Presentation on theme: "Fluid Mechanics 2 The Bernoulli Equation"— Presentation transcript:
1Fluid Mechanics 2 The Bernoulli Equation CEVE 101Fluid Mechanics 2 The Bernoulli EquationDr. Phil Bedient
2FLUID DYNAMICS THE BERNOULLI EQUATION The laws of Statics that we have learned cannot solve Dynamic Problems. There is no way to solve for the flow rate, or Q. Therefore, we need a new dynamic approach to Fluid Mechanics.
3The Bernoulli Equation By assuming that fluid motion is governed only by pressure and gravity forces, applying Newton’s second law, F = ma, leads us to the Bernoulli Equation.P/g + V2/2g + z = constant along a streamline (P=pressure g =specific weight V=velocity g=gravity z=elevation)A streamline is the path of one particle of water. Therefore, at any two points along a streamline, the Bernoulli equation can be applied and, using a set of engineering assumptions, unknown flows and pressures can easily be solved for.
4The Bernoulli Equation (unit of L) At any two points on a streamline:P1/g + V12/2g + z1 = P2/g + V22/2g + z212
5Determine the difference in pressure between points 1 and 2 A Simple Bernoulli Example V2Zg = gairDetermine the difference in pressure between points 1 and 2Assume a coordinate system fixed to the bike (from this system, the bicycle is stationary, and the world moves past it). Therefore, the air is moving at the speed of the bicycle. Thus, V2 = Velocity of the BikerHint: Point 1 is called a stagnation point, because the air particle along that streamline, when it hits the biker’s face, has a zero velocity (see next slide)
6Stagnation PointsOn any body in a flowing fluid, there is a stagnation point. Some fluid flows over and some under the body. The dividing line (the stagnation streamline) terminates at the stagnation point. The Velocity decreases as the fluid approaches the stagnation point. The pressure at the stagnation point is the pressure obtained when a flowing fluid is decelerated to zero speed by a frictionless process
7Apply Bernoulli from 1 to 2 V2Zg = gairPoint 1 = Point 2P1/gair + V12/2g + z1 = P2/gair + V22/2g + z2Knowing the z1 = z2 and that V1= 0, we can simplify the equationP1/gair = P2/gair + V22/2gP1 – P2 = ( V22/2g ) gair
8A Simple Bernoulli Example If Lance Armstrong is traveling at 20 ft/s, what pressure does he feel on his face if the gair= lbs/ft3?We can assume P2 = 0 because it is only atmospheric pressureP1 = ( V22/2g )(gair) = P1 = ((20 ft/s)2/(2(32.2 ft/s2)) x lbs/ft3P1 =.475 lbs/ft2Converting to lbs/in2 (psi)P1 = psi (gage pressure)If the biker’s face has a surface area of 60 inchesHe feels a force of x 60 = .198 lbs
9Bernoulli Assumptions There are three main variables in the Bernoulli Equation Pressure – Velocity – ElevationTo simplify problems, assumptions are often made to eliminate one or more variablesKey Assumption # 1Velocity = 0Imagine a swimming pool with a small 1 cm hole on the floor of the pool. If you apply the Bernoulli equation at the surface, and at the hole, we assume that the volume exiting through the hole is trivial compared to the total volume of the pool, and therefore the Velocity of a water particle at the surface can be assumed to be zero
10Bernoulli Assumptions Key Assumption # 2Pressure = 0Whenever the only pressure acting on a point is the standard atmospheric pressure, then the pressure at that point can be assumed to be zero because every point in the system is subject to that same pressure. Therefore, for any free surface or free jet, pressure at that point can be assumed to be zero.
