GATE Chemical Engineering Mass Transfer Flashcards

24 sample cards from the Mass Transfer deck

Sampled from the end of the deck, so these are different cards from the ones shown on the syllabus page.

1. What is an azeotrope, and why can't ordinary distillation separate it?

   An azeotrope is a mixture whose vapour and liquid have the same composition at boiling ($y_A = x_A$, $\alpha=1$). Since no enrichment occurs, ordinary distillation cannot cross the azeotropic composition; special methods (extractive/azeotropic distillation, pressure swing) are needed.

2. Write the McCabe–Thiele rectifying-section operating line equation.

   $$y_{n+1} = \frac{R}{R+1}\,x_n + \frac{x_D}{R+1}$$ where $R = L/D$ is the reflux ratio and $x_D$ the distillate composition. Slope $= R/(R+1)$, intercept $= x_D/(R+1)$.

3. Define the q-line and give its slope in McCabe–Thiele analysis.

   The q-line locates the intersection of the operating lines based on feed thermal condition $q$ (fraction of feed that is liquid). Its equation is $$y = \frac{q}{q-1}x - \frac{x_F}{q-1}$$ with slope $q/(q-1)$. For saturated liquid feed $q=1$ (vertical line); for saturated vapour feed $q=0$.

4. What happens at total reflux, and which equation gives the minimum number of stages?

   At total reflux $R\to\infty$, no product is withdrawn, the operating lines coincide with the diagonal, and the number of stages is minimum. Fenske equation: $$N_{min} = \frac{\ln\!\left[\dfrac{(x_D/(1-x_D))}{(x_W/(1-x_W))}\right]}{\ln \alpha} - 1$$

5. What is the minimum reflux ratio $R_{min}$ and its operational significance?

   $R_{min}$ is the reflux ratio at which separation requires an infinite number of stages (pinch point where operating line touches the equilibrium curve). Real columns operate at $R$ typically $1.2$–$1.5\times R_{min}$ to balance capital and operating cost.

6. State Raoult's law and Henry's law for vapour–liquid equilibrium.

   Raoult's law (ideal solutions): $p_A = x_A\,P_A^{sat}$. Henry's law (dilute solute): $p_A = H\,x_A$. Raoult applies near $x_A\to1$; Henry applies near $x_A\to0$.

7. What is gas absorption, and on what equilibrium relation is its design based?

   Absorption is the transfer of a soluble gas component (solute) from a gas stream into a contacting liquid solvent. Design is based on solubility equilibrium, commonly Henry's law $y = m\,x$ (or $p_A = H x_A$). Stripping is the reverse operation.

8. Write the Kremser equation form for the number of theoretical stages in dilute gas absorption.

   $$N = \frac{\ln\!\left[\dfrac{x_{N+1}-y_1/m}{x_0 - y_1/m}\left(1-\dfrac{1}{A}\right)+\dfrac{1}{A}\right]}{\ln A}$$ where the absorption factor $A = \dfrac{L}{mG}$.

9. Define the absorption factor $A$ and explain its effect on absorber performance.

   $$A = \frac{L}{m\,G}$$ where $L$, $G$ are liquid and gas molar flows and $m$ the equilibrium slope. $A>1$ favours absorption (high recovery achievable); $A<1$ limits the fraction of solute that can be absorbed regardless of stages. The minimum liquid rate corresponds to $A=1$ at the pinch.

10. What is leaching, and name its main process variables.

    Leaching (solid–liquid extraction) is the removal of a soluble constituent from a solid by contacting it with a liquid solvent. Key variables: particle size, solvent choice, temperature, contact time, and degree of agitation. Smaller particles and higher temperature speed up leaching.

11. In leaching, what is the 'underflow' and 'overflow', and what is the ideal-stage assumption?

    Overflow is the clear solution (solvent + dissolved solute) leaving a stage; underflow is the inert solid carrying adhered solution. In an ideal stage the overflow and the solution retained in the underflow have the same composition (equilibrium), and the solute is fully dissolved.

12. Define liquid–liquid extraction and the distribution (partition) coefficient.

    Liquid–liquid extraction transfers a solute from one liquid (feed/raffinate) to an immiscible solvent (extract) phase. The distribution coefficient is $$K_D = \frac{y^{*}\ (\text{solute in extract})}{x\ (\text{solute in raffinate})}$$ A high $K_D$ favours extraction.

