5%20Bernoulli%27s%20Equation1

5%20Bernoulli%27s%20Equation1 - CHG 2312 CHG Fluid Flow...

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Unformatted text preview: CHG 2312 CHG Fluid Flow Bernoulli's Equation Outline 1. 2. 3. Energy balance: Steady, incompressible flow Friction heating Bernoulli equation 3.1. Zero flow (fluid static) 3.2. Head form of Bernoulli equation Diffusers (gradual expansion) 5. Sudden expansion 6. Torricelli equation 7. Fluid flow measurement a) Pitot tube b) Pitot-static tube c) Venturi meter d) Orifice meter d) Orific te e) Rotameter 8. Cavitation 8. 9. Unsteady flow 10. Non-uniform flows 1. Energy balance: Steady, incompressible flow Bernoulli’s Equation: dW ⎛P V2 ⎞ ⎛ P V2 ⎞ dQ ⎜ u + + gz + ⎟ − ⎜ u + + gz + ⎟ = − n.f. − ⎜ ρ 2 ⎟in ⎜ ρ 2 ⎟out dm dm ⎝ ⎠⎝ ⎠ ⎛P V 2 ⎞ dWn.f. ⎛ dQ ⎞ ⎟= − ⎜ ∆u − ∆⎜ + gz + ⎟ ⎟ ⎜ρ 2⎠ dm ⎝ dm ⎠ ⎝ University of Ottawa, CHG 2312, P. Mehrani 3 2. Friction heating Examples in solids: smoking brakes, saw cutting wood, etc. Temperature increases in fluid due to friction heating are less, why? o Am ount of frictional work per unit m ass is generally less than in examples of solids above. (in solids energy is concentrated in small volume, whereas in fluid it is spread over a larger volume). o Heat capacity of liquids is generally greater than that of solids. University of Ottawa, CHG 2312, P. Mehrani 4 2. Friction heating For constant-density materials only: ∆u = friction h eating dQ + dm dm ∆u − dQ =F dm F is the friction heating per unit mass and if it exist, it will be positive. Usually referred to as friction loss! 5 University of Ottawa, CHG 2312, P. Mehrani 2. Friction heating Incompressible flow vs. incompressible fluid: o Incompressible flow: density changes unimportant. o All steady flows of liquid and most steady flows of gasses at low velocities are considered incompressible. o Some unsteady flows of liquids and all steady flows of gases at high velocities may not be considered incompressible. P1-Patm V2 (ft/s,calc) V2 (ft/s, actual) T2 (oF) 0.01 35 35 67.9 0.1 110 111 67.0 0.3 190 191 65.0 0.6 267 268 62.0 1.0 340 343 58.2 2.0 466 476 49.1 5.0 678 713 25.6 University of Ottawa, CHG 2312, P. Mehrani 6 3. Bernoulli’s equation: ⎛P V 2 ⎞ dWn.f. ⎛ dQ ⎞ ⎟= − ⎜ ∆u − ∆⎜ + gz + ⎟ ⎟ ⎜ρ 2⎠ dm dm ⎠ ⎝ ⎝ ⎛P V 2 ⎞ dW n. f . ⎟= ∆ ⎜ + gz + −F ⎜ρ 2⎟ dm ⎝ ⎠ F is a positive number for all real flows. Referred to as the mechanical energy balance. 7 University of Ottawa, CHG 2312, P. Mehrani 3.1. Zero flow (fluid statics) z Bernoullli’s equation l ⎛P V 2 ⎞ dWn. f . ⎟= ∆⎜ + gz + −F ⎜ρ 2⎟ dm ⎝ ⎠ 2 1 Fluid at rest: ⎛P ⎞ ∆⎜ ⎜ ρ + gz ⎟ = 0 ⎟ ⎝ ⎠ lim ∆z→ 0 1 ρ ∆P = − g∆z ∆P dP = = − ρg ∆z dz University of Ottawa, CHG 2312, P. Mehrani 8 3.2 Head form of Bernoulli’s equation Divide both sides of Bernoulli's equation by gravitational constant (g): ⎛P V2 ∆⎜ ⎜ ρg + z + 2 g ⎝ ⎞ dW n. f . F ⎟= − ⎟ gdm g ⎠ Every term has the dimension of