# Fluid Mechanics 04.

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Fluid Mechanics 04

Buoyant Force A buoyant force is defined as the upward
force that is produced on a body that is totally or partially submerged in a fluid. Fb= ϒ * Vd Vd is the displacement fluid volume

If the body is totally submerged, the displaced volume is the volume of the body. If a body is partially submerged, the displaced volume is the portion of the volume that is submerged. For a fluid of uniform density, the line of action of the buoyant force passes through the centroid of the displaced volume.

G : center of gravity C : center of buoyancy M : metacenter

Example A metal part (object 2) is hanging by a thin cord from a floating wood block (object 1). The wood block has a specific gravity S1 = 0.3 and dimensions of 50 × 50 × 10 mm. The metal part has a volume of 6600 mm3. Find the mass m2 of the metal part and the tension T in the cord.

Solution Body (1) Fb1 = T + W1 Fb1 = ϒw * Vd1 Vd1 = 0.05*0.05*0.0075
Vd1 = m^3 Fb1 = 9800*Vd1=0.184 N W1 = ϒb1*Vb1 Vb1 = 0.05*0.05*0.01 Vb1 = W1 = 9800*0.3*Vb1= N

T = Fb1-W1 = – = 0.11 N Body (2) W2 = Fb2+T T = 0.11 N Fb2 = ϒw*Vb2 = 9800* = N W2 = = N m2 = W2/g = kg = 17.8 g

Fluid Motion Flow pattern is consisted of stream lines.
The local velocity vector is tangent to the streamline at every point along the line.

Uniform and Non uniform
Uniform Flow the velocity does not change along a fluid path. Non uniform Flow the velocity changes along a fluid path.

In a steady flow the velocity at a given point on a fluid path does not change with time Unsteady Flow In a unsteady flow the velocity at a given point on a fluid path changes with time

Laminar and Turbulent Laminar flow fluid layers move smoothly with respect to each other. (Low Velocity) Turbulent flow is an unsteady flow characterized by intense cross-stream mixing.(High Velocity)

1-D, 2-D, 3-D

Acceleration In this section we will give a general expression to acceleration. 𝑎= 𝑑𝑉 𝑑𝑡 But 𝑉=𝑉 𝑠,𝑡 ∗𝑒𝑡 Then 𝑎= 𝑑𝑉(𝑠,𝑡) 𝑑𝑡 ∗𝑒𝑡+𝑉∗ 𝑑𝑒𝑡 𝑑𝑡 𝑑𝑒 𝑑𝑡 = 𝑉 𝑟 ∗𝑒𝑛 𝑎𝑡= 𝑑𝑉(𝑠,𝑡) 𝑑𝑡 , 𝑎𝑛= 𝑉 2 𝑟

𝑑𝑉(𝑠,𝑡) 𝑑𝑡 = 𝑑𝑉 𝑑𝑠 ∗ 𝑑𝑠 𝑑𝑡 +( 𝑑𝑉 𝑑𝑡 ) = 𝑉∗ 𝑑𝑉 𝑑𝑠 + 𝑑𝑉 𝑑𝑡
𝑎𝑡=𝑉∗ 𝑑𝑉 𝑑𝑠 + 𝑑𝑉 𝑑𝑡

Euler's Equation 𝑃∗ ∆𝐴− 𝑃+∆𝑃 ∗∆𝐴−∆𝑊𝑙=𝑚 ∗𝑎𝑙 𝑚= ρ ∗ ∆𝐴 ∗ ∆𝑙
∆𝑊𝑙= ∆𝑊∗ sin α , sin α= ∆𝑧 ∆𝑙 ∆𝑊=𝑚∗𝑔 −∆𝑃∗∆𝐴 − ρ∗∆𝐴∗∆𝑙∗𝑔∗ ∆𝑧 ∆𝑙 = ρ∗∆𝐴∗∆𝑙 ∗𝑎𝑙 ∆𝑃 ∆𝑙 +ɣ ∗ ∆𝑧 ∆𝑙 =−ρ∗𝑎𝑙

Example A column water in a vertical tube is being accelerated by a piston in the vertical direction at 100 m/s2. The depth of the water column is 10 cm. Find the gage pressure on the piston. The water density is 1000 kg/m3.

Solution 𝑑(𝑃+ɣ∗𝑧) 𝑑𝑙 = −ρ∗𝑎𝑙 1 2 𝑑(𝑃+ɣ∗𝑧) = −ρ∗𝑎𝑙∗ 1 2 𝑑𝑙
𝑑(𝑃+ɣ∗𝑧) 𝑑𝑙 = −ρ∗𝑎𝑙 1 2 𝑑(𝑃+ɣ∗𝑧) = −ρ∗𝑎𝑙∗ 1 2 𝑑𝑙 𝑃2−𝑃1 +ɣ∗ 𝑧2−𝑧1 =− ρ∗𝑎𝑙∗(𝑧2−𝑧1) P2= Patm = Zero gauge, z2-z1=0.1m (-P1)+(1000*9.8*0.1)=-1000*100*0.1 P1 = Pa

Example The tank on a trailer truck is filled completely with gasoline, which has a specific weight of 42 lbf/ft3 (6.60 kN/m3). The truck is decelerating at a rate of 10 ft/s2 (3.05 m/s2). (a) If the tank on the trailer is 20 ft (6.1 m) long and if the pressure at the top rear end of the tank is atmospheric, what is the pressure at the top front? (b) If the tank is 6 ft (1.83 m) high, what is the maximum pressure in the tank?

Solution 𝑑 𝑃+ɣ𝑧 =−ρ∗𝑎𝑙∗𝑑𝑙 𝑃2−𝑃1 +ɣ∗ 𝑧2∗𝑧1 =−ρ∗𝑎𝑙∗(𝐿2−𝐿1)
Where z2-z1=Zero, L2-L1=6.1m , P1=Patm= Zero gauge 𝑃2=− ∗−3.05∗6.1=12530 Pa 𝑃𝑚𝑎𝑥=𝑃2+ ρ∗𝑔∗ℎ= ∗ ∗1.83 = Pa