Chapter 1- Introduction to Hydraulics

ChapterChapter 11Chapter Chapter 1 1

Introduction to Hydraulics Introduction to Hydraulics EngineeringEngineering

Dr. Tewfik MahdiDr. Tewfik [email protected]@sharjah.ac.ae

Offi E tOffi E t 29432943Office Ext.: Office Ext.: 29432943

IntroductionIntroduction The word "hydraulics" originates from the Greek word The word "hydraulics" originates from the Greek word

((hydraulikoshydraulikos) meaning Study of pipe. in applied ) meaning Study of pipe. in applied (( yy ) g y p p pp) g y p p ppengineering the study of water or other fluids at rest or in engineering the study of water or other fluids at rest or in motion.motion.

HydraulicHydraulic topics range through most science and topics range through most science and engineering disciplines. In water resources engineering it engineering disciplines. In water resources engineering it covers concepts such as dam design river works floodcovers concepts such as dam design river works floodcovers concepts such as dam design, river works, flood covers concepts such as dam design, river works, flood control, sediment and erosion control, pressurized pipe control, sediment and erosion control, pressurized pipe system, hydropower, flow measurement and free surface system, hydropower, flow measurement and free surface hydraulics such as occurring in rivers, canals, lakes, hydraulics such as occurring in rivers, canals, lakes, estuaries and seas.estuaries and seas.

ObjectivesObjectivesObjectivesObjectives

Fundamentals of fluid mechanicsFundamentals of fluid mechanics Fundamentals of fluid mechanicsFundamentals of fluid mechanics Basic hydraulic system componentsBasic hydraulic system components

D i f W t Di t ib ti S tD i f W t Di t ib ti S tDesign of Water Distribution SystemsDesign of Water Distribution Systems TurbomachinesTurbomachinesDesign of free surface canals (i.e., storm Design of free surface canals (i.e., storm

water systems)water systems)y )y )

Fundamentals of Fluid MechanicsFundamentals of Fluid MechanicsA A fluidfluid is a substance that flows under the action of is a substance that flows under the action of

shearing forces If a fluid is at rest the forces on itshearing forces If a fluid is at rest the forces on itshearing forces. If a fluid is at rest the forces on it shearing forces. If a fluid is at rest the forces on it are in balance.are in balance.

AA i fl id th t i ibl Ai fl id th t i ibl A li idli id ii A A gasgas is a fluid that is compressible. A is a fluid that is compressible. A liquidliquid is a is a fluid that is hard to compress or incompressible. fluid that is hard to compress or incompressible.

A A free surfacefree surface is formed as a boundary between a is formed as a boundary between a liquid and a gas above it (e.g. water and air).liquid and a gas above it (e.g. water and air).

Basic of Fluid PropertiesBasic of Fluid Properties Pressure and Force Pressure and Force :: Pressure Pressure

P P is defined as the amount of is defined as the amount of force force FF exerted on a unit areaexerted on a unit area AAff b t (i fl id)b t (i fl id) A

FP Force, {F}={M.L.T-2} ={N}

Pressure= {P}={M.L.T-2} /{L2}2of a of a substance (i.e., fluid). substance (i.e., fluid). A ={N}/m2}={Pa}

1pa= 105 Nm-2 1psi =6895Pa DensityDensity (rho) is the amount(rho) is the amount m {}={M/L3} DensityDensity (rho) is the amount (rho) is the amount

of of MassMass per per Unit Volume Unit Volume of of fluid:fluid: V

m {}={M/L3} , SI: {Kg/m3} or Slugs/ft3 in the GB system

Specific weight (gamma) also known as Unit Weight is defined as the amount of V

W {KN/m{KN/m33} or } or lb/ftlb/ft33 in GB systemin GB systemdefined as the amount of

weight of fluid per unit volume of the fluid:

gV

. yy

CCSg

w

s

w

s00 4@4@

Specific Gravity (Sg) is the ratio of the weight of the body to the weight of an equal volume of standard substance

HYDROSTATIC PRESSUREHYDROSTATIC PRESSURE The pressure at any given point of a nonThe pressure at any given point of a non--moving (static) fluid moving (static) fluid

is called the is called the hydrostatic pressurehydrostatic pressure Pressure is function of the fluid properties and directly Pressure is function of the fluid properties and directly

proportional to the height of the proportional to the height of the fluid column above the fluid column above the areaarea of concern. of concern.

hp . is the specific weighth is the height of the water columngSI Unit for pressre: Pa, kPa, Psi, Bar, N/m2

F: is the hydrostatic forceA.hF cg.

