 The Cessna Citation executive jet weighs 67 kilonewtons and has a wing area of 32 square meters. It cruises at 10 kilometers standard altitude with a lift coefficient of 0.21 and a drag coefficient of 0.015. So the lift coefficient is to the force of lift, the same one that the drag coefficient is to the drag force. It's gonna be one half times the coefficient of lift times density times area times velocity squared. This will be one half times the coefficient of drag times density times area times velocity squared. And in these equations or aircrafts, that area is actually the chord length of the airfoil which is the distance from the leading edge to the trailing edge of the airfoil multiplied by the length of the wing itself and it's the same for both the lift and drag. In this particular circumstance, it isn't the perspective from the forward moving air. That is special to how the lift and drag coefficients are made for airfoils. Anyway, I want us to determine the cruising velocity of this aircraft when pointed horizontally in miles per hour. The power output required to maintain cruise velocity, the Mach number of the aircraft at cruise velocity. And if the atmospheric turbulence caused the aircraft to have to ascend to an altitude of 15 kilometers, how would that affect the answers to A, B, and C? So we start by recognizing the statics problem. In our aircraft, which I will draw as a circle, we have a sum of forces in the x direction and a sum of forces in the y direction for this to be flying at a constant velocity horizontally. I'm saying the forces in the x direction have to equal zero and the forces in the y direction have to equal zero. Therefore, the drag force is going to be equal to the thrust force and the lift force is going to be equal to the weight. So the fact that I know the executive weight is 67 kN means that I can say the force of lift is also 67 kN. So we'll start with this equation and think about what we know. In this equation, I know the force of lift because I know the weight. I know one half. I was given the coefficient of lift. I can look up the density at this altitude. The area is given 32 square meters. I can solve for velocity. So the velocity is going to be the square root of 2 times the force of lift divided by the coefficient of lift times density times area. So that's 2 times 67 kN divided by our coefficient of lift was 0.21. We are using a density at an altitude of 10 km. So if we go into table A6, 10 km is going to have a density of 0.4125 kg for cubic meter. And then our area of effect is 32 square meters for both the lift and drag. So far, a kN is 1000 N. And a N is a kilogram meter per second squared. So if I were to do no other unit conversion, I would be left with an answer in meters per second after I evaluate the square root. Since we're probably going to use this velocity in other parts of the calculation, let's leave it in meters per second and then convert to what the answer actually is asked to be in. So we start with 2 times 67 times 1000 divided by 0.21 times 0.4125 times 32, which when taken to the square root gives me a velocity of 219 meters per second. And that is going to be useful for other parts of the calculations, but the problem actually asked for an answer in miles per hour, I believe. Yes. So we start with the recognition that a mile is 5,280 feet. There are 3.2808 feet per one meter and 3600 seconds per hour. So if I take that result and I multiply it by 3.2808 times 3600 divided by 5,280, we get 492 miles an hour. Then part B asks us to figure out what the power output required to maintain cruise velocity is going to be. Well, we know power is going to be work per unit time and work is going to be the integral of force with respect to displacement. So if my force is constant, which it's going to be because the thrust force is constant if I'm traveling at a constant velocity, this comes out and I'm left with force times dx dt, which is force times velocity. We know the velocity, we just figured it out, which force do we use? That's right, it's the force of thrust. We don't know the force of thrust, but we do know it has to equal the drag force and we can calculate the drag force because we know the drag coefficient. So I'm going to take one half times the coefficient of drag times the density times the area times velocity squared. That's going to give me the drag force which is equal to the force of thrust and then I'm going to multiply that by velocity. So if I combine those two calculations symbolically, that'll leave me with one half times the coefficient of drag times density times area times velocity cubed. The coefficient of drag was given as 0.015. We are using the same density as earlier, 0.4125 and then our area is the same because that's how the coefficient of lift and coefficient of drag are generated for an airfoil. And then our velocity was 219.865 meters per second and I'm going to cube everything. And naturally I want this answer in horsepower so I have to convert from metric to imperial once again. For that I will start by recognizing that a horsepower can be expressed as 0.7457 kilowatts or 550 feet pound-force per