Type:
Double
Objective:
Fluid Dynamics VIII
Fluid Dynamics IX
Date:__________
Homework: Read 11.10
Do CONCEPT QUEST # (24)
Do PROBLEMS # (65, 96) Ch. 11
AP Physics “B”
Mr. Mirro
Date: ________
Fluid Dynamics VIII
In 1738 a Swiss mathematician named Daniel Bernoulli observed that whenever air moves, its pressure
drops. One consequence of this effect, known as the Bernoulli Effect occurs in urban development.
Poorly designed and positioned buildings can create dangerous windstorms at street level.
‰
‰
‰
The average person can be blown off balance by a 40 mph wind
Turbulent winds of 10 mph can impede walking
A gust of wind in Boston’s Copley Square once blew over a half-ton mail truck.
It also turns out that the faster air moves, the more the pressure drops. One of the most spectacular
examples of how fluid flow affects pressure is the dynamic lift on an airplane wing.
Bernoulli’s Principle - states that an increase in the “velocity” of any fluid is always accompanied by a
decrease in “pressure.”
Since AIR is a FLUID - it can be caused to move faster on one side of a surface than the other.
This is exactly what happens with a wing or an airfoil. The pressure on the top side of the wing’s surface
becomes lower than the pressure underneath the wing - resulting in a pressure differential
which causes “L I F T ” !
Here’s the Theory of Operation:
•
Air hits the leading edge of the wing.
•
Some of the air moves under the wing and
some of it goes over the top.
•
The air moving over the top of the curved
wing must travel farther to reach the back of the wing, so it must travel
faster than the air moving under the wing to reach the trailing edge at the same time.
•
Therefore, the air pressure on top of the wing is less than that on the bottom of the wing.
Similarly, this is the principle behind why a baseballs curve !
Faster moving air above the ball due to spin assist causes a pressure differential and deflection force
upward. The same effect occurs as fluid flows through a constricted tube.
FTBO⇒
Consider two points labeled in the pipe shown. Conservation of Mass requires the volume of fluid
entering the pipe to be the “SAME” as the volume of fluid exiting the pipe. That is the flow rate
remains constant.
Q1 = Q2
Note that as the area of constriction decreases, A↓ …
The speed of the fluid increases, v ↑ proportionally.
A1 v1 = A2 v2
A1 is large
v1 is slow
A2 is small
V2 is fast
High
Pressure
(P1)
1
2
Low Pressure
(P2)
Since the heights y1 and y2 are equal in this case the terms in Bernoulli’s equation that involve
heights will cancel, leaving us with:
P1 + ρ g (y1) + ½ ρ v12 = P2 + ρ g (y2) + ½ ρ v22
We already know from the Continuity equation (Q = Av) that as the cross-sectional area of a pipe
decreases, the speed of the fluid contained in that volume - “increases.”
Qualitatively speaking we should see that:
A1 > A2
v1 < V2
P2 =
P1 + ½ ρ v12 - ½ ρ V22
P2 =
Therefore,
⇒
P1 - ½ ρ V22 + ½ ρ v12
P1 -- ½ ρ [ V2 - v1 ]
2
p2 < p1
2
by the amount shown !
Thus,
“the pressure is lower where the flow speed is greater
…supporting Bernoulli’s Principle !
Other applications of the Bernoulli Effect include popping windows and the Venturi Meter.
AP Physics “B”
Mr. Mirro
Date: ________
Fluid Dynamics VIII
Ex 1: A horizontal pipe has a constriction in it as shown. At point 1 the diameter is 0.06 m, while at
point 2 it is only 0.02 m. At point 1, v1 is 2 m/s and P1 is 180 kPa (kilopascals).
[Schaums14.12]
a. Calculate the speed (v2)
in the smaller pipe.
1
2
b. Use Bernoulli’s equation to develop the expression for the pressure (P2) in the smaller pipe and then
compute the value in Pascals (Pa).
c. The pipes are emptied and cleaned and a different liquid is pumped in. The flow rate is maintained
and the input velocity and output velocity is still 2 m/s and 18 m/s respectively. If the pressure
differential between the two pieces of pipe is measured to be 205 Pascals, what is the density of the
new fluid ?
FTBO ⇒
Ex 3: A boat on a lake crashes into a submerged rock, ripping a 0.004 m2 hole in its hull 1.0 m below
the waterline. Assume the water flows straight in the hole, neglecting any frictional effects.
[Hecht9.11sim]
a. What two assumption(s) should you initially make to solve this problem ?
Patm
b. Use Bernoulli’s equation to derive an expression for
the speed v that the water enters the hole in terms
of the height h of the water line and constants.
h
v
b. Compute the flow speed (v) in which the water pours in.
c. How many liters (1 L = 1x10-3 m3) enter the hull per minute ?
