Calculating skin friction drag PDF

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AERO 309 Cal cul at i ngski nf r i ct i ondr ag We're going to go over an example of calculating skin friction drag. We are going to do two three different example or three cases. We are going to do fully laminar. Fully turbulent. And then one call with transition. Be partly a laminar part turbulent. Now each three, each of these three cases can have practical applications with fully laminar. It would be more for much smaller aircrafts like a small UAV airplane. So small and slow flight. You could end up with cases where the boundary layer is for the most part fully laminar or mostly lamer. You could have fully turbulent cases. That's a good approximation for full size aircraft or airplanes that are flying at higher speeds. You can consider most of the boundary layer will be turbulence. So, you can do a very good approximation by assuming the whole boundaryless turbulence. And then you have with transition which is actually more real flow that usually start off with an area near the leading edge that is laminar. And then at some point boundary layer will transition to a turbulent boundary layer. So that happens at both low and high speed small and large airplanes that many times will have a real Flows will have a transition from laminar to turbulent boundary layer. Now if you read through the discussion on this, you’ll notice you'll know that I'm predicting when something changes from Laminar to turbulent is not easy. And it depends a lot on the flow itself. How much turbulences are in the air? Sound can have a problem on the surface of the body of which boundary layer, they are going. And a lot about the shape of the shape of the body just on what pressure distributions are having over the body here doing so many different things can cause a different kind of boundary to transition from laminar to turbulent. So, in these three cases we're going to deal with the same flat, plate wings. I'd say it has a cord. Five centimeters. But a span of one meter of free stream velocity of one hundred and twenty meters per second. And over sea level. Standard sea level. So, air density one point two two five. KG per meter cubed. An air of viscosity B one point seventy-nine times ten to the negative five. Kilograms per meter second. It's always good in problems engineering aerodynamic problems. You draw a picture exactly to scale. Gives you an idea. And that's supposed to be a rectangle. The way of oncoming flow at 120 meters per second. This is the leading edge. So, five centimeters two equals zero point zero five meters. And this band is one meter very thin. Well we'll learn later. This has a very high aspect ratio. Wings. So, let's look at fully laminar cases. Fully laminar. And we are wanting to skin friction drag so we are going for. Skin friction. Remember skin friction. Q Infinity’s s. ZF NCF. Is one point three to eight divided by square root. Sort of the real center of the wing. And the nurses these are Laminar equation's rails no length will be Ro V. L. divided by new and just for completeness the dynamic pressure member is one 1/2 Ro. V squared. So, let's calculate the rules over first we’re going to use it quite a bit. So, a Ro. v. or up there Rova L Romeo. Let's enter in what we know. Sea level one point two two five. Losses 120 meters per second length. It is the chord length so zero point is the distance going in the direction of the flow. So not span which is one meter but five centimeters or point zero point zero five meters. Where well in direction of flow. And viscosity is one point seven eight nine times up with four point one one times into the fifth. And remember brindled number is non dimensional. You put in all the dimensions cereal. See that they all cancel out. So that's the four hundred eleven thousand. Is our room somewhere here. So, let's now calculate Kerferd total skin friction coefficient on point three to eight hired by the rules. Square root of the real summer. We have the rental summer there which needed to add in one point three to eight square root of four point one one time. Stenseth fifth. And the calculation you end up with zero point zero zero two It’s. Now that we have the skin friction coefficient you can calculate the drag. Now return discussion of skin confection kill or skin friction drag that this equation right here equation up here its confection drags. We're using s and that is just the wing

