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Unformatted text preview: Ch. 10 OpenChannel Flow
Previous internal flow analyses have considered only closed conduits where the
fluid typically fills the entire conduit and may be either a liquid or a gas.
This chapter considers only partially filled channels of liquid flow referred to as
openchannel flow.
OpenChannel Flow: Flow of a liquid in a conduit with a free surface.
Openchannel flow analysis basically results in the balance of gravity and friction
forces.
One Dimensional Approximation
While openchannel flow can, in general, be very complex ( three dimensional and
transient), one common approximation in basic analyses is the
OneD Approximation:
The flow at any local cross
section can be treated as
uniform and at most varies
only in the principal flow
direction. This results in the following equations.
Conservation of Mass (for ρ = constant)
Q = V(x) A(x) = constant
Energy Equation
2 2 V1
V
+ Z1 = 2 + Z 2 + h f
2g
2g  v v X1  eText Main Menu  Textbook Table of Contents  Study Guide The equation in this form is written between two points ( 1 – 2 ) on the free surface
of the flow. Note that along the free surface, the pressure is a constant, is equal to
local atmospheric pressure, and does not contribute to the analysis with the energy
equation.
The friction head loss hf is analogous to the head loss term in duct flow, Ch. VI,
and can be represented by
where P = wetted perimeter 2 x − x Vavg
hf = f 2 1
D h 2g Dh = hydraulic diameter = 4A
P Note: One of the most commonly used formulas uses the hydraulic radius: 1
A
R h = Dh =
4
P
Flow Classification by Depth Variation
The most common classification method is by rate of change of freesurface depth.
The classes are summarized as
1. Uniform flow (constant depth and slope)
2. Varied flow
a. Gradually varied (onedimensional)
b. Rapidly varied (multidimensional)
Flow Classification by Froude Number: Surface Wave Speed
A second classification method is by the dimensionless Froude number, which is a
dimensionless surface wave speed. For a rectangular or very wide channel we
have Fr =
and V
V
=
1/2
co (gy) where y is the water depth co = the speed of a surface wave as the wave height approaches zero. There are three flow regimes of incompressible flow. These have analogous flow
regimes in compressible flow as shown below:  v v X2  eText Main Menu  Textbook Table of Contents  Study Guide Incompressible Flow
Fr < 1
Fr = 1
Fr > 1 Compressible Flow subcritical flow
critical flow
supercritical flow Ma < 1
Ma = 1
Ma > 1 subsonic flow
sonic flow
supersonic flow Hydraulic Jump
Analogous to a normal shock in compressible flow, a hydraulic jump provides a
mechanism by which an incompressible flow, once having accelerated to the
supercritical regime, can return to subcritical flow. This is illustrated by the
following figure. Fig. 10.5 Flow under a sluice gate accelerates from subcritical to critical to
supercritical and then jumps back to subcritical flow.
The critical depth Q
yc = 2 b g 1/3 is an important parameter in open channel flow and is used to determine the local flow regime (Sec. 10.4).
Uniform Flow; the Chezy Formula
Uniform flow 1. Occurs in long straight runs of constant slope
2. The velocity is constant with V = Vo
3. Slope is constant with So = tan θ  v v X3  eText Main Menu  Textbook Table of Contents  Study Guide From the energy equation with V1 = V2 = Vo, we have h f = Z1 − Z 2 = S o L
Since the flow is fully developed, we can write from Ch. VI
1/2 2 L Vo
hf = f
Dh 2g 8g
1/2 1/2
Vo = Rh So
f and 1/2 8g
For fully developed, uniform flow, the quantity is a constant
f
and can be denoted by C. The equations for velocity and flow rate
thus become Vo = C R h So
1/2 1/2 Q = CA R h So
1/2 and 1/2 1/2 The quantity C is called the Chezy coefficient, and varies from 60 ft /s for small
1/2
1/2
rough channels to 160 ft /s for large rough channels (30 to 90 m /s in SI).