The pressure at a given depth in a continuous, static body of liquid is constant.

p1 p3 p1 = p2 = p3

4 4 September September 20122012 66 Dr. Tarek MerabteneDr. Tarek Merabtene

p2

Law of Conservation of MassF di i l i iblF di i l i ibl t d Fl RTTRTT For one dimensional incompressibleFor one dimensional incompressible steady Flow RTT RTT reduces to: Outflow = Inflow. At constant density the reduces to: Outflow = Inflow. At constant density the equations is known as theequations is known as the continuity or flow ratecontinuity or flow rateequations is known as theequations is known as the continuity or flow rate continuity or flow rate Equation:Equation:

nnnn

nnnn1i

outii1i

inii1i

out1i

in )A.(v)A.(vorQQ

1i

outiii1i

iniii1i

out1i

in )A.v.()A.v.(ormm

v v is called the is called the flow average velocityflow average velocity : : vvavav=Q/A= =Q/A= 11//AA ((vv.n)..n).dAdA For pipes of constant diameter and incompressible For pipes of constant diameter and incompressible

flow average velocity flow average velocity vvavgavg stays the same down the stays the same down the g yg y avg avg yypipe, even if the velocity profile changespipe, even if the velocity profile changes

Fluid Dynamics: The Bernoulli Equation To solve for the flow rate (Q) we need dynamic approach to Fluid Mechanics Assuming that fluid motion is governed only by pressure and gravity forces, applying Newtons second law, F = ma, leads us to the Bernoulli Equation.

P/ + V2/2g + z along a streamline.2gvz

p 211

1 Cst(P=pressure =specific weight V=velocity g=gravity z=elevation)

A streamline is the path of one particle of water Therefore at any two points

2g

A streamline is the path of one particle of water. Therefore, at any two points along a streamline, the Bernoulli equation can be applied to evaluate unknown flows and pressures using a set of engineering assumptions.

At any two points on a streamline: 1 222

2gvz

p2gvz

p 22

211

1

Reynolds Number defines the flow regimeReynolds Number defines the flow regimeLaminar Laminar vsvs Turbulent flowTurbulent flow

Critical Reynolds number (ReCritical Reynolds number (Recrcr) for flow in a ) for flow in a round piperound pipeDefinition of Reynolds number

D.D..Re vv round piperound pipe

Re < Re < 2300 2300 laminarlaminar2300 2300 Re Re 4000 4000 transitional transitional Re > Re > 4000 4000 turbulentturbulent

y

Note that these values are approximate.Note that these values are approximate. For a given application, RFor a given application, Ree depends upondepends uponv: velocityD di Pipe roughnessPipe roughness

VibrationsVibrations Upstream fluctuations, disturbances (valves, Upstream fluctuations, disturbances (valves,

elbows, etc. that may disturb the flow)elbows, etc. that may disturb the flow)

D: diameter: density: kinematic viscosity : Dynamic viscosity , y ), y ) : Dynamic viscosity

Friction and Minor Losses in pipeFriction and Minor Losses in pipe Total head loss Total head loss hhT T in a system (i.e., the pipe sections) is in a system (i.e., the pipe sections) is

comprised of friction losses (comprised of friction losses (hhLL) and the minor losses () and the minor losses (hhmm) ) (in the hydraulics components): (in the hydraulics components): hhT T = h= hL L + h+ hmm

vv 2j2i i KLfh

ff: friction factor (friction loss coefficient) K: Minor loss coefficient: friction factor (friction loss coefficient) K: Minor loss coefficient

2g2gji

)components(jj

)pipe(i i

iiT KD

fh

ff: friction factor (friction loss coefficient), K: Minor loss coefficient , : friction factor (friction loss coefficient), K: Minor loss coefficient , D: diameter [m], L: length [m], v: velocity [m/s]D: diameter [m], L: length [m], v: velocity [m/s]