area. So, reality we are just looking at half of the wing something called top. So, when you look at the top surface right now D.F. Q infinity FCF plug in our values one half. Q is one half Roe V squared and one-half Roe one point two two five, V is 120. Square area which will be the core of zero point zero five times span which is one times r. Coefficient we just found zero point zero zero two o seven. That gives us a drag of zero point nine one three Newtons. Now we want the full wing. Versus top and bottom. So total. It's conversion back will be two times zero point nine one three. So total. Yes, skin friction drag is one point eight three nudes. And this is. Or the fully laminar case. Now let's look at the Foley turbulent case. So. Only turbulence. So again, we have the exact same equation for. Friction drag infinity S.F. with this time just as fully turbulent. Total skin friction drag is different. Equation for its zero point zero seven four divided by Reynoldsburg and zero point two from earlier are rentals No it's going to be the same because it's the same wing. It's four point one one time tenth to the fifth. So, let's calculate. Our total skin friction coefficient. Clear Point 074 divided by Four point one one fifth zero-point two power and you plug that in. And you end up with. Zero point zero zero five five eight. And from her the other one was zero zero two o seven. So, this makes sense. This is. Larger than Laminar. As we would expect. So, let’s calculate the drive. Now remember. We use wing area really calculating half of the drag. So, the top surface. Infinity SCF. It's like everything in Kantha Today’s dynamic pressure so it's one half ro one point two to five V squared. Hundred and twenty squared s zero point zero five. The cord span was one NCF with zero point zero zero five five eight. Do that calculation. Double checking my own work. We have two point four six Newtons, which was. It’s pretty much pretty much bigger than the point nine one three Newtons that we calculational fully laminar but total. It's top and bottom so two times that. We get total. Skin friction drag for a turbulent case fully turbulent case is four point nine two Newtons and this is fully turbulent. Now we have one last case, and this is with transition. This one's a little more complicated. It's a. With transition and here I'm going to specify exactly where the transition takes takes place. So, let’s say. And this occurs Downstream. From the leadingedge stream of the leading edge. Let's draw a picture here to show what I'm talking about. So, we have. The infinity is under 20 meters per second. That's leading edge. And I'm gonna blow up the wing. So that's a cut. That's a cut. So, from here to here is zero point zero five meters to the. And let's say this is a full 30 percent or so. For the first 30 percent going to have laminar for the last 70 percent. We're going to have turbulent boundary layer. So how do we find. The total skin friction or how did we find out how to find, It’s friction coefficient. So, now in reality it's very complicated. But we have a method here it's a pretty good approximation. So, we won't say that the dry total skin friction drags. We'll be the total skin friction drag when it is 100 percent turbulent. Minus. The skin friction drags of if we just said we just calculated the first 30 percent of the wing. Turbulent but then we're going to add back in. If. Just calculating the drag for just the first 30 percent when it's laminar. So, let's look at this picture wise. Let's say. Or doing here is we have a flat plate. And this is zero point zero five meters so we're looking at it edgewise. Now calculate the full. Turbulent boundary layer or subtract then. Thanks. We'll just do the turbulence for the first This term turbulent, we're going to add back an. Oh, laminar case. And it's a little hard to see but there's a draw in the picture that this laminar is smaller than it’s turbulent. And where we end up with. As for the first 30 percent we get laminar. We’re going to jump up and for the last 70 percent. We get turbulent. So, Laminar. Turbulent. And this is a pretty good approximation. Now going to require some extra work but we are already. Have. One hundred percent turbulent case because we just calculate that this is the Foley. Turbulent case that we did and that was if remembered for the top. D.F. hundred percent turbulent. We found that the drag was two point four six Newtons. So, we need to calculate the drag for the first 30 percent with for the turbulent. For the turbulent boundary layer and also a laminar boundary layer. First, we need to figure out what the length of the core length is for 30 percent Page 2 of 3

downstream. So. 30 percent downstream. So, you want to call this El 30. Which will be 30 percent. So, zero point three times or four chord length which was zero point zero five meters, and we end up with zero point zero one five meters. Just for completeness just to remember that zero one hundred percent length was zero point zero five meters. So, we're right near the Reynold's number at this point though. It's a real somber tone. 30. Roe v. L. Thirty. I knew at one point two two five losses a hundred and twenty meters per second. Now our length the zero point zero one five meters. Scores changed change one point seven eight nine times to the negative five. Just checking my work. And we end up with one point two three times into the fifth. So, one hundred twenty-three thousand is our real number. Now let's calculate the infection for the 30 percent when it's laminar. So, look at laminar. Part of it or I call it. CF. say L 30 sub trips and get a little confusing here but it's going to be one point three to eight. We're square root of the Reynolds number one laminar kaso round number at the 30 percent location. So, one point three to eight by one point two three square root of one point two three tenths of the fifth and that gives us a skin friction coefficient of zero point zero zero three seven nine. So. Or that into the dry calculations of the drag. Conviction dragged for 30 percent laminar. Q infinity D. S. It's CF L 30 so. Q hasn’t changed so one half. Roe one point two two five the Hundred and twenty squared area has changed though her role in looking at the first 30 percent. So, zero point zero one five times span is 30 percent. See. And this is span which has not changed it's still one. Aren't your new. Coefficient Friction zero point three seven nine. And that gives us zero-point five o one Newton’s. Now finally let's do the turbulent case for the first 30 percent. So, we have turbulent. Yes, four or 30 percent use a turbulent boundary layer equation zero point zero seven four square not square root a Reynolds number. Zero point two zero point zero seven for my numbers. One hundred and twenty-three thousand. So, one point two three times. Raise that to zero point two. We just made that up. Your car will clear. And we have zero point zero zero seven one zero. So, looking at just the top surface. Top Drag. Three 30. Again, this is. 30 percent. Court area so one half. Roe one point two two five velocities under 20. Square that area zero point zero one five times one. And or new. Turbulent skin friction coefficient false conviction zero point zero zero seven one zero. That gives us. Zero point nine three nine. Now we have everything we need. So, let's look at. Top surface. Drag ghouls. A hundred percent turbulent case. Minus the 30 percent. The first 30 percent turbulent case. Then add back first 30 percent laminar case. Our values. We have two point four six. Turbulent. For the first 30 percent. Was zero point nine. Thirty-nine Newtons and Laminar was zero point five. One. And gives value of 2.0 to news. So total, be two times that for the top and bottom surfaces gives us four-point O four. Newtons Gockley the fully laminar fully turbulent a one with transition. So just to summarize. We have. Well we laminar. We end up with top bottom. One point eight three Newtons. Only turbulent. In the drag skin friction drag was four point nine two Newton's. And then with 30 percent transition. First 30 percent was laminar last 70 percent was. Turbulent. We end up with four point 0 four Newtons and could see that the kind of this magnitude here makes somewhat sense for does make sense an order of which is smallest to largest fully laminar its smallest fully turbulent is largest laminar smallest. The largest is fully turbulent to expect. If you had both a mixture that would fall in between. And it does.

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