1. Home
  2. College physics
  3. Fluid dynamics and its biological
  4. The most general applications

12.3 The most general applications of bernoulli’s equation

<< Chapter < Page
  College physics     Page 1 / 2
Chapter >> Page >
  • Calculate using Torricelli’s theorem.
  • Calculate power in fluid flow.

Torricelli’s theorem

[link] shows water gushing from a large tube through a dam. What is its speed as it emerges? Interestingly, if resistance is negligible, the speed is just what it would be if the water fell a distance h size 12{h} {} from the surface of the reservoir; the water’s speed is independent of the size of the opening. Let us check this out. Bernoulli’s equation must be used since the depth is not constant. We consider water flowing from the surface (point 1) to the tube’s outlet (point 2). Bernoulli’s equation as stated in previously is

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} } } {}

Both P 1 size 12{P rSub { size 8{1} } } {} and P 2 size 12{P rSub { size 8{2} } } {} equal atmospheric pressure ( P 1 size 12{P rSub { size 8{1} } } {} is atmospheric pressure because it is the pressure at the top of the reservoir. P 2 size 12{P rSub { size 8{2} } } {} must be atmospheric pressure, since the emerging water is surrounded by the atmosphere and cannot have a pressure different from atmospheric pressure.) and subtract out of the equation, leaving

1 2 ρv 1 2 + ρ gh 1 = 1 2 ρv 2 2 + ρ gh 2 . size 12{ { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{1} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{1} } = { { size 8{1} } over { size 8{2} } } ρv rSub { size 8{2} } rSup { size 8{2} } +ρ ital "gh" rSub { size 8{2} } } {}

Solving this equation for v 2 2 size 12{v rSub { size 8{2} } rSup { size 8{2} } } {} , noting that the density ρ cancels (because the fluid is incompressible), yields

v 2 2 = v 1 2 + 2 g ( h 1 − h 2 ) . size 12{v rSub { size 8{2} } rSup { size 8{2} } =v rSub { size 8{1} } rSup { size 8{2} } +2g \( h rSub { size 8{1} } - h rSub { size 8{2} } \) } {}

We let h = h 1 − h 2 size 12{h=h rSub { size 8{1} } - h rSub { size 8{2} } } {} ; the equation then becomes

v 2 2 = v 1 2 + 2 gh size 12{v rSub { size 8{2} } rSup { size 8{2} } =v rSub { size 8{1} } rSup { size 8{2} } +2 ital "gh"} {}

where h size 12{h} {} is the height dropped by the water. This is simply a kinematic equation for any object falling a distance h size 12{h} {} with negligible resistance. In fluids, this last equation is called Torricelli’s theorem . Note that the result is independent of the velocity’s direction, just as we found when applying conservation of energy to falling objects.

Part a of the figure shows a photograph of a dam with water gushing from a large tube at the base of a dam. Part b shows the schematic diagram for the flow of water in a reservoir. The reservoir is shown in the form of a triangular section with a horizontal opening along the base little near to the base. The water is shown to flow through the horizontal opening near the base. The height which it falls is shown as h two. The pressure and velocity of water at this point are P two and v two. The height to which the water can fall if it falls from a height h above the opening is given by h 2. The pressure and velocity of water at this point are P one and v one.
(a) Water gushes from the base of the Studen Kladenetz dam in Bulgaria. (credit: Kiril Kapustin; http://www.ImagesFromBulgaria.com) (b) In the absence of significant resistance, water flows from the reservoir with the same speed it would have if it fell the distance h size 12{h} {} without friction. This is an example of Torricelli’s theorem.
Figure shows a fire engine that is stationed next to a tall building. A floor of the building ten meters above the ground has caught fire. The flames are shown coming out. A fire man has reached close to the fire caught area using a ladder and is spraying water on the fire using a hose attached to the fire engine.
Pressure in the nozzle of this fire hose is less than at ground level for two reasons: the water has to go uphill to get to the nozzle, and speed increases in the nozzle. In spite of its lowered pressure, the water can exert a large force on anything it strikes, by virtue of its kinetic energy. Pressure in the water stream becomes equal to atmospheric pressure once it emerges into the air.

All preceding applications of Bernoulli’s equation involved simplifying conditions, such as constant height or constant pressure. The next example is a more general application of Bernoulli’s equation in which pressure, velocity, and height all change. (See [link] .)

Calculating pressure: a fire hose nozzle

Fire hoses used in major structure fires have inside diameters of 6.40 cm. Suppose such a hose carries a flow of 40.0 L/s starting at a gauge pressure of 1 . 62 × 10 6 N/m 2 size 12{1 "." "62" times "10" rSup { size 8{6} } `"N/m" rSup { size 8{2} } } {} . The hose goes 10.0 m up a ladder to a nozzle having an inside diameter of 3.00 cm. Assuming negligible resistance, what is the pressure in the nozzle?

Strategy

Here we must use Bernoulli’s equation to solve for the pressure, since depth is not constant.

