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

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 × 10 ³ kg m ^{- 3}
Viscosity of glycerine, η = 0.83 Pa s
Volume of glycerine flowing per sec : 

\(V =\frac{ M }{ P} \)

\(= \frac{4.0 × 10 ^{- 3} }{ 1.3 × 10^ 3} \)

= 3.08 × 10 ^{- 6} m3 s^-1
According to Poiseville’s formula, we have the relation for the rate of flow : 

\(V = \frac{πpr4 }{ 8ηl}\)
Where, p is the pressure difference between the two ends of the tube

\(∴ P = \frac{V8ηl }{ πr^4 }\)

\(=\frac{ 3.08 × 10 - 6 × 8 × 0.83 × 1.5 }{π × (0.01)^4 }\)

= 9.8 × 10² Pa
Reynolds’ number is given by the relation : 

\(R = \frac{4ρV }{πdη }\)

\(=\frac{ 4 × 1.3 × 10 3 × 3.08 × 10 ^{- 6} }{ π × (0.02) × 0.83 }= 0.3\)
Reynolds’ number is about 0.3. Hence, the flow is laminar.