Basic Hydraulics Rabi H. Mohtar ABE 325
The river continues on its way to the sea, broken the wheel of the mill or not. Khalil Gibran The forces on moving body of fluid mass are: 1. Inertial due to mass (ρ density) 2. Gravitational associated with specific gravity 3. Viscous related to viscosity (μ) 4. Pressure to water/fluid • Steady flow vs. transient flow • Uniform vs. varied flow • Pipeflow velocity changes with CS area • Open channel velocity “V” changes with slope (s), CS area (A) and roughness (n)
Shear stress:velocity relationship dv τμ= dy
Pipeflow Open Channel
VD2 ρ inertial force v speed of flow R == Fr#= = e μ viscous force gy celerity (speed of wave)
Re <2000 laminar flow Fr#>1 supercritical
Re >4500 turbulent Fr#<1 subcritical
Mass conservation:
Q=A1V1=A2V2
Energy conservation:
Momentum conservation: Examples
AV=AV11 2 2
AV11 V=2 A2
In the above network, what are the flows from the two reservoirs:
∑=Qi 0 Q2 reservoirs + 800 + 450 – (50 + 500 + 1000 + 250 + 200 + 300+ 150 + 400 + 100)=0 Q2 reservoirs = 1700 gpm
Energy conservation 1 1. kinetic energy mass flowing at a certain velocity = mv2 2 2. potential energy mass of a body situated at a distance above a datum = Zg 3. energy due to pressure P*A=Force*L=energy PAL PAL P Unit mass ≡ = ρAL ρ
Bernoulli’s Energy Equations: Energy = constant 1P v2 + +gz=constant 2 ρ vP2 + +z=constant 2g ρg
4. Mechanical energy (hm) pump (+ve) or turbine (-ve) add or remove energy
Friction forces or losses: 10.29 n2 L Manning: h = f D5.33 Darcy Weisbach: L(ft) Q2 h (ft)=f f D(ft) 2gA2 1.852 ⎛⎞Q 1.852 K ⎜⎟ 10.44QL n ⎝⎠C Hazen-Williams = 1.852 4.87 = KQ or hf = 4.87 CDHW D \ Bernoulli’s can be written now between any 2 sections c to d (see image below) PV22 P V 11+Z + ±h = 2 +Z + 2 +h γ 1m2L2g γ 2g
∑ h=0f (around a loop) Flow will split such that flow and will have the same friction loss.
Momentum Conservation → ∑=∑F momentum fluxes leaving - ∑ momentum entering the system →→ ∑ F=Qρ ( V21 - V ) 2 Hydraulic Grade Line (HGL) =Z++PP Energy Line (EL) =Z+ V γγ2g 2 v2g 2 2 V h v2g f 2g 2 1 h V P w 2g γ 1 p P γ So p 1 γ
γ Z S Z o So
P γ
hf=hw=So for open channel
Hazen William Friction Formula 1.852 ⎛⎞Q K ⎜⎟ C h = ⎝⎠ f D4.87 m h = friction f 100 m pipe length C = Hazen – William (H-W) resistance coefficient D = diameter in mm Q = flow rate in l sec K = 1.22 * 1012 for SI units 1 C unlike n (Manning coef. ) is Inversely proportional to roughness (friction) α Pipe C values - New coated steel 120 aluminum w/ - Asbestos-cement 140 - Plastic 150
Minor losses Friction due to Bend, Valves, other fittings v2 hK= fL2g K = coefficient of head losses Valves of KL for each of the components are available in the literature.
Orifice flow Q = CA 2gh C = 0.6 - 0.63 for sharp-edged orifices = 1 for smooth orifice A = area of orifice h = head Uses of this in open-channel in pipe flow
Open Channel Hydraulics: Continuity Equation: Q = AV Q = flow rate cfs or cms V = flow velocity mps or fps A = area of flow ft2 or m2
Velocity profile
Energy equation VPVP22 1122+++=++++ YZYZh 22gg112212γγL − velocity elevation pressure=0 in open channel kinetic energy potential energy Y = flow depth Z = channel elevation (bottom) P = pressure g = unit weight of water hL = energy loss if hL = negligible 2 v ++=Yzconstant 2g if Z=0 (datum = channel bottom) v2 +=Y constant=E=specific energy for rectangular channel at depth y 2g q=vy q=flow per unit width 2 q +=y E 2gy2 specific energy diagram dE critical depth flow with min E, i.e. yc is = 0 dy dE 2 =−2q +10 = dy 2gy3
3 2 or Yqgc = / or v =1 gyc v = Froude number gyc F = 1 critical flow F>1 supercritical flow F<1 subcritical flow
VArea For non-rectangular channel F= dh=hydraulic depth= gdh top width
Life had stunned my eyes and left them so confused, they wanted to keep looking. Danté
Manning Equation 2 1 R 3 S 2 V = n n = surface roughness coefficient R = hydraulic radius (m) = Area/wetted perimeter S = slope of energy line º slope of bed V = velocity in m/sec ft Insert a factor 3 3.281= 1.486 m To convert to ft/sec 2 1 1.486R3 S 2 v = n substitute A Q=AV and R= then P 5 1 c A3 S 2 Q= 2 nP 3 c = 1 m3/sec c = 1.486 ft3/sec A = m2(ft2) P = m(ft)
S, Q, and n are solved explicitly 2 Q.np 3 S= 5 cA 3
5 1 cA3 S 2 n = 2 QP 3 A & P require an implicit solution, similarly flow depth, top width. This requires trial and error or Newton Raphson/regula falsi Design Velocity – Table 7.2 of Soil and Water Conservation Engineering, Schwab 4th ed. 0.8 – 2.5 m/sec Roughness (n) 0.04 - 0.2 for small channels
