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Fluid Mechanics 2 The Bernoulli Equation Dr. Phil Bedient CEVE 101.

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Presentation on theme: "Fluid Mechanics 2 The Bernoulli Equation Dr. Phil Bedient CEVE 101."— Presentation transcript:

1 Fluid Mechanics 2 The Bernoulli Equation Dr. Phil Bedient CEVE 101

2 FLUID 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.

3 The 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/  + V 2 /2g + z = constant along a streamline (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 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.

4 The Bernoulli Equation (unit of L) At any two points on a streamline: P 1 /  + V 1 2 /2g + z 1 = P 2 /  + V 2 2 /2g + z 2 1 2

5 A Simple Bernoulli Example  V 2 Z  =  air Determine the difference in pressure between points 1 and 2 Assume 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, V 2 = Velocity of the Biker Hint: 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)

6 Stagnation Points On 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

7 Apply Bernoulli from 1 to 2  V 2 Z Point 1 = Point 2 P 1 /  air + V 1 2 /2g + z 1 = P 2 /  air + V 2 2 /2g + z 2 Knowing the z 1 = z 2 and that V 1 = 0, we can simplify the equation P 1 /  air = P 2 /  air + V 2 2 /2g P 1 – P 2 = ( V 2 2 /2g )  air  =  air

8 A Simple Bernoulli Example If Lance Armstrong is traveling at 20 ft/s, what pressure does he feel on his face if the  air =.0765 lbs/ft 3 ? We can assume P 2 = 0 because it is only atmospheric pressure P 1 = ( V 2 2 /2g )(  air ) = P 1 = ((20 ft/s) 2 /(2(32.2 ft/s 2 )) x.0765 lbs/ft 3 P 1 =.475 lbs/ft 2 Converting to lbs/in 2 (psi) P 1 =.0033 psi (gage pressure) If the biker’s face has a surface area of 60 inches He feels a force of.0033 x 60 =.198 lbs

9 Bernoulli Assumptions Key Assumption # 1 Velocity = 0 Imagine 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 There are three main variables in the Bernoulli Equation Pressure – Velocity – Elevation To simplify problems, assumptions are often made to eliminate one or more variables

10 Bernoulli Assumptions Key Assumption # 2 Pressure = 0 Whenever 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.

11 Bernoulli Assumptions Key Assumption # 3 The Continuity Equation In cases where one or both of the previous assumptions do not apply, then we might need to use the continuity equation to solve the problem A 1 V 1 =A 2 V 2 Which satisfies that inflow and outflow are equal at any section

12 Bernoulli Example Problem: Free Jets What is the Flow Rate at point 2? What is the velocity at point 3? γ H2O Part 1: Apply Bernoulli’s eqn between points 1 and 2 P 1 /  H2O + V 1 2 /2g + h = P 2 /  H20 + V 2 2 /2g + 0 simplifies to h = V 2 2 /2g  solving for V V = √(2gh) Q = VA or Q = A 2 √(2gh) 0 A2A2 Givens and Assumptions: Because the tank is so large, we assume V 1 = 0 (Vol out <<< Vol tank ) The tank is open at both ends, thus P 1 = P 2 = P 3 = atm  P 1 and P 2 and P 3 = 0

13 1 2 3 γ H2O Z = 0 A2A2 Bernoulli Example Problem: Free Jets Part 2: Find V 3 ? Apply Bernoulli’s eq from pt 1 to pt 3 P 1 /  H2O + V 1 2 /2g + h = P 3 /  H20 + V 3 2 /2g – H Simplify to  h + H = V 3 2 /2g Solving for V  V 3 = √ ( 2g ( h + H ))

14 The 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 Velocity  1 A 1 V 1 (mass inflow rate) =  2 A 2 V 2 ( mass outflow rate) If the fluid at both points is the same, then the density drops out, and you get the continuity equation: A 1 V 1 =A 2 V 2 Therefore If A 2 V 1 Thus, water exiting a nozzle has a higher velocity Q 1 = A 1 V 1 A1A1 V 1 -> Q 2 = A 2 V 2 A 1 V 1 = A 2 V 2 A 2 V 2 ->

15 Free Jets The 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

16 The Energy Line and the Hydraulic Grade Line Looking at the Bernoulli equation again: P/  + V 2 /2g + z = constant on a streamline This constant is called the total head (energy), H Because energy is assumed to be conserved, at any point along the streamline, the total head is always constant Each term in the Bernoulli equation is a type of head. P/  = Pressure Head V 2 /2g = Velocity Head Z = elevation head These three heads summed equals H = total energy Next we will look at this graphically…

