- #1
rcummings89
- 19
- 0
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 + [itex]\frac{1}{2}[/itex]ρairv12 = P2 + [itex]\frac{1}{2}[/itex]ρairv22 Where P1 = 0 gage. Then P2 = [itex]\frac{1}{2}[/itex]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 [itex]\frac{1}{2}[/itex]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?
Thanks in advance!
"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 + [itex]\frac{1}{2}[/itex]ρairv12 = P2 + [itex]\frac{1}{2}[/itex]ρairv22 Where P1 = 0 gage. Then P2 = [itex]\frac{1}{2}[/itex]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 [itex]\frac{1}{2}[/itex]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?
Thanks in advance!