Fundamentals of Hydraulics: Flow, Spring/Summer 2010, Vol

PUBLISHED BY THE NATIONAL ENVIRONMENTAL SERVICES CENTER

FundamentalsBy Zane Satterfield, of P. E. ,Hydraulics:NESC Engineering Scientist Flow

Summary Hydraulics is the branch of engineering that focuses on the practical problems of collecting, storing, measuring, trans- porting, controlling, and using water and other liquids. This Tech Brief—the second in a two-part series—provides basic information about hydraulic problems and will focus on calculating flow rates in water or wastewater conveyance (distribution and collection) systems. The first Tech Brief in this series discussed various aspects of pressure.

Why is understanding hydraulics important? Flow Liquid in motion produces forces and pressure whenever its In pipes under pressure or in open channels under velocity or flow direction changes. Knowing pipe pressure the force of gravity, the volume of water flowing past and flow at certain points along the pipe’s path can help any given point in the pipe or channel per unit time determine the necessary pipe size and capacity, as well as is called the flow rate or discharge (Q). what pipe material would work best in given situations. In the U.S., flow rate may be expressed as cubic feet Understanding hydraulics can help systems decide what 3 pipe-flow rates related to size and material are necessary to per second (ft /s or cfs), gallons per minute (GPM), or transport water in an efficient manner.A key rationale for million gallons per day (MGD). An approximate but understanding hydraulics is that it can save money for convenient conversion to remember is 1 MGD = 1.55 small systems, especially those that can’t afford engineering cfs = 700 GPM. services or may not have an engineer on retainer. In the metric system, the unit for flow rate is cubic meters per second (m3/s). The term liters per second Units and Variables (L/s) is also used for relatively small flow rates. Understanding that water has a unit weight is important. Another expression for flow rate is megaliters per day 3 6 System designers need this information to calculate other (ML/d), So 1 m = 1000 L and 1 ML = 10 L. variables. The physical and hydraulic behavior of waste- Flow rate is expressed by the formula: water is similar to that of clean water, and there is gener- ally no difference in the design or analysis of systems Q = A x V involving these two liquids. Both are considered to be Where Q = flow rate or discharge in gpm, mgd or cfs incompressible liquids because their volume does not (ft3/s) in U. S. m3/s, ML/d or L/s in SI change significantly with changing pressure. — metric units Water or wastewater has a unit weight of 62.4 pounds A = cross-sectional flow area usually in ft2 per cubic foot (62.4 lb/ft3); so, one cubic foot (1ft x 1ft x in U. S. units m2 in SI metric units 1ft) of water weighs 62.4 pounds. There are 7.48 gallons — in a cubic foot, with each gallon weighing approximately V = velocity of flow usually in ft/s in 8.34 pounds, so 7.48 gal x 8.34 lbs/gal = 62.4 pounds. U. S. units — m/s in SI metric units In the metric system, the term for weight (force due to gravity) is a Newton (N), and the unit weight of water is Flow in Pipes under Pressure 9,800 Newtons per cubic meter (9,800 N/m3). More appro- When water flows in a pipe, friction occurs between priately, this is expressed as 9.8 kilonewtons per cubic the flowing water and the pipe wall, and between the meter (9.8 kN/m3), where the prefix kilo stands for 1,000. layers of water moving at different velocities in the pipe due to the viscosity of the water. The flow velocity is zero at the pipe wall and maximum along the all of our centerline of the pipe. Velocity of flow means the aver- Download age velocity over the cross section of the flow. at www.nesc.wvu.edu/techbrief.cfmTech Briefs DWFSOM150 two

