FE Rapid Preparation for the Civil Fundamentals of Engineering Exam CIVIL REVIEW MANUAL Michael R. Lindeburg, PE Professional Publications, Inc. • Belmont, California 10 Hydraulics . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . .. . . . .. . . . .. . . . .. . . 1. Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2. Conservation Laws . . . . . . . . . . . . . . . . . . . . . . 3. Steady Incompressible Flow in Pipes and Conduits . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4. Flow in Noncircular Conduits . . . . . . . . . . . . 5. Open Channel and Partial-Area Pipe Flow . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6. The Impulse-Momentum Principle . . . . . . . . 10-3 10-3 10-4 10-7 1. INTRODUCTION . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . In a general sense, hydraulics is the study of the practical laws of fluid flow and resistance in pipes and open channels. Hydraulic formulas are often developed from experimentation, empirical factors, and curve fitting, without an attempt to justify why the fluid behaves the way it does. 2. CONSERVATION LAWS m2 m m1/2/s m m m – N – m/s2 m m Ns – – m kg kg/s – Pa kgm/s m3/sm m3/s m – s m m/s m m . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . Equation 10.1 Through Eq. 10.3: Continuity Equation A1 v1 ¼ A2 v2 Q ¼ Av _ ¼ Q ¼ Av m 10:1 10:2 10:3 Description Fluid mass is always conserved in fluid systems, regardless of the pipeline complexity, orientation of the flow, and fluid. This single concept is often sufficient to solve simple fluid problems. _2 _1¼m m When applied to fluid flow, the conservation of mass law is known as the continuity equation. 1 A1 v1 ¼ 2 A2 v2 If the fluid is incompressible, then 1 = 2. Equation 10.1, then, is the continuity equation for incompressible flow. Volumetric flow rate, Q, is defined as the product of cross-sectional area and velocity, as shown in Eq. 10.2. From Eq. 10.1 and Eq. 10.2, it follows that Q1 ¼ Q2 Symbols kinetic energy correction factor specific weight density Subscripts c critical h hydraulic H hydraulic – N/m3 kg/m3 Various units are used for volumetric flow rate. MGD (millions of gallons per day) and MGPCD (millions of gallons per capita day) are units commonly used in municipal water works problems. MMSCFD (millions of standard cubic feet per day) may be used to express gas flows. Calculation of flow rates is often complicated by the interdependence between flow rate and friction loss. P P I * w w w . p p i 2 p a s s . c o m Hydraulics/ Hydrologic Sys. Nomenclature A area B channel width C Hazen-Williams coefficient d depth D diameter E specific energy f friction factor F force Fr Froude number g acceleration of gravity, 9.81 h head hf head loss due to friction I impulse k1 constant K constant L length m mass _ m mass flow rate n Manning’s roughness coefficient p pressure P momentum q flow per unit width Q flow rate RH hydraulic radius S slope t time T width at surface v velocity y depth z height above datum 10-1 10-1 10-2 F E C I V I L R E V I E W M A N U A L Each affects the other, so many pipe flow problems must be solved iteratively. Usually, a reasonable friction factor is assumed and used to calculate an initial flow rate. The flow rate establishes the flow velocity, from which a revised friction factor can be determined. Example Example The diameter of a water pipe gradually changes from 5 cm at the entrance, point A, to 15 cm at the exit, point B. The exit is 5 m higher than the entrance. The pressure is 700 kPa at the entrance and 664 kPa at the exit. Friction between the water and the pipe walls is negligible. The water density is 1000 kg/m3. An incompressible fluid flows through a pipe with an inner diameter of 10 cm at a velocity of 4 m/s. The pipe contracts to an inner diameter of 8 cm. What is most nearly the velocity of the fluid in the narrower pipe? B 15 cm (A) 4.7 m/s (B) 5.0 m/s 5m (C) 5.8 m/s (D) 6.3 m/s A Solution 5 cm The cross-sectional areas of the two pipes are Hydraulics/ Hydrologic Sys. A1 ¼ pD 21 pð10 cmÞ2 ¼ ¼ 78:54 cm2 4 4 A2 ¼ pD 22 pð8 cmÞ2 ¼ ¼ 50:27 cm2 4 4 What is most nearly the rate of discharge at the exit? (A) 0.0035 