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TL2101 Mekanika Fluida I

TL2101 Mekanika Fluida I. Benno Rahardyan. Pertemuan 7. 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.

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TL2101 Mekanika Fluida I

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  1. TL2101 Mekanika Fluida I Benno Rahardyan Pertemuan 7

  2. FLUID DYNAMICSTHE 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/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.

  4. Bernoulli Example Problem : Free Jets 2 A small cylindrical tank is filled with water, and then emptied through a small orifice at the bottom. Case 1 What is the flow rate Q exiting through the hole when the tank is full? Case 2 What is the flow rate Q exiting through the hole when the tank is half full? -Hint- The Continuity Equation is needed R=1’ R=1’ Assumptions Psurf = Pout = 0 Because it’s a small tank, Vsurf ≠ 0 γH20=62.4 lbs/ft3 4’ R=.5’ R=.5’ 2’ Q? Q? Case 1 Case 2

  5. Free Jets 2 Case 1 Apply Bernoulli’s Equation at the Surface and at the Outlet: 0 + Vsurf2/2g + 4 = 0 + Vout2/2g + 0 With two unknowns, we need another equation : The Continuity Equation AsurfVsurf=AoutVout p(1)2 x Vsurf = p(.5)2 x Vout  Vsurf=.25Vout R=1’ R=1’ Substituting back into the Bernoulli Equation  (.25Vout)2/2g + 4 = Vout2/2g Solving for Vout if g = 32.2 ft/s2 Vout = .257 ft/s Qout = AV = .202 ft3/s (cfs) γH20=62.4 lbs/ft3 4’ R=.5’ R=.5’ 2’ Q? Q? Case 1 Case 2

  6. Bernoulli Example Problem : Free Jets 2 Case 2 Bernoulli’s Equation at the Surface and at the Outlet is changed: 0 + Vsurf2/2g + 2 = 0 + Vout2/2g + 0 Continuity eqn remains the same. Substituting back into the Bernoulli Equation  (.25Vout)2/2g + 2 = Vout2/2g Solving for Vout if g = 32.2 ft/s2 Vout = .182 ft/s Qout = AV = .143 cfs Note that velocity is less in Case 2 R=1’ R=1’ γH20=62.4 lbs/ft3 4’ R=.5’ R=.5’ 2’ Q? Q? Case 1 Case 2

  7. 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 the Glen Canyon Dam in Colorado

  8. The Energy Line and the Hydraulic Grade Line Looking at the Bernoulli equation again: P/γ + V2/2g + 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/γ = Pressure Head V2/2g = Velocity Head Z = elevation head These three heads, summed together, will always equal H Next we will look at this graphically…

  9. The Energy Line and the Hydraulic Grade Line Lets first understand this drawing: Measures the Total Head 1: Static Pressure Tap Measures the sum of the elevation head and the pressure Head. 2: Pilot 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 Measures the Static Pressure 1 2 1 2 EL V2/2g HGL Q P/γ Z

  10. The 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 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 EL V2/2g V2/2g HGL P/γ 2 Q P/γ Z 1 Z

  11. The Complete 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 Assumptions and Hints: P1 and P4 = 0 --- V3 = V4 same diameter tube We must work backwards to solve this problem 1 γH2O= 62.4 lbs/ft3 R = .5’ 4’ R = .25’ 2 3 4 1’

  12. Point 1: Pressure Head : Only atmospheric  P1/γ = 0 Velocity Head : In a large tank, V1 = 0  V12/2g = 0 Elevation Head : Z1 = 4’ 1 γH2O= 62.4 lbs/ft3 4’ R = .5’ R = .25’ 2 3 4 1’

  13. Point 4: Apply the Bernoulli equation between 1 and 4 0 + 0 + 4 = 0 + V42/2(32.2) + 1 V4 = 13.9 ft/s Pressure Head : Only atmospheric  P4/γ = 0 Velocity Head : V42/2g = 3’ Elevation Head : Z4 = 1’ 1 γH2O= 62.4 lbs/ft3 4’ R = .5’ R = .25’ 2 3 4 1’

