t2

TOPIC T2 - FLOW IN PIPES AND CHANNELS

AUTUMN 2002

1. Pipe flow 1.1 Introduction 1.2 Governing equations for circular pipes 1.3 Laminar flow 1.4 Turbulent flow 1.5 Expressions for the friction factor f 1.6 Frictional losses in pipes 1.7 Other losses 1.8 Energy and hydraulic grade lines 1.9 Pipe networks 2. Open-channel flow 2.1 Uniform flow 2.1.1 Hydraulic radius and the drag law 2.1.2 Friction factors 2.1.3 Chézy and Manning’s formulae 2.1.4 Uniform flow calculations 2.1.5 Optimal shape of cross-section 2.2 Introduction to non-uniform flow 2.2.1 The Froude number – subcritical and supercritical flow 2.2.2 Frictionless flow over an obstacle 2.2.3 Gradually-varying flow 2.2.4 Rapidly-varying flow – the hydraulic jump Appendix

Hydraulics 2

T2-1

David Apsley

1. PIPE FLOW 1.1 Introduction The flow of water, oil and gas in pipelines is of immense importance to engineers. An understanding of the relationship between flow rate and head loss is vital in designing reservoir and distribution systems. Laminar and Turbulent Flow In 1883, Osborne Reynolds’s pipe-flow experiments demonstrated the occurrence of two regimes of flow - laminar or turbulent – according to the Reynolds number. For pipes, the conventional choice of length and velocity scales gives VD Re ≡ (1) ν where V = average velocity (= flow rate ÷ area) D = diameter ν = kinematic viscosity (= µ/ρ)

laminar

turbulent

For smooth-walled pipes the critical Reynolds number for transition is generally accepted to be Recrit ≈ 2300 (2) but in commercial pipes transition from intermittent to fully-turbulent flow typically occurs over a range 2000 < Re < 4000.

Development Length At inflow, the flow velocity is often nearly uniform. Boundary layers develop on the side walls because of frictional drag. These grow with distance to merge at the centre of the pipe. Beyond this distance the velocity profile becomes fully-developed (i.e., doesn’t change any further with distance). Typical correlations for this development length are: Ldevel 0.06 Re (laminar) (3) = 1/6 (turbulent) D î 4.4 Re The kinematic viscosity of air and water is such that most pipe and duct flows in civil engineering have high Reynolds numbers and are fully turbulent. The development length can usually be ignored as a function of total length. However, there are some duct flows – notably laboratory wind tunnels – where one would prefer the development length to be much longer than the working section. Exercise. νwater = 1.0×10-6 m2 s-1. Calculate the Reynolds numbers for average velocity 0.5 m s-1 in pipes of inside diameter 12 mm and 0.3 m. Estimate the development length in each case. Answer. Re = 6000, 1.5×105. Ldevel = 0.23 m, 9.6 m. Hydraulics 2

T2-2

David Apsley

1.2 Governing Equations For Circular Pipes Fully-developed pipe flow is determined by a balance between: - pressure - weight - shear stress We shall see that the net effect of pressure and weight can be combined as a piezometric pressure p* = p + ρgz (4) Changes in piezometric pressure are conveniently expressed as head losses hf via − ∆p * or ∆p * = −ρgh f = hf ρg

(5)

For a circular pipe of radius R, consider the component of forces along the pipe axis for a cylindrical element of radius r < R and length δl.

δl R

p

τ

r

mg pressure forces: weight (component): viscous forces:

p+δp

θ

p (πr 2 ) − ( p + δp )(πr 2 ) mg sin θ − τ(2πrδl )

Note that: (i) p is the average pressure over a cross-section; for circular pipes this is equal to the centreline pressure, with equal and opposite hydrostatic variation above and below; (ii) for this analysis a stress direction is assumed such that τ is positive. Balancing forces:






− δp (πr 2 ) + ρ(πr 2 δl ) g sin θ − τ(2πrδl ) = 0 pressure

weight

shear stress

δz , δl δp δz τ − − ρg −2 =0 r δl δl

Dividing by πr2 δl, and noting that sin θ = −

Hydraulics 2

T2-3

David Apsley

Hence, 1 dp * τ=− r 2 dl

(6)

Note that pressure only appears in the combination p* = p + ρgz, the piezometric pressure. Equation (6) determines the size of pressure gradient required to drive the flow against frictional forces. Note that: (i) p* is independent of radius (since there is no acceleration across the pipe); dp * (ii) is independent of distance l (since τ is independent of l in fully-developed flow). dl Hence, we may write, using G for pressure Gradient and hf for the head loss over a finite length L: dp * − ∆p * ρgh f = = , constant (7) G=− dl L L Then the equation governing behaviour of shear stress τ across the pipe is τ = 12 Gr

(8)

Shear stress varies linearly from 0 at the centre to a maximum at the walls. In particular the wall shear stress is given by GD τw = (9) 4 where D = 2R is the pipe diameter.

Equations (8) and (9) apply in all fully-developed pipe flow, whether laminar or turbulent. For laminar flow we can use it to establish an actual velocity profile, because we can write τ in terms of ∂u/∂r (section 1.3). For turbulent flow a velocity profile is unobtainable, but gross parameters (flow rate, head loss, ...) may be obtained if τw can be related (empirically) to the average velocity V by means of a friction factor (section 1.4).

Hydraulics 2

T2-4

David Apsley

1.3 Laminar Flow Laminar, pressure-driven flow through a circular pipe is often called Poiseuille flow or Hagen1-Poiseuille2 flow. It is one of a very small class of simple flows for which exact solutions of the Navier-Stokes equations can be written down analytically. Others include: • Plane Poiseuille flow – pressure-driven flow between parallel plane walls; • Couette3 flow – flow between parallel plane walls, driven by the movement of one wall; • flow between concentric rotating cylinders. In each case a critical Reynolds number exists, beyond which the laminar flow solution becomes unstable to small disturbances and the flow becomes turbulent. In fully-developed pipe flow, shear stresses balance pressure and weight: τ = 12 Gr where dp * ρgh f = G=− dl L is the pressure gradient.

(8)´

(10)

Viscous forces act in flows with velocity gradients. Across any interface the drag of upper fluid on lower fluid is equal and opposite to the drag of lower fluid on upper. The direction of the drag force is such as to reduce the velocity difference. The magnitude of the drag force is given by: stress (force per unit area) = viscosity × velocity gradient With the sign convention adopted for τ here this gives (for the force on the inner fluid): ∂u −τ=µ ∂r ∂u 1G =− ⇒ r ∂r 2µ This integrates to give G 2 u=− r + constant 4µ Applying the no-slip condition at the wall (u = 0 on r = R) the final velocity solution is G (R2 − r2 ) u= (11) 4µ The shape of the velocity profile is parabolic (quadratic in r).

