = B/2 P = 2y+B = 2B 9 = 2B , B = 4.5 m, y = 2.25cm A = 4.5 ×2.25 = 10.125 m2
R =10.125
9= 1.125m
Q =1
0.210.125 × (1.125)2 3 1
1500
12
= 14.148m3 /s
∴ change in Q = 14.148 -13.8558 = 0.29242 m3 /s
E310/1 198 WE (4) Water is to flow in a channel at 12.5 m3 / s with a mean velocity of 1.25 m/s .Calculate the economical cross section of a). rectangular b). Triangular c). Trapezoidal and d). circular section. Which of these have least perimeter and maximum perimeter a). Economical rectangular channel
y = B/2, A = B× y = 2y × y = 2y2 ×
A =12.51.25
= 10 = 2y2 , ∴ y =102
= 5 = 2.236
So B = 2×y = 2 × 2.236 = 4.47 2m p = B+2y = 8.944 m b). Triangular section to be economical θ = 450
and A = 12
× 2y × y = y 2
10 = y2 y = 3.162m p = (2 × 2)y = 8.943m
c). Trapezoidal section to be economical side slopeθ =600 with horizontal
Z = 13
A= (B + 2zy) + B
2
y = (B + zy)y
(B + 2zy)2
= y 1 + z 2 = y 1+13
=2y
3
B + 2 ×13
y
2
=2y
3
B =23
y
So A = 23
y +13
y = 3y 2
10 = 3y2
y =10
3
12
= 2.4m
B =23
× 2.4 = 2.775m
p = 2y 1 + z2
+ B = 2y 1 +13
+23
y = 2 3y = 8.3m
d. Circular section is economical for a given cross section when Q = c A3 p s is maximum when p is minimum i.e
E310/1 199
P = dθ A= r 2
2(θ − sinθ ) =
d 2
8(θ − sinθ)
d =8A
θ − sinθ
∴ p =d2θ
=8A
(θ − sinθ )×θ
For P to be minimum, dp dθ = 0
θ − sinθ −θ
2 θ − sinθ(1− cosθ ) = 0
2(θ − sinθ ) − θ(1 − cosθ) = 0θ = Π
ie economical section is semi circle.
d=8A
θ − sinθ=
8 × 103.142 − 0
= 5.045m
p=rθ =5.045
2× 3.142 = 7.926m
∴The least perimeter is of circular one and highest perimeter is rectangular & triangular Worked Example (5) A power channel of Trapezoidal section has to be excavated through hard clay at the least cost. Determine the dimensions of the channel given Q = 14 m3 /s. Bed slope 1 in1500 and Manning’s n = 0.02
Side slope for best section is 600 with horizontal. So Z = 13
Trapezoidal channel to be economical we have
b + 2zd
2= y z 2 +1
b + 2 ×13
y
2= y y 3
ie b = 2∫3
y
and A = y(b+zy) = 3y2
Q = Aϑ = 14 = 3y 2 ×1
0.02y2
2 3 12500
1 2
∴y = 2.604m
b = 2∫3
y = 3.007m
Worked Example 6) A lined channel ( n = 0.014 ) is of trapezoidal section with one side vertical and the other on a side slope of 1.5 H to 1 v . If the channel is to deliver 9 m3 /s on a slope of
E310/1 200 0.0002 , find the efficient cross section which requires minimum lining. Find the corresponding mean velocity. Sol
figure
A = B +zy2
y
B = A y −zy2
and p = B + y + y Z 2 +1
=Ay
−zy2
+ y + y 1+ z2
=Ay
+ 1 −Z2
+ 1 + z 2
y
For p to be minimum, dpdy = 0
ie dp dy =−Ay2 + 1−
z2
+ 1+ z2
= 0
A = y2 (1− 1 −1.52
+ 1 +1.52
= 2.0528 y2
B =Ay
−zy2
=2.0528y2
y−
1.5 × y2
= 1.3028y
p = 1.3028 y +y+ 3.25y = 4.1056y
R = Ap =
2.05284.1056
y = 0.5y
Q =1n
AR2 3s 1 2
9 =1
0.0142.0528y 2 .(0.5y)2 3 (0.0002)
12
y = 2.062 m B = 2.687 m So A = 2.0528 y2 = 8.7282 m2
∴ Velocity v = QA
=9
8.7282= 1.031m / s
SUMMARY 1. Kinetic energy correction factor
α =ϑ 3dA
A∫
Av3 > 1, for turbulent flow
α = 1.03 - 1.36 and momentum correction factor
E310/1 201
β =ϑ 2
A∫ dA
Av 2 > 1
β for turbulent flow = 1.01 - 1.12 2. Chezy’s equation is v = C RS
C - Chezy’s coefficient R -Hydraulic mean radius S - bed slope for uniform flow bed slope = water surface slope = Energy slope
Manning’s formula v = 1n
R2 3S 1 2
n - Manning’s roughness coefficient and c = 1n
R1 6
