Question

Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1.0 cm. If the amount of glycerine collected per second at one end is 4.0 × 10–3 kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine = 1.3 × 103 kg m–3 and viscosity of glycerine = 0.83 Pa s). [You may also like to check if the assumption of laminar flow in the tube is correct].​

Answer 1

Explanation:

Length of the horizontal tube, l = 1.5 m

Radius of the tube, r = 1 cm = 0.01 m

Diameter of the tube, d = 2r = 0.02 m

Glycerine is flowing at a rate of 4.0 × 10–3 kg s–1.

M = 4.0 × 10–3 kg s–1

Density of glycerine, ρ = 1.3 × 103 kg m–3

Viscosity of glycerine, η = 0.83 Pa s

Volume of glycerine flowing per sec:

V = M / ρ

= 4 × 10-3 / (1.3 × 103) = 3.08 × 10-6 m3s-1

According to Poiseville’s formula, we have the relation for the rate of flow:

V = πpr4 / 8ηl

Where, p is the pressure difference between the two ends of the tube

∴ p = V8ηl / πr4

= 3.08 × 10-6 × 8 × 0.83 × 1.5 / [ π × (0.01)4 ]

= 9.8 × 102 Pa

Reynolds’ number is given by the relation:

R = 4pV / πdη

= 4 × 1.3 × 103 × 3.08 × 10-6 / ( π × 0.02 × 0.83)

= 0.3

Reynolds’ numbe