2009

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20. Rörströmning I (6.1-6.4, 6.6)

Energiförluster vid rörströmningLokala energiförlusterSeriekopplade rörledningarParallellkopplade rörledningar

Övningstal: J11 12 J18

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Övningstal: J11-12, J18

PIPE FLOWFlow of water, oil and gas in pipes is of immense importance in civil engineering:

Distribution of water from source to consumers (private, municipal process industries)municipal, process industries)Transport of waste water and storm water to recipient via treatment plantTransport of oil and gas from source to refineries (oil) or into distribution networks (gas) via pipelines

Some data from Sweden:

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• Average water consumption: 330 liters/(person and day)• Purchase cost (“Anskaffningsvärde”) for water and waste

water pipes: 250 billion SEK (250 miljarder kr)• Length of all water pipes put together: 67.000 km

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TWO FACTORS OF IMPORTANCE IN DESIGN OF PIPES

1) Hydraulic transport capacity of the pipe

In a pressurized system the hydraulic transport capacity is a function of the fall of pressure along the pipe. The fall of pressure is caused by energy losses in the pipe:- Energy losses due to friction due to shear stresses along

pipe walls - Local losses that arises at pipe bends, valves, enlargements,

contractions, etc

2) St th f i

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2) Strength of pipeUsually determined on basis of high and low pressures in conjunction with flow changes (closing of valve or pump stop)

(total energi)

(trycknivå)

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Energy Losses In Pipe Flow

Energy equation:

2 2V V

The objective is to determine a relation between energy losses and

1 1 2 21 22 2

p V p Vz z hlossesg gγ γ

+ + = + + + ∑

h h hlosses friction local∑ = + ∑

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The objective is to determine a relation between energy losses andmean velocity in a pipe:

hfriction = f(V) and hlocal = f’(V)

Energy losses due to friction

Calculated using Darcy – Weisbach’s formula (Eq. 6.12)(allmänna friktionsformeln för både laminärt och turbulent flöde; ):

hf – energy loss due to friction over a distance, L (m), along the pipef – pipe friction factor [f=f(Re), ”Pipe wall roughness”];

2 2

5 2

162 2f f

L V L Qh f or h fD g D gπ

= =

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Fig. 6.10 – Moody diagram, laminar flow → f = 64/Re; (Re = VD/ν)D – Pipe diameter (m)V – average velocity in the pipe (m/s)Q – flowrate in the pipe (m3/s)

roughness = råhet

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J11:Calculate the smallest reliable flowrate that can be pumped through this pipeline. D = 25 mm, f = 0.020, L = 2 x 45 m, Vertical distances are 7.5 m and 15 mrespectively. Assume atmospheric pressure 101.3 kPa.

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Local energy lossesMinor head losses in pipelines occur at pipe bends, valves (“ventiler”), enlargement and contraction of pipe sections, junctions (“knutpunkter”) etc.

In long pipelines these local head losses are often minor in comparison with energy losses due to friction and may be neglected.

In short pipes, however, they may be greater than frictional losses and should be accounted for.

Local losses usually result from abrupt changes in velocity leading t dd f ti hi h t t f th fl

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to eddy formation which extract energy from the mean flow.

Increase of velocity is associated with small head (energy) losses and decrease of velocity with large head losses

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Local energy losses (cont.)Usually it is possible to write local energy losses in pipe flow using

the following formula:

hlocal = local energy loss

2

2

Vh Klocal local g

= ⋅

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Klocal = local loss coefficient (different for different types of losses)

(Förlust koefficient Klocal är ofta bestämd empiriskt)

V2/(2g) = kinetic energy (hastighetshöjd)

LOCAL ENERGY LOSS - ENLARGEMENT

V2V1 V2

Loss coefficient K for sudden enlargement (V=V ):

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D2/D1 1.5 2.0 2.5 5 10

KL 0.31 0.56 0.71 0.92 0.98

Loss coefficient, KL, for sudden enlargement (V=V1):

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ENERGY LOSS FOR OUTFLOW IN RESERVOIR

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LOCAL ENERGY LOSS - CONTRACTION

Loss coefficient for sudden contractionsudden contraction(Franzini and Finnemore, 1997, V = V2):

