An air plane wing is designed so that the sped of air across the top of the wing is 450 m s⁻¹, the speed of air below the wing is 410 m s⁻¹. What is the pressure difference between the top and bottom of the wing? (density of air = 1.29 Kg m³) (Ans: 22 kPa)



Data Given:


Velocity at top of the wing  = v₁ = 450 m s⁻¹

Velocity below of the wing  = v₂ = 410 m s⁻¹

value of g = 9.8 m s⁻² 

Density of air = ⍴ = 1.29 kg m⁻³


To Find:


Pressure difference  = P₁ - P₂ = ?


Solution:


By using the formula Bernoulli's equation


P₁ + `\frac {1}{2}`⍴v₁²  =  P₂ + `\frac {1}{2}`⍴v₂² 

P₁ - P₂  = `\frac {1}{2}`⍴v₂² - `\frac {1}{2}`⍴v₁²  

P₁ - P₂  = `\frac {1}{2}`⍴ (v₂² - v₁²)


By putting values

P₁ - P₂  = `\frac {1}{2}` ｘ 1.29 kg m⁻³ ((450 m s⁻¹)²- (410 m s⁻¹)²) 

P₁ - P₂  = 0.645 kg m⁻³ (202500 m² s⁻² - 168100 m² s⁻²) 

P₁ - P₂  = 0.645 kg m⁻³ (34400m² s⁻²)

P₁ - P₂  = 22188  kg m⁻¹ s⁻²
 

As Pa = kg m⁻¹ s⁻²

P₁ - P₂  = 22188  Pa

P₁ - P₂  = 22.188 ｘ10³  Pa

P₁ - P₂ = 22 kPa -----------------Ans.


************************************

Shortcut Links For 


1. Website for School and College Level Physics   
2. Website for School and College Level Mathematics  
3. Website for Single National Curriculum Pakistan - All Subjects Notes 

© 2022-Onwards by Academic Skills and Knowledge (ASK)     

Note:  Write me in the comment box below for any query and also Share this information with your class-fellows and friends.