6Fluid2_flow

# 6Fluid2_flow - Fluid Flow Simplifying assumptions for flow...

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1 Fluid Flow Simplifying assumptions for flow of an ideal fluid : – steady (not turbulent) – incompressible (density does not change appreciably with pressure) – non-viscous – irrotational (no vortices, whirlpools, etc.) v velocity is tangent to streamlines Fluid doesn’t cross streamlines

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Equation of Continuity: Δm 1 /Δt = ρΔV 1 /Δt Δm 2 /Δt = ρΔV 2 /Δt Δx 1 Δx 2 ΔV 1 = A 1 Δx 1 ΔV 2 = A 2 Δx 2 Δm 1 = Δm 2 mass going in = mass coming out (incompressible so ρ = constant) ΔV 1 = ΔV 2 A 1 Δx 1 = A 2 Δx 2 divide by Δt to get: Q = A 1 v 1 = A 2 v 2 volume/sec = flow rate is position independent = ΔV/Δt
d 1 = 9.6 cm (hose) d 2 = 2.5 cm (nozzle) Example: Fire Hose v 1 = 1.3 m/s v 2 = ? A 1 v 1 = A 2 v 2 so π(0.048) 2 (1.3) = π(0.0125) 2 v 2 v 2 = 19.17 m/s r 1 = (9.6/2)/100 = 0.048m r 2 = (2.5/2)/100 = 0.0125m

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Example: Blood flow and mass conservation Given: the aorta radius is 1.0 cm and blood speed there is 30 cm/s. If capillary size is r cap = 4x10 -4 cm [i.e., 4 microns (μm)] and speed is 600 times slower (i.e., 0.05 cm/s), how many capillaries must there be in the body? ΔV aorta / Δt = ΔV cap / Δt A aorta v aorta = A cap (total) v cap A cap (total) = N A cap (individual) = Nπ(4x10 -4 cm) 2 π(1cm) 2 (30 cm/s) = Nπ16x10 -8 cm 2 (0.05cm/s) N = 30/[16(0.05)] x 10 8 ~ 4 x 10 9
Fluids in Motion Bernoulli’s Principle: velocity high means pressure low

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Bernoullis’ equations: Work-Energy theorem for fluids A fluid element of volume ΔV moves from pipe 1 to pipe 2. The pressure P 1 does positive work and the pressure P 2 does negative work. F 1 Δx 1 – F 2 Δx 2 = net work done so W net = ΔKE = ½m 2 v 2 2 - ½m 1 v 1 2 m 1 = m 2 and divide by ΔV to get: P 1 – P 2 = ½ρv 2 2 - ½ρv 1 2 or P 1 + ½ρv 1 2 = P 2 + ½ρv 2 2 v changes ! FΔx = PAΔx = PΔV P 1 ΔV – P 2 ΔV = ½ m 2 v 2 2 – ½ m 1 v 1 2
Now do the same thing with gravity (same area pipe) Work done by gravity is negative: W g = -mg(y 2 – y 1 ) Work done by P 1 and P 2 is positive: W p = P 1 ΔV – P 2 ΔV higher pressure lower pressure g No KE change (v = constant and flow rate = constant) W g + W P = 0 or when divide by ΔV: ρgy 1 – ρgy 2 + P 1 – P 2 = 0 ρgy 1 + P 1 = ρgy 2 + P 2 (constant area) v constant ! A 1 v 1 = A 2 v 2 A is constant so v is constant

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Now put both gravity and area change together to find the general equation: General Bernoulli equation when both vary: P 1 + ρgy 1 + ½ρv 1 2 = P 2 + ρgy 2 + ½ρv 2 2 fluid motion multiply by ΔV: F 1 Δx 1 + mgy 1 + ½mv 1 2 = F 2 Δx 2 + mgy 2 + ½mv 2 2 Constant area: ρgy 1 + P 1 = ρgy 2 + P 2 Constant height: P 1 + ½ρv 1 2 = P 2 + ½ρv 2 2
Torricelli’s Theorem (equal pressures) General equation when both vary: P 1 + ρgy 1 + ½ρv 1 2 = P 2 + ρgy 2 + ½ρv 2 2 P 1 = P 2 v 1 = 0 ρgy 1 = ρgy 2 + ½ρv 2 2 v 2 = √2gh note, reversed “1” and “2” notation from that of the book

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Venturi Tube: No gravity term Constant height: P 1 + ½ρv 1 2 = P 2 + ½ρv 2 2 Direct connection between the velocity and pressure.
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## This note was uploaded on 01/02/2012 for the course PHYS 1149 taught by Professor Bala during the Fall '11 term at Northeastern.

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6Fluid2_flow - Fluid Flow Simplifying assumptions for flow...

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