CIV2263 â Water Systems (Part 1)
CIV2263 – Water Systems (Part 1) Contents 1. Fluid Properties.………………………………………………………………………………………………………………………………………….1 2. Hydrostatics..………………………………………………………………………………………………………………………………………………3 2.1 Pressure Diagram..….….…………………………………………………………………………………………………………………………4 2.2 Pressure Measures….….…………………………………………………………………………………………………………………………5 2.3 Hydrostatic Forces.….….…………………………………………………………………………………………………………………………7 3. Fluid Kinematics….………………………………………………………………………………….……………………………………….……..…..13 4. Fluid Dynamics………...……..………………………….………………….………………………….………………………………………….……14 4.1 Free Jets……………………….………………………………………….…………………………………..………………………………….……17 4.1 Momentum Equation..….………………………………………….…………………………………..………………………………….……20 5. Pipe Flow…………………...……………..…………..…………………….………………………………….….………………………………………23 5.1 Reynolds Number..….….…………………………………………………………………………………………………………………………23 5.2 Laminar Flow….…………..…………………………………………………………………………………………………………………………26 5.3 Shear Stress………...….….…………………………………………………………………………………………………………………………28 6. Turbulent Flow…………………………………………………………..………………………….……….…..………………………………………29 6.1 Minor Losses………..….….…..……………………………………………………………………………………………………………………31 7. Pumps………….……….….………………………………………………………………………………….…..………………………………………..35 8. Long Pipes.…………………………….………………………………………….………………………………..……………………………………..38
Fluid Properties ▪ ▪
Identify key fluid properties used in the analysis of fluid behaviour Understand the concepts of density, compressibility, and viscosity
Continuum Hypothesis A fluid = a substance that deforms continuously when acting on by a shearing stress of any magnitude i.e something that flows (can be water, oil or gas) Bcas water has large no. of particles we consider it to be made up of fluid particles that interact with each other & their surroundings (these fluid particles fill the whole water body) *Note that water temperature varies with volume
Water Properties
Compressibility: assume water is incompressible = constant = 1000 kg/m3 Bcas only a tiny amount of change in vol is observed when heavily loading water it is ignored *Note gases are compressible Viscosity: Viscosity is related to the ‘fluidity’ of a fluid (force water exerts when you try change its velocity) In a fluid in motion, two adjacent particles exchange a pressure force (normal to the surface contact) & a shear stress (like internal friction) viscosity results in resistance to shear deformation via cohesion & interaction between fluid molecules
Fluid particles attached to bottom plate don’t move, velocity of particle then changes linearly from fixed plate to free plate Moving top plate w/ certain force generates shear stress in fluid () = force/area which is proportional to change of velocity (dynamic viscosity) This relationship is valid for Newtonian fluids i.e water, air & oil
Hydrostatics ▪ ▪ ▪
Understand the concept of pressure Determine the pressure in any location of a fluid at rest Understand the principles of manometers
Pressure Pressure (p) = normal force per unit of area [N/m2] at a given point acting on a given plane w/in the fluid mass of interest *Note: Pressure doesn’t change w/ orientation for fluids at rest
Pressure is constant on horizontal planes Pressure only changes vertically Pressure at surface is = in all containers
Pressure Diagram In static conditions Pressure at bottom (p.1): Sum of all forces acting on control volume = 0 m.w can be written in terms of density & vol. (Elevation & air pressure is usually given can find p.1)
Pressure at a generic point:
Hydrostatic Law In a fluid at rest the sum of the elevation plus the pressure head is constant: elevation (z) + pressure head (density/specific weight) = piezometric head (constant)
Linear Pressure Diagram
P.a = atmospheric pressure Pressure linearly w/ depth Pressure due to weight of water above certain elevation
Absolute & Gage Pressure We will use gage pressure in CIV2263 Gage pressure (pg) = absolute pressure – atmospheric pressure We consider the red line as our pressure value (not the green)
Gage Pressure Choose an elevation at atmospheric pressure as a reference! ▪ p>0 when the pressure is larger than the atmospheric pressure ▪ p SR3 because of inherent instability in two channel systems Unstable Channel If below Sv < SR1 we have unstable channel & no adjustment mechanism exists. In long term, equilibrium can be achieved by adjusting valley slope
