Bernoulli's equation enables you to express the relationship between a fluid substance's velocity, pressure and height at different points along its flow. It doesn't matter whether the fluid is air flowing through an air duct or water moving along a pipe. In the Bernoulli equation, p + 1/2dv^2 + dgh = C, p is pressure, d represents the fluid's density and v equals its velocity. The letter g stands for the gravitational constant and h is the fluid's elevation. C, the constant, lets you know that the sum of a fluid's static pressure and dynamic pressure, multiplied by the fluid's velocity squared, is constant at all points along the flow. Here, we'll see how the Bernoulli equation works by calculating the pressure at one point in an air duct when you know the pressure at another point.

Use the Bernoulli equation to solve other types of fluid flow problems. For instance, you might want to calculate the pressure at a point in a pipe where liquid flows. Ensure that you accurately determine the liquid's density so that you can plug it into the equation correctly. If one end of a pipe is higher than the other, don't remove dgh1 and dhg2 from the equation because those represent the water's potential energy at different heights.

You can also rearrange the Bernoulli equation to compute a fluid's velocity at one point if you know pressure at two points and the velocity at one of those points.

Write the following equations:

p1 + (1/2)dv1^2 + dgh1 = Constant p2 + (1/2)dv2^2 + dgh2 = Constant

The first one defines fluid flow at one point where pressure is p1, velocity is v1 and height is h1. The second equation defines the fluid flow at another point where pressure is p2. Velocity and height at that point are v2 and h2. Because these equations equal the same constant, we can combine them to create one equation, as seen below:

p1 + (1/2)dv1^2 + dgh1 = p2 + (1/2)dv2^2 + dgh2

Remove dgh1 and dgh2 from both sides of the equation because acceleration due to gravity and height do not change in this example. The equation appears as shown below after the adjustment:

p1 + (1/2)dv1^2 = p2 + (1/2)dv2^2

Define some sample property values. Assume that the pressure p1 at one point is 1.2 x 10^5 N/m^2 and the air velocity at that point is 20 m/sec. Also, assume that the air velocity at a second point is 30 m/sec. The density of air, d, is 1.2 kg/m^3. Rearrange the equation to solve for p2, the unknown pressure, and the equation appears as shown:

p2 = p1 - 1/2d(v2^2 - v1^2)

Replace the variables with actual values to get the following equation:

p2 = 1.2 x 10^5 N/m^2 - (1/2) (1.2 kg/m^3) (900 m^2/sec^2 - 400 m^2/sec^2)

Simplify the equation to obtain the following:

p2 = 1.2 x 10^5 N/m^2 - 300 kg/m per sec^2

Because 1 N equals 1 kg per m/sec^2, update the equation as seen below:

p2 = 1.2 x 10^5 N/m^2 - 300 N/m^2

Solve the equation for p2 to get 1.197 x 10^5 N/m^2.

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