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

Bernoulli’s equation is a form of the conservation of energy principle. Note that the second and third terms are the kinetic and potential energy with m size 12{m} {} replaced by ρ size 12{ρ} {} . In fact, each term in the equation has units of energy per unit volume. We can prove this for the second term by substituting ρ = m / V size 12{ρ=m/V} {} into it and gathering terms:

1 2 ρv 2 = 1 2 mv 2 V = KE V . size 12{ { {1} over {2} } ρv rSup { size 8{2} } = { { { {1} over {2} } ital "mv" rSup { size 8{2} } } over {V} } = { {"KE"} over {V} } "."} {}

So 1 2 ρv 2 size 12{ { { size 8{1} } over { size 8{2} } } ρv rSup { size 8{2} } } {} is the kinetic energy per unit volume. Making the same substitution into the third term in the equation, we find

ρ gh = mgh V = PE g V , size 12{ρ ital "gh"= { { ital "mgh"} over {V} } = { {"PE" rSub { size 8{"g"} } } over {V} } "."} {}

so ρ gh size 12{ρ ital "gh"} {} is the gravitational potential energy per unit volume. Note that pressure P size 12{P} {} has units of energy per unit volume, too. Since P = F / A size 12{P=F/A} {} , its units are N/m 2 size 12{"N/m" rSup { size 8{2} } } {} . If we multiply these by m/m, we obtain N m/m 3 = J/m 3 size 12{N cdot "m/m" rSup { size 8{3} } ="J/m" rSup { size 8{3} } } {} , or energy per unit volume. Bernoulli’s equation is, in fact, just a convenient statement of conservation of energy for an incompressible fluid in the absence of friction.

Making connections: conservation of energy

Conservation of energy applied to fluid flow produces Bernoulli’s equation. The net work done by the fluid’s pressure results in changes in the fluid’s KE size 12{"KE"} {} and PE g size 12{"PE" rSub { size 8{g} } } {} per unit volume. If other forms of energy are involved in fluid flow, Bernoulli’s equation can be modified to take these forms into account. Such forms of energy include thermal energy dissipated because of fluid viscosity.

The general form of Bernoulli’s equation has three terms in it, and it is broadly applicable. To understand it better, we will look at a number of specific situations that simplify and illustrate its use and meaning.

Bernoulli’s equation for static fluids

Let us first consider the very simple situation where the fluid is static—that is, v 1 = v 2 = 0 size 12{v rSub { size 8{1} } =v rSub { size 8{2} } =0} {} . Bernoulli’s equation in that case is

P 1 + ρ gh 1 = P 2 + ρ gh 2 . size 12{P rSub { size 8{1} } +ρ ital "gh" rSub { size 8{1} } =P rSub { size 8{2} } +ρ ital "gh" rSub { size 8{2} } "."} {}

We can further simplify the equation by taking h 2 = 0 size 12{h rSub { size 8{2} } =0} {} (we can always choose some height to be zero, just as we often have done for other situations involving the gravitational force, and take all other heights to be relative to this). In that case, we get

P 2 = P 1 + ρ gh 1 . size 12{P rSub { size 8{2} } =P rSub { size 8{1} } +ρ ital "gh" rSub { size 8{1} } "."} {}

This equation tells us that, in static fluids, pressure increases with depth. As we go from point 1 to point 2 in the fluid, the depth increases by h 1 size 12{h rSub { size 8{1} } } {} , and consequently, P 2 size 12{P rSub { size 8{2} } } {} is greater than P 1 size 12{P rSub { size 8{1} } } {} by an amount ρ gh 1 size 12{ρ ital "gh" rSub { size 8{1} } } {} . In the very simplest case, P 1 size 12{P rSub { size 8{1} } } {} is zero at the top of the fluid, and we get the familiar relationship P = ρ gh size 12{P=ρ ital "gh"} {} . (Recall that P = ρgh size 12{P=hρg} {} and Δ PE g = mgh . size 12{Δ"PE" rSub { size 8{g} } = ital "mgh"} {} ) Bernoulli’s equation includes the fact that the pressure due to the weight of a fluid is ρ gh size 12{ρ ital "gh"} {} . Although we introduce Bernoulli’s equation for fluid flow, it includes much of what we studied for static fluids in the preceding chapter.

Bernoulli’s principle—bernoulli’s equation at constant depth

Another important situation is one in which the fluid moves but its depth is constant—that is, h 1 = h 2 size 12{h rSub { size 8{1} } =h rSub { size 8{2} } } {} . Under that condition, Bernoulli’s equation becomes

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} } "." } {}

Situations in which fluid flows at a constant depth are so important that this equation is often called Bernoulli’s principle    . It is Bernoulli’s equation for fluids at constant depth. (Note again that this applies to a small volume of fluid as we follow it along its path.) As we have just discussed, pressure drops as speed increases in a moving fluid. We can see this from Bernoulli’s principle. For example, if v 2 size 12{v rSub { size 8{2} } } {} is greater than v 1 size 12{v rSub { size 8{1} } } {} in the equation, then P 2 size 12{P rSub { size 8{2} } } {} must be less than P 1 size 12{P rSub { size 8{1} } } {} for the equality to hold.

Questions & Answers

how do they get the third part x = (32)5/4
kinnecy Reply
can someone help me with some logarithmic and exponential equations.
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ninjadapaul
20/(×-6^2)
Salomon
okay, so you have 6 raised to the power of 2. what is that part of your answer
ninjadapaul
I don't understand what the A with approx sign and the boxed x mean
ninjadapaul
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
ninjadapaul
oops. ignore that.
ninjadapaul
so you not have an equal sign anywhere in the original equation?
ninjadapaul
Commplementary angles
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The answer is neither. The function, 2 = 0 cannot exist. Hence, the function is undefined.
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Differences Between Laspeyres and Paasche Indices
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No. 7x -4y is simplified from 4x + (3y + 3x) -7y
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Need to simplify the expresin. 3/7 (x+y)-1/7 (x-1)=
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. After 3 months on a diet, Lisa had lost 12% of her original weight. She lost 21 pounds. What was Lisa's original weight?
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I start with an easy one. carbon nanotubes woven into a long filament like a string
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Yes, Nanotechnology has a very fast field of applications and their is always something new to do with it...
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AMJAD
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Stotaw
In this morden time nanotechnology used in many field . 1-Electronics-manufacturad IC ,RAM,MRAM,solar panel etc 2-Helth and Medical-Nanomedicine,Drug Dilivery for cancer treatment etc 3- Atomobile -MEMS, Coating on car etc. and may other field for details you can check at Google
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At high concentrations (>0.01 M), the relation between absorptivity coefficient and absorbance is no longer linear. This is due to the electrostatic interactions between the quantum dots in close proximity. If the concentration of the solution is high, another effect that is seen is the scattering of light from the large number of quantum dots. This assumption only works at low concentrations of the analyte. Presence of stray light.
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the Beer law works very well for dilute solutions but fails for very high concentrations. why?
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Source:  OpenStax, College physics (engineering physics 2, tuas). OpenStax CNX. May 08, 2014 Download for free at http://legacy.cnx.org/content/col11649/1.2
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