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jeka94
3 years ago
5

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).
Physics
1 answer:
Alenkasestr [34]3 years ago
6 0

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.

<u>Using the equation of motion:</u>

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}

<u>For Reynold's no. we have the relation as:</u>

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

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

Re=1992.24

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