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D2/D1 0.0 0.1 0.2 0.3 0.4 0.5 0.6 0.7 0.8 0.9 1.0

KL 0.50 0.45 0.42 0.39 0.36 0.33 0.28 0.22 0.15 0.06 0.00

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Head loss coefficient for different types of pipe entrancesIn/utloppstyper från reservoir

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Head loss at smooth pipe bends

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Loss coefficients at right angle bends

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Pipe systems – pipes in series

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Solution

Energy equation ⇒ Total head, H = Δz = hf1 + hf2 + Σhlocal

Continuity equation ⇒ Q = Q1 = Q2

J12:Water is flowing. Calculate the gage reading when V300 is 2.4 m/s. (NOTE El. = elevation)

2222

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Pipe systems – pipes in parallel

Solution

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Energy equation ⇒ hf1 + Σhlocal,1 = hf2 + Σhlocal,2

Continuity equation ⇒ Q = Q1 + Q2

(elevation z in reservoirs same for both pipes, velocity V in reservoirs

equal to zero or otherwise same for both pipes)

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2009

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21. Rörströmning II (6.4, 7.1-7.4)

TrereservoirproblemetKvasistationär rörströmning

Övningstal: J22-23

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Pipe systems – branched pipe systems

Solution3 Possible flow situations:1) From reservoir 1 and 2 to reservoir 32) From reservoir 1 to reservoir 2 and 33) From reservoir 1 to 3 (Q = 0)

As HJ (HJ is total head at J) is initiallyunknown, a method of solution is as follows:

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3) From reservoir 1 to 3 (Q2 = 0)For the situation as shown:Energy equation ⇒

HJ = PJ/w + zJ + V2J/2g

hf1 + Σhlocal,1 = z1 – HJhf2 + Σhlocal,2 = z2 – HJhf3 + Σhlocal,3 = HJ – z3

Continuity equation ⇒ Q3 = Q1 + Q2

1) Guess HJ2) Calculate Q1, Q2, and Q33) If Q1 + Q2 = Q3, then the solution is

correct4) If Q1 + Q2 ≠ Q3, then return to 1).

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Reservoir Elevation (m) Pipe length (m)

A 90 3200

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B 60 4800

C 30 6800

Reservoir Elevation (m) Pipe length (m)

Diameter (mm)

A 150 1600 300

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A 150 1600 300

B 120 1600 200

C 90 2400 150

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Non-stationary Pipe Flow -Outflow From Reservoir Under Varying Pressure Level

When water flows under varying pressure levels the outflow from the reservoirwill vary accordingly.

In the figure V representsthe volume in the reservoirat a certain time. There is

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also an inflow, Qi, to and an outflow, Qo, from the reservoir.

Non-stationary Pipe Flow, Cont.

Volume change in the reservoir during a small time interval dt, can be expressed as:dV = (Qi – Qo) · dt and dV = As · dz →A · dz = (Q – Q ) · dt where both inflow and outflow can vary in timeAs dz (Qi Qo) dt where both inflow and outflow can vary in time.The outflow can normally be determined by the energy equation that gives outflow as a function of z. For example outflow through a hole: Qo = Ahole · CD · (2gz)1/2 (Eqn. 5.12)If time is to be estimated that changes the water level from z1 to z2integration of dt = (As /(Qi – Qo)) dz gives:

t = 1∫z2 (A /(Qi – Q )) dz

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t z1∫ (As /(Qi Qo)) dz

This expression can be derivated if Qi = 0 or if Qi = constant and Qo canbe re-written as a function of z. Qo can be determined by the energyequation during short time periods assuming stationarity. If water levelchanges quickly an acceleration term has to be included though.

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Example – unsteady pipe flow

The open wedge-shaped tank in the figurebelow has a length of 5 m perpendicularto the sketch It is drained through a 75to the sketch. It is drained through a 75mm diameter pipe, 3.5 m long whosedischarge end is at elevation zero. Thecoefficient of loss at the pipe entrance is0.5, the total of the bend loss coefficientsis 0.2, and the friction factor is f = 0.018.Find the time required to lower the watersurface in the tank from elevation 3 m to

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surface in the tank from elevation 3 m to1.5 m. Assume that the accelerationeffects in the pipe are negligible.