17 The Energy Line and the Hydraulic Grade Line Q Measures the static pressure Pitot measures the total head 1 Z P/  V 2 /2g EL HGL 2 1: Static Pressure Tap Measures the sum of the elevation head and the pressure Head. 2: Pitot Tube Measures the Total Head EL : Energy Line Total Head along a system HGL : Hydraulic Grade line Sum of the elevation and the pressure heads along a system

18 The Energy Line and the Hydraulic Grade Line Q Z P/  V 2 /2g EL HGL 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. V 2 /2g P/  Z 1 2 Point 1: Majority of energy stored in the water is in the Pressure Head Point 2: Majority of energy stored in the water is in the elevation head If the tube was symmetrical, then the velocity would be constant, and the HGL would be level

19 Tank Example Solve for the Pressure Head, Velocity Head, and Elevation Head at each point, and then plot the Energy Line and the Hydraulic Grade Line ’ 4’ Assumptions and Hints: P 1 and P 4 = V 3 = V 4 same diameter tube We must work backwards to solve this problem R =.5’ R =.25’

20 ’ 4’ Point 1: Pressure Head : Only atmospheric  P 1 /  = 0 Velocity Head : In a large tank, V 1 = 0  V 1 2 /2g = 0 Elevation Head : Z 1 = 4’ R =.5’ R =.25’

21 ’ 4’ γ H2O = 62.4 lbs/ft 3 Point 4: Apply the Bernoulli equation between 1 and = 0 + V 4 2 /2(32.2) + 1 V 4 = 13.9 ft/s Pressure Head : Only atmospheric  P 4 /  = 0 Velocity Head : V 4 2 /2g = 3’ Elevation Head : Z 4 = 1’ R =.5’ R =.25’

22 ’ 4’ Point 3: Apply the Bernoulli equation between 3 and 4 (V 3 =V 4 ) P 3 / = P 3 = 0 Pressure Head : P 3 /  = 0 Velocity Head : V 3 2 /2g = 3’ Elevation Head : Z 3 = 1’ R =.5’ R =.25’

23 ’ 4’ Point 2: Apply the Bernoulli equation between 2 and 3 P 2 / V 2 2 /2(32.2) + 1 = Apply the Continuity Equation ( .5 2 )V 2 = ( .25 2 )x13.9  V 2 = ft/s P 2 / /2(32.2) + 1 = 4  P 2 = lbs/ft 2 R =.5’ R =.25’ Pressure Head : P 2 /  = 2.81’ Velocity Head : V 2 2 /2g =.19’ Elevation Head : Z 2 = 1’

24 Plotting the EL and HGL Energy Line = Sum of the Pressure, Velocity and Elevation heads Hydraulic Grade Line = Sum of the Pressure and Velocity heads EL HGL Z=1’ V 2 /2g=3’ Z=4’ P/  =2.81’ V 2 /2g=.19’

25 Pipe Flow and Open Channel Flow Pipe Flow and Open Channel Flow CEVE 101

26 Open Channel Flow Uniform 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 channel Manning’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.

27 Open Channel Flow Manning’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.

28 Manning’s Equation Terms in the Manning’s equation: V = Channel Velocity A = Cross sectional area of the channel P = Wetted perimeter of the channel R = Hydraulic Radius = A/P S = Slope of the channel bottom (ft/ft or m/m) n = Manning’s roughness coefficient (.015,.045,.12) Y n = Normal depth (depth of uniform flow) Area Wetted Perimeter YnYn Y X Slope = S = Y/X

29 Manning’s Equation V = (1/n)R 2/3 √(S) for the metric system V = (1.49/n)R 2/3 √(S) for the English system Q = A(k/n)R 2/3 √(S) k is either 1 or 1.49 Y n is not directly a part of Manning’s equation. However, A and R depend on Y n. 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 Channel Area = A = B x Y n Wetted Perimeter = P = B + 2Y n Hydraulic Radius = A/P = R = BY n /(B+2Y n ) B YnYn

30 Simple Manning’s Example A rectangular open concrete (n=0.015) channel is to be designed to carry a flow of 2.28 m 3 /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. 2 m YnYn First, find A, P and R A = 2Y n P = 2 + 2Y n R = 2Y n /(2 + 2Y n ) Next, apply Manning’s equation Q = A(1/n)R 2/3 √(S)  2.28 = (2Y n )x(1/0.015) * (2Y n /(2 + 2Yn)) 2/3 * √(0.006) Solving for Y n with Goal Seek Y n = 0.47 meters

31 The 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? 25’ B=35’ a = 20° Concrete Lined n = Slope S = ft/ft A, P and R for Trapezoidal Channels B YnYn θ A = Y n (B + Y n cot a) P = B + (2Y n /sin a ) R = (Y n (B + Y n cot a)) / (B + (2Y n /sin a))

