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Making connections: take-home investigation with two strips of paper

For a good illustration of Bernoulli’s principle, make two strips of paper, each about 15 cm long and 4 cm wide. Hold the small end of one strip up to your lips and let it drape over your finger. Blow across the paper. What happens? Now hold two strips of paper up to your lips, separated by your fingers. Blow between the strips. What happens?

Velocity measurement

[link] shows two devices that measure fluid velocity based on Bernoulli’s principle. The manometer in [link] (a) is connected to two tubes that are small enough not to appreciably disturb the flow. The tube facing the oncoming fluid creates a dead spot having zero velocity ( v 1 = 0 size 12{v rSub { size 8{1} } =0} {} ) in front of it, while fluid passing the other tube has velocity v 2 size 12{v rSub { size 8{2} } } {} . This means that Bernoulli’s principle as stated in P 1 + 1 2 ρv 1 2 = P 2 + 1 2 ρv 2 2 size 12{P rSub { size 8{1} } + { {1} over {2} } ρv rSub { size 8{1} } "" lSup { size 8{2} } =P rSub { size 8{2} } + { {1} over {2} } ρv rSub { size 8{2} } "" lSup { size 8{2} } } {} becomes

P 1 = P 2 + 1 2 ρv 2 2 . size 12{P rSub { size 8{1} } =P rSub { size 8{2} } + { {1} over {2} } ρv rSub { size 8{2} } "" lSup { size 8{2} } "." } {}
Part a of the figure shows a picture of a wing. It is in the form of an aerofoil. One side of the wing is broader and the other end tapers. The direction of the air is shown as lines along the length of the wing. The direction of the air below the wing is shown as flowing along the length of the wing. The pressure exerted by the air given by P b is upward. The direction of the air on the top or front part of the wing is shown as flowing along the length of the wing. The pressure exerted by the air is given by P f, and it acts downward. Part b of the figure shows a boat with a sail. The direction of the sail is almost across the boat. The direction of the air in the sail is shown by lines on the front and back sides of the sail. The air currents on the front exert a pressure P front toward the sail, and air currents on the back sides of sail exert a pressure P back again toward the sail.
(a) The Bernoulli principle helps explain lift generated by a wing. (b) Sails use the same technique to generate part of their thrust.

Thus pressure P 2 size 12{P rSub { size 8{2} } } {} over the second opening is reduced by 1 2 ρv 2 2 size 12{ { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } } {} , and so the fluid in the manometer rises by h on the side connected to the second opening, where

h 1 2 ρv 2 2 . size 12{h prop { {1} over {2} } ρv rSub { size 8{2} } rSup { size 8{2} } "."} {}

(Recall that the symbol size 12{ prop } {} means “proportional to.”) Solving for v 2 size 12{v rSub { size 8{2} } } {} , we see that

v 2 h . size 12{v rSub { size 8{2} } prop sqrt {h} "."} {}

[link] (b) shows a version of this device that is in common use for measuring various fluid velocities; such devices are frequently used as air speed indicators in aircraft.

Part a shows a U-shaped manometer tube connected to ends of two tubes which are placed close together. Tube one is open on the end and shows a velocity v one equals zero at the end. Tube two has an opening on the side and shows a velocity v two across the opening. The level of fluid in the U-shaped tube is more on the right side than on the left. The difference in height is shown by h. Part b of the figure shows a velocity measuring device a pitot tube. Two coaxial tubes, one broader outside and other narrow inside are connected to a U-shaped tube. The U-shaped tube is also narrow at one end and broader at the other. The narrow end of the U-shaped tube is connected to the narrow inner tube and the broader end of the U-shaped tube is connected to the broader outer tube. The tube one has an opening at one of its edges and the velocity of the fluid at the end is v one equals zero. Tube two has an opening on the side and shows a velocity v two across the opening. The level of fluid in the U-shaped tube is more on the right side than on the left. The difference in height is shown by h.
Measurement of fluid speed based on Bernoulli’s principle. (a) A manometer is connected to two tubes that are close together and small enough not to disturb the flow. Tube 1 is open at the end facing the flow. A dead spot having zero speed is created there. Tube 2 has an opening on the side, and so the fluid has a speed v across the opening; thus, pressure there drops. The difference in pressure at the manometer is 1 2 ρv 2 2 size 12{ { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } } {} , and so h is proportional to 1 2 ρv 2 2 size 12{ { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } } {} . (b) This type of velocity measuring device is a Prandtl tube, also known as a pitot tube.

