Question

Calculate the Reynolds number for an oil gusher that shoots crude oil 25.0 m into the air through a pipe with a 0.100-m diameter. The vertical pipe is 50 m long. Take the density of the oil to be 900 kg/m3 and its viscosity to be 1.00 (N/m2)·s (or 1.00 Pa·s).

Answer 1

Answer:

$Re=1992.24$

Explanation:

Given:

vertical height of oil coming out of pipe, $h=25\ m$

diameter of pipe, $d=0.1\ m$

length of pipe, $l=50\ m$

density of oil, $\rho = 900\ kg.m^{-3}$

viscosity of oil, $\mu=1\ Pa.s$

Now, since the oil is being shot verically upwards it will have some initial velocity and will have zero final velocity at the top.

Using the equation of motion:

$v^2=u^2-2gh$

where:

v = final velocity

u = initial velocity

Putting the respective values:

$0^2=u^2-2\times 9.8\times 25$

$u=22.136\ m.s^{-1}$

For Reynold's no. we have the relation as:

$Re=\frac{\rho.u.d}{\mu}$

$Re=\frac{900\times 22.136\times 0.1}{1}$

$Re=1992.24$