Open Channel Flow
2/4/2016
OPEN CHANNELS (OPEN CHANNEL FLOW AND HYDRAULIC MACHINERY) UNIT – I
Rambabu Palaka, Assistant Professor
BVRIT
Learning Objectives 1. Types of Channels 2. Types of Flows 3. Velocity Distribution 4. Discharge through Open Channels 5. Most Economical Sections
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Learning Objectives 6. Specific Energy and Specific Energy Curves 7. Hydraulic Jump (RVF) 8. Gradually Varied Flow (GVF)
Types of Channels Open channel flow is a flow which has a free surface and flows due to gravity. Pipes not flowing full also fall into the category of open channel flow In open channels, the flow is driven by the slope of the channel rather than the pressure
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Types of Channels Open channel flow is a flow which has a free surface and flows due to gravity. Pipes not flowing full also fall into the category of open channel flow In open channels, the flow is driven by the slope of the channel rather than the pressure
Types of Flows 1. Steady and Unsteady Flow 2. Uniform and Non-uniform Flow 3. Laminar and Turbulent Flow 4. Sub-critical, Critical and Super-critical Flow
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
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1. Steady and Unsteady Flow Steady flow happens if the conditions (flow rate, velocity, depth etc) do not change with time. The flow is unsteady if the depth is changes with time
2. Uniform and Non-uniform Flow 1. Steady and Unsteady Flow the velocity of If for a given length of channel, flow, depth of flow, slope of theFlow channel and cross 2. Uniform and Non-uniform section remain constant, the flow is said to be Uniform The flow is Non-uniform, if velocity, depth, slope and cross section is not constant
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
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2. Non-uniform Flow 1. Steady Unsteady Flow Types of and Non-uniform Flow 1. Uniform Gradually Varied Flow (GVF) 2. and Non-uniform Flow
If the depth of the flow in a channel changes gradually over a length of the channel.
2. Rapidly Varied Flow (RVF) If the depth of the flow in a channel changes abruptly over a small length of channel
Types of Flows 1. Steady and Unsteady Flow 2. Uniform and Non-uniform Flow
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
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3. Laminar and Turbulent Flow Both laminar and turbulent flow can occur in open channels 1. Steady and Unsteady Flow depending on the Reynolds number (Re)
2. Uniform and Non-uniform Flow Re = ρVR/µ 3. Laminar and Turbulent Flow
Where, ρ = density of water = 1000 kg/m3 µ = dynamic viscosity R = Hydraulic Mean Depth = Area / Wetted Perimeter
Types of Flows 1. Steady and Unsteady Flow 2. Uniform and Non-uniform Flow 3. Laminar and Turbulent Flow
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
4. Sub-critical, Critical and Super-critical Flow
Types of Flows
1. Steady and Unsteady Flow 2. Uniform and Non-uniform Flow 3. Laminar and Turbulent Flow 4. Sub-critical, Critical and Super-critical Flow
Types of Flows 1. Steady and Unsteady Flow 2. Uniform and Non-uniform Flow 3. Laminar and Turbulent Flow 4. Sub-critical, Critical and Super-critical Flow
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Velocity Distribution Velocity is always vary across channel because of friction along the boundary The maximum velocity usually found just below the surface
Velocity Distribution Velocity is always vary across channel because of friction along the boundary The maximum velocity usually found just below the surface
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Discharge through Open Channels 1. Chezy’s C 2. Manning’s N 3. Bazin’s Formula 4. Kutter’s Formula
Forces acting on the water between sections 1-1 & 2-2 Discharge through Open Channels 1. Component of weight of Water = W sin i 2. Friction Resistance = f P L V2
1. Chezy’s C
2. Manning’s N 3. Bazin’s Formula 4. Kutter’s Formula
Rambabu Palaka, Assistant Professor, BVRIT
