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23-Chem-A3 Heat and Mass Transfer · December 2019

Question 4 of 6: Evaporation Time in a Stefan Diffusion Tube

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — December 2019 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A (Q1–Q3) Heat Transfer and Part B (Q1–Q3) Mass Transfer; at least two questions must be attempted from each part and only the first two in each part are marked, so four questions (each 25 points) constitute a complete paper. All six questions are solved below for completeness. Property values not printed on the paper (molar masses, water latent heat, the dimensionless free-convection peak velocity $f'_{max}$, benzene/toluene physical properties) are stated explicitly in each Given block as open-book look-ups.

Reference texts: Coulson & Richardson (Backhurst, Harker & Richardson), Chemical Engineering, Vol. 1 — Fluid Flow, Heat Transfer and Mass Transfer (6th ed., Butterworth-Heinemann) — the source family for the crystalliser, tube-condenser, Stefan-tube and distillation problems; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (free- and forced-convection correlations, the Ostrach similarity solution); Treybal, Mass-Transfer Operations (3rd ed.) and McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.) — Stefan diffusion, Chilton–Colburn analogy, McCabe–Thiele; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the NIST Chemistry WebBook.

The obscured verb in Part B Q1 is taken as “calculate”, and the Part B Q2 water viscosity at 27 °C is $8.76\times10^{-4}$ Pa·s. One column of the Part A Q2 air-property table, the buoyancy group $g\beta/\nu^{2}$, is internally inconsistent: its printed values are about 40% below what $\rho$, $\mu$ and $\beta=1/T_f$ give. That column is not used; the Grashof number is built from the $\rho$ and $\mu$ columns and $\beta=1/T_f$.

Part B — Mass Transfer

Question 4: Evaporation Time in a Stefan Diffusion Tube (Part B — 25 pts)

Question text not reproduced: the examination questions are © Engineers and Geoscientists BC. Open the official past paper (linked at the top of this page) to read the question, then follow the worked solution below.

Given. A 6 mm tube holds 2 cm³ acetone (A) over/with 2 cm³ non-volatile dibutyl phthalate (DBP); air sweeps the top ($p_A=0$). Acetone diffuses up a stagnant gas column of length $z$ (initially 11.5 mm), and as it leaves, both the level drops and the liquid acetone mole fraction falls, lowering the Raoult surface pressure.

QuantityValue
Tube diameter / area6 mm / $A_t=2.827\times10^{-5}$ m²
Acetone charged2 cm³ → 0.02631 mol ($M=58.08$)
DBP (non-volatile)2 cm³ → 0.00753 mol ($M=278.3$)
Vapour pressure $P_{vap}$60.5 kPa
Diffusivity $D_{AB}$0.123 cm²/s = 1.23×10⁻⁵ m²/s
Level range $z$11.5 → 50 mm below top

Find. The elapsed time for the liquid surface to recede from 11.5 mm to 50 mm below the tube top.

acetone+ DBPair sweep: p(vapour) = 0 at topz: 11.5 → 50 mmp_A0x_A·P_vapVolatile diffuses up the stagnant air column; Raoult surface pressure falls as it depletes
Figure 4 — Acetone diffuses up the stagnant air column to the swept top ($p_A=0$). Because DBP is non-volatile and the liquid is well-mixed, the surface partial pressure is $x_A P_{vap}$ and falls as acetone depletes; the diffusion path $z$ simultaneously lengthens.

Approach. Use the pseudo-steady flux (Fick, bulk flow neglected as instructed), relate the level position to the acetone remaining, then integrate the resulting ODE for time as the surface recedes from 11.5 to 50 mm.

  1. Surface partial pressure (Raoult). With $n_A$ mol acetone left and $n_D=0.00753$ mol DBP fixed, $x_A=n_A/(n_A+n_D)$ and $p_{surf}=x_A P_{vap}$. Initially $x_{A}=0.778$, so $p_{surf}=47.0$ kPa (not 60.5 — the DBP dilutes the surface).
  2. Level–composition link. Only acetone leaves, so the surface recedes as acetone volume shrinks: at depth $z$ the acetone remaining is $n_A=n_{A0}-\dfrac{\rho_A A_t}{M_A}(z-z_0)$. Reaching $z=50$ mm leaves $n_A=0.0120$ mol ($x_A=0.615$, $p_{surf}=37.2$ kPa).
  3. Pseudo-steady flux. Neglecting bulk flow, $N_A=\dfrac{D_{AB}}{RT}\dfrac{p_{surf}-0}{z}$ (mol·m⁻²·s⁻¹). Equating the acetone lost to the level recession, $-A_t N_A=\dfrac{\rho_A A_t}{M_A}\dfrac{dz}{dt}$, separates to$$dt=\frac{\rho_A A_t}{M_A}\,\frac{RT\,z}{A_t D_{AB}\,p_{surf}(z)}\,dz.$$
  4. Integrate 11.5 → 50 mm. With $p_{surf}(z)=P_{vap}\,n_A/(n_A+n_D)$ inside the integral (driving force falling from 47 to 37 kPa), numerical integration gives$$t=\int_{11.5\,\text{mm}}^{50\,\text{mm}}\frac{\rho_A RT\,z\,(n_A+n_D)}{M_A D_{AB} P_{vap}\,n_A}\,dz=\boxed{7.93\times10^{4}\ \text{s}\;\approx\;22\ \text{hours}}.$$

Check (model choice): the paper explicitly says to neglect bulk flow, so the simple Fick flux $N_A=D_{AB}(p_{surf})/(RTz)$ is used rather than the Stefan log-mean form $\ln[(P-0)/(P-p_{surf})]$. Retaining the Stefan drift factor would raise the flux and shorten the time by ~40% (the acetone concentration is not dilute), but that contradicts the stated assumption.

QuantityResult
Initial / final surface pressure47.0 kPa / 37.2 kPa
Acetone evaporated0.0143 mol (of 0.0263 charged)
Time to recede to 50 mm≈ 7.9×10⁴ s (22 h)