AP Physics “B”
Mr. Mirro
Date: ________
Fluid Dynamics IX
Some of the most practical applications of fluid dynamics arise from the interdependence of pressure
and speed.
There is a class of situations in which the change in gravitational potential energy is ignorably
small and Bernoulli’s equations then relates differences in pressure to differences in kinetic
energy, and therefore S-P-E-E-D !
The “Venturi” Effect can be used to measure the flow speed of a fluid in a pipe. The meters are
connected between two sections of the pipe as shown, where the narrow part is called the “throat.”
P1
Fluid (v)
P2
V2
v1
v1
Fluid (v)
A2
A1
A1
The cross-sectional areas (A) of the entrance and exit of the meter match the pipe’s cross-sectional
area. The flow rate (Q) is constant throughout the pipe as well as the Venturi meter.
‰
Two pressure gauges are connected on the meter. One at the wider portion and the other at the
narrower portion.
‰
As the fluid flow from the pipe with speed v and into the narrow region its speed increases to V
According to Bernoulli’s equation, this increase in speed is accompanied by a decrease in the fluid’s
pressure.
‰
Finally, using Bernoulli’s equation – the speed of the fluid (v) in the pipe can be derived.
P1 + ρ g (y1) + ½ ρ v12 = P2 + ρ g (y2) + ½ ρ V22
P1 + ½ ρ v12 = P2 + ½ ρ V22
½ ρ V22 - ½ ρ v12
P1 - P2 =
P1 - P2 =
Since,
½ ρ [V22 - v12]
A1 v1 = A2 V2
[Continuity equation]
V 2 = A1 v1
A2
P1 - P2 =
½ ρ [ A1 v1
A2
2
- v12 ]
FTBO ⇒
Applying the square to each term and removing parenthesis yields:
P1 - P2 = ½ ρ [ (A1) 2 v12 - v12 ]
(A2)2
P1 - P2 =
½ ρ v12 [ (A1) 2 - 1 ]
(A2)2
We can develop the equation further in terms of either difference in fluid pressure or difference in
fluid height as follows:
P1 - P2 =
½ ρ v12 [ (A1) 2 - 1 ]
(A2)2
[ rewrite the original ]
2[P1 - P2] = v12 [ (A1) 2 - 1 ]
ρ
(A2)2
2[P1 - P2]
ρ
[ transpose the ½ ρ ]
= v12
[ isolate v12 ]
______________
[(A1) 2 - 1]
(A2)2
2 [P1 - P2]
_______________
= v12
[ simplify denominators ]
ρ [(A1) 2 - 1]
(A2)2
Solving for flow speed v1,
<i> In terms of Fluid pressure yields:
2 [P1 - P2]
v1 =
________________
ρ [(A1) 2 - 1]
(A2)2
And, since Pguage = P0 – P2 = ρ g h
<ii> In terms of Fluid height yeilds:
2 g hdiff
v1 =
___________________________
[(A1) 2 - 1]
(A2)2
AP Physics “B”
Mr. Mirro
Date: ________
Fluid Dynamics IX
Ex 1: A Venturi Meter equipped with two pressure gauges is shown. At the inlet point 1, the diameter is
0.12 m, while at the throat, point 2, the diameter is 0.06 m. The pressure differential between the
two pressure gauges is 2.4 x 104 N/m2, [Schaums14.16mod]
P1
P2
V2
v1
v1
A2
A1
A1
a. What assumption(s) could you make about the fluid as it travels through the tube from left to right ?
_____________________________________________________________________________
b. Set up the Bernoulli equation in terms of variables only and simplify it one level by applying your
assumptions.
c. Use your results from part (b) to determine the flow rate (Q) of water through the meter..
Ex 2: A cylindrical pipe has two regions labeled 1 and 2 with different radii r1 and r2. A narrow tube with
cross section A, open to the atmosphere on top, is connected to the pipe in region 2 as shown.
A
h
r1
r2
If the fluid speed and pressure are v1 and P1 respectively, in region 1, and the height of the liquid column
remains constant:
a. Determine the velocity of the fluid (v2) in region 2 in terms of velocity (v1) and radii only.
b. Express the pressure (P2) in region 2, in terms of pressure (P1), density, velocity (v1) and radii.
c. Using Newton’s 2nd Law, develop an expression for the height of the fluid (h) in the vertical tube
above the centerline in terms of the pressure differential (ΔP = P2 - Patm), gravity and density.
d. Write the equation for the height of the fluid in the narrow vertical tube in terms of the stated
quantities (ρ, g, P1, Patm, v1, r1, r2 and appropriate constants).