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!)

Respuesta :

The pipe must drop in height by 13.101 m in order to achieve this pressure.

Bernoulli stated that at whatever point along the fluid flow, the total amount of pressure, kinetic energy and potential energy per unit volume is the same.

P + 1/2 ρv² + ρ* g* h = constant

Given that, v = 1 m/s

P₁ = 101300 Pa

P₂ = 202600 Pa

ρ ethanol = 789 kg/m³

g = 9.8 m/s²

Now, let us write the bernoulli equation,

ρ*g*h₁ + P₁ = ρ*g*h₂ + P₂

P₁ - P₂ =  ρ*g*h₂ - ρ*g*h₁

( h₂ - h₁) = (P₁ - P₂)/ρ*g

Δh = (P₁ - P₂)/ρ*g

Δh = -101300 / ( 789 * 9.8) = -13.101 m

Minus sign indicates that height must be reduced.

To know more about bernoulli equation:

https://brainly.com/question/17074630

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