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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. how far must the pipe drop in height in order to achieve this pressure? assume the velocity does not change. (hint: use the bernoulli equation. the density of ethanol is 789 kg/m3 and gravity g is 9.8 m/s2. pay attention to units!)



Answer :

Δh = -13.101 m Far must the pipe drop in height in order to achieve this pressure.

13,101 m must be removed from the height.

Bernoulli went on to say that at whatever point along the fluid flow, the total amount of pressure, kinetic energy, and potential energy per volume is the same.

P+ 1/2 ρv² + ρgh = constant

Given

v = 1 m/s ⇒ constant

P₁ = 101300 Pa (1 atm)

P₂ = 202600 Pa (2 atm)

ρ ethanol = 789 kg/m3

g = 9.8 m/s2

Required

The height(lower level)

Solution

The Bernoulli equation becomes when v₁=v₂ constant

ρgh₁+P₁=ρgh₂+P₂

Solve for the value (height difference):

P₁-P₂=ρgh₂-ρgh₁

P₁-P₂=ρg(h₂-h₁)

(P₁-P₂)/(ρg) = Δh

- 101300 kg/ms² / 789 kg/m³ × 9.8 m/s² = Δh

Δh = -13.101 m

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