Mass Transfer and Separation Processes

Batch drying time with constant and falling rate periods

Batch drying is conventionally described by two periods separated by the critical moisture content X_c. In the constant-rate period, the surface stays wet with unbound water, so the rate R_c is set by external heat and mass transfer and does not depend on the moisture content. For a drying area A and a dry solid mass L_s per unit area, the constant-rate time is t_1 = \frac{L_s(X_0-X_c)}{R_c}, where all moisture contents are on a dry basis (kg water per kg dry solid). Once the moisture falls below X_c, the rate decreases. When the falling-rate curve is linear in X, the rate is R = R_c\frac{X-X^*}{X_c-X^*}, where X^* is the equilibrium moisture content. Integrating dX/dt = -R/L_s gives t_2 = \frac{L_s(X_c-X^*)}{R_c}\ln\frac{X_c-X^*}{X_f-X^*}, which reduces to t_2 = \frac{L_sX_c}{R_c}\ln\frac{X_c}{X_f} when X^*=0. For example, with L_s = 10 kg m^{-2}, X_0 = 0.4, X_c = 0.15 and R_c = 2.5 kg m^{-2} h^{-1}, the constant-rate time is t_1 = 1 h. When the falling-rate period is described only by the time over which the rate decays to a fraction of its initial value, the moisture removed in that interval equals the average of the two end rates multiplied by the time, so the final moisture follows directly from the removal. When the drying area is given per kilogram of dry solid, the product AR_c is expressed as a rate per kilogram of dry solid, and the same period formulas apply without change. Moisture data given on a dry-solid basis must be used consistently with the mass of dry solid. For a wet mass W at dry-basis moisture X_0, the dry solid is W_s = W/(1+X_0), and the constant-rate and falling-rate times are then computed on that basis.

Binary diffusive flux in liquids (Fick vs Stefan)

Diffusion in a binary mixture is described by the flux of A relative to the molar average motion, N_A = -cD_{AB}\frac{dx_A}{dz} + x_A(N_A+N_B). Two simplified forms are used in steady problems. For equimolar counter-diffusion, N_B = -N_A, and the flux is N_A = \frac{cD_{AB}}{\Delta z}(x_{A1}-x_{A2}). For diffusion of A through a stagnant B, N_B = 0, and the flux is N_A = \frac{cD_{AB}}{\Delta z}\ln\frac{1-x_{A2}}{1-x_{A1}}. Writing x_{Bm} = \frac{x_{B2}-x_{B1}}{\ln(x_{B2}/x_{B1})} gives N_A = \frac{cD_{AB}}{\Delta z}\frac{x_{A1}-x_{A2}}{x_{Bm}}, so the stagnant-B flux exceeds the equimolar flux by the factor 1/x_{Bm}, since x_{Bm} is less than one. The percentage by which the equimolar flux falls short of the stagnant flux is \left(1-\frac{N_{EMD}}{N_{UMD}}\right)\times 100. The ratio of the two fluxes is exactly 1/x_{Bm}, so for a concentrated mixture the stagnant-B flux is noticeably larger, while in dilute solutions x_{Bm} approaches one and the two fluxes nearly coincide. For example, with x_{A1}=0.6 and x_{A2}=0.2, x_{Bm}=0.577, so the stagnant flux is about 1.73 times the equimolar flux. For equimolar transfer across a slab, the molar flow rate is the flux multiplied by the area normal to the direction of diffusion, and the path length is the thickness in that direction.

