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I would like some help with the explicit math steps to go from equation 2 to 3. These equations are presented in a paper that I am reading. I will show where these equations came from and my attempt further below. The paper reads:

The volumetric isothermal flow rate of nitrogen, which behaves as an ideal gas, from a storage tank at pressure $P_o$ is:

$$\frac{-V_tP'_o(t)}{P(x,t)}=\frac{-k_lA(1+b/P(x,t))}{\mu}\frac{\partial P(x,t)}{\partial x} \tag{1}$$

The subscript $o$ in these equations refers to conditions just upstream of the inlet face of the sample.

If Eqn. (1) is integrated with respect to lenth and divided by $1/2(P_l-P_o)$, it becomes

$$\frac{-2V_t \mu P'_o(t)L}{k_lA(P_o-P_l)}=P_l+P_o+2b \tag{2}$$

Pressures in all equations above have been absolute pressures, expressed in atmospheres and permeability in darcies. If we now switch to gauge pressure (psig), and express permeability in millidarcies, Eqn. (2) becomes (since $P_l=0$ psig):

$$\frac{-V_t P'_o(t)}{P_o(t)}=\frac{k_lA}{2000 \times 14.696 \mu L}(P_o(t)+2P_a+2b) \tag{3}$$

Note: I will be using the notation $P_L$ not $P_l$ as the author did above

Some background information about these equations. The equations are being used to describe a test flowing gas through a rock.

The test setup consists of a tank and pressure transducer that can be pressurized with nitrogen. A rock-core holder is attached to the tank, separated by a quick opening valve. To perform a run, the tank is charged with nitrogen to an initial pressure. If the valve at the bottom of the tank is opened, nitrogen will flow (axially) through the core (a right-cylindrical rock sample) and the pressure in the tank will decline -- rapidly at first, then more and more slowly. The volumetric rate of nitrogen flow at the inlet face of the core can be derived from the ideal gas law, since the compressibility factor (deviation factor) is unity for nitrogen at low pressure and room temperature.

The Derivation to get to Eqn. (1):

The volumetric isothermal flow rate of nitrogen, which behaves as an ideal gas, from a storage tank at pressure $P_o$ is

$$q_o(t)=\frac{M}{\rho_o(t)}\frac{-dn}{dt} = \frac{-MV_t}{\rho_o(t)RT}\frac{dP_o}{dt} \tag{a}$$

where:

$q_o(t)=$ the volumetric flow rate at the inlet face of the core at time $t$

$M=$ molecular weight, g/mol

$\rho_o(t)=$ gas density at the inlet face of the core at time $t$

$n=$ the number of moles of nitrogen in the reservoir tank

$t=$ time

$V_t=$ the volume of the nitrogen tank, which remains constant

$R=$ universal gas law constant

$T=$ absolute temperature

$P_o=$ the absolute pressure of nitrogen gas in the reservoir tank/inlet core face

Density is given by:

$$\rho_o(t)=\frac{MP_o(t)}{RT} \tag{b}$$

Therefore,

$$q_o(t)=\frac{-V_t}{P_o(t)}\frac{dP_o}{dt} \tag{c}$$

Assume for the moment that at any instant in time the mass velocity throughout the length of the core is constant. (This is not rigorously true.) As the nitrogen flows through the core it expands, such that

$$q(x,t)=\frac{q_o(t)P_o(t)}{P(x,t)}=\frac{-V_t}{P(x,t)}\frac{dP_o}{dt} \tag{d}$$

where,

$x=$ distance in core (=0 at inlet end and L at outlet end)

Klinkenberg's relationship

$$k(x,t)=k_l(1+b/P(x,t)) \tag{e}$$

where,

$k=$ apparent permeability, darcy

$k_l=$ Klinkenberg, or "liquid" permeability, darcy

$b=$ Klinkeberg slip factor, psi

Darcy's equation for one-dimentional flow

$$q=\frac{-kA}{\mu}\frac{dP}{dx} \tag{f}$$

where,

$A=$ cross-sectional area of core (normal to direction of flow)

$\mu =$ nitrogen viscosity (=0.0177 cp at 23 deg C)

Substituting Eqn. (d) and Klinkenberg's relationship, Eqn. (e), into the Darcy equation for one-dimensional flow, Eqn. (f), yields Eqn. (1):

$$\frac{-V_tP'_o(t)}{P(x,t)}=\frac{-k_lA(1+b/P(x,t))}{\mu}\frac{\partial P(x,t)}{\partial x} $$

My attempt to derive Eqn. (3)