4. Economical section is that when maximum discharge can pass through it for a given cross sectional area ie Q = Ac A pS is maximum when p is minimum for given A 5. For rectangular cross section it is economical when y = B/2, R = y/2
6. For Trapezoidal section it is economical when, for a given cross section A, with side slope constant
12
(Top width) = side slope
R = y/2 or when semicircle of radius y and mid point of top width as centre, will be tangential to sides and bottom. When bottom width constant , side slope variable, for given cross section, it is economical when
B y =1− z 2
zand A =
y2
zwhen depth y = constant,
it is economical when side slope θ = 600 with horizontal.
7. Triangular section is economical when the included angle is 900 , ie hole slope is 450 with vertical. 8. Circular section is economical when y = 0.95D for maximum Q R = 0.29D
y = 0.81D for maximum velocity R = 0.3D
and for given cross section A it is economical when it is semicircle Answers to SAQ
9.1 A trapezoidal channel base width 8 m and bed slope 1 in 400, carries water at 12 m3/s and side slopes are 1:1 , compute the normal depth. Take n = 0.025 [0.85 m ] 9.2 A trapezoidal channel has a bed width of 2.0 m side slope of 1.25 horizontal : 1 vertical and carries discharge of 9 m3 / s at a depth of ;2.0m . Calculate the average velocity and bed slope of the channel . Take n = 0.015 [ v = 1.0 m/s , s0 = 2.0554×10-4]9.3 A rectangular channel 3.0 m wide had a badly damaged living whose Manning’s ‘n’ was estimated as 0.025. The lining was repairedand it has now an n = 0.014. If the depth of flow remains the same as 1.3 m as before the repair, estimate the new discharge [Q = 4.894 m3 / s] 9.4 What diameter of a semi circular channel will have the same discharge as a rectangular channel of width 2.0m and depth 1.2 m ? Assume the bed slope & Manning’s n are the same for both the sections [ D = 2.396 m] 9.5 A triangular channel has a vertex angle of 75 0 and a longitudinal slope of 0.001 If mannigs n = 0.015 estimate the normal depth for a discharge of 250 lps in this channel [yn = 0.668m] 9.6 A rectangular channel [ n = 0.02] is 5.0 m wide and 0.9 m deep and has slope of 1 in 1600. If the channel had been designed to be of efficient rectangular section for the same wetted perimeter, what additional discharge could it carry [ ∆Q = 2.211
m3 / s] 9.7 A trapezoidal channel has one side vertical and the other side has a slope of 1.5 H ; 1V . This carries a discharge of 15 m3 /s with velocity of 1.5 m/s . Calculate the dimensions of an efficient section of this shape and also the bottom slope necessary to achieve this discharge [ n = 0.0130] [ y = 2.207n, B = 2.876m, s0 = 3.335 × 10-4 ] 9.8 Deter mine the dimensions of a concrete lined ( n= 0.014) trapezoidal channel of most efficient proportions to carry a discharge of 10.0m 3/s. The bed slope of the channel is 0.005 [ y = 1.25 m, B = 1.444m, side slope Z = 0.5773] 9.9 Determine the efficient section and bed slope of a trapezoidal channel ( n = 0.025) designed to carry 15 m3 /s of flow. To prevent scouring the velocity is to be 1.0 m/s and the side slope of channel are 1V : 2H [y = 2.463m, B = 1.163m, s0 = 4.735 × 10-4]
* * *
FLUID MECHANICS & HYDRAULIC MACHINERY
UNIT X
OPEN CHANNEL FLOW
E310/1 203 Aims:
The aims of this unit are to define specific energy, specific force and to derive conditions
for critical depth to compute critical depth and to apply this concept to channel
transistors.
