At a given point on a horizontal streamline in flowing air, the static pressure is â2.0 psi (i.e., a vacuum) and the velocity is 150 ft/s. determine the pressure at a stagnation
1 answer:
At a point on the streamline, Bernoulli's equation is p/ρ + v²/(2g) = constant where p = pressure v = velocity ρ = density of air, 0.075 lb/ft³ (standard conditions) g = 32 ft/s² Point 1: p₁ = 2.0 lb/in² = 2*144 = 288 lb/ft² v₁ = 150 ft/s Point 2 (stagnation): At the stagnation point, the velocity is zero. The density remains constant. Let p₂ = pressure at the stagnation point. Then, p₂ = ρ(p₁/ρ + v₁²/(2g)) p₂ = (288 lb/ft²) + [(0.075 lb/ft³)*(150 ft/s)²]/[2*(32 ft/s²) = 314.37 lb/ft² = 314.37/144 = 2.18 lb/in² Answer: 2.2 psi
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