The total energy equation with head loss becomes:The total energy equation with head loss becomes:

2gv

2gv

2gv

2gv 2j

222ii

)components(jj

)pipe(i i

ii

ii KD

LfPzP

z hhpp: pump head (energy) in meter (if any): pump head (energy) in meter (if any)

gggg )components(j)pipe(i i

Energy Equation in a pipe with friction and Energy Equation in a pipe with friction and Minor losses, Pump and turbineMinor losses, Pump and turbine, p, p

When a piping system involves pumps and/or turbines, When a piping system involves pumps and/or turbines, pump and turbine head must be included in the energy pump and turbine head must be included in the energy

titiequationequation

The useful head of the pump (hThe useful head of the pump (h ) or the head) or the head The useful head of the pump (hThe useful head of the pump (hpump,upump,u) or the head ) or the head extracted by the turbine (hextracted by the turbine (hturbine,eturbine,e), are functions of volume ), are functions of volume flow rate, i.e., they are not constants.flow rate, i.e., they are not constants.

Operating point of system is where the system is inOperating point of system is where the system is in Operating point of system is where the system is in Operating point of system is where the system is in balance, e.g., where pump head is equal to the head balance, e.g., where pump head is equal to the head losses.losses.

Simple Piping SystemSimple Piping System Two general Two general types:types:

Pi i iPi i i Pipes in seriesPipes in series Volume flow rate is constantVolume flow rate is constant Head loss is the summationHead loss is the summationHead loss is the summation Head loss is the summation

of partsof parts Pipes in parallelPipes in parallel

Volume flow rate is the sum Volume flow rate is the sum of the componentsof the components

Pressure loss across allPressure loss across allPressure loss across all Pressure loss across all branches is the samebranches is the same

Energy (EGL) and Hydraulic Grade Line (HGL)Energy (EGL) and Hydraulic Grade Line (HGL) The fall of the EGL reflects the energy losses in the system EGL drops slowly due toThe fall of the EGL reflects the energy losses in the system. EGL drops slowly due to friction losses and it drops sharply due to a major loss (a valve or transition) or due to work extraction (to a turbine). The EGL can rise only if there is work addition (as from a pump). HGL is obtained as EGL minus the velocity head V2/2g above a datum.

The EGL and HGL slope downward in the direction of the flow due to the head loss in the pipe. A sudden change occurs in the HGL and the EGL whenever a loss occurs due to a sudden geometry change as represented At points where the HGL passes through g y g p p p gthe centerline of the pipe, the pressure is zero. If the pipe lies above the HGL, there is a vacuum in the pipe, a condition that is often avoided, if possible, in the design of piping systems; an exception would be in the design of a siphon.

Computational model of the Computational model of the friction factorfriction factor fffriction factor friction factor ff

Moody chart was developed for circular pipes, but can be used Moody chart was developed for circular pipes, but can be used for nonfor non--circular pipes using hydraulic diametercircular pipes using hydraulic diameterfor nonfor non circular pipes using hydraulic diametercircular pipes using hydraulic diameter

Colebrook equation is a curveColebrook equation is a curve--fit of the data which is convenient fit of the data which is convenient for computations.for computations.

Both Moody chart and Colebrook equation are accurate to Both Moody chart and Colebrook equation are accurate to 1515% % d t h i i t l fitti f d td t h i i t l fitti f d t

Implicit equation for f which can be solved using simple numerical model

due to roughness size, experimental error, curve fitting of data, due to roughness size, experimental error, curve fitting of data, etc.etc.

Modified Energy EquationModified Energy Equation The total kinetic energy at the section using averageThe total kinetic energy at the section using average The total kinetic energy at the section using average The total kinetic energy at the section using average

velocity is given by velocity is given by VV33AA. Therefore, the correction factor . Therefore, the correction factor is ( is (AA vv33dA)dA)/(V/(V33A) or A) or 11/A/AAA((vv/V)/V)33dAdA. .

If we have a defined velocity distribution function we can If we have a defined velocity distribution function we can find the correction factor analytically. find the correction factor analytically.