Substituting the volumetric flowrate relation and Klinkenberg's relationship into the Darcy equation for one-dimensional flow, Eqn. 16, yields:

\begin{equation} -\frac{V_t}{P_s(x,t)}\left(\frac{dP_{s,o}}{dt}\right)=-\frac{k_lA(1+\frac{b}{P_s(x,t)})}{\mu}\frac{\partial{P_s(x,t)}}{\partial{x}}\end{equation}

where,

$P_{s,o}=$ absolute pressure measured at the inlet face of the core sample

$P_g(x,t)=$ gauge pressure in the core that varies with both distance and time, psig

$P_a=$ atmospheric pressure, psia

Recall differential notation:

$$\frac{dP_{s,o}}{dt}=P'_{s,o}(t)$$

Separate variables and integrate pressure with respect to the length of the core. This will sum all the infintesimal changes in pressure over the length of the core.

\begin{align} \frac{V_t \mu P'_{s,o}(t)}{k_lA} \int_{o}^{L}dx &=\int_{P_s(o,t)}^{P_s(L,t)}(P_s(x,t)+b) \ dP_s(x,t) \\ \nonumber\\ &=\frac{P_s(x,t)^2}{2}\Biggr|_{P_s(o,t)}^{P_s(L,t)}+bP_s\Biggr|_{P_s(o,t)}^{P_s(L,t)} \end{align}

let $P_s(L,t)=P_{s,L}$ and $P_s(o,t)=P_{s,o}$

\begin{align} &=\frac{P_{s,L}^2}{2}-\frac{P_{s,o}^2}{2}+P_{s,L}b-P_{s,o}b \\ \nonumber\\ &=\frac{1}{2}(P_{s,L}-P_{s,o})(P_{s,L}+P_{s,o}+2b) \\ \nonumber\\ \frac{V_t \mu P'_{s,o}(t)L}{k_lA}&=\frac{1}{2}(P_{s,L}-P_{s,o})(P_{s,L}+P_{s,o}+2b) \\ \nonumber\\ \frac{-1}{-1} \times \frac{2 V_t \mu P'_{s,o}(t)L}{k_lA(P_{s,L}-P_{s,o})}&=P_{s,L}+P_{s,o}+2b \\ \nonumber\\ \frac{-2 V_t \mu P'_{s,o}(t)L}{k_lA(P_{s,o}-P_{s,L})}&=P_{s,L}+P_{s,o}+2b \end{align}

Since $P_{s,o}$ is the only pressure changing with respect to time, the equation should be written as:

\begin{equation} \frac{-2 V_t \mu P'_{s,o}(t)L}{k_lA(P_{s,o}(t)-P_{s,L})}=P_{s,L}+P_{s,o}(t)+2b\end{equation}

This is the same equation as Eqn. (2) written by the author.

I'm repeating the paper's verbage here: Pressures in all equations above have been absolute pressures expressed in atmospheres and permeability in darcies. If we now switch to gauge pressure (psig), and express permeability in millidarcies, the equation becomes (since $P_l=0$ psig (I will be using $P_L=0$) and $k \text{[mD]}=\frac{k\text{[D]}}{1000}$):

Recall that absolute pressure, $P_s=P_g+P_a$, where $P_a$ is a constant. Therefore, when integrating absolute pressure plus Klinkenberg's coefficient, $(P_s(x,t)+b)$, we get:

$$\int (P_s(x,t)+b) \ dP_s=\int P_g(x,t) \ dP_g + \int P_a \ dP_s + \int b \ dP_s $$

*Question: how should I notate my bounds of integration? Specifically for the $\int P_g(x,t) \ dP_g$ term?

\begin{align} \int_{P_o}^{P_L}(P_s+b)&=\frac{P_{g,L}^2}{2}-\frac{P_{g,o}^2}{2}+P_LP_a-P_oP_a+P_Lb-P_ob \\ \nonumber\\ &=\frac{1}{2}(P_{g,L}-P_{g,o})(P_{g,L}+P_{g,o}+2P_a+2b) \end{align}

Therefore,

\begin{equation} \frac{-2 V_t \mu P'_{s,o}(t)L}{k_lA(P_{g,o}(t)-P_{g,L})}=(P_{g,L}+P_{g,o}(t)+2P_a+2b) \end{equation}

\begin{align} \frac{-2 V_t \mu P'_{s,o}(t)L}{\left(\frac{k_l}{1000}\right)A(P_{g,o}(t)-0)}&=(0+P_{g,o}(t)+2P_a+2b) \\ \nonumber\\ \frac{-2000 V_t \mu P'_{s,o}(t)L}{k_lA(P_{g,o}(t))}&=P_{g,o}(t)+2P_a+2b \end{align}

*Question:Somehow the paper went from Eqn. (2) to Eqn. (3), using the fact that 1 atm is equal to 14.696 psia. How does dividing atmospheres by absolute pressure give you gauge pressure?:

\begin{equation} \frac{-V_t P'_o(t)}{P_o(t)}=\frac{k_lA}{2000 \times 14.696 \mu L}(P_o(t)+2P_a+2b)\end{equation}

I'm not sure how to notate the $P'_o$ term on the left hand side of the equation with regards to subscripts $s$ or $g$. I know that ultimately it needs to be gauge pressure but in my math steps I haven't done anything to that term. I have the same issue with the $P_o(t)$ terms.