Objectives:
1. To define specific energy and explain meaning of critical depth by means of specific energy curve.
2. To derive conditions for critical depth for (a). given discharge and (b) given
specific energy
3. To define momentum in open channel flow and obtain an expression for specific
force
4. To derive a condition for critical state of flow for (a) given discharge and (b)
given specific force.
5. To compute critical depth
6. To analyse critical flow in rectangular channels.
7. To apply the above to solve practical problems.
10.1 Introduction:
In the design of transitions in open channels, specific energy and critical depth concept is
very much necessary. Similarly the energy description below spillways, hydraulic jump is
utilised. So the conditions required for the foundation of hydraulic jump can be obtained
from specific force concept. In addition to it there are many practical applications of
hydraulic jump. So this unit is very useful for civil engineers in solving open channel
problems.
10.2 Specific energy critical depth
The concept of specific energy was first introduced by Bakhmeteff in 1912 It is defined
as total energy of flow per unit weight of water which is measured with respect to the
channel bed as datum. Thus datum head in Bernoulli’s total head is zero.
ie specific energy E = y +ϑ 2
2g, But V =
QA
so E= y +Q2
2gA2
E310/1 204 Thus for a given Q in a prismatic channel, specific energy is function of depth of flow
only. So the relationship between specific energy for a given Q, and cross sectional area
A, and depth of flow is as shown in the figure.
Since E = y + v 2
2g , the curve of specific energy is asymptotic to 450 line through
origin and x axis as shown in figure. In the figure it is seen that specific energy is
minimum at a depth of yc . This depth is called as critical depth. So critical depth ‘yc’ in
open channel is defined as that depth at which the specific energy is minimum for given Q
and A. Velocity at this depth is called as critical velocity.
When the depth of flow in the channel is increased velocity decreases for the same
discharge in same channel or vice versa. When depth of flow is decreased the velocity
increased. When depth of flow is more than the critical depth yc , the flow is called as
subcritical flow or tranquil flow and when the depth is less than critical depth the flow is
called as supercritical flow or rapid flow. Thus for any given specific energy E , there are
two possible depths y1, and y2 which are called as subcritical depth and y2 is called as
super critical depth. These depths are called as alternate depths.
For given Q the minimum specific energy can be obtained by differentiating E
with respect to y as follows
E = y +Q 2
2gA2
dEdy
= 1 +Q 2
2g(−2)
1A3
dAdy
= 0
But as shown in figure dA = Tdy
So dA/dy =T
substituting those in the above equation, we have
1=Q 2TgA3 or
Q 2
g=
A3
Tis the condition for critical flow.
E310/1 205 But v =
QA
hydraulic depth D = AT
Then we have Q2
A2
×
1g
=AT
v 2
g= D
v2
gD= 1 or
vgD
= 1
A at critical depth, Fr , Froude number = 1
So for sub critical flow Fr <1 Q D is more and v is less
for super critical flow Fr >1 D is less and v is more
Similarly for a given specific energy, we have
Q2
2gA2 = E − y (A)
Q = A 2g(E − y)
Thus for given E and A, Q is function of y only . So the relationship between Q and y is
obtained as shown in figure.
ie Q is maximum at a depth called as critical depth, which can be obtained by
differentiating Q w.r.to y as follows
dθdg
= 2g(E − y)dAdy
+12
A1
2g(E − y)2g(−1) = 0
2g(E-y)T-Ag = 0
2(E-y) = A/T
E310/1 206
But Q2
gA 2 = 2(E − y) from equation (A)
ieQ2
gA 2 =AT
or Q 2
g=
A3
Tis the condition for critical flow at which Q is maximum, which is same
as the previous equation, for critical depth.
