Prototype: Diameter of pipe $D_{p}$=1.2m, Viscosity$\mu_{p}$=0.9paise

Discharge, $Q_{p}=1000 liters/s=1m^{3}/s$, specific gravity Sp=0.9

$P_{p}=specific gravity=s_{m}=1,\mu=0.01$paise

In case of flow the dynamic similarity will be applicable. This dynamic similarity will be applicable. This dynamic similarity will be obtained when the Reynold's number both for thr rprototype and model are equal i.e,$(p_{e})_{p}=(R_{e})_{m}$

$\frac{P_{p},v_{p},D_{p}}{\mu_{p}}=\frac{p_{m},v_{m},D_{m}}{\mu_{m}}$

$\frac{v_{m}}{v_{p}}=\frac{\mu_{p}}{\mu_{m}}\times \frac{\mu_{m}}{\mu_{p}}\times \frac{D_{p}}{D_{m}}=\frac{900}{1000}\times \frac{0.01}{0.9}\times \frac{1.2}{0.1}$

$v_{m}=0.12v_{p}$

Velocity of flow in prototype pipe

$v_{p}=\frac{Rate \ of \ discharge}{Area \ of \ pipe A_{p}}$ = $\frac{QP}{(\frac{\pi}{4})D_{p}^{2}}$

$=\frac{4}{\pi}\times \frac{1}{(1.2)^{2}}=0.8842$m/s

$v_{m}=0.1time v_{p}=0.12\times 0.8842=0.1061 m/s$

Rate of flow in the pipe of model $Q_{m}$

$Q_{m}=A_{m}.v_{m}=\frac{\pi}{4}(D^{2}_{m})\times v_{m}$

$\frac{\pi}{4}\times (0.1)^{2}\times 0.1061=8.333\times 10^{-4}m^{3}/s$