13. Define the selectivity (separation factor) $\beta$ in liquid–liquid extraction.

    $$\beta = \frac{K_{D,\text{solute}}}{K_{D,\text{carrier}}} = \frac{(y_A/x_A)}{(y_B/x_B)}$$ It measures how preferentially the solvent extracts the solute relative to the carrier; $\beta>1$ is required for a feasible separation, and larger $\beta$ means easier separation.

14. Name the four periods/regions seen on a typical batch drying-rate curve.

    (1) Initial adjustment (warm-up) period; (2) Constant-rate period; (3) First falling-rate period; (4) Second falling-rate period. Drying proceeds from high to low moisture content through these regions.

15. What controls the drying rate in the constant-rate period, and what surface temperature prevails?

    In the constant-rate period the surface is saturated with free moisture; drying is controlled by external heat/mass transfer (gas-film), and the wetted surface stays at the wet-bulb temperature of the drying air. Rate $\propto k_y(H_s - H)$.

16. Define critical moisture content and equilibrium moisture content in drying.

    Critical moisture content $X_c$ is the moisture at which drying changes from the constant-rate to the falling-rate period. Equilibrium moisture content $X^{*}$ is the moisture the solid retains in equilibrium with the air; it is the lower limit (free moisture $= X - X^{*}$).

17. Define absolute (specific) humidity and relative humidity (relative saturation).

    Absolute humidity: $$H = \frac{M_w}{M_{air}}\cdot\frac{p_A}{P - p_A} = 0.622\,\frac{p_A}{P-p_A}\ \text{(kg water/kg dry air)}$$ Relative humidity: $$\%RH = \frac{p_A}{p_A^{sat}}\times 100$$ the ratio of actual to saturation partial pressure at the same temperature.

18. Distinguish the dry-bulb, wet-bulb, and dew-point temperatures.

    Dry-bulb: actual air temperature. Wet-bulb: steady temperature of a wetted surface under adiabatic evaporative cooling. Dew point: temperature at which the air becomes saturated (condensation begins) on cooling at constant humidity. For unsaturated air: dry-bulb > wet-bulb > dew point.

19. Why does the wet-bulb temperature approximately equal the adiabatic saturation temperature for the air–water system?

    Because the Lewis number $Le = \dfrac{h}{k_y c_s} \approx 1$ for the air–water system, the ratio of heat- to mass-transfer coefficients matches the humid heat, making the wet-bulb temperature nearly coincide with the adiabatic saturation temperature. This is generally not true for other vapour–gas systems.

20. Differentiate adsorption from absorption, and physisorption from chemisorption.

    Adsorption is a surface phenomenon (solute accumulates on a solid surface); absorption is bulk uptake throughout a phase. Physisorption: weak van der Waals forces, low heat of adsorption ($<\sim 40\ \text{kJ/mol}$), reversible, multilayer; chemisorption: chemical bonds, high heat ($>\sim 80\ \text{kJ/mol}$), often irreversible, monolayer only.

21. Write the Langmuir adsorption isotherm and state its key assumptions.

    $$q = \frac{q_m\,K\,C}{1 + K\,C}$$ where $q$ is amount adsorbed, $q_m$ the monolayer capacity, $K$ the equilibrium constant. Assumptions: monolayer coverage, energetically uniform sites, no interaction between adsorbed molecules. At low $C$, $q\propto C$; at high $C$, $q\to q_m$.

22. Write the Freundlich adsorption isotherm and note when it applies.

    $$q = K_F\,C^{1/n}$$ an empirical isotherm (with $1/n<1$) describing adsorption on heterogeneous surfaces over moderate concentration ranges. A log–log plot of $q$ vs $C$ is linear with slope $1/n$.

23. In a fixed-bed adsorber, define the breakthrough curve and breakthrough point.

    The breakthrough curve is the plot of effluent concentration $C/C_0$ vs time at the bed outlet as the mass-transfer zone moves through the bed. The breakthrough point is the time/volume at which the effluent concentration reaches a set maximum allowable value, signalling the bed must be regenerated.

24. Why must absorption equilibrium and operating lines be analysed with mole ratios (Y, X) for concentrated streams rather than mole fractions?

    Because in concentrated systems the total gas and liquid molar flows change along the column as solute transfers. Using solute-free mole ratios $Y = \dfrac{y}{1-y}$, $X = \dfrac{x}{1-x}$ with carrier flows $G_s$, $L_s$ keeps the operating line straight: $$G_s(Y_1 - Y_2) = L_s(X_1 - X_2)$$

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