The above curve can be obtained when water is controlled by a gate into a prismatic
channel. When the gate is closed discharge into channel is Q = 0 and depth of flow v/s of
the gate is maximum. As the gate is gradually opened, the depth of flow v/s of the gate
decreases and discharge is gradually increased, till it reaches a maximum value at which
the depth of flow v/s of gate is critical. Further opening of the gate will have no effect
over the flow and thus top portion of the curve is obtained ;and the bottom portion is
imaginary.
Similarly d/s of the gate Q = 0, y = 0 and depth of flow gradually increases till it
reaches a value at which the Q is maximum then this depth is called critical depth. After
words opening of gate will have no effect over the flow. Thus lower portion of the curve
is obtained and upper portion of the curve is imaginary.
From the curve it is seen for any discharge there are two possible depths, y1 and y2 . These
depths y1 is called as subcritical and y2 is called as super critical depth. These two depths
are called as alternate depths.
10.3 Critical flow in rectangular channel
consider a constant flow in a rectangular channel of B wide.
In this case specific energy E = y +v2
2g
Let q be the discharge per metre width
Then ϑ =qy
E310/1 207
So E = y +q2
2gy2
and for min E, dEdy = 0
ie dEdy
= 1 +q 2
2g(−2)y3 = 0
q2
gy 3 = 1
q2
g= yc
3
or yc = q g23 which is called as critical depth Thus at minimum specific energy the
flow is critical.
ie q2
(y2 )gy= 1
ϑ 2
gy= 1
ϑgy
= 1
ie Froude no Fr =1
suppose for E constant we have
q2
2gy2 = (E − y)
or q = y 2g(E − y)
for maximum discharge dq/dy = 0
ie dqdy
= 2g E − y +y
2 E − y(−1)
= 0
2(E-y) - y = 0
But from eqn (A) q2
gy 2 = 2(E − y)
∴q 2
gy2 = y
q2
g= yc
3 which is same as before
E310/1 208 So in rectangular channel , for constant discharge specific energy is minimum or for
constant specific energy discharge is maximum. For these two conditions the flow is
critical which is given by
q2
g= yc
3
Further 2(E-y) -y =0
(E-y) = y/2
E = y + y 2 =32
yc
ie critical depth yc =23
L5
At this depth velocity is critical velocity and bed slope is critical bed slope.
The above condition can be obtained by
Q 2
g=
A3
T
for rectangular channel T = B
∴Q 2
A2( )g =AB
= y
Q2
B2 y2( )g = y
q2
g= y3c which is derived as above
10.4 Momentum of flow in open channel-specific force
Consider a small reach of a channel as shown in figure
The forces that are acting on the fluid are
1. hydrostatic forces p1 and p2 where
P1 = wAz1 and P2 =wAz2
E310/1 209 z1 & z 2 = Centre of Gravity of area of cross section A1 & A2 from force surface
respectively
2. Water weight component W sin θ in the direction of flow where W = water weight in
the reach.
3. Frictional resistance due to surface in contact with water.
By Newton’s 2ns law, we have
∑f = rate of change of momentum per second
∴P1 -P2 +W sin θ -Ff =wQg
(v2 − v1 )
Here as the reach is small, we can neglect Ff =0 also as the slope is small, we can also
neglect the weight component in the direction of flow.
So p1 - p2 =wqg
(v2 − v1 )
wA1z1 − wAz2 =wqg
(v2 − v1 )
or qv1
g+ Az1 =
qv2
g+ Az2 = F is constant
ie qvg
+ Az is called as specific force which is constant . This specific force of the
water at any cross section is the force per unit specific weight of water.
