Solids Drying with Air Recycle: 23-Chem-A1 Process Balances and Chemical Thermodynamics, December 2017, Q1

Question 1: Solids Drying with Air Recycle (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. Wet solid enters at 15 wt% water and leaves at 7 wt% water. Air humidities (kg moisture per kg dry air): fresh $H_F=0.01$, recycle $H_R=0.10$, mixed air into the drier $H_M=0.03$. The recycle is drawn from the air leaving the drier, so exit-air humidity $=H_R=0.10$. Basis: 100 kg wet solid.

Quantity	Value
Wet solid fed (basis)	100 kg
— bone-dry solid (85%)	85 kg
— water in feed (15%)	15 kg
Product water content	7 wt%
Fresh / mixed / recycle humidity	0.01 / 0.03 / 0.10

Find. (a) water evaporated; (b) dry air in the fresh stream $F$; (c) dry air in the recycle stream $R$ — all per 100 kg wet solid.

Figure 1 — Fresh air is blended with recycle at the mixing tee, passes through the drier picking up evaporated water, and the exit air is split into recycle and exhaust. The recycle and exhaust share the exit humidity of 0.10.

Approach. A water balance on the solid fixes the evaporation; a humidity balance on the mixing tee relates $F$ and $R$; a moisture balance across the drier (dry air $F+R$ gains $H_M\!\to\!H_R$) closes the system.

Water removed from the solid (part a). The bone-dry solid (85 kg) is conserved. In the 7% product it is 93% of the mass, so the product $=85/0.93=91.40$ kg carrying $0.07\times91.40=6.40$ kg water. The evaporated water is the feed water minus the product water: $$w_{\text{evap}}=15-6.40=\boxed{8.60\ \text{kg water per 100 kg wet solid}}.$$
Humidity balance on the mixing tee. Fresh dry air $F$ (at 0.01) and recycle dry air $R$ (at 0.10) blend to the mixed dry air $F+R$ at 0.03. Balancing moisture: $$0.01F+0.10R=0.03(F+R)\;\Rightarrow\;0.07R=0.02F\;\Rightarrow\;F=3.5\,R.$$ The tee alone fixes the fresh-to-recycle ratio without any reference to the drier.
Moisture balance across the drier. The same dry air $(F+R)$ enters at 0.03 and leaves at 0.10, having absorbed the 8.60 kg of evaporated water: $$(F+R)(0.10-0.03)=8.60\;\Rightarrow\;F+R=\frac{8.60}{0.07}=122.9\ \text{kg dry air}.$$
Solve for the two air streams (parts b, c). Substituting $F=3.5R$ into $F+R=4.5R=122.9$: $$R=\frac{122.9}{4.5}=\boxed{27.3\ \text{kg dry air (recycle)}},\qquad F=3.5R=\boxed{95.6\ \text{kg dry air (fresh)}}.$$

Check: fresh air brings in $0.01(95.6)=0.96$ kg and recycle brings $0.10(27.3)=2.73$ kg into the tee, giving 3.69 kg on 122.9 kg dry air $=0.030$ — the specified mixed humidity, confirming closure.

Quantity	Result (per 100 kg wet solid)
(a) Water evaporated	8.60 kg
(b) Dry air in fresh stream	95.6 kg
(c) Dry air in recycle stream	27.3 kg

Concept & Theory

Topic. Steady-state moisture (humidity) balances on a drier with an air recycle loop.

Key relations. Solid-phase water conservation via the bone-dry tie component; humidity ratio $H=\text{kg water}/\text{kg dry air}$; tee balance $H_F F+H_R R=H_M(F+R)$; drier balance $(F+R)(H_{\text{out}}-H_{\text{in}})=w_{\text{evap}}$.

Why this works. Dry air and bone-dry solid are tie (inert) components that pass through unchanged, so each balance is written on the conserved carrier. Because recycle is drawn from the exit air, the exit humidity equals the recycle humidity, which supplies the extra relation needed to close the loop.

Common pitfall. Using total (wet) air instead of the dry-air basis in the humidity balances, or forgetting that the exit-air humidity is 0.10 (the recycle value), not some new unknown.

Source. Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes §8.4 (humidity, drying) and §4.5–4.7 (recycle balances).

Question 1 of 6: Solids Drying with Air Recycle

Question 1: Solids Drying with Air Recycle (Part A — equal value)