where W = density x volume = w (AL) = wAL Equate both Forces: f P L V2 = wAL sin i
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Open Channel Flow
2/4/2016
Chezy’s Formula, V=
w f
V= C mi
A sin i → 1 P
A = m = Hydraulic Radius → 2 P w = C = Chezy' s Constant → 3 f
Chezy’s Formula,
V= C mi
substitute 2& V = w AEqn. sin i → 1 3 in Eqn. 1, f
P
VA == m C = Hydraulic m. sin i Radius → 2 P forwsmall values of i, sin i = tan i = i = C = Chezy' s Constant → 3
f ∴V = C m. i
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
1. Manning’s N Chezy’s formula can also be used with Manning's Roughness Coefficient
C = (1/n) R1/6 where R = Hydraulic Radius n = Manning’s Roughness Coefficient
2. Bazin’s Formula Chezy’s formula can also be used with Bazins’ Formula 1. Manning’s N
C= 157.6 1.81 + k m
2. Bazin’s Formula where k = Bazin’s constant m = Hydraulic Radius
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Chezy’s Formula,
V= C mi
1. Manning’s N 2. Bazin’s Formula
3. Kutter’s Formula Chezy’s formula can also be used with Kutters’ Formula
1. Manning’s N
1 23 + 0.00155 + 2. Bazin’s Formula N C= 0.00155 N 3. Kutter’s Formula 1 + 23 + m i where N = Kutter’s constant m = Hydraulic Radius, i = Slope of the bed
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Chezy’s Formula,
V= C mi
1. Manning’s N 2. Bazin’s Formula 3. Kutter’s Formula
Problems 1. Find the velocity of flow and rate of flow of water through a rectangular channel of 6 m wide and 3 m deep, when it is running full. The channel is having bed slope as 1 in 2000. Take Chezy’s constant C = 55 2. Find slope of the bed of a rectangular channel of width 5m when depth of water is 2 m and rate of flow is given as 20 m3/s. Take Chezy’s constant, C = 50
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Problems 3. Find the discharge through a trapezoidal channel of 8 m wide and side slopes of 1 horizontal to 3 vertical. The depth of flow is 2.4 m and Chezy’s constant C = 55. The slope of bed of the channel is 1 in 4000 4. Find diameter of a circular sewer pipe which is laid at a slope of 1 in 8000 and carries a discharge of 800 litres/s when flowing half full. Take Manning’s N = 0.020
Problems 5. Find the discharge through a channel show in fig. 16.5. Take the value of Chezy’s constant C = 55. The slope of bed of the channel is 1 in 2000
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Most Economical Sections 1. Cost of construction should be minimum 2. Discharge should be maximum Types of channels based on shape: 1. Rectangular 2. Trapezoidal 3. Circular
Most Economical Sections 1. Cost of construction should be minimum Q = A V = A C m i 2. Discharge should be maximum
1 Types of channels based on shape: Q = K where K = A C A i 1. Rectangular P 2. Trapezoidal If P is minimum, Q will be maximum 3. Circular
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Rectangular Section for most economical section, P should be minimum dP
=0
d(d)
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
A →1 d A P = b + 2d = + 2d → 2 d for most economical seciton, P should be minimum
Rectangular Section
A = bd ⇒ b =
for most economical section,
A d + 2d dP A d = 0 ⇒P−should =0⇒ + 2 = 0be ⇒ minimum A = 2d 2 ⇒ bd = 2d 2 2 d (d ) d (d ) d b = 2d or d = b/2 dP m =
A P
=
bd b + 2d
=
2d
2
2d + 2d
Rambabu Palaka, Assistant Professor, BVRIT
=
d
=0
d(d)
2
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Open Channel Flow
2/4/2016
Trapezoidal Section for most economical section, P should be minimum dP
=0
d(d)
A − nd → 1 d A P = b + 2d n 2 + 1 = − nd + 2d n 2 + 1 → 2 d for most economical section, for most economical seciton, P should be minimum
Trapezoidal Section
A = (b + nd)d ⇒ b =
A be minimum P should d − nd + 2d n 2 + 1 dP d = 0 ⇒ b + 2nd = d n 2 + 1 =0⇒ dP d(d) d(d) 2 =0 d d(d) m = and θ = 60 0 2
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Circular Section d for Max. Velocity,
A P =0 dθ
3 A d P for Max. Discharge, =0 dθ
sin 2 θ )→ 1 2 P = 2R θ → 2 A = R 2 (θ -
Circular Section