Since it was stated that $P_L=0$ psig (because the pressure of the outlet end of the core is at 1 atmosphere), shouldn't $P_a$ also be equated to zero?

Any help will be much appreciated with these matters.

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  • $\begingroup$ Are $P_{s,o}$ and $P_{s,0}$ different? $\endgroup$ – Enredanrestos Aug 7 '14 at 14:25
  • $\begingroup$ @Enredanrestos no, they are not. both define the inlet face of the core. would it be more clear and correct to stick with one subscript? I also did the same for L and l - they both designate the outlet end of the core. I can edit the post to keep the subscripts consistent if you like. $\endgroup$ – Armadillo Aug 7 '14 at 17:41
  • $\begingroup$ Indeed, it would clarify. $\endgroup$ – Enredanrestos Aug 7 '14 at 18:10
  • $\begingroup$ @Enredanrestos I don't know if you were automatically updated, but I have made the subscripts consistent. $\endgroup$ – Armadillo Aug 7 '14 at 19:36
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how should I notate my bounds of integration?

I guess you mean for this integral, passing to gauge pressure \begin{align} \int_{P_s(o,t)}^{P_s(L,t)}(P_s(x,t)+b) \ dP_s(x,t). \end{align} Since $P_s=P_g+P_a$, all you need to do is that change of variables. Besides, $dP_s=dP_g$ because $P_a$ is a constant. Also, there is no need to say $dP_s(x,t)$, since these are dummy variables. \begin{align} \int_{P_s(o,t)}^{P_s(L,t)}(P_s+b) \ dP_s&=\int_{P_s(o,t)}^{P_s(L,t)}(P_g+P_a+b)dP_s\\ &=\int_{P_s(o,t)}^{P_s(L,t)}(P_g+P_a)dP_s+(P_{s,L}-P_{s,o}) b\\ &=\int_{P_s(o,t)}^{P_s(L,t)}P_g dP_s+(P_{s,L}-P_{s,o}) (b+P_a)\\ &=\int_{P_g(o,t)}^{0}P_g dP_g-P_g (b+P_a)\\ &=-\frac{P_g^2}{2}-P_g (b+P_a)=-\frac{P_g}{2}(P_g+2P_a+2b),\\ \end{align} where in the last line $P_g$ is $P_{g,o}(t)$. Note that $P_{s,L}-P_{s,o}=-P_g$.

How does dividing atmospheres by absolute pressure give you gauge pressure?

Here I think you are confused. psi, psig, atm, and Pa are all units of pressure. That is why you can add gauge pressure to atmospheric pressure in the first place. The problem is that pressures expressed in these units are not proportional since they have different zero, which is the same problem when working between the different temperature scales (farenheit, celsius, and Kelvin). My advice is to make every algebraic and mathematical manipulation without units, or using a single system like SI, and only translate to more practical units at the end.

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  • $\begingroup$ this is great!, and thank you for helping me out with now two of my problems! \begin{align} \frac{V_t\mu P'_{s,o}(t)L}{k_lA}&=-\frac{P_g}{2}(P_g+2P_a+2b)\\ \frac{-2V_t\mu P'_{s,o}(t)L}{k_lA}&=P_{g,o}(t)(P_{g,o}(t)+2P_a+2b)\\ \frac{-2V_t\mu P'_{s,o}(t)L}{k_lAP_{g,o}(t)}&=P_{g,o}(t)+2P_a+2b\end{align} the change in absolute pressure w.r.t. time is the same as the change in gauge pressure, therefore $P'_{s,o}(t)=P'_{g,o}(t)$. The RHS is divided by 14.696 to convert pressure units from psig to atm? Again, thanks for your time! $\endgroup$ – Armadillo Aug 8 '14 at 1:10
  • $\begingroup$ @jakemcgregor Indeed, since the difference between gauge and absolute pressure is a constant, the derivatives are the same. All that is left is the conversion factor between psi (or psig) and atm. At this point, I would like to give you a small advice for you to get more feedback: since your questions have been mostly mathematical, try to streamline more the notation and jargon. You put a lot of effort in your posts, but perhaps instead of writing everything, try to make them as concise as possible. Just an advice. $\endgroup$ – Enredanrestos Aug 8 '14 at 2:22

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