Question

The rate of flow of glycerine of density $ 1.25\times {{10}^{3}}kg{{m}^{-3}} $ through the conical section of a pipe if the radii of its ends are $0.1\, m$ and $0.04 \,m$ and the pressure drop across its length $ 10\text{ }N{{m}^{-2}} $ is

• A. $ 6.93\times {{10}^{-4}}{{m}^{3}}{{s}^{-1}} $
• B. $ 7.8\times {{10}^{-4}}{{m}^{3}}{{s}^{-1}} $
• C. $ 10.4\times {{10}^{-5}}{{m}^{3}}{{s}^{-1}} $
• D. $ 14.5\times {{10}^{-5}}{{m}^{3}}{{s}^{-1}} $

Answer 1

Answer: (A)

From Bernoulli's theorem
$p_{1}+\frac{1}{2} \rho v_{1}^{2}=P_{2} \frac{1}{2} \rho v_{2}^{2}$
$\therefore p_{1}-p_{2}=\frac{1}{2} \rho\left(v_{2}^{2}-v_{1}^{2}\right) $
$\therefore 10=\frac{1}{2} \times 1.25 \times 10^{3}\left(v_{2}^{2}-v_{1}^{2}\right) $
$\therefore v_{2}^{2}-v_{1}^{2}=\frac{10 \times 2}{1.25 \times 10^{3}}=16 \times 10^{-3}\,\,\,...$(i)
Also, from equation of continuity
$=A_{1} v_{1}=A_{2} v_{2}$
Or $ \pi r_{1}^{2} v_{1}=\pi r_{2}^{2} v_{2} $
$\therefore \frac{v_{1}}{v_{2}}=\left[\frac{r_{2}}{r_{1}}\right]^{2}=\frac{0.04}{0.1}=0.4 $
ie, $ v_{1}=0.4 v_{2} ....$ (ii)
Substituting this value in Eq. (i)
$v_{2}^{2}-\left(0.4 v_{2}\right)^{2}=16 \times 10^{-3}$
$\Rightarrow v_{2}=1.38 \times 10^{-1}=0.138\, ms ^{-1}$
Rate of flow of glycerine $v=A_{2} v_{2}$
$=\pi r_{2}^{2} v_{2}$
$=6.93 \times 10^{-4} \,m ^{3}\, s ^{-1}$