1.5 Bed Forms ▪ The bed forms are flow-induced ▪ Bed forms directly affect the flow resistance ▪ Computation of the river stage & flow velocity relies on the determination of bed form roughness
Bars - Point bars (little beaches at elbow of meanders) - Alternating Bars tend to be distributed periodically along a channel w/ bars near alternate channel banks Ripples - Sinusodial shapes, 30mm height Dunes - Larger than ripples but smaller than bars - Longitudinal shape is out of phase w/ the water surface dunes cause large eddies to occur & causes turbulence @ surface Transition - Low amplitude ripples or dunes and flat areas. - Transition stage occurs at a higher flow intensity than dunes Antidunes - Bed forms which exist in phase w/ water surface waves (water surface total shaped by dunes) - Free surface waves have larger amplitude than sand waves & grow w/ velocity & Fr. No. eventually becoming unstable & breaking in the upstream direction Chutes & Pools - Comprise large elongated mounds of sediment forming chutes w/in which the flow is supercritical and connected by pools w/ flow that may be supercritical or subcritical i.e Hydraulic jumps - Occurs at relatively large slopes & sediment discharges
1.4.Spillway Design Procedure
1. Select crest elevation (W) 2. Choose crest width (B) 3. Determine Design Flow rate (Qd) from risk analysis & flood routing 4. For Qd determine h from q Can change B & Qd until we get reasonable result 5. Choose chute toe elevation (Za) Then can find Z 6. Calculate terminal velocity (vt) & y1 W/ vt can find y1 7. Calculate length of spillway, Lr = length of energy dissipator Note: Different spillways have different eqns. 8. Calculate downstream depth y2 Use alternate depth equation 9. Compare JHRL & TWRL so 𝐽𝐻𝑅𝐿 = 𝑍𝑎 + 𝑦2 JHRL = Jump height rating level TWRL = Tail-water rating level Need y2 > tail water river level (TWRL) so JHRL = TWRL If JHRL TWRL: Need to alter crest elevation, crest width, or elevation of stilling basic bed & re-calc E.g.
E.g.
A) Want max discharge (Q) capacity of spillway - As q = Q/B - so Q = q*B = 1475 m3/s need to evacuate 1475m3/s in extreme events
b) Terminal velocity vt = sqrt[2g(Z – h/2)] Z = From toe to top of water vt = 19.6m/s
c) Calcukate JHRL - Find y1 = Q / b*v = 1.232m - Then final alternate depth y2 using y1 & FR1 - so JHRL = Chute toe level + y2
e) Select type of stilling basin - Knowing Fr1 = 7.906 & q = 12.3 - Select appropriate basin recommend SAF
d) Find height of broad-crested weir to tail-water level to match JHRL - Need to design weir so we have 6.704m in spilling basin - Calculate E2 we know y2, Q & width (b) - Over weir we want Ec Need 3.734m of energy to go over weir Have 6.875m of energy Difference is height of weir
Fluid Properties ▪ ▪
Identify key fluid properties used in the analysis of fluid behaviour Understand the concepts of density, compressibility, and viscosity
Continuum Hypothesis A fluid = a substance that deforms continuously when acting on by a shearing stress of any magnitude i.e something that flows (can be water, oil or gas) Bcas water has large no. of particles we consider it to be made up of fluid particles that interact with each other & their surroundings (these fluid particles fill the whole water body) *Note that water temperature varies with volume
Water Properties
Compressibility: assume water is incompressible = constant = 1000 kg/m3 Bcas only a tiny amount of change in vol is observed when heavily loading water it is ignored *Note gases are compressible Viscosity: Viscosity is related to the ‘fluidity’ of a fluid (force water exerts when you try change its velocity) In a fluid in motion, two adjacent particles exchange a pressure force (normal to the surface contact) & a shear stress (like internal friction) viscosity results in resistance to shear deformation via cohesion & interaction between fluid molecules
Fluid particles attached to bottom plate don’t move, velocity of particle then changes linearly from fixed plate to free plate Moving top plate w/ certain force generates shear stress in fluid () = force/area which is proportional to change of velocity (dynamic viscosity) This relationship is valid for Newtonian fluids i.e water, air & oil