second. So I'm going to say one horsepower is let's go with the kilowatt conversion this particular time and a kilowatt is a thousand watts and a watt is a joule per second and a joule is a Newton meter and a Newton is a kilogram meter per second squared. Newtons cancel Newtons, joules cancel joules, watts cancel watts, kilowatts cancel kilowatts. So then cubic meters and square meters is going to cancel cubic meters, meters and meters. Second squared is going to cancel two of the seconds. Second squared and seconds are going to cancel cubic seconds. Kilograms cancels kilograms leaving me with just horsepower. So 0.5 times 0.015 times 0.4125 times 32 times 219.865 and just to be as accurate as possible let's actually just grab that value you know because eight decimal places will really jazz up that result. And then I'm taking that entire quantity divided by, don't need the two because I multiply by one half 0.7457 times 1000 and I get 1411 horsepower. Then part C I wanted to know the Mach number. So the Mach number is probably a non-dimensional parameter that you have heard of. The Mach number represents the velocity divided by the speed of sound at those conditions. The speed of sound at an elevation of 10 kilometers can be calculated by using this approximation that you will probably use if you take aerodynamics. But for now, like everything else we do, we're going to look it up in a table. So on table A6 we have the speed of sound at that altitude which at 10 kilometers is 299.5 meters per second. And speed of sound varies based on the ambient conditions. So we had to be specific in regard to where we were grabbing those properties or which value we were grabbing. Then our velocity is 491.82 miles per hour or 219.865 to our 19.865 meters per second divided by 299.5 meters per second. And again, I will grab the actual velocity just for all those extra decimal places that we don't need divided by 299.5 that gives us a Mach number of 0.734. So as we go up in elevation, the speed of sound will decrease. So even if we're traveling at the same speed, our Mach number will increase as well. Speaking of which, Part D asks if we went to 15 kilometers, how would that affect A, B, and C? So at 15 kilometers, our density is now 0.1935. Furthermore, our speed of sound is going to be a little bit different. It is now going to be 295.1, so not particularly different. And then with those two numbers, I can recalculate A, B, and C. So I will do that by writing out the velocity, the work, and the Mach number. And we can just rerun our calculations. So first thing we did was calculate a velocity. And that velocity, when the density is changed to 0.1935, gives us 321.016 meters per second, which we had then converted to miles an hour. So if I repeat that process here, I'm going to have a velocity of 718 miles per hour. So this same aircraft design, if pointed horizontally, would have to be traveling 718 miles per hour in order to maintain flight at 15 kilometers. Even though it only climbed by 5,000 meters, it had to double its speed in order to maintain that cruising altitude. Next calculation was the power required. That power required had been 14.11. So we're just going to go in here and we're going to change the density and velocity. So the velocity I'll change first. Instead of 219.8648, we are using 321. And then I go change the density as well from 0.4125 to 0.1935 and I get 2060. So the engine is going to have to output 2060 horsepower in order to maintain flight at that altitude. And lastly, if we take our velocity in meters per second and divide it by 295.1, we will be traveling at a supersonic speed. So that's how the actual numerical results compare. We could have looked at these equations and seen how changing the density is going to a change. The velocity we could have plugged in this equation symbolically and simplified it down to a proportion with the densities. But in this particular case, it happened to make just as much sense just to actually calculate the numbers. Do you think this aircraft can travel at 15 kilometers? At least horizontally? Probably not. Probably not made to travel faster than the speed of sound. Which makes the question, why do we care about Mach number? Why is that an important quantity in fluid mechanics? Well, it's because our assumption of incompressible flow becomes less accurate the faster the fluid is moving. At higher velocities, especially relative to the speed of sound, that compressibility of the fluid becomes more important. The speed of sound seems arbitrary, but the speed of sound is just the speed of a pressure wave in the fluid. When you change the pressure, it's going to manifest forward at a rate that is the speed of sound. It just happens to be that sound is a pressure wave that we have immediate context for. So as we make changes to a property of the fluid, those changes propagate forward at the speed of sound. That's why the speed of sound is important for our analysis. At this point in our analysis, if we were to continue, we have to start talking about compressible flow. But that's outside the scope of what we're doing in this class. So we'll leave it here.