00:01
Welcome to this numerary tutorial.
00:03
If we get started with part one of this problem, we can first start by solving for the wing aspect ratio, which is b squared divided by s, so the wing span squared divided into the plan form wing area.
00:30
And then we have the e, which is actually the curve d.
00:34
This is the span efficiency.
00:35
C factor specified at 0 .85, then the c sub -do is a zero -lift drag coefficient.
00:47
So zero -lift implies the force lift is equal to the force weight of the aircraft.
00:54
So the clean zero -lift drag coefficient in this particular part of the problem is specified at 0 .033.
01:07
So the next step is to solve for the l over d or the lift to drag ratio max.
01:14
So when we enter in the specified data, we compute a lift to drag ratio of 14 .55 to 1.
01:25
And it's important to filter out the information and apply what is relevant.
01:33
So in this case, cruise speed, coefficient of lift max, the takeoff 1 .2 v .s and the v land 1 .3 vs are not applicable in this particular problem.
01:53
In fact, the entire problem set, because the projected frontal area, the a sub -f, is not specified.
02:06
So we have the vlo, the velocity lift -off speed, which is a function of the vs, 2 times the weight, divided into the row air density times the wing area times a cl max so we need to apply this to the a half and that is not specified so it's actually easier to apply the l over d max formula so the next estimation actually an estimation to make is approximately a 6 % fuel mass consumption from, move this up a little bit, to 6 % fuel mass consumption estimate from takeoff to climb, and then entering initial cruise speed and altitude.
03:19
So the boeing 737, dress -specific fuel consumption, if you look up the reference data on this particular airframe and its power plant, either.
03:33
The cfm -56 or the ge cf6 turbofans.
03:42
There's two of them on board the 737, and usually there's about a 6 % fuel consumption en masse from the point the airplane takes off, rolls down into its v1, and then rotates the v2, climbs, and then has to drop its throttle, and then climb to its cruise altitude.
04:16
So it is important to mention, even though the cruise speed is not relevant to the problem that the 833 km per hour cruise speed can only be attained at the 36 ,000 feet cruise altitude because the density of the air and the temperature as well as the pressure all stable.
04:46
Facilitate the cruise zero lift drag coefficient.
04:56
So that would not be possible until that cruise altitude of about 10 ,976 meters is reached.
05:09
So that is a value or metric to make reference to in order to infer if it's a relevant piece of data to use to see.
05:22
Solve the problem.
05:23
So another slight variation is that the gravitational acceleration constant at 10 ,976 meters msl pressure altitude will decrease just a tiny bit to 9 .76 meters per second squared.
05:41
So the next step is to compute the estimated weight of the aircraft right when it enters initial cruise altitude of 10 ,976 meters pressure altitude msl or 36 ,000 feet msl altitude.
06:04
So we have to subtract the 6 % fuel burn from the time the aircraft takes off to the time it reaches the cruise altitude of 10 ,976 meters or 36 ,000 feet.
06:19
So this verse is accomplished by computing the aircraft mass after the 6 % fuel consumption.
06:32
So 68 ,560 kilograms from its initial takeoff weight of 70 ,000 kilograms takeoff mass.
06:44
So move up here.
06:52
So we continue on and then solve for the aircraft.
06:59
Weight, which is mass times gravitational acceleration, and then we compute 6669 ,146 newtons force weight, right when the aircraft enters initial cruise altitude, 36 ,000 feet, or 10 ,976 meters, msl, pressure altitude.
07:33
Okay, so next we have the 600 ,000.
07:39
169 ,146 newtons force weight and then we divide this into the lift to calculated lift to drag ratio of 14 .55 newton's force lift per one newton force drag and this then computes to a force drag at clean cruise speed of 45 ,9809809.
08:16
0 .4 newtons force drag or expressed as 46 kilonutons of drag force.
08:42
So the final answer for part one of this problem is the clean cruise speed force drag is equal to 46 kilonutons or 10 ,337 pounds force drag.
09:10
All right, moving on to part two of the problem.
09:15
We have to first establish a zero lift drag coefficient summation.
09:26
So this is accomplished by applying which zero lift drag coefficients are applicable to the takeoff mode of the airplane.
09:37
So the takeoff mode is the zero lift v1 speed just prior to rotation and climb to lift off and then approaching the v2 speed, which is the climb speed, so that would not be relevant.
09:54
So we're trying to understand what question or part two is asking, and it's asking for the takeoff drag at zero lift drag hole efficient, the zero lift drag hole efficiency.
10:13
So what is pertinent to the takeoff mode, it is the zero lift drag hole efficient.
10:18
With landing gear and zero lift drag coefficient flaps take off.
10:26
So we have to add 0 .02 plus 0 .058 to yield 0 .078.
10:35
So the zero left drag coefficient is a constant...