Question

Calculate rate a flow of glycerin of density 1.25*10^3 kg/m^3 through the conical section of a horizontal pipe if the radii of its ends are 0.1 m and 0.04 m and pressure drop across its length is 10 N/m

Answer 1

According to Bernoulli's theorem , the pressure of drop is given as
∆P = (ρ/2) [v1² - v2²]
also , A1v1 = A2v2
where A , v and ρ shows cross section area , velocity and density of liquid.

A2 = π(0.1)² , A1 = π(0.04)²
now, A1v1 = A2v2
(0.04)² v1 = (0.1)²v2
25v2= 4v1 ------(1)

now, ∆P = (ρ/2)[v1² - 4/25 v1²]
= (ρ/2) [21/25]v1²

now, here pressure drop across it's end is 10N/m² , e.g., ∆P = 10N/m² and ρ = 1.25 × 10³ Kg/m³

so, 10 = (1.25 × 10³/2 ) × (21/25) v1²
10 = 625 × 21/25 v1²
v1² = 10/25 × 21
v1 = √{2/105} m/s

Answer 2

hope it will help you out