Single-Effect Caustic-Soda Evaporator Area: 23-Chem-A2 Unit Operations and Separation Processes, May 2016, Q5

Question B2: Single-Effect Caustic-Soda Evaporator Area (25 marks)

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. Single-effect evaporator.

Quantity	Value
Feed $F$	4500 kg/hr @ 10% NaOH
Product	41% NaOH
Vapour-space pressure	13 kPa
Steam temperature	390 K (116.85 °C)
$U$	1.25 kW/m²·K
Liquid head above surface	1.2 m (ρ = 1390 kg/m³)
Boiling-point rise	30 K
Feed temperature	291 K (18 °C)

Find. The heating-surface area $A$.

Figure B2 — Feed is concentrated from 10% to 41% NaOH; the coil sees a true driving force reduced by both the boiling-point rise and the hydrostatic head at the heating surface.

Approach. A solute (NaOH) balance fixes the product and vapour rates; the true boiling temperature is the saturation temperature at 13 kPa raised by the boiling-point rise and by the hydrostatic head at the coil; an enthalpy balance gives the duty $Q$, and $A=Q/(U\,\Delta T)$.

Mass balances. Solute balance $F x_F=P x_P$ gives $$P=\frac{4500(0.10)}{0.41}=1098\ \mathrm{kg/hr},\qquad V=F-P=3402\ \mathrm{kg/hr}.$$
Saturation and true boiling temperatures. At 13 kPa the saturation temperature is $T_{sat}=51.1$ °C. The liquid boils at the surface at $T_b=51.1+30=81.1$ °C (boiling-point rise). At the coil, 1.2 m of 1390 kg/m³ liquid adds a hydrostatic head of $\rho g h=1390(9.81)(1.2)=16.4$ kPa, so the solution there is at $T_{sat}(29.4\,\mathrm{kPa})+30=68.7+30=98.7$ °C.
True temperature driving force. Referencing the steam (116.85 °C) to the boiling solution at the heating surface, $$\Delta T=116.85-98.7=18.15\ \mathrm{K}.$$ Ignoring the head would overstate $\Delta T$ (and understate the area).
Heat duty (enthalpy balance). Heating the feed from 18 °C to the vapour-space boiling point 81.1 °C and vaporizing $V$ at $\lambda_v\approx2380$ kJ/kg: $$Q=F c_{p,F}(T_b-T_F)+V\lambda_v=4500(4.0)(63.1)+3402(2380)=1.14\times10^6+8.10\times10^6\ \mathrm{kJ/hr}.$$ Dividing by 3600, $Q=2565$ kW.
Heating-surface area. $$A=\frac{Q}{U\,\Delta T}=\frac{2565}{1.25(18.15)}=\;\boxed{113\ \mathrm{m^2}}.$$

Quantity	Result
Product / vapour rate	1098 / 3402 kg/hr
Boiling T (surface, with head)	98.7 °C
True $\Delta T$	18.15 K
Heat duty $Q$	≈ 2565 kW
Heating-surface area $A$	≈ 113 m²

Check (data): saturation temperatures (51.1 °C at 13 kPa, 68.7 °C at 29.4 kPa) and the vapour latent heat (≈2380 kJ/kg at the vapour-space condition) are steam-table values; the feed sensible heat uses $c_{p,feed}=4.0$ and neglects heat of dilution (not given). Using $c_{p,product}$ or including dilution shifts $Q$ by a few percent.

Concept & Theory

Topic. Single-effect evaporator sizing with boiling-point rise and hydrostatic-head elevation.

Key relations. Solute balance $Fx_F=Px_P$; $V=F-P$; true $\Delta T=T_{steam}-(T_{sat}+\text{BPR}+\text{head elevation})$; $Q=Fc_p(T_b-T_F)+V\lambda$; $A=Q/U\Delta T$.

Why this works. The coil can only transfer heat across the temperature gap between condensing steam and the actual boiling liquid it touches. Two effects raise that liquid's boiling point above the pure-water value: the dissolved solute (boiling-point rise) and the pressure of the liquid column above the coil (hydrostatic head). Subtracting both from the steam temperature gives the true, smaller driving force that governs area.

Common pitfall. Sizing on $T_{steam}-T_{sat}$ and forgetting the boiling-point rise and hydrostatic head — this can under-size the area by 30–50%. Also mixing absolute (K) and Celsius scales in the sensible term (differences are identical, but reference points must be consistent).

Source. Coulson & Richardson, Chemical Engineering Vol. 2 ch. 14 (evaporation); Geankoplis ch. 8.

Question 5 of 6: Single-Effect Caustic-Soda Evaporator Area

Question B2: Single-Effect Caustic-Soda Evaporator Area (25 marks)