Module 6 Notes
Contents
1. Page 11 of 22 CE 3372 Water Systems Design FALL 2013 Substitute in values 432016 ft x 2 4 A1471b ft r 1 5 4 Fanchore 1 94slug f 25 85 14 15 7 96ft s 18 Solving for the anchor force one obtains Fanchor 59431b To conclude the forces on fittings are analyzed by momentum Remember that the pressures matter as does gravity in the vertical direction Generally one solves for the forces of the anchor on the control volume and then uses these forces is later parts of the design to design foundations to resist the hydraulic forces 1 2 2 Cavitation Cavitation occurs in flowing liquids when the flow passes through a zone of low pressure near the vapor pressure of the liquid Small bubbles of vapor are formed than when returned to a liquid release substantial energy that damages the pipe and equipment EGL BOL LC BGL a i A f M n fiu omes C VOS b Grade change a Venturi U c Power outage to pump Figure 8 Typical locations of cavitation in hydraulic systems The cavitation in pumps is problematic and is the reason the engineer makes the Net Positive Suction Head available N PSH computations Page 12 of 22 CE 3372 Water Systems Design FALL 2013 Good hydraulic design practice is to avoid negative gage pressures in a system and where necessary keep these negative pressures greater than 10 feet of head If the pressure head gets at all close to 332. 0 58 a Class B bedding b Class C bedding c Class D bedding load factor 1 9 load factor 1 5 load factor 1 1 Figure 10 Different trench bedding conditions from CITE The load on the pipe per unit length is W CYsoi B where ani is the unit weight of the soil B is the trench width and C is a coefficient that is a function of fill height and soil type Page 15 of 22 CE 3372 Water Systems Design FALL 2013 Figure 11 is a chart that relates the C factor to different soil types and depth to width ratios Once the load is computed then the engineer can select a pipe material and thickness to resist the anticipated load Observe that changing the backfill can have an effect of performance by lowering the C factor i e backfill in a saturated clay with cement stabilized pea gravel in a deep trench could reduce the load factor by nearly one allowing a thinner wall pipe to serve 5 0 4 0 2 0 Saturated UE topsoil Sand and gravel 0 3 l l PETET eee OE Od L 20 304050 100 200 300 HiB Figure 11 Load factors for trench design from CITE The typical formula is called Marston s general formula for loads on buried objects CITE If the pipe is rigid the formula is W C Ysoil B 23 where B is the trench width The coefficient C is taken from a soil description and either a tabulation or a chart such as Figure 11 If the pipe is flexible the formula is modified W C Ysoil B Ba
3. 1 3 3 Concrete Concrete pipe is common in civil engineering both as a pressure pipe and as unpressurized drainage pipes culverts sewers etc When used unpressurized corrosion is an issue especially with sewage where microbial induced concrete corrosion MICC can destroy the pipe crown in as few as two years Concrete is a rigid material Figure 14 is a typical table of strengths for different bedding conditions Page 18 of 22 CE 3372 Water Systems Design FALL 2013 Table 3 Mechanical properties of pipes fittings and accessories Type of castin Minimum tensile Minimum ring crush Maximum Brinell and material 9 strength strength hardness MPa MPa HB Pipes grey cast iron 200 350 260 spheroidal graphite cast iron 420 230 Fittings and accessories grey cast iron 150 260 spheroidal graphite cast iron 420 250 1 Other types of cast iron shall satisfy the criteria laid down for grey cast iron 2 Tensile and ring crush strength for other products see annex A 3 332 MPa for nominal sizes equal to or greater than DN 250 Figure 13 Ductile iron properties from CITE 1 3 4 Plastic Plastic PVC pipes are flexible and the concept of crush is somewhat exaggerated Surely PVC can be crushed but the failure is considerably more ductile than iron or concrete Figure 15 is a photograph of PVC pipe in a compression test frame Observe the large deformation of the material befor