The Manning Roughness Correlation
The friction factor f in the Chezy equations can be obtained from the Moody
chart of Ch. VI. However, since most flows can be considered fully rough, it is
appropriate to use Eqn 6.64:
fully rough flow: 3.7Dh f ≈ 2.0 log ε −2 However, most engineers use a simple correlation by Robert Manning:
S.I. Units Vo (m/s ) ≈ B.G. Units Vo (ft/s ) ≈ α [R (m)]
nh α 2/3 [R (ft )]
nh 2/3 S1/2
o S1/2
o  v v X4  eText Main Menu  Textbook Table of Contents  Study Guide where n is a roughness parameter given in Table 10.1 and is the same in both
systems of units and α is a dimensional constant equal to 1.0 in S.I. units and 1.486
in B.G. units. The volume flow rate is then given by Uniform flow Q = Vo A ≈ α
n A R 2/3 S1/2
h
o Table 10.1 Experimental Values for Manning’s n Factor  v v X5  eText Main Menu  Textbook Table of Contents  Study Guide Example 10.1
Given:
Rectangular channel,
finished concrete, slope = 0.5˚
water depth: y = 4 ft, width: b = 8 ft
Find:
3
Volume flow rate (ft /s) 4 ft crosssection 8 ft θ For the given conditions: n = 0.012 A = b y = (8 ft) (4 ft) = 32 ft 2 2 A 32 ft
Rh = =
= 2 ft
P 16 ft So = tan 0.5˚ = 0.0873
P = b + 2 y = 8 + 2 (4) = 16 ft D h = 4 R h = 8 ft Using Manning’s formula in BG units, we obtain for the flow rate Q≈ 1.486
(32 ft 2 )(2 ft )2/3 (0.00873)1.2 ≈ 590 ft 3/s ans.
0.012 Alternative Problem
The previous uniform problem can also be formulated where the volume flow rate
Q is given and the fluid depth is unknown. For these conditions, the same basic
equations are used and the area A and hydraulic radius Rh are expressed in terms
of the unknown water depth yn.
The solution is then obtained using iterative or systematic trial and error techniques
that are available in several math analysis/ math solver packages such as EES
(provided with the text) or Mathcad ®.  v v X6  eText Main Menu  Textbook Table of Contents  Study Guide Uniform Flow in a Partly Full, Circular Pipe
Fig. 10.6 shows a partly full, circular
pipe with uniform flow. Since
frictional resistance increases with
wetted perimeter, but volume flow
rate increases with cross sectional
flow area,
the maximum velocity and flow rate
occur before the pipe is completely
full.
For this condition, the geometric
properties of the flow are given by the
equations below.
Fig. 10.6 Uniform Flow in a Partly Full,
Circular Channel sin2θ 2
A=R θ− 2 P = 2 Rθ Rh = R sin2θ 1 − 2θ 2 The previous Manning formulas are used to predict Vo and Q for uniform flow
when the above expressions are substituted for A, P, and Rh. α R sin2θ Vo ≈ 1 − n 2 2θ 2 /3
1/2 So Q = Vo R 2 sin2θ θ− 2 These equations have respective maxima for Vo and Q given by  v v X7  eText Main Menu  Textbook Table of Contents  Study Guide α Vmax = 0.718 R 2/3 S1/2
o n Q max = 2.129 α
n at θ = 128.73Þ and y = 0.813 D R 8/3 S1/2
o at θ = 151.21Þ and y = 0.938 D Efficient Uniform Flow Channels
A common problem in channel flow is
that of finding the most efficient lowresistance sections for given conditions.
This is typically obtained by maximizing
Rh for a given area and flow rate. This
is the same as minimizing the wetted
perimeter.
Note: Minimizing the wetted perimeter for a given flow should minimize the
frictional pressure drop per unit length for a given flow.
It is shown in the text that for constant value of area A and α = cot θ, the
minimum value of wetted perimeter is obtained for [ A = y 2 2 (1 + α 2 ) − α
1/ 2 P = 4 y (1 + α ) 2 1/ 2 − 2α y Rh = 1
y
2 Note: For any trapezoid angle, the most efficient cross section occurs when the
hydraulic radius is onehalf the depth.
For the special case of a rectangle (a = 0, q = 90˚), the most efficient cross section
occurs with A = 2y 2 P = 4y Rh = 1
y
2 b = 2y  v v X8  eText Main Menu  Textbook Table of Contents  Study Guide Best Trapezoid Angle
The general equations listed previously are valid for any value of α. For a given,
fixed value of area A and depth y the best trapezoid angle is given by
α = cot θ = 1
3 θ = 60 o or 1/2 Example 10.3
What are the best dimensions for a rectangular brick channel designed to carry 5
3
m /s of water in uniform flow with So = 0.001?
2 Taking n = 0.015 from Table 10.1, A = 2 y , and Rh = 1/2 y ; Manning’s
formula is written as
2/3
1.0
2 1 or 5 m / s =
(2 y ) 2 y (0.001)1/ 2
0.015 1.0
2/3 1/2
Q≈
A R h So
n 3 This can be solved to obtain y 8/ 3 = 1.882 m 8/3 or y = 1.27 m The corresponding area and width are A = 2 y 2 = 3.21 m2 and b= A
= 2.53 m
y Note: The text compares these results with those for two other geometries having
the same area.  v v X9  eText Main Menu  Textbook Table of Contents  Study Guide Specific Energy: Critical Depth
2 One useful parameter in channel
flow is the specific energy E,
where y is the local water depth. V
E = y+
2g Defining a flow per unit channel
width as q = Q/b we write q
E = y+
2 g y2 2 Fig. 10.8b is a plot of
the water depth y vs.
the specific energy E.
