# Aerospace Finding velocity of air at a point on a wing

1. Jan 8, 2010

### lp2789

Hi all, Could somebody point me in the right direction with a question that I am stuck on, thanks.

An F-16 fighter jet is flying at 220 km/h as it commences to land. The
atmospheric pressure and density are 101.3 kN/m2 and 1.226 kg/m3
respectively. At a point on the upper surface of the wing, the pressure is
measured equal to 98.6 kN/m2. Using Bernoulli’s equation, find the
velocity of the flow at this point of the wing and the total pressure acting
on the aircraft.

My first thought was to use the 1d continuity form of the bernoulli equation,
P1-P1 = 0.5p(V2^2 - V1^2), where P1 and V1 are atmospheric pressure and aircraft velocity and P2 and V2 are the respective velocity and pressure at the point on the wing. Is this correct?
thanks

2. Jan 8, 2010

### Astronuc

Staff Emeritus
P1 - P2 = 0.5p(V22 - V12), then one is OK.

It's better to start with p1/ρ + V12/2 = p2/ρ + V22/2, then group the variables into unknown and known.

3. Jan 8, 2010

### lp2789

Thanks, but how would I go about finding the total pressure acting on the aircraft?

4. Jan 8, 2010

### mugaliens

Atmospheric. You have the two pressures and the density. All you need do is solve for velocity.

5. Jan 8, 2010

### minger

By definition the total pressure is the pressure that the fluid would have if it were isentropically brought to rest. So, you were given a pressure and calculated a velocity. Convert that velocity into pressure. If the velocites are high, you may want to consider using compressible flow equations rather than Bernoulli's.

6. Jan 8, 2010

### Cyrus

More aptly, if Ma > 0.3, then account for compressibility.

7. Jan 8, 2010

### minger

Damn, it's like you're following me around with poop on a stick swinging it around anytime I make a post.

8. Jan 8, 2010

### Brian_C

This isn't a homework board.

9. Jan 8, 2010

### Cyrus

Hahahhaahha. It's because I love you. This is getting awkward. I think I'll go back to stalking you.

Really though, your post was good. I was just giving a precise criteria rather than 'if your velocity is fast'.

Last edited: Jan 8, 2010
10. Jan 8, 2010

### minger

Well Ma>0.3 is purely subjective and is up to interpretation. Maybe you're at Mach 0.2 and just want a little more accuracy. Who are we to determine when compressibility effects start to become a factor in his problem? Huh Huh??

By the way, I can see you in the tree and please put your pants back on.

11. Jan 8, 2010

### Cyrus

No its not, see andersons book for a plot of the change in density vs mach number. For Ma < 0.3 the variation in density is less than 5%. This is the accepted definition of low speed (incompressible) aerodynamics.

Last edited: Jan 8, 2010
12. Jan 8, 2010

### minger

You mean:
$$\frac{\rho}{\rho_t} = \left(1+\frac{\gamma-1}{2}M^2\right)^{\frac{-1}{\gamma-1}}$$
It looks to me like a lot of constants, multiplied by an exponential functions of Mach number.

The fact of the matter is what is 5%? Why is 5% important? It is similar to boundary layer theory. Why is the boundary layer defined when the velocity is 0.99 the freestream velocity. Why not 0.999? Why not 0.98?

The fact of the matter is that these are simply arbitrary values. A reference standard set such that people can talk about the same thing. Furthermore, is it wrong to use compressible flow equations for Mach<0.3? Absolutely not. In fact, you'll get a more accurate answer everytime. By how much? Well, you said at Ma=0.3 we have 5% density error. Is 5% acceptable?

13. Jan 9, 2010

### Cyrus

Ugh, stop saying "The fact of the matter" I hate that phrase. But anyways, I never said you did not have to use compressibility equations below Ma < 0.3, it is that the added complexity is will only give you an increase in your answer, at best 5%. :tongue2:

14. Jan 9, 2010

### mugaliens

Precisely. Cyrus, it's a rule of thumb, and one that's commonly accepted. But it's not an ironclad rule.

15. Jan 9, 2010

### Cyrus

I wasn't implying that it was. I was just giving a number rather than a vagary of "when the velocity is fast enough". What is: "fast enough"? Why is it "fast enough"? These are answered in rule I stated.

16. Jan 9, 2010

### minger

That's just something that I think needs to be addressed on a per-problem basis rather than simply based on a rule-of-thumb.

The fact of the matter is that each engineer needs to address the problem before hand. The first step in any problem is making assumptions. If one is OK living with errors up to 5%, then by all means save your company the time by doing a truncated analysis.

The point I was trying to make is simply that arbitrary rules of thumb should be taken with a grain of salt and each problem we solve is usually unique and often times necessitates a unqiue set of inputs and methodology.

17. Jan 9, 2010

### Cyrus

Of course.

Damn you................damnnnn youuuu!!

Absolutely.