4. Page 14 of 22 CE 3372 Water Systems Design FALL 2013 If the equivalent deflection is converted into a stress i e via a stress strain relation the result is o EaAT 22 where F is the elastic modulus of the material Most real systems use expansion fittings joints that move slightly to accommodate thermal changes although long plastic sewer pipes are often welded solvent or ultrasonic and do not have expansion fittings Thermal issues would be important in above ground large diameter pipe systems and in systems that carry heated liquids e g crude oil pipelines 1 2 5 External Loading Most pipes are placed in a trench and buried These pipes need to be able to resist internal loads pressure possible thermal loads temperature and the external load of the backfill and any additional overburden anticipated Trench and backfill design is largely empirical because the systems are complex to ana lyze Figure 10 shows a pipe with three different bedding conditions these are tabulated in CITE The load on the pipe is equal to the weight of the soil above the pipe plus any added weight above the soil less the shear force like a skin friction between the trench wall and the backfill Lightly compacted backfill Loose Carefully backfill X i L compacted tote 8 fet backfill 1 6 in min a Compacted crushed dC stone or AB B Im m Tier pea gravel tnl
5. feet of head cavitation is almost assured The choice of 10 feet is somewhat arbitrary but is intended to serve as a guideline with a reasonable factor of safety The conditions need to be checked at all operating conditions changes in flow start up and shut down ie If the hydraulic grade line is at an elevation 33 feet below the pipe system cavitation will occur Page 13 of 22 CE 3372 Water Systems Design FALL 2013 1 2 3 Internal Pressure Thin walled pipes are examined using a hoop tension analysis pDL a Figure 9 Thin wall pipe free body diagram A force balance on the ring section in Figure 9 produces the following formula for internal pipe stress pD dal 19 where t is the pipe thickness Thick walled pipes consider that the internal stress is non uniform and the formula be comes o 1 NE r2 ri p 20 2 2 Pri To 2 2 Once the internal stress is known based on geometry and pressure then materials that do not yield under such stresses can be selected 1 2 4 Temperature Stress and Strain Temperature stresses develop when temperature changes occur in pipes often between instal lation and service but also in service If a pipe is restrained and subjected to a temperature change then the pipe is subjected to an equivalent longitudinal deflection of AL LaAT 21 The term a is the coefficient of thermal expansion and is tabulated for different materi als
6. for negative pressure pipelines and thermal forces as temperature of the surroundings change 1 2 1 Forces on Bends and Transitions Momentum is used to calculate the forces in bends and transitions These forces are com puted to design thrust blocks and connections to keep a pipeline fixed in space instead of flopping around like a garden hose with a nozzle For a bend the momentum equation has at least two components if the plane of the bend is perpendicular or parallel to the gravitational acceleration direction three otherwise The momentum equations are Sod a Viz gt Fy pQ Vay Viy 9 25 F PQ Vaz Viz The subscripts 2 and 1 refer to the downstream and upstream locations flow from 1 to 2 Usually the third equation includes a body force grabity in the force summation Example Figure 4 depicts a section of pipe with a direction change The horizontal bend is 30 The 1 meter diameter pipe carries water at 3m per second The pressure in the bend is approximately 75 kPa gage and the volume of the bend is 1 8m The bend itself metal weights 4kN What forces are applied to the bend by the anchor thrust block to hold the bend in place Assume the pipe walls carry no force along their axis i e expansion joints that cannot transmit substantial longitudinal load Page 8 of 22 CE 3372 Water Systems Design FALL 2013 Concrete anchor Expansion joints a Plan view b Elevation view F