1

G.L.H. Hagen, German engineer who, in 1839, measured water flow in long brass pipes and reported that there appeared to be two regimes of flow. 2 J.L.M Poiseuille (1799-1869), French physician who was interested in flow in blood vessels. 3 M.F.A. Couette (1858-1943)

Hydraulics 2

T2-5

David Apsley

Exercise. Find, from the velocity distribution given above, (a) the centreline velocity; (b) the average velocity V; (c) the flow rate in terms of head loss and pipe diameter; 1 (d) the momentum factor β = ∫ ρu 2 dA (= factor by which the actual ρV 2 A momentum flux varies from that assuming uniform velocity for the same mass flow rate). (e) the wall shear stress; τ (f) the friction factor, defined as f = 1 w 2 as a function of Re. 2 ρV

Answers. (a) u0 = u(0) = (b) V = (c) Q =

GR 2 4µ

GR 2 1 = 2 u0 8µ πD 4 ρgh f 128µ

L

(d) β = 4/3 (e) τ w = − µ (f) f =

∂u = 12 GR ∂r r = R

16 Re

The exercise demonstrates some of the key qualitative aspects of laminar pipe flow: • the average velocity is only half the maximum velocity (in turbulent flow they would be nearly equal); • the discharge varies as (diameter)4; • the friction factor varies as Re-1.

Hydraulics 2

T2-6

David Apsley

1.4 Turbulent Flow In turbulent flow one is usually interested in time-averaged or mean quantities. The shear stress τ represents the mean rate of transport of momentum across any interface and is dominated by turbulent fluctuations rather than molecular viscosity. There is no longer a clear-cut relationship between stress τ and velocity gradient ∂u/∂r. Hence, to relate flow rate to head loss we require an empirical relation connecting wall shear stress and average velocity. The problem is to relate: head loss hf discharge Q pipe diameter D Other factors involved are the surface roughness height (ks) and the viscosity of the fluid (ν). The main elements of the analysis are: (i) For uniform flow, wall friction balances weight and pressure forces; (ii) A friction factor f is used to relate wall stress to average velocity; (iii) Combining these two, pipe-flow calculations then depend on: - a head-loss equation; - an expression for the friction factor. Step (i) Consider the length of pipe shown.

L τw

p area

D θ

p+∆p θ

               mg

The balance of forces along the axis in fully-developed flow gives: − (τ w × perimeter × L) − (∆p × area ) + (ρ × area × L × g ) sin θ = 0 wall drag

pressure

weight

Dividing by area × L and noting that L sin θ = − ∆z , perimeter ∆p ρg∆z − τw × − − =0 area L L or perimeter − ∆( p + ρgz ) τw × = area L ρg × (head loss ) = L

Hydraulics 2

T2-7

David Apsley

Hence,

hf L

=

τ w perimeter × ρg area

(12)

The quantity Sf = hf/L is called the friction slope – the (downward) slope of the energy grade line – see later. The quantity area ÷ perimeter is called the hydraulic radius. Step (ii) The friction factor f is defined as the ratio of wall shear stress to dynamic pressure: τ (13) f =1 w2 2 ρV

Step (iii) Combining (12) and (13) and noting that, for round pipes, perimeter 2 πR 2 4 = = = area πR 2 R D we have the head-loss equation for circular pipes: DARCY4-WEISBACH5 EQUATION h f 2 fV 2 = L gD

(14)

It remains to specify f for a given pipe flow. Methods for doing so are discussed in section 1.5 and lead to the Colebrook formula. The main problem is that f depends on both the relative roughness of the pipe (ks/D) and the flow velocity itself (through the Reynolds number Re ≡ VD/ν): either an iterative solution or a chart-based solution is usually required. The discharge Q can be related to the average velocity and pipe diameter via πD 2 Q =V 4 At high Reynolds numbers f tends to a constant for any particular pipe. Thus: V 2 Q2 hf ∝ ∝ 5 (turbulent) D D Compare: V Q h f ∝ 2 ∝ 4 (laminar) D D Very important note. Both Dr Lane-Serff and myself favour the above definition of f, which coincides with the definition of skin-friction coefficient in aerodynamics. However, what we refer to as f, many engineers refer to as λ/4 or, unfortunately, as f/4. Be very wary of the definition. You should be able to distinguish it from the expression for f in laminar flow: 16/Re in our notation; 64/Re with the alternative definition.

4 5

Henri Darcy (1803-1858), French engineer; conducted experiments on pipe flow. Julius Weisbach, German professor who, in 1850, published the first modern textbook on hydrodynamics.

Hydraulics 2

T2-8

David Apsley

1.5 Expressions for the Friction Factor f Smooth Pipes •

Laminar flow (theory)

•

Turbulent flow (Blasius6)

16 Re 0.079 f = Re 0.25

f =

Rough Pipes Nikuradse (1933) used sand grains to roughen pipe surfaces. He defined a relative roughness ks/D, where ks is the sand grain size and D the diameter of the pipe. His experimental curves for friction factor showed 5 regions: 1. laminar flow (Re < Recrit 2000; relative roughness irrelevant) 2. transition (2000 < Re < 4000) 3. smooth turbulence (limiting line for all ks/D; f a function of Reynolds number only) 4. rough turbulence (f a function of relative roughness only) 5. transitional turbulence (f a function of both Re and ks/D)



In the two limiting cases, 3 and 4, Prandtl7 and Von Kármán8 gave, respectively: Re f 1 • Smooth turbulence: = 4.0 log10 1.26 f •

Rough turbulence:

1 f

= 4.0 log10

3.71D ks

Unfortunately, in practice, many pipes of interest lie in the 5th regime - both roughness and Reynolds number are important, so that the friction factor is not a constant for any particular pipe. Colebrook (1939) combined smooth and rough turbulence laws into a single formula, the Colebrook (or sometimes, Colebrook-White) transition formula: COLEBROOK FORMULA  k 1.26  1 = −4.0 log10  s + (15)  3.71D Re f  f   This remains the principal formula used as the basis of establishing frictional losses in pipes. The main difficulty in using it is that it is implicit – f appears on both sides of the equation and an iterative solution is usually required (although it is a straightforward computerprogramming exercise – try it!). Colebrook (1939) and Moody (1944) also gathered data to establish effective roughness for commercial pipe materials (where the pattern of surface roughness may be very different to that in the artificially-roughened surfaces of Nikuradse). Typical values of ks are given in the Appendix. 6

P.R.H. Blasius (1883-1970); first postgraduate student of Prandtl; also derived a famous equation for the velocity profile in a laminar, flat-plate boundary layer. 7 Ludwig Prandtl (1875-1953); German engineer; introduced boundary-layer theory. 8 Theodore von Kármán (1881-1963); Hungarian mathematician and aeronautical engineer; gave his name to the double row of vortices shed from a 2-d bluff body and now known as a Kármán vortex street

Hydraulics 2

T2-9

David Apsley

Explicit Approximations to the Colebrook Formula Moody (1944), Barr (1975) and Haaland (1983) have all produced explicit approximations to the solution of (15), accurate to within a few percent for realistic ranges of Reynolds number – see the references in Massey (1998) and White (1999). Charts The two most popular graphical solutions of (15) are: • the Moody chart ( f versus Re for various values of ks/D) • Hydraulic Research Station Charts for the Hydraulic Design of Channels and Pipes (a set of charts – one for each value of ks – giving hydraulic quantities D, V, Q, Sf on the same diagram.