PAGE OF FOUR al .Hazen-Williams Coefficient for Different Types ofPipe Table 1. hr Q=flowrateinm Where Q =0.28xCD commonly usedistheHazen-Williams equation: hydraulics todothis,butoneofthosemost in severalformulas are the system.There andflowsthroughout the headlosses,pressures, existing pipenetworks,youwillneedtocalculate mainsortoanalyze pipelines orsewageforce beabletodesignnewwaterdistribution To alongthepathofflow. drop tinuous pressure loss inthesystem.Thisenergy lossisacon- toflowcausesenergy The frictionalresistance W V Tin Steel Plastic Lead Glass Galvanized Iron Copper Concrete orconcretelined Cast Iron Brick Sewer Brass Asbestos Cement itrified Clay(good condition) ood Stave(average condition) Riveted New unlined Coal-tar enamellined Centrifugally spun Wooden forms Steel forms 40 yr. old 30 yr. old 20 yr. old 10 yr. old New, unlined dimensionless S =slopeofhydraulicgradeline, (in) inchesU.S.units D =pipediameterin(m)SIunitsor 1. based onpipematerialsuchasTable (dimensionless) canbefoundintables coefficient C =piperoughness cient accuracy. both SIunitsandU.S.withsuffi- 0.28 =constantwhichcanbeusedfor gpm inU.S. Pipe Materials in feetormeters Wher or meters S =h Wher 2.63 e Listhelengthofpipe e h L /L L x S is headlossinfeet 0.54 3 /s inSImetricor T e c h

B r i e f

• Fundamentals ofHydraulics: Flow 130-140 140-150 145-150 140-150 130-140 130-140 C 107-113 1 89-100 10-140 HW 64-83 75-90 120 130 140 120 135 120 140 130 100 140 110 Now compute the value of S, where S=h Now computethevalueofS,where pipe tobeC=140. finds thepiper Solution: headisadequate. giving thatthestorageandpressure to seeiftheexistingpipingwillhandletheirdemands the flowrateinpipeusingHazen-William equation head lossof10feetper1,000pipeline.Determine ing 12-inchdiameternew,plasticpipecarrieswaterwitha industrialparksiteandneeds1,800gpm.Theexist- entire existing waterlinesinplace.Acompanywantstousethe Your watersystemservicesavacantindustrialparkwith Example 1 thr open channelflow,exceptwhenthewaterispumped orsanitarysewersisalso channel flow.Flowinstorm orriverflowisopen losses ingravityflowaswell.Stream as withpr But, themoving force. or gravityflow.Gravityprovides it iscalledopenchannel face exposedtotheatmosphere, When waterflowsinapipeorchannelwithfree sur- (Not underPressure) Gravity FlowinPipesorOpenChannels new facility. pr The calculationshowsthattheexisting12-inchpipewill Now applyingtheHazen-Williams equation The diameterisgivenininchesandstays wetted perimeter. Thesizeofthechannel orpipe,aswell sur invert elevations alongthepipeline.The lengthofwetted objective ofsewerdesign is toestablishappropriate the bottomofpipewall iscalledtheinvert.Onebasic and The topoftheinsidepipe walliscalledthecrown, istheslopeofchannelbottom. water surface conditions,theslopeof Under steady,uniform-flow constant inthelengthofchannelbeinganalyzed. also are flowarea andcross-sectional of thewatersurface flow means that the slope Uniform time. with constant is flow.Steadyflowmeansthatthedischarge steady uniform orsanitarysewersystemsinvolveacondition called storm calculations inthedesignoranalysisof routine Most ovide more thanther ovide more ough thepipeunderpr face on the pipe or stream cross sectionis calledthe cross face onthepipe orstream Q =2,246gpm Q =0.28x140689.040.08317 get onethatdoes. If yoursdoesnot, Q =0.28x14012 Most calculatorshaveay Q =0.28xCD D =12in S =0.010 S =10ft/1,000ft essur From Table 1,theHazen-Williams Table equation From oughness coefficient (C)foranew,plastic oughness coefficient e flow, there are frictionlossesorener are e flow,there , Spring/Summer 2010, Vol. 10,Issue1 2.63 equir 2.63 x S essur ed amountneededbythe x 0.010 0.54 x e (a force main). e (aforce button. 0.54 L /L gy as its slope and wetted perimeter, are important factors Example 2 in relation to its discharge capacity. A storm sewer pipe is going to be extended to an existing 3-feet-wide by 2-feet-deep concrete The common formula for solving open channel flow prob- lined channel. The extended storm sewer pipe lems is called Manning’s formula and is written as: has a peak discharge of 20 cfs (ft3/s). A 6- Q = 1.0 or 1.5/n x A x R2/3 x S1/2 inch freeboard—the difference between the top of the water and the container it is in—is 3 Where Q = channel discharge capacity in (m /s) SI needed in the channel. Will the channel 3 units or (ft /s) U.S. units handle the flow with 6-inch freeboard? 1.0 = constant for SI metric units A rectangular drainage channel is 3-feet wide and 1.5 = constant for U.S. units is lined with concrete. The bottom of the channel drops in elevation at a rate of 0.5 feet per 100 n = channel roughness coefficient (Manning’s n) feet. What is the discharge from the channel dimensionless when the water depth is flowing at 1.5 feet that gives us a 6-inch freeboard? Use Manning’s n of A = cross-sectional flow area (not the cross- 0.013 for a concrete-lined channel. section of pipe or channel) in m2 SI units or ft2 U.S. units Solution: Because the data is in U.S. units, use the constant of 1.5 in the Manning’s equation. R = hydraulic radius of the channel in (m) SI units or (ft) U.S. units The cross-sectional flow area (A) in the channel is: S = slope of the channel bottom, dimensionless A = 3.0 ft wide x 1.5 ft water depth From the variables above the hydraulic radius of a channel R, is defined as the ratio of the cross-sectional flow A = 4.5 ft2 area A to the wetted perimeter P. In formula form: The wetted perimeter P is: R = A/P P = 1.5 ft (side) + 3.0 ft (bottom) + 1.5 ft Where R = hydraulic radius of the channel in (m) SI (side) units or (ft) U.S. units P = 6.0 ft A = cross-sectional flow area (not the cross-

