Water, with a density of rho = 1145 kg/m 3, flows in a horizontal pipe. In one segment of the pipe, the flow speed is v 1 = 7.33 m/s. In a second segment, the flow speed is v 2 = 1.57 m/s. What is the difference between the pressure in the second segment (P 2) and the pressure in the first segment (P 1) ?

the pressure difference will be ΔP = P₂ - P₁ = 29348.64 Pa

Explanation:

from Bernoulli's equation

P₁ + ρgh₁ + 1/2ρv₁² = P₂ + ρgh₂+1/2ρv₂²

where P = pressure, ρ = density, g = gravity, h = height, v=flow speed and 1 and 2 denote first and second segment respectively

then since the pipe is horizontal there is no difference in height (h=h₁=h₂), thus

P₁ + 1/2ρv₁² = P₂ + 1/2ρv₂²

the pressure difference ΔP = P₂ - P₁ will be

ΔP = P₂ - P₁ = 1/2ρv₁² - 1/2ρv₂² = 1/2*ρ * (v₁² - v₂²)

replacing values

ΔP = 1/2*ρ * (v₁² - v₂²) = 1/2 * 1145 kg/m³ * [ (7.33 m/s) ² - (1.57 m/s) ²] = 29348.64 Pa

ΔP = 29348.64 Pa