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23-Chem-A3 Heat and Mass Transfer · December 2016

Question 2 of 7: Evaporative Cooling from an Open Bowl

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

Notes on this paper

Format: open-book, three-hour paper; seven questions in three parts (Part A Q1–2, Part B Q3–4, Part C Q5–7). A candidate answers one from A, one from B and two from C (four questions, equal value). All seven are solved here as a complete study resource.

Reference texts: R.E. Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — film theory, wetted-wall and flat-plate convective mass transfer, packed-tower absorption, distillation; Coulson & Richardson, Chemical Engineering Vol. 1 (6th ed.) and Vol. 2 (5th ed., Richardson, Harker & Backhurst) — diffusion, drying/evaporation, absorption with reaction, adsorption, membrane separations; C.J. Geankoplis, Transport Processes and Separation Process Principles (4th ed.) — Sherwood-number correlations, McCabe–Thiele with side streams; property data from Perry’s Chemical Engineers’ Handbook (9th ed.).

Question 2: Evaporative Cooling from an Open Bowl (Part A — equal value)

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. An open bowl, $d=0.30$ m ($A=0.0707$ m²), water at $T=350$ K, atmospheric $P_T=101.3$ kPa. Vapour diffuses through a 1 mm molecular-diffusion film; strong air currents keep the bulk partial pressure at essentially zero. Thermal mass = 10 kg water-equivalent.

QuantityValue
Surface vapour pressure $p_s$41.8 kPa
Diffusivity $D$ / film $L$$0.2\times10^{-4}$ m²/s / 1 mm
Latent heat $\lambda$2318 kJ/kg
Water equivalent $m$, $C_p$10 kg, 4.187 kJ/kg·K

Find. the rate of temperature fall $dT/dt$ of the water caused by evaporation.

well-mixed water, 350 Kwater equivalent 10 kgfree surface, d = 0.30 m1 mm gas filmvapour swept away by air currents (p∞ ≈ 0)Evaporation removes latent heat → the pool cools
Figure 2 — Vapour leaves the 350 K surface at $p_s=41.8$ kPa and is swept away by air currents ($p_\infty\approx0$); the latent heat carried off cools the well-mixed pool.

Approach. Compute the Stefan evaporation flux with a zero bulk partial pressure, turn it into a mass rate, multiply by the latent heat, and divide by the thermal mass.

  1. Total molar concentration in the film. $C_T=\dfrac{P_T}{RT}=\dfrac{101.3\times10^{3}}{8.314(350)}=34.8$ mol/m³ (equivalently $\tfrac{1}{22.4}\!\times\!\tfrac{273}{350}=0.0348$ kmol/m³).
  2. Stefan flux, bulk partial pressure zero. $$N_A=\frac{D}{L}\,C_T\ln\!\frac{P_T}{P_T-p_s}=\frac{0.2\times10^{-4}}{10^{-3}}(34.8)\ln\!\frac{101.3}{101.3-41.8}=\boxed{0.370\ \text{mol/m}^2\text{s}}.$$
  3. Mass evaporation rate. $\dot m=N_A\,A\,M=0.370(0.0707)(0.018015)=4.72\times10^{-4}$ kg/s (1.70 kg/h).
  4. Heat removed and cooling rate. The evaporative duty is $\dot Q=\dot m\,\lambda=4.72\times10^{-4}(2318\times10^{3})=1.09\ \text{kW}$, which cools the thermal mass at $$\frac{dT}{dt}=\frac{\dot Q}{mC_p}=\frac{1.09\times10^{3}}{10(4187)}=\boxed{0.026\ \text{K/s}\approx 1.57\ \text{K/min}}.$$
QuantityResult
Evaporation flux $N_A$0.370 mol/m²·s
Evaporation rate $\dot m$$4.72\times10^{-4}$ kg/s (1.70 kg/h)
Evaporative duty $\dot Q$1.09 kW
Rate of cooling $dT/dt$0.026 K/s (1.57 K/min)