Question

The rate of flow of glycerine of density $1.25 \times 10^{3}\, kgm ^{-3}$ 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 $10 \,Nm ^{-2}$ 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 Bernoullis theorem
$p_{1}+\frac{1}{2} \rho v_{1}^{2}=p_{2}+\frac{1}{\alpha} \rho v_{2}^{2}$
$\therefore p_{1}-p_{2}=\frac{1}{2} \rho\left(v_{2}^{2}-v_{1}^{2}\right)$
$\therefore 10=\frac{1}{2} \times 1.25 \times 10^{3}\left(v_{2}^{2}-v_{1}^{2}\right)$
$\therefore v_{2}^{2}-v_{1}^{2}=\frac{10 \times 2}{1.25 \times 10^{3}}$
$=16 \times 10^{-3}$ ...(i)
Also from equation of continuity
$=A_{1} v_{1}=A_{2} v_{2} $
$\pi r_{1}^{2} v_{1}=\pi r_{2}^{2} v_{2}$
$\therefore \frac{v_{1}}{v_{2}}=\left[\frac{r_{2}}{r_{1}}\right]^{2}=\frac{0.04}{0.1}=0.4$
$v_{1}=0.4\, v_{2}$ ... (ii)
Substituting this value in Eq. (i)
$v_{2}^{2}-\left(0.4\, v_{2}\right)^{2}=16 \times 10^{-3}$
$ v_{2}=1.38 \times 10^{-1}=0.138\, ms ^{-1}$
Rate of flow of glycrine $v=A_{2} v_{2}$
$=\pi r_{2}^{2} v_{2}$
$=6.93 \times 10^{-4} \,m^{3}\, s^{-1}$