For inviscid flows it is For inviscid flows it is 11, for laminar flow it is , for laminar flow it is 2 2 and for and for turbulent flows it varies between turbulent flows it varies between 11..03 03 and and 11..3 3 with Reynolds with Reynolds Number.Number.Number. Number.

Thus, the real flows calculation using Bernoulli Equation Thus, the real flows calculation using Bernoulli Equation must be written as:must be written as:

2gvK

DLf

2gvPz

2gvPz

22

jj

i i

ii

22

22

221

2 pipe in components :j;pipeintionssec:iggg ji i

2

Assignment Assignment 11 Your are called to design a water slide as shown in the figure to be installed in a swimming Your are called to design a water slide as shown in the figure to be installed in a swimming

pool. Your client insist on the satisfaction and safety of his customers (pool. Your client insist on the satisfaction and safety of his customers (i.ei.e, ensure their , ensure their happiness but without burning their bottoms). Your researcher ends to recommend a happiness but without burning their bottoms). Your researcher ends to recommend a continuous water flow of continuous water flow of 11..39 39 L/s (about L/s (about 22 22 gal/min) down the slide. To be on a safe side you gal/min) down the slide. To be on a safe side you decide to use an decide to use an 8080%%--efficient pump under the slide, submerged efficient pump under the slide, submerged 1 1 m below the water surface m below the water surface to feeds a to feeds a 55--mm--long, long, 44--cmcm--diameter hose, of roughness diameter hose, of roughness 00..008 008 cm, to the slide. The hose cm, to the slide. The hose discharges the water at the top of the slide, discharges the water at the top of the slide, 4 4 m above the water surface, as a free jet. Ignore m above the water surface, as a free jet. Ignore minor losses and assume = minor losses and assume = 11..0606. Find the brake horsepower needed to drive the pump. For . Find the brake horsepower needed to drive the pump. For water take = water take = 998 998 kg/mkg/m3 3 and = and = 00..001 001 kg/kg/mmss. Write the steady. Write the steady--flow energy equation from the flow energy equation from the water surface (water surface (11) to the outlet () to the outlet (22) at the top of the slide: ) at the top of the slide:

Assignment Assignment 22 A 200mm Diameter pipeline of 5000m long and effective roughness 0.03 mm, A 200mm Diameter pipeline of 5000m long and effective roughness 0.03 mm, p p g g ,p p g g ,

delivers water between two reservoirs where the minimum difference of water level delivers water between two reservoirs where the minimum difference of water level is 40 m.is 40 m.

(a)(a) Determine the steady discharge between the two reservoirs (include friction losses Determine the steady discharge between the two reservoirs (include friction losses d th i l t th t d it f th t k )d th i l t th t d it f th t k )and the minor losses at the entrance and exit of the tanks).and the minor losses at the entrance and exit of the tanks).

Ans. 43.52 l/sAns. 43.52 l/s

(a)(a) If the discharge is to be increased to 50 l/s without increase in gross head, If the discharge is to be increased to 50 l/s without increase in gross head, determine the length of a 200 mm diameter pipe of effective roughness 0.015 mm to determine the length of a 200 mm diameter pipe of effective roughness 0.015 mm to be fitted in parallel. Consider only friction losses.be fitted in parallel. Consider only friction losses.

Ans 1645mAns 1645mAns. 1645mAns. 1645m

Assignment Assignment 33 A four pumpA four pump--turbine units of pumped storage hydroturbine units of pumped storage hydro--electric scheme are each to electric scheme are each to p pp p p p g yp p g y

supplied by a highsupplied by a high--pressure pipeline of length pressure pipeline of length 2000 2000 m. The minimum water head m. The minimum water head level between the upper and lower reservoirs is level between the upper and lower reservoirs is 310 310 m and the maximum is m and the maximum is 340340m.m.

The upper reservoirs has a usable volume of The upper reservoirs has a usable volume of 33..2525xx101066 mm33 which could be released which could be released t th t bi i i i i d ft th t bi i i i i d f 44 hhto the turbines in a minimum period of to the turbines in a minimum period of 4 4 hours.hours.

