Equations in OCF
Equations in OCF
Equations in OCF
PART 2
EXPECTED LEARNING
OUTCOMES
1
CONTINUITY EQUATION
ENERGY EQUATION
Specific Energy
Critical Flows
MOMENTUM EQUATION
Hydraulics Jump
CONTINUITY EQUATION
Commonly used in the analysis of flowing fluids
Developed from the law conservation of mass
A2
V2
A1
V1
Q = A1V1 =A2V2
where, A is the area
of cross section and V
is the mean velocity.
In
steady flow, the discharge, passing any sections
along the channel must be the same.
Q = A1V1 = A2V2 = AiVi
V2
H z y cos
2g
Elevation
Pressure
Head
Velocity
Head
SPECIFIC ENERGY
SPECIFIC ENERGY
Specific Energy, E in OCF is,
Specific energy (E) = Pressure head (y) + Velocity
head (V2/2g)
V2
E y cos
2g
Or
If the slope is small ( =1) and the flow is uniform
flow ( =1), the equation becomes,
2
V
E y
2g
V2
E y
2g
Q2
E y
2gA 2
EXAMPLE
1. Rectangular channel. B = 2.5 m, Q = 6.48 m3/s, E
= 1.5 m. Find alternate depths and the
corresponding Froude numbers.
Given,
CRITICAL
FLOWS
Subcritical Flow
Critical Flow
Supercritical Flow
dy
y
A(y)
T(y)
dA=T dy
dy
y
A(y)
T(y)
dA=T dy
dy
y
To get Ec,
let dE/dy =0.
Thus, the remaining terms will be in critical
conditions.
Q 2 dAc
0 1 3
gAc dyc
A(y)
T(y)
dA=T dy
dy
y
A(y)
And Froude
Number,
Q 2Tc
1
3
gAc
F
Ac
Q2
g
Tc
V
gA
T
OOPS
How
?
y = yc
E = Ec
A = Ac
F=1
Q Tc
Ac
Q
1 or
3
gAc
g
Tc
RECTANGULAR SECTION
For
a rectangular section, and
by equation
T=B
TRIANGULAR SECTION
For
2my
Hence,
The specific energy at critical depth
y
Therefore,
Ec 1.25 yc
1
m
In a sloping channel
Flow from a reservoir to a channel
Flow flowing towards a free overfall channel
Flow above the hump
Flow in a contracted section
EXAMPLE
0.5
(yc=1.828m,
E1
Energy line
E2
y1
yc
y2
No change to:
y1
y2 = y2
E1 = E2 + z
Case II z = zmax
No change to:
y1
y2 = yc
E1 = Ec + zmax
Case
III
z > zmax
Choking
condition
y1 changes (y1>
y1)
y2 = yc
E1
Energy line
E2
y1
yc
y2
E1
Energy line
E2
z < zmax
No change to:
y1
y2 = y2
E1 = E2 + z
Case II
z = zmax
No change to:
y1
y2 = yc
E1 = Ec + zmax
y1 changes
(y1< y1)
y2 = yc
E1
Energy line
E2
MOMENTUM
EQUATION
MOMENTUM EQUATION
The commonly used momentum equation for open
channel flow problems is the linear-momentum
equation
It is used when:
Impossible to estimate head loss, hL (cannot use
energy equation)
The internal flow is complicated
Concerns with internal forces between two sections
Forces acting on a
control volume are as
follows,
1. Pressure forces, Fp
2. Friction forces along
the bed of channel, Ff
3. Body force (weight of
the fluid) in the
longitudinal direction,
Wx
By
the linear-momentum
equation in the longitudinal
direction for a steady flow
discharge of Q,
In which
and
Figure 1.17 A control volume bounded by
sections 1 and 2, the boundary and a surface
HYDRAULICS JUMP
Momentum equation can be applied to analyze the
hydraulic jump.
What is a hydraulic jump?
It is a sudden transition from a rapid flow
(supercritical) to slow flow (subcritical).
Why momentum equation?
There is a significance amount of energy loss in
the hydraulic jump. Therefore, the energy
equation is not suitable to analyze the situation.
Hydraulic Jump
EXAMPLE
Figure below shows a hydraulic jump in a horizontal
apron aided by a two-dimensional block on the
apron. Obtain an expression for the drag per unit
length of the block. (assume a frictionless, horizontal
channel and hydrostatic pressure distribution at
section 1 and 2)
V
2
y2
V1
P1
y1
1
P2
Fd
2
OOPS
What is
apron?of concrete or
A ground covering
other material used to protect
the underlying earth from water
erosion.
SOLUTION
Forces
are:
by
* Can assume or equals to 1 when the channels are straight, prismatic and uniform flow or GVF takes place
TODAYS OUTCOMES
1
CONTINUITY EQUATION
ENERGY EQUATION
Specific Energy
Critical Flows
MOMENTUM EQUATION
Hydraulics Jump
WAIT
(5%) QUIZ 1
7th June 2016
13.02.06
2.05pm