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This question stems from gas flow through pipelines, quoting this paper (link):

For the calculation of the $z$ factor one needs the average pressure $p_{av}$. An obvious choice is the arithmetic average $$\tag{1} p_{ava}=\frac{p_b+p_e}{2}$$

However the pressure over the pipeline is not linear but quadratic. If the influence of the $z$-factor is ignored, the pressure over the pipeline is $$\tag{2} p(x)=\sqrt{p_b^2-cx}$$ with $$\tag{3} cL=p_b^2-p_e^2$$ The average pressure over the piple line $p_{avl}=$ $$\tag{4} \frac{1}{L}\int_0^L \sqrt{p_b^2-cx}\ dx=\frac{-2}{3cL}(p_b^2-cx)^{\frac{3}{2}}|_0^L=\frac{2}{3cL}(p_b^3-p_e^3)$$ Elimination of $cL$ gives $$\tag{5} p_{avl}=\frac{2}{3}\frac{(p_b^3-p_e^3)}{(p_b^2-p_e^2)}$$

$p_b$=upstream pressure; $p_e$=downstream pressure; $L$=length of pipe; $x$ is distance along length of pipe; $c$=?(author did not state)

Questions:

  1. How does one derive Eqn. (2)? (Since the $z$-factor is ignored, I guess one would start with the ideal gas law,$pV=nRT$? I guess this derivation also assumes isothermal flow.)
  2. I guess to answer the first question, one needs to know what the $c$ variable is...what is it?
  3. If one were to not ignore the influence of the $z$-factor, how would one derive the average pressure over the length of pipe? (assuming isothermal flow)
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  • $\begingroup$ Do you know how to factor the numerator and denominator of Eqn. 5 so that you can reduce this fraction to lowest terms? $\endgroup$ Commented Jul 12, 2017 at 11:24
  • $\begingroup$ @ChesterMiller $p_b^3-p_e^3=(p_b-p_e)(p_e^2+p_b^2+p_ep_b)$ and $p_b^2-p_e^2=(p_e-p_b)(p_e+p_b)$ therefore $\frac{(p_e^2+p_b^2+p_ep_b)}{(p_e+p_b)}$ $\endgroup$
    – Armadillo
    Commented Jul 12, 2017 at 13:57
  • $\begingroup$ @ChesterMiller however, I have also seen the equation factored as: $$\frac{2}{3}(p_e+p_b-\frac{p_ep_b}{p_e+p_b})$$ which I am not sure how to get $\endgroup$
    – Armadillo
    Commented Jul 12, 2017 at 14:02
  • $\begingroup$ Reduce to common denominator. $\endgroup$ Commented Jul 12, 2017 at 14:52
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    $\begingroup$ @ChesterMiller From $$\frac{(p_e^2+p_b^2+p_ep_b)}{(p_e+p_b)}$$ Add $(p_ep_b-p_ep_b)=0$ to the numerator, $$\frac{(p_e^2+2p_ep_b+p_b^2-p_ep_b)}{(p_e+p_b)}$$ which can be factored to $$\frac{(p_e+p_b)^2}{(p_e+p_b)}-\frac{p_ep_b}{(p_e+p_b)}=p_e+p_b-\frac{p_ep_b}{(p_e+p_b)}$$ $\endgroup$
    – Armadillo
    Commented Jul 12, 2017 at 19:15

1 Answer 1

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1 and 2. To get the constant c, you treat the flow as fully developed locally, calculate the Reynolds number, represent the density in terms of the pressure (using the ideal gas law), determine the friction factor, and determine the local shear stress at the wall. This all then gives you the gradient of $ p^2$. You then integrate along the pipe.

You are correct, that the derivation assumes isothermal flow.

  1. Without ignoring the z factor, you would use a similar procedure, but the problem would have to be done iteratively. You could guess the inlet pressure, then integrate the local pressure gradient numerically to the outlet end of the pipe. Then you would check to make sure that the outlet pressure matched the known outlet pressure. If not, you would adjust the inlet pressure and iterate until it did.
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  • $\begingroup$ If I make no assumption about the change in pressure as a function of position along the flow, e.g. 'it is linear or it is quadratic', how does one come to Eqn (2)? In another paper [link][1] the authors state: the average pressure will be determined through $(dp/dx\propto 1/p)$, $$p_{avg}=\frac{\int_1^2p\ dx}{\int_1^2\ dx}=\frac{\int_1^2p^2\ dp}{\int_1^2p\ dp}=\frac{2}{3}\left[p_1+p_2-\frac{p_1p_2}{p_1+p_2}\right]$$ [1]: scielo.br/pdf/jbsmse/v29n3/a05v29n3.pdf $\endgroup$
    – Armadillo
    Commented Jul 12, 2017 at 14:19
  • $\begingroup$ 1) How to conceptually come to grips with the statement: $(dp/dx\propto 1/p)$? 2) How to arrive at the first integral: $\frac{\int_1^2p\ dx}{\int_1^2\ dx}$ from $(dp/dx\propto 1/p)$? 3) How to arrive to the second integral, $\frac{\int_1^2p^2\ dp}{\int_1^2p\ dp}$, from $\frac{\int_1^2p\ dx}{\int_1^2\ dx}$? $\endgroup$
    – Armadillo
    Commented Jul 12, 2017 at 14:19
  • $\begingroup$ Are you aware that both results are mathematically the identical? See my comment following your original post. $\endgroup$ Commented Jul 12, 2017 at 14:51
  • $\begingroup$ I can arrive to Eqn (2), $p(x)=\sqrt{p_b^2-cx}$, by combining $Re=\frac{\rho u \delta}{\mu}$, $\rho=\frac{pM}{RT}$, and $f=\frac{8\tau_w}{\rho u^2}$? $\endgroup$
    – Armadillo
    Commented Jul 12, 2017 at 20:42
  • $\begingroup$ Yes, ... and the ideal gas law. $\endgroup$ Commented Jul 13, 2017 at 2:17

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