Hydraulics 2

T2-10

David Apsley

1.6 Frictional Losses in Pipes Pipeline systems are subject to two sorts of losses: - frictional losses, contributing a slow change of head over a large distance; - local losses due to abrupt changes in geometry; e.g. pipe junctions, valves, ... . Each type of loss can be represented by a change in head. In this section we deal with cases where frictional losses predominate and minor losses can be ignored. Such is the case for many large-scale reservoir and distribution systems. The General Problem

hf

L

Given two parameters from diameter (D), discharge (Q) and head loss (hf), find the third. Other contributory parameters are the surface roughness (ks) and the kinematic viscosity (ν). The problem is solved by the simultaneous solution of: • head loss from the Darcy-Weisbach equation: h f 2 fV 2 = L gD • friction factor from the Colebrook formula:  k 1.26 1 = −4.0 log10  s +  3.71D Re f f 

(14)´    

(15)´

In most problems it is necessary to solve (15) iteratively. The exception is the calculation of discharge Q when head loss hf and diameter D are known, because Re f = ( D/ν) fV 2 , and the combination fV2 can be found from (14). In this special case (Type I below), f can be found directly from (15):  k 1.26ν  1 = −4.0 log10  s + (16)  3.71D D fV 2  f   Knowledge of both fV2 and f allows one to find the average velocity V and thence the discharge Q.

Hydraulics 2

T2-11

David Apsley

Type 1 (flow-rate problem): diameter (D) and head loss (hf) known; find the discharge (Q) Example. A pipeline 10 km long, 300 mm diameter and with roughness 0.03 mm, conveys water from a reservoir (top water level 850 m above datum) to a water treatment plant (inlet water level 700 m above datum). Assuming that the reservoir remains full, estimate the discharge. Take ν = 1.13×10-6 m2 s-1. Solution: Since D (= 0.3 m) and hf (= 150 m) are known, the Darcy-Weisbach headloss equation enables us to find fV2: hf gDh f 9.81 × 0.3 × 150 2 fV 2 = 0.02207 m 2 s −2 = ⇒ fV 2 = = 2L 2 × 10000 L gD Hence, rewriting the Colebrook formula,  k 1 1.26  = −4.0 log10  s +  3.71D Re f  f    k 1.26ν  = −4.0 log10  s +  3.71D D fV 2    −5  3 × 10 1.26 × 1.13 × 10 −6    + = −4.0 log10   3 . 71 0 . 3 × 0 . 3 0 . 02207   = 16.92 Hence, 1 f = = 0.003493 16.92 2 Knowledge of both fV2 and f gives

V=

fV 2 f

=

0.02207 0.003493

= 2.514 m s -1

Finally, the discharge may be computed as velocity × area: πD 2 π × 0.3 2 Q =V = 2.514 × = 0.178 m 3 s -1 4 4

Hydraulics 2

T2-12

David Apsley



Type 2 (head loss problem) – diameter (D) and discharge (Q) known; find the head loss (hf) Example. The known outflow from a branch of a distribution system is 30 s-1. The pipe diameter is 150 mm, length 500 m and roughness estimated at 0.06 mm. Find the head loss in the pipe. Solution: To find the head loss (14) we require the friction factor. This is found by an iterative solution of the Colebrook equation. The average velocity is given by Q 0.03 = = 1.698 m s -1 V = 2 πD / 4 π × 0.15 2 / 4 The Reynolds number is then VD 1.698 × 0.15 = Re = = 225 400 ν 1.13 × 10 −6 Substituting values for ks (= 6×10-5 m), D (= 0.15 m) and Re (= 225 400) in the Colebrook formula (15), one obtains the iterative relation  1 5.590 × 10 −6  = −4.0 log10 1.078 × 10 −4 +   f f   or, rearranging for f, 1 f = 2  5.590 × 10 −6  −4 16.0log10 (1.078 × 10 + ) f   Starting from an initial guess, successive values of f are substituted into the RHS and new values of f evaluated. Initial guess: f = 0.01 First iteration f → 0.00436 Second iteration f → 0.00453 Third iteration f → 0.00452 Fourth iteration f → 0.00452 The value of f can then be substituted in the Darcy-Weisbach equation (14) to derive the head loss: 2 fV 2 L 2 × 0.00452 × 1.698 2 × 500 = 8.9 m = hf = 9.81 × 0.15 gD

Hydraulics 2

T2-13

David Apsley

Type 3 (sizing problem) – discharge (Q) and available head (hf) known; find the necessary diameter (D). This sort of problem is probably better dealt with using the Hydraulics Research Station Charts! However, an appropriate procedure is to write D in terms of f and iterate for f. Note that expressions involving V must be replaced by Q/(πD2/4), and at intermediate stages it is convenient to have expressions for Re and ks/D ready in terms of D.



Example. A discharge of 400 s-1 is to be conveyed from a headworks at 1050 m above datum to a treatment plant at 1000 m above datum. The pipeline length is 5 km. Estimate the required diameter, assuming that ks = 0.03 mm. Solution: Before iterating, try to write D in terms of f and expressions for Re and ks/D in terms of D. By Darcy-Weisbach: 2 hf 2 fV 2 2 f  Q  32Q 2 f = = =   L gD gD  πD 2 / 4  π 2 gD 5 Rearranging for D: 1/ 5

 32Q 2 L   f 1/ 5 D= 2  π gh  f   3 -1 Substituting values of Q (= 0.4 m s ), hf (= 50 m), L (= 5000 m), ks (= 3×10-5 m), together with g = 9.81 m s-2 gives a working expression (with D in metres): D = 1.395 f 1 / 5 For the Reynolds number, with ν = 1.13×10-6 m2 s-1: VD  Q  D  4Q  1 4.507 × 10 5 = Re = =  =  2 D ν  πD / 4  ν  πν  D and for the relative roughness, ks 8.086 × 10 −6 = 3.71D D We then proceed to iterate the Colebrook equation for f:  k 1 1.26  1 or f = = −4.0 log10  s + 2   f   3.71D Re f  ks 1.26  ) 16 log10 ( + 3 . 71 D Re f   5 Guess: f = 0.01 ⇒ D = 0.5554 m, Re = 8.115×10 , ks/(3.71D) = 1.456×10-5 Iter.1: f → 0.003057 ⇒ D = 0.4382 m, Re = 1.029×106, ks/(3.71D) = 1.845×10-5 Iter. 2: f → 0.003241 ⇒ D = 0.4433 m, Re = 1.017×106, ks/(3.71D) = 1.824×10-5 Iter. 3: f → 0.003231 ⇒ D = 0.4430 m, Re = 1.017×106, ks/(3.71D) = 1.825×10-5 Iter. 4: f → 0.003232 ⇒ D = 0.4431 m The requisite diameter is 0.44 m. In practice (as anyone who does DIY knows) commercial pipes are only made with certain standard diameters and the next available larger diameter should be chosen.

