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 $4.0\times {10}^{–3}kg{s}^{–1}$ , what is the pressure difference between the two ends of the tube? (Density of glycerine =$1.3\times {10}^{3}kg{m}^{–3}$ 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].

Answer 1

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 $4.0\times {10}^{–3}kg{s}^{–1}.$
$M=4.0\times {10}^{–3}kg{s}^{–1}$
Density of glycerine, $\rho =1.3\times {10}^{3}kg{m}^{–3}$
Viscosity of glycerine, $\eta =0.83Pas$
Volume of glycerine flowing per sec:
$V=\frac{M}{\rho }=\frac{4.0\times {10}^{-3}}{1.3\times {10}^{-3}}=3.08\times {10}^{-6}{m}^3{s}^{-1}$
According to Poiseville’s formula, we have the relation for the rate of flow:
$V=\frac{\pi p{r}^4}{8\eta l}$
Where, p is the pressure difference between the two ends of the tube
$\therefore p=\frac{V8\eta l}{\pi r^4}$
$=\frac{3.08\times {10}^{-6}\times 8\times 0.83\times 1.5}{\pi \times (0.01{)}^4}=9.8\times {10}^{2}Pa$
Reynolds’ number is given by the relation:
$R=\frac{4\rho V}{\pi d\eta }=\frac{4\times 1.3\times {10}^3\times 3.08\times {10}^{-6}}{\pi \times \left(0.02\right)\times 0.83}=0.3$
Reynolds’ number is about 0.3.
Hence, the flow is laminar.