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MR. SURRETTE
VAN NUYS HIGH SCHOOL
CHAPTER 9: FLUID MECHANICS
CLASS NOTES
FLUIDS
A fluid is a substance whose molecules move freely past one another and has the tendency to assume the
shape of its container. In general, liquids and gases are fluids.
DENSITY
The density  of a substance is mass per unit volume. It has units of kilograms per cubic meter (or
grams per cubic centimeter) in the metric system:
=m/V
PRESSURE
Pressure is force exerted per unit area:
p=F/A
Pressure [Pascals] = Force [Newtons] / Area [m2]
BASIC LAW OF FLUID PRESSURE
The pressure at the bottom of a fluid can be expressed as:
p = gh
( = density of fluid)
(h = depth of fluid)
(g = 9.8 m/s2)
The pressure at any point in a fluid depends only on its density and its depth. It acts equally in all
directions.
ABSOLUTE PRESSURE
The absolute pressure, p, at a depth, h, below the surface of a liquid which is open to the atmosphere is
greater than atmospheric pressure, Pa, by an amount that depends on the depth below the surface:
p = pa + gh
( = density of the liquid)
Example 1. What is the total pressure at the bottom of a 5 m deep swimming pool? (Note the pressure
contribution from the atmosphere is 1.01 x 105 N/m2, the density of water is 103 kg/m3, and g = 9.8 m/s2)
1A.
(1) p = pa + gh
(2) p = (1.01 x 105 N/m2)+(1.0 x 103 kg/m3)(9.8 m/s2)(5 m)
(3) p = 1.5 x 105 N/m2
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
PASCAL’S PRINCIPLE
Pressure applied to an enclosed fluid is transmitted undiminished to every portion of the fluid and the
walls of the containing vessel:
F1 / A1 = F2 / A2
PASCAL’S PRINCIPLE
Example 2. A hydraulic lift raises a 2000 kg automobile when a 500 N force is applied to the smaller
piston. If the smaller piston has an area of 10 cm2, what is the cross-sectional area of the larger piston?
2A.
(1) F1/A1 = F2/A2
(2) A2 = (F2A1)/F1
(3) F2 = w = mg
(4) A2 = (mgA1)/F1
(5) A2 = (2000 kg)(9.8 m/s2)(1.0 x 10-3 m2) / 500 N
(6) A2 = 3.92 x 10-2 m
ARCHIMEDE’S PRINCIPLE
Archimede’s Principle is the law of buoyancy. When an object is submerged in a fluid (completely
or partially), there exists an upward force on the object that is equal to the weight of the fluid that is
displaced by the object.
ARCHIMEDE’S PRINCIPLE
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
BUOYANT FORCE
One common expression for the upward buoyant force is:
FB = Vg
(FB = buoyant force)
 = density of liquid)
(V = volume of submerged object)
(g = 9.8 m/s2)
BUOYANT FORCE
Another expression for buoyant force is:
FB = w – w’
(w = actual weight of object
w’ = apparent weight of submerged object)
DENSITY OF OBJECTS
The density of a submerged object can be determined:
object = liquid / w – w’
Example 3. An object of mass 0.5 kg is suspended from a scale and submerged in a liquid. What is the
buoyant force that the fluid exerts if the reading on the scale is 3.0 N?
3A.
(1) w = mg
(2) w = (0.5 kg)(9.8 m/s2)
(3) w = 4.9 N
(4) FB = w – w’
(5) FB = 4.9 N – 3.0 N
(6) FB = 1.9 N
FLOW RATE
The product Av is called the flow rate. The flow rate at any point along a pipe carrying an
incompressible fluid is constant:
A1v1 = A2v2
Example 4. If the flow rate of a liquid going through a 2.00 cm radius pipe is measured at
0.8 x 10-3 m/s, what is the average fluid velocity
in the pipe?
4A.
(1) Flow rate = Av
(2) v = Flow rate / A
(3) v = Flow rate / r2
(4) v = 0.8 x 10-3 m/s / (2.0 x 10-2 m)2
(5) v = 0.64 m/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
TORRICELLI’S THEOREM
The speed of efflux is the same as the speed a body would acquire in falling freely through a height h:
v = (2gh)1/2
TORRICELLI’S THEOREM
Torricelli’s Theorem can be analyzed another way: Point 1 is at the surface of water in a tank and Point
2 is the position of the hole. y1 is the vertical distance from the ground to Point 1 and y2 is the vertical
distance from the ground to Point 2:
v = (2g(y1-y2))1/2
(Notice that (y1 – y2) = h = fluid depth)
TORRICELLI’S THEOREM
Example 5. A liquid filled tank has a hole on its vertical surface just above the bottom edge. If the
surface of the liquid is 0.4 m above the hole, at what speed will the stream of liquid emerge from the
hole?
Question 5.
5A.
(1) v = (2gh)1/2
(2) v = [(2)(9.8 m/s2)(0.4 m)]1/2
(3) v = 2.8 m/s
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
BERNOULLI’S EQUATION
Bernoulli’s equation relates to the Work Energy Theorem and to the Conservation of Energy. It is
derived from Newtonian mechanics:
p1 + ½ v12 + gy1 = p2 + ½ v22 + gy2
(p1 = initial pressure, p2 = final pressure)
 = density of fluid, g = 9.8 m/s2)
(v1 = initial velocity, v2 = final velocity)
(y1 = initial height, y2 = final height)
Example 6. Water (density = 1 x 103 kg/m3) is flowing through a pipe whose radius is 0.04 m with a
speed of 15 m/s. This same pipe goes up to the second floor of the building, 3 m higher, and the
pressure remains unchanged. What is the cross-sectional area of the pipe on the second floor?
6A.
(1) First Equation:
p1 + ½ v12 + gy1 = p2 + ½ v22 + gy2
(2) ½ v12 + gy1 = ½ v22 + gy2
(3) v12 = v22 + 2gy2
(4) v2 = [(v1)2 – 2gy2]1/2
(5) v2 = [(15 m/s)2 – 2(9.8 m/s2)(3 m)]1/2
(6) v2 = 12.9 m/s
(7) Second Equation: A1v1 = A2v2
(8) A2 = A1v1 / v2
(9) Third Equation: A1 = r2
(10) A2 = (r2)v1 / v2
(11) A2 = (0.04 m)2(15 m/s) / (12.9 m/s)
(12) A2 = 5.84 x 10-3 m2
BERNOULLI’S PRINCIPLE
Bernoulli’s principle states that swiftly moving fluids exert less pressure than slowly moving fluids.
APPLICATION OF BERNOULLI’S PRINCIPLE
CROSS-SECTION OF WING (AIR FOIL)

APPLICATION OF BERNOULLI’S PRINCIPLE
P = P1 – P2. When the force exerted by P exceeds the weight of the airplane, the airplane is
provided lift and flies.
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PHYSICS
MR. SURRETTE
VAN NUYS HIGH SCHOOL
BERNOULLI’S TUBE
The velocity of a fluid at the constricted end of a pipe is greater than the velocity at the wider ends if
steady flow is maintained.
BERNOULLI’S TUBE
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PHYSICS