VISCOSITY

Definition

Immerse two parallel plates in an incompressible fluid which sticks to the surface of the plates (Fig 1).

When a force F is applied to the upper plate and it begins to move to the right while the lower plate is fixed in place, the layer of fluid attached to the upper plate will acquire the same velocity as the plate while the layer next to the lower plate will remain static. The layers in between, are dragged by the upper layers while being slowed down by the lower ones. The result is that the velocity of the layers, as indicated by the horizontal lines, decreases as the distance from the moving plate along the Y axis increases. The property of the fluid which determines that reduction in velocity is its viscosity. For many fluids it is found that the ratio of the shear stress and the shear strain remains constant for a relatively wide range of values of both shear stress and strain. This ratio is a definition of viscosity. Since, in this figure, shear stress is the applied force divided by the area of the plate ( f/A ) and the shear strain is the gradient of the velocity with respect to the Y axis,  

                 f /A = 0 (dV_X /dY)      Eq.  1

where 0 is the viscosity. The units of viscosity are, then,

                (N @ m^-2 ) / (m @  s^-1 @ m^-1 ) = N @ s @ m^-2 = Pa @ s.

When water flows through a pipe, or blood through a vessel, the phenomenon is exactly the same, although the equations look differently because of the different geometry of the system. If the system of interest consists instead on solid particles moving within a viscous fluid, as the diffusion of molecules or of ions in a solution, we find the same situation: the viscosity of the fluid offers resistance against the movement of these particles. All of these cases respond to the same physical principles and, therefore, can be described by almost identical equations. Later on, the concept of flux will be introduced and, in the derivation of its equation, the notions discussed regarding viscosity will play an important role. It will be seen that the same concept can be applied to the flow of a liquid, or to molecules, or to electric charge, the only difference being that the units used would be, respectively, those of volume, moles or coulombs.  At this point we need to expand our discussion of viscosity and its effects.

Movement in a Viscous Fluid

Suppose you are observing a solution of a substance in water and that you find that the movement of the solute molecules shows, superimposed on an apparently random motion, a drift in a preferred direction. As a result, there is a flow of solute from one region of the solution to another. If the solution were in an electric field and the solute particles electrically charged, we would logically consider that an electrical force is responsible for that drift. Similarly, if a concentration difference existed in different regions of the solution, we could make the same assumption even when the reality of a body force whose existence depends on a concentration difference is not evident. This assumption allows us to express the quantitative relationship between the force experienced by each particle ( f ) and their motion in terms of Newton’s second law, which applies to both cases. We will assume that, when time is zero, the force is applied and the particles have zero velocity in the direction of the drift, but receive an acceleration a = f / m.  A semi quantitative examination of this situation indicates that, as time elapses, the velocity of the particles will increase due to that acceleration, but that velocity will generate a drag force due to friction (f _D ) which opposes f. Consequently, the acceleration will be reduced as the net force      (f - f_D ) diminishes. As long as f is larger than  f_D, the particles will be accelerated and their velocity increases, but with an increasingly smaller acceleration. Since the velocity increases, so will the drag force and, therefore, the acceleration will decrease until it becomes zero. Then, the velocity will remain constant.

This examination provides some insight into the nature of the phenomenon, but a fully satisfactory understanding of it is achieved only by applying quantitative reasoning. As its result, we will derive equations that describe the change in time of the velocity.

                  f = m @ a = m @ (dv/dt)                          Eq.  2

where v is the velocity of a particle in the direction of the drift and m is the mass of each particle. It is clear that all particles will not have the same velocity; so, the velocity we are deriving is the mean of the velocities shown by the large population of molecules. The movement of a particle in a viscous fluid will generate a drag force ( f _D ) directed opposite to the particles velocity and proportional to that velocity.

                 f _D =  " @ v                                        Eq   3

 Therefore, Newton’s second law may now be expressed in terms of a net force which is the difference between the applied force f and the drag force f _D. So, 

                (f - f _D ) / m = a = dv/dt                     Eq   4

But, in applying Equation 1 to this situation, it is possible to show that, for newtonian fluids, " is proportional to 0  ( " = k @ 0 in newton s per m). k is a constant that depends on the shape of the particles and has units of meters. Making the corresponding substitutions and rearranging terms,

                dv/dt + (k 0/m) @ v = f / m                 Eq.   5

This is a first degree linear differential equation whose solution is

                        Eq.   6

In which C is a constant of integration. Integrating the exponents of e and making the substitution k 0/ m = $, which is a rate constant,

