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Watershed Sciences 6900FLUVIAL HYDRAULICS & ECOHYDRAULICS
FORCES & FLOW CLASSIFICATION + ENERGY & MOMENTUM PRINCIPLES
Joe Wheaton
Slides by Wheaton et al. (2009-2014) are licensed under a Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
TODAY’S PLAN A
I. Force Classification
II. Overall Force Balance
III. Take-Homes from Summary of Force Magnitudes
IV. Reynolds Number as a Force Ratio
V. Froude Number as a Force Ratio
I. Froude Number & Hydraulic Jumps
FORCES & FLOW CLASSIFCATION
From Dingman (2009)
WE’VE ALREADY DEALT WITH FORCES… BUT:
• Understanding the relative magnitudes of all forces in different situations gives us a basis for ignoring them… (i.e. simplifying assumptions)
• To keep comparison simple, consider:
Wide, rectangular channel (Y=R) with constant width (W1=W2=W) but
spatially variable depth
𝐹𝐷 = 𝐹𝑅 𝐹 = 𝑚 ∙ 𝑎
From Dingman (2008), Chapter 7
TODAY’S PLAN A
I. Force Classification
II. Overall Force Balance
III. Take-Homes from Summary of Force Magnitudes
IV. Reynolds Number as a Force Ratio
V. Froude Number as a Force Ratio
I. Froude Number & Hydraulic Jumps
FORCES & FLOW CLASSIFCATION
From Dingman (2008)
TODAY’S PLAN A
I. Force Classification
II. Overall Force Balance
III. Take-Homes from Summary of Force Magnitudes
IV. Reynolds Number as a Force Ratio
V. Froude Number as a Force Ratio
I. Froude Number & Hydraulic Jumps
FORCES & FLOW CLASSIFCATION
What did Dingman Mean by Force Balance?
From Dingman (2008), Chapter 7
RECALL… 1D FORM OF NAVIER-STOKES
• Conservation of Momentum… in 1D:
𝑚 ∙ 𝑎 = 𝐹
𝜌𝜕𝑈𝑥𝜕𝑡+ 𝑈𝑥𝜕𝑈𝑥𝜕𝑥= −𝜌𝑔
𝜕𝑧
𝜕𝑥−𝜕𝑃
𝜕𝑥+ 𝐹𝑣𝑖𝑠𝑐𝑥
𝑄 = 𝑊 ∙ 𝑌1 ∙ 𝑈1 = 𝑊 ∙ 𝑌2 ∙ 𝑈2
Wide, rectangular channel (Y=R) with
constant width (W1=W2=W) but
spatially variable depth
SO IN TERMS OF a= 𝐹
𝑚CLASSIFICATION
𝜌𝜕𝑈𝑥𝜕𝑡+ 𝑈𝑥𝜕𝑈𝑥𝜕𝑥= −𝜌𝑔
𝜕𝑧
𝜕𝑥−𝜕𝑃
𝜕𝑥+ 𝐹𝑣𝑖𝑠𝑐𝑥
𝜕𝑈𝑥𝜕𝑡+ 𝑈𝑥𝜕𝑈𝑥𝜕𝑥= −𝑔𝜕𝑧
𝜕𝑥−1
𝜌∙𝜕𝑃
𝜕𝑥+ 𝐹𝑣𝑖𝑠𝑐𝑥𝜌
𝑎𝑡 + 𝑎𝑋 = −𝑎𝐺 − 𝑎𝑃 + 𝑎𝑉
1D Momentum Equation…
Rearranged in terms of our
classification…
Matched to Table 7.1
From Dingman (2008), Chapter 7
RECALL: DEFINITIONS OF UNIFORM & STEADY FLOW
• Uniform Flow: if the element velocity at any instant is constant along a streamline the flow is uniform (i.e. convective acceleration 𝑑𝑢 𝑑𝑥 = 0),.
• Steady Flow: if the element velocity 𝑢 at any
given point on a streamline does not change with time, the flow is steady (i.e. local acceleration 𝑑𝑢 𝑑𝑡 = 0) .
