Question

Glycerine flows steadily through a horizontal tube of length 1.5 m and radius 1 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? $\left({\rho }_{\text{Glycerine }}=1.3\times 10^{3}kg{m}^{-3}\right)\left({\eta }_{\text{Glycerine }}=0.83Pa.s\right)$

• A. 975 Pa
• B. 1025 Pa
• C. 1050 Pa
• D. 1200 Pa

Answer 1

Answer: (A)

Using Poisuelle's formula, $\frac{V}{t}=\frac{\pi P{r}^4}{8\cdot \eta \cdot l}$
$\frac{V}{t}=\frac{M}{\rho \cdot t}=\frac{4\times 10^{-3}kg{s}^{-1}}{1.3\times 10^{3}kg{m}^{-3}}$
Now, $P=\left(\frac{V}{t}\right)\cdot \left(\frac{8\eta l}{\pi r^4}\right)$
$=\left(\frac{4\times 10^{-3}kg{s}^{-1}}{1.3\times 10^{3}kg{m}^{-3}}\right)\cdot \left(\frac{8\times 0.83Pas(1.5m)}{3.14\times {\left(10^{-2}m\right)}^4}\right)$
$=975Pa$