Hydrostatics ▪ ▪ ▪
Understand the concept of pressure Determine the pressure in any location of a fluid at rest Understand the principles of manometers
Pressure Pressure (p) = normal force per unit of area [N/m2] at a given point acting on a given plane w/in the fluid mass of interest *Note: Pressure doesn’t change w/ orientation for fluids at rest
Pressure is constant on horizontal planes Pressure only changes vertically Pressure at surface is = in all containers
Pressure Diagram In static conditions Pressure at bottom (p.1): Sum of all forces acting on control volume = 0 m.w can be written in terms of density & vol. (Elevation & air pressure is usually given can find p.1)
Pressure at a generic point:
Hydrostatic Law In a fluid at rest the sum of the elevation plus the pressure head is constant: elevation (z) + pressure head (density/specific weight) = piezometric head (constant)
Linear Pressure Diagram
P.a = atmospheric pressure Pressure linearly w/ depth Pressure due to weight of water above certain elevation
Absolute & Gage Pressure We will use gage pressure in CIV2263 Gage pressure (pg) = absolute pressure – atmospheric pressure We consider the red line as our pressure value (not the green)
Gage Pressure Choose an elevation at atmospheric pressure as a reference! ▪ p>0 when the pressure is larger than the atmospheric pressure ▪ p SR3 because of inherent instability in two channel systems Unstable Channel If below Sv < SR1 we have unstable channel & no adjustment mechanism exists. In long term, equilibrium can be achieved by adjusting valley slope
1.5 Bed Forms ▪ The bed forms are flow-induced ▪ Bed forms directly affect the flow resistance ▪ Computation of the river stage & flow velocity relies on the determination of bed form roughness
Bars - Point bars (little beaches at elbow of meanders) - Alternating Bars tend to be distributed periodically along a channel w/ bars near alternate channel banks Ripples - Sinusodial shapes, 30mm height Dunes - Larger than ripples but smaller than bars - Longitudinal shape is out of phase w/ the water surface dunes cause large eddies to occur & causes turbulence @ surface Transition - Low amplitude ripples or dunes and flat areas. - Transition stage occurs at a higher flow intensity than dunes Antidunes - Bed forms which exist in phase w/ water surface waves (water surface total shaped by dunes) - Free surface waves have larger amplitude than sand waves & grow w/ velocity & Fr. No. eventually becoming unstable & breaking in the upstream direction Chutes & Pools - Comprise large elongated mounds of sediment forming chutes w/in which the flow is supercritical and connected by pools w/ flow that may be supercritical or subcritical i.e Hydraulic jumps - Occurs at relatively large slopes & sediment discharges
1.4.Spillway Design Procedure
1. Select crest elevation (W) 2. Choose crest width (B) 3. Determine Design Flow rate (Qd) from risk analysis & flood routing 4. For Qd determine h from q Can change B & Qd until we get reasonable result 5. Choose chute toe elevation (Za) Then can find Z 6. Calculate terminal velocity (vt) & y1 W/ vt can find y1 7. Calculate length of spillway, Lr = length of energy dissipator Note: Different spillways have different eqns. 8. Calculate downstream depth y2 Use alternate depth equation 9. Compare JHRL & TWRL so 𝐽𝐻𝑅𝐿 = 𝑍𝑎 + 𝑦2 JHRL = Jump height rating level TWRL = Tail-water rating level Need y2 > tail water river level (TWRL) so JHRL = TWRL If JHRL TWRL: Need to alter crest elevation, crest width, or elevation of stilling basic bed & re-calc E.g.
E.g.
A) Want max discharge (Q) capacity of spillway - As q = Q/B - so Q = q*B = 1475 m3/s need to evacuate 1475m3/s in extreme events
b) Terminal velocity vt = sqrt[2g(Z – h/2)] Z = From toe to top of water vt = 19.6m/s
c) Calcukate JHRL - Find y1 = Q / b*v = 1.232m - Then final alternate depth y2 using y1 & FR1 - so JHRL = Chute toe level + y2
e) Select type of stilling basin - Knowing Fr1 = 7.906 & q = 12.3 - Select appropriate basin recommend SAF
d) Find height of broad-crested weir to tail-water level to match JHRL - Need to design weir so we have 6.704m in spilling basin - Calculate E2 we know y2, Q & width (b) - Over weir we want Ec Need 3.734m of energy to go over weir Have 6.875m of energy Difference is height of weir
