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

Answer: 9.8 × 10² Pa

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:

= 3.08 × 10^–6 m³ s^–1

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

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

= 9.8 × 10² Pa

Reynolds’ number is given by the relation:

Reynolds’ number is about 0.3. Hence, the flow is laminar.