The water depth for
which E is a minimum
is referred to as the
critical depth yc. Fig. 10.8 Specific Energy Illustration q2 y = yc = g Emin occurs at 1/3 Q2 = 2 b g E min = The value of Emin is given by 1/3 3
y
2c At this value of minimum energy and minimum depth we can write Vc = (g y c ) = Co
1/2 and Fr = 1  v v X10  eText Main Menu  Textbook Table of Contents  Study Guide Depending on the value of Emin and V, one of several flow conditions can exist.
For a given flow, if E < Emin No solution is possible E = Emin Flow is critical, y = yc, V = Vc E > Emin , V < Vc Flow is subcritical, y > yc ,disturbances can
propagate upstream as well as downstream E > Emin , V > Vc Flow is supercritical, y < yc , disturbances can
only propagate downstream within a wave angle
given by C
(g y )
µ = sin o = sin1
V
V 1/ 2 1 Nonrectangular Channels
For flows where the local channel width varies with depth y, critical values can be
expressed as bo Q 2 Ac = g 1/3 and Q g Ac Vc =
= A c bo 1/2 where bo = channel width at the free surface.
These equations must be solved iteratively to determine the critical area Ac and
critical velocity Vc.
For critical channel flow that is also moving with constant depth (yc), the slope
corresponds to a critical slope Sc given by  v v X11  eText Main Menu  Textbook Table of Contents  Study Guide 2 n gAc
Sc = 2
α b o Rh, c α = 1. for S I units and 2.208 for B. G. units and Example 10.5
Given: a 50˚, triangular channel has a
3
flow rate of Q = 16 m /s.
Compute: (a) yc, (b) Vc,
(c) Sc for n = 0.018
a. For the given geometry, we have
P = 2 ( y csc 50˚) A = 2[y (1/2 y cot 50˚)] Rh = A/P = y/2 cos 50˚ bo = 2 ( y cot 50˚) For critical flow, we can write g Ac = bo Q
3 2 or g(yc cot 50Þ = (2 yc cot 50Þ Q
)
)
2 yc = 2.37 m 2 ans. b. With yc, we compute
Pc = 6.18 m 2 Ac = 4.70 m bo,c = 3.97 m
3 The critical velocity is now Q 16 m / s
Vc = =
= 3.41 m / s
A 4.70 m ans. c. With n = 0.018, we compute the critical slope as g n2 P
9.81(0.018)2 (6.18)
Sc = 2
=
= 0.0542
α b o R1/3 1.0 2 (3.97) (0.76)1 /3
h  v v X12  eText Main Menu  Textbook Table of Contents  Study Guide Frictionless Flow over a Bump
Frictionless flow over a bump
provides a second interesting
analogy, that of compressible
gas flow in a nozzle.
The flow can either increase or
decrease in depth depending on
whether the initial flow is
subcritical or supercritical.
The height of the bump can also
change the results of the
downstream flow.
Fig. 10.9 Frictionless, 2D flow over a bump
Writing the continuity and energy equations for two dimensional, frictionless flow
between sections 1 and 2 in Fig. 10.10, we have V1 y1 = V2 y2 2 and 2 V1
V
+ y1 = 2 + y2 + ∆ h
2g
2g Eliminating V2, we obtain
2 2 2 Vy
V
y − E2 y + 1 1 = 0 where E2 = 1 + y1 − ∆ h
2g
2g
3
2 2
2 The problem has the following solutions depending on the initial flow condition
and the height of the jump:  v v X13  eText Main Menu  Textbook Table of Contents  Study Guide Key Points:
1. The specific energy E2 is exactly ∆h less than the approach energy E1.
2. Point 2 will lie on the same leg of the curve as point 1.
3. For Fr < 1, subcritical
approach The water level will decrease at the bump. Flow
at point 2 will be subcritical. 4. For Fr > 1, supercritical
approach The water level will increase at the bump. Flow at
point 2 will be supercritical. 5. For bump height equal to
∆hmax = E1  Ec Flow at the crest will be exactly critical (Fr =1). 6. For ∆h > ∆hmax No physically correct, frictionless solutions are
possible. Instead, the channel will choke and
typically result in a hydraulic jump. Flow under a Sluice Gate
A sluice gate is a bottom opening in a wall as shown below in Fig. 10.10a. For
free discharge through the gap, the flow smoothly accelerates to critical flow near
the gap and the supercritical flow downstream. Fig. 10.10 Flow under a sluice gate
This is analogous to the compressible flow through a convergingdiverging nozzle.