From Dingman (2008) – Chapter 4& 6
THREE MAIN OPEN CHANNEL FLOW TYPES
1. Steady Uniform Flow
2. Steady Nonuniform Flow
3. Unsteady Nonuniform Flow
𝑎𝑡 + 𝑎𝑋 = −𝑎𝐺 − 𝑎𝑃 + 𝑎𝑉𝑎𝐺 + 𝑎𝑃 − 𝑎𝑉 + 𝑎𝑇 + 𝑎𝑋 + 𝑎𝑡 = 0
𝐹𝐷 = 𝐹𝑅
• Downstream forces (+): 𝐹𝐷• Upstream Forces (-): 𝐹𝑅
𝑎𝐺 − 𝑎𝑉 + 𝑎𝑇 = 0 EQ 7.2
EQ 7.3
EQ 7.4
𝑎𝐺 + 𝑎𝑃 − 𝑎𝑉 + 𝑎𝑇 + 𝑎𝑋 = 0
• Steady/Unsteady (Time)• Uniform/Nonuniform (Space)
𝑎𝐺 + 𝑎𝑃 − 𝑎𝑉 + 𝑎𝑇 + 𝑎𝑋 + 𝑎𝑡 = 0
𝑎𝐺 + 𝑎𝑃 − 𝑎𝑉 + 𝑎𝑇 = 𝑎𝑋
𝑎𝐺 + 𝑎𝑃 − 𝑎𝑉 + 𝑎𝑇 = 𝑎𝑋 + 𝑎𝑡
SO WHAT IS NAVIER STOKES?
• Unsteady or Steady?
• Uniform or Nonuniform?
𝜌𝜕𝑈𝑥𝜕𝑡+ 𝑈𝑥𝜕𝑈𝑥𝜕𝑥= −𝜌𝑔
𝜕𝑧
𝜕𝑥−𝜕𝑃
𝜕𝑥+ 𝐹𝑣𝑖𝑠𝑐𝑥
𝑎𝐺 + 𝑎𝑃 − 𝑎𝑉 + 𝑎𝑇 = 𝑎𝑋 + 𝑎𝑡
𝜕𝑈𝑥𝜕𝑡+ 𝑈𝑥𝜕𝑈𝑥𝜕𝑥= −𝑔𝜕𝑧
𝜕𝑥−1
𝜌∙𝜕𝑃
𝜕𝑥+ 𝐹𝑣𝑖𝑠𝑐𝑥𝜌
𝑎𝑡 + 𝑎𝑋 = −𝑎𝐺 − 𝑎𝑃 + 𝑎𝑉
“The difference between the driving & resisting forces is acceleration.”
𝐹𝐷 ≠ 𝐹𝑅
𝐹𝐷 − 𝐹𝑅 = 𝑎EQ 7.4
TODAY’S PLAN A
I. Force Classification
II. Overall Force Balance
III. Take-Homes from Summary of Force Magnitudes
IV. Reynolds Number as a Force Ratio
V. Froude Number as a Force Ratio
I. Froude Number & Hydraulic Jumps
FORCES & FLOW CLASSIFCATION
RANGE OF FORCE VALUES
• Downstream forces (+): 𝐹𝐷• Upstream Forces (-): 𝐹𝑅
• Body Forces
• Surface Forces
TODAY’S PLAN A
I. Force Classification
II. Overall Force Balance
III. Take-Homes from Summary of Force Magnitudes
IV. Reynolds Number as a Force Ratio
V. Froude Number as a Force Ratio
I. Froude Number & Hydraulic Jumps
FORCES & FLOW CLASSIFCATION
From Dingman (2008)
REYNOLDS NUMBER
• Can also be shown that 𝑅𝑒 ∝𝑎𝑇
𝑎𝑉(ratio of turbulent
forces to viscous forces):
• So… is it fair to say that Reynolds number is sort of a ratio of driving forces to resisting forces?
• No! Why?
𝑅𝑒 ≡𝜌∙𝑌∙𝑈
𝜇=𝑌∙𝑈
𝜈Ratio of ‘inertial forces’ to viscous forces
𝑎𝑇𝑎𝑉=
Ω2 ∙ 𝜌 ∙ 𝑈2
𝑌3 ∙ 𝜇 ∙ 𝑈𝑌2
=Ω2 ∙ 𝜌 ∙ 𝑈 ∙ 𝑌
3 ∙ 𝜇=Ω2 ∙ 𝑈 ∙ 𝑌
3 ∙ 𝜐=Ω2
3∙ 𝑅𝑒
TODAY’S PLAN A
I. Force Classification
II. Overall Force Balance
III. Take-Homes from Summary of Force Magnitudes
IV. Reynolds Number as a Force Ratio
V. Froude Number as a Force Ratio
I. Froude Number & Hydraulic Jumps
FORCES & FLOW CLASSIFCATION
From Dingman (2008)
TODAY’S PLAN A
I. Force Classification
II. Overall Force Balance
III. Take-Homes from Summary of Force Magnitudes
IV. Reynolds Number as a Force Ratio
V. Froude Number as a Force Ratio
I. Froude Number & Hydraulic Jumps
FORCES & FLOW CLASSIFCATION
From Dingman (2008)
FROUDE NUMBER
• What are Dimensions?
