Video: Finding the Airflow Speeds Required for Generating Particular Lift Forces

The wings of an aircraft are required to produce 1.00 kN of lift per square meter of wing, counting the average area of the top and bottom wing surfaces as wing area. Lift is provided by the differential flow of air around the wing surface and depends on the density and speed of air over the wing surfaces. Use a value of 1.29 kg/mยณ for the density of air at sea level and assume that the speed of airflow over the lower wing surface is equal to the speed of the aircraft. At takeoff, an aircraft has a speed of 60.0 m/s. At what speed must air move over the upper wing surface for the required lift for the aircraft to be produced? At an altitude where the density of air is 25.0% of the sea level air density the aircraft has a speed of 245 m/s. At what speed must air move over the upper wing surface for the required lift for the aircraft to be produced?

10:10

Video Transcript

The wings of an aircraft are required to produce 1.00 kilonewtons of lift per square metre of wing, counting the average area of the top and bottom wing surfaces as wing area. Lift is provided by the differential flow of air around the wing surface and depends on the density and speed of air over the wing surfaces. Use a value of 1.29 kilograms per metre cubed for the density of air at sea level and assume that the speed of airflow over the lower wing surface is equal to the speed of the aircraft. At takeoff, an aircraft has a speed of 60.0 metres per second. At what speed must air move over the upper wing surface for the required lift for the aircraft to be produced? At an altitude where the density of air is 25.0 percent of the sea level air density the aircraft has a speed of 245 metres per second. At what speed must air move over the upper wing surface for the required lift for the aircraft to be produced?

We have a two-part problem here and weโ€™ve been given a lot of information to set up the scenario. Letโ€™s start off by recalling the critical information that was given on the previous screen introducing the problem. First, we were told that in order for an airplane to lift up, 1.00 kilonewtons of lift per square metre of wing is required. We can write that as capital ๐‘ƒ sub ๐‘™ standing for pressure of lift is equal to 1.00 kilonewtons per square metre.

Next, we were told on the previous screen that the density of air at sea level is 1.29 kilograms per metre cubed. We will write that as the Greek letter ๐œŒ. And throughout this problem, weโ€™ll assume that value is an exact value. On this screen, weโ€™re given two different scenarios, both asking what air speed is necessary over the upper wing surface in order for the required lift for the aircraft to be produced.

Letโ€™s begin by looking at the first part of this problem and weโ€™ll draw a diagram to clarify the scenario. We show here a picture of one of the wings of the aircraft looking at it end on. The air that approaches the wing moves over the top and the bottom of it. And weโ€™ll establish two separate locations: one right above the wing that weโ€™ll call location one and the second spot right below the wing that weโ€™ll call location two.

In this part one of our problem, weโ€™re told that the aircraft has a speed of 60.0 metres per second, which is the speed of the air at location two. So weโ€™ll write that down and label that speed ๐‘‰ sub two. Recall that weโ€™ve also been told the pressure or the lift force required per unit area of the wing so that the aircraft produces its required lift; thatโ€™s 1.00 kilonewtons per square metre of wing. The density of air at this altitude, ๐œŒ, is 1.29 kilograms per cubic metre.

And this first part of the problem asks us to solve for ๐‘‰ sub one, the speed of the air over the top part of the wing required in order to produce a sustainable lift force. Letโ€™s begin solving for ๐‘‰ one by recognizing that this is a situation where Bernoulliโ€™s equation applies.

Bernoulliโ€™s equation tells us that as it relates to points one and two on our diagram, the pressure at point one plus one-half times ๐œŒ times the speed at point one squared plus ๐œŒ times ๐‘” times the altitude at point one is equal to those seen terms at point two. So this is the canonical form of Bernoulliโ€™s equation.

