# Thermodynamics/Fluids Homework Problem

1. Jan 29, 2014

### rcummings89

Hello, the problem statement reads thus:

"Consider a low-speed flow of air over an airplane wing at standard sea level conditions; the free-stream velocity far ahead of the wind [point 1] is 100 mi/h [44.7m/s]. The flow accelerates over the wing reaching a maximum velocity of 150 mi/h [67.1 m/s] at some point on the wing [point 2]. What is the percentage pressure change between this point and the free stream?" (I inserted the brackets)

I have never come across the term (to the best of my knowledge) "percentage pressure change" in fluids or thermo so I'm a little unsure if I am doing this right.

My first instinct was to use Bernoulli's Eqn., neglecting height differences:

P1 + $\frac{1}{2}$ρairv12 = P2 + $\frac{1}{2}$ρairv22 Where P1 = 0 gage. Then P2 = $\frac{1}{2}$ρair(v12 - v22) = -1.508 kPa

But it mentions the percentage change, so my next idea was to find the pressure at points 1 & 2 using just $\frac{1}{2}$ρairvn2 and obtains P1 = 1.203 kPa, and P2 = 2.707 kPa;

Then I interpreted the percentage change as

(P2-P1)/P1 * 100 = 125%.

Am I in the right direction or completely lost?

2. Jan 29, 2014

### LoopInt

You found that P2 is higher than P1, and this is a bad sign. If you increase the velocity, pressure should lower.
In your problem you can consider P1 to be atmospheric since its at sea level. Then you plug in the values for v1 and v2 and density, and find P2.

Then you plug in p1 and p2 you've found on the percentage formula you wrote and it should work.
I got -1,48%. Makes sense that its negative, because the pressure at P2 should be lower than P1.

I'm new to fluid mechanics. Maybe someone can verify that its correct.