08 May 2025
Compressible Capillary Flow and Rankine's Method
For an incompressible liquid, the volume rate is the same at every cross-section. A gas expands as its pressure falls, so its local volume rate changes even though the mass rate remains constant.
Isothermal flow of a gas through a capillary
Consider steady, laminar flow through a capillary of radius $a$ and length $l$. Let the absolute pressures at the inlet and outlet be $p_1$ and $p_2$, with $p_1>p_2$. At position $x$, write the pressure as $p(x)$ and the local volume rate as $Q(x)$.
Mass conservation requires
\[\dot m=\rho Q=\text{constant}.\]For an ideal gas flowing isothermally, $\rho/p$ is constant. Consequently,
\[pQ=C,\]where $C$ is constant along the tube.
Apply the local Poiseuille relation to an element $dx$:
\[Q=-\frac{\pi a^4}{8\eta}\frac{dp}{dx}.\]Substituting $Q=C/p$ gives
\[\frac{C}{p}=-\frac{\pi a^4}{8\eta}\frac{dp}{dx},\]or
\[C\,dx=-\frac{\pi a^4}{8\eta}p\,dp.\]Integrating from $(0,p_1)$ to $(l,p_2)$,
\[\begin{aligned} Cl &=-\frac{\pi a^4}{8\eta}\int_{p_1}^{p_2}p\,dp\\ &=\frac{\pi a^4}{16\eta}(p_1^2-p_2^2). \end{aligned}\]Therefore
\[\boxed{ C=pQ =\frac{\pi a^4(p_1^2-p_2^2)}{16\eta l} }.\]If $Q_2$ is measured at the outlet pressure $p_2$, then $C=p_2Q_2$ and
\[\boxed{ Q_2=\frac{\pi a^4(p_1^2-p_2^2)}{16\eta lp_2} }, \qquad \boxed{ \eta=\frac{\pi a^4(p_1^2-p_2^2)}{16lp_2Q_2} }.\]Both $p_1$ and $p_2$ must be absolute pressures. A manometer reading supplies only their difference, so atmospheric pressure must be included before using this equation.
Let
\[p_m=\frac{p_1+p_2}{2}, \qquad \Delta p=p_1-p_2.\]The volume rate referred to the mean pressure is $Q_m=C/p_m$. Since $p_1^2-p_2^2=2p_m\Delta p$,
\[\boxed{Q_m=\frac{\pi a^4\Delta p}{8\eta l}}.\]Thus the familiar incompressible form is recovered when the gas rate is stated at the mean pressure. If the outlet rate $Q_2$ is used and $\Delta p\ll p_2$, then
\[Q_2=\frac{\pi a^4\Delta p}{8\eta l} \left(1+\frac{\Delta p}{2p_2}\right) \simeq\frac{\pi a^4\Delta p}{8\eta l}.\]Rankine’s gas viscometer
Rankine’s apparatus is a closed glass loop. One limb is a fine capillary and the other is a wider, nearly uniform fall tube. A mercury pellet seals the wide limb and acts as a moving piston.
Let the capillary have radius $a$ and length $l$, the fall tube have cross-sectional area $A$, and the pellet have mass $m$. When the instrument is vertical, the pellet descends and forces gas from its lower side through the capillary to its upper side. Neglecting pellet-wall resistance and the weight of the gas, force balance on the pellet gives the nearly constant pressure difference
\[mg=A\Delta p_R, \qquad \boxed{\Delta p_R=p_H-p_L=\frac{mg}{A}}.\]The closed-loop gas is compressible, so the swept volume divided by time is not, in general, the gas volume rate at one fixed pressure. Let $V$ be the total gas volume, neglecting the small capillary volume, and let $P$ be the uniform pressure when the apparatus is horizontal. At an instant during the fall, let $V_L$ be the volume above the pellet and $V_H=V-V_L$ the volume below it. Isothermal conservation of the enclosed gas gives
\[p_LV_L+p_HV_H=PV, \qquad p_H-p_L=\Delta p_R.\]Solving these equations,
\[p_L=P-\Delta p_R+\frac{\Delta p_R}{V}V_L, \qquad p_H=P+\frac{\Delta p_R}{V}V_L.\]Thus both chamber pressures change as the pellet moves, although their difference stays constant. The pressure-volume throughput through the capillary is
\[\mathcal C=p_LQ_L=p_HQ_H =\frac{\pi a^4(p_H^2-p_L^2)}{16\eta l}.\]Meanwhile, the pressure-volume content above the pellet increases at the rate
\[\mathcal C=\frac{d(p_LV_L)}{dt} =\left(P-\Delta p_R+\frac{2\Delta p_RV_L}{V}\right) \frac{dV_L}{dt}.\]Using $p_H^2-p_L^2=\Delta p_R(p_H+p_L)$ therefore gives
\[dt=\frac{16\eta l}{\pi a^4\Delta p_R} \frac{P-\Delta p_R+2\Delta p_RV_L/V} {2P-\Delta p_R+2\Delta p_RV_L/V}\,dV_L.\]If the timed fall changes $V_L$ from $V_a$ to $V_b$, define
\[\mathcal I=(V_b-V_a) -\frac{PV}{2\Delta p_R} \ln\!\left[ \frac{2P-\Delta p_R+2\Delta p_RV_b/V} {2P-\Delta p_R+2\Delta p_RV_a/V} \right].\]The exact ideal-gas result is then
\[\boxed{ \eta=\frac{\pi a^4\Delta p_R\,t}{16l\mathcal I} }.\]For $\Delta p_R\ll P$, $\mathcal I\simeq(V_b-V_a)/2$. If the calibrated swept volume is $\Omega=V_b-V_a=As$, this reduces to the commonly quoted approximation
\[\boxed{ \eta\simeq\frac{\pi a^4mgt}{8lA\Omega} \qquad(\Delta p_R\ll P) }.\]In practice, mercury surface tension and contact with the fall tube alter the effective driving pressure. The same instrument is therefore often calibrated with a reference gas of known viscosity $\eta_0$. With the same pellet, geometry, marks, equilibrium pressure and temperature, all factors except $\eta$ and $t$ are unchanged, so
\[\boxed{\frac{\eta}{\eta_0}=\frac{t}{t_0}}.\]The instrument is inverted to return the pellet above the first mark and repeat the timing. The editable apparatus diagram is available as a TikZ file.
The capillary-pressure integration, exact Rankine integral and small-$\Delta p_R/P$ limit are checked in the Maxima worksheet; every displayed residual is zero.
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