Question

Crude oil having density and viscosity of $850\ \text{kg/m}^3$ and $2\times10^{-3}\ \text{Pa}\!\cdot\!\text{s}$, respectively, is flowing at an average velocity of $0.35\ \text{m/s}$ through a horizontal capillary tube. The inside diameter and length of the capillary tube are $2.5\times10^{-3}\ \text{m}$ and $0.30\ \text{m}$, respectively. The Fanning friction factor is $f=\dfrac{16}{\mathrm{Re}}$. The pressure drop across the capillary tube is ________ Pa (rounded to one decimal place).

Hint:

For laminar flow in a tube: $f=\dfrac{16}{\mathrm{Re}}$ and with Fanning $f$, $\Delta P = 2 f \dfrac{L}{D}\rho v^2$.

Answer 1

- Reynolds number: $\mathrm{Re}=\dfrac{\rho v D}{\mu}=\dfrac{850\times0.35\times2.5\times10^{-3}}{2\times10^{-3}}=371.875$.
- Fanning factor: $f=\dfrac{16}{\mathrm{Re}}=\dfrac{16}{371.875}=0.043025$.
- Using Darcy–Weisbach with Fanning $f$: $\Delta P = 2 f \dfrac{L}{D}\,\rho v^2$.
$\Rightarrow \Delta P = 2(0.043025)\left(\dfrac{0.30}{2.5\times10^{-3}}\right) (850)(0.35)^2 = 1075.2\ \text{Pa}$.