Question

Pressure gauge indicate that the pressure at point 1 is 10kgPa greater than the pressure at point 2. The cross sectional area at point 2 is 1/5th the cross sectional area at point 1. Determine the speed of the air at point 2

Answer 1

P₁ - is pressure which pressure gauge indicates at the point 1

P₂ - is pressure which pressure gauge indicates at the point 2

$\Delta$P = 10kP

P₁ = P₂ - $\Delta$P

S₂ = S₁ / 5; S₁ = 5S₂

T = const in pressure gauge

according to ideal gas law

PV = const

P₁S₁H = P₂V₂H -> (P₂ - $\Delta$P) *5S₂ = P₂ S₂

P₂ = 1.25P₁ = 12500Pa

v = $\sqrt{3P/\rho}$ = >

$\rho$ = 1.29kg/m³ - for air

v = $\sqrt{3 * 12500/1.29}$ = 170m/s