For a free discharge, we can neglect friction. Since this flow has no bump (∆h = 0)
and E1 = E2, we can write  v v X14  eText Main Menu  Textbook Table of Contents  Study Guide 2 V1 2 2 V12 y1
y −
+ y1 y2 +
=0 2g 2g
3
2 This equation has the following possible solutions.
Subcritical upstream flow
and low to moderate
tailwater (downstream
water level) One positive, real solution. Supercritical flow at y2
with the same specific energy E2 = E1. Flow rate
varies as y2/y1. Maximum flow is obtained for
y2/y1 = 2/3. Subcritical upstream flow
and high tailwater The sluice gate is drowned or partially drowned
(analogous to a choked condition in compressible
flow). Energy dissipation will occur downstream in
the form of a hydraulic jump and the flow
downstream will be subcritical. The Hydraulic Jump
The hydraulic jump is an irreversible,
frictional dissipation of energy which
provides a mechanism for supercritical flow to transition (jump) to
subcritical flow analogous to a
normal shock in compressible flow. The development of the theory is equivalent to that for a strong fixed wave (Sec
10.1) and is summarized for a hydraulic jump in the following section.  v v X15  eText Main Menu  Textbook Table of Contents  Study Guide Theory for a Hydraulic Jump
If we apply the continuity and momentum equations between points 1 and 2 across
a hydraulic jump, we obtain
1 /2
2 y2
= − 1 + (1 + 8 Fr12 )
y1 which can be solved for y2. V2 = We obtain V2 from continuity: V1 y1
y2 The dissipation head loss is obtained from the energy equation as V12 V22 h f = E1 − E2 = + y1 − + y2 2 g 2g or (y − y1 )
=2 3 hf 4 y1 y2 Key points:
1. Since the dissipation loss must be positive, y2 must be > y1.
2. The initial Froude number Fr1 must be > 1 (supercritical flow).
3. The downstream flow must be subcritical and V2 < V1. Example 10.7
3 Water flows in a wide channel at q = 10 m /(s m) and y1 = 1.25 m. If the flow
undergoes a hydraulic jump, compute: (a) y2, (b) V2, (c) Fr2, (d) hf, (e) the
percentage dissipation, (f) power dissipated/unit width, and (g) temperature rise.  v v X16  eText Main Menu  Textbook Table of Contents  Study Guide a. The upstream velocity is 3
q 10 m /(s ⋅ m)
V1 = =
= 8.0 m / s
y1
1.25 m Fr1 = The upstream Froude number is 8.0
V1
1/ 2 =
1/ 2 = 2.285
(g y1 ) [9.81 (1.25)] This is a weak jump and y2 is given by 2 y2
2 1/ 2
= − 1 + ( + 8 (2.285) ) = 5.54
1
y1
and y2 = 1 / 2 y1 (5.54) = 3.46 m
V2 = b. The downstream velocity is V1 y1 8.0 (1.25)
=
= 2.89 m / s
y2
3.46 c. The downstream Froude number is Fr2 = 2.89
V2
1/ 2 =
1/ 2 = 0.496
9.81 (3.46)]
(g y2 ) [ and Fr2 is subcritical as expected.
d. The dissipation loss is given by (y − y1 )
=2 3 hf 4 y1 y2 = (3.46 − 1.25)3 = 0.625 m
4 (3.46) (1.25) e. The percentage dissipation is the ratio of hf/E1.
2 2 8.0
V
E1 = 1 + y1 = 1.25 +
= 4.51 m
2g
2 (9.81)  v v X17  eText Main Menu  Textbook Table of Contents  Study Guide The percentage loss is thus given by % Loss = hf
E1 100 = 0.625
100 = 14%
4.51 f. The power dissipated per unit width is
3 3 Power = ρ Q g hf = 9800 M/m *10 m /(s m) * 0.625 m = 61.3 kw/m
g. Using Cp = 4200 J/kg K, the temperature rise is given by
&
Power dissipated = mCp ∆ T or
61,300 W/m = 10,000 kg/s m * 4200 J/kg K* ∆T
∆T = 0.0015˚K
negligible temperature rise  v v X18  eText Main Menu  Textbook Table of Contents  Study Guide ...
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This note was uploaded on 10/27/2009 for the course MAE 101a taught by Professor Sakar during the Spring '08 term at UCSD.
 Spring '08
 Sakar

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