length [m] From left: pressure head, gravity head, velocity head, pump or turbine head and friction head loss. University of Ottawa, CHG 2312, P. Mehrani 9 4. Diffusers (gradual expansion) One of the ways to slow down a fluid: gradual expansion. of he slow own luid radua xpansion. U niv ersity of Ottawa, CHG 2312, P. Mehrani 10 4. Diffusers 4. ffuser ⎛P V 2 ⎞ dW n. f . ⎟= ∆ ⎜ + gz + −F ⎜ρ 2⎟ dm ⎝ ⎠ P2 − P1 V 2 2 − V 12 + = −F ρ 2 P2 − P1 = ρ V 12 2 V 2 = V1 A1 A2 ⎛ A2 ⎞ ⎜ 1 − 1 ⎟ − ρF 2 ⎜ A2 ⎟ ⎝ ⎠ 11 U niv ersity of Ottawa, CHG 2312, P. Mehrani 5. Sudden expansions A2 A1 P2 − P1 = ρ V12 2 − ρF Separation regions: eddies University of Ottawa, CHG 2312, P. Mehrani 12 6. Torricelli’s equation Tank-draining problem z 1 h In addition to assumptions for Bernoulli: o V2 >> V1 (V1 ≈ 0), o P1 ≈ P2 = Patm o W=F=0 o steady-state flow, i.e. constant tank level g ( z 2 − z1 ) + 2 V22 = 0 & z 2 − z1 = − h 2 V2 = (2gh)1 2 University of Ottawa, CHG 2312, P. Mehrani 13 7. Fluid flow measurement a) Pitot Tube P 2 − P 1 V 12 − = −F ρ 2 University of Ottawa, CHG 2312, P. Mehrani 14 7. Fluid flow measurement B.E. Equation: P2 − P1 ρ V 12 − = −F 2 Inside the tube the fluid is not moving: P 2 = P a t m + ρ g (h 1 + h 2 ) From fluid statics: P1 = P a t m + ρ g h 2 Velocity at point 1: V1 = (2 g h 1 + 2F ) 1 2 V1 ≈ (2 gh1 ) University of Ottawa, CHG 2312, P. Mehrani 1 2 15 7. Fluid flow measurement b) Pitot-Static Tube ⎛ 2∆P⎞ V1 = ⎜ ⎟ ⎝ρ⎠ 1 2 University of Ottawa, CHG 2312, P. Mehrani 16 7. Fluid flow measurement Pitot vs. Pitot-Static Tube V1 ≈ (2gh1 ) ⎛ 2∆P⎞ V1 = ⎜ ⎟ ⎝ρ⎠ 1 2 1 2 17 University of Ottawa, CHG 2312, P. Mehrani 7. Fluid flow measurement c) Venturi Meter Venturi eter 2 − V12 P2 − P1 V2 + ρ 2 =0 ⎡ 2( P − P ) ⎤ 1 2 ⎢ ⎥ ρ ⎢ ⎥ V2 = 2 ⎢⎛ A ⎞⎥ ⎢ ⎜1 − 2 ⎟ ⎥ 2 ⎜ A1 ⎟ ⎥ ⎢⎝ ⎠⎦ ⎣ University of Ottawa, CHG 2312, P. Mehrani 1 2 18 7. Fluid flow measurement University of Ottawa, CHG 2312, P. Mehrani 19 7. Fluid flow measurement Differences exist between calculated and actual! (actual < calc.) Introduce coefficient of discharge Cv ⎡ 2(P − P ) ⎤ 1 2 ⎢ ⎥ ρ ⎢ ⎥ V2 = CV 2 ⎢⎛ A ⎞⎥ ⎢ ⎜1 − 2 ⎟ ⎥ 2 ⎜ A1 ⎟ ⎥ ⎢⎝ ⎠⎦ ⎣ 1 2 University of Ottawa, CHG 2312, P. Mehrani 20 7. Fluid flow measurement Cv depends only on Reynolds number Re = 1. 2. 3. 4. 5. V1 D1 ρ µ Use V2 from equation with no Cv Us Compute Reynolds number Read Cv from graph Calculate new V2 Repeat steps 2, 3 and 4 until convergence achieved. maximum minimum average 21 University of Ottawa, CHG 2312, P. Mehrani 7. Fluid flow measurement maximum minimum Cv average Re =V1 D1 ρ/µ University of Ottawa, CHG 2312, P. Mehrani 22 Fluid flow measurement Orifice Meter ⎡ 2( P1 − P2 ) ρ ⎤ ⎥ V2 = Cv ⎢ 2 2 ⎢ (1 − A2 A1 ) ⎥ ⎣ ⎦ 1 2 University of Ottawa, CHG 2312, P. Mehrani 20 7. Fluid flow measurement Cv Re =V2 D2 ρ/µ University of Ottawa, CHG 2312, P. Mehrani 24 7. Fluid flow measurement Orifice vs. Venturi University of Ottawa, CHG 2312, P. Mehrani 25 