Further integration followed by division by e^{$  t} yields

                                            Eq. 7

Imposing the initial condition v = 0 at t = 0 to find the value of C, and substituting J = 1/ $,

                  Eq   8

where J = m / k 0 is the time constant of the process. In an exponential of this form (1 - e ^{- t/J}), J is a measure of how long it takes for the value of the expression in parenthesis to approach 1; the larger the value of J, the longer it takes. Since this exponential approaches that value asymptotically, in practice, a value of t sufficiently larger than J is taken. For example, when J is one second, after five seconds have elapsed the value of the quantity in parentheses is 0.993. These numbers are too large for the problem at hand, but, if they were applicable, it would mean that in five seconds the velocity would have reached a value of 0.993 f / k 0.

It would be interesting to find a similar equation that describes the growth in time of the drag force f _D.   This can be done easily by substituting v in Eq. 3 with its value in Eq. 7, with the result

Another approach would be to take the derivative with respect to t of Eq. 3

      d f_D /dt = " @d v/dt = k 0 dv /dt = k 0 a

       df_D = [k 0 (f - f_D) / m] dt

      -df_D /(f_D - f) = (k 0 / m)dt = $ dt

       df_D /(f_D -f) = - $ dt

and, integrating,

         ln (f_D -f) = - $ t + C

taking the antilog,

                f_D -f = e ^{-$ t} @ e^C = c’ @ e ^{-$ t}

to which we impose the initial condition f_D = 0 at t = 0 to obtain   c’ = -f

so        f_D -f = -f e ^{-$ t}        and       f_D = f (1 - e ^{-$ t} ).

.If this last equation is compared with Eq. 7, it will be seen that the only difference is the value of the constant, which in Eq. 7 has the same units as velocity while here it has the units of force. The quantities in parenthesis, which determine the rate of change, are identical. At time zero, both the velocity and the drag force are zero, since the quantity in parenthesis is zero. At that time, the net applied force is equal to f because, although the acceleration is high, the velocity (a @ t) is still zero. But, as time passes, the velocity grows because the particles are being accelerated, f _D = k 0 v    increases and  the acceleration   ( a = [f - f _D]/ m )  is reduced. In that manner, the rate at which the velocity grows becomes smaller and, consequently so does the rate at which f_D becomes smaller.

We could arrive at a similar result without using calculus but the derivation is less rigorous and the results are not the same. We start by stating that the velocity of the particles is equal to the acceleration times time

                and     

Now multiply both numerator and denominator of the right side by m / k 0

 in which it is clear that when  t >> m / k0  the denominator becomes essentially equal to t, and v = f / k 0. Although the result of this derivation also shows that v progressively approaches a constant value, this last equation represents a hyperbolic curve similar to the one obtained for the Michaelis-Menten model for enzyme kinetics, whereas Eq. 8 represents an exponential curve and more accurately describes the changes in v as time elapses.

Generalized Flux

When particles of a given mass immersed in a viscous fluid are exposed to a gravitational field      ( force per unit mass) an acceleration is imposed to them and, in time, the acceleration is reduced to zero and the velocity reaches a constant value. At that point the applied force and the drag force are equal in intensity and opposite in direction:

                - (dU/dx) @ m - k 0 v = 0                        Eq. 9

where U is a gravitational potential and its derivative with respect to x is the potential gradient. Given the definition of gradient, this assumes that x is taken in the direction in which the slope of U with respect to distance is the largest; in other words, straight down. The negative sign in front of it indicates that the direction of the force induced by the field on the mass is opposite to the direction of that slope. Solving for v,

                v = - (m / k 0) @ (dU/dx)                        Eq.  10

in which the constant of proportionality between the gradient and the flux (m / k 0 in this case), represented by u, has the dimensions of mass times velocity per unit force and is inversely proportional to the viscosity of the medium and, therefore, to the resistance the medium opposes to the movement. For this reason it has been called mobility. Up to this point, v may be considered either the velocity of a single particle or the mean velocity of a population of particles. Upon defining flux in the next paragraphs, we will use only the latter, the mean velocity of a large population of particles.

