Question 1 of 6: Part A: Steam Temperature for Maximum Boiling Rate
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: open-book, three-hour exam; Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6); all problems 25 points, at least two attempted per part. Every problem is worked in full below.
Reference texts: J. M. Coulson & J. F. Richardson, Chemical Engineering Vols. 1&2 (Coulson & Richardson) — boiling curves, tube-bank convection, gas absorption and column sizing; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — free-convection correlations and the gas-property tables reproduced in the paper; J. R. Welty, C. E. Wicks & R. E. Wilson, Fundamentals of Momentum, Heat and Mass Transfer — the laminar vertical-plate local Nusselt relation; R. E. Treybal, Mass-Transfer Operations (McGraw-Hill) — stagnant-film diffusion, differential (Rayleigh) distillation and packed-tower flooding; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.).
Question 1 — Part A: Steam Temperature for Maximum Boiling Rate (25 points)
Given. A wall separates condensing steam (outside) from an organic liquid boiling at $T_b=340$ K (inside). Wall: thickness $x=3$ mm, $k=42$ W/m·K. Steam-side film coefficient $h_o=11$ kW/m²K (constant). The boiling coefficient $h_b$ varies with the wall-to-liquid difference $\Delta T_b$:
$\Delta T_b$ (K)
22.2
27.8
33.3
36.1
38.9
41.7
44.4
50.0
$h_b$ (kW/m²K)
4.43
5.91
7.38
7.30
6.81
6.36
5.73
4.54
Find. The steam temperature $T_{steam}$ that maximises the evaporation rate, i.e. the boiling heat flux $q=h_b\,\Delta T_b$.
Figure 1 — Series path steam→wall→boiling liquid (left), and the boiling curve $q=h_b\Delta T_b$ (right), which has an interior maximum near $\Delta T_b=42$ K. Maximum evaporation means operating at that peak flux, then finding the steam temperature that drives it.
Approach. The evaporation rate follows the boiling heat flux $q=h_b\Delta T_b$; tabulate that product to locate its maximum, then add the wall and steam-film temperature drops carried by that same flux to the liquid temperature to get the required steam temperature.
Boiling heat flux at each point. Evaporation rate $\propto q=h_b\,\Delta T_b$. Multiplying each pair (in kW/m²):
$$q:\;98.3,\;164.3,\;245.8,\;263.5,\;264.9,\;265.2,\;254.4,\;227.0.$$
The flux rises, peaks, then falls — the classic nucleate-boiling curve with a maximum (the critical heat flux).
Locate the maximum. The largest product is at $\Delta T_b=41.7$ K, where $h_b=6.36$ kW/m²K:
$$q_{max}=h_b\,\Delta T_b=(6.36)(41.7)=\boxed{265.2\ \text{kW/m}^2}.$$
Operating anywhere past this point (higher $\Delta T_b$) actually reduces the flux, so this peak is the maximum evaporation condition.
Temperature drop across the wall. The same flux conducts through the 3-mm wall:
$$\Delta T_{wall}=\frac{q\,x}{k}=\frac{(265{,}200)(0.003)}{42}=18.9\ \text{K}.$$
Temperature drop across the steam film. Through the condensing-steam film,
$$\Delta T_{film}=\frac{q}{h_o}=\frac{265{,}200}{11{,}000}=24.1\ \text{K}.$$
Assemble the steam temperature. Working outward from the boiling liquid, the steam must sit above the liquid by the boiling drop plus the wall and film drops:
$$T_{steam}=T_b+\Delta T_b+\Delta T_{wall}+\Delta T_{film}=340+41.7+18.9+24.1,$$
$$\boxed{T_{steam}\approx 424.8\ \text{K}\;(\approx 425\ \text{K}\;\text{or}\;152\ ^\circ\text{C}).}$$
Raising the steam above this only increases $\Delta T_b$ past 41.7 K, which drops the flux — so 425 K is the optimum, not merely a lower bound.