7. Fluid flow measurement e) Rotameter e) o The float rises within the tube to a location where the float weight, drag oat ight, dr force, and buoyancy orce forces balance each orces other. University of Ottawa, CHG 2312, P. Mehrani 26 7. Fluid flow measurement Force balance: Fgravity + F pressure from above = Fbuoyancy + F pressure from bottom 0= π 6 3 D o ρ ball g + P3 π 4 2 Do − π 6 3 Do ρ fluid g − P1 π 4 2 Do From B.E: ⎛V 2 ⎛V 2 V 2 ⎞ P − P2 = ρ fluid ⎜ 2 − 1 ⎟ = ρ fluid ⎜ 2 1 ⎜2 ⎟ ⎜2 2⎠ ⎝ ⎝ V2 = 2 ⎞⎛ A2 ⎞ ⎟ ⎟⎜1 − ⎟⎜ A 2 ⎟ 1⎠ ⎠⎝ 4 Do g ρ ball − ρ fluid • ρ fluid 3 Q = A 2V 2 University of Ottawa, CHG 2312, P. Mehrani 27 7. Fluid flow measurement e) Rotameter e) o The float rises within the tube to a location where the float weight, drag oat ight, dr force, and buoyancy orce forces balance each orces other. University of Ottawa, CHG 2312, P. Mehrani 28 7. Fluid flow measurement For gas flow rate measurements: o From B.E: ⎛ ρball − ρ fluid V∝ ⎜ ⎜ ρ fluid ⎝ o ⎞ ⎟ ⎟ ⎠ ⇒ Q1 ρ1 = Qo ρo Mass flow rate is constant through a system: Q1 ρ1 = Qo ρ o 29 University of Ottawa, CHG 2312, P. Mehrani 7. Fluid flow measurement For finding the process flow rate from the rotameter scale reading: Q 2 = Q0 T2 P2 P0 P1 M 0 T0T1 M 1 For setting the gas flow rate at the rotameter to give a desired process gas flow rate: Accessories Q0 = Q 2 P2 T2 T0T1 M 1 P0 P1 M 0 Process conditions: Q2, T2, ρ2, M2 Rotameter conditions: Q1, T1, ρ1, M1 Reference (standard) conditions: Q0, T0, ρ0, M0 University of Ottawa, CHG 2312, P. Mehrani 30 8. Cavitation (negative absolute pressures) Bernoulli’s equation sometimes predicts negative absolute pressures. o In gas systems: likely indicates very high gas velocities, thus Bernoulli’s equation does not apply (see Chap. 8). o In liquid systems: Occurs in very rare conditions. When the absolute pressure on a liquid approaches the vapour pressure of the liquid, liquid boils (two-phase flow). Therefore, the calculated flow is physically impossible. University of Ottawa, CHG 2312, P. Mehrani 31 8. Cavitation (negative absolute pressures) Examples: University of Ottawa, CHG 2312, P. Mehrani 32 8. Cavitation (negative absolute pressures) University of Ottawa, CHG 2312, P. Mehrani 33 9. Unsteady flows Bernoullli’s equation can be applied to some unsteady flows, l if the changes in flow rate are slow enough to be ignored. Most problems in which unsteady flow cannot be neglected involve suddenly stopping the flow with a valve or starting the flow from rest. These problems are more easily solved using the momentum balance. University of Ottawa, CHG 2312, P. Mehrani 34 10. Non-uniform flows Normally ignore difference in velocity perpendicular to flow. Cannot ignore for shallow gravity-driven flows, e.g., weirs in distillation columns, etc. Sieve plates Bubble caps University of Ottawa, CHG 2312, P. Mehrani 35 ...
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