Summary

  • Bernoulli’s equation states that the sum on each side of the following equation is constant, or the same at any two points in an incompressible frictionless fluid:
    P 1 + 1 2 ρv 1 2 + ρ gh 1 = P 2 + 1 2 ρv 2 2 + ρ gh 2 . size 12{P rSub { size 8{1} } + { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{1} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{1} } =P rSub { size 8{2} } + { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{2} } } {}
  • Bernoulli’s principle is Bernoulli’s equation applied to situations in which depth is constant. The terms involving depth (or height h ) subtract out, yielding
    P 1 + 1 2 ρv 1 2 = P 2 + 1 2 ρv 2 2 . size 12{P rSub { size 8{1} } + { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{1} } rSup { size 8{2} } =P rSub { size 8{2} } + { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } } {}
  • Bernoulli’s principle has many applications, including entrainment, wings and sails, and velocity measurement.

Conceptual questions

You can squirt water a considerably greater distance by placing your thumb over the end of a garden hose and then releasing, than by leaving it completely uncovered. Explain how this works.

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Water is shot nearly vertically upward in a decorative fountain and the stream is observed to broaden as it rises. Conversely, a stream of water falling straight down from a faucet narrows. Explain why, and discuss whether surface tension enhances or reduces the effect in each case.

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Look back to [link] . Answer the following two questions. Why is P o size 12{P rSub { size 8{o} } } {} less than atmospheric? Why is P o size 12{P rSub { size 8{o} } } {} greater than P i size 12{P rSub { size 8{i} } } {} ?