Hydraulics 2

T2-14

David Apsley

1.7 Other Losses Frictional losses in pipes are determined by the Darcy-Weisbach formula: 2 fV 2 L hf = g D Other “minor” losses occur at abrupt non-uniformities such as - entry and exit - sudden changes in pipe radius - bends - valves These losses are usually quantified in terms of a loss coefficient K, the ratio of head loss hL to dynamic head (V2/2g): V2 hL = K (17) 2g

Typical values of K are given for entry/exit and for commercial pipe fittings in the tables below (from Massey, 1998). For pictures of the various valve types see White’s textbook. Pipe friction L K = 4f D (This is where the alternative definition of friction factor, λ = 4f, comes from) Entry/exit losses Configuration Bell-mouthed entry Abrupt entry Protruding entry Bell-mouthed exit Abrupt enlargement Exit to atmosphere*

K 0 0.5 1.0 0.2 0.5 1.0

Commercial pipe fittings (very approximate – consult manufacturer’s literature) Fitting K Globe valve 10 Gate valve – wide open 0.2 Gate valve – ½ open 5.6 Pump foot valve 1.5 0.9 90° elbow 0.4 45° elbow side outlet of T-junction 1.8

Minor losses are a “one-off” loss, occurring at a single point. Frictional losses hf are proportional to the length of pipe L. In the grand scheme of things, hf usually dominates and, for long pipelines, both minor losses (and non-uniformities in f due to the development length) are often ignored.

Hydraulics 2

T2-15

David Apsley

The energy equation (“Bernoulli’s equation with losses”) for a pipeline can then be expressed as: change in head = head input (pumps) - frictional losses - minor losses V2 p ) = ∆( + z + − − ∑ hin ∑ hf ∑ hL 2g ρg

* Exit to Atmosphere Applying Bernoulli’s equation with losses: V2 V2 p p ( 2 + z 2 + 2 ) − ( 1 + z1 + 1 ) = − h f ρg 2g ρg 2g V1 is 0. Strictly the exit velocity V2 is part of the total head at section 2. In practice, it is convenient to transfer it to the RHS as a minor loss in piezometric head: p V2 ∆( + z ) = −h f − ρg 2g



exit loss

The extra term on the RHS is equivalent to a minor loss with loss coefficient 1.0.

Hydraulics 2

T2-16

David Apsley

1.8 Energy and Hydraulic Grade Lines Energy grade lines and hydraulic grade lines are graphical means of portraying the energy changes in, for example, reservoir/pipeline systems. In what follows, p is the gauge pressure (i.e. pressure relative to atmospheric) and, unless otherwise specified, z is the pipe elevation. For pipelines three elevations may be drawn: pipe centreline: z hydraulic grade line (HGL):

p +z ρg

energy grade line (EGL):

p V2 +z+ 2g ρg

Illustrations energy hydrau

Pipe friction only

reservoir pipelin

grade l ine lic gra de line

V2/2g p/ρg

e

reservoir

entry loss EGL HGL

Pipe friction with minor losses (exaggerated), including change in pipe diameter.

exit loss

pipelin e

Pumped system EGL HGL

pipeline

Hydraulics 2

T2-17

pump David Apsley

Energy Grade Line • Shows the change in total head along the pipeline. - starts at level of water in supply reservoir; - tiny drops reflect entry loss, exit loss or minor loss - steady downward slope reflecting pipe friction (slope change if pipe radius changes) - large drops (turbines) or rises (pumps). Again we note that non-frictional losses can often be ignored. • The EGL always lies a distance V2/2g above the HGL. For uniform pipes, the two are parallel. • The EGL represents, at any station, the maximum height to which water may be delivered.

Hydraulic Grade Line • Shows the change in piezometric head along the pipeline. • For pipe flow the HGL lies a distance p/ρg above the pipe centreline. Thus, the difference between pipe elevation and hydraulic grade line gives the static pressure. If the HGL drops below pipe elevation this means negative gauge pressures (i.e. less than atmospheric). This is generally undesirable since: (i) extraneous matter may be sucked into the pipe through any leaks; (ii) for large negative gauge pressures, dissolved gases may come out of solution and cause water hammer. A hydraulic grade line more than patm/ρg (about 10 m of water) below the pipeline is impossible. • The HGL is the height to which the liquid would rise in a piezometer tube. • For open-channel flows, pressure is atmospheric (i.e. p = 0) at the surface: the HGL is then the height of the free surface.

Hydraulics 2

T2-18

David Apsley

Example. Two reservoirs, the water levels in which are at elevations 180 m and 150 m respectively are connected by a pipe 3000 m long, 600 mm diameter and friction factor 0.00625. The elevation of the ground along the line of the pipeline is given in the table below. Distance (m) 0 150 300 1800 3000 Elevation (m) 175 165 190 140 147 Assuming a rounded inlet and an abrupt outlet calculate the discharge. Find the maximum depth of the pipeline below ground if the absolute pressure therein is not to fall below 3 m of water. Draw to scale on graph paper, the ground level, pipeline level and energy and hydraulic grade lines for a suitable pipeline. Take atmospheric pressure to be 10 m of water. Solution. The losses are as follows (average velocity V in m s-1): 2 fV 2 2 × 0.00625 × 3000 2 Friction: L = V hf = 9.81 × 0.6 gD Entry loss: hL (entry) = 0

= 6.371 V 2

V2 ) = 0.025 V 2 2g ∴ Total head loss = (6.371 + 0.025)V 2 = 6.396V 2 Exit loss:

hL (exit ) = 0.5(

Total head loss = difference in water levels of reservoirs. Hence 6.396 V 2 = 30

⇒ V = The discharge is therefore πD 2 Q =V 4

30 6.396

= 2.166 × π ×

= 2.166 m s -1 0.6 2 4

= 0.612 m 3 s −1

It is useful at this stage to calculate also the dynamic head (V2/2g = 0.239 m) and the exit loss (½×V2/2g = 0.120 m). The energy grade line can now be drawn (see below) – it starts at the water level in the first reservoir and descends with uniform slope to a height 150.12 m (i.e. allowing for the exit loss) at the second reservoir. The hydraulic grade line can then be drawn a distance V2/2g = 0.239 m below the energy grade line. The start and end coordinates give it an equation z HGL = 179.761 − 0.0096 x The ground elevation is now marked on the graph from the data given in the equation. We are now asked to ensure that the absolute pressure does not fall below 3 m of water: in other words (since atmospheric pressure is equivalent to 10 m of water) that

