Equation 1 represents an improvement over the intial assumption of constant mass velocity (i.e., $\rho v=\rho_o v_o$). We can now improve Eqn. (2) to get to Eqn. (3). My question is: what are the steps to obtain Eqn. (3)?

Equation 1 is: $$\rho v = \rho_o v_o(1+\delta f[c,g]) \tag{1}$$

where $\rho$ is gas density, $v$ is gas velocity. The subscript $o$ denotes that the variable is evaluated at the inlet face of the core. $(1+\delta f[c,g])$ is the gas expansion corection factor.

Equation 2 is: $$q(x,t)=\frac{q_o(t)P_o(t)}{P(x,t)}=\frac{-V_t}{P(x,t)}\left(\frac{dP_o}{dt}\right) \tag{2}$$

Equation 3 is: $$q(x,t)=\frac{-V_t (1+\delta f[c,g])}{P(x,t)}\left(\frac{dP_o}{dt}\right) \tag{3}$$

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 volumetric flow rate at any position downstream from the inlet face of the core can be calculated with correction factors that account for variable mass flow rate with position at any instant in time. Hence the use of the correction factor used in Eqn. (1).

My steps taken to obtain Eqn. (2):

The volumetric isothermal flow rate of nitrogen as a function of time, while the nitrogen is assumed to behave as an ideal gas, from a storage tank at pressure $P_o$ is:

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

The subscript $o$ in these equations refers to conditions just upstream of the inlet face of the sample. Since the only measuring instrument in place is a pressure transducer, we need to find the relationships between gas density and moles of gas versus pressure. Doing so we can re-write our volumetric flow rate equation in terms of pressure.

The Ideal Gas Law for the gas in the tank can be written as:

$$ P_oV_t=nRT \tag{b}$$


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

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

$R=$ universal gas law constant

$T=$ absolute temperature

Equation (b) can be rearranged to solve for the number of moles of nitrogen just upstream of the core as:

$$ n=\frac{P_oV_t}{RT} \tag{c} $$

Equation (c) can be substituted into Eqn. (a):

$$ q_o(t)=\frac{-MV_t}{\rho_o(t)RT}\left(\frac{dP_o}{dt}\right) \tag{d}$$

The density of gas just upstream of the core as a function of time is defined as:

$$ \rho_o(t)=\frac{m(t)}{V_t} \tag{e} $$


$m(t)=$ mass of gas just upstream of the core as a function of time

$V_t=$ tank volume

From the Ideal Gas Law, Eqn. (b), we can relate the density of the gas just upstream of the core in Eqn. (d) to the pressure just upstream of the core as:


where $n=\frac{m(t)}{M}$. Substituting this relation in to Eqn. (b) and solving for pressure:

$$P_oV_t=\frac{m(t)}{M}RT$$ $$ P_o=\frac{m(t)}{V_t}\frac{RT}{M}\tag{f}$$

substituting Eqn. (e) into Eqn. (f) and then rearranging for solve for density:

$$ P_o=\rho_o \frac{RT}{M} \rightarrow \rho_o=\frac{MP_o}{RT}\tag{g}$$

Substitute Eqn. (g) into Eqn. (d) and solve for the volumetric flow rate of nitrogen based on the pressure just upstream of the core, as a function of time:

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

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, hence the correction factor in Eqn. (1).) As the nitrogen flows through the core it expands as a function of pressure change throughout the position of the core; the volumetric flow rate will be proportional to the ratio of inlet pressure and pressure along the core and as a function of time:

$$ \Delta q \propto \frac{P_o(t)}{P(x,t)}\tag{i}$$

Therefore, the volumetric flowrate throughout the core as a function of position and time can be related by equations h and i:

\begin{equation} q(x,t)=\frac{q_o(t)P_o(t)}{P(x,t)}=-\frac{V_t}{P(x,t)}\left(\frac{dP_o}{dt}\right)\end{equation}

So this is how I got to Eqn. (2). Any help with how to get to Eqn. (3) using Eqn. (1) would be much appreciated.

  • $\begingroup$ Are you sure you didn't make a typo in the subscripts or something? Eq. (2) and Eq. (3) look completely the same other than the gas expansion correction factor. If they are indeed the same equation, equating the two, would lead to a gas expansion correction factor of 1. My guess is either one of them has a different location (and thus subscript). If this is the case, then it might be possible to relate the two. $\endgroup$
    – ROIMaison
    Commented Jul 25, 2014 at 7:16
  • $\begingroup$ @ROIMaison I'm not sure, I'm following a paper and prior to getting to Eqn. (3) it states: above derivation (Eqn. (2)) assumed a constant mass velocity, or equivalently that $\frac{\partial(\rho v)}{\partial x}=0$, but the continuity equation for one-dimensional flow requires that $\frac{\partial (\rho v)}{\partial x}=-\phi \frac{\partial \rho}{\partial t}$ and since pressure (hence, density) is changing with time, the continuity equation was violated. Then it goes on to present Eqn. (3) without showing the explicit steps. $\endgroup$
    – Armadillo
    Commented Jul 25, 2014 at 13:39
  • $\begingroup$ What is $\phi$? $\endgroup$ Commented Jul 25, 2014 at 14:05
  • $\begingroup$ @Enredanrestos $\phi$ represent porosity -- the void (empty) space within the rock. The pores are interconnected and the gas flows through this interstitial space. $\endgroup$
    – Armadillo
    Commented Jul 25, 2014 at 14:11
  • $\begingroup$ So it is constant cross-section ($A$ say). Therefore $q=\rho v A\phi/M$ right? $\endgroup$ Commented Jul 25, 2014 at 14:22

1 Answer 1


Mass conservation assuming stationary conditions: $$\rho(x,t)v(x,t)=\rho(0,t)v(0,t):=\rho_0(t)v_0(t),$$ or equivalently $\rho(x,t)q(x,t)=\rho_0(t)q_0(t)$ since $q=vA$, with $A$ a constant cross section. Therefore $$q(x,t)=\frac{\rho_0(t)q_0(t)}{\rho(x,t)}.$$ Then, we use the initial conditions $$q_0(t)=-\frac{V}{\rho_0(t)}\frac{d \rho_0(t)}{dt},$$ and you get the the first result. Of course, $P=K\rho$ with $K=RT/M$, a constant so it does not matter.

The second part I think it is very simple, you depart a little bit from the stationary assumptions $$(1+\delta f)\rho_0(t)v(0,t)=\rho(x,t)v(x,t),$$ and work your way replacing again $q=vA$, with $A$ a constant cross section: $$q(x,t)=\frac{\rho_0(t)q_0(t)(1+\delta f)}{\rho(x,t)}.$$ Using the same initial condition for $q_0(t)$ you should derive Eq.(3).

  • $\begingroup$ I was able to make it work out and make sense to me, thank you for your help! $\endgroup$
    – Armadillo
    Commented Jul 25, 2014 at 15:40

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