Model velocity and flow for oil pipeline (0.04 poise)

Flow in a pressurised pipe is dominated by viscous forces, so the Reynolds model law is used: $Re_m = Re_p$.

Given

Prototype: $D_p = 2$ m, $Q_p = 5\ \text{m}^3/\text{s}$, oil of s.g. 0.92, so $\rho_p = 920\ \text{kg/m}^3$, $\mu_p = 0.04$ poise $= 0.004\ \text{N s/m}^2$.

Model: $D_m = 0.2$ m, water $\rho_m = 1000\ \text{kg/m}^3$, $\mu_m = 0.01$ poise $= 0.001\ \text{N s/m}^2$.

Prototype velocity

$$V_p = \frac{Q_p}{\frac\pi4D_p^2} = \frac{5}{\frac\pi4(2)^2} = 1.5915\ \text{m/s}$$

Model velocity

$$\frac{\rho_mV_mD_m}{\mu_m} = \frac{\rho_pV_pD_p}{\mu_p} \;\Rightarrow\; V_m = V_p\,\frac{\rho_p}{\rho_m}\,\frac{D_p}{D_m}\,\frac{\mu_m}{\mu_p}$$ $$V_m = 1.5915\times\frac{920}{1000}\times\frac{2}{0.2}\times\frac{0.001}{0.004} = 3.66\ \text{m/s}$$

Model discharge

$$Q_m = \frac\pi4D_m^2V_m = \frac\pi4(0.2)^2(3.6606) = 0.115\ \text{m}^3/\text{s}$$

Answer: model velocity $\approx 3.66$ m/s; model discharge $\approx 0.115\ \text{m}^3/\text{s}$ (115 L/s).