Hydraulics 2

T2-19

David Apsley

the pipeline is not more than 7 m above the hydraulic grade line. The most significant problem occurs at x = 300 m, where the maximum pipeline height can be z pipeline = z HGL + 7 = 179.761 − 0.0096 × 300 + 7 = 183.8 m Since the ground level here is 190 m, the pipeline depth below ground must be 190 − 183.8 = 6.2 m Since unnecessary excavation is undesirable, the pipeline is laid at ground level, except where it must be lowered to satisfy pressure constraints. A suitable pipeline is marked on the diagram. 190

excavation necessary 180

7m

175

EGL HGL

165

velocity head = 0.239 m

pe pi

exit loss = 0.12 m

lin e 150 147

140

0 150 300

Hydraulics 2

1800

T2-20

3000 m

David Apsley

1.9 Pipe Networks For all pipe combinations the following basic principles apply: (1) mass conservation at junctions (total flow in = total flow out) (2) there can be only one value of head at any one point (3) each pipe must satisfy its individual head-loss vs discharge relation There is a close analogy with electrical networks: head H potential V discharge Q current I However, the hydraulic equivalent of Ohm’s law is usually non-linear: head loss ∆H = kQ2 potential difference ∆V = RI In pipe-network calculations, the discharge coefficient k is usually taken as a constant for each pipe (although, in reality, it will vary slightly with Reynolds number).

Pipes in Series Q1 = Q2 same flow in each pipe ∆H = ∆H1+∆H2 total head loss equals sum of individual losses Pipes in Parallel ∆H1 = ∆H2 Q = Q1+Q2

2

1

same head loss across each pipe total flow is sum of individual flows

1 2

Branched Pipes – Single Junction Heads HA, HB and HC are known (from the levels of water in the reservoirs). The head at J must be found (iteratively) to satisfy: A (i) the loss equation (∆H = kQ2) for each pipe; in this example: 2 B H A − H J = k AJ Q AJ

J ?

2 H B − H J = k BJ QBJ

2 H J − H C = k JC Q JC C (ii) continuity at junction J: Q AJ + QBJ = Q JC Note that the direction of flow in pipe BJ cannot be established a priori but must emerge as part of the solution.

Solution procedure: (1) Guess HJ (2) Calculate flow rates in all pipes (from head differences) (3) Calculate flow into and flow out of J (4) If necessary, adjust HJ to reduce any flow imbalance and repeat from (2) (If the direction of flow in pipe BJ is not obvious then a good initial guess is to set HJ = HB so that there is initially no flow in this pipe. The first flow rate calculation will then establish whether HJ should be lowered or raised.

Hydraulics 2

T2-21

David Apsley

Example. Reservoirs A, B and C have constant water levels of 150, 120 and 90 m respectively. A 30 cm pipe, 1600 m long, leaves A and runs to O at elevation 137 m. Here it divides into a 20 cm pipe, 1600 m long, leading to B and a 15 cm pipe, 2400 m long, leading to C. Assuming f = 0.005 in all pipes, calculate the flow in each pipe. Calculate also the reading of a Bourdon pressure gauge attached to the junction O. 150 m

A 120 m

0m 160 m L= =0.2 D

O m 00 24 5 m L= =0.1 D

B

L= 1 D=0600 m .3 m

90 m

C Solution. First we need the head-loss law (i.e. hf as a function of Q) for each pipe. In terms of the friction factor, 2 fV 2 Q and L hf = V = gD πD 2 / 4 32 fQ 2 L L ⇒ h f = 2 5 = 1.653 × 10 −3 5 Q 2 D π gD Substituting length L and diameter D for each pipe, H A − HO 2 Q AO = Pipe AO: h f = 1088Q AO or 1088 HO − HB 2 Pipe OB: h f = 8265QOB or QOB = 8265 HO − HC 2 QOC = Pipe OC: h f = 52243QOC or 52243 (i) Guess HO = 140 m. This gives QAO = 95.9×10-3 m3 s-1, QOB = 49.2×10-3 m3 s-1, QOC = 30.9×10-3 m3 s-1. Total flow into junction O = QAO = 95.9×10-3 m3 s-1 Total flow out of junction O = QOB + QOC = 80.1×10-3 m3 s-1 There is 15.8×10-3 m3 s-1 more flow into the junction than out of it, so HO needs to be increased. (ii) Guess HO = 145 m. This gives QAO = 67.8×10-3 m3 s-1, QOB = 55.0×10-3 m3 s-1, QOC = 32.4×10-3 m3 s-1. Total flow into junction O = QAO = 67.8×10-3 m3 s-1 Total flow out of junction O = QOB + QOC = 87.4×10-3 m3 s-1 There is 19.6×10-3 m3 s-1 more flow out of the junction than into it, so HO needs to be decreased. Hydraulics 2

T2-22

David Apsley

(iii) With two benchmarks straddling the solution we can interpolate between them and get a better guess. Guess 15.8 H O = 140 + × (145 − 140) = 142.2 m 15.8 + 19.6 This gives QAO = 84.7×10-3 m3 s-1, QOB = 51.8×10-3 m3 s-1, QOC = 31.6×10-3 m3 s-1. Total flow into junction O = QAO = 84.7×10-3 m3 s-1 Total flow out of junction O = QOB + QOC = 83.4×10-3 m3 s-1 There is 1.3×10-3 m3 s-1 more flow into the junction than out of it, so HO needs to be increased. (iv) This is probably close enough, but a further interpolation gives a new guess 1 .3 H O = 142.2 + × (145 − 142.2) = 142.4 m 1.3 + 19.6 This gives QAO = 83.6×10-3 m3 s-1, QOB = 52.1×10-3 m3 s-1, QOC = 31.7×10-3 m3 s-1. Total flow into junction O = QAO = 83.6×10-3 m3 s-1 Total flow out of junction O = QOB + QOC = 83.8×10-3 m3 s-1 The total mass flow error at the junction is now 0.2×10-3 m3 s-1 or about 0 .2 × 100 = 0.24% 83.6 A Bourdon gauge measures absolute pressure. Applying Bernoulli’s equation with losses between, for example, A and O,  p  p V2 V2  = H A − H O = 7.6 m   −  +z+ +z+ 2 g  A  ρg 2 g  O  ρg Also, pA = patm, VA = 0 and Q AO 83.6 × 10 −3 = VO = = 1.183 m s −1 2 2 πD / 4 π × 0.3 / 4 Hence, p O = p atm + ρg ( z A − z O ) − 12 ρVO2 − ρg ( H A − H O ) = p atm + 1000 × 9.81 × 13 − 0.5 × 1000 × 1.183 2 − 1000 × 9.81 × 7.6 = 101325 + 52274 = 153600 Pa Answer: 1.54 bar.

