Question

Oil of density 920 kg/m${}^3$ and dynamic viscosity 0.3 Pa-s flows in a pipe of diameter 15 cm at a velocity of 2 m/s. What would be the velocity of flow of water in a 1 cm diameter pipe, to make the two flows dynamically similary? The density and viscosity of water can be taken as a 998 kg/m${}^3$ and 1.31 $\times$ 10${}^{-3}$ Pa-s respectively.

• A. 0.08 m/s
• B. 0.1 m/s
• C. 0.12 m/s
• D. 0.8 m/s

Answer 1

The correct option is C 0.12 m/s

Reynold's similarity law is applicable (pipe flow).

$(Re{)}_m=\frac{V_m{d}_m}{v_m}=(Re{)}_P=\frac{V_p{d}_p}{v_P}$

$V_r=\frac{V_m}{V_p}=\frac{d_p{v}_m}{d_m{v}_p}=\frac{d_p{\mu }_m}{{\rho }_m}\times \frac{{\rho }_p}{d_m{\mu }_p}$

$\Rightarrow \frac{V_m}{V_p}=\frac{{\mu }_m}{{\mu }_p}\times \frac{1}{\left(\frac{L_m}{L_p}\right)\left(\frac{{\rho }_m}{{\rho }_p}\right)}$

Referring to oil with a subscript p and water with a suffix m.

$\frac{V_m}{V_p}=\frac{1.31\times {10}^{-3}}{0.3}\times \frac{1}{\frac{1}{15}\times \frac{998}{920}}=0.0604$

$\therefore V_m=$ Velocity of water $=V_p\times 0.0604=2\times 0.06=0.125m/s$