23-Chem-A5 Chemical Plant Design and Economics · December 2015
Question 6 of 6: Evaporator Heat-Transfer Area
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
National Exams — December 2015 — 04-Chem-A5 Chemical Plant Design and Economics. Three-hour, closed-book exam; one two-sided aid sheet and an approved calculator permitted. Six equally weighted (20-mark) questions are posed and the candidate answers any five; only the first five are marked. All six are answered below for completeness. Question 1 is a conceptual flowsheet-synthesis question (hydrodealkylation of toluene to benzene) answered with a process flow sheet and organised prose; questions 2, 3 and 6 are numerical (capacity-scaled and index-escalated plant cost, yield-improvement rate of return, and evaporator heat-transfer area); questions 4 and 5 are qualitative essays on materials selection against the common corrosion mechanisms and on process-hazard classification.
Reference texts: M.S. Peters, K.D. Timmerhaus & R.E. West, Plant Design and Economics for Chemical Engineers (5th ed., McGraw-Hill) — the exam's named primary text (cost estimation Ch. 6, interest and profitability Ch. 7–10, materials of construction Ch. 12, plant safety and loss prevention Ch. 3); J.M. Douglas, Conceptual Design of Chemical Processes (McGraw-Hill) — the hydrodealkylation (HDA) flowsheet-synthesis case study used in Question 1; R. Turton et al., Analysis, Synthesis, and Design of Chemical Processes (4th ed., Prentice Hall) — flowsheet synthesis and equipment cost correlations; AIChE, Dow’s Fire & Explosion Index Hazard Classification Guide (7th ed.) — the process-hazard checklist behind Question 5; supporting Canadian practice from CCOHS/WHMIS 2015, the Canadian Environmental Protection Act (CEPA), and CSA/ASME materials and pressure-vessel codes.
Question 6: Evaporator Heat-Transfer Area (20 marks)
Given. Feed $\dot m_F = 4000$ kg/h; water removed $=40\%$ of feed; steam $T_s=120$ °C; boiling $T_b=100$ °C; latent heat $\lambda = 2200$ kJ/kg; overall coefficient $U$ read from the heat-transfer nomograph (boiling aqueous / steam condensing).
Find. The required heat-transfer area $A$.
Figure 6.1 — Single-effect evaporator. Apple juice (4000 kg/h) is fed to a vacuum evaporator boiling at 100 °C; LP steam at 120 °C condenses to supply the latent heat, driving off 1600 kg/h of water vapour and leaving concentrated juice.
Approach. Take the feed as essentially aqueous (composition not given), so the duty is the latent heat needed to evaporate the removed water; read the overall coefficient off the nomograph for the boiling-aqueous / steam-condensing pairing, then size the area from $Q = U A\,\Delta T$.
Water evaporated.
$$\dot m_v = 0.40\times4000 = 1600\ \text{kg/h}$$
Heat duty. Latent heat only (feed and product near the same temperature):
$$Q = \dot m_v\,\lambda = 1600\ \tfrac{\text{kg}}{\text{h}}\times2200\ \tfrac{\text{kJ}}{\text{kg}} = 3.52\times10^{6}\ \text{kJ/h} = \boxed{978\ \text{kW}}$$
Overall coefficient from the nomograph. Join the process-fluid scale to the service-fluid scale and read $U$ where the line crosses the middle diagonal. The printed brackets put boiling aqueous at about $2300\text{--}2700$ and steam condensing at $4000$ and above (the bracket starts exactly on the 4000 tick and runs to the end of the
scale, $\approx4800$). The chart is simply the two film resistances in series,
$$\frac{1}{U} = \frac{1}{h_{\text{proc}}} + \frac{1}{h_{\text{serv}}} \;\Rightarrow\; U = \left(\frac{1}{2500}+\frac{1}{4400}\right)^{-1} \approx \boxed{1600\ \text{W m}^{-2}\text{K}^{-1}}$$
with the bracket corners giving a band of about $1450\text{--}1750$.
Driving force and area. With steam at 120 °C and boiling at 100 °C, $\Delta T = 20$ K, so
$$A = \frac{Q}{U\,\Delta T} = \frac{977{,}800\ \text{W}}{1600\times20} = \boxed{30.6\ \text{m}^2}$$
Quantity
Value
Water evaporated
$1600$ kg/h
Heat duty $Q$
$\approx978$ kW
Overall coefficient $U$ (nomograph)
$\approx1600$ W m$^{-2}$K$^{-1}$
Temperature driving force $\Delta T$
$20$ K
Heat-transfer area $A$
$\approx31$ m$^2$
Check: the area scales inversely with the read coefficient, so the nomograph band $U=1450\text{--}1750$ gives $A\approx28\text{--}34$ m$^2$ — a $\approx31$ m$^2$ central estimate. A common slip is to pick $U\approx2000$: with steam condensing at $\le4800$ that would need a process-side film coefficient near $3500$, well above the printed
boiling aqueous bracket, so it is not a reading the chart supports. Treating the juice as aqueous is the key assumption; a real fruit juice has a small boiling-point elevation and a lower $U$ once fouling is allowed for — Sinnott's table of typical overall coefficients puts a steam/aqueous evaporator at $1000\text{--}1500$, i.e. $33\text{--}49$ m$^2$, so the clean-surface
nomograph figure is the optimistic end and a design would carry the larger area. Sensible-heat pre-warming of the feed is neglected because feed and product sit near 100 °C.