A R sin 2 θ A = (θ )→ 3 d P 2θ 2 P for Max. Velocity, =0 dm for max. velocity, = 0 ⇒ θ = 128 0 45 ' , d = 0.81D,d θm = 0.3D dθ 3 A A3 A Q = AC m i = AC i =C i , C and i are constants d P P P Max. Discharge, =0 A 3for dθ d P = 0 ⇒ θ = 154 0 , d = 0.95D for max. discharge, dθ m =
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Problems 1. A trapezoidal channel has side slopes of 1 horizontal and 2 vertical and the slope of the bed is 1 in 1500. The area of cross section is 40m2. Find dimensions of the most economical section. Determine discharge if C=50 Hint: Equate Half of Top Width = Side Slope (condition 1) and find b in terms of d Substitute b value in Area and find d Find m = d/2 (condition 2) Find V and Q
Problems 1. A trapezoidal channel has side slopes of 1 horizontal and 2 vertical and the slope of the bed is 1 in 1500. The area of cross section is 40m2. Find dimensions of the most economical section. Determine discharge if C=50
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Problems 1. A trapezoidal channel has side slopes of 1 horizontal and 2 vertical and the slope of the bed is 1 in 1500. The area of cross section is 40m2. Find dimensions of the most economical section. Determine discharge if C=50
Problems 2. A rectangular channel of width 4 m is having a bed slope of 1 in 1500. Find the maximum discharge through the channel. Take C=50 3. The rate of flow of water through a circular channel of diameter 0.6m is 150 litres/s. Find the slope of the bed of the channel for maximum velocity. Take C=50
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Non-uniform Flow In Non-uniform flow, velocity varies at each section of the channel and the Energy Line is not parallel to the bed of the channel. This can be caused by 1. Differences in depth of channel and 2. Differences in width of channel. 3. Differences in the nature of bed 4. Differences in slope of channel and 5. Obstruction in the direction of flow
Specific Energy v2 Total Energy of flowing fluid, E = z + h + 2g where z = Height of bottom of channel above datus, If the channel bottom is taken as datum, v2 Es = h + which is called as Specific 2g
Rambabu Palaka, Assistant Professor, BVRIT
Energy
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Open Channel Flow
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Specific Energy Q=AV⇒V=
Q Q = A bh
If discharge per unit width , q = V=
Q = constant b
Q q = bh h 2
2 q V ∴ Es = h + =h+ 2g 2g h 2
Specific Energy Potential Energy (h)
dE for Critical Depth,
Modified Equation to plot Specific Energy Curve
=0 dh
where, E = h +
q
2
Es= h + q2/2gh2
2g h 2 1
q2 3 hc = g
3
⇒ hc =
Q subsitute value q =
2
g
2 3 ⇒ h c .g = q → 1
bh. v =
b ⇒ Vc =
q
b
= h c V c in Eqn. 1
g hc
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Specific Energy Potential Energy (h)
dE
Minimum Energy in terms of Critical Depth; E = for Critical Depth,Specific =0 dh 2 when specific energy is minimum, Depth ofq2flow q Es = h + /2gh2 is critical where, E = h + 1 2g h 2
h+
q
2
2g h 2
2 q2 3 3 q E = hc + substitute h c = or h c = q 2 32g 23 q 2 g g 2 3 hc = = ⇒ ⇒ . g = → 1 q hc hc hc 3g g
1
q
2
E min = h c +
hc
3h c h = hc + c = bh.2v 2 2
Q subsitute value q = 2g =h c b b ⇒ orVch c=
= h c V c in Eqn. 1
2 E min 3
=g h c
Alternate Depths 1 & 2 Hydraulic Jump
Specific Energy Curve
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Problems 1. The specific energy for a 3 m wide channel is to be 3 kg-m/kg. What would be the max. possible discharge 2. The discharge of water through a rectangular channel of width 6 m, is 18 m3/s when depth of flow of water is 2 m. Calculate: i) Specific Energy ii) Critical Depth iii) Critical Velocity iv) Minimum Energy 3. The specific energy for a 5 m wide rectangular channel is to be 4 Nm/N. If the rate of flow of water through the channel us 20 m3/s, determine the alternate depths of flow.
Hydraulic Jump
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Hydraulic Jump The hydraulic jump is defined as the rise of water level, which takes place due to transformation of the unstable shooting flow (super-critical) to the stable streaming flow (sub-critical). When hydraulic jump occurs, a loss of energy due to eddy formation and turbulence flow occurs.