7. in a similar fashion and design guidance is provided in Manuals of Practice and from manufacturer tests Page 17 of 22 CE 3372 Water Systems Design FALL 2013 1 8 Pressure Pipe Materials The common pressure pipe materials are steel concrete ductile iron plastic ABS and PVC Less common but still available are vitrified clay and asbestos cement Lead wood stone and brick pipes are obsolete and except for a historical installation would not be used Lead would be unavailable in the US 1 3 1 Steel Steel pipe in civil engineering use is medium carbon content Stainless would be unusual in a pressure pipe system but is available Typical recomendations are the use of a tensile stress for design of steel pipe to be equal to 1 2 the yield stress factor of safety of at least 2 Corrosion is an issue and coatings are used to protect the pipe as is cathodic protection Steel would probably be considered a quasi flexible pipe and backfill would be critical 1 3 2 Ductile Iron Ductile grey iron is common in civil engineering It is used for force mains in sewage as well as water supply lines Joints are important and the material can be coated with epoxies for corrosion control It is a rigid material and can tolerate poor backfill conditions somewhat better than flexible materials Figure 13 lists some important material properties Even though tolerant of poor backfill good backfill design should be used if at all possible
8. s method Starting with the network equations they are rewritten into a functional form suitable for a Newton s type approach A Q Q D f Q 0 5 Recall from Newton s method that d sea ge lag En 6 thus the extension to the pipeline case is Qi Qr I QAE 7 where J Q is the Jacobian of the coefficient matrix A evaluated at Q Although a bit cluttered here is the formula for a single update step with the matrix demand vector and the solution vector in their proper places U Qr QJ A Q4 Qe DJ 8 The Jacobian of the pipeline model is a matrix with the following properties 1 The partition of the matrix that corresponds to the node formulas is identical to the original coefficient matrix it will be comprised of 0 or 1 in the same pattern at the equivalent partition of the A matrix 2 The partition of the matrix that corresponds to the loop formulas will consist of values that are twice the values of the coefficients in the original coefficient matrix at any supplied value of Qx In the current example the Jacobian would look like the following array columns and rows are abbreviated to fit the page Page 5 of 22 CE 3372 Water Systems Design FALL 2013 1 0 0 0 0 1 1 0 0 0 0 0 1 dese i 1 2f K Qil 0 0 e 0 0 0 2fK Qj 2fK Qs 2fK Qio 0 0 0 0 JA U U In this document the pipeline solution is a true Newton s method because analytical Jacobi
9. 24 Page 16 of 22 CE 3372 Water Systems Design FALL 2013 where D is the diameter of the conduit If the trench is wide compared to the pipe say two pipe diameters then the formula will over estimate the applied load and more extensive than here geotechnical considerations are applied These formulas estimate only the soil load added live loads trucks railroads etc are added to these applied loads to examine the anticipated load on the buried pipe 1 2 6 Rigid Pipe Strength Rigid pipe strength is determined using some variation of a three point bearing test as depicted in Figure 12 possibly to failure of the testing machine has the capacity Like a concrete compression test the instrument records loading rate and deformations Load Pipe 2D 1 in min Bearing strips Figure 12 Three point bearing test The test is identical to tests you performed in the materials laboratory The results of such tests are tabulated by manufacturers and used by the design engineer to select materials Such strength values are used with a factor of safety to support the external loads as expressed in Equation 25 mu Lia dae P load Lsafe 25 Paten where Lsafe is the safe design load per unit length L3 454 is the material strength from a 3 edge test manufacturer supplied Fj is the load bedding factor from Figure 10 and Frafety is the factor of safety an expression of uncertainty Flexible pipes are treated
10. CE 3372 Water Systems Design FALL 2013 1 Network Theory This article presents the theory that is employed to perform pipe network calculations essentially it is a look under the hood of EPA NET The important point is that one could make computations by hand if needed the quasi linearization approach is general and the computation method can be extended to other kinds of systems electric circuits non linear structures logistics Pipe networks like single path pipelines are analyzed for head losses in order to size pumps determine demand management strategies and ensure minimum pressures in the system Conceptually the same principles are used for steady flow systems conservation of mass and energy with momentum