P1 + ½ρv1² + ρgy1 = P2 + ½ρv2² + ρgy2
where ρ is the fluid's density. This equation holds for fluids in a variety of physical circumstances, including the fluid flowing through the pipe in the diagram below.

The pipe changes diameter, so the fluid's speed changes. The pipe's height also changes. Both of these changes affect the fluid pressure according to Bernoulli's equation.
In this exercise, we'll only change the speed of the experimental fluid, which will be air, and so the 3rd terms on both sides of the equation can be dropped:
P1 + ½ρairv1² =
P2 + ½ρairv2²
P1 - P2 =
½ρair(v2² - v1²)
(equation 1)
The air will be moving through a pipe with two widths, as in the figure below

If the fluid doesn't change density as it moves through the 2 parts of the pipe, then the following equation of continuity also holds:
A1v1 = A2v2
where A1 and A2 are the cross-sectional areas of the two widths of the pipe. We can measure the inner diameters of the pipe sections and calculate these areas. This will give us the following relationship between the velocities:
v2 = (A1/A2)v1
So then we can substitute for v2 in equation 1:
P1 - P2 =
½ρairv1²{(A1/A2) -
1}
(equation 2)
Equation 2 will allow us to calculate v1, the velocity through pipe diameter 1. We'll measure the pressure difference between the two pipe diameters (P1 - P2) using an apparatus depicted in the figure below:
Water is in the center loop of the narrow tube. The difference in height between the surfaces is due to the difference in pressure in the two pipes. In fact, Bernoulli's equation can be used here too in order to determine the difference in pressure, because at the two surfaces of the water, the pressures are related by:
P1 + ρwgy1 =
P2 + ρwgy2
P1 - P2 =
ρwg(y2 - y1)
P1 - P2 =
ρwgh
So, using this in equation 2 we get:
ρwgh =
½ρairv1²{(A1/A2) -
1}
(equation 3)
So, when the air is running through the pipes, we can measure h and then solve for v1.
This calculation of v1 will be compared with a more direct determination of its value. We'll measure the rate that the air is flowing through the pipe by running it into a garbage bag of known volume, timing how long it takes for the bag to fill with air, and then computing what v1 must be to fill it in that amount of time. The following will hold:
where V is the volume of the garbage bag and t is the time to fill it.
You can determine the volume of the bag by filling it and measuring it dimensions. When filled, its shape will approximate a cylinder. You can measure the circumference and length of this cylinder and calculate its volume, since V = πr²L is the volume of a cylinder with height L.
After determining v1 in this way, find the percent difference between this value and the one calculated from equation 3.
Now reverse the flow of the air and do the experiment again. Compare the percent difference in the first trial with that in the second trial.