2
$\begingroup$

I'm going to ask my question, explain my problem, show my work, and then re-state my question.

Question: how is my solution for velocity different than another solution that I have found in literature?

Explanation of my problem: I have a (hyperbolic? or is it quadratic?) equation relating pressure drop and velocity, $v$: $$\tag{1} -(dp/dx-\rho g)=\frac{\mu}{kk_{rf}}v+\beta\rho v^2$$ I want to solve this equation for the velocity variable $v$.

My solution: I have performed the "complete the square" method, e.g., rearranging Eqn (1) $$\tag{2} \beta\rho v^2+\frac{\mu}{kk_{rf}}v=-(dp/dx-\rho g)$$ dividing through by $\beta\rho$ $$\tag{3} v^2+\frac{1}{\beta\rho}\frac{\mu}{kk_{rf}}v=\frac{-(dp/dx-\rho g)}{\beta\rho}$$ completing the square by adding $\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2$ to both sides: $$\tag{4} v^2+\frac{\mu}{\beta\rho kk_{rf}}v+\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2=\frac{-(dp/dx-\rho g)}{\beta\rho}+\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2$$ simplifying the LHS: $$\tag{5} \left(v+\frac{\mu}{2\beta\rho kk_{rf}}\right)^2=\frac{-(dp/dx-\rho g)}{\beta\rho}+\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2$$ taking the squareroot of both sides: $$\tag{6} v+\frac{\mu}{2\beta\rho kk_{rf}}=\pm \sqrt{\frac{-(dp/dx-\rho g)}{\beta\rho}+\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2}$$ and solving for $v$: $$\tag{7} v=-\frac{\mu}{2\beta\rho kk_{rf}}\pm \sqrt{\frac{-(dp/dx-\rho g)}{\beta\rho}+\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2}$$ For the equation to make sense I would take the positive root: $$\tag{8} v=-\frac{\mu}{2\beta\rho kk_{rf}}+\sqrt{\frac{-(dp/dx-\rho g)}{\beta\rho}+\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2}$$

Back to the question: The other solution I have found in literature is this: $$\tag{9} v=\frac{1}{2k\rho\beta}\left[-\frac{\mu}{k_{rf}}+\sqrt{\left(\frac{\mu}{k_{rf}}\right)^2-4k^2\rho\beta\left(dp/dx+\rho g\right)}\right]$$ Are Equations 8 and 9 equivalent? If so, how? (Also, is Eqn 1 a hyperbolic or quadratic equation?)

$\endgroup$
  • $\begingroup$ I found my solution, but I will leave this question up so that someone may answer if they wish. $\endgroup$ – Armadillo Dec 12 '17 at 18:59
0
$\begingroup$

$$v=\frac{1}{2k\rho\beta}\left[-\frac{\mu}{k_{rf}}+\sqrt{\left(\frac{\mu}{k_{rf}}\right)^2-4k^2\rho\beta\left(dp/dx+\rho g\right)}\right]$$

$$v=-\frac{\mu}{2\beta\rho kk_{rf}}+\frac{1}{2k\rho\beta}\sqrt{\left(\frac{\mu}{k_{rf}}\right)^2-4k^2\rho\beta\left(dp/dx+\rho g\right)}$$

$$v=-\frac{\mu}{2\beta\rho kk_{rf}}+\sqrt{\frac{1}{4k^2\rho^2\beta^2}\Bigg(\left(\frac{\mu}{k_{rf}}\right)^2-4k^2\rho\beta\left(dp/dx+\rho g\right)\Bigg)}$$

Distribute within the radical to acquire

$$v=-\frac{\mu}{2\beta\rho kk_{rf}}+\sqrt{\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2-\frac{(dp/dx-\rho g)}{\beta\rho}}$$

Rearrange to get

$$v=-\frac{\mu}{2\beta\rho kk_{rf}}+\sqrt{\frac{-(dp/dx-\rho g)}{\beta\rho}+\left(\frac{\mu}{2\beta\rho kk_{rf}}\right)^2}$$ as needed.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.