In a prismatic channel for given discharge this specific force is function of depth ‘y’ . So
the relationship between specific force F and depth y is as shown in figure.
ie when the specific force is minimum the depth is called as critical depth. So for specific
force to be minimum we have dFdy
= 0
ie dFdy
= (−1)q2
gA2dAdy
+d Az( )
dy= 0
E310/1 210 where
dAdy
= T
and d(A z) u is the change in moment of Area A about force surface for a change in
depth dy which can be calculated as given below.
d(Az ) = A z + dy( ) + Tdy dy2
− Az = Ady +
T(dy)2
2
∴d(Az ) = Ady by neglecting small terms.
substituting these in above equation we have
q2
gA 2 T =Adydy
= A
q2
g=
A3
Tis the condition for critical flow at which F is minimum and the
depth of flow is critical depth yc
Further q2
A2 ×1g
=AT
= D
v2
gD= 1 or
vgD
= 1 = Fr
is at critical depth Froude number is 1 if Fr is <1, the flow is called as sub critical at which
depth of flow is more and velocity is less
If Fr is >1, the flow is called as super critical flow at which the depth of flow is less and
velocity is more.
Thus when flow passes from supercritical to sub critical through a critical depth, then
hydraulic jump force as shown in figure in which there is turbulent mixing of the fluid.
Due to this mixing lot of energy is dissipated.
So this concept is used for energy dissipation below spillways etc.
From the specific force diagram it is seen that for the same specific force these are two
possible depths y1 and y2.
E310/1 211 The depth y1 is called as super critical depth and the depth y2 is called as subcritical depth.
so these depths for which specific force is same is called as initial depth (y1) and
sequential depth (y2).
Now from specific force we have
qgA
F Az2
= −
or q = gA(F − Az )
so for a given specific force q is maximum when dqdy
= 0
ie ( )dqdy
gA F Az
F AzdAdy
Ad Azdy
=−
× − +−
=
12
1 0( )
( )
ie (F-A z)T = Ad(Az )
dy= A
Adydy
= A2
But (F-A z) =q2
gA
q2
gAT = A2
or q2
g=
A3
Tis the condition for critical flow at which depth is critical and q is
maximum
Thus for given specific energy or specific force, the discharge is maximum when the flow
is critical.
10.5 Computation of critical flow.
When the depth of flow is equal to critical depth, then the flow is called as critical flow.
This can be known by
q2
g=
A3
T
c
qg
= AAT
= zc
where zc is called as section factor for critical flow.
Thus for a given discharge zc is function of depth ‘y’ only.
So for different depths, the relationship between Z and y is shown in figure, where Z =
E310/1 212
AAT
, for different depths, Z can be called as A A T and plotted as shown in figure.
Now for a given discharging
Zc =qq
so from the graph, for this section factor Zc we have yc read. which is critical depth. Or
for a given depth of flow, the critical section factor Zc is known from the graph from
which
Zc =qc
g, the critical discharge ‘qc ’ can be calculated.
For a given discharge the slope of bed can be adjusted such that the uniform depth is equal
to critical depth yc . Or for a given depth of flow, the discharge and slope can be adjusted
such that the flow is critical . Then this slope is called as critical slope sc.
Thus when the flow is in critical state we have.
1. E = specific energy is minimum for a given discharge
2. F = specific force is minimum for a given discharge
3. Discharge is maximum for a given E or F
4. Froude number Fr = 1
5. For rectangular channels
yc =23
E or yc =q2
g
13
and ν c
2
2g=
12
yc or Ec = yc +ϑ c
2
2g= yc +
12
yc =32
yc
ie Ec = 32 yc
Identify the following as True or False
E310/1 213 SAQ 1. For critical flow at a given discharge Froude Number = 1 and E is minimum
SAQ 2. Discharge per unit width is 1m3/s. Find the critical depth in rectangular channel.
SAQ 3.. For rectangular channel critical depth is 1m. Find the maximum discharge
SAQ 4. For rectangular channel E is 1.5 m find the critical depth
SAQ 5 For a discharge 1m3 /s per metre width the critical depth is 0.465 m Find Froude
number.
SAQ 6. When qyc
= gyc , the flow is critical.
Worked Example (1) A rectangular channel with a bed slope of 1 in 200, carries a
discharge of 10 m3 / s , Find the critical depth if the width of channel is 2m.
q = 102
= 5m3 / s / m
yc =q2
g
13
=52
9.81
13
= 1.368 m
Worked Example (2) If the specific energy of flow in 5m wide rectangular
channel is 1.5 m, then calculate the maximum discharge
yc =23
E =23
×32
= 1m
but yc =q2
g
13
or qmax = gyc3( )
12 = 3.13m 3 / s / m
Worked Example (3) A trapezoidal channel with 5 m bottom width and 2H to 1v
side slope carries a discharge of 9.81 m3 / s. Find the critical depth.
