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Published byElfrieda Hines Modified over 9 years ago
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Fluid Flow conservation and continuity § 12.4
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Volume Flow Rate Volume per time through an imaginary surface perpendicular to the velocity dV/dtunits: m 3 /s
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Volume Flow Rate dV/dt = v·A if v is constant over A = ∫v·dA if it is not
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Flow Continuity Constant mass flow for a closed system 1A1v1 = 2A2v21A1v1 = 2A2v2 dm dt dm dt = 12
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Flow Continuity For an incompressible fluid: constant 1 A 1 v 1 = 2 A 2 v 2 dm dt dm dt = 12 dV dt dV dt = 12 A 1 v 1 = A 2 v 2
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Poll Question Where is the velocity greatest in this stream of incompressible fluid? A.Here. B.Here. C.Same for both. D.Can’t tell.
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Quick Question Where is the density greatest in this stream of incompressible fluid? A.Here. B.Here. C.Same for both. D.Can’t tell.
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Poll Question Where is the volume flow rate greatest in this stream of incompressible fluid? A.Here. B.Here. C.Same for both. D.Can’t tell.
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Bernoulli’s Equation Energy in fluid flow § 12.5
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Incompressible Fluid Continuity condition: constant volume flow rate dV 1 = dV 2 v 1 A 1 = v 2 A 2
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Poll Question Where is the kinetic energy of a parcel greatest in this stream of incompressible fluid? A.Here. B.Here. C.Same for both. D.Can’t tell.
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Changing Cross-Section Fluid speed varies –Faster where narrow, slower where wide Kinetic energy changes Work is done!
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Ideal Fluid No internal friction (viscosity) No non-conservative work!
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Poll Question Where would the pressure be greatest if the fluid were stationary? A.Here. B.Here. C.Same for both. D.Can’t tell.
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Work-Energy Theorem W net = K K 1 + W net = K 2 K 1 + U 1 + W other = K 2 + U 2 W other = K + U dW = dK + dU
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Work done by Pressure dW = F·ds Work done on fluid at bottom: dW 1 = p 1 A 1 ·ds 1 Work done on fluid at top: dW 2 = –p 2 A 2 ·ds 2 = (p 1 – p 2 )dV Total work done on fluid : dW = p 1 A 1 ·ds 1 –p 2 A 2 ·ds 2
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Kinetic Energy Change Steady between “end caps” Lower cap: dK 1 = 1/2 dmv 1 2 Upper cap: dK 2 = 1/2 dmv 2 2 dm = dV dK = 1/2 dV (v 2 2 –v 1 2 )
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Potential Energy Change Steady between “end caps” Lower cap: dU 1 = dmgy 1 Upper cap: dU 2 = dmgy 2 dm = dV dU = gdV (y 2 –y 1 )
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Put It All Together dW = dK + dU (p 1 – p 2 )dV = 1/2 dV (v 2 2 –v 1 2 ) + gdV (y 2 –y 1 ) (p 1 – p 2 ) = 1/2 (v 2 2 –v 1 2 ) + g(y 2 –y 1 ) p 1 + 1/2 v 1 2 + gy 1 = p 2 + 1/2 v 2 2 + gy 2 – This is a conservation equation – Strictly valid only for incompressible, inviscid fluid
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What Does It Mean? Faster flow lower pressure Maximum pressure when static pV is energy
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Example problem A bullet punctures an open water tank, creating a hole that is a distance h below the water level. How fast does water emerge from the hole?
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Torricelli’s Theorem p 1 + 1/2 v 1 2 + gy 1 = p 2 + 1/2 v 2 2 + gy 2 1/2 v 2 2 = g(y 2 –y 1 ) + (p 2 –p 1 ) – 1/2 v 1 2 1/2 v 2 2 = gh v 2 2 = 2gh h v v 2 = 2gh look familiar? 1 2
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Example problem Water emerges from a downward-facing tap with a diameter of 2.0 cm at a flow rate of 40 L/min. As the water falls, it accelerates downward and the stream becomes thinner. What is the diameter of the stream after it has fallen a distance y from the tap?
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