We have mentioned previously (XXXXX) that a flux of something is the amount that flows in one second through an area of one square meter perpendicular to the direction of the flow. Since the field we are considering is a gravitational field with units of newtons per kilogram, it is necessary to express the units of mobility as  kg @ m/ (N @ s),  the flux of particles in kg / s @ m². But, if all particles have the same mass, the flux could also be expressed in terms of the number of particles that cross one meter square in one second. In addition, we have also derived in an earlier section that this flux (here represented by j ) is equal to the concentration of these particles in the suspension times the mean velocity of the particles

                j = c @ v

in which concentration means mass of particles per unit volume or, alternatively, number of particles per unit volume. Substitute the value of v from Eq. 10

                j = - (m / k 0) @c @ (dU/dx)                     Eq. 11

For this kind of flux and of gradient, the units of the mobility are: kg @ (N s m^-1)^-1, those of concentration are: kg / m^-3, those of the gradient are: N /kg, and those of flux: kg /s m^-2.

When the process we are observing is diffusion, the gravitational field is mostly irrelevant unless we were talking about relatively large molecules exposed to a large gravitational field in a high speed centrifuge. On the other hand, the existence of a concentration gradient in the solution is the reason why diffusion takes place. In fact, an early study of diffusion by Fick led him to a model from which he established his first law of diffusion:

                j_{i =}  - D (dc_i /dx)

in which D is called diffusion coefficient. In the general model we are using here, the driving force for the transport of matter is derived from the negative gradient of a potential, in other words a field. Free energy is the potential energy function whose change is used to identify spontaneous processes in which changes in entropy become significant. Therefore, it makes sense to use the negative of its gradient as the force that drives the directed movement in diffusion. The contribution of a solute to the free energy of a solution is given by

                G_i = n_i G^o_i + n_i RT ln c_i 

where n_i is the total number of moles of the solute i present in the solution. Its gradient is

                d G_i /dx = n_i RT (1/c_i) (dc_i /dx).           Eq.  12

The units are a bit different in this case because the potential has units of joules per mole and the field is expressed in newtons per mole. This potential is called chemical potential, represented by the symbol : and defined as

We also need to redefine concentration as number of moles per unit volume. So, in terms of chemical potential, Eq. 12 may be rewritten as

                d :_i /dx = RT 1/c_i (dc_i /dx)

Before we proceed to substitute this in Eq. 10, we need to justify these changes in units. The units of this gradient are no longer newtons per kilogram, but newtons per mole, and the unit used to measure the amount of matter on which the field acts to generate the driving force is the mole. In other words, the force that drives the movement is equal to the negative gradient of the appropriate potential function  times the amount of matter that moves when exposed to that field, and that amount of matter will be expressed in units such that when multiplied times newtons per kilogram will yield newtons, or that when multiplied times newtons per mole will yield newtons, or that when multiplied times newtons per cubic meter will yield newtons. There is nothing mysterious about this. In the previous situation, where particles with a mass moved in a gravitational field,  we defined the flux as kg / s @ m² . In other words, we selected a number of particles whose mass added up to a convenient unit, the kilogram. This definition of flux determines that  the field and the mobility must use the same unit of mass. When the field is derived from a gradient of chemical potential, we select N_A as the number of particles, which represents a mole, because that gradient is expressed in N / mole.

Equations 9 and 10 need to be rewritten as follows

                - (d:_i /dx) @ n_i - k 0 v = 0                      Eq.  9'

                v = - n_i / k 0 @ (d:_i /dx)                         Eq.  10'

Using this expression for v we get,

                j_i  = - (n_i  / k 0) @ c_i @ d :_i /dx = - (n_i  / k 0) @ c_i  @ RT (1/c_i ) @ (dc_i /dx)

The units here are:

moles @ s^-1 @ m^-2 =  [(moles @ (N s m^-1)^-1 ] @ [moles @ m^-3] @ [N m (mole)^-1] @ [m³ mol^-1 mol m^-3   m  ^-1]

          Flux                       Mobility           Concentration            RT                  1 /c_i @ (dc_i /dx)

                j_i = - (n_i  / k 0) @ RT (dc_i /dx) = u_i RT  (dc_i /dx)                                   Eq.   13

This needs to be integrated

The units here are:

      moles @ s^-1 @ m^-2 = [(moles @ m @ (N s )^-1 ] @ [N m (moles)^-1] @[ moles  @ m^-3  @ m^-1 ]

                                            mobility                          RT                        dc_i /dx

If the particles moving are ions and we are interested in studying the movement of charge, 

Develop for charge and volume. Improve the chart for hydraulic, diffusional and electric fluxes.

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