Mechanical Properties of Fluids

Question : 13 of 31

Marks: +1, -0

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 \times 10^{-3}\,\text{kg}\,\text{s}^{-1}

, what is the pressure difference between the two ends of the tube? (Density of glycerine

= 1.3 \times 10^{3}\,\text{kg}\,\text{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:

Here,

l=1.5\ \text{m} ; r=1\ \text{cm}=0.01\ \text{m} ; \eta=0.83\ \text{Pa}\ \text{s}

mass of the glycerine flowing per sec,

m=4.0 \times 10^{-3}\,\text{kg}\,\text{s}^{-1}

density of glycerine,

\rho=1.3 \times 10^{3}\,\text{kg}\,\text{m}^{-3}

Therefore, volume of glycerine flowing per second,

V=\frac{m}{\rho}

=\frac{4.0 \times 10^{-3}}{1.3 \times 10^{3}}

=3.077 \times 10^{-6}\,\text{m}^{3}\,\text{s}^{-1}

\text{Now},\ V=\frac{\pi p r^{4}}{8 \eta l}

p=\frac{8 V \eta l}{\pi r^{4}}

=\frac{8 \times 3.077 \times 10^{-6} \times 0.83 \times 1.5}{\pi \times (0.01)^{4}}

=975.37\,\text{Pa}

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