__Question 10.13__:
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 × 10^{3}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].

__Solution__:*l*= 1.5 m

Radius of the tube,

*r*= 1 cm = 0.01 m

Diameter of the tube,

*d*= 2

*r*= 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 = M /

*ρ*

= 4 X 10

^{-3}/ (1.3 X 10

^{3}) = 3.08 X 10

^{-6}m

^{3}s

^{-1}

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

V = πpr

^{4}/ 8

*η*l

Where,

*p*is the pressure difference between the two ends of the tube

∴ p = V8

*η*l / πr

^{4}

= 3.08 X 10

^{-6}X 8 X 0.83 X 1.5 / [ π X (0.01)

^{4}]

= 9.8 × 10

^{2}Pa

Reynolds’ number is given by the relation:

R = 4

*p*V / πd

*η*

= 4 X 1.3 X 10

^{3}X 3.08 X 10

^{-6}/ ( π X 0.02 X 0.83)

= 0.3

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

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