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22. Rörströmning III (6.5)

Laminär och turbulent strömningTurbulensmodellerTurbulent rörströmningFriktionskoefficient, Moody´s diagramIcke-cirkulära rör

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Övningstal: J4, J6-7

Reynold’s 2nd experiment – Laminar vs. Turbulent flow

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Laminar flow: hfriction ~ VTurbulent flow: hfriction ~ V2

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l = mixing length

l

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2

32f

Lh VDμ⋅ ⋅

= ⋅ (Laminar flow: Poiseuille´s equation, 6.11)

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2

0

0

2

f

f

w DLh

w RrR

τ

τ τ

⋅⋅ ⋅

=⋅

= ⋅(Laminar AND turbulent flow)

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Friction in laminar pipe flowPoiseuille´s equation:

hf = 32νLV / gD2 (Eq. 6.11)hf 32νLV / gD (Eq. 6.11)

Darcy-Weisbach equation: (Eq. 6.12)

hf = fLV2 / 2gD (also for turbulent pipe flow)

Equaling gives (laminar flow)

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Equaling gives (laminar flow)32 ν LV / gD2 = f LV2 / 2gD

(Re = DV/ ν) → f = 64/Re (Eq. 6.13)

Models for turbulent shear stress τ

Boussinesq´s modelτ = ε (du/dy) where ε is eddy viscosity( y) y y

(laminar τ = μ(du/dy))

Prandtl´s model (for pipe flow)τ = ρ l2 (du/dy)2 where l is mixing length and function of y,

close to wall l = κ y where κ = turbulenceconstant ( 0 4)

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constant (~0.4)

Equaling the two models gives:ε = ρ l2 (du/dy) (Pa · s)

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Nikuradse’s experimentsA major contribution on determining the friction factor as a function of Reynolds number, Re, and pipe roughness

A series of experiments where friction factor and velocity distribution were determined for various Reynolds numbers

In the experiments, pipes were artificially roughened by stickinguniform sand grains to smooth pipes

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ff

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a) δL >> kS: f = f(y0), y0 = y0(δL), δL = δL(Re) ⇒ f = f(Re)

b) δL ≈ kS: f = f(y0), y0 = y0(δL, kS), ⇒ f = f(Re, kS/D)

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) L S (y0), y0 y0( , S), ( , S )

c) δL << kS: f = f(y0), y0 = y0(kS) ⇒ f = f(kS)

Thickness of laminar sublayerLaminar sublayer thickness, δL (from measurements):

4L v

νδ = where ν = kinematic viscosity

Friction velocity, v*, can be determined from:

*v

20 0

*4

2 8fL L V fh f v V

gD D gτ τρ ρ

= = ⇒ = =

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R = radiusτ = τ0 r/R r = distance from pipe wall

τ0 = shear stress at pipe wall

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Example – laminar sublayerWater is flowing in a 100 mm pipe with an average velocity g p p g yof 1 m/s. Pipe friction factor is 0.02 and kinematic viscosity,ν = 1·10-6 m2/s.

* 0.05 /8

4

fv V m s

ν

= =

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*

4 0.08L mvνδ = =

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Formulas Used To Determine The Friction Factor

Prandtl’s and von Karman’s semi-empirical laws

For smooth pipes Re1 2log( )2.51

ff= Eq. 6.16

For rough pipes

Colebrook-White transition formula:

This formula is applicable to the whole turbulent region fori l i (it ti l ti )

1 3.72log( )/Sk Df

=

1 2.512log( )3.7 Re

SkDf f

= − +

Eq. 6.17

Eq. 6.18

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commercial pipes (iterative solution)

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NON-CIRCULAR PIPES – THE HYDRAULIC RADIUS

Although the majority of pipes have circular cross-sections, there aresome cases where one has to consider flow in rectangular or othertypes of non-circular pipes.yp p p

Head loss calculations for non-circular pipe sections are done using the hydraulic radius concept

The hydraulic radius, Rh, is defined as the area, A, of the pipe sectiondivided by the “wetted perimeter” (circumference), P.

