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Fluid Mechanics 05 Rotating Flow In the rotating flow βl = βr and the acceleration will be only in the normal direction. π −π£ 2 − π + Ι£π§ = ρ ∗ ππ π Where π£ = π€ ∗ π π π + Ι£π§ = ρ ∗ π€ 2 ∗ π ππ ρ ∗ π€2 ∗ π2 π + Ι£π§ − = Const 2 Example A cylindrical tank of liquid shown in the figure is rotating as a solid body at a rate of 4 rad/s. The tank diameter is 0.5 m. The line AA depicts the liquid surface before rotation, and the line A′ A′ shows the surface profile after rotation has been established. Find the elevation difference between the liquid at the center and the wall during rotation. Solution ρ1 ∗ π€12 ∗ π12 π1 + Ι£1π§1 − 2 ρ2 ∗ π€22 ∗ π22 = π2 + Ι£π§2 − 2 Where P1=P2=Patm=Zero gauge, r1=0, r2=0.25 m, w=4 rad/s, ρ=1000Kg/m^3, g=9.81m/s^2 π€ 2 ∗ π22 42 ∗ 0.252 π§2 − π§1 = = = 0.051π 2∗π 2 ∗ 9.81 Bernoulli Equation We apply Bernoulli equation of the tangent component of acceleration. π − π + Ι£π§ = ρ ∗ ππ‘ ππ ππ£ ππ£ Where ππ‘ = π£ ∗ + ππ ππ‘ Assume the flow is steady then π£2 ππ£ ππ£ π( 2 ) = 0 π‘βππ ππ‘ = π£ ∗ = ππ‘ ππ ππ π π π£2 − π + Ι£π§ = ρ ∗ ( ) ππ ππ 2 π π£2 π + Ι£π§ + ρ ∗ =0 ππ 2 π£2 π + Ι£π§ + ρ ∗ = Const 2 π π£2 +π§+ = ππππ π‘ Ι£ 2π Example Piezometric tubes are tapped into a venturi section as shown in the figure. The liquid is incompressible. The upstream piezometric head is 1 m, and the piezometric head at the throat is 0.5 m. The velocity in the throat section is twice large as in the approach section. Find the velocity in the throat section. Solution π£12 π£22 β1 + = β2 + 2∗π 2∗π Where v2=2*v1, h1=1m, h2=0.5m π£12 (2 ∗ π£1)2 1+ = 0.5 + 2 ∗ 9.81 2 ∗ 9.81 3 ∗ π£12 0.5 = 2 ∗ 9.81 v1 = 1.808m/s , v2=3.62m/s Example A open tank filled with water and drains through a port at the bottom of the tank. The elevation of the water in the tank is 10 m above the drain. The drain port is at atmospheric pressure. Find the velocity of the liquid in the drain port. Solution π1 π£12 π2 π£22 + π§1 + ρ ∗ = + π§2 + ρ ∗ Ι£ 2π Ι£ 2π Where P1=P2=Patm, v1=zero, z1-z2=10m 10= 1000 ∗ π£22 2∗9.81 Then π£2 = 10 ∗ 2 ∗ 9.81 = 14π/π Rate of flow Discharge (Volume flow rate):General equation π = π£ ∗ ππ΄ Note that we take only the velocity component normal to the area. Laminar flow eqn π = π£ππ£ ∗ π΄ Where vav is the average velocity Mass flow rate π. = ρ ∗ π£ ∗ ππ΄ For constant density across the flow π. = ρ π£ ∗ ππ΄ π. = ρ ∗ π Example The water velocity in the channel shown in the accompanying figure has a distribution across the vertical section equal to u/umax = (y/d)1/2. What is the discharge in the channel if the water is 2 m deep (d = 2 m), the channel is 5 m wide, and the maximum velocity is 3 m/s? Solution π¦ 1/2 π’ = π’πππ₯ ∗ ( ) π π= π= π’πππ₯ ∗ π= π ∗ ππ΄ 1 1/2 ( ) *π¦1/2 π 1 1 2 3∗ 2 Q = 10.606 ∗ ∗ 5 ∗ ππ¦ ∗5∗ 1 π¦2 ∗ ππ¦ π¦1/2 ∗ ππ¦ 2 Q = 10.606 ∗ ∗ π¦ 3/2 3 3 2 π3 Q = 10.606 ∗ ∗ 2 2 = 20 3 π Example A jet of water discharges into an open tank, and water leaves the tank through an orifice in the bottom at a rate of 0.003 m3/s. If the crosssectional area of the jet is 0.0025 m2 where the velocity of water is 7 m/s, at what rate is water accumulating in (or evacuating from) the tank? Solution π3 πππ = π΄1 ∗ π£1 = 0.0025 ∗ 7 = 0.0175 π π3 πππ’π‘ = 0.003 π π3 πππ − πππ’π‘ = 0.0175 − 0.003 = 0.0145 π πΎπ . π = ρ ∗ πππ − πππ’π‘ = 14.5 π Example A 10 cm jet of water issues from a 1 m diameter tank. Assume that the velocity in the jet is 2πβ m/s where h is the elevation of the water surface above the outlet jet. How long will it take for the water surface in the tank to drop from h0 = 2 m to hf = 0.50 m? Solution πππ’π‘ = ππ π£ ∗ ππ΄ = ππ‘ ππ π΄π‘ ∗ πβ ππ‘ = = πππ’π‘ 2 ∗ π ∗ β ∗ π΄π Π π΄π‘ = ∗ π·π‘ 2 = 0.785 π2 4 Π π΄π = ∗ π·π2 = 0.00785 π2 4 π‘ β π΄π‘ πβ ππ‘ = ∗ 2 ∗ π ∗ π΄π 0 β0 β π‘= π‘= 2 ∗ π΄π‘ 2 ∗ π ∗ π΄π 2 ∗ 0.785 ∗ 2 ∗ 9.81 ∗ 0.00785 t= 31.9 s β0 − β ∗ 2 − 0.5 Example Water with a density of 1000 kg/m3 flows through a vertical venturimeter as shown. A pressure gage is connected across two taps in the pipe (station 1) and the throat (station 2). The area ratio Athroat/Apipe is 0.5. The velocity in the pipe is 10 m/s. Find the pressure difference recorded by the pressure gage. Solution π1 = π2 = π΄1 ∗ π£1 = π΄2 ∗ π£2 π΄1 π£2 π£2 = =2= π΄2 π£1 10 20π π£2 = π π£12 π£22 β1 + = β2 + 2∗π 2∗π π£22 − π£12 202 − 102 β1 − β2 = = = 15.3π 2∗π 2 ∗ 9.8 π1 − π2 = ρ ∗ π ∗ β1 − β2 = 150πππ Rotation and Irrotation flow Assume 2-D flow (x,y) β θ = + θπ΄ 2 Π β = + θπ΅ − θπ΄ 2 Π (θπ΄ + θπ΅) θ= + 4 2 The rotational rate (θΚΉπ΄ + θΚΉπ΅) θΚΉ = 2 If θΚΉ=0 the flow is irrotational. The rotational rate in terms of velocity. 1 ππ€ ππ£ Ωπ₯ = ∗ ( − ) 2 ππ¦ ππ§ 1 ππ’ ππ€ Ωπ¦ = ∗ ( − ) 2 ππ§ ππ₯ 1 ππ£ ππ’ Ωπ§ = ∗ ( − ) 2 ππ₯ ππ¦ Vorticity (w) w=2*Ω
