 The main cross cable between towers of a coastal suspension bridge is 60 centimeters in diameter and 90 meters long. Determine A, if the flow conditions are laminar, and B, the drag force on this cable. Well first of all, I didn't give us a velocity, so let's add that in. So if we had a 50 mile an hour average wind, what would the load be and is that flow going to be laminar around the cable or not? And for our purposes here, we are assuming that the wind is coming at the cable from the side and with that we can calculate a Reynolds number. Here the characteristic length is going to be the diameter. We are considering air and we don't know the conditions, so we could say 20 degrees Celsius and approximately one atmosphere, at which point we can jump into table A2. And for air at 20 degrees Celsius, a density of 1.2 kilograms per cubic meter and a kinematic viscosity of 1.5 times 10 to the negative fifth. So with a velocity of 50 miles an hour, a diameter of 60 centimeters and a kinematic viscosity of 1.5 times 10 to the negative fifth meter squared per second, we will have to convert some units to get this into a unitless proportion. So I'll start with one mile is 5,280 feet and then one meter is 3.2808 feet is 3600 seconds, miles cancels miles, seconds cancel seconds, hours cancels hours. I have centimeters and meters in the numerator and I have square meters and feet in the denominator. Feet cancels feet. So I need to throw in 100 centimeters per meter and then centimeters will cancel centimeters, meters squared will cancel meters and meters. Now with our handy-dandy calculator, if you would cooperate calculator, I would start by turning on awesome, thank you. 50 times 60 times 5 to 80 divided by 1.5 e to the negative fifth times 3.2808 times 3600 times 100. We have a Reynolds number of 89 excuse me, 894,091. Like every other time, we're going to compare our Reynolds number to the critical Reynolds number. The critical Reynolds number for internal flow was 2300, the critical Reynolds number we used for flat plate theory was 5e5, but this situation is neither a flat plate nor is it internal flow. So since we have a different situation, we have to look at some actual data. For that, let's go into the textbook, specifically section 7.6. In 7.6, we have some data corresponding to some actual flow conditions. Specifically on 7.16a, we have the coefficient of drag corresponding to Reynolds numbers for a couple of different types of shapes. This curve at the top is a smooth circular cylinder in cross flow, which is what I'm assuming my cable is. Now where on this blue line do you think we have laminar flow? Yeah, it's definitely this first curve. Right about here, we start the transition to turbulence, and right about here, we assume it is turbulent. So I would say anything greater than 2, maybe 3 times 10 to the 5th is going to be turbulent. Anything less than 10 to the 4th is definitely laminar. Anything between the two is transitioning from laminar to turbulent. Our Reynolds number was 8.9 times 10 to the 5th, which is going to be closer to 10 to the 6th, especially considering that this is a logarithmic scale. Therefore, I will confidently say this is turbulent flow. So in answer to part A, we can pretty confidently say this is not laminar. The question specifically asked if it was laminar, and if we had ended up in a situation where we had a Reynolds number in, say, this area here, even there we would still answer no, even though it isn't necessarily turbulent. It's definitely not laminar, or at least completely laminar anymore. Now on to part B. For part B, I want us to determine the drag force, which is going to be 1 half times the coefficient of drag times the density times the area times velocity squared. The coefficient of drag will come from this figure. Again, we're assuming smooth circular cylinder, and we're saying that the length is so much longer than the diameter that it is essentially an infinite relation. It's a very long relative to its diameter. I mean, if we wanted to, we could calculate that proportion 90 over 5, excuse me, 90 over 0.6. We have 150, and our two options here are a proportion of 5, or a proportion of infinity. We're going to say it's closer to infinity than it is to 5, which makes sense if you don't think about it too hard. So I'm going to say my coefficient of drag is maybe 0.2, 0.25. I mean, remember with logarithmic scales like this, we're going to jump about halfway for the first increment. So I'm going to say about 0.2. With that number of 0.2, I have the coefficient of drag, I looked up the density of air. We know the velocity is 50 miles per hour. The area we're going to use is the diameter times the length, because we're looking at the approximately rectangular shape. It's a beautiful rectangle. I know what you're thinking. And that is 60 centimeters tall and 90 centimeters long, excuse me, 90 meters long. So we're using 60 centimeters times 90 meters for the effective area for drag. So if we called the coefficient of drag about 0.2, and the density of air was 1.2 kilograms per cubic meter, and my area is 60 centimeters multiplied by 90 meters, and my velocity is 50 miles an hour squared. So I square the 50 square the miles square the hours. And I don't believe that, oh, I do have a unit. So we're looking for units of newtons. A newton is a kilogram meter per second squared. One mile is 5,280 feet square everything. There are 3.2808 feet in one meter square everything. And then one hour squared is 3,600 squared second squared. And then at this point, second squared cancels second squared, hour squared cancels hour squared, mile squared cancels mile squared, feet squared cancels feet squared, kilograms cancels kilograms, 100 centimeters is one meter centimeters, cancel centimeters. And then I have meters, meters squared, and meters canceling cubic meters and meters. And that leaves me with newtons. So calculator, if you would please, that's 0.5 times 0.2 times 1.2 times 60 times 90, times 50 times 50 times 5,280 times 5,280 times 1 over 32, excuse me, 3.2 808 times 3.2808 times 3,600 times 3,600 times 100. And let's double check that that math looks like it actually carried over correctly, even though I kind of arbitrarily chose some random parentheses. So 3,237.56 newtons. Now how confident are we in that number? Well, keep in mind that we are reading a coefficient of drag off of a chart with a lot of wiggle room. If we had 0.3 instead of 0.2, that would increase this drag by 50%. And that's difficult to see. The demarcation between 0.2 and 0.3 is going to be pretty difficult on a chart like this. The best thing to do would be to get the actual experimental data and build the correlation ourselves. But this is what we have for the moment.