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Rh = A/P

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NON-CIRCULAR PIPES – THE HYDRAULIC RADIUS (cont.)

For a circular pipe: Rh = area/wet perimeter = (πD2/4)/(πD) = D/4; or D = 4Rh

This value for the diameter may be substituted into the Darcy-Weisbachequation, Reynold’s number, and relative roughness:

Using these equations, the head loss for non-circular cross-sections may be determined, using an equivalent diameter for a non-circular pipe.

)2 (4, Re ,4 2 4

h sf hH

kV R kL V sh fR g D R

ρμ

= = =

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The hydraulic radius approach works well for turbulent flow, but not for laminar flow.

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2009

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25. Rörströmning IV (11.5-11.7)

PumptyperPumpsystemSerie- och parallellkopplade pumpar

Övningstal: J27-28, J30

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Pump types – centrifugal pump

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Pump types – axial flow pumps

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Rotating movement by centrifugal impeller or propeller ⇒Pressure increases over the pump

Pressure increases over pump: (pout – pin) / ρg = Hp = pump head

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Specific energy consumption pump = (w Q Hp)/(3600 Q η) kWh/m3

η = power output/power input

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A pump is characterized by a so-called pump curve

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η = power output/power input

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Calculation of flowrate and pressure in a pump system

Water is pumped from reservoir A toreservoir B

What will the flowrate, Qp, be if the pipecharacteristics, L, D, ks are known as wellas static head, Δz, and pump curve?

The pressure increase over pump (energysupply from pump to water) shouldachieve two things with respect to lifting

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achieve two things with respect to liftingwater from reservoir A to B: 1) overcome geometric height, Δz2) overcome head losses hf1 + hf2

The hydraulic characteristics for thepipe system, Hsyst, is obtained fromthe energy equation ⇒

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L QH z h z fsyst losses D gA=Δ +Σ =Δ +

(local losses neglected in this case)

Hsyst states how much energy that is needed to transport 1 kg of waterfrom A to B

Hp states how much energy thepump can provide to the water

2gA

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pump can provide to the water

When the pump is introduced in thepipe system the flowrate and pumphead will adjust so that Hsyst = Hp

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J28:Water is pumped between two reservoirs with the same water surface elevation zo. Total pipe length is L = 2500 m, diameter D = 0 1 m and equivalent sand

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33

= 0.1 m and equivalent sand roughness k = 0.0001 m. The pump characteristics is given by the figure. What is the maximum permissible distance x from the upstream reservoir to the suction side of the pump, if the pressure must not be less than

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atmospheric. The local losses may be neglected. The temperature is 20°C.

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PARALLEL PUMPING

To deal with cases where you have pumps operating in parallel you canconsider them as being replaced by a fictive equivalent pump with a pump curve obtained by horizontal addition of the single pumps pump curves

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PUMPS IN SERIES

To deal with cases where you have pumps operating in series you can considerthem as being replaced by a fictive equivalent pump with a pump curveobtained by vertical addition of the single pumps pump curves

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EXAMPLES OF SYSTEM CURVES

1) Two pumps operating in parallel

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2) Heat pump systems

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3) Increase of natural flow rate

Without pump: 2 2

2 2 02 2syst

L Q L Qz f H z fD gA D gA

−Δ = ⇒ = Δ + =

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4) Flow control using a valve

2

2( )2syst valve

L QH z K fD gA

= Δ + +

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5) Time-varying reservoir surface

2

2 (Δz varies)2syst

L QH z fD gA

= Δ +

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6) Flow regulation using speed-adjustable pumps

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Discharge Q (m3/s) 0.020 0.030 0.040 0.050

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Head (m) 55 47 35 20

Total efficiency (%) 78 80 72 60

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J30:Water (20°C) is pumped between two reservoirs through two identical, parallel pipes each with a diameter of 0.2 m, length 1000 m, and equivalent sand roughness of 4⋅10-4 m.

a) What flow is expected through the pump?b) How much energy (kWh) is needed to pump 1 m3 of water? The efficiency ofb) How much energy (kWh) is needed to pump 1 m of water? The efficiency of the pump h = 0.75

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