Countercurrent absorption operating line

In a countercurrent absorber, the solute balance is written with mole ratios on a solute-free basis, Y = y/(1-y) and X = x/(1-x), because the carrier gas and solvent flows remain constant along the column. The operating line is then straight, G'(Y_{in}-Y_{out}) = L'(X_{out}-X_{in}), and it joins the two ends of the column. When the equilibrium relation is y^* = mx and the solvent enters pure, the minimum liquid-to-gas ratio is set by the pinch at the gas inlet, (L'/G')_{min} = \frac{Y_{in}-Y_{out}}{X^*_{out}}, where X^*_{out} is the liquid in equilibrium with the entering gas. Operating at a multiple of this minimum, the exit liquid ratio is obtained from the balance; for example, twice the minimum ratio halves the exit liquid composition relative to the equilibrium limit. For dilute systems with a linear equilibrium line Y^* = mX and a linear operating line, define the stripping factor S = mG/L. The gas-side transfer units are then N_{OG} = \frac{1}{1-S}\ln\left[(1-S)\frac{y_b-mx_t}{y_t-mx_t}+S\right], where the subscripts b and t denote the bottom and top of the absorber. For example, with S = 0.5 and y_b/y_t = 20 at x_t = 0, N_{OG} = 2\ln 10.5 \approx 4.7. The packing height is Z = H_{tOG}N_{tOG}, where N_{tOG} is the integral \int dy/(y-y^*) between the column ends. For a linear equilibrium line the integral has the same closed form as the expression above, so the height follows from the transfer-unit height and the transfer-unit count.

Reverse osmosis transport model

Reverse osmosis forces solvent through a semipermeable membrane against the osmotic pressure of the solution. At equilibrium, the solvent chemical potential is the same on both sides of the membrane, and for a dilute solution this is expressed as P_1-\pi_1 = P_2-\pi_2. Osmotic pressure therefore opposes the hydraulic pressure, so it enters with a negative sign rather than being added to it, and equality of either pressure alone is not the equilibrium condition. In the solution-diffusion description, the water flux is proportional to the net driving pressure, J_w = A(\Delta P-\Delta\pi), where A is the water permeability, while solute passage is driven by the concentration difference across the membrane. Concentration polarization is the accumulation of rejected solute in a boundary layer at the membrane surface. The higher surface concentration raises the osmotic pressure at the membrane, which reduces the net driving force and lowers the water flux, and it also increases the solute passing through the membrane, so the observed solute rejection is reduced. For example, if \Delta P = 40 bar and \Delta\pi = 25 bar, the net driving pressure is 15 bar, and polarization that raises the effective \Delta\pi to 30 bar cuts the flux by one-third.

Countercurrent stripping operating line

In stripping, a solute moves from a liquid into a gas that enters clean. With liquid entering at the top at composition x_2 and leaving at the bottom at x_1, the solute balance gives the operating line y = \frac{L}{G}(x-x_1), which passes through the bottom end where pure gas enters. Equilibrium is written as x^* = y/m, so the operating line in terms of liquid composition is x^* = \frac{L}{mG}(x-x_1). The number of overall liquid-phase transfer units is N_{OL} = \int_{x_1}^{x_2}\frac{dx}{x-x^*}. A useful special case is L/G = m: the driving force x-x^* then equals the constant x_1 at every point, and the integral gives N_{OL} = (x_2-x_1)/x_1. For a linear equilibrium line and pure stripping gas, the minimum number of ideal stages is found from the Kremser relation for stripping, \phi = \frac{S^{N+1}-S}{S^{N+1}-1}, where S = mV/L is the stripping factor and \phi = \frac{x_{in}-x_{out}}{x_{in}-y_{in}/m} is the fraction of solute removed from the liquid. The solved value of N is rounded up to the next integer to give the number of stages.

Binary distillation material balance

Material balances on a binary column use the overall balance F = D + B and the component balance Fz_F = Dx_D + Bx_B. Together they fix the distillate and bottoms flows for given compositions. For example, with F = 100 kmol/h, z_F = 0.5, x_D = 0.95 and x_B = 0.05, the distillate is D = F(z_F-x_B)/(x_D-x_B) = 50 kmol/h. For a saturated liquid feed under constant molal overflow, the vapour flow is the same above and below the feed, and the stripping section carries liquid L' = L + F with vapour V' = V. The stripping operating line is y = \frac{L'}{V'}x - \frac{Bx_B}{V'}, which passes through the point (x_B, x_B). Its intercept fixes V' in terms of B and x_B, and the rectifying line y = \frac{R}{R+1}x + \frac{x_D}{R+1}, with R = L/D, then gives the reflux ratio from the same material balance. The minimum reflux is set by the pinch at the feed. For a saturated liquid feed the q-line is vertical at x = z_F, and R_{min} satisfies \frac{R_{min}}{R_{min}+1} = \frac{x_D - y^*_F}{x_D - z_F}, where y^*_F is the equilibrium vapour at the feed. Under constant molal overflow the vapour rate is V = (R+1)D, so the minimum reboiler duty is Q = V\lambda, with \lambda the molar latent heat.

