Density of a liquid in a horizontal pipe

[Joe Cruz]

1) Alright, so the problem reads, "The liquid in a horizontal pipe of diameter d1 before a constriction is at pressure P1 and flowing with the speed v1, and then in the constriction with diameter d2 the gauge measures pressure P2. What is the density of the liquid?"

2) I'm assuming the equations I'd need would be the equation of continuity, A₁v₁ = A₂v₂, and Bernoulli's equation.

3) So my first step was to figure out what v₂ was by using the equation of continuity which gave me
v₂ = v₁d₁² / d₂²

The second step was to set up Bernoulli's equation, P₁ + ρv₁²/2 + ρgh = P₂ + ρv₂²/2 + ρgh. Since the pipe is horizontal ρgh cancels out on both sides leaving P₁ + ρv₁²/2 = P₂ + ρv₂²/2. Then I began the process of isolating ρ.

ρ(v₁²/2 - v₂²/2) = P₂ - P₁

ρ = (P₂ - P₁) / (v₁²/2 - v₂²/2)

ρ = (P₂ - P₁) / [v₁²/2 - (v₁²d₁⁴)/ (2d₂⁴)]

ρ = (P₂ - P₁) / [(v₁²d₂⁴ - v₁²d₁⁴) / 2d₂⁴]

And finally I arrived at my final simplified solution;

ρ = [2d₂⁴ (P₂ - P₁)] / [v₁² (d₂⁴ - d₁⁴)]

I'm not all that confident in my math for this particular problem. Any input would be greatly appreciated as an exam approaches and I want to make sure I have a good understanding of this particular type of problem.

 

[mfb]

Looks right.

As d₂<d₁, both numerator and denominator will be negative. You can change that by swapping the sign in both of them, but it does not change the result.