11The Continuity Equation Bernoulli AssumptionsKey Assumption # 3The Continuity EquationIn cases where one or both of the previous assumptions do not apply, then we might need to use the continuity equation to solve the problemA1V1=A2V2Which satisfies that inflow and outflow are equal at any section
12Bernoulli Example Problem: Free Jets What is the Flow Rate at point 2? What is the velocity at point 3?Givens and Assumptions: Because the tank is so large, we assume V1 = 0 (Volout <<< Voltank) The tank is open at both ends, thus P1 = P2 = P3 = atm P1 and P2 and P3= 0Part 1:Apply Bernoulli’s eqn between points 1 and 2P1/gH2O + V12/2g + h = P2/gH20 + V22/2g + 0simplifies toh = V22/2g solving for V V = √(2gh)Q = VA or Q = A2√(2gh)1γH2O2A23
13Bernoulli Example Problem: Free Jets Part 2: Find V3?Apply Bernoulli’s eq from pt 1 to pt 3P1/gH2O + V12/2g + h = P3/gH20 + V32/2g – HSimplify to h + H = V32/2gSolving for V V3 = √( 2g ( h + H ))1γH2O2Z = 0A23
14The Continuity Equation Why does a hose with a nozzle shoot water further?Conservation of Mass: In a confined system, all of the mass that enters the system, must also exit the system at the same time.Flow rate = Q = Area x Velocityr1A1V1(mass inflow rate) = r2A2V2( mass outflow rate)If the fluid at both points is the same, then the density drops out, and you get the continuity equation: A1V1 =A2V2Therefore If A2 < A1 then V2 > V1Thus, water exiting a nozzle has a higher velocityV1 ->A1A2 V2 ->Q2 = A2V2Q1 = A1V1A1V1 = A2V2
15Free JetsThe velocity of a jet of water is clearly related to the depth of water above the hole. The greater the depth, the higher the velocity. Similar behavior can be seen as water flows at a very high velocity from the reservoir behind a large dam such as Hoover Dam
16The Energy Line and the Hydraulic Grade Line Looking at the Bernoulli equation again:P/g + V2/2g + z = constant on a streamline This constant is called the total head (energy), HBecause energy is assumed to be conserved, at any point along the streamline, the total head is always constantEach term in the Bernoulli equation is a type of head.P/g = Pressure HeadV2/2g = Velocity HeadZ = elevation headThese three heads summed equals H = total energyNext we will look at this graphically…
17The Energy Line and the Hydraulic Grade Line Measures the static pressurePitot measures the total head1: Static Pressure TapMeasures the sum of the elevation head and the pressure Head.2: Pitot TubeMeasures the Total HeadEL : Energy LineTotal Head along a systemHGL : Hydraulic Grade lineSum of the elevation and the pressure heads along a system12ELV2/2gHGLQP/gZ
18The Energy Line and the Hydraulic Grade Line Understanding the graphical approach of Energy Line and the Hydraulic Grade line is key to understanding what forces are supplying the energy that water holds.Point 1:Majority of energy stored in the water is in the Pressure HeadPoint 2:Majority of energy stored in the water is in the elevation headIf the tube was symmetrical, then the velocity would be constant, and the HGL would be levelELV2/2gV2/2gHGLP/g2QP/ gZ1Z
19Tank ExampleSolve for the Pressure Head, Velocity Head, and Elevation Head at each point, and then plot the Energy Line and the Hydraulic Grade LineAssumptions and Hints:P1 and P4 = V3 = V4 same diameter tubeWe must work backwards to solve this problem1R = .5’4’R = .25’2341’
20Pressure Head : Only atmospheric P1/g = 0 Point 1:Pressure Head : Only atmospheric P1/g = 0Velocity Head : In a large tank, V1 = 0 V12/2g = 0Elevation Head : Z1 = 4’14’R = .5’R = .25’2341’
21Pressure Head : Only atmospheric P4/ g = 0 Point 4:Apply the Bernoulli equation between 1 and = 0 + V42/2(32.2) + 1V4 = 13.9 ft/sPressure Head : Only atmospheric P4/ g = 0Velocity Head : V42/2g = 3’Elevation Head : Z4 = 1’1γH2O= 62.4 lbs/ft34’R = .5’R = .25’2341’
22Point 3:Apply the Bernoulli equation between 3 and 4 (V3=V4) P3/ =P3 = 0Pressure Head : P3/g = 0Velocity Head : V32/2g = 3’Elevation Head : Z3 = 1’14’R = .5’R = .25’2341’
23Apply the Continuity Equation Point 2:Apply the Bernoulli equation between 2 and P2/ V22/2(32.2) + 1 =Apply the Continuity Equation(P.52)V2 = (P.252)x13.9 V2 = ft/sP2/ /2(32.2) + 1 = 4 P2 = lbs/ft2Pressure Head :P2/ g = 2.81’Velocity Head :V22/2g = .19’Elevation Head : Z2 = 1’14’R = .5’R = .25’2341’
24Plotting the EL and HGLEnergy Line = Sum of the Pressure, Velocity and Elevation headsHydraulic Grade Line = Sum of the Pressure and Velocity headsV2/2g=.19’ELP/ g =2.81’V2/2g=3’V2/2g=3’Z=4’HGLZ=1’Z=1’Z=1’
25Pipe Flow and Open Channel Flow CEVE 101Pipe Flow and Open Channel Flow
26Open Channel FlowUniform Open Channel Flow is the hydraulic condition in which the water depth and the channel cross section do not change over some reach of the channelManning’s Equation was developed to relate flow and channel geometry to water depth. Knowing Q in a channel, one can solve for the water depth Y. Knowing the maximum allowable depth Y, one can solve for Q.