section of pipe or channel) in m2 SI units or The hydraulic radius R is: 2 ft U.S. units R = A/P

P = wetted perimeter in (m) SI units or ft U.S. units R = 4.5 ft2 / 6.0 ft The roughness coefficient (n), often referred to as R = 0.75 ft Manning’s n, depends on the material and age of the pipe or lined channel and on topographic features for a The slope S is: natural streambed. It can range from a value of 0.01 for S = rise (drop)/run a smooth, clay pipe to 0.1 for a small natural stream. A value of n commonly assumed for concrete pipes or con- S = 0.5 ft / 100 ft crete lined channels is 0.013. S = 0.005 Table 2. Manning’s Roughness Coefficient, n. Manning’s n Using Manning’s equation we can calculate Q Type of Pipe Min. Max. discharge: Glass, brass, or copper 0.009 0.013

Smooth cement surface 0.010 0.013 Q = (1.5/n) x A x R2/3 x S1/2 or (0.5) Wood-stave 0.010 0.013 It is very helpful if your calculator Vitrified sewer pipe 0.010 0.017 has a yx button Cast-Iron 0.011 0.015 Q = (1.5/0.013) x 4.5 ft2 x 0.752/3 ft x

1/2 P

Concrete, precast 0.011 0.015 0.005 AGE OF Cement mortar surfaces 0.011 0.015 Q = 115.38 x 4.5 ft2 x 0.8255 ft x Common-clay drainage tile 0.011 0.017 0.07071 three

