1. Home
  2. Physics 101
  3. Fluids
  4. The most general applications

10.10 The most general applications of bernoulli’s equation  (Page 43/2)

<< Chapter < Page
  Physics 101     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.

Questions & Answers

the definition for anatomy and physiology
Watta Reply
what is microbiology
Agebe Reply
What is a cell
Odelana Reply
what is cell
Mohammed
how does Neisseria cause meningitis
Nyibol Reply
what is microbiologist
Muhammad Reply
what is errata
Muhammad
is the branch of biology that deals with the study of microorganisms.
Ntefuni Reply
What is microbiology
Mercy Reply
studies of microbes
Louisiaste
when we takee the specimen which lumbar,spin,
Ziyad Reply
How bacteria create energy to survive?
Muhamad Reply
Bacteria doesn't produce energy they are dependent upon their substrate in case of lack of nutrients they are able to make spores which helps them to sustain in harsh environments
_Adnan
But not all bacteria make spores, l mean Eukaryotic cells have Mitochondria which acts as powerhouse for them, since bacteria don't have it, what is the substitution for it?
Muhamad
they make spores
Louisiaste
what is sporadic nd endemic, epidemic
Aminu Reply
the significance of food webs for disease transmission
Abreham
food webs brings about an infection as an individual depends on number of diseased foods or carriers dully.
Mark
explain assimilatory nitrate reduction
Esinniobiwa Reply
Assimilatory nitrate reduction is a process that occurs in some microorganisms, such as bacteria and archaea, in which nitrate (NO3-) is reduced to nitrite (NO2-), and then further reduced to ammonia (NH3).
Elkana
This process is called assimilatory nitrate reduction because the nitrogen that is produced is incorporated in the cells of microorganisms where it can be used in the synthesis of amino acids and other nitrogen products
Elkana
Examples of thermophilic organisms
Shu Reply
Give Examples of thermophilic organisms
Shu
advantages of normal Flora to the host
Micheal Reply
Prevent foreign microbes to the host
Abubakar
they provide healthier benefits to their hosts
ayesha
They are friends to host only when Host immune system is strong and become enemies when the host immune system is weakened . very bad relationship!
Mark
what is cell
faisal Reply
cell is the smallest unit of life
Fauziya
cell is the smallest unit of life
Akanni
ok
Innocent
cell is the structural and functional unit of life
Hasan
is the fundamental units of Life
Musa
what are emergency diseases
Micheal Reply
There are nothing like emergency disease but there are some common medical emergency which can occur simultaneously like Bleeding,heart attack,Breathing difficulties,severe pain heart stock.Hope you will get my point .Have a nice day ❣️
_Adnan
define infection ,prevention and control
Innocent
I think infection prevention and control is the avoidance of all things we do that gives out break of infections and promotion of health practices that promote life
Lubega
Heyy Lubega hussein where are u from?
_Adnan
en français
Adama
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, Physics 101. OpenStax CNX. Jan 07, 2013 Download for free at http://legacy.cnx.org/content/col11479/1.1
Google Play and the Google Play logo are trademarks of Google Inc.

Notification Switch

Would you like to follow the 'Physics 101' conversation and receive update notifications?

Ask