Retardance Class Class Length of grass Good stand Fair stand >30" A B 11 - 24" B C 6 – 10" C D 2 - 6" D D <2" E E To find “n”
Orifice flow equation Q=AC 2gh Q=ft3 / s design flow A=cross sectional Area ft2
C=constant = 0.6 g=gravitational acceleration 32 ft/s2 h=height of water above orifice in ft
Example: Runoff = 1.5 in Terrace = 600 ft long with spacing of 100 ft If h=4 ft (ridge height) Find the diameter (D) for 48h max time of storage. Q=1.5 in ∗∗∗ 12 in 600 100 = 7500 ft3 ft q = 7500 ft33 /8 hr/3600 s/h=0.043 ft / s 9 A== 0.0045 ft2 C2gL π D2 A== D=0.9 in=23 mm 4
Weir and Orifice Flow Equation from computer applications in hydraulic engineering Orifice flow equation across 1-2 (1 inside flow, 2 orifice location) PV22 P V 11++=++Z+ H 2 Z+ 2 H γγ1G2L2g 2g
Z122 =Z ;V= 2g(H-H L ) V2 1 ==0; Q=AV 2g(H-H ) 2g 2L P 2 ==0; Q=AV A 2g(H-H ) γ 2L P 1 = H; Q=CA 2gH γ
Weir equation:
V2L = 2g(H-H ) AV= Q=LH 2g(H-H ) 2L C=weir coefficient that includes headloss and 2g 3 Q=CLH 2
Steady, uniform flow Design criteria for open channel: 1. flow velocity is not serious scouring or sedimentation occurs 2. sufficient capacity to carry the design runoff 3. adequate depth 4. stable sideslopes
Manning’s formula: 1.49 2 1 v= R3 S 2 n v = average velocity (ft/sec) n = roughness coefficient, measure of channel lining resistance R = hydraulic radius (ft=area/wetted perimeter) S = hydraulic gradient (ft/ft) = headloss due to friction divided by channel length or sinθ, θ=angle between channel bottom as horizontal = slope of channel R is a measure of slope resistance of a channel 2 3 1.49⎛⎞ a 1 q= a⎜⎟ S 2 np⎝⎠ q is determined using design flow for T a and P are typically unknown “n” varies from 0.01 – 0.1. 0.04 is typical of grass-lined channels Θ Most efficient section is close to a circular b=2d tan 2 d=depth b=bottom width θ=side slope angle with the horizontal
1. solve quadratic equation root 2. trial and error variation of “n” and other uncertainties 1. stage and discharge
2. size and shape of channel affects R and V 3. vegetal cover seasonal changes in vegetal cover and the resistance due to flow depth
4. channel irregularities. They may change by 0.005 and therefore affect is at small values. 5. seasonal changes ice, vegetal, obstruction all change n n values table are available in all books. Table 6.1
Flow classification: Steady: does not vary in time Non-steady: varies in time Uniform: does not vary in space Non-uniform: varies in space Steady flow: 1. uniform 2. non-uniform a. gradually varied b. rapidly varie Unsteady flow 1. unsteady uniform flow (rare) 2. unsteady varied flow (unsteady flow) a. gradually varied b. rapidly varied
Open channel canal designs 1.49 2 1 v= R3 S 2 n Types of open channel problems encountered: 1. Z, S, b, d, characteristic of channel Find Q Solution direct from Manning’s Equation 2. Z, S, b, n, Q known Find y Solution A, P are functions of y Substitute all parameters (polynomial) Solve for y by trial and error 3. Z, S, Q, soil type (erodible) find b, d? Vmax in erodible channel Solution: Pick Vmax, n Manning’s equation; solve for “R” 2 equations, 2 unknowns quadratic equation with only one acceptable solution 4. Vegetated channel (n=f(d)) usually Q is known (storm), S, grass type and management known. Find depth of flow d, D, & T, t Solution: a. Find retardance class for cut and uncut grass b. Select Vmax for mowed c. Determine A from continuity d. From Vmax, slope determine R trial or error Manning’s or graphical solution for retardance class Fig. 4.15 (Soil and Water Conservation 4th ed., Schwab) e. 2 equations and 2 unknowns, solve for t and d f. the above solutions carries the flow during short grass period. How about when grass is tall? Add depth to carry capacity when grass is long. Pick D, determine T and R and A g. use retardance class chart, R, S to find v. h. Check Q for V and A i. If not adequate increase D until satisfied.
Channel Example Given: Short grass S = 0.5% = 0.005 3 Q = 50 ft /s d 1 F.B.(Free Board) = 6" 10 d = ? and total depth (D) 6 ft Z=10 n = 0.05 A = bd + Zd2 P = b + 21dZ2 + d(ft) A(ft2) P(ft) R(ft) V(ft/s) q(ft3/s) 1 16 26.1 0.61 1.52 24.3 Too shallow 2 52 46.2 1.13 2.28 118.6 Too deep 1.5 31.5 36.15 0.87 1.92 60.1 Too deep 1.4 28 34.14 0.82 1.85 51.7 Close enough