 In some speed ranges, when a flow passes over a cylinder in cross-flow, vortices are shed and carried downstream with the flow. The situation is sketched below. We have a uniform flow coming in from the left with a velocity of v. There's a cylinder shedding vortices. The shedding frequency has been measured for a range of velocity as given in the table. All these data were obtained for air flowing over a cylinder of diameter, 10mm, and the tube length was very long compared to its diameter. The shedding frequency has been shown to depend on velocity of the flow, the diameter of the cylinder, the density, and viscosity of the fluid. The following data were gathered from an experimental setup using air with density 1.225 kg per cubic meter and a viscosity of 1.8 x 10 to the negative fifth kilograms per meter second. The first thing I want us to do is to use a Buckingham Pi analysis to determine which dimensionless variables govern the physical situation and then to try to predict the shedding frequency for a cylinder of a smaller diameter with a velocity of 36.75 m per second. So the first step in this process is going to be to list all the physical variables involved. We know that the shedding frequency is a function of velocity of the flow, diameter of the cylinder, density of the fluid, and viscosity of the fluid. So when we start our Buckingham Pi analysis, we are going to list those five variables. The next step is going to be to identify the dimensions of each of those variables. I know frequency is going to be one over time. I know velocity is going to be length over time. I know diameter is going to be length. I know density is going to be mass per length cubed and viscosity is going to be mass over length times time. Step three is to determine a J value, the number of repeated variables that would generate dimensionless Pi groups. My general rule on that is to start with a J value of the number of physical dimensions present. I have three, so I will start with a J value of three. If I can't make it work, I'll reduce J by one and start over. Next, I get to identify which of these five variables should be the three repeating variables. Generally speaking, I want to avoid viscosity and I want to leave my dependent variable if I can. So I will make velocity, diameter, and density my repeating variables. Step five is to generate the Pi groups one at a time, one for each of the non-repeating parameters. Since I have five total variables, three are repeating, that means two are non-repeating, therefore I should have two Pi groups. Pi group one would be generated using the first non-repeating parameter. And if you've been following along with the previous two example problems, you probably have a pretty good idea of how to proceed with this. Why don't you pause the video here, try to generate the first Pi group on your own, and you can see if you were correct. So I started by using exponents a, b, and c, and tried to determine values for a, b, and c that would produce a non-dimensional number on the left. For that, I like to set up my dimensions as three separate equations, and then use three equations and three unknowns to solve for them. It's very similar to balancing a chemical reaction. My mass equation is zero on the left, mass appears only once on the right, and that's with c. For length I have zero on the left, and then a plus b minus three times c on the right. For time I have zero on the left, and then negative one minus a on the right. That gives me an a value of negative one, a b value of positive one, and a c value of zero. Therefore, my first Pi group is n times diameter over velocity. For the second Pi group, I have essentially the same setup as the previous problem. I have viscosity as a non-repeating parameter, then velocity to the e, excuse me, the d power, diameter to the e power, and density to the f power. So for my second Pi group, I get viscosity divided by velocity times density times diameter, which remember out of convention of the Reynolds number, we typically write as density times velocity times diameter divided by viscosity. That will produce positive numbers that have a positive correlation with what we're doing more often than not. So as a result of my Buckingham Pi theorem analysis, I can write out shedding frequency times diameter divided by velocity is a function of density times velocity times diameter divided by viscosity. Using that information, I will generate a row for Pi group one and Pi group two for each of the observations in my table. So I'm going to be calculating a Pi group one and a Pi group two value for each of my five observations using a density of 1.225 kilograms per cubic meter, a viscosity of 1.8 times 10 to the negative fifth kilograms per meter second, and a diameter of 10 millimeters. Remember that I want my two Pi groups to produce dimensionless parameters, which means that I need all the units to cancel, which means if I'm using values of kilograms, meters and seconds, I need the diameter to be a meter as well. So instead of 10 millimeters, I will use 0.01 meters. And now we can get to calculating. So for the first observation, I had a shedding frequency of 30.87 and I'm multiplying by a diameter of 0.01 and dividing by a velocity of 1.47. I get to 0.21. For observation number two, I have 61.95 times 0.01 divided by 2.95 and I get 0.21. Then for observation number three, I get 92.61 times 0.01 divided by 4.41 and I get 0.21. For observation four, I have 123.48 times 0.01 divided by 5.88 and I get 0.21. Then for observation number five, I have 154.35 times 0.01 divided by 7.35 and I get 0.21. For Pi group number two, observation one, I have 0.01 multiplied by the velocity of 1.47 times the density of 1.225 divided by the viscosity of 1.8 times 10 to the negative fifth. So for observation number one, I have 1000.42. For observation number two, I'm changing my velocity and nothing else. So instead of 1.47, I have 2.95. So I get 2007.64. For observation number three, and just for convenience, I'm going to wrap this in parentheses and then multiply by the velocity at the end. So 4.41, I get 3001.25, like this little neater. For observation number four, multiply by 5.88, I get 4001.67 and then 5002.08. For observation number five. So looking at these data, we are looking for similarity and what we can say with a reasonable amount of confidence is that if the Reynolds number, if Pi group two, is between 1000 and 5000, then Pi group one is going to be about 0.21. That's our conclusion. So for part A of the problem, we were just creating the dimensionless variables. For part B, we are trying to predict the shedding frequency with a diameter of 1mm and a velocity of 36.75. With this new diameter and velocity, the first thing we should do is check for similarity. So I'm going to calculate a Pi group two. It's going to be diameter times velocity times density divided by viscosity. We are using the same air, so we have the same diameter, excuse me, we have the same density and the same viscosity. But instead of a diameter of 0.01 meters, we are using 0.001 meters. And for our velocity, we are using 36.75. So I get a Pi group two of 2501. My Pi group two value at part B is less than 5000 and greater than 1000. So it's within the range we established with our earlier five data points. As a result of that, we can be relatively confident that similarity exists. So we can say at this data point, n times diameter divided by velocity is probably going to be 0.21. Therefore, n would be 0.21 times velocity divided by diameter. So 0.21 times our velocity, which was 36.75 divided by 0.001. So the shedding frequency is likely to be about 7,717.5. But note that we could only do that because we had similarity. If we had a situation where our diameter was much bigger, maybe something like 100 millimeters, and our velocity was much bigger, maybe something like, if we had a situation where our diameter was much bigger, maybe 100 millimeters. And let's use about the same velocity. Oh, I don't know, 30-ish. With that, our new Reynolds number, our new Pi group two, is going to be 204,000. 204,000 is much, much bigger than 5,000. So no similarity is present, or rather, not enough similarity is present to be confident in our prediction for a shedding frequency.