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# Bernoulli's equation derivation part 1

Bernoulli's equation is an equation from fluid mechanics that describes the relationship between pressure, velocity, and height in an ideal, incompressible fluid. Learn how to derive Bernoulli’s equation by looking at the example of the flow of fluid through a pipe, using the law of conservation of energy to explain how various factors (such as pressure, area, velocity, and height) influence the system. Created by Sal Khan.

## Want to join the conversation?

• Isn't the work in putted into the fluid manifested in the form of kinetic energy? Why do we need to have two separate terms indicating KE and W? I can understand that KE1 + PE1 = KE2 + PE2 , by law of conservation of energy.... So, why do we need a work term? Please help.
• Bernoulli's expression (ρ+ρgh+1/2ρv^2) states that for an incompressible fluid not experiencing friction forces (low viscosity), the sum of the static pressure and dynamic pressure will be constant within a closed container.

The term 1/2ρv^2 looks similar to Kinetic energy and is actually known as "dynamic pressure", which is the pressure associated with the movement of the fluid.

The term ρgh looks a lot like the expression for gravitational potential of an object. It is the pressure associated with the mass of a fluid sitting above some position of depth.

The term P is derived from what Sal refers to as Work. Specifically, it is derived from "energy density".
We know that Pressure = Force/Area (N/m^2); if we manipulate the formula by multiply force and area by meters: N/m^2 x m/m; we get N·m/m^3; which is actually J/m^3 (energy density). A system at a higher pressure cotain a greater density of energy than those systems at lower pressure.

Recall that adding P + ρgh gives us the static pressure (looks just like the absolute pressure formula).

Ultimately, Bernoulli's principle says more energy dedicated towards fluid movement (higher 1/2ρv^2 value) means less energy dedicated towards fluid pressure (lower P + ρgh values). The opposite can be true too, less movement means more static pressure.

Sorry for the long post, hoped it helped.
• At , Sal divides Force by area than multiplies by area. Why is he multiplying by area? Aren't we just looking for pressure which is simply Force/Area?
• Because when F is divided by Area it also needs to be multiplied by it as it is being introduced by us and hence to maintain the equality in the equation
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• Maybe this is a really stupid question but wouldn't the Kinetic Energy of the liquid at the start of the system be a result of/induced by the work inputted? So adding up the work inputted along with the KE of the liquid being inputted would give too large a total result for total energy at start of system wouldn't it?

Or have I lost the plot lol..
• The kinetic energy will be induced by a force (and not the work done). Since for a mass to have kinetic energy, it needs to have a velocity, but if the initial velocity is zero, we would need to accelerate the mass--and so we require a force. Here the forces are mg on the two sections of water (which gives us the potential energy for the two sections of water we are considering here) and we also need to apply a force at the bottom of the pipe to make sure the water reaches the top (like a pump).

Sal has just written Work done by all forces = change in kinetic energy in a different way.
• how is it possible for liquid to go up?
• By definition it can not become more dense so it has to go wherever is available when the force on one end becomes greater than the resistance on the other end
• this question may not be related here but here it goes. viscosity is inversely proportional to temperature i.e if temp increases viscosity decreases and if temp decreases viscosity increases. but lava is a very hot liquid. and also it has a very high viscosity. but it violates the above law of temperature viscosity relationship . how is this explained??
• That relation has to do with the change in viscosity of a single substance at different temperatures. As lava increases in temperature it decreases in viscosity so the law is not violated.
• If the pipe narrows in the middle, wouldn't this equation be obsolete? I am thinking about when I kink a hose, the pressure is not the same at the delivery end as the production end, so should there not be a caveat that this equation only works if the pipe does not narrow in the middle?
• Your question makes alot of sense. So let us consider a hose with water flowing through it. Say somewhere around the middle of the hose, you kink it. Have you noticed that at the kink, from the source side, the presence of a very hard bulge, tight would be a better word. This bulge is solely due to a large increase in pressure. This pressure develops because of all that water trying to fit into a tiny opening. On the other side of the kink, the water would come out at an increased velocity, but much reduced volume. This is in accordance with the equation.
Then one would have to keep in mind that, a hose is an enclosed pipe of small diameter rarely ever placed in a linear fashion, this water exiting the kink, doesn't really have much space to 'be fast'(couldn't think of a better word.), its speed would be reduced due to collisions with hose walls (I hope you can visualize this.) And the water to reach the output end, would require more water to raise it to the end of the hose(like filling a glass of water, you need more water to reach its brim). And so we find a reduced pressure instead of an increased pressure. This is not due to inadequacy of the given equation but due to the 'bendy' nature and length of a hose.
• why here we add extra term as work in conservation of energy?
• Is it possible to re-record this topic again by Sal, the quality is so bad...and the Hindi version doesn't suit...please...