𝐹𝑟 ≡𝑈
𝑔 ∙ 𝑌 12
Ratio of velocity to
Celerity (𝐶𝑔𝑤 = 𝑔 ∙ 𝑌)
[LT-1]
[LT-2 L]1/2
[LT-2 L]1/2 =LT-1
WHAT DOES THIS RATIO MEAN PHYSICALLY?
• If 𝐹𝑟 < 1, flow is subcritical
– Meaning that the celerity is > then the velocity; ∴ waves can propagate upstream
• If 𝐹𝑟 = 1, flow is critical
– Meaning that the celerity is = velocity (rare transitional point)
• If 𝐹𝑟 > 1, flow is supercritical
– Meaning that the celerity is < then the velocity; ∴ waves cannot propagate
upstream
𝐹𝑟 ≡𝑈
𝑔 ∙ 𝑌 12
THE HYDRAULIC JUMP
• Transition from supercritical flow to subcritical flow is a hydraulic jump
• For a hydraulic jump to occur, there has to be a transition from subcritical to supercritical flow!
• Characterized by:
– Development of large-scale turbulence
– Surface waves
– Spray
– Energy dissipation
– Air entrainment
– River can not maintain supercritical state over too long a distance…
WHY CAN YOU SURF IN A RIVER?
• Explain it in terms of Froude Number…
• Which way is water going? (kayakers can’t answer!)
TODAY’S PLAN B
I. Energy Principle in 1D Flows
I. The Energy Equation (§ 8.1.1)
II. Specific Energy (§ 8.1.2)
ENERGY & MOMENTUM PRINCIPLES (Applied to 1D flows)
From Chanson (2004)
TODAY’S PLAN
I. Energy Principle in 1D Flows
I. The Energy Equation (§ 8.1.1)
II. Specific Energy (§ 8.1.2)
ENERGY & MOMENTUM PRINCIPLES (Applied to 1D flows)
From Dingman (2008), Chapter 8
TOTAL ENERGY 𝒉 (FROM §4.4)
• The total energy ℎ of an element is the sum of its potential energy ℎ𝑃𝐸 & its kinetic energy ℎ𝐾𝐸:
• Recall that ℎ𝑃𝐸 is the sum of gravitational potential energy ℎ𝐺 and pressure potential energy ℎ𝑃 :
• ∴ total energy ℎ is:
From Dingman (2008), Chapter 8
ℎ = ℎ𝑃𝐸 + ℎ𝐾𝐸
ℎ𝑃𝐸 = ℎ𝐺 + ℎ𝑃
ℎ = ℎ𝐺 + ℎ𝑃 + ℎ𝐾𝐸
EQ 8.1
EQ 8.2
Energy quantities are expressed as energy
[F L] divided by weight [F], which is
called head [L]
DEFINITON DIAGRAM FOR ENERGY IN 1D
• Used for derivation of the macroscopic one-dimensional energy equation
• Flow through reach is steady
From Dingman (2008), Chapter 8
TODAY’S PLAN B
I. Energy Principle in 1D Flows
I. The Energy Equation (§ 8.1.1)
II. Specific Energy (§ 8.1.2)
ENERGY & MOMENTUM PRINCIPLES (Applied to 1D flows)
From Dingman (2008), Chapter 8
TOTAL MECHANICAL ENERGY AT A CROSS SECTION
• Based on what Ryan showed us, using the channel bottom as a reference point, at cross section 𝑖 we can
