Double-Effect Forward-Feed Evaporator Capacity: 23-Chem-A2 Unit Operations and Separation Processes, December 2019, Q6

Question B3: Double-Effect Forward-Feed Evaporator Capacity (25 points)

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.

Quantity	Value
Configuration	double-effect, forward feed
Areas / coefficients	$A_1=A_2=10$ m²; $U_1=2.8$, $U_2=1.7$ kW/m²·K
Steam / condenser	375 kPa ($T_s=141.3$ °C) / 13.5 kPa ($T_\text{cond}=51.5$ °C)
Boiling-point rise	3 K (in the concentrated 2nd effect)
Feed	10% solids at 310 K (36.85 °C)
Product	50% solids
$\lambda$ / $c_p$	2330 kJ/kg / 4.1868 kJ/kg·K (printed as “kJ/kg”; per-kelvin intended)

Find. The maximum feed rate $F$ the two effects can process (i.e. the feed that exactly loads both 10-m² surfaces).

Figure B3 — Forward feed: steam drives effect 1, whose vapour $V_1$ heats effect 2. The available overall $\Delta T$ (steam to condenser, less the 3-K BPR) splits between the effects in inverse proportion to $UA$; the feed is sized so both 10-m² surfaces are exactly loaded.

Approach. Read the steam and condenser saturation temperatures, deduct the BPR, then split the overall $\Delta T$ between the effects by solving the coupled rate/energy equations for the boiling temperature $T_1$ that makes the total evaporation equal the 80% demanded by the 10% → 50% concentration change.

Terminal temperatures. At 375 kPa, $T_s=141.3$ °C; at 13.5 kPa, interpolation of the steam table gives $T_\text{cond}=51.5$ °C. With the 3-K rise, the second effect boils at $T_2=51.5+3=54.5$ °C, so the total available driving force is $T_s-T_2=86.8$ K.
Evaporation demand. Concentrating 10% → 50% solids removes a fraction $1-0.10/0.50=0.80$ of the feed as vapour, so $V_1+V_2=0.80\,F$.
Effect-1 balance. Heat into effect 1, $U_1A_1(T_s-T_1)$, heats the feed from 310 K to $T_1$ and boils off $V_1$ (whose vapour then heats effect 2 with $V_1\lambda=U_2A_2(T_1-T_2)$): $$F=\frac{U_1A_1(T_s-T_1)-U_2A_2(T_1-T_2)}{c_p(T_1-T_f)},\qquad V_1=\frac{U_2A_2(T_1-T_2)}{\lambda}.$$
Effect-2 balance. The liquor entering effect 2 flashes from $T_1$ to $T_2$, adding to the vapour raised there: $$V_2=V_1+(F-V_1)\frac{c_p(T_1-T_2)}{\lambda}.$$
Solve for $T_1$. Iterating until $V_1+V_2=0.80F$ gives $$T_1=102.7\ \text{°C}\ (\Delta T_1=38.6,\ \Delta T_2=48.2\ \text{K}),\quad V_1=0.352,\ V_2=0.403\ \text{kg/s}.$$ The corresponding maximum feed is $$\boxed{F_{\max}=0.944\ \text{kg/s}}.$$

Quantity	Value
Effect temperatures $T_1$ / $T_2$	102.7 °C / 54.5 °C
$\Delta T_1$ / $\Delta T_2$	38.6 K / 48.2 K
Vapour $V_1$ / $V_2$	0.352 / 0.403 kg/s
Total evaporation	0.755 kg/s ( = 0.80 F )
Maximum feed rate $F$	≈ 0.944 kg/s

Concept & Theory

Topic. Multiple-effect evaporator capacity by the effect-by-effect area-matching method.

Key relations. Rate equations $Q_i=U_iA_i\Delta T_i$; vapour coupling $V_1\lambda=Q_2$; concentration constraint $\sum V=F(1-x_f/x_p)$; $\sum\Delta T_i=T_s-T_\text{cond}-\text{BPR}$.

Why this works. With both areas fixed at 10 m², the temperature split is not free: each effect must pass the same heat it receives, so the $\Delta T$'s distribute inversely to the $UA$ products. The single unknown boiling temperature $T_1$ is pinned by requiring the combined evaporation to match the mass-balance demand set by the 10% → 50% concentration change; that closes the problem and yields the largest feed the surfaces can handle.

Common pitfall. Forgetting the boiling-point rise (it subtracts from the useful $\Delta T$), or neglecting the feed-preheat and inter-effect flash sensible terms, which shift the required steam and the capacity by several percent.

Source. Coulson & Richardson Vol. 2 ch. 14 (evaporation); McCabe ch. 16; Geankoplis ch. 8.

Question 6 of 6: Double-Effect Forward-Feed Evaporator Capacity

Question B3: Double-Effect Forward-Feed Evaporator Capacity (25 points)