Packed column diameter at flooding velocity

The diameter of a packed column is set from the volumetric vapour flow and a design superficial velocity u. The vapour volumetric flow is the molar flow multiplied by the molecular weight and divided by the vapour density. The design velocity is a fraction of the flooding velocity u_F, commonly between 50 and 80 percent. The cross-sectional area is A = Q_V/u, and the diameter is D = \sqrt{4A/\pi}. For example, 0.2 kmol/s of vapour with molecular weight 30 and density 1.5 kg m^{-3} gives Q_V = 4 m^3/s; at 60 percent of a flooding velocity of 0.2 m/s the area is 33.3 m^2 and the diameter is about 6.5 m. The flooding velocity is read from a generalized correlation whose abscissa is \frac{G_L}{G_V}\sqrt{\rho_V/\rho_L} and whose ordinate is u_F\sqrt{\rho_V/(\rho_L-\rho_V)}, with a family of curves for different packings. Both packings share the same density terms, so the ratio of their flooding velocities equals the ratio of their chart ordinates at the same abscissa. The allowable vapour mass velocity is the design fraction times u_F\rho_V, so the ratio of allowable vapour mass velocities between two packings follows directly from that ratio.

Batch distillation Rayleigh equation

In a simple batch still with liquid holdup L and light-component mole fraction x, the differential material balance gives dL/L = dx/(y-x), which integrates to the Rayleigh equation \ln\frac{L_0}{L} = \int_{x_f}^{x_0}\frac{dx}{y-x}. For a constant relative volatility, y = \frac{\alpha x}{1+(\alpha-1)x}, and the integral has the closed form \ln\frac{L_0}{L} = F(x_0)-F(x_f), where F(x) = \frac{1}{\alpha-1}\left[\ln x - \alpha\ln(1-x)\right]. Because the holdup and the composition change together, the final composition is found by combining this equation with the balance on the light component remaining in the still, usually by iteration. The vapour leaving the still at any instant is in equilibrium with the liquid, so the distillate collected over the batch has a cumulative average composition. That average equals the total light component in the distillate divided by the total distillate, where the light component in the distillate is the amount charged minus the amount left in the still. If a fixed fraction of the charged light component is recovered, the amount remaining in the still, Wx_W, is known, and the still composition and the distillate amount follow from the Rayleigh relation together with this balance.

Ternary liquid-liquid extraction single stage

Ternary liquid-liquid systems are plotted on a triangular diagram whose corners are the pure components. The binodal curve bounds the two-phase region, and each tie-line joins the two equilibrium phases, the extract and the raffinate. Mixing feed with solvent produces a mixing point on the straight line joining the two feed points, and the solvent-to-feed ratio is given by the lever rule along that line. The solvent demand is at its minimum when the mixing point lies on the boundary of the two-phase region at the end of a tie-line, so the phases formed are that tie-line's ends. The solute is normally more soluble in the solvent-rich phase, so along any tie-line the extract end carries more solute than the raffinate end. A mixing point lying on a tie-line splits into exactly those two end phases, with the amounts given by the lever rule across that tie-line.