m3/s (B) 0.0064 m3/s (C) 0.010 m3/s (D) 0.018 m3/s Use Eq. 10.1. A1 v1 ¼ A2 v2 Solution m ð78:54 cm2 Þ 4 A v s v2 ¼ 1 1 ¼ 50:27 cm2 A2 ¼ 6:25 m=s ð6:3 m=sÞ The answer is (D). .............................................................................................................................. Equation 10.4 and Eq. 10.5: Bernoulli Equation Use the Bernoulli equation, Eq. 10.5, to find the velocity at the exit. v2 p2 v22 p þ þ z2 ¼ 1 þ 1 þ z1 2g 2g 10:4 v2 p2 v22 p þ þ z2g ¼ 1 þ 1 þ z1g 2 2 10:5 Description The Bernoulli equation, also known as the field equation or the energy equation, is an energy conservation equation that is valid for incompressible, frictionless flow. The Bernoulli equation states that the total energy of a fluid flowing without friction losses in a pipe is constant. The total energy possessed by the fluid is the sum of its pressure, kinetic, and potential energies. In other words, the Bernoulli equation states that the total head at any two points is the same. P P I * w w w . p p i 2 p a s s . c o m First, find the relationship between the entrance and exit velocities, v1 and v2, respectively. From Eq. 10.1 and substituting the pipe area equation, A1 v1 ¼ A2 v2 2 2 pD1 pD2 v1 ¼ v2 4 4 ! ! D 22 ð15 cmÞ2 v2 v2 ¼ ¼ 9v2 v1 ¼ D 21 ð5 cmÞ2 2 v2 ð9v2 Þ p2 v22 p p þ þ z2g ¼ 1 þ 1 þ z1g ¼ 1 þ þ z 1g 2 2 2 sﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ p2 p1 gðz 2 z 1 Þ þ v2 ¼ 49 49 vﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ ﬃ u u ð664 kPa 700 kPaÞ 1000 Pa u u kPa u kg u ð49Þ 1000 3 ¼u m u u u m 9:81 2 ð5 mÞ t s þ 49 ¼ 0:516 m=s H Y D R A U L I C S Multiply the velocity at the exit with the cross-sectional area to get the rate of flow. pD 22 v2 4 0 1 2 B pð15 cmÞ C m ¼@ 2 A 0:516 s cm ð4Þ 100 m ¼ 0:0091 m3 =s ð0:010 m3 =sÞ 10-3 Solution Use the energy equation, Eq. 10.7. Take point 1 at the reservoir surface and point 2 at the pipe outlet. Q ¼ A 2 v2 ¼ v2 p v2 p1 þ z 1 þ 1 ¼ 2 þ z 2 þ 2 þ hf g 2g g 2g The pressure is atmospheric at the reservoir and the outlet, so p1 = p2. The velocity at the reservoir surface is v1 ≈ 0 m/s, so the equation reduces to z1 ¼ z2 þ The answer is (C). v22 þ hf 2g Solve for the velocity at the pipe outlet, v2. 3. STEADY INCOMPRESSIBLE FLOW IN PIPES AND CONDUITS . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . Equation 10.6 and Eq. 10.7: Extended Field Equation pﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ 2gðz 1 z 2 hf Þ rﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃﬃ m ¼ ð2Þ 9:81 2 ð200 m 180 m 18 mÞ s v2 ¼ v22 p1 p þ z 1 þ ¼ 2 þ z 2 þ þ hf 2g 2g 10:6 v2 p v2 p1 þ z 1 þ 1 ¼ 2 þ z 2 þ 2 þ hf g 2g g 2g 10:7 Description The extended field (or energy) equation, also known as the steady-flow energy equation, for steady incompressible flow is shown in Eq. 10.6 and Eq. 10.7. Hydraulics/ Hydrologic Sys. ¼ 6:26 m=s v21 The flow rate out of the pipe outlet is 2 pD Q ¼ v2 A ¼ v2 4 ! 2 m pð1 mÞ ¼ 6:26 s 4 ¼ 4:92 m3 =s ð4:9 m3 =sÞ The answer is (A). The head loss due to friction is denoted by the symbol hf. .............................................................................................................................. If the cross-sectional area of the pipe is the same at points 1 and 2, then v1 = v2 and v21 =2g ¼ v22 =2g. If the elevation of the pipe is the same at points 1 and 2, then z1 = z2. When analyzing discharge from reservoirs and large tanks, it is common to use gauge pressures, so that p1 = 0 at the surface. In addition, since the surface elevation changes slowly (or not at all) when drawing from a large tank or reservoir, v1 = 0. p1 p2 ¼ hf ¼ ghf Equation 10.8: Pressure Drop 10:8 Description For a pipe of constant cross-sectional area and constant elevation, the pressure change (pressure drop) from one point to another is given by Eq. 10.8. 4. FLOW IN NONCIRCULAR CONDUITS . