Answer to Question #166298 in Mechanics | Relativity for Ronalyn Sacman

Question

Through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of 101300 Pa. The refinery needs the ethanol to be at a pressure of 2 atm (202600 Pa) on a lower level. The density of ethanol is 789 kg/m3 and gravity g is 9.8 m/s2. How far must the pipe drop in height in order to achieve this pressure? Assume the velocity does not change.

Answer 1

From the Bernoulli's equation:

$p_1+\frac{\rho v_1^2}{2}+\rho gh_1=p_2+\frac{\rho v_2^2}{2}+\rho gh_2,$

where $p_1$ and $p_2$ - the pressures at the levels h₁ and h₂ respectively;

$v_1=v_2=v$ - the velocity at the same levels;

$\rho$ - the density of ethanol;

g - the gravitational acceleration.

So,

$h_2-h_1=\frac{p_1-p_2}{\rho g}=\frac{202600-101300}{789\cdot 9.8}=13.1\space m.$

Answer: 13.1 m.