Wind blows over the top of a 10.0 m tall chimney at 12.7 m/s. The air at the bottom of the chimney is stationary. What is the difference in air pressure between the top and bottom of the chimney?

Answer 1

This question is an illustration of Bernoulli's equation:

So, plug in your known values

P₁ + 0 + 0 = P₂ + 1/2*p(12.7 m/s)² + p*(10 m²/s²)*10.0 m

assuming standard sea-level conditions: p = 1.225 kg/m³ 

ΔP = 1/2*1.225 kg/m³ *(12.7 m/s)² + 1.225 kg/m³ *(10 m²/s²)*10.0 m

ΔP = 221 kg*s^-2*m^-1 = 221 Pa

So, the pressure at the bottom of the chimney is 221 Pa greater than at the top.

Answer 2

This question is based on Bernoulli's Principle

That is Sum of Kinetic Energy , Potential energy and Pressure energy is conserved

1/2 mV² + mgh + P* Volume = constant ( KE = 1/2 mV² ,G PE = mgh & Pressure Energy = Pressure * Volume)

Dividing by m

1/2 V² + gh + P/d = constant

Where d = density of air

1/2( 0) +g( 0 ) + P1/d = 1/2 ( 12.7) ² + g(10 ) +P2/d

(P1 - P2 ) /d = 1/2( 12.7 * 12.7) +9.8*10

(P1 - P2 )/d = 80.65 + 98= 178.65

Pressure Difference = d( 178.65) = 230.45 Pa

As d = 1.29Kg/m3 =Density of air