Branched Pipe – Multiple Junctions This is considerably more complex as more than one unknown head must be adjusted. These sort of problems will not be covered in this course (although there is one such problem on the example sheet). General algorithms – the loop method and the nodal method – are discussed in Chadwick and Morfett (1998). They are based on making estimates for the small increments δQ or δH respectively in pipes or junctions and coupling these increments by a requirement to satisfy continuity and head.

Hydraulics 2

T2-23

David Apsley

2. OPEN-CHANNEL FLOW Flow in open channels (rivers, canals, guttering, ...) and partially-full conduits (e.g. sewers) is characterised by the presence of a free surface, at which pressure is atmospheric. Since the (gauge) pressure is zero at the free surface, the hydraulic grade line (p/ρg + z) coincides with the free surface. The energy grade line lies a distance V 2 /2 g above it. Open-channel flows are driven principally by gravitational forces. Flow must occur when the free surface is tilted because the (hydrostatic) pressure is greater under a greater depth of water and hence there is a net horizontal force. In this course we shall study: (i) uniform flow – weight component along the channel exactly balances bed friction; (ii) non-uniform flow – introduction only; more advanced study in Hydraulics 3.

2.1 Uniform Flow The flow is uniform if the velocity profile does not change along the channel.

p=p a

tm

h hf L Steady uniform flow is called normal flow and the depth of water the normal depth. In normal flow the bed slope, hydraulic grade line and energy grade line are all parallel; i.e. geometric slope S0 = friction slope Sf hf dz b =− dx L In uniform flow the component of weight down the slope exactly balances bed friction.

Hydraulics 2

T2-24

David Apsley

2.1.1 Hydraulic Radius and the Drag Law In both open channels and partially-full pipes, wall friction only occurs along the wetted perimeter.

A P

Let A be the cross-sectional area occupied by fluid and P the wetted perimeter.

L

For steady, uniform flow, the component of weight down the slope balances wall friction: (ρAL) g sin θ = τ w PL where τw is the average wall friction. Hence, A τ w = ρg ( ) sin θ P

mg

τw

θ

The quantity Rh ≡

area A = P wetted perimeter

(18)

is called the hydraulic radius (or, sometimes, the hydraulic mean depth, m). Hence, for normal flow, τ w = ρgRh S

(19)

where S (= drop ÷ length) is the slope; note that tanθ ≈ sinθ for small angles. Examples. (i) For a circular pipe running full, A πR 2 1 = R = 14 D Rh = = (20) P 2 πR 2 i.e. for a full circular pipe, the hydraulic radius is half the geometric radius. (Sorry folks, this is just one of those things!). As a result, it is common to define a hydraulic diameter Dh by D h = 4 Rh (ii)

For a very wide channel of uniform depth h, side walls make negligible contribution to the wetted perimeter and hence Rh = h i.e. Rh is equal to the depth of flow.

To progress we need some expression for the average wall stress τw.

Hydraulics 2

T2-25

David Apsley

2.1.2 Friction Factors One technique for finding τw is to extend friction-factor theory to partially-full pipes, taking into account only the drag on the wetted perimeter. As before, the friction factor is defined as the ratio of wall shear stress to dynamic pressure: τ f =1 w2 2 ρV Hence, substituting the normal-flow relationship (19) for τw: ρgR S 2 gRh S f = 1 h2 = V2 2 ρV Hence hf 2 fV 2 fV 2 slope S (≡ ) = = (21) 2 gRh L gDh But this is just the Darcy-Weisbach equation with D replaced by Dh = 4Rh. The friction factor may be obtained from the Colebrook transition formula, but with Reynolds number D V 4 RhV Re = h = ν ν The procedure for solving problems is similar to that for full pipes, but with the added problem of having to determine the level of fluid in the conduit as part of the solution.

2.1.3 Chézy and Manning’s Formulae The friction-factor method is untenable for many open conduits - in particular, natural water courses – because the shear stress is not constant around the wetted perimeter. For practical problems, engineers tend to use simpler empirical formulae due to Chézy9 and Manning10. The formula for the slope (21) can be rearranged as one for velocity: Chézy’s formula: V = C Rh S

(22)

C ( = 2 g/f ) is called Chézy’s coefficient, which varies with channel roughness and (to a smaller extent) hydraulic radius. The most popular correlation for C is that of Manning who proposed C = R h1 / 6 /n , in terms of a roughness-dependent coefficient n. Combined with Chézy’s formula (22), this yields: MANNING’S FORMULA 1 V = Rh2 / 3 S 1 / 2 (23) n Very important: both Chézy’s C and Manning’s n are dimensional and depend on the units used. Typical values of n in metre-second units are given in the Appendix. Typical figures for artificially-lined channels and natural water courses are about 0.015 and 0.03 respectively. 9

Antoine Chézy (1718-1798); French engineer who carried out experiments on the Seine and on the Courpalet Canal in 1769. 10 Robert Manning (1816-1897); Irish engineer.

Hydraulics 2

T2-26

David Apsley

2.1.4 Uniform Flow Calculations Assuming that the channel slope, shape and lining material are known, there are two main classes of problem: (1) Determine the discharge (Q) given the depth (h): - calculate the area A and wetted perimeter P from geometry - calculate the hydraulic radius Rh =

A P

- calculate the average velocity from Manning’s formula: V =

1 2 / 3 1/ 2 Rh S n

- calculate the discharge as velocity × area: Q = VA (2) Determine the depth (h) given the discharge (Q): - follow the steps in (1) above to write algebraic expressions for, successively, A, P, Rh, V and Q in terms of depth h - invert the Q(h) relation graphically or numerically.

Hydraulics 2

T2-27

David Apsley

Example. A smooth concrete-lined channel has trapezoidal cross-section with base width 6m and sides of slope 1V:2H. If the bed slope is 1 in 500 and the normal depth is 2 m calculate the discharge. Solution. We are given slope S = 0.002. From the Appendix, Manning’s n is 0.012.