Solution

Bernoulli’s equation states

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} } } {}

where the subscripts 1 and 2 refer to the initial conditions at ground level and the final conditions inside the nozzle, respectively. We must first find the speeds v 1 size 12{v rSub { size 8{1} } } {} and v 2 size 12{v rSub { size 8{2} } } {} . Since Q = A 1 v 1 size 12{Q=A rSub { size 8{1} } v"" lSub { size 8{1} } } {} , we get

v 1 = Q A 1 = 40 . 0 × 10 − 3 m 3 /s π ( 3 . 20 × 10 − 2 m ) 2 = 12 . 4 m/s . size 12{v rSub { size 8{1} } = { {Q} over {A rSub { size 8{1} } } } = { {"40" "." 0 times "10" rSup { size 8{ - 3} } " m" rSup { size 8{3} } "/s"} over {π \( 3 "." "20" times "10" rSup { size 8{ - 2} } " m" \) rSup { size 8{2} } } } ="12" "." 4" m/s"} {}

Similarly, we find

v 2 = 56.6 m/s . size 12{v rSub { size 8{2} } ="56" "." 6" m/s"} {}

(This rather large speed is helpful in reaching the fire.) Now, taking h 1 size 12{h rSub { size 8{1} } } {} to be zero, we solve Bernoulli’s equation for P 2 size 12{P rSub { size 8{2} } } {} :

P 2 = P 1 + 1 2 ρ v 1 2 − v 2 2 − ρ gh 2 . size 12{P rSub { size 8{2} } =P rSub { size 8{1} } + { {1} over {2} } ρ \( v rSub { size 8{1} rSup { size 8{2} } } - v rSub { size 8{2} rSup { size 8{2} } } \) - ρ ital "gh" rSub { size 8{2} } } {}

Substituting known values yields

P 2 = 1 . 62 × 10 6 N/m 2 + 1 2 ( 1000 kg/m 3 ) ( 12 . 4 m/s ) 2 − ( 56 . 6 m/s ) 2 − ( 1000 kg/m 3 ) ( 9 . 80 m/s 2 ) ( 10 . 0 m ) = 0 . size 12{P rSub { size 8{2} } =1 "." "62" times "10" rSup { size 8{6} } " N/m" rSup { size 8{2} } + { {1} over {2} } \( "1000"" kg/m" rSup { size 8{3} } \) left [ \( "12" "." 4" m/s" \) rSup { size 8{2} } - \( "56" "." 6" m/s" \) rSup { size 8{2} } right ] - \( "1000"" kg/m" rSup { size 8{3} } \) \( 9 "." 8" m/s" rSup { size 8{2} } \) \( "10" "." 0" m" \) =0} {}

Discussion

This value is a gauge pressure, since the initial pressure was given as a gauge pressure. Thus the nozzle pressure equals atmospheric pressure, as it must because the water exits into the atmosphere without changes in its conditions.

Got questions? Get instant answers now!

Questions & Answers

A golfer on a fairway is 70 m away from the green, which sits below the level of the fairway by 20 m. If the golfer hits the ball at an angle of 40° with an initial speed of 20 m/s, how close to the green does she come?
Aislinn Reply
cm
tijani
what is titration
John Reply
what is physics
Siyaka Reply
A mouse of mass 200 g falls 100 m down a vertical mine shaft and lands at the bottom with a speed of 8.0 m/s. During its fall, how much work is done on the mouse by air resistance
Jude Reply
Can you compute that for me. Ty
Jude
what is the dimension formula of energy?
David Reply
what is viscosity?
David
what is inorganic
emma Reply
what is chemistry
Youesf Reply
what is inorganic
emma
Chemistry is a branch of science that deals with the study of matter,it composition,it structure and the changes it undergoes
Adjei
please, I'm a physics student and I need help in physics
Adjanou
chemistry could also be understood like the sexual attraction/repulsion of the male and female elements. the reaction varies depending on the energy differences of each given gender. + masculine -female.
Pedro
A ball is thrown straight up.it passes a 2.0m high window 7.50 m off the ground on it path up and takes 1.30 s to go past the window.what was the ball initial velocity
Krampah Reply
2. A sled plus passenger with total mass 50 kg is pulled 20 m across the snow (0.20) at constant velocity by a force directed 25° above the horizontal. Calculate (a) the work of the applied force, (b) the work of friction, and (c) the total work.
Sahid Reply
you have been hired as an espert witness in a court case involving an automobile accident. the accident involved car A of mass 1500kg which crashed into stationary car B of mass 1100kg. the driver of car A applied his brakes 15 m before he skidded and crashed into car B. after the collision, car A s
Samuel Reply
can someone explain to me, an ignorant high school student, why the trend of the graph doesn't follow the fact that the higher frequency a sound wave is, the more power it is, hence, making me think the phons output would follow this general trend?
Joseph Reply
Nevermind i just realied that the graph is the phons output for a person with normal hearing and not just the phons output of the sound waves power, I should read the entire thing next time
Joseph
Follow up question, does anyone know where I can find a graph that accuretly depicts the actual relative "power" output of sound over its frequency instead of just humans hearing
Joseph
"Generation of electrical energy from sound energy | IEEE Conference Publication | IEEE Xplore" ***ieeexplore.ieee.org/document/7150687?reload=true
Ryan
what's motion
Maurice Reply
what are the types of wave
Maurice
answer
Magreth
progressive wave
Magreth
hello friend how are you
Muhammad Reply
fine, how about you?
Mohammed
hi
Mujahid
A string is 3.00 m long with a mass of 5.00 g. The string is held taut with a tension of 500.00 N applied to the string. A pulse is sent down the string. How long does it take the pulse to travel the 3.00 m of the string?
yasuo Reply
Who can show me the full solution in this problem?
Reofrir Reply
Got questions? Join the online conversation and get instant answers!
Jobilize.com Reply
<< Chapter < Page Page > Chapter >>

Read also:

Get Jobilize Job Search Mobile App in your pocket Now!

Get it on Google Play Download on the App Store Now




Source:  OpenStax, College physics. OpenStax CNX. Jul 27, 2015 Download for free at http://legacy.cnx.org/content/col11406/1.9
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'College physics' conversation and receive update notifications?

Ask