 Let's consider an example of our flow continuity equation. In this case, we have a hose and a nozzle on the end of the hose. Many of you probably have a garden hose and you have some sort of sprayer on the end that takes the flow of the hose and adjusts it to make it spray differently. Usually it comes out at a higher speed. Well, let's look at this particular example here. Here I have a hose that has a diameter of 10 centimeters. It's actually a fairly substantial hose that's about four inches, okay? And we're shrinking it to a diameter of two and a half centimeters, which is only about an inch. So maybe this is something like a fire hose and we're trying to actually spray. And in this case, the flow in the hose is given here as three gallons per minute. And notice we're assuming continuity that the flow in, q over here on this side, the flow in is the same as the flow out, okay? So we assume some continuity for these two things. And if we do that, we know q in equals q out equals q. It's the same flow in and out. But we can also use our relationship that q is equal to v a, where v is the velocity of the flow. And a is the cross-sectional area of each of the sections. Here we have the diameter of each of these sections. But in order for this diameter to be useful, let's go ahead and calculate the area of both sections, okay? So the area, and usually this is the diameter that's given here, is usually the inside diameter, often abbreviated id, the inside diameter of the hose would be the one that we'd be interested in. So we're going to presume that this is an inside diameter in this particular case. Let's go ahead and find area one, which we're going to know it is pi r squared or pi d1 squared over four. And similarly for area two, pi r squared or pi d2 squared over four, okay? So if we do that, we get values of, we plug in the 10 centimeters here, or 5 centimeters squared, and we get a value of 78.54 square centimeters. And over here, if we do similar, we get an area here of 4.91 square centimeters. Notice a substantial reduction in the area between the two, okay? The question asks, how fast is the water speed inside the hose, here where it's flowing in, and outside the nozzle as it exits the hose? What's the difference in the speed between the two water things? Well, in this case, we can manipulate this equation where our flow is the velocity times the area and solve for speed. In other words, the velocity equals the flow divided by the area. And that'll be true for the velocity inside the hose, which we're calling one, and the velocity outside the hose, which we're calling two, position one, position two. So in this case, velocity one is equal to our flow, which is in gallons per minute. Hmm. Well, we might actually have to do something to change our units here. Let's go ahead and convert our gallons per minute into a metric unit. So if I have three gallons per minute, three gallons per minute, and I want to make a conversion, I know, for example, that there are 3.79 liters in one gallon, and there are 1,000 cubic centimeters in one liter. And generally, we want to do flow rates in meters per second. So we're going to go ahead and convert the minutes by representing one minute is equal to 60 seconds. In that particular case, we're able to cancel out gallons. Leaders, we end up with our result in centimeters cubed. We cancel out minutes in centimeters cubed per second. And if we do the math there, we end up with the value of 189.5 centimeters cubed per second. That's a little easier to work with since our numbers down here, our areas down here are in square centimeters. So now I plug that value in 189.5 cubic centimeters per second. I divide it by a one here, which is the 78.54 square centimeters. Notice I'll have some units that will cancel out the square centimeters and two of the three cubic centimeters. And that will give me an answer in centimeters per second. Which is what we desire. When I do so, I get a value for the first one here of 2.41 centimeters per second. So that means the average speed of the water in the hose is moving about a little under an inch every second. However, when I spray the hose, the outside speed is going to use the same flow rate, centimeter cubed per second. But now we use the much smaller 4.91 centimeters squared. Again, we get the same cancellation of units. And our final velocity value there is 38.6 centimeters per second. Notice this is a much more substantial speed. We've multiplied it by about 18 or 19 times, actually probably more like 15 times or so. But you can see that now we have a spray that the water is moving much faster. 38.6 centimeters per second is something like about that big moving every second as opposed to moving something about that size every second. So the reduction of the nozzle size changes the cross-sectional area and that changes the speed at which the water comes out even though it's coming in and out at the same overall volumetric flow rate.