Terms
Density
Specific Gravity
Pressure
Gauge Pressure
Absolute Pressure
Pascal’s Principle
*In pressure problems…the force is
always perpendicular to the surface area
 = M/V
S.G. = sample/ water
P = F/A
P = gh
Pabs = gh + Patm
Pressure applied to a
confined fluid is transmitted
throughout the entire fluid
and acts in all directions
Pressure Examples
Suction Cups
Remove air under the cup
Outside air pressure pushes against cup
Difference in air pressure…”suction”
Straws
Reduce pressure at top of straw
Greater pressure on bottom
Net upward push…liquid rises
Question: What is the longest straw you could use?
…more Terms
Archimedes Principle
the buoyant force on a body
immersed in a fluid is equal to
the weight of the fluid displaced
by the object*
A
F1
h
FB
F B = F2 – F 1
F2
Buoyant Force
Acts upward…because F2 is
greater due to “h2”
*FB = FgV = mFg (V = Ah)
The buoyant force occurs because the pressure in a fluid increases with depth
…more Buoyancy
FB = w’
The buoyant force is equal to the weight of the liquid displaced by the object…
…the buoyant force is equal to the weight of the body of fluid whose
volume equals the volume of the original submerged object
Objects appear to weigh less when submerged in a fluid
Statue example: pg. 284
Applying Archimedes Principle
scale reads 13.4 kg
scale reads 14.7 kg
FT
F’T
FB
m
w
m
FT = Fg (w = mg)
FT = w
w
F’T + FB = w
w’ = m’g = F’T scale reading based on effective weight
F’T = w’ = w – FB
Therefore FB = w – w’
finishing up calculations…
FB = FgV
so (w - w’) = FB = FgV
set a ratio: w
= samplegV
(w – w’) FgV
S.G. = sample = w
water
(w – w’)
Specific Gravity definition
Fluid Flow
Streamline (laminar flow) - smooth
Turbulent Flow – Eddy Currents
Fluid Terms
Viscosity – internal friction
Syrup is more viscous than water
Flow Rate (f = Av)
Mass Flow Rate = Δm / Δt = ( ΔV) / Δt = Av
Equation of Continuity
(Av)1 = (Av)2 since  is constant in a fluid…
**in = out (Av)1 = (Av)2**
*a large cross-sectional area results in a slower fluid velocity
Bernoulli’s Equation
1. The fluid is incompressible
2. The fluid’s viscosity is negligible
3. The flow is streamline
Bernoulli’s Principle:
where the velocity of a fluid is high…the pressure is low
where the velocity of a fluid is low…the pressure is high
*less fluid pressure results in greater fluid acceleration*
P1 + ½ v12 + gy1 = P2 + ½ v22 + gy2
P + ½ v12 + gy = Constant
P + gy + ½ v2
Examples
1.
A pipe of non-uniform diameter carries water. At one
point in the pipe, the radius is 2 cm and the flow speed
is 6 m/s.
a) What is the flow rate?
b) What is the flow speed at a point where the pipe
constricts to a radius of 1 cm?
a) f = Av = πr2v = π(0.02m)2(6 m/s) = 0.0075 m3/s
b) v α 1/A & A α r2 so if A decreases by a factor of 4…v increases by a
factor of 4…4 * (6 m/s) = 24 m/s
Examples
2. If the diameter of a pipe increases from 4cm to 12 cm,
what will happen to the flow speed?
A = πr2 = π(1/2 d)2 = ¼ πd2…so if d increases by a factor of 3…flow speed
decreases by a factor of 9.
Examples
3. What does Bernoulli’s Equation tell us about a fluid at
rest in a container open to the atmosphere?
Because the fluid in the tank is at rest: v1 & v2 = 0 m/s
P1 + gy1 = P2 + gy2
since P1 = P atm…
P2 = Patm + g(y1 – y2) = Patm + gh…
which is the formula for Absolute Pressure!!!
Examples
4. In the figure below, a pump forces water at a constant
flow rate through a pipe whose cross-sectioanl area, A,
gradually decreases: at the exit point, A has decreased
to 1/3 its value. If y=60cm and the flow speed of the
water at point 1 is 1 m/s, what is the gauge pressure at
point 1?
exit
P1 + ½ v12 + gy1 = P2 + ½ v22 + gy2
y
y1 = 0
P1 – Patm = gy2 + ½ v2 - ½ v1
2
Point 1
2
P1 – Patm = gy2 + ½ (3v1 - ½ v1
)2
2
Pump
P1 – Patm = (gy2 + 4v12)
P1 – Patm = (1000 kg/m3)[(10 m/s)(0.6 m) + 4(1 m/s)2]
Pg = 10000 Pa
Examples
5. The side of an above-ground pool is punctured, and
water gushes out through the hole. If the total depth of
the pool is 2.5 m, and the puncture is 1 m above ground
level, what is the speed of the water gushing out?
gy1 = ½ v22 + gy2
v2 = 2g(y1-y1) = 2gh = 30 = 5.5 m/s