32 The Trapezoidal Channel 25’ 35’ Θ = 20° Concrete Lined n = Slope S = ft/ft A = Y n (B + Y n cot a) A = 25( cot(20)) = 2592 ft 2 P = B + (2Y n /sin a ) P = 35 + (2 x 25/sin(20)) = ft R = 2592’ / 181.2’ = 14.3 ft

33 The Trapezoidal Channel 25’ 35’ Θ = 20° Concrete Lined n = Slope S = ft/ft Q for Bayou = A(1.49/n)R 2/3 √(S) Q = 2592 x (1.49 /.015) (14.3) 2/3 √(.0003) Q = Max allowable Flow = 26,300 cfs

34 Manning’s Over Different Terrains 3’ 5’ Grass n=.03 Concrete n=.015 Grass n=.03 Estimate 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. S =.005 ft/ft

35 Manning’s Over Grass 3’ 5’ Grass n=.03 Concrete n=.015 Grass n=.03 The Grassy portions: For each section: A = 5’ x 3’ = 15 ft 2 P = 5’ + 3’ = 8 ft R = 15 ft 2 /8 ft = 1.88 ft Q = 15(1.49/.03)1.88 2/3 √(.005) Q = 15(1.49/.03)1.88 2/3 √(.005) Q = cfs per section  For both sections… Q = 2 x = cfs S =.005 ft/ft

36 Manning’s Over Concrete 3’ 5’ Grass n=.03 Concrete n=.015 Grass n=.03 The Concrete section A = 5’ x 6’ = 30 ft 2 P = 5’ + 3’ + 3’= 11 ft R = 30 ft 2 /11 ft = 2.72 ft Q = 30(1.49/.015)2.72 2/3 √(.005) Q = 30(1.49/.015)2.72 2/3 √(.005) Q = cfs For the entire channel… Q = = 540 cfs S =.005 ft/ft

37 Pipe 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. P 1 /  + V 1 2 /2g + z 1 = P 2 /  + V 2 2 /2g + z 2 + H maj + H min H maj Energy line with no losses Energy line with major losses 12

38 Major Losses Major 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. H maj = f (L/D)(V2/2g) H maj Energy line with no losses Energy line with major losses 12

39 Minor Losses Unlike 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, K L to go with it. Minor Losses

40 Major and Minor Losses Major Losses: H maj = f (L/D)(V2/2g) f = friction factor L = pipe length D = pipe diameter V = Velocity g = gravity Minor Losses: H min = K L (V 2 /2g) K l = sum of loss coefficients V = Velocity g = gravity When solving problems, the loss terms are added to the system at the second analysis point P 1 /  + V 1 2 /2g + z 1 = P 2 /  + V 2 2 /2g + z 2 + H maj + H min

41 Loss Coefficients

42 Pipe Flow Example 1 2 Z 2 = 130 m 130 m 7 m 60 m r/D = 2 Z 1 = ?  oil = 8.82 kN/m 3 f =.035 If 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. K out =1 r/D = 0

43 Pipe Flow Example 1 2 Z 2 = 130 m 130 m 7 m 60 m r/D = 2 Z 1 = ? K out =1 r/D = 0 Apply Bernoulli’s equation between points 1 and 2: Assumptions: P 1 = P 2 = Atmospheric = 0 V 1 = V 2 = 0 (large tank) Z 1 = m + H maj + H min H maj = (f L V 2 )/(D 2g)=(.035 x 197m * (1.58m/s) 2 )/(.15 x 2 x 9.8m/s 2) H maj = 5.85m

44 Pipe Flow Example 1 2 Z 2 = 130 m 130 m 7 m 60 m r/D = 2 Z 1 = ? K out =1 r/D = Z 1 = m m + H min H min = 2K bend V 2 /2g + K ent V 2 /2g + K out V 2 /2g From Loss Coefficient table: K bend = 0.19 K ent = 0.5 K out = 1 H min = (0.19x ) * ( /2*9.8) H min = 0.24 m

45 Pipe Flow Example 1 2 Z 2 = 130 m 130 m 7 m 60 m r/D = 2 Z 1 = ? K out =1 r/D = Z 1 = m + H maj + H min Z 1 = m m m Z 1 = meters

46 Stormwater Mgt Model (SWWM) Most advanced model ever written for dynamic hydraulic routing Most advanced model ever written for dynamic hydraulic routing Solves complex equations for pipe flow with consideration of tailwater at outlet Solves complex equations for pipe flow with consideration of tailwater at outlet New graphical user interface for easy input and presentation of results New graphical user interface for easy input and presentation of results Will allow for direct evaluation of flood control options under various conditions Will allow for direct evaluation of flood control options under various conditions

47 SWMM Input Bayou Level Inlets to Pipes Pipe Elevations and Sizes Junction Locations Rainfall Pattern

48 SWMM Output Backflow at Outlet High Bayou Level Flooding Areas Pipe at Capacity

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