Hydraulic Jump The most typical cases for the location of hydraulic jump are: 1. Below control structures like weir, sluice are used in the channel 2. when any obstruction is found in the channel, 3. when a sharp change in the channel slope takes place. 4. At the toe of a spillway dam
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Hydraulic Jump d1 + d2 = − 2
d2=−
d2 =
d1
d1
2 +
4
2q
2 → interms of q
g d1
2 2v 1 d 1 + → interms of V 1 4 g1
d1
+
2
2
d1
2 1 + 8 F e − 1 → interms of F e 2
Hydraulic Jump Loss of Energy 2: d2 = −
d1
+
d1
+
2q
2 → interms of q
[gdd 1− d ]3 2 1 h L = E1 − E 22= 2 d1 d1 2v 14 dd 11 d→2interms + + d2=− 2 4 g 2
4
of V 1
1
Length of jump = 5 to 7 times of (d 2 − d1) d 2
d2 =
1
1+ 8 F
− 1 → interms of F e
e Hydrualic 2 Jump = d 2 − d1
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Problems 1. The depth of flow of water, at a certain section of a rectangular channel of 2 m wide is 0.3 m. The discharge through the channel is 1.5 m3/s. Determine whether a hydraulic jump will occur, and if so, find its height and loss of energy per kg of water. 2. A sluice gate discharges water into a horizontal rectangular channel with a velocity of 10 m/s and depth of flow of 1 m. Determine the depth of flow after jump and consequent loss in total head.
Gradually Varied Flow (GVF)
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Gradually Varied Flow (GVF) In GVF, depth and velocity vary slowly, and the free surface is stable The GVF is classified based on the channel slope, and the magnitude of flow depth. Steep Slope (S): So > Sc or h < hc Critical Slope (C): So = Sc or h = hc Mild Slope (M): So < Sc or h > hc Horizontal Slope (H): So = 0 Adverse Slope(A): So = Negative where So : the slope of the channel bed, Sc : the critical slope that sustains a given discharge as uniform flow at the critical depth (hc).
Gradually Varied Flow (GVF) In GVF, depth and velocity vary slowly, and the free surface is stable The GVF is classified based on the channel slope, and the magnitude of flow depth. Steep Slope (S): So > Sc or h < hc Critical Slope (C): So = Sc or h = hc Mild Slope (M): So < Sc or h > hc Horizontal Slope (H): So = 0 Adverse Slope(A): So = Negative where So : the slope of the channel bed, Sc : the critical slope that sustains a given discharge as uniform flow at the critical depth (hc).
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Flow Profiles The surface curves of water are called flow profiles (or water surface profiles). Depending upon the zone and the slope of the bed, the water profiles are classified into 13 types as follows: 1. Mild slope curves M1, M2, M3 2. Steep slope curves S1, S2, S3 3. Critical slope curves C1, C2, C3 4. Horizontal slope curves H2, H3 5. Averse slope curves A2, A3 In all these curves, the letter indicates the slope type and the subscript indicates the zone. For example S2 curve occurs in the zone 2 of the steep slope
Normal Depth Line
Flow Profiles in Mild slope Critical Depth Line
Flow Profiles in Steep slope
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Flow Profiles in Critical slope
Flow Profiles in Horizontal slope
Flow Profiles in Adverse slope
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
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Gradually Varied Flow (GVF) Equation of GVF : ib − ie
dh =
1 −
dx
dh = dx
[
V
2
1− (F e)
2
→ in terms of Velocity
gh
ib−ie
Sc or ib Energy Line Slope So or ie Bed Slope
h1 h2
→ in terms of Fe
]
L =
ib - ie
dh where
E 2 - E1
represents the variation of water depth along the bottom of the channel dx
Gradually Varied Flow (GVF) IfEquation dh/dx =of0,GVF Free: Surface of water is parallel to the bed of channel ib − ie
dh =
→ in terms of Velocity
If dx dh/dx > 0, Depth increases in the 2 V direction1of− water flow (Back Water Curve)
Sc or ib Energy Line Slope So or ie Bed Slope
h1
gh
h2
If dh/dx < 0, Depth of water decreases in dh direction i b − i eof flow (Dropdown Curve) the =
dx
[
1− (F e)
2
→ in terms of Fe
]
E 2 - E1 ib - ie
dh where
L =
represents the variation of water depth along the bottom of the channel dx
Rambabu Palaka, Assistant Professor, BVRIT
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Open Channel Flow
2/4/2016
Problems 1. Find the rate of change of depth of water in a rectangular channel of 10 m wide and 1.5 m deep, when water is flowing with a velocity of 1 m/s. The flow of water through the channel of bed slope in 1 in 4000, is regulated in such a way that energy line is having a slope of 0.00004 2. Find the slope of the free water surface in a rectangular channel of width 20 m, having depth of flow 5 m. The discharge through the channel is 50 m3/s. The bed of channel is having a slope of 1 in 4000. Take C=60
Reference Chapter 16 A Textbook of Fluid Mechanics and Hydraulic Machines Dr. R. K. Bansal
Laxmi Publications
Rambabu Palaka, Assistant Professor, BVRIT
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