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Learning objectives

By the end of this section, you will be able to:

  • Calculate using Torricelli's theorem.
  • Calculate power in fluid flow.

The information presented in this section supports the following AP® learning objectives and science practices:

  • 5.B.10.2 The student is able to use Bernoulli's equation and/or the relationship between force and pressure to make calculations related to a moving fluid. (S.P. 2.2)
  • 5.B.10.3 The student is able to use Bernoulli's equation and the continuity equation to make calculations related to a moving fluid. (S.P. 2.2)

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.

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Questions & Answers

If a prism is fully imersed in water then the ray of light will normally dispersed or their is any difference?
Anurag Reply
the same behavior thru the prism out or in water bud abbot
Ju
If this will experimented with a hollow(vaccum) prism in water then what will be result ?
Anurag
What was the previous far point of a patient who had laser correction that reduced the power of her eye by 7.00 D, producing a normal distant vision power of 50.0 D for her?
Jaydie Reply
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
What is the far point of a person whose eyes have a relaxed power of 50.5 D?
Jaydie
A young woman with normal distant vision has a 10.0% ability to accommodate (that is, increase) the power of her eyes. What is the closest object she can see clearly?
Jaydie
29/20 ? maybes
Ju
In what ways does physics affect the society both positively or negatively
Princewill Reply
how can I read physics...am finding it difficult to understand...pls help
rerry Reply
try to read several books on phy don't just rely one. some authors explain better than other.
Ju
And don't forget to check out YouTube videos on the subject. Videos offer a different visual way to learn easier.
Ju
hope that helps
Ju
I have a exam on 12 february
David Reply
what is velocity
Jiti
the speed of something in a given direction.
Ju
what is a magnitude in physics
Jiti Reply
Propose a force standard different from the example of a stretched spring discussed in the text. Your standard must be capable of producing the same force repeatedly.
Giovani Reply
What is meant by dielectric charge?
It's Reply
what happens to the size of charge if the dielectric is changed?
Brhanu Reply
omega= omega not +alpha t derivation
Provakar Reply
u have to derivate it respected to time ...and as w is the angular velocity uu will relace it with "thita × time""
Abrar
do to be peaceful with any body
Brhanu Reply
the angle subtended at the center of sphere of radius r in steradian is equal to 4 pi how?
Saeed Reply
if for diatonic gas Cv =5R/2 then gamma is equal to 7/5 how?
Saeed
define variable velocity
Ali Reply
displacement in easy way.
Mubashir Reply
binding energy per nucleon
Poonam Reply

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Source:  OpenStax, College physics for ap® courses. OpenStax CNX. Nov 04, 2016 Download for free at https://legacy.cnx.org/content/col11844/1.14
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