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A liquid of density 1390 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.63 m/s and the pipe diameter d 1 is 10.1 cm . At Location 2, the pipe diameter d 2 is 15.3 cm . At Location 1, the pipe is Δ y = 8.85 m higher than it is at Location 2. Ignoring viscosity, calculate the difference Δ P between the fluid pressure at Location 2 and the fluid pressure at Location 1.

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davidlcastiblanco

Answer:

The difference of power is

ΔP = 172.767 kPa

Explanation:

ρ = 1390 kg / m³

v = 9.63 m/s

d₁ = 10.1 cm , d₂ = 15.3 cm

Δz = 8.85 m

To find the difference ΔP between the fluid pressure at locations 2 and the fluid pressure at location 1

ΔP = ρ * g * Z + ¹/₂ * ρ * v² * ( 1 - (d₁ / d₂)⁴ )

ΔP = 1390 kg / m³ * 9.8 m/s² * 8.85 m + 0.5 * 1390 kg / m³ *(9.63 m /s)² * (1 - (0.101 m / 0.153 m )⁴ )

ΔP = 172.767 x 10 ³ Pa

ΔP = 172.767 kPa

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