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].
The Poiseuille’s law states the relation:
Where,
‘p’ is the difference in the pressure between the two ends.
‘r’ is the radius of the tube.
‘ɳ’ is the viscosity of the fluid.
‘l’ is the horizontal length of the tube.
Given,
Length of the tube, l = 1.5 m
Radius of the tube, r = 1 cm=0.01 m
Mass of glycerine flowing per second, M = 4.0 × 10-3 kg/s
Density of the glycerine, ρ = 1.3 × 103 kg/m3
Viscosity of the glycerine, ɳ = 0.83 Pa s
Volume of glycerine flowing per second is given as,
⇒V= 
⇒ V=3.08 × 10-6 m3/sec
Applying Poiseuille’s law we get the relation,
⇒ 
⇒ p=9.8 × 102 Pa
To check whether the assumption of laminar flow in the tube is correct we use the Reynolds’s number relation given as,
Where,
and if the Reynolds number is less than 2000, the flow is laminar.
Where,
‘ρ’ is the density of the fluid.
‘V’ is the volume of fluid flowing per sec
‘d’ is the diameter of the pipe
‘ɳ’ is the viscosity of the fluid
Thus,
⇒ 
⇒ R =0.3
Since, the Reynolds’s number is less than 2000. Hence, the flow is laminar.
The pressure difference between the two ends of the tube is 9.8 × 102 Pa