27Open Channel FlowManning’s equation is only accurate for cases where the cross sections of a stream or channel are uniform. Manning’s equation works accurately for man made channels, but for natural streams and rivers, it can only be used as an approximation.
28Manning’s Equation Terms in the Manning’s equation: V = Channel VelocityA = Cross sectional area of the channelP = Wetted perimeter of the channelR = Hydraulic Radius = A/PS = Slope of the channel bottom (ft/ft or m/m)n = Manning’s roughness coefficient (.015, .045, .12)Yn = Normal depth (depth of uniform flow)AreaYnYXWetted PerimeterSlope = S = Y/X
29For a rectangular Channel Manning’s EquationV = (1/n)R2/3√(S) for the metric systemV = (1.49/n)R2/3√(S) for the English systemQ = A(k/n)R2/3√(S) k is either 1 or 1.49Yn is not directly a part of Manning’s equation. However, A and R depend on Yn. Therefore, the first step to solving any Manning’s equation problem, is to solve for the geometry’s cross sectional area and wetted perimeter:For a rectangular ChannelArea = A = B x YnWetted Perimeter = P = B + 2YnHydraulic Radius = A/P = R = BYn/(B+2Yn)YnB
30Simple Manning’s Example Next, apply Manning’s equation A rectangular open concrete (n=0.015) channel is to be designed to carry a flow of 2.28 m3/s. The slope is m/m and the bottom width of the channel is 2 meters Determine the normal depth that will occur in this channel.First, find A, P and RA = 2Yn P = 2 + 2Yn R = 2Yn/(2 + 2Yn)Next, apply Manning’s equationQ = A(1/n)R2/3√(S) 2.28 = (2Yn)x(1/0.015) * (2Yn/(2 + 2Yn))2/3 * √(0.006)Solving for Yn with Goal SeekYn = 0.47 metersYn2 m
31The Trapezoidal Channel House flooding occurs along Brays Bayou when water overtops the banks. What flow is allowable in Brays Bayou if it has the geometry shown below?Slope S = ft/ft25’a = 20°Concrete Lined n = 0.015B=35’A, P and R for Trapezoidal ChannelsA = Yn(B + Yn cot a)P = B + (2Yn/sin a )R = (Yn(B + Yn cot a)) / (B + (2Yn/sin a))YnθB
32The Trapezoidal Channel Slope S = ft/ft25’Θ = 20°Concrete Lined n = 0.01535’A = Yn(B + Yn cot a)A = 25( cot(20)) = 2592 ft2P = B + (2Yn/sin a )P = 35 + (2 x 25/sin(20)) = ftR = 2592’ / 181.2’ = 14.3 ft
33The Trapezoidal Channel Slope S = ft/ft25’Θ = 20°Concrete Lined n = 0.01535’Q for Bayou = A(1.49/n)R2/3√(S)Q = 2592 x (1.49 / .015) (14.3)2/3 √(.0003)Q = Max allowable Flow = 26,300 cfs
34Manning’s Over Different Terrains S = .005 ft/ft5’5’5’3’3’Grass n=.03Grass n=.03Concrete n=.015Estimate the flow rate for the above channel?Hint:Treat each different portion of the channel separately. You must find an A, R, P and Q for each section of the channel that has a different n coefficient. Neglect dotted line segments.