2m 6m

4m

Break the trapezoidal section into rectangular and triangular elements to obtain: area, A = 6 × 2 + 2 × 12 × 4 × 2 = 20 m 2 wetted perimeter, P = 6 + 2 × 4 2 + 2 2 = 14.94 m Hence, the hydraulic radius is A 20 Rh = = = 1.339 m P 14.94 Substituting into Manning’s formula we can obtain the average velocity: 1 2 / 3 1/ 2 1.339 2 / 3 × 0.0021 / 2 V = Rh S = = 4.527 m s −1 n 0.012 The discharge is Q = VA = 4.527 × 20 = 90.5 m 3 s -1 Example. For the channel above, if the discharge is 40 m3 s-1, what is the normal depth? Solution. This time we need to leave all quantities as functions of height h. area, A = 6h + 2 × 12 × 2h × h = 6h + 2h 2 wetted perimeter, P = 6 + 2 × (2h) 2 + h 2

hydraulic radius, Rh =

=6+2 5 h

A 6h + 2 h 2 = P 6 + 2 5h

1 bulk velocity, V = Rh2 / 3 S 1 / 2 n discharge, Q = VA = 3.728

Rh2 / 3 0.0021 / 2 = 0.012

 6 h + 2h 2   = 3.728  + h 6 2 5  

2/3

( 6h + 2 h 2 ) 5 / 3

( 6 + 2 5h) 2 / 3 We can now try a few values (h in m, Q in m3 s-1): h=2 Q =90.52 h=1 Q =24.91 h = 1.23 Q = 36.28 (latest guess for h based on interpolation to get Q = 40) h = 1.31 Q = 40.73 h = 1.30 Q = 40.16 Answer: h = 1.30 m. Note: evaluating Q(h) is an ideal application for Microsoft Excel – try it!

Hydraulics 2

T2-28

David Apsley

2.1.5 Optimal Shape of Cross-Section The most hydraulically-efficient shape of channel is the one which can pass the greatest discharge for any given area or, equivalently, the smallest area for a given discharge. From Manning’s formula and the corresponding expression for discharge we see that this occurs for the minimum hydraulic radius or, equivalently, for the minimum wetted perimeter. For channel shapes that do not taper inwards, maximum hydraulic efficiency is achieved with a semi-circular cross section. However, hydraulic efficiency is not the only consideration and one must also consider, for example, fabrication costs, excavation and, for loose granular linings, the maximum slope of the sides. Many applications favour trapezoidal channels. General expressions for A, P and Rh for important channel shapes are given in the table below. rectangle

trapezoid

h

α

bh

wetted perimeter P

b + 2h

hydraulic radius Rh

h 1 + 2 h/ b

h

θ R

b

b

area A

circle

h2 bh + tan α 2h b+ sin α b + h/ tan α h b + 2h/ sin α

h

R 2 (θ − 12 sin 2θ) 2 Rθ

R  sin 2θ   1 − 2θ  2

Trapezoidal Channels For a trapezoidal channel we have, from the table in the previous section: h2 h A = bh + Area: tan α α 2h Wetted perimeter: P =b+ b sin α What depth of flow and what angle of side should we design for to achieve maximum hydraulic efficiency? To minimise the wetted perimeter for maximum hydraulic efficiency, we substitute for b in terms of the fixed area A: 2h A h 2h A 2 1 ) P =b+ )+ (24) =( − = + h( − sin α h tan α sin α h sin α tan α To minimise P with respect to water depth we set

Hydraulics 2

T2-29

David Apsley

∂P A 2 1 )=0 ≡− 2 +( − sin α tan α ∂h h and, on substituting the bracketed term into the expression (24) for P, we obtain 2A P= h Hence the hydraulic radius is A h Rh ≡ = P 2 In other words, for maximum hydraulic efficiency, a trapezoidal channel should be so proportioned that its hydraulic radius is half the depth of flow.

Similarly, to minimise P with respect to the angle of slope of the sides, α, we set ∂P −2 1 h ≡ h( 2 cos α + sec 2 α) = (1 − 2 cos α) = 0 2 ∂α sin α tan α sin 2 α This occurs when cos α = ½. i.e. the most efficient side angle for a trapezoidal channel is 60°. Taking the results for h and α together, it is straightforward to show that the most efficient trapezoidal channel shape is half a regular hexagon.

Circular Ducts In similar fashion it can be shown that the maximum discharge for a circular duct actually occurs when the duct is not full – in fact for a depth about 94% of the diameter (Exercise. Prove this; also try to explain in words why you might expect this).

Hydraulics 2

T2-30

David Apsley

2.2 Introduction to Non-Uniform Flow (Optional) This is only a brief introduction to the phenomena encountered: in particular, • slowly-varying flow (energy methods appropriate) - subcritical and supercritical flow (Froude number and critical depth) - flow over a surface bump - gradually-varying-flow equation • rapidly-varying flow (mechanical energy lost) - hydraulic jumps Non-uniform flow with a free surface will be considered in much greater detail in Hydraulics 3, where you will also have the opportunity to study wave motion.

2.2.1 The Froude Number – Subcritical and Supercritical Flow Flow in open channels is driven by gravitational forces. If the flow has characteristic length scale L and velocity scale U then the ratio of “inertial forces” (mass × acceleration) to gravitational forces (mass × g) is of order acceleration U (U/L) U 2 ≈ = g g gL The square root of this quantity is called the Froude number Fr. Since the characteristic length scale involved in these flows is typically (but not always) a depth h, the Froude number is usually written U Fr = (25) gh A flow with Fr < 1 is called subcritical or tranquil. A flow with Fr > 1 is called supercritical or rapid. Since Fr determines the ratio of inertial to gravitational forces, one might expect very different behaviour according as Fr < 1 or Fr > 1 and this is indeed the case.

A second interpretation comes from noting that the speed of long waves in a channel of depth h is given by c = gh . Then the Froude number is the ratio flow velocity Fr = wave speed If Fr > 1 then the flow is moving faster than any disturbance can propagate upstream. Hence the upstream flow cannot be changed by downstream conditions. If, on the other hand, Fr < 1 then circumstances downstream (e.g. weirs) can control the upstream flow.

Hydraulics 2

T2-31

David Apsley

2.2.2 Frictionless Flow Over an Obstacle Assumptions: steady, frictionless and quasi-1-dimensional. At any streamwise location let u be the velocity, h the depth of water and zb the bed height relative to some horizontal datum. The analysis comes from a combination of energy and continuity.

h

u

zb

(a) Energy Since the free surface is a streamline at constant pressure p = 0 and height h + zb, Bernoulli’s theorem (in “head” form) gives total head = zb + h + 12 u 2 /g = constant (26) Define specific energy E as the head relative to the bed of the channel; i.e.

E =h+

u2 2g

(27)

Then the total head equation says E + z b = constant zb increases ⇔ E decreases (b) Continuity For quasi-1-d flow,

(28)

uh = q (flow rate per unit width; constant)

Combining energy and continuity, velocity u can be eliminated to give 1 q2 E =h+ 2 gh 2

(29)

water depth

h

1

subcritical (Fr1)

∆z b

E(h)

Ec

specific energy

The graph of E vs h has the shape shown, with a minimum at the critical point (Ec,hc) given (do the differentiation and prove it) by

Hydraulics 2

T2-32

David Apsley

hc Ec

 q2  =    g  = 23 hc

1/ 3

(30)

(i)

For any given flow rate q, Ec is the minimum possible specific energy.

