Question

A liquid of density 1290 kg/m 3 flows steadily through a pipe of varying diameter and height. At Location 1 along the pipe, the flow speed is 9.91 m/s and the pipe diameter d 1 is 10.5 cm . At Location 2, the pipe diameter d 2 is 16.7 cm . At Location 1, the pipe is Δ y = 9.01 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.

Answer 1

Answer:

114 kPa

Explanation:

By Bernoulli's equation when a fluid flows steadily through a pipe:

P + ρ*g*y + v² = constant in the pipe, where P is the pressure, ρ is the density of the fluid, g is the gravity acceleration (9.8 m/s²), y is the high, and v the velocity.

By the continuity equation, the liquid flow must be constant in the pipe, and then:

A1*v1 = A2*v2

Where A is the area, v is the velocity, 1 is the point 1, and 2 the point 2 in the pipe. The are is the circle area: π*(d/2)². So:

π*(0.105/2)²*9.91 = π*(0.167/2)²*v2

0.007v2 = 0.027

v2 = 3.9 m/s

Then:

P1 + ρ*g*y1 + v1² = P2 + ρ*g*y2 + v2²

ρ*g*y1 - ρ*g*y2 + v1² - v2² = P2 - P1

ρ*g*Δy + v1² - v2² = ΔP

ΔP = 1290*9.8*9.01 + 9.91² - 3.9²

ΔP = 113,987.42 Pa

ΔP = 114 kPa