write the integrated gravitational head (AKA elevation head), 𝐻𝐺𝑖:
• Where 𝑍𝑖 is the elevation of the channel bottom, and the integrated pressure head 𝐻𝑃𝑖 is:
• Kinetic energy head 𝐻𝐾𝐸 for a fluid element with velocity 𝑢 is given by:
From Dingman (2008), Chapter 8
𝐻𝐺𝑖 = 𝑍𝑖 EQ 8.3
𝐻𝑃𝑖 = 𝑌𝑖 ∙ 𝑐𝑜𝑠𝜃0 EQ 8.4
ℎ𝐾𝐸 =𝑢2
2 ∙ 𝑔EQ 8.5
VELOCITY COEFFICIENT 𝜶 FOR ENERGY
• Because velocity varies from point to point n a cross section, we need a fudge factor 𝛼𝑖 to account for this
variation, so the velocity head 𝐻𝐾𝐸𝑖 is:
ℎ𝐾𝐸𝑖 =𝛼𝑖 ∙ 𝑈𝑖
2
2 ∙ 𝑔
EQ 8.6
From Dingman (2008), Chapter 8
A CONVENIENT OUTCOME OF BOX 8.2
Velocity Head 𝐻𝐾𝐸 tends to be at
least an order of magnitude smaller than pressure head 𝐻𝑃
𝛼 ≈ 1.3 good enough typically…
From Dingman (2008), Chapter 8
TOTAL HEAD 𝑯𝒊 AT A CROSS SECTION 𝒊
• ‘The total mechanical energy-per-weight, or total head 𝐻𝑖, at cross-section 𝑖, is the sum of the gravitational, pressure and velocity heads:
𝐻𝑖 = 𝐻𝐺𝑖 +𝐻𝑃𝑖 +𝐻𝐾𝐸𝑖ℎ = ℎ𝐺 + ℎ𝑃 + ℎ𝐾𝐸
𝐻𝑖 = 𝑍𝑖 + 𝑌𝑖 ∙ 𝑐𝑜𝑠𝜃0 +𝛼𝑖 ∙ 𝑈𝑖
2
2 ∙ 𝑔
From Dingman (2008), Chapter 8
EQ 8.6
But, this is just at one cross section 𝒊… What
about for this diagram?
APPLY 𝑯𝒊 BETWEEN TWO CROSS SECTIONS
• What is the change in in cross-sectional integrated energy from an upstream section (𝑖 = 𝑈) to a downstream section (𝑖 = 𝐷)?
• Where ∆𝐻 is the energy lost
(converted to head) per weigh of fluid, or head loss.
From Dingman (2008), Chapter 8
𝑍𝑈 + 𝑌𝑈 ∙ 𝑐𝑜𝑠𝜃0 +𝛼𝑈∙𝑈𝑈
2
2∙𝑔=𝑍𝐷 + 𝑌𝐷 ∙ 𝑐𝑜𝑠𝜃0 +
𝛼𝐷∙𝑈𝐷2
2∙𝑔+ ∆𝐻
EQ 8.8b
𝐻𝑖 = 𝑍𝑖 + 𝑌𝑖 ∙ 𝑐𝑜𝑠𝜃0 +𝛼𝑖 ∙ 𝑈𝑖
2
2 ∙ 𝑔 EQ 8.6
𝐻𝐺𝑈 + 𝐻𝑃𝑈 + 𝐻𝐾𝐸𝑈 = 𝐻𝐺𝐷 + 𝐻𝑃𝐷 + 𝐻𝐾𝐸𝐷 + ∆𝐻EQ 8.8a
This is the ENERGY EQUATION!
THE ENERGY EQUATION APPLIES WHERE?
• Depth can vary!
• Assumed that pressure distribution was hydrostatic (i.e. streamlines no significantly curved)
• We call this steady gradually varied flow
• This is focus of Chapter 9
• We’ll use it in flume on Thursday!
𝑍𝑈 + 𝑌𝑈 ∙ 𝑐𝑜𝑠𝜃0 +𝛼𝑈∙𝑈𝑈
2
2∙𝑔=𝑍𝐷 + 𝑌𝐷 ∙ 𝑐𝑜𝑠𝜃0 +
𝛼𝐷∙𝑈𝐷2
2∙𝑔+ ∆𝐻
EQ 8.8b
SOME IMPORTANT IMAGINARY LINES
• The line representing the total potential energy from section to section is called the piezometric head line
𝑆𝐸 ≡ −𝐻𝐷−𝐻𝑈
∆𝑋=∆𝐻
∆𝑋= EQ 8.9
• The line representing the total head from section to section is called the energy grade line
• The slope 𝑆𝐸 of the energy grade line is the energy slope:
WOULD THE ENERGY EQUATION APPLY HERE?
• Hint, energy loss..