Fenske equation at total reflux

At total reflux no product is withdrawn, so the operating lines coincide with the diagonal, y_{n+1} = x_n, and each equilibrium stage is stepped directly from the liquid composition. The Fenske equation gives the minimum number of stages for this condition, \frac{x_D/(1-x_D)}{x_B/(1-x_B)} = \alpha^{N}, where N is the number of ideal stages including the reboiler and \alpha is the average relative volatility. Taking the logarithm gives N_{min} = \frac{\ln\left[\frac{x_D}{1-x_D}\cdot\frac{1-x_B}{x_B}\right]}{\ln\alpha}. For example, with \alpha = 3, N = 3 and x_B = 0.05, the distillate composition is x_D \approx 0.59. In a two-stage case, stepping from the reboiler liquid through the equilibrium vapour of each stage gives the distillate composition directly, and the Fenske result agrees with this stage-by-stage calculation.

Convective mass transfer from evaporating surface

When a volatile component evaporates from a surface into a flowing gas, the flux at the interface is N_A = k_c(C_{Ai}-C_{Ab}), where the interface concentration is fixed by the equilibrium vapour concentration at the surface. Within the boundary layer the concentration falls from its interface value toward the bulk value. When the gas velocity varies across the layer, the stream composition that matters for mass balances is the flow-averaged concentration, C_{A,m} = \frac{\int v_z C_A\,dy}{\int v_z\,dy}, which weights each layer by the local flow. For a linear concentration profile C_A = C_{Ai}(1-y/\delta) and a velocity that varies as v_z = 2y, the ratio of the flow-averaged to the interface concentration is 1/3, independent of the layer thickness. A velocity that increases from the surface makes the slow near-wall layer contribute less, so the flow-averaged value is lower than the arithmetic average of the profile. Because the ratio depends only on the profile shape, it is a dimensionless quantity that can be evaluated from the profile alone.

Mass transfer rate to spherical droplets

For a stationary sphere of diameter d in a stagnant medium with pure molecular diffusion, the Sherwood number takes its limiting value Sh = k_c d/D = 2, so k_c = 2D/d. The mass transfer rate is W = k_c(\pi d^2)\Delta C = 2\pi D d\,\Delta C, which is proportional to the diameter rather than to its square, because the coefficient falls as the diameter grows. Doubling the diameter therefore doubles the transfer rate for a given concentration difference. For a moving droplet the correlation Sh = 2 + 0.6\,Re^{1/2}Sc^{1/3} applies, and the convective term is added to the stagnant limit of two.

Reflux ratio optimization and total annual cost

The total annualized cost of a distillation column combines the annualized capital cost of the column with the annual energy cost of the reboiler and condenser. As the reflux ratio increases above its minimum, fewer stages are needed, so the capital cost falls, but the vapour and liquid loads rise, so the energy cost increases approximately in proportion to the reflux ratio. Near the minimum reflux ratio, the number of stages grows without bound, so the capital cost rises steeply. For reflux ratios far above the minimum, the energy cost dominates and the total cost increases roughly linearly. The total annualized cost therefore passes through a minimum at an intermediate reflux ratio, commonly within a range of about 1.1 to 1.5 times the minimum. Because the stage requirement diverges at the minimum, the total cost rises most steeply as the reflux ratio approaches the minimum from above, not as it decreases.

Overall mass transfer coefficient for liquid–liquid extraction

In liquid–liquid mass transfer the two-film model describes the flux through the interface. Let the distribution coefficient be defined by the interface equilibrium y_A^i = m\,y_B^i. The flux is N = k_A(y_A-y_A^i) = k_B(y_B^i-y_B), and eliminating the interface compositions gives an overall coefficient K_A based on phase A, with \frac{1}{K_A} = \frac{1}{k_A} + \frac{m}{k_B} and N = K_A(y_A - m\,y_B). For example, with k_A = 50, k_B = 100 and m = 2, 1/K_A = 0.02 + 0.02, so K_A = 25. The interfacial compositions are obtained by applying the equilibrium relation at the interface together with the two flux expressions. When the individual coefficients are unknown, a measured flux and the bulk compositions allow the individual coefficient in one phase to be deduced once the ratio of the coefficients and the interface equilibrium are known.