used to determine head losses The following brief description of the arithmetic behind such analyses and the algorithm common to most if not all network analysis programs is accomplished using a simple example Keep in mind that the network is simply an extension of the branched pipe example Illustrative Example d LE a ndoa Day gan B zf 636 Dry riso Dry cc Te oer Figure 1 Pipe network for illustrative example with supply and demands identified Pipe dimensions and diameters are also depicted Figure 1 is a sketch of the example problem that will be used The network supply and Page 1 of 22 CE 3372 Water Systems Design FALL 2013 demand node are annotated with the discha
11. Q Q D 3 Finding a Solution to the Network Equations The network equations include the node equations and the loop equations The set of equations are solved simultaneously for pipe discharge and then these results are used to determine system pressures or other nodal quantities The system for a pipeline network happens to be a quadratic system of equations therefore non linear and therefore some adaptation of linear solvers is used At the correct solution the following matrix vector system is true A Q Q D 4 Non linear because of the Q Q term These equations are statements of conservation of mass These equations are statements of conservation of energy and independence of path Page 4 of 22 CE 3372 Water Systems Design FALL 2013 In this expression A is a function of Q therefore the coefficient matrix is a function of the solution vector a non linear system If the system were a univariate non linear equation we would proceed by subtracting the right hand side from the left hand side and re expressing the entire relationship as a function of the solution variable that is supposed to equal 0 Newton Raphson Method Theory The Newton Raphson method extends the classical Newton s method to vector valued func tions of vector arguments The first derivative of Newton s method is replaced by the Jaco bian of the function but otherwise the method is for all purposes identical to the univariate Newton
12. S i eie odd 3sazs e Bul 70WH ous L see th 0 5 Wo om so D L ED LSOS8S7S C LESGPLIL C AICLOLCO SGPIS LZEL9vCE 0 zverteer 0 SrLoEHLOC 26986 C L exo gr O ECS L aag o30 sace sare maa ed K mae oars gt Sem Jreetzoc e SIRSE c OER Z ran cep Z RCC S FES lecerweze 0 laen UgL28 BZ v 1iGA CC8 U L L b FES ERL Paran ERE FES PILL EDER LL WEBTISZs peo 10 11 ls riai PES Panai Tj S400 swg leuren SLC 0 5100 s0 102 SC 9 II SZ LRE C BASS LRR 7 ELpress sbBBLGG Fr re 8658195 1 Esprit SKRBpSLSS 4 srn e e rm EGE OGL coe oor n L U L 3 D 8 8 dal ae ees E e Tet 2593 E DNJIH S a 57 apt Figure 3 Pipe network for illustrative example with loops pipes and nodes labeled Page 7 of 22 CE 3372 Water Systems Design FALL 2013 1 1 Readings 1 http en wikipedia org wiki Newton s method 2 http en wikipedia org wiki Pipe network analysis 1 2 Forces and Stresses on Pipes Pipes are structural elements that are subjected to internal and external forces The in ternal forces are pressure and momentum transfer when direction changes and cavitation when the liquid pressure is so low that the liquid phase changes switchces between gas and liquid rapidly The external forces are the restraining forces to counteract momentum changes crushing forces of loads outside the pipe including air pressure
13. an values are used If a numerical method to approximate the derivatives is used it would be called a quasi Newton method As an algorithm the engineer would supply a guess for Qg compute the update value used this just computed value as the new guess and repeat the computation until the computed vector is relatively unchanging Typically even with a poor first guess the solution can be found in z 2 x rank A This method is the basis of nearly all network models it is even used in groundwater hy draulics and surface water networks The method with some effort can be extended to transient systems Spreadsheet Example Figure 3 is an image of a spreadsheet model to implement such calculations The purpose here is to illustrate a bit of the layout the spreadsheet itself required manual recalculation and iteration counter and reasonably intricate updating and linear algebra operations All the pieces needed are part of a spreadsheet system and an actual spreadsheet will be illustrated in class n a transient solver there would be a set of iterations per time step hence the method is used over and over to evolve forward in time In a transient case analytical derivatives would be extremely desirable but if geometry changes as in open channel cases the programs usually sacrifice speed and use numerical approximation of the Jacobian at each time step Page 6 of 22 FALL 2013 CE 3372 Water Systems Design F out i
