Question

Through a refinery, fuel ethanol is flowing in a pipe at a velocity of 1 m/s and a pressure of 101300 Pa (1 atm). The refinery needs the ethanol to be at a pressure of 200,000 Pa on a lower level. How far must the pipe drop in height in order to achieve this pressure? Assume the velocity does not change. (The density of ethanol is 789 kg/m^3 and gravity g is 9.8 m/s^2.

Answer 1

According to bernaulli theorm - when liquid flows in a pipe then sum of kinetic enrgy per unit volume, pressure energy and potential energy per unit volume is constant.

i.e -

1/2 ρv^2 + ρgH1 + P1 = 1/2 ρ v^2 + ρgH2 + P2

Since P1 = 101300 pa (N /m ^2)

P2 = 200000 pa

And ρ = density of liquid in kg/m^3

Since velocity at both point lower and upper is, same so kinetic enrgy per unit volume will be same. And the equation becomes -

ρgH1 + P1 = P2 + ρgH2

ρg ( H1 - H2) = P2 - P1

(H1 - H2) = (200000 - 101300) / 9.8×789

( H1 -H2) = 12.76 m

This is required drop in height to achieve this pressure.