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
we’ve been given a lot of information to set up the scenario. Let’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 𝑃 sub
𝑙 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 𝜌. And throughout this problem, we’ll
assume that value is an exact value. On this screen, we’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.
Let’s begin by looking at the first
part of this problem and we’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 we’ll establish two separate
locations: one right above the wing that we’ll call location one and the second spot
right below the wing that we’ll call location two.
In this part one of our problem,
we’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 we’ll write that down and label
that speed 𝑉 sub two. Recall that we’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; that’s 1.00 kilonewtons per square metre of
wing. The density of air at this
altitude, 𝜌, is 1.29 kilograms per cubic metre.
And this first part of the problem
asks us to solve for 𝑉 sub one, the speed of the air over the top part of the wing
required in order to produce a sustainable lift force. Let’s begin solving for 𝑉 one by
recognizing that this is a situation where Bernoulli’s equation applies.
Bernoulli’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 𝜌 times the speed at point one squared plus 𝜌 times 𝑔 times the
altitude at point one is equal to those seen terms at point two. So this is the canonical form of
Now, in our scenario, if we write
out this equation for our situation, we’ll see that not all of these terms stay. For example, look at the two terms
in this equation that involve heights: ℎ one and ℎ 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 airplane’s altitude. Therefore, 𝜌 𝑔 ℎ one is
effectively equal to 𝜌 𝑔 ℎ two, meaning that we can subtract this term from both
sides of our equation and cancel it out. That simplifies our equation.
Now, let’s begin to solve for 𝑉
sub one by rearranging, which can first involve subtracting 𝑃 sub one from both
sides cancelling that term out on the left side, and then multiplying both sides by
two divided by 𝜌. Cancelling the two and 𝜌 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 𝑉 sub one.
Looking at this cleaned up version
of the solution for 𝑉 sub one, we can begin to plug in our numbers. Looking at these first two terms 𝑃
sub two and 𝑃 sub one, we don’t know either one particularly, but we do know that
𝑃 sub two must be 1.00 kilonewtons per metre squared greater than 𝑃 sub one in
order for the aircraft lift to be sustained.
Continuing on, 𝜌 the density of
air at sea level we’re given as 1.29 kilograms per metre cubed and 𝑉 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 𝑉 sub one of 71.8 metres per
second. That’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 we’re told is
different is that 𝑉 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 we’re told about
is that 𝜌, 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 let’s call this new air density
𝜌 sub 𝑎 for 𝜌 at altitude. 𝜌 sub 𝑎 is equal to 25.0 percent
of 𝜌. And when we calculate that out,
25.0 percent of 1.29 kilograms per metre cubed equals 0.323 kilograms per metre
Now with these two new pieces of
information, we want to again solve for 𝑉 sub one, the speed of air as it moves
over the upper surface of the wing. We’ll again use Bernoulli’s
equation, we’ll again cancel out the terms involving height because the height
differences are negligibly small, and once more we’ll rearrange that resulting
equation to solve for 𝑉 sub one. Let’s take that equation for 𝑉 sub
one we see on screen now and modify it in light of the two new pieces of
information; namely, the new speed for 𝑉 two and the new air density, 𝜌 sub
In this expression, everywhere we
see a 1.29 kilograms per metre cubed, we’ll replace that with 𝜌 sub 𝑎, 0.232
kilograms per metre cubed. Having done that, now let’s replace
each instance of 𝑉 sub two, which originally was 60.0 metres per second, with our
updated value of 245 metres per second. With these updates, we’re now ready
to solve for 𝑉 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 𝑉 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.