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . Example An open reservoir with a water surface level at an elevation of 200 m drains through a 1 m diameter pipe with the outlet at an elevation of 180 m. The pipe outlet discharges to atmospheric pressure. The total head losses in the pipe and fittings are 18 m. Assume steady incompressible flow. The flow rate from the outlet is most nearly 3 (A) 4.9 m /s 3 (B) 6.3 m /s (C) 31 m3/s (D) 39 m3/s Equation 10.9: Hydraulic Radius RH ¼ cross-sectional area D H ¼ wetted perimeter 4 10:9 Description The hydraulic radius is defined as the area in flow divided by the wetted perimeter. The area in flow is the cross-sectional area of the fluid flowing. When a fluid is flowing under pressure in a pipe (i.e., pressure flow), the area in flow will be the internal P P I * w w w . p p i 2 p a s s . c o m 10-4 F E C I V I L R E V I E W M A N U A L area of the pipe. However, the fluid may not completely fill the pipe and may flow simply because of a sloped surface (i.e., gravity flow or open channel flow). The wetted perimeter is the length of the line representing the interface between the fluid and the pipe or channel. It does not include the free surface length (i.e., the interface between fluid and atmosphere). For a circular pipe flowing completely full, the area in flow is pR2. The wetted perimeter is the entire circumference, 2pR. The hydraulic radius in this case is half the radius of the pipe. RH ¼ What is most nearly the hydraulic radius of the flow? (A) 1.5 cm (B) 2.5 cm (C) 3.0 cm (D) 5.0 cm Solution The hydraulic radius is cross-sectional area wetted perimeter 3 ð8 cmÞ ð12 cmÞ 4 ¼ 3 ð8 cmÞ þ 12 cm ð2Þ 4 ¼ 3:0 cm RH ¼ pR R D ¼ ¼ 4 2pR 2 2 The hydraulic radius of a pipe flowing half full is also R/2, since the flow area and wetted perimeter are both halved. Hydraulics/ Hydrologic Sys. Many fluid, thermodynamic, and heat transfer processes are dependent on the physical length of an object. The general name for this controlling variable is characteristic dimension. The characteristic dimension in evaluating fluid flow is the hydraulic diameter (also known as the equivalent diameter), DH. The hydraulic diameter for a full-flowing circular pipe is simply its inside diameter. If the hydraulic radius of a noncircular duct is known, it can be used to calculate the hydraulic diameter. DH ¼ 4RH ¼ 4 area in flow wetted perimeter The frictional energy loss by a fluid flowing in a rectangular, annular, or other noncircular duct can be calculated from the Darcy equation by using the hydraulic diameter, DH, in place of the diameter, D. The friction factor, f, is determined in any of the conventional manners. Example The 8 cm 12 cm rectangular flume shown is filled to three-quarters of its height. The answer is (C). 5. OPEN CHANNEL AND PARTIAL-AREA PIPE FLOW . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . .. . . . . . . . An open channel is a fluid passageway that allows part of the fluid to be exposed to the atmosphere. This type of channel includes natural waterways, canals, culverts, flumes, and pipes flowing under the influence of gravity (as opposed to pressure conduits, which always flow full). A reach is a straight section of open channel with uniform shape, depth, slope, and flow quantity. There are difficulties in evaluating open channel flow. The unlimited geometric cross sections and variations in roughness have contributed to a relatively small number of scientific observations upon which to estimate the required coefficients and exponents. Therefore, the analysis of open channel flow is more empirical and less exact than that of pressure conduit flow. This lack of precision, however, is more than offset by the percentage error in runoff calculations that generally precede the channel calculations. .............................................................................................................................. Equation 10.10: Manning’s Equation 2=3 v ¼ ðK =nÞRH S 1=2 10:10 Values 8 cm SI units K=1 customary U.S. units K = 1.486 concrete n ≈ 0.013 Description 12 cm P P I * w w w . p p i 2 p a s s . c o m Manning’s equation has typically been used to estimate the velocity of flow in any open channel. It depends on the hydraulic radius, RH, the slope of the energy grade line, S, and a dimensionless Manning’s roughness coefficient, n. A conversion constant, K, modifies the equation

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