(ii)

Conversely, for any given specific energy, the maximum discharge occurs at the critical depth.

(iii)

For any value E greater than Ec there are two possible depths with the same E: - a shallow (h < hc), high-speed flow – supercritical (Fr > 1) - a deep (h < hc), low speed flow – subcritical (Fr < 1) These are called alternative depths.

(iv)

hc (the critical depth) and Ec depend on the flow rate q. 1/ 3

(v)

(vi)

(vii)

1/ 3

h  gh   g  Since q = uh, one has, for any given depth h, = h 2 2  =  2  . Hence the hc u  u h  ratio of depth to critical depth is determined by the local Froude number number: h u 1 = 2/3 , Fr = (31) hc Fr gh If the bed height zb increases, the specific energy E must decrease by the same amount to retain the same total head – equation (28). Changes in E and h are related by the constant-q curve above. • supercritical: [1] → [2]; h increases ⇒ u decreases ⇒ absolute water level increases over bump; • subcritical: [1]′ → [2]′; h decreases ⇒ u increases ⇒ absolute water level decreases over bump. This may be counter-intuitive!

supercritical

subcritical

From the E-h curve, the only way in which a flow can pass from subcritical (low speed, large depth) to supercritical (high speed, small depth) is if h = hc (i.e. Froude number Fr = 1) at the top of the obstacle.

(viii) E cannot be reduced below Ec without the assumptions breaking down. In practice, if the obstacle height zb is too large, the flow will “back up” upstream, so as to obtain a subcritical flow there. At the top of the obstacle, Fr = 1 and the flow changes smoothly from subcritical to supercritical. The obstacle will then be controlling the flow.

Hydraulics 2

T2-33

David Apsley

2.2.3 Gradually-Varying Flow Small changes in water depth often accompany changes in bed friction, slope or channel cross-section. The friction slope Sf = loss of total head per unit length due to friction = downward slope of energy grade line hf = L The geometric slope S0 = (downward) slope of the channel bed dz =− b dx

h(x)

zb(x)

slope S 0

For uniform flow: - the friction slope Sf and bed slope S0 are the same; - the depth and velocity of water are constant along the channel; - this is common in artificial conduits, but rare in natural water courses. For non-uniform flow: - the friction slope Sf and bed slope S0 are different; - the depth and velocity of water vary along the channel; In both cases Sf is typically correlated with average velocity and hydraulic radius by Manning’s formula which can be inverted to give: S f = n 2V 2 /Rh4 / 3 where n is related to the bed roughness. Let h be the depth of water and zb the bed height above some horizontal datum. Over a small length δx, Sf is approximately constant, so total head loss = − S f × length

⇒ δ( h + z b +

V2 ) = − S f δx 2g

1 δV 2 δh δz b + = −S f ⇒ + δx δx 2 g δx Letting δx → 0 and writing S0 = –dzb/dx = (downward) bed slope, dh 1 dV 2 + = S0 − S f dx 2 g dx If the channel is of uniform width b, then V = Q/hb and dV 2 Q 2 d 1 Q 2 dh V 2 dh = 2 = −2 ( 2 ) = −2 2 3 dx h dx b dx h b h dx Hence,

V2 dh (1 − ) = S0 − S f gh dx (The analysis can be extended to non-uniform width – with the same result – if h is interpreted as the mean depth.) The final result is the

Hydraulics 2

T2-34

David Apsley

GRADUALLY-VARYING FLOW EQUATION dh S 0 − S f = dx 1 − Fr 2 where dz = bed slope S0 = − b dx S f = τ w / ρgRh = friction slope V = Froude number Fr = gh

(32)

You will be offered plenty of chance to solve this numerically next year. For the present we simply notice that the gross behaviour – h increasing or decreasing – depends on two factors: Slope: S0 > Sf or S0 < Sf Froude number: Fr < 1 (subcritical) or Fr > 1 (supercritical) This gives a table showing how the water depth will change along the channel: S0 > Sf S0 < Sf Subcritical, Fr < 1 increases decreases Supercritical, Fr > 1 decreases increases

Hydraulics 2

T2-35

David Apsley

2.2.4 Rapidly-Varying Flow – the Hydraulic Jump A hydraulic jump is an abrupt change from a shallow high-speed (supercritical, Fr>1) flow to a deep low-speed (subcritical, Fr 1, it follows that Fr1 > 1 and Fr2 < 1; i.e. the upstream flow is supercritical and the downstream flow subcritical. Fr22 ≡

(38) can be rearranged as a quadratic for h2/ h1:

Hydraulics 2

T2-37

David Apsley

2

 h2   h2    +   − 2Fr12 = 0  h1   h1  with solution (the negative root is impossible): h2 1 = (−1 + 1 + 8Fr12 ) h1 2 h1 and h2 are called conjugate depths.

(40)

Summary A hydraulic jump is a rapid transition from supercritical to subcritical flow with consequent loss of mechanical energy. The ratio of conjugate depths is determined by the upstream Froude number (equation (40)) and the loss of energy by equation (37).

Hydraulics 2

T2-38

David Apsley

Appendix Material Riveted steel Concrete Wood stave Cast iron Galvanised iron Asphalted cast iron Commercial steel or wrought iron Drawn tubing Glass

ks (mm) 0.9-9.0 0.3-3.0 0.18-0.9 0.26 0.15 0.12 0.046 0.0015 0 (smooth)

Table 1. Typical roughness for commercial pipes (from White, 1999).

Artificial lined channels: Glass Brass Steel, smooth painted riveted Cast iron Concrete, finished unfinished Planed wood Clay tile Brickwork Asphalt Corrugated metal Rubble masonry Excavated earth channels: Clean Gravelly Weedy Stony, cobbles Natural channels: Clean and straight Sluggish, deep pools Major rivers Floodplains: Pasture, farmland Light brush Heavy brush Trees

n (metre-second units)

ks (mm)

0.01 0.011 0.012 0.014 0.015 0.013 0.012 0.014 0.012 0.014 0.015 0.016 0.022 0.025

0.3 0.6 1.0 2.4 3.7 1.6 1.0 2.4 1.0 2.4 3.7 5.4 37 80

0.022 0.025 0.03 0.035

37 80 240 500

0.03 0.04 0.035

240 900 500

0.035 0.05 0.075 0.15

500 2000 5000 ?

Table 2. Typical values of Manning’s n and equivalent sand roughness (from White, 1999).

Hydraulics 2

T2-39

David Apsley