Mechanical Properties of Fluids

Question : 13

Total: 31

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 4.0×10–3kg s–1, what is the pressure difference between the two ends of the tube? (Density of glycerine =1.3×103kg 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 m;r=1 cm=0.01 m;η=0.83 Pa s
mass of the glycerine flowing per sec,
m=4.0×10−3 kgs−1
density of glycerine, ρ=1.3×103 kg m−3
Therefore, volume of glycerine flowing per second,
V=

4.0×10−3

1.3×103

=3.077 ×10−6m3 s−1
‌Now,V=

πpr4

8 ηl

or p=

8V ηl

πr4

8×3.077×10−6×0.83×1.5

π×(0.01)4

=975.37Pa

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