Question

A brine solution with a density of 1230 kg/m3 moves through a constricted pipe in steady, ideal flow. At the lower point shown in the figure below, the pressure is P1 = 1.95 × 104 Pa, and the pipe diameter is 5.00 cm. At another point y = 0.35 m higher, the pressure is P2 1.00 104 Pa and the pipe diameter is 2.50 cm. P2 (a) Find the speed of flow (in m/s) in the lower section m/s (b) Find the speed of flow (in m/s) in the upper section. m/s Find the volume flow rate (in m3/s) through the pipe. (c) m2/s

Answer 1

Gravitational acceleration = g = 9.81 m/s²

Density of the fluid = $\rho$ = 1230 kg/m³

Pressure at point 1 = P₁ = 1.95 x 10⁴ Pa = 19500 Pa

Diameter of pipe at point 1 = D₁ = 5 cm = 0.05 m

Speed of flow at point 1 = V₁

Area of flow at point 1 = A₁

A₁ = $\pi$D₁²/4

Pressure at point 2 = P₂ = 1 x 10⁴ Pa = 10000 Pa

Diameter of pipe at point 2 = D₂ = 2.5 cm = 0.025 m

Speed of flow at point 2 = V₂

Height of point 2 compared to point 1 = y = 0.35 m

Area of flow at point 2 = A₂

A₂ = $\pi$D₂²/4

By continuity equation,

A₁V₁ = A₂V₂

($\pi$D₁²/4)V₁ = ($\pi$D₂²/4)V₂

D₁²V₁ = D₂²V₂

(0.05)²V₁ = (0.025)²V₂

V₂ = 4V₁

By bernoulli's equation,

$P_{1} + \frac{\rho V_{1}^{2}}{2} = P_{2} + \frac{\rho V_{2}^{2}}{2} + \rho gy$

19500+ 1230) 230)(41+ (1230) (9.81) (0.35)

V₁ = 0.756 m/s

V₂ = 4V₁

V₂ = 4(0.756)

V₂ = 3.024 m/s

Volume flow rate = Q

Q = A₁V₁

Q = $\pi$D₁²V₁/4

Q = $\pi$(0.05)²(0.756)/4

Q = 1.484 x 10^-3 m³/s

a) Speed of flow in the lower section = 0.756 m/s

b) Speed of flow in the higher section = 3.024 m/s

c) Volumetric flow rate = 1.484 x 10^-3 m³/s