9Fluid2_flow - 11/14/2010 Fluid Flow Example: Fire Hose •...

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11/14/2010 1 Fluid Flow • Simplifying assumptions for flow of an ideal fluid : – steady (not turbulent) – incompressible (density does not change appreciably with pressure) –non - v i scou s – irrotational (no vortices, whirlpools, etc.) 1 v velocity is tangent to streamlines Fluid doesn’t cross streamlines Equation of Continuity: Δ m 1 / Δ t= ρΔ V 1 / Δ t Δ m 2 / Δ t= ρΔ V 2 / Δ t Δ m 1 = Δ m 2 mass going in = mass coming out (incompressible so ρ = constant) Δ V 1 = Δ V 2 Δ x 1 Δ x 2 Δ V 1 = A 1 Δ x 1 Δ V 2 = A 2 Δ x 2 A 1 Δ x 1 = A 2 Δ x 2 divide by Δ t to get: Q = A 1 v 1 = A 2 v 2 volume/sec = flow rate is position independent = Δ V/ Δ t 2 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 3 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 ( μ m)] and speed is 600 times slower (i.e., 0.05 cm/s), how many capillaries must there be in the body? Δ V aorta / Δ t = Δ V cap / Δ t A aorta v aorta = A cap (total) v cap A cap (total) = N A cap (individual) = N π (4x10 -4 cm) 2 π (1cm) 2 (30 cm/s) = N π 16x10 -8 cm 2 (0.05cm/s)
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11/14/2010 2 Fluids in Motion Bernoulli’s Principle: velocity high means pressure low 5 Bernoullis’ equations: Work-Energy theorem for fluids A fluid element of volume Δ V moves from pipe 1 to pipe 2. The pressure P 1 does positive work and the pressure P 2 does negative work. F 1 Δ x 1 –F 2 Δ x 2 = net work done so W net = Δ KE = ½m 2 v 2 2 -½m 1 v 1 2 m 1 = m 2 and divide by Δ V to get: P 1 –P 2 = ½ ρ v 2 2 ρ v 1 2 or P 1 + ½ ρ v 1 2 = P 2 + ½ ρ v 2 2 v changes ! F Δ x = PA Δ x= P Δ V P 1 Δ V – P 2 Δ V = ½ m 2 v 2 2 –½ ±m 1 v 1 2 6 Now do the same thing with gravity (same area pipe) higher pressure lower pressure g v constant ! A 1 v 1 = A 2 v 2 A is constant so v is constan 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 Δ V – P 2 Δ V No KE change (v = constant and flow rate = constant) W g + W P = 0 or when divide by Δ V: ρ gy 1 ρ gy 2 + P 1 –P 2 = 0 ρ gy 1 + P 1 = ρ gy 2 + P 2 (constant area) constant so v is constant 7 Now put both gravity and area change together to find the general equation: General Bernoulli equation when both fluid motion Constant area: ρ gy 1 + P 1 = ρ gy 2 + P 2 Constant height: P 1 + ½ ρ v 1 2 = P 2 + ½ ρ v 2 2 General Bernoulli equation when both vary: P 1 + ρ gy 1 + ½ ρ v 1 2 = P 2 + ρ gy 2 + ½ ρ v 2 fluid motion multiply by Δ V: F 1 Δ x 1 + mgy 1 + ½mv 1 2 = F 2 Δ x 2 + mgy 2 + ½mv 2 2
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11/14/2010 3 Torricelli’s Theorem (equal pressures) General equation when both vary: P 1 + ρ gy 1 + ½ ρ v 1 2 = P 2 + ρ gy
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9Fluid2_flow - 11/14/2010 Fluid Flow Example: Fire Hose •...

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