14. anual version 5 0 Tech nical Report EPA 600 R 05 040 U S Environmental Protection Agency National Risk Management Research Laboratory Cincinnati OH 45268 Page 22 of 22
15. be used so the Reynolds number component is superfluous The head loss in any pipe is Hioss f K Q Q 2 Write the mass balance for each node and head loss for each loop This step builds the equation system the matrix below has two partitions the upper partition corresponds to the nodal equations and the lower partition to the loop equations Notice that the lower partition will change value for any change in the discharges 1 0 0 0 0 1 0 0 0 0 1 1 0 1 0 0 0 0 0 0 0 1 1 0 0 0 0 0 0 0 0 0 U U 1 1 1 0 0 0 0 0 0 1 i1 0 0 0 1 0 0 0 0 0 0 0 1 i 0 0 0 0 0 0 0 0 0 1 1 1 0 U 1 U U U U U U 1 fKIQi 0 0 fK Q fK Qs fK Qs 0 0 0 0 0 jK Q fK Q4 fK Qi 0 0 0 0 fK Qs fKlQso 0 0 0 0 fK lQ 0 Klg fK Qs fK Qu 0 The convention used here is that flow into a node is algebraically positive while flow away from a node is negative The sign convention in the loop partition is that if the loop traverse is the same direction as the assumed flow then the sign convention is a positive loss while a negative loss is computed for the opposite situation The array above is a coefficient matrix dependent on discharge This array is represented here as A Q The discharges are written as the vector Q Page 3 of 22 cOooccccoco occ CE 3372 Water Systems Design FALL 2013 The demand is written as the vector D ene eme uds c occ So the resulting system of equations are A
16. e failure If the material is buried and backfilled correctly the deformation would be opposed by the compression of backfill at the springline of the pipe Page 19 of 22 CE 3372 Water Systems Design CONCRETE PIPE CULVERT CRUSHING STRENGTH LBS PER LIN FT ULTIMATE STRENGTH OR CLASS METHOD A BEDDING DIAMETER DIAMETER MAXIMUM HEIGHT OF COVER IN FEET UN SQ FT IN Inr w M als a lw lM r N o Oja j jjs 147 2t 30 14 2T 30 2r 30 2T 30 2v 30 2r 30 2r 30 102 2r 30 108 Heights of cover shown in toble are for finished construction To protect pipe during construction minimum heights of cover prior to allowing construction traffic to cross installation are to be 2 or 3 0 whichever is greater This cover shall extend the full length of the pipe culvert The approach fill ramp is to extend a minimum of 10 Dia 3 on each side of the culvert or to the intersection with a cut Minimum finished height of cover to be Dia or 2 0 whichever is greater except pipe under entrances and median crossovers where a 9 min willbe permitted 2 CONCRETE PIPE CLASS TABLE FOR H 20 LIVE LOAD VIRGINIA DEPARTMENT OF TRANSPORTATION Figure 15 Three point bearing test on flexible pipe FALL 2013 Sheet_1 of 17 Page 20 of 22 CE 3372 Water Systems Design FALL 2013 General Physical Properties of PVC Pipe Value Test Method GENERAL Cell Classification 12454 ASTM D1784 Maxim
17. igure 4 Pipe with direction change The physical restraint that holds the pipe is called a thrust block Solution Apply the momentum equations to the control volume between the expansion joints as depicted in Figure 5 Figure 5 Control volume FBD of pipe bend z axis is perpendicular with plane of the figure First the x component of the momentum equation is pi A1 paAacos 30 Finrust e 1000kg m 3m s Q A2 cos 30 Q A1 10 The two pressures are equal as is are the cross sectional areas so the equation reduces o 75 000Pa A 1 cos 30 Finrust o 1000kg m 3m s Q A cos 30 1 11 Page 9 of 22 CE 3372 Water Systems Design FALL 2013 The cross sectional area is A m D 4 0 785m so solving for the force of the thrust block gives Pe 19 000Pa 0 785m 1 0 866 1000kg m 3n s 3 31 3 82 m s 9 420N 12 Thus the thrust block pushes to the left on the pipe Repeating a similar analysis for the y component of momentum is Fihrusty 75 000Pa 0 785m sin 30 1000kg m 3m s 3 825in 30 m s d 170N 13 Finally the z component is considered to find the vertical force the thrust block msut provide to hold the system in place against its weight F Wrend W water ye ees 1000kg m 3m s 0 14 Solving for the anchor force produces Fanchor z Weend Wwater 4000N 1 8m 9800N m 21 600N 15 Recall these a