A = (B+zy)y
T = (B+2zy)
Now q2
g=
A3
Tfor critical depth.
(9.81)2
9.81= 9.81 =
(b + 2yc )yc( )3
(b + 4yc )
9.81 13 (b + 4yc )
13 = (b + 2yc )yc
or yc = 0.59
E310/1 214 for different depths y = 0.1, 0.2, 0.3 , 0.4, 0.5, 0.6, 0.7. 0.8 find Z = A A T and plot a
curve between z & y as shown in figure
Then find zc =qg
= =9 819 81
9 81..
.
for this value from the graph find yc
10.6 Transitions
A transition is a small portion of a channel of varying cross section connecting two
channel sections. It may be sudden or gradual. It may be contracted or expanded.
It may be obtained either by
a. reducing and expanding bed width or
b. raising or lowering bed
c. by varying both bed width and bed elevation.
These transitions are necessary for measuring devices, to change the velocity to create
hydraulic jump to dissipate energy and to economise cost of hydraulic structures like
aquident etc.
The purpose of a transition is to minimise the energy loss by changing the hydraulic
conditions gradually .
10.6.1 Transitions with reduction in width
In this case the specific energy will be constant for a given discharge in the channel.
When the width is reduced the uniform depth y1 at (1) is reduce to y2 at (2) in subcritical
flow as shown in figure, as the velocity at (2) is increased. This reduction in depth at (2)
continues till it reaches a critical depth at which the discharge per unit width is maximum.
This critical depth yc 2 is less than y2..
If the width at (2) is further reduced beyond the critical flow, the depth at (1) is increased
changing the specific energy E1 at (1) to E11 such that the new depth yc2
1 at (2) will be
the critical depth for the new specific energy E11. The new critical depth yc2
1 is greater
than yc2.
E310/1 215
Similarly if the flow is supercritical flow the depth y2 at (2) is more than y1 at (1) for the
given E1 . As the width goes an reduced the depth at (2) ‘y2’ goes on increased, till it
reaches a critical condition at which the discharge qmax per unit width at (2) is maximum
and the depth yc2. If further the width is reduced beyond this critical flow, the head at (1)
is lowered to increase the specific energy E1 such that the new depth it (2) would be
critical depth yc2
1 for the new specific energy. E11 and the new critical depth Y1
c2 is more
than yc2.
Thus for a given discharge and specific energy, there is a limit for a reduction in width at
which the flow is critical and beyond this reduction in width causes a change in u/s depth
Here E1 = E2 = Ec =32
yc
and yc =q2
2
g
13
q2 = q/B2
10.6.2 Transition with raise in bed.
In this for the same discharge, the specific energy varies, when the bed is raised by ∆z as
shown in figure
E310/1 216
Here E1 = E2 + ∆z
Hence for a given q1 E reaches a minimum as bed gradually is raised. At this minimum
specific energy the raise in bed ie ∆Zmax is maximum possible. This can be obtained by
E1 = Ec + ∆z max
Ec =32
yc
and yc =q2
g
1 3
As in the previous section here also in case of subcritical flow the depth at (2) goes on
decreases till it reaches critical value yc2which is less than y2 which is less than y1, and
in
case of supercritical the depth ‘y2’ at (2) goes on increases till it reaches a critical value
yc2as shown in figure.
If further raise in bed causes the y, depth to raise or fall as the flow is subcritical
or supercritical.
E310/1 217 In case of both reduction in width and raise in bed is analysed by the same procedure.
Identify the following as True or False
SAQ 7 By changing hydraulic conditions the transition reduces loss of energy.
SAQ 8 For a given discharge in a channel the critical depth in the transition is same both
in subcritical and supercritical flows.
SAQ 9 The minimum width or maximum raise in bed of a channel corresponds to critical
flow.
