Question

Calculate the rate of flow of glycerine of density. 1.25xx10^(3) kg//m^(3) through the conical section of a pipe. If the radii of its ends are 1.0m and 0.04m and the pressure drop across its length is 10N//m^(2).

Answer 1

Thus the rate of flow of glycerin is Q = 6.43 × 10^−4 m^3/s

Explanation:

From continuity equation:

A1v1 = A2v2

or, v1v2 = A2A1 − πr2^2 / πr1^2 = (r2 / r1)^2

= (0.04 / 0.1)^2 = 4 / 25   -------(1)

From Bernouli's equation:

p1 + 1 / 2ρv1^2 = p2 + 1 / 2ρv2^2

v2^2 − v1^2 = 2(p1 − p2) / ρ

v2^2 − v1^2 = 2 × 10 / 1.25 × 10^3

= 1.6 × 10^−2 m^2/s^2    -------(2)

Solving Equations, (1) ans (2), we get v2 ≈ 0.128 m/s

Rate of volume flow through the tube

Q = A2v2 = (πr2^2)v2  

Q = π(0.04^2)(0.128)

Q  = 6.43 × 10^−4 m^3/s

Thus the rate of flow of glycerin is Q  = 6.43 × 10^−4 m^3/s