18. re all vector components so direction up down left right matters So the resultant force the block must be able to transmit to the bend is Finrust 9 420i 35 170j 21 600jN Once the forces are understood then we could design the block foundation recall we need to include the dead weight of the block itself Transitions are analyzed in nearly identical fashion again an example will illustrate the analysis Example Water flows through the pipe diameter reducer shown in Figure 6 at a rate of 25 cubic feet per second The minor loss coefficient or this fitting is 0 20 based on velocity in the smaller pipe What longitudinal force from a thrust block is required to hold the reducer in place Upstream pressure is 30 psi gage Page 10 of 22 CE 3372 Water Systems Design FALL 2013 Figure 6 Pipe reducer transition diagram Solution First sketch a control volume FBD to identify momentum terms and such as in Figure 7 Again we assume the pipes are connected by zero force joints so the external force is applied to only the reducer Control surface gt NE r Figure 7 Control volume FBD of transition Next use the energy equation to find the downstream pressure pa MW po Vs Vj to y m IL 030 16 y 29 2g 2g Substituting in numerical values and solving for p we obtain na 4147lb ft note the units Now we apply the momentum equation to find the required restraining force
19. rge in this case 10 ft s Sketch the network The first step in analysis is to sketch the network E podes B f oges 3 x npe Jo Figure 2 Pipe network for illustrative example with loops pipes and nodes labeled Check geometry A unique solution exists only if the sum of the node count and loop count is equal to the pipe count In the current example this sum is 11 but the drawing shows only 10 pipes A modeling trick is to add a fictitious pipe at either a supply of demand point to satisfy this geometric requirement The particulars of this imaginary pipe are irrelevant as flow in this pipe should vanish at the solution Prepare f K Re Tables The next step is to prepare tables for use in the head loss equations In these notes the TA single demand is unusual but helps with clarity in the example Page 2 of 22 CE 3372 Water Systems Design FALL 2013 Darcy Weisbach formula is used for head loss thus the relevant equations for any particular pipe are 1 The head loss coefficient just the constant part to be multiplied by Q Q to obtain loss for the pipe 4p S 1 gD 2 The Reynolds number coefficient to be multiplied by Q to obtain the pipe Reynolds number for determination of friction factors He BL Q uD 3 An the friction factor table if variable factors are to be used typically the Colebrook White formula is used but table look up is also valid and fast In this example fixed values will
20. um Service Temperature 140 F Color white dark gray Water Absorption increase 24hrs 25 C 05 ASTM D570 Hardness Rockwell 110 120 ASTM D785 Poisson s Ratio 8 73 F 410 Hazen Williams Factor C2150 Mechanical Physical Properties of PVC Pipe Value Test Method MECHANICAL Specific Gravity g cu cm 1 40 t 02 ASTM D792 Tensile Strength psi 73 F 7 450 ASTM D638 Modulus of Elasticity psi 73 F Tensile Modulus 420 000 ASTM D638 Flexural Strength psi 73 F 14 450 ASTM D790 Compressive Strength psi 73 F 9 600 ASTM D695 Izod Impact ft lb in 73 F 275 ASTM D256 Figure 16 Plastic PVC properties from CITE Page 21 of 22 CE 3372 Water Systems Design FALL 2013 References Chin D A 2006 Water Resources Engineering Prentice Hall Gironas J L A Roesner and J Davis 2009 Storm Water Management Model appli cations manual Technical Report EPA 600 R 09 077 U S Environmental Protection Agency National Risk Management Research Laboratory Cincinnati OH 45268 NCEES 2008 Fundamentals of Engineering Supplied Reference Handbook 8th ed 280 Seneca Creek Road Clemson SC 29631 National Council of Examiners for Engineering and Surveying ISBN 978 1 932613 37 7 Rossman L 2000 EPANET 2 users manual Technical Report EPA 600 R 00 057 U S Environmental Protection Agency National Risk Management Research Laboratory Cincinnati OH 45268 Rossman L 2009 Storm Water Management Model user s m
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