PART II Continuity and Conservation of Mass • Some applications of fluid mechanics • Flow through a pipe: • . Conservation. of mass for steady state (no storage) says m in = m out . m m _1A1Vm1 = _ 2A2Vm2 • For incompressible fluids, density does not changes (_ 1 = _ 2) so A1Vm1 = A2Vm2 = Q Fluid Mechanics – Continuity Equation Bernoulli’s equation • The equation of continuity states that for an Y incompressible fluid flowing in a tube of varying 2 A V cross-sectional area (A), the mass flow rate is the 2 2 same everywhere in the tube: . m1= m2 _ 1A1V1 = _ 2A2V2 _1A1V1= _2 A2V2 • Generally, the density stays constant and then it's Y For incompressible flow simply the flow rate (Av) that is constant. 1 A1 V1 A1V1= A2V2 Assume steady flow, V parallel to streamlines & no viscosity 1 Bernoulli Equation – energy Bernoulli Equation – work • Consider energy terms for steady flow: • Consider work done on the system is Force x distance • We write terms for KE and PE at each point • We write terms for force in terms of Pressure and area Y Y 2 2 Wi = FiVi dt =PiViAi dt A2 V2 A2 V2 E = KE + PE i i i Note ViAi dt = mi/_i 1 2 E1 = 2 m&1V1 + gm&1 y1 W1 = P1m&1 / ρ1 2 Y 1 Y 1 E2 = 2 m&2V2 + gm&2 y2 1 W2 = −P2m&2 / ρ2 A1 V1 A1 V1 As the fluid moves, work is being done by the external Now we set up an energy balance on the system. forces to keep the flow moving. For steady flow, the work Conservation of energy requires that the change in done must equal the change in mechanical energy. energy equals the work done on the system. Bernoulli equation- energy balance Forms of the Bernoulli equation Energy accumulation = _Energy – Total work • Most common forms: 1 2 1 2 0 = (E2-E1) – (W1+W2) i.e. no accumulation at steady state P1 + 2 ρV1 = P2 + 2 ρV2 + gρΔh Or W +W = E -E Subs terms gives: 1 2 2 1 PS1 + PV1 = PS 2 + PV 2 + ΔPht & & P1m1 P2m2 1 2 1 2 − = ( 2 m&V + gm& y ) − ( 2 m&V + gm&y ) The above forms assume no losses within the volume… ρ ρ 2 2 2 2 1 1 1 1 1 2 If losses occur we can write: Pm& P m& 1 1 1 m&V 2 gm&y 2 2 1 m&V 2 gm& y + 2 1 1 + 1 1 = + 2 2 2 + 2 2 PS1 + PV1 = PS 2 + PV 2 + losses + ΔPht ρ1 ρ2 And if we can ignore changes in height: For incompressible steady flow m&1 = m&2 and ρ1 = ρ2 = ρ 1 2 1 2 PS1 + PV1 = PS 2 + PV 2 + losses Key eqn P1 + 2 ρV1 + gρy1 = P2 + 2 ρV2 + gρy2 2 Application of Bernoulli Equation Bernoulli Equation for a venturi Daniel Bernoulli developed the most important equation in fluid • A venturi measures flow rate in a duct using a pressure hydraulics in 1738. this equation assumes constant density, difference. Starting with the Bernoulli eqn from before: irrotational flow, and velocity is derived from velocity potential: PS1 + PV1 = PS 2 + PV 2 + losses + ΔPht • Because there is no change in height and a well designed venturi will have small losses (<~2%) We can simplify this to: PS1 + PS 2 = PV 2 + PV1 or − ΔPS = ΔPV • Applying the continuity condition (incompressible flow) to get: 2(P − P ) V S1 S 2 1 = 2 A2 ρ1− 2 A1 Venturi Meter VP Measurement upstream downstream SUCTION SIDE, INLET PRESSURE SIDE, OUTLET Ps <0, PT <0, PV >0 Ps >0, PT >0, PV >0 P1 P2 2(P − P ) F V C S1 S 2 A 1 = e 2 N A2 ρ1− 2 A1 • Discharge Coefficient Ce corrects for losses = f(Re) PT - Ps = (Ps + PV)- Ps = PV 3 Pitot tube END HERE The static and Pitot tube are often combined into the one-piece Pitot- static tube. Total pressure port V = 4005√Pv V = velocity in fpm PV = velocity pressure in “wg Static pressure port 5.12 4.