SAQ10 If the specific energy in the channel is 10 m and the critical energy is 9.8 m, the
maximum raise in bed is 0.2 m
SAQ11 If the width is reduced or bed is raised beyond the critical flow, then the v/s depth
will be altered.
SAQ12 A rectangular channel 2.5 m wide, the specific energy is 10m . Then when the
width is reduced till critical flow is obtained the critical specific energy is 10m
SAQ13 The specific energy in a channel is 10.2 m. The maximum raise in bed is 0.2 m.
Then the critical depth is 0.67m.
SAQ 14 Critical velocity head is 12
critical depth
Worked Example (4) A 10 m wide is rectangular channel carries a discharge of 30 m3/s
with 1.66 m depth of flow. Calculate the maximum raise in bed and the corresponding
fall in water surface. If the raise is increased by 10% of the width at critical flow, find the
rise in water bed U/s.
q = 3010
= 3m 3 / s / m
ϑ 1 =3
1.66= 1.8m ϑ 1
2
2g = 0.166m
∴ E1 = y1 +ϑ1
2
2g= 1.66 + 0.166 = 1.826m
and yc = q 2
g
13
=32
9.81
13
= 0.97m
Ec =32
yc = 1.457m
∴ ∆zmax = E1 - Ec = 1.826 -1.457 = 0.36m
Drop in water level = y1 - (∆zmax + yc ) =vc
2
2g−
ϑ12
2g
E310/1 218 = 1.66-(0.36+0.97) = 0.32m
or0.97
2− 0.166 = 0.32m
raise in bed level =- 0.1 ×0.36 = 0.036
New raise in bed is 0.396 m
New specific energy U/s is E11 = Ec +∆z = 1.457+0.396 = 1.853m
E11 = 1.853 = y1
1 +32
y11( )2
× 2 × 9.81
y11 = 1.7 m
Worked Example (5) A rectangular channel 10m wide carries a discharge of
10m3/s with a depth of 0.8 m Find the minimum contraction in bed. If the bed width is
further reduced by 10%. What would be the rise in water level U/s.
v1 =qy1
=1
0.8= 1.25m / s
v12 / 2g = 0.0796 m
E1 = y1 + v12 /2g = 0.8 + 0.0796 = 0.8796 m
E1= E2 = Ec
∴ yc = 2 3 Ec =23
× 0.8796 = 0.5814 = q22 g( )
13
∴q2 = yc3g = 0.58643 × 9.81 = 1.4m3 / s / m
vc2 2g =
12
yc =0.5864
2= 0.2932m
vc = 2.398 m/s
10 =Bcyc Vc = Bc ×0.5864×2.398
∴ Bc = 7.11m
or q2 B2 = Q
∴B2 =Qq
m2
1014
714= =.
.
New width = 7.11 × 0.9 = 6.399m
q21 =
100.399
= y113
g = 1.562
∴ yc1 = 0.629m
Ec1 = E1
1 = 3 2 × yc1 = 0.9436m
E310/1 219
E11 = y1+v1
2/2g = 0.9436 = y 1 + q2
y12 2g
= y11 +
1y1
2 ×19.62
y11 = 0.86 m
Worked Example (6) A rectangular channel 5 m wide carries a discharge of 10 m3
/s at a depth of 1.5 m. If the bed width is reduced to 3m and a hump[ is constructed to
create critical flow determine the same.
ϑ 1 =qy1
=2
1.5= 1.53m / s
ϑ 12
2g = 0.0906m
E1 = y1 + ϑ 12 2g = 1.681m
q2 =10B2
=103
= 3.33m
yc =q2
2
g
13
=3.332
9.81
13
= 1.04m
E1 = ∆z+yc + vc2 / 2g
= ∆z+ yc +1/2 yc
1.681 = ∆z +1.5 × 1.04
∴∆z = 0.0545m
10.7 Summary
1. Specific energy E = y + ϑ 2
2gis the total energy with r.to bed as datum
2. The condition for critical flow is a prismatic channel is Q2 /g = A3/T where T =
top width of flow both for q constant or E is constant
At this critical flow , E is minimum when Q is constant and Q is maximum for