• Okay.. If it does apply, what would the energy grade line look like for a hydraulic jump?
TODAY’S PLAN B
I. Energy Principle in 1D Flows
I. The Energy Equation (§ 8.1.1)
II. Specific Energy (§ 8.1.2)
ENERGY & MOMENTUM PRINCIPLES (Applied to 1D flows)
From Dingman (2008), Chapter 8
SPECIFIC ENERGY 𝑯𝑺𝒊
• Similar to 𝐻𝑖, just that datum is changed so that mechanical energy is measured with respect to the channel bottom instead of horizontal datum
• Upshot is that elevation head 𝐻𝐺𝑖 → 0 in 𝐻𝑖 = 𝐻𝐺𝑖 +𝐻𝑃𝑖 +𝐻𝐾𝐸𝑖 , such that:
𝐻𝑆𝑖 = 𝐻𝑃𝑖 +𝐻𝐾𝐸𝑖
𝐻𝑆𝑖 = 𝑌𝑖 ∙ 𝑐𝑜𝑠𝜃0 +𝛼𝑖 ∙ 𝑈𝑖
2
2 ∙ 𝑔EQ 8.10
SPECIFIC HEAD APPLICATION…
• Consider a flow of discharge 𝑄 in a channel of constant width 𝑊:
• Which can be substituted into our definition of specific
energy (𝐻𝑆𝑖 = 𝑌𝑖 ∙ 𝑐𝑜𝑠𝜃0 +𝛼𝑖∙𝑈𝑖
2
2∙𝑔):
• If 𝑄 and width 𝑊 are constant, specific head 𝐻𝑆 only depends on flow depth 𝑌. However, 𝐻𝑆 is a function of both 𝑌 and 𝑌2, which means it can be solved with two different positive values of depth 𝑌!
𝑄 = 𝑊 ∙ 𝑌𝑖 ∙ 𝑈𝑖
𝐻𝑆 = 𝑌 +𝛼𝑖 ∙ 𝑄
2
2 ∙ 𝑔 ∙ 𝑊2 ∙ 𝑌2 EQ 8.12
From Dingman (2008), Chapter 8
ALTERNATE DEPTHS VS. CRITICAL DEPTH
• Fancy way of saying there are two depth 𝑌 solutions for every specific head 𝐻𝑆, except at 𝐻𝑆𝑚𝑖𝑛 , which is the
critical depth (i.e. 𝐻𝑆𝑚𝑖𝑛 = 𝑌𝐶).
• To find 𝑌𝐶, find where derivative of 𝐻𝑆 is
equal to zero:
𝐻𝑆 = 𝑌 +𝛼𝑖 ∙ 𝑄
2
2 ∙ 𝑔 ∙ 𝑊2 ∙ 𝑌2
𝑑𝐻𝑆𝑑𝑌= 1 −
𝑄2
𝑔 ∙ 𝑊2 ∙ 𝑌3= 0
EQ 8.12
EQ 8.13
𝑌𝐶 =𝑄2
𝑔 ∙ 𝑊2
1 3
EQ 8.14
𝑌𝐶 =𝑊 ∙ 𝑌𝐶 ∙ 𝑈𝐶
2
𝑔 ∙ 𝑊2
1 3
=𝑈𝐶2
𝑔EQ 8.15
WHY IS IT CRITICAL DEPTH?
• If 𝑌𝐶 = 1 =𝑈𝐶2
𝑔, notice the similarity with
the Froude Number 𝐹𝑟:
• The minimum value of 𝐻𝑆 occurs when 𝐹𝑟2 = 1 (i.e. when 𝐹𝑟 = 1) or when flow is critical!
• ddd
𝑌𝐶 =𝑈𝐶2
𝑔 EQ 8.15
𝐹𝑟 ≡𝑈
𝑔 ∙ 𝑌 12
TODAY’S PLAN B
I. Energy Principle in 1D Flows
I. The Energy Equation (§ 8.1.1)
II. Specific Energy (§ 8.1.2)
ENERGY & MOMENTUM PRINCIPLES (Applied to 1D flows)
From Dingman (2008), Chapter 8
THIS WEEK’S LAB
• We will do a complete lab assignment and calculations for a hydraulic jump produced by a sluice gate
• You’ll get to put the energy equation to use
• We will then play with different configurations of the flume to induce different hydraulics. We will play with:
– Width constrictions
– Different spillways
– Altering Q & S