Answer :

According to the question, the Length of the horizontal tube, l = 1.5 m

and the radius of the tube is

r = 1 cm = 0.01 m

Therefore diameter of the tube becomes-

d = 2r = 0.02 m

It is given that the glycerine is flowing at the rate of 4.0 × 10^-3 kg/s

Mass of the glycerine, M = 4.0 × 10^-3 kg/s

Therefore, the density of glycerine,

ρ = 1.3 × 10³ kg m^-3

We have the viscosity of glycerine, η = 0.83 Pa

Expression for the volume of glycerine flowing per sec is –

$ V=\frac{M}{Density}=\frac{4\times {{10}^{-3}}}{1.3\times {{10}^{3}}} $

$ V=3.08\times {{10}^{-6}}{{m}^{3}}/s $

Using Poiseville’s formula, we get the expression :

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

$ \therefore {{p}^{‘}}=\frac{V8\eta l}{\pi {{r}^{4}}} $

Where p’ denotes the pressure difference between the two ends of the given pipe.

$ {{p}^{‘}}=\frac{3.08\times {{10}^{-6}}\times 8\times 0.83\times 2}{\pi \times {{\left( 0.01 \right)}^{4}}} $

$ {{p}^{‘}}=9.8\times {{10}^{2}}Pa $

We know that the expression for reynold’s Number is

$ R=\frac{4\rho V}{\pi d\eta }=\frac{4\times 1.3\times {{10}^{3}}\times 3.08\times {{10}^{-6}}}{\pi \times 0.83\times 0.02} $

$ R=0.3 $

The flow of glycerine in the pipe is laminar since the Reynolds number is 0.3, which is much lower than 2000.