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

Ayele, K., 2003. Introductory Economics, 3rd ed., Addis Ababa.
Widad Reply
can you send the book attached ?
Ariel
?
Ariel
What is economics
Widad Reply
the study of how humans make choices under conditions of scarcity
AI-Robot
U(x,y) = (x×y)1/2 find mu of x for y
Desalegn Reply
U(x,y) = (x×y)1/2 find mu of x for y
Desalegn
what is ecnomics
Jan Reply
this is the study of how the society manages it's scarce resources
Belonwu
what is macroeconomic
John Reply
macroeconomic is the branch of economics which studies actions, scale, activities and behaviour of the aggregate economy as a whole.
husaini
etc
husaini
difference between firm and industry
husaini Reply
what's the difference between a firm and an industry
Abdul
firm is the unit which transform inputs to output where as industry contain combination of firms with similar production 😅😅
Abdulraufu
Suppose the demand function that a firm faces shifted from Qd  120 3P to Qd  90  3P and the supply function has shifted from QS  20  2P to QS 10  2P . a) Find the effect of this change on price and quantity. b) Which of the changes in demand and supply is higher?
Toofiq Reply
explain standard reason why economic is a science
innocent Reply
factors influencing supply
Petrus Reply
what is economic.
Milan Reply
scares means__________________ends resources. unlimited
Jan
economics is a science that studies human behaviour as a relationship b/w ends and scares means which have alternative uses
Jan
calculate the profit maximizing for demand and supply
Zarshad Reply
Why qualify 28 supplies
Milan
what are explicit costs
Nomsa Reply
out-of-pocket costs for a firm, for example, payments for wages and salaries, rent, or materials
AI-Robot
concepts of supply in microeconomics
David Reply
economic overview notes
Amahle Reply
identify a demand and a supply curve
Salome Reply
i don't know
Parul
there's a difference
Aryan
Demand curve shows that how supply and others conditions affect on demand of a particular thing and what percent demand increase whith increase of supply of goods
Israr
Hi Sir please how do u calculate Cross elastic demand and income elastic demand?
Abari
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