# Deriving Bernoullis equation

To derive Bernoulli’s equation, we apply the work–energy theorem to the fluid in a section of a flow tube. In Fig. 12.23 we consider the element of fluid that at some initial time lies between the two cross sections a and c. The speeds at the lower and upper ends are v1 and v2. In a small time interval dt, the fluid that is initially at a moves to b, a distance ds1 = v1 dt, and the fluid that is initially at c moves to d, a distance ds2 = v2 dt. The cross-sectional areas at the two ends are A1 and A2, as shown. The fluid is incompressible; hence by the continuity equation, Eq. (12.10), the volume of fluid dV passing any cross section during time dt is the same. That is, dV = A1 ds1 = A2 ds2.

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Let’s compute the work done on this fluid element during dt. If there is negligible internal friction in the fluid (i.e., no viscosity), the only nongravitational forces that do work on the element are due to the pressure of the surrounding fluid. The pressures at the two ends are p1 and p2; the force on the cross section at a is p1A1, and the force at c is p2A2. The net work dW done on the element by the surrounding fluid during this displacement is therefore

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The term p2A2ds2 has a negative sign because the force at c opposes the displacement of the fluid.

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The work dW is due to forces other than the conservative force of gravity, so it equals the change in the total mechanical energy (kinetic energy plus gravitational potential energy) associated with the fluid element. The mechanical energy for the fluid between sections b and c does not change. At the beginning of dt the fluid between a and b has volume A1 ds1, mass rA1 ds1, and kinetic energy 1/ 2 r(A1 ds12v1 ). At the end of dt the fluid between c and d has kinetic energy 1/ 2 r1A2 ds2v2 2. The net change in kinetic energy dK during time dt is

What about the change in gravitational potential energy? At the beginning
of time interval dt, the potential energy for the mass between a and b is
dm gy1 = r dV gy1. At the end of dt, the potential energy for the mass between c
and d is dm gy2 = r dV gy2. The net change in potential energy dU during dt is

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Combining Eqs. (12.13), (12.14), and (12.15) in the energy equation dW = dK + dU, we obtain

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This is Bernoulli’s equation. It states that the work done on a unit volume of fluid by the surrounding fluid is equal to the sum of the changes in kinetic and potential energies per unit volume that occur during the flow. We may also interpret Eq. (12.16) in terms of pressures. The first term on the right is the pressure difference associated with the change of speed of the fluid. The second term on the right is the additional pressure difference caused by the weight of the fluid and the difference in elevation of the two ends

We can also express Eq. (12.16) in a more convenient form as

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Subscripts 1 and 2 refer to any two points along the flow tube, so we can write

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Note that when the fluid is not moving (so v1 = v2 = 0), Eq. (12.17) reduces to
the pressure relationship we derived for a fluid at rest, Eq. (12.5).

Caution bernoulli’s equation applies in certain situations only We stress again that
Bernoulli’s equation is valid for only incompressible, steady flow of a fluid with no internal
friction (no viscosity). It’s a simple equation, but don’t be tempted to use it in situations
Demo in which it doesn’t apply!

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Ans: According to the continuity equation, the speed of fluid flow can vary along the paths of the fluid. The pressure can also vary; it depends on height as in the static situation (see Section 12.2), and it also depends on the speed of flow. We can derive an important relationship called Bernoulli’s equation, view more..
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Ans: The mass of a moving fluid doesn’t change as it flows. This leads to an important relationship called the continuity equation view more..
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Ans: We are now ready to consider motion of a fluid. Fluid flow can be extremely complex, as shown by the currents in river rapids or the swirling flames of a campfire. But we can represent some situations by relatively simple idealized models. An ideal fluid is a fluid that is incompressible (that is, its density cannot change) and has no internal friction (called viscosity). view more..
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Ans: To derive Bernoulli’s equation, we apply the work–energy theorem to the fluid in a section of a flow tube. In Fig. 12.23 we consider the element of fluid that at some initial time lies between the two cross sections a and c. The speeds at the lower and upper ends are v1 and v2. In a small time interval dt, the fluid that is initially at a moves to b, a distance ds1 = v1 dt, and the fluid that is initially at c moves to d, a distance ds2 = v2 dt. The cross-sectional areas at the two ends are A1 and A2, as shown. The fluid is incompressible; hence by the continuity equation, Eq. (12.10), the volume of fluid dV passing any cross section during time dt is the same. That is, dV = A1 ds1 = A2 ds2. view more..
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Ans: HERE ARE SOME EXAMPLES TO DEAL WITH view more..
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Ans: Viscosity is internal friction in a fluid. Viscous forces oppose the motion of one portion of a fluid relative to another. Viscosity is the reason it takes effort to paddle a canoe through calm water, but it is also the reason the paddle works. Viscous effects are important in the flow of fluids in pipes, the flow of blood, the lubrication of engine parts, and many other situations view more..
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Ans: When the speed of a flowing fluid exceeds a certain critical value, the flow is no longer laminar. Instead, the flow pattern becomes extremely irregular and complex, and it changes continuously with time; there is no steady-state pattern. This irregular, chaotic flow is called turbulence view more..
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Ans: SUMMARY OF EVERY TOPIC OF FLUID MECHANISM. view more..
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Ans: Every particle of matter in the universe attracts every other particle with a force that is directly proportional to the product of the masses of the particles and inversely proportional to the square of the distance between them. view more..
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Ans: We have stated the law of gravitation in terms of the interaction between two particles. It turns out that the gravitational interaction of any two bodies having spherically symmetric mass distributions view more..
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Ans: To determine the value of the gravitational constant G, we have to measure the gravitational force between two bodies of known masses m1 and m2 at a known distance r. The force is extremely small for bodies that are small enough to be brought into the laboratory, but it can be measured with an instrument called a torsion balance, which Sir Henry Cavendish used in 1798 to determine G. view more..
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Ans: HERE ARE SOME SOLVED EXAMPLES TO CLEAR YOUR CONCEPTS view more..
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Ans: gravitational forces are negligible between ordinary household-sized objects but very substantial between objects that are the size of stars. Indeed, gravitation is the most important force on the scale of planets, stars, and galaxies view more..
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Ans: We defined the weight of a body in Section 4.4 as the attractive gravitational force exerted on it by the earth. We can now broaden our definition and say that the weight of a body is the total gravitational force exerted on the body by all other bodies in the universe view more..
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Ans: When we first introduced gravitational potential energy in Section 7.1, we assumed that the earth’s gravitational force on a body of mass m doesn’t depend on the body’s height. This led to the expression U = mgy view more..
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Ans: As a final note, let’s show that when we are close to the earth’s surface, Eq. (13.9) reduces to the familiar U = mgy view more..
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Ans: Artificial satellites orbiting the earth are a familiar part of technology But how do they stay in orbit, and what determines the properties of their orbits? We can use Newton’s laws and the law of gravitation to provide the answers. In the next section we’ll analyze the motion of planets in the same way. view more..

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