35Manning’s Over Grass The Grassy portions: S = .005 ft/ft5’5’5’3’3’Grass n=.03Grass n=.03Concrete n=.015The Grassy portions:For each section: A = 5’ x 3’ = 15 ft2 P = 5’ + 3’ = 8 ft R = 15 ft2/8 ft = 1.88 ftQ = 15(1.49/.03)1.882/3√(.005) Q = cfs per section For both sections…Q = 2 x = cfs
36Manning’s Over Concrete S = .005 ft/ft5’5’5’3’3’Grass n=.03Grass n=.03Concrete n=.015The Concrete section A = 5’ x 6’ = 30 ft2 P = 5’ + 3’ + 3’= 11 ft R = 30 ft2/11 ft = 2.72 ftQ = 30(1.49/.015)2.722/3√(.005) Q = cfsFor the entire channel…Q = = 540 cfs
37Pipe Flow and the Energy Equation For pipe flow, the Bernoulli equation alone is not sufficient. Friction loss along the pipe, and momentum loss through diameter changes and corners take head (energy) out of a system that theoretically conserves energy. Therefore, to correctly calculate the flow and pressures in pipe systems, the Bernoulli Equation must be modified.P1/g + V12/2g + z1 =P2/g + V22/2g + z2 + Hmaj + HminEnergy line with no lossesHmajEnergy line with major losses12
38Major LossesMajor losses occur over the entire pipe, as the friction of the fluid over the pipe walls removes energy from the system. Each type of pipe as a friction factor, f, associated with it.Hmaj = f (L/D)(V2/2g)Energy line with no lossesHmajEnergy line with major losses12
39Minor LossesUnlike major losses, minor losses do not occur over the length of the pipe, but only at points of momentum loss. Since Minor losses occur at unique points along a pipe, to find the total minor loss throughout a pipe, sum all of the minor losses along the pipe. Each type of bend, or narrowing has a loss coefficient, KL to go with it.Minor Losses
40Major and Minor Losses Major Losses: Minor Losses: Hmaj = f (L/D)(V2/2g)f = friction factor L = pipe length D = pipe diameterV = Velocity g = gravityMinor Losses:Hmin = KL(V2/2g)Kl = sum of loss coefficients V = Velocity g = gravityWhen solving problems, the loss terms are added to the system at the second analysis pointP1/g + V12/2g + z1 =P2/g + V22/2g + z2 + Hmaj + Hmin
42Pipe Flow Example goil= 8.82 kN/m3 1Z1 = ?2Z2 = 130 m60 mKout=17 mr/D = 0130 mr/D = 2If oil flows from the upper to lower reservoir at a velocity of 1.58 m/s in the D= 15 cm smooth pipe, what is the elevation of the oil surface in the upper reservoir?Include major losses along the pipe, and the minor losses associated with the entrance, the two bends, and the outlet.
43Hmaj = (f L V2)/(D 2g)=(.035 x 197m * (1.58m/s)2)/(.15 x 2 x 9.8m/s2) Pipe Flow Example1Z1 = ?2Z2 = 130 m60 mKout=17 mr/D = 0130 mr/D = 2Apply Bernoulli’s equation between points 1 and 2: Assumptions: P1 = P2 = Atmospheric = V1 = V2 = 0 (large tank)Z1 = m + Hmaj + HminHmaj = (f L V2)/(D 2g)=(.035 x 197m * (1.58m/s)2)/(.15 x 2 x 9.8m/s2)Hmaj= 5.85m
44Pipe Flow Example 0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + Hmin 2Z2 = 130 m60 mKout=17 mr/D = 0130 mr/D = 2Z1 = m m + HminHmin= 2KbendV2/2g + KentV2/2g + KoutV2/2gFrom Loss Coefficient table: Kbend = Kent = Kout = 1Hmin = (0.19x ) * (1.582/2*9.8)Hmin = 0.24 m
45Pipe Flow Example 0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + 0.24m Kout=17 mr/D = 0130 mr/D = 2Z1 = m + Hmaj + HminZ1 = m m mZ1 = meters
46Stormwater Mgt Model (SWWM) Most advanced model ever written for dynamic hydraulic routingSolves complex equations for pipe flow with consideration of tailwater at outletNew graphical user interface for easy input and presentation of resultsWill allow for direct evaluation of flood control options under various conditions
47SWMM Input Rainfall Pattern Inlets to Pipes Pipe Elevations and Sizes Junction LocationsBayou Level
48SWMM Output Flooding Areas High Bayou Level Pipe at Capacity Backflow at Outlet