Now, in our scenario, if we write out this equation for our situation, weโ€™ll see that not all of these terms stay. For example, look at the two terms in this equation that involve heights: โ„Ž one and โ„Ž two. When we consider the height of the plane above ground and the relative difference in altitude between points one and two on the top and bottom side of the wing, we can assume that that height difference is negligibly small compared to the airplaneโ€™s altitude. Therefore, ๐œŒ ๐‘” โ„Ž one is effectively equal to ๐œŒ ๐‘” โ„Ž two, meaning that we can subtract this term from both sides of our equation and cancel it out. That simplifies our equation.

Now, letโ€™s begin to solve for ๐‘‰ sub one by rearranging, which can first involve subtracting ๐‘ƒ sub one from both sides cancelling that term out on the left side, and then multiplying both sides by two divided by ๐œŒ. Cancelling the two and ๐œŒ on the left-hand side of our equation, finally we can take the square root of both sides, which leads to cancelling the squared factor and square root on the left side, leaving us simply with ๐‘‰ sub one.

Looking at this cleaned up version of the solution for ๐‘‰ sub one, we can begin to plug in our numbers. Looking at these first two terms ๐‘ƒ sub two and ๐‘ƒ sub one, we donโ€™t know either one particularly, but we do know that ๐‘ƒ sub two must be 1.00 kilonewtons per metre squared greater than ๐‘ƒ sub one in order for the aircraft lift to be sustained.

Continuing on, ๐œŒ the density of air at sea level weโ€™re given as 1.29 kilograms per metre cubed and ๐‘‰ sub two the speed of air on the underside of the wing is given as 60.0 metres per second. With all these numbers entered in as shown, the last thing we do before entering all of this on our calculator is to convert the 1.00 kilonewtons to 1.00 times 10 to the third newtons.

Now when we enter all these numbers in, we find a speed over the top part of the wing ๐‘‰ sub one of 71.8 metres per second. Thatโ€™s how fast air must be flowing over the top part of the wing compared to 60.0 metres per second under the bottom part of the wing to generate the necessary lift.

Now, we can move on to part two of our problem, which involves a very similar scenario, except two of the initial conditions have been changed. The first condition weโ€™re told is different is that ๐‘‰ sub two the speed of air under the underside of the wing is now no longer 60.0 metres per second, but 245 metres per second.

The other change weโ€™re told about is that ๐œŒ, the density of the air, is no longer 1.29 kilograms per metre cubed, but actually has decreased to a value of 25.0 percent of that original value. This decrease in air density is due to a gain in altitude. So letโ€™s call this new air density ๐œŒ sub ๐‘Ž for ๐œŒ at altitude. ๐œŒ sub ๐‘Ž is equal to 25.0 percent of ๐œŒ. And when we calculate that out, 25.0 percent of 1.29 kilograms per metre cubed equals 0.323 kilograms per metre cubed.

Now with these two new pieces of information, we want to again solve for ๐‘‰ sub one, the speed of air as it moves over the upper surface of the wing. Weโ€™ll again use Bernoulliโ€™s equation, weโ€™ll again cancel out the terms involving height because the height differences are negligibly small, and once more weโ€™ll rearrange that resulting equation to solve for ๐‘‰ sub one. Letโ€™s take that equation for ๐‘‰ sub one we see on screen now and modify it in light of the two new pieces of information; namely, the new speed for ๐‘‰ two and the new air density, ๐œŒ sub ๐‘Ž.

In this expression, everywhere we see a 1.29 kilograms per metre cubed, weโ€™ll replace that with ๐œŒ sub ๐‘Ž, 0.232 kilograms per metre cubed. Having done that, now letโ€™s replace each instance of ๐‘‰ sub two, which originally was 60.0 metres per second, with our updated value of 245 metres per second. With these updates, weโ€™re now ready to solve for ๐‘‰ sub one in this second part of the problem. Everything else in the equation including the lift force required to keep the airplane in the air is the same.

When we enter these updated values into our calculator to solve for ๐‘‰ sub one, we find a result of 257 metres per second. Under these updated conditions, this is the minimum air speed over the upper surface of the wing required to meet the conditions of lift for the aircraft.

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