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

Given, the radius and length of the pipe is 1.0 cm and 1.5 m respectively, the mass flow rate of glycerin is 4.0× 10 −3   kg/s , and the density and viscosity of glycerin is 1.3× 10 3   kg/ m 3 and 0.83 Pa⋅s respectively.

The flow rate of glycerin is given by the equation,

V= M ρ

Substituting the values in the above equation, we get:

V= 4.0× 10 −3   kg/s 1.3× 10 3   kg/ m 3 =3.08× 10 −6   m 3 /s

From Poiseuille’s equation,

V= πp r 4 8ηl p= 8Vηl π r 4

Substituting the values in the above equation, we get:

p= 8( 3.08× 10 −6   m 3 /s )( 0.83 Pa⋅s )( 1.5 m ) π ( 1.0× 10 −2  m ) 4 =9.8× 10 2  Pa

Reynolds’ number is given by the equation,

R= 4ρV πdη = 4ρV π( 2r )η

Substituting the values in the above equation, we get:

R= 4( 1.3× 10 3   kg/ m 3 )( 3.08× 10 −6   m 3 /s ) π( 2×1.0× 10 −2  m )( 0.83 Pa⋅s ) =0.3

Since the Reynolds’ number of the flow is 0.3, the flow is laminar.