  14. Point 3: Apply the Bernoulli equation between 3 and 4 (V3=V4) P3/62.4 + 3 + 1 = 0 + 3 + 1 P3 = 0 Pressure Head : P3/γ = 0 Velocity Head : V32/2g = 3’ Elevation Head : Z3 = 1’ 1 γH2O= 62.4 lbs/ft3 4’ R = .5’ R = .25’ 2 3 4 1’

  15. Point 2: Apply the Bernoulli equation between 2 and 3 P2/62.4 + V22/2(32.2) + 1 = 0 + 3 + 1 Apply the Continuity Equation (Π.52)V2 = (Π.252)x13.9  V2 = 3.475 ft/s P2/62.4 + 3.4752/2(32.2) + 1 = 4  P2 = 175.5 lbs/ft2 Pressure Head : P2/γ = 2.81’ Velocity Head : V22/2g = .19’ Elevation Head : Z2 = 1’ 1 γH2O= 62.4 lbs/ft3 4’ R = .5’ R = .25’ 2 3 4 1’

  16. 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 V2/2g=.19’ EL P/γ =2.81’ V2/2g=3’ V2/2g=3’ Z=4’ HGL Z=1’ Z=1’ Z=1’

  17. 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. P1/γ + V12/2g + z1 = P2/γ + V22/2g + z2 + Hmaj + Hmin Major losses: Hmaj 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. Energy line with no losses Hmaj Energy line with major losses 1 2

  18. Pipe Flow and the Energy Equation Minor Losses : Hmin Momentum losses in Pipe diameter changes and in pipe bends are called 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, KL to go with it. Minor Losses

  19. Major and Minor Losses Major Losses: Hmaj = f x (L/D)(V2/2g) f = friction factor L = pipe length D = pipe diameter V = Velocity g = gravity Minor Losses: Hmin = KL(V2/2g) Kl = sum of loss coefficients V = Velocity g = gravity When solving problems, the loss terms are added to the system at the second point P1/γ + V12/2g + z1 = P2/γ + V22/2g + z2 + Hmaj + Hmin

  20. Loss Coefficients Use this table to find loss coefficients:

  21. Pipe Flow Example γoil= 8.82 kN/m3 f = .035 1 Z1 = ? 2 Z2 = 130 m 60 m Kout=1 7 m r/D = 0 130 m r/D = 2 If oil flows from the upper to lower reservoir at a velocity of 1.58 m/s in the 15 cm diameter 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.

  22. Pipe Flow Example γoil= 8.82 kN/m3 f = .035 1 Z1 = ? 2 Z2 = 130 m 60 m Kout=1 7 m r/D = 0 130 m r/D = 2 Apply Bernoulli’s equation between points 1 and 2:Assumptions: P1 = P2 = Atmospheric = 0 V1 = V2 = 0 (large tank) 0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin Hmaj = (fxLxV2)/(Dx2g)=(.035 x 197m x (1.58m/s)2)/(.15 x 2 x 9.8m/s2) Hmaj= 5.85m

  23. Pipe Flow Example γoil= 8.82 kN/m3 f = .035 1 Z1 = ? 2 Z2 = 130 m 60 m Kout=1 7 m r/D = 0 130 m r/D = 2 0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + Hmin Hmin= 2KbendV2/2g + KentV2/2g + KoutV2/2g From Loss Coefficient table: Kbend = 0.19 Kent = 0.5 Kout = 1 Hmin = (0.19x2 + 0.5 + 1) x (1.582/2x9.8) Hmin = 0.24 m

  24. Pipe Flow Example γoil= 8.82 kN/m3 f = .035 1 Z1 = ? 2 Z2 = 130 m 60 m Kout=1 7 m r/D = 0 130 m r/D = 2 0 + 0 + Z1 = 0 + 0 + 130m + Hmaj + Hmin 0 + 0 + Z1 = 0 + 0 + 130m + 5.85m + 0.24m Z1 = 136.09 meters

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