Max. Power output required/turbine = Max. Power output required/turbine = 110 110 MWMW TurboTurbo--Generator Efficiency = Generator Efficiency = 80 80 %%

Eff ti h f i liEff ti h f i li 00 66 Effective roughness of pipeline = Effective roughness of pipeline = 00..6 6 mmmmTaking Minor losses in the pipeline, power station and draft tube to be Taking Minor losses in the pipeline, power station and draft tube to be 33..0 0 m:m:(1)(1) Determine the minimum diameter of the pipeline to enable the maximum specified Determine the minimum diameter of the pipeline to enable the maximum specified

power to be generatedpower to be generated Ans :Ans : 22 66mm 22 77 mmpower to be generated. power to be generated. Ans.: Ans.: 22..66mm--22..7 7 mm(2)(2) Determine the pressure head to be developed by the pumpDetermine the pressure head to be developed by the pump--turbines when reversed turbines when reversed

to act in the pumping mode to deliver a total volume of to act in the pumping mode to deliver a total volume of 33..2525xx101066 mm33 to the upper to the upper reservoir uniformly duringreservoir uniformly during 66 hours in the offhours in the off--peak period.peak period. Ans.:Ans.: 365365mm--370370 mmreservoir uniformly during reservoir uniformly during 6 6 hours in the offhours in the off peak period. peak period. Ans.: Ans.: 365365mm 370 370 mm

Assignment 5Assignment 5 A reservoir A reservoir AA delivers water to a reservoir delivers water to a reservoir BB through two uniform pipelines AJ:JB of through two uniform pipelines AJ:JB of g p pg p p

diameters diameters 300 300 mm and mm and 200 200 mm respectively. Just upstream of the node J (change mm respectively. Just upstream of the node J (change in section) a control discharge of in section) a control discharge of 30 30 l/s is taken off.l/s is taken off.

Length AJ=Length AJ=3000 3000 m; JB=m; JB=4000 4000 m ; effective roughness of both pipes m ; effective roughness of both pipes 00..015 015 mm; and mm; and h d t l lh d t l l 2525head water level =head water level =25 25 m.m.

Determine the discharge to B (neglecting losses at node Determine the discharge to B (neglecting losses at node J)J)..Ans. QAns. Q11==6262..5 5 l/s, Ql/s, Q22==3232..5 5 l/sl/s

Assignment Assignment 44 A pump delivers a steady flow of water (,) from a large tank to two other higherA pump delivers a steady flow of water (,) from a large tank to two other higher--p p y ( ) g gp p y ( ) g g

elevation tanks, as shown. The same pipe of diameter d and roughness is used elevation tanks, as shown. The same pipe of diameter d and roughness is used throughout. All minor losses except through the valve are neglected, and the partiallythroughout. All minor losses except through the valve are neglected, and the partially--closed valve has a loss coefficient Kclosed valve has a loss coefficient Kvalvevalve. Turbulent flow may be assumed with all kinetic . Turbulent flow may be assumed with all kinetic energy flux correction coefficients equal toenergy flux correction coefficients equal to 11..0606. The pump net head H is a known. The pump net head H is a knownenergy flux correction coefficients equal to energy flux correction coefficients equal to 11..0606. The pump net head H is a known . The pump net head H is a known function of Qfunction of QAA and hence also of Vand hence also of VAA = Q= QAA/A/Apipepipe. Subscript . Subscript JJ refers to the junction point at refers to the junction point at the node J where branch A splits into B and C. Pipe length Lthe node J where branch A splits into B and C. Pipe length LCC is much longer than Lis much longer than LBB. It . It is desired to predict the pressure at is desired to predict the pressure at JJ, the three pipe velocities and friction factors, and , the three pipe velocities and friction factors, and the pump head Thus there arethe pump head Thus there are 88 variables: H Vvariables: H V VV VV ff ff ff pp Write down theWrite down thethe pump head. Thus there are the pump head. Thus there are 8 8 variables: H, Vvariables: H, VAA, V, VBB, V, VCC, f, fAA, f, fBB, f, fCC, p, pJJ. Write down the . Write down the eight equations needed to resolve this problem, but do not solve for now, since an eight equations needed to resolve this problem, but do not solve for now, since an elaborate iteration procedure, or an equation solver would be required.elaborate iteration procedure, or an equation solver would be required.