Wrought Iron 0.012 0.017 FOUR Q = 30.3 cfs (ft3/s) Brick with cement mortar 0.012 0.017 Riveted-steel 0.017 0.020 The concrete lined channel will handle the extended storm sewer with a peak discharge of Cement rubble surfaces 0.017 0.030 20 cfs (ft3/s) with plenty of freeboard. Corrugated metal storm drain 0.020 0.024 NATIONAL ENVIRONMENTAL SERVICES CENTER Circular Pipes Flowing Full In U.S. units, use the constant of 1.5 in Manning’s equation for a full flowing pipe: Most sanitary and storm sewer systems are built with circular pipe. Along with the roughness coefficient, the Q = (1.5/n) x (π x D2)/4) x (D/4)2/3 x S1/2 or 0.5 other important factors in design or analysis are the flow velocity (V), pipe diameter (D), pipe slope (S), and how Q = (1.5/0.022) x (π x 22)/4) x (2/4)2/3 x 0.021/2 much the pipe can handle flow full (max discharge Q). Q = 68.18 x 3.1415 x 0.6299 x 0.1414 In a circular pipe carrying water so that the pipe is just Q = 19.08 cfs (ft3/s) full to the crown (but still under atmospheric pressure and gravity flow), the flow area (A) would be the pipe The combined sewer overflow is 14 cfs (ft3/s). This equation cross-sectional area (π D2)/4, the area of the pipe A = illustrates that the calculated flow for the 24-inch corru- (π x D2)/4. The wetted perimeter (P) is the perimeter of gated metal pipe flowing full at two-percent slope will handle the pipe, or circle or P = π x D. Because the hydraulic 19.08 cfs (ft3/s). The existing 24-inch pipe will work. radius (R) is defined as A divided by P, (R = A/P), we get: R = A/P Closing 2 Hydraulics can be complex when designing or analyzing R = ((π x D )/4) / (π x D) entire collection or distribution systems. Knowing the cor- Or written as rect formulas to calculate forces or pressure is the first step in having a properly functioning system. Pipe size, tank (π x D2) 1 storage, and location are all key in getting water to system R = x customers. If any situation should come up that the water 4(πx D) system staff does not feel comfortable analyzing, they This can be reduced to: should consult a licensed engineer. R = D/4 Several software and spreadsheet programs are available to assist in the hydraulics of system design. However, the data For circular pipes flowing full, the Manning’s formula is: they generate is only as good as the information that is Q = ((1.0 or 1.5)/n) x ((π x D2)/4) x (D/4)2/3 x S1/2 or 0.5 input. Knowing the fundamentals of hydraulics can help users get the most out of these programs. For a given value n, only the pipe diameter Note: The first part of this series, “Fundamentals of Hydraulics: and slope are needed to solve for discharge (Q) Pressure,” is available on the National Environmental Services Center in a circular pipe flowing full. website at www.nesc.wvu.edu/pdf/dw/publications/ontap/ To help in analysis there are charts or nomo- 2010_tb/hydraulics_pressure_DWFSOM147.pdf. graphs for the Manning’s equation just as with the Hazen-Williams equation. References Example 3 Hwang, Ned H. C. and Robert J. Houghtalen. 1996. Fundamentals of Hydraulic Engineering Systems, Third Edition. New Jersey: A combined sewer overflow is calculated to be Prentice-Hall, Inc. 14 cfs (ft3/s). The existing discharge point needs to be redirected under a road. A 24-inch Lindeburg, Michael R. 1999. Civil Engineering Reference Manual corrugated metal pipe is on hand in the for the PE Exam. Seventh Edition. Belmont, CA: Professional supply yard. Will this pipe work? Publications, Inc.

Calculate the discharge of a 24-inch corrugated Nathanson, Jerry A. 1997. Basic Environmental Technology, Water metal pipe flowing full at a slope of two percent. Supply, Waste Management and Pollution Control. Second From Table 2, use the Manning’s roughness Edition. New Jersey: Prentice-Hall, Inc. coefficient n. The table has a range for corrugated metal storm drain from 0.020 min for NESC Engineering Scientist Zane Satterfield is a licensed professional new condition to 0.024 for older condition. engineer and water operator who previously worked for the West Virginia Because the condition is not given, use the Bureau of Public Health, Environmental Engineering Division. middle range of 0.022, or to generate a worst- case condition, use 0.024. However, this exam- For a complete list of Tech Briefs, visit the NESC website at ple uses the middle range of 0.022, n = 0.022. www.nesc.wvu.edu/techbrief.cfm. You may download Tech Briefs for free from the site or you may order them for a nominal cost by calling FOUR The diameter is given at 24 inches but needs (800) 624-8301 or by sending and email to [email protected]. to be in feet; there are 12 inches in a foot so, four 24 in/12 in per foot = 2 ft (D = 2 ft). AGE OF P The slope is given at two percent. To get this into decimal form, 2% /100 = 0.02. (S = 0.02). An Equal Opportunity/Affirmative Action Institution

Published by The National Environmental Services Center at West Virginia University, P.O. Box 6064, Morgantown, WV 26506-6064 • Fundamentals of Hydraulics: Flow, Spring/Summer 2010, Vol. 10, Issue 1 Tech Brief