Chapter 14
Fluids
Examples & Problems
Devices that use Chapter 14:
Ships (buoyancy)
Hydraulic Lifts (Pascal)
Barometer (Pascal)
Airplanes (Bernoulli)
gasoline automatic shut-off valves (Bernoulli)
Incompressibile Fliuds
Aerodynamics: For slow air speeds (v < 100m/s), air can be
considered incompressible fluid.
Liquids
Then
the Continuity Equation and Bernoulli Equation hold.
1
Continuity Equation
∆V = A ∆x = A v ∆t
∆V = A1v1∆t = A2v2∆t
A1v1 = A2v2 (equation of continuity)
thus
Av = constant (volume flow rate)
A1v1 = A2v2 = constant
Incompressible fluids:
liquids
air -- for slow air speeds (v < 100m/s).
Bernoulli’s equation
Bernoulli’s equation is based on
Conservation of Mechanical Energy
=
Conserved
Potential energy per volume
Kinetic energy per volume
Pressure
p + ½ ρv2 + ρgy = constant
p1 + ½ ρv12 + ρgy1 = p2 + ½ ρv22 + ρgy2 = constant
2
Problem 14-29
FB
A block of wood floats in water with
2/3 of its volume submerged and in oil
with 9/10 of its volume submerged.
mg
A) Find the density of the wood.
ρwaterVwater g = ρwoodVwood g where Vwater is the water displaced

V
 2
ρ wood = ρ water  water  = ρ water   = 667 kg/m 3
 3
 Vwood 
B) Find the density of the oil.
ρoilVoil g = ρwoodVwood g where Voil is the oil displaced

V
 10 
 2  10 
ρoil = ρ wood  wood  = ρ wood   = ρ water    = 741 kg/m 3
9
 3  9 
 Voil 
Flight
3
Airfoils
from Anderson, Introduction to Flight, 3rd ed.1978
Airfoils
top airstream moves faster than
the lower airstream.
Consider
vlower = 70m/s
plower = patm
ρair = 1.21kg/m3
(160 mph -- takeoff speed)
The shape of the airfoil causes vupper = 120m/s.
Find ∆pupper = plower − pupper
4
Airfoils
Consider
vlower = 70m/s
plower = patm
ρair = 1.21kg/m3
The shape of the airfoil causes vupper = 120m/s.
Find ∆pupper = plower − pupper
p1 + ½ ρv12 + ρgy1 = p2 + ½ ρv22 + ρgy2 = constant
pupper + ½ ρvupper2 = plower + ½ ρvlower2
1
1
2
∆p = p lower − p upper = ρair v 2upper − ρair v lower
2
2
1
1
2
= ρair v 2upper − v lower
=  (1.21) 120 2 − 70 2 = 5,748N/m 2
2
 2
(
)
(
)
Airfoils
wings 30m x 10m x 2
= 600m2
5
Airfoils
Consider
vlower = 70m/s, vupper = 120m/s
plower = patm
ρair = 1.21kg/m3
wings 30m x 10m x 2
= 600m2
1
1
2
∆p = p lower − p upper = ρair v 2upper − ρair v lower
2
2
1
1
2
= ρair v 2upper − v lower
=  (1.21) 120 2 − 70 2 = 5,748N/m 2
2
 2
(
)
(
)
=> 5,748 x 600 = 3,450,000 N of lift!!! (=776,250 lbs = 388 tons)
Daily Quiz, October 29, 2004
What caused the beachball to “float” in air?
1) Bernoulli Effect
2) Pascal’s Principle
3) Archimedes Principle
4) Other equation/effect
5) I thought it sank??!?
6
Daily Quiz, October 29, 2004
pressure reduced
What caused the beachball to “float” in air?
1) Bernoulli Effect
2) Pascal’s Principle
3) Archimedes Principle
4) Other equation/effect
5) I thought it sank??!?
Problem 14-42
Two streams join to become one river.
The dimensions are:
w1 = 8.2m, d1 = 3.4m, v1 = 2.3m/s
w2 = 6.8m, d2 = 3.2m, v2 = 2.6m/s
wf = 10.5m, df = ?, vf = 2.9m/s
Continuity Equation
Volume flowing into junction = Volume flowing out of junction
A1v1 + A2v2
A1v1 = (8.2)(3.4)(2.3)m3/s
A2v2 =
+
(6.8)(3.2)(2.6)m3/s
=
Afvf
=
constant
= Afvf = (10.5)(df)(2.9)m3/s
(64.124 + 56.576) m3/s = (30.450)(df)m2/s => df = 3.964m
7
Problem 14-53
D = 15m
radius of hole = 0.02m
d = 6.0m
A) Find the frictional force needed by the plug.
pressure = Force divided by Area
Problem 14-53
D = 15m
radius of hole = 0.02m
d = 6.0m
B) Find the water volume exiting in 3.0h.
p + ½ ρv2 + ρgy = constant
8
Problem 14-54
v1 = 15m/s
d1 = 5.0 cm
d2 = 3.0 cm
A) Find the volume of water exiting in 10 minutes.
Problem 14-54
v1 = 15m/s
d1 = 5.0 cm
d2 = 3.0 cm
B) Find the speed v2 in the larger pipe section.
9
Problem 14-54
v1 = 15m/s
d1 = 5.0 cm
d2 = 3.0 cm
C) Find the pressure p2 in the larger pipe section.
Problem 14-58
r1 = 2.00R, r2 = 3.00R
vR = 0.50m/s
Find net work done on 0.400m3 of water as it passes through the pipe.
A1v1 = A2v2 = constant
10