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Questions & Answers

Why is there no 2nd harmonic in the classical electron orbit?
Shree Reply
how to reform magnet after been demagneted
Inuwa Reply
A petrol engine has a output of 20 kilowatts and uses 4.5 kg of fuel for each hour of running. The energy given out when 1 kg of petrol is burnt is 4.8 × 10 to the power of 7 Joules. a) What is the energy output of the engine every hour? b) What is the energy input of the engine every hour?
Morris Reply
what is the error during taking work done of a body..
Aliyu Reply
what kind of error do you think? and work is held by which force?
Daniela
I am now in this group
smart
theory,laws,principles and what-a-view are not defined. why? you
Douglas Reply
A simple pendulum is used in a physics laboratory experiment to obtain an experimental value for the gravitational acceleration, g . A student measures the length of the pendulum to be 0.510 meters, displaces it 10 o from the equilibrium position, and releases it. Using a s
Emmanuel Reply
so what question are you passing across... sir?
Olalekan
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed?
Emmanuel Reply
54 joule
babar
how?
rakesh
Reduce that two body problem into one body problem. Apply potential and k. E formula to get total energy of the system
rakesh
i dont think dere is any potential energy... by d virtue of no height present
Olalekan
there is compressed energy,dats only potential energy na?
rakesh
yes.. but... how will u approach that question without The Height in the question?
Olalekan
Can you explain how you get 54J?
Emmanuel
Because mine is 36J
Emmanuel
got 36J too
Douglas
OK the answer is 54J Babar is correct
Emmanuel
Conservation of Momentum
Emmanuel
woow i see.. can you give the formula for this
joshua
Two masses of 2 kg and 4 kg are held with a compressed spring between them. If the masses are released, the spring will push them away from each other. If the smaller mass moves off with a velocity of 6m/s, what is the stored energy in the spring when it is compressed? Asume there is no external force.
Emmanuel Reply
Please help!
Emmanuel
please help find dy/dx 2x-y/x+y
Inuwa
By using the Quotient Rule dy/dx = 3y/(x +y)²
Emmanuel
3y/(x+y)²
Emmanuel
may be by using MC^2=MC^2 and Total energy=kinetic energy +potential energy so 1st find kinetic energy and den find potential energy which is stored energy
rakesh
i think i m correct
rakesh
But how?
Emmanuel
3y/(x+y)²
Douglas
what's the big bang?
kwame Reply
yes what is it?
LamaBbake
it is the explanation of how the universe began
Zainab
yes
Ana
explain
Chinagorom
in
Chinagorom
it is a theory on how the universe began. to understand more I would suggest researching the topic online.
david
thanks guys
kwame
if a force of 12N is applied to load of 200g what us the work done
Joshua Reply
We can seek accelation first
Nancy
we are given f=12 m=200g which is 0.2kg now from 2nd law of newton a= f/m=60m/s*2 work done=force applied x displacement cos (theta) w= 12x60 =720nm/s*2
Mudang
this very interesting question very complicated for me, í need urgent help. 1,two buses A and B travel along the same road in the same direction from Harper city (asume They both started from the same point) to Monrovia. if bus A maintains a Speedy of 60km/h and bus B a Speedy of 75km/h, how many
mohammed
hours Will it take bus B to overtake bus A assuming bus B starts One hour after bus A started. what is the distance travelled by the buses when They meet?.
mohammed
pls í need help
mohammed
4000 work is done
Ana
speed=distance /time distance=speed/time
Ana
now use this formula
Ana
what's the answer then
Julius
great Mudang
Kossi
please Ana explain 4000 ?
babar
hey mudang there is a product of force and acceleration not force and displacement
babar
@Mohammed answer is 0.8hours or 48mins
Douglas
nice
A.d
its not possible
Olalekan
í want the working procedure
mohammed
the answer is given but how Will One arrive at it. the answers are 4hours and 300m.
mohammed
physics is the science that studies the non living nature
isidor Reply
ancient greek language physis = nature
isidor
what is phyacs
technical Reply
if i am going to start studying physics where should i start?
BRIAN Reply
I think from kinematics
Nancy
You can find physics books at the library or online. That's how I started.
Chelsea
And yes, kinematics is usually where you can begin.
Chelsea
study basic algebra and calculus and can start from classical mechanics
Mudang
yes think so but dimension is the best starting point
Obed
3 formula's of equations of motion
benjamin Reply
vf=vi+at........1 s=vit+1/2(at)2 vf2=vi2+2as
Ana
solve the formula's please
benjamin
those are the three .. what you wanna solve ?
Nihrantz
For first equation simply integrate formula of acceleration in the limit v and u
Tripti
For second itegrate velocity formula by ising first equation
Tripti
similarly for 3 one integrate acceleration again by multiplying and dividing term ds
Tripti
any methods can take to solve this eqtions
a=vf-vi/t vf-vi=at vf=vi+at......1
Ana
suppose a body starts with an initial velocity vi and travels with uniform acceleration a for a period of time t.the distance covered by a body in this time is "s" and its final velocity becomes vf
Ana
what is the question dear
Zeeshan
average velocity=(vi+vf)/2 distance travelled=average velocity ×time therefore s=vi+vf/2×t from the first equation of motion ,we have vf =vi+at s=[vi+(vi+at)]/2×t s=(2vi+at)/2×t s=bit+1/2at2
Ana
find the distance
Ana
how
Zeeshan
Two speakers are arranged so that sound waves with the same frequency are produced and radiated through a room. An interference pattern is created. Calculate the distance between the two speakers?
Hayne Reply
How can we calculate without any information?
Amir
I think the formulae used for this question is lambda=(ax)/D
Amir
Practice Key Terms 2

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Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
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