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 = 0.00001875 m^3
Fb1 = 9800*Vd1=0.184 N
W1 = ϒb1*Vb1
Vb1 = 0.05*0.05*0.01
Vb1 = 0.000025
W1 = 9800*0.3*Vb1=0.0735 N
T = Fb1-W1 = 0.184 – 0.0735 = 0.11 N
Body (2)
W2 = Fb2+T
T = 0.11 N
Fb2 = ϒw*Vb2 = 9800*0.0000066 = 0.0647 N
W2 = 0.0647+0.11 = 0.1747 N
m2 = W2/g = 0.0178 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.
Steady and Unsteady
Steady Flow
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.
𝑑𝑉
𝑎=
𝑑𝑡
But 𝑉 = 𝑉 𝑠, 𝑡 ∗ 𝑒𝑡
Then 𝑎 =
𝑑𝑒
𝑑𝑡
𝑑𝑉(𝑠,𝑡)
𝑑𝑡
𝑉
= 𝑟 ∗ 𝑒𝑛
𝑑𝑉(𝑠,𝑡)
𝑎𝑡 =
𝑑𝑡
∗ 𝑒𝑡 + 𝑉 ∗
, 𝑎𝑛 =
𝑉2
𝑟
𝑑𝑒𝑡
𝑑𝑡
𝑑𝑉(𝑠,𝑡)
𝑑𝑡
=
𝑑𝑉
𝑑𝑠
∗
𝑑𝑠
𝑑𝑡
+
𝑑𝑉
( )
𝑑𝑡
𝑑𝑉
𝑑𝑠
=𝑉∗
𝑑𝑉 𝑑𝑉
𝑎𝑡 = 𝑉 ∗
+
𝑑𝑠
𝑑𝑡
+
𝑑𝑉
𝑑𝑡
Euler's Equation
𝑃 ∗ ∆𝐴 − 𝑃 + ∆𝑃 ∗ ∆𝐴 − ∆𝑊𝑙 = 𝑚 ∗ 𝑎𝑙
𝑚 = ρ ∗ ∆𝐴 ∗ ∆𝑙
∆𝑧
∆𝑊𝑙 = ∆𝑊 ∗ sin α, sin α =
∆𝑙
∆𝑊 = 𝑚 ∗ 𝑔
∆𝑧
−∆𝑃 ∗ ∆𝐴 − ρ ∗ ∆𝐴 ∗ ∆𝑙 ∗ 𝑔 ∗ = ρ ∗ ∆𝐴
∆𝑙
∗ ∆𝑙 ∗ 𝑎𝑙
∆𝑃
∆𝑧
+ɣ ∗
= −ρ ∗ 𝑎𝑙
∆𝑙
∆𝑙
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)
= −ρ ∗ 𝑎𝑙
𝑑𝑙
2
𝑑(𝑃 + ɣ ∗ 𝑧) =−ρ
1
2
𝑑𝑙
1
∗ 𝑎𝑙 ∗
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 = 10980 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
6600
𝑃2 = −
∗ −3.05 ∗ 6.1 = 12530 Pa
9.81
𝑃𝑚𝑎𝑥 = 𝑃2 + ρ ∗ 𝑔 ∗ ℎ = 12530 + (6600