Mass Transfer Coefficient Mass transfer coefficients - simplified method to describe complex boundary condition involving flow and diffusion. Mass transfer from a surface to a fluid
Mass Transfer Coefficient Flux from surface into the fluid: -Brackets denote a spatial average. -S is the surface area in the direction of the flux (plane of constant y = 0 in this case). Now, we define an empirical equation describing the flux: kf unit = m/s. kf is not an intrinsic property of the system. Equating these two relations:
Mass Transfer Coefficient Non-dimensionalization: Results: A new parameter: Sherwood number
Mass Transfer Coefficient • Determine Sh experimentally for different system. (Table 7.5) • Estimate flux using the Sh number. • Flux at wall for flat plate=
Estimation of Flux in the Presence of Convection In many mass transfer problems with convection, the concentration gradient is confined to a thin “boundary layer” of thickness M (M <<R) near the surface. To estimate this mass transfer boundary layer we need to first consider fluid boundary layers.
Boundary Layer Prandtl, 1905
Boundary Layer • High Reynolds Number Flow. • Two length scales • 1. Far from surface, viscous forces are unimportant • and inertial forces dominate. • 2. Near the surface, viscous forces are comparable • to inertial forces
Boundary Layer Approach: Perform scaling for two dimensional flow for a boundary layer of thickness in y direction and a length scale L in the x direction. Derive the “boundary layer” equations Examine approximate solutions to obtain boundary layer thickness and shear stress Apply to mass transfer boundary layers Estimate mass transfer coefficients
Conservation of mass and Navier-Stokes equation for 2-D High Re laminar flow over a plate Conservation of mass X-component Y-component Length scales - in y direction L in the x direction; expect << L Velocity scales - U0 in x direction V in y direction; expect V << U0 Will deduce magnitude of , V and scaling for pressure from an order of magnitude analysis
Scaling Conservation of mass becomes: Velocity gradient in x and y direction are proportional. Thus the following terms must balance to maintain validity of the conservation of mass or So, we now have a relation for V and can use it in the conservation of linear momentum to find .
x component of the conservation of linear momentum Can neglect underlined term since 2 << L Simplifying and rearranging Since viscous and inertial forces are equally important in the boundary layer
Rearranging and solving foryields where ReL = UoL/µ This is the correct scaling for the boundary layer. Almost all subsequent efforts are involved in finding the numerical coefficients to make the relation exact. Since the pressure cannot be neglected either, P2/µUoL must be of order 1. This leads to: This result should not be too surprising since we had used it before.
Use the assumption that 2/L2<<1. The y component of the conservation of linear momentum yields That is, the pressure varies only in the direction of flow (P = P(x) only). The pressure in the boundary layer at a given x position is the same inside and outside the boundary layer. To find this pressure and the x dependence of pressure, apply Bernoulli’s equation along a horizontal streamline far above the boundary layer.
Take the x derivative and rearrange to obtain an expression for the pressure gradient For a flat plate, U(x) = U0 and the pressure gradient is zero.
The boundary layer equations Conservation of mass x-component Options: Solve numerically, or approximate by integrating along the length L. By integrating, derive the von Karman integral momentum equation for boundary layer of thickness .
Approach • Assume an expression for vx that satisfies the boundary conditions y = 0, vx = 0 y = , vx = U(x) Depending on the expression used, additional criteria can be used such as the first and second derivatives must be continuous at y = , vx = U(x) • Insert into equations and solve for . • Compute shear stress
Example, Flow over a flat plate Models hemodialysis membranes for many geometries, since curvature can be neglected since is small. A uniform flow field with velocity Uo approaches a flat plate of length L which is oriented in the direction of flow. For this flow, U(x) = Uo. The von Karman Momentum integral reduces to: The simplest expression that can be used and that simplifies the boundary conditions is
Inserting this expression into the momentum integral yields So, now we have a first order ODE for . We just need an initial which is that at x = 0, = 0. This yields the following results
Comparing different expressions for the velocity, let Conclusion: Even relatively simple approximations work well
Now, let’s consider the conservation relations when we have boundary flows. • The first issue that confronts us is that there are now two length scales in the y direction, the momentum boundary layer thickness M and the concentration boundary layer thickness C. This is important because: • If M << C, the velocity profile is essentially uniform as the concentration changes from 0 to C0. • If C << M, the velocity profile is linear as the concentration changes from 0 to C0. • If C ~ M,the concentration and momentum boundary layers are of the same thickness
In the non-dimensionalization of the conservation relation for solute transport, the concentration is a function of Re and Sc. The Schmidt number controls the relative importance of momentum and diffusive transport. For solutes in water, Sc =/Dij ranges from 103 for small solutes to 103 for proteins. Thus, convective transport is much greater than diffusive transport and the solute concentration gradient is confined in a narrow region near the surface. C << M Linear velocity profile,
From the conservation of mass and linear profile for vx For solute transport, the dimensionless equation is: Since C << L, the underlined terms can be neglected.
Since diffusive and convective transport are equally important, the non-dimensionalization indicates that Using this result, we can approximate the flux as: