Continuity Equation Compressible Vs Incompressible Flow

Continuity

In most of the cases we will consider, the matter of the fluid is neither created or destroyed. We call this continuity, which means the amount of matter stays the same. This has some interesting consequences for the flow of fluid through a pipe. The rate at which fluid goes into the pipe must be equal to the rate at which fluid comes out. There isn't anywhere inside the pipe where fluid accumulates or where new fluid is generated.

We have to treat gases (and compressible fluids) differently from incompressible fluids (such as water). First consider the harder situation -- a compressible gas. For a pipe with area A, the rate at which matter flows into the pipe is $I = ρAv$, where $I$ is the matter flow rate (kg⋅m³/s), $ρ$ is the density of the gas, $A$ is the cross sectional area in $m^2$, and $v$ is the fluid velocity in m/s. Let's consider a specific example where the pipe changes its cross section.

If the system is a continual flow with nothing changing inside, the mass of gas flowing into and out of the pipe must be the same in each instant of time. The volume going in during the time $Δt$ is $A_1Δx_1 = A_1v_1Δt$ and the volume going out is $A_2 Δx_2 = A_2 v_2 Δt$. So setting the amount of matter going in equal to the amount going out gives the continuity equation:

$$\rho_1 A_1v_1 = \rho_2 A_2v_2$$

This is actually pretty complicated since lots of things can change.

The situation simplifies considerably when we switch to considering an incompressible fluid. In that case, the density doesn't change and the volumes going in and out have to be the same. Therefore, the volumetric currents must be equal. If we introduce $Q$ where $I = \rho Q$, so $Q = Av$ (see the webpage Quantifying fluid flow):

$$Q_1 = Q_2$$

$$A_1 v_1 = A_2 v_2$$

This tells us that the ratio $v_2/v_1$ is equal to $A_1/A_2$. Since $A_2 < A_1$ in our example, this means the flow velocity out of the pipe's narrow end will be faster than the flow velocity into the pipe's broader end. This makes sense. In order to keep a constant volume of fluid moving through the pipe, when the pipe's area gets smaller, the fluid has to move faster. This relation is called the continuity equation for incompressible fluids.

Internal flow can be used by organisms to draw fluids into their body, such as by breathing, or to transport fluids around the body, as in blood circulation. In moving these flows around the body, should they be fast or slow? In some situations, it is important to get the flow quickly to where it needs to go, say from the heart to the brain, in which case fast flow is useful. However, once it gets there, it might help to slow the flow down, in order to transfer oxygen or nutrients from the flow to the surrounding tissues.

Biological implications

In order to slow down the flow velocity in the small blood capillaries, there must be enough of them to increase the overall cross sectional area. The total capillary area, across all of the small vessels, must be greater than that of the aorta which delivers the heart's flow. Otherwise, the flow will move just as fast as it does in the aorta.

$$Q_{aorta} = A_{aorta} v_{aorta} = Q_{capillaries} = A_{capillaries} v_{capillaries}$$

Since it can be easily measured that $v_{capillaries} << v_{aorta}$ , it must therefore be true that $A_{capillaries} >> A_{aorta}$.

Since the blood splits up and goes through all the capillaries at once, we have to add up the cross sectional area of all the capilaries.

Some data for the dog circulatory system suggests that the total cross sectional area of the smallest capillaries is 600 cm², compared to the 2 cm² of the large aorta, an increase of 300 fold.* As a result of this increase in cross sectional area, the flow velocity in the capillaries will be 300 times slower than the flow in the aorta. Flow velocities in our capillaries have been measured to be 0.7 mm/s. Since typical capillaries have a radius of just 3 μm (compared to the 8 mm radius of the aorta), the capillary radius times the flow velocity is 7 x 10^-4 m/s * 3 x 10^-6 m = 2.1 x 10^-9 m²/s. This is quite close to oxygen's diffusion coefficient of 1.8 x 10^-9 m2/s, suggesting that flow speed is quite well matched to the diffusion of oxygen to the surrounding tissues in the small capillaries.

We should point out that in spite of organisms using flow to get fluids where they need to go, once the fluid gets there, the transfer of nutrients and waste products must still occur by diffusion. This is the result of the boundary layer which surrounds every surface. The thickness of the boundary layer can be impacted by the flow characteristics, being thicker for slower flows and thinner for faster flows.

* Steven Vogel, Comparative Biomechanics: Life's Physical World (Princeton U. Press, 2003) chapters 6, 7, 9, 10. table 10.1

Workout: The continuity equation

Joe Redish and Karen Carleton 10/26/11

mcquaigopithought.blogspot.com

Source: https://www.compadre.org/nexusph/course/The_continuity_equation

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