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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

what is temperature
Adeleye Reply
temperature is the measure of degree of hotness or coldness of a body. measured in kelvin
a characteristic which tells hotness or coldness of a body
Average kinetic energy of an object
average kinetic energy of the particles in an object
Mass of air bubble in material medium is negative. why?
Hrithik Reply
a car move 6m. what is the acceleration?
Umaru Reply
depends how long
What is the simplest explanation on the difference of principle, law and a theory
Kym Reply
how did the value of gravitational constant came give me the explanation
Varun Reply
how did the value of gravitational constant 6.67×10°-11Nm2kg-2
A steel ball is dropped onto a hard floor from a height of 1.50 m and rebounds to a height of 1.45 m. (a) Calculate its velocity just before it strikes the floor.
Kris Reply
0.5m* mate.
0.05 I meant.
Guess your solution is correct considering the ball fall from 1.5m height initially.
How can we compare different combinations of capacitors?
Prakash Reply
find the dimension of acceleration if it's unit is ms-2
Happiness Reply
b=-2 ,a =1
M^0 L^1T^-2
what is botany
it is a branch of science which deal with the study of plants animals and environment
what is work
Sunday Reply
a boy moving with an initial velocity of 2m\s and finally canes to rest with a velocity of 3m\s square at times 10se calculate it acceleration
6.6 lol 😁😁
show ur work
sorry..the answer is -10
your question is wrong
If the boy is coming to rest then how the hell will his final velocity be 3 it'll be zero
re-write the question
men i -10 isn't correct.
using v=u + at
ya..1/10 is very correct..
how did the value 6.67×10°-11Nm2kg2 came tell me please
Work is the product of force and distance
what is longitudinal wave
Badmus Reply
A longitudinal wave is wave which moves parallel or along the direction of propagation.
longitudinal wave in liquid is square root of bulk of modulus by density of liquid
Is British mathematical units the same as the United States units?(like inches, cm, ext.)
Nina Reply
We use SI units: kg, m etc but the US sometimes refer to inches etc as British units even though we no longer use them.
Thanks, just what I needed to know.
What is the advantage of a diffraction grating over a double slit in dispersing light into a spectrum?
Uditha Reply
can I ask questions?
Boniface Reply
hello guys
when you will ask the question
anybody can ask here
is free energy possible with magnets?
you could construct an aparatus that might have a slightly higher 'energy profit' than energy used, but you would havw to maintain the machine, and most likely keep it in a vacuum, for no air resistance, and cool it, so chances are quite slim.
calculate the force, p, required to just make a 6kg object move along the horizontal surface where the coefficient of friction is 0.25
Yes ask
if a man travel 7km 30degree east of North then 10km east find the resultant displacement
Ajali Reply
disagree. Displacement is the hypotenuse length of the final position to the starting position. Find x,y components of each leg of journey to determine final position, then use final components to calculate the displacement.
1.The giant star Betelgeuse emits radiant energy at a rate of 10exponent4 times greater than our sun, where as it surface temperature is only half (2900k) that of our sun. Estimate the radius of Betelgeuse assuming e=1, the sun's radius is s=7*10exponent8metres
James Reply
2. A ceramic teapot (e=0.20) and a shiny one (e=0.10), each hold 0.25 l of at 95degrees. A. Estimate the temperature rate of heat loss from each B. Estimate the temperature drop after 30mins for each. Consider only radiation and assume the surrounding at 20degrees
Practice Key Terms 2

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