Air Dehumidification with a Bypass: 23-Chem-A1 Process Balances and Chemical Thermodynamics, May 2018, Q2

Question 2: Air Dehumidification with a Bypass (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. Entering (and bypass) air: 320 K, saturated. Dehumidifier exit: 285 K, saturated. Target humidity of the reheated product: $Y_f=0.03$ kg water/kg dry air (a maximum, taken as the design value). $P=101.3$ kPa; $p_w^{\text{sat}}(320)=10.6$ kPa, $p_w^{\text{sat}}(285)=1.4$ kPa. Psychrometric ratio $M_w/M_{\text{air}}=18.02/28.97=0.622$; $R=8.314$ kPa·L/mol·K.

Stream	Humidity $Y$ (kg/kg dry air)
Entering / bypass air (320 K, sat)	$Y_1=\dfrac{0.622\,(10.6)}{101.3-10.6}=0.0727$
Dehumidifier exit (285 K, sat)	$Y_2=\dfrac{0.622\,(1.4)}{101.3-1.4}=0.00872$
Final reheated air (target)	$Y_f=0.0300$

Find. (a) bypass dry air per kg dry air through the dehumidifier; (b) water condensed per 100 m³ fed to the dehumidifier; (c) final-air volume per 100 m³ fed.

Figure 2 — The 320 K saturated air is split: one part is cooled to 285 K (dropping water as condensate), the other bypasses. Recombining the dry (285 K) stream with the humid bypass and reheating to 320 K sets the product humidity at 0.03.

Approach. A moisture balance on the mixing junction (dry stream $Y_2$ + bypass $Y_1$ → product $Y_f$) gives the bypass ratio; the ideal-gas law converts the 100 m³ feed to a dry-air mass, and the humidity drop $Y_1\!-\!Y_2$ gives the condensate; the total dry air and its water content then set the final volume at 320 K.

Bypass ratio from the mixing balance (part a). Let $x$ be kg bypass dry air per kg dry air through the dehumidifier. Water in equals water out at the mixer: $$1\cdot Y_2+x\cdot Y_1=(1+x)\,Y_f\;\Rightarrow\;x=\frac{Y_f-Y_2}{Y_1-Y_f}=\frac{0.0300-0.00872}{0.0727-0.0300}=\boxed{0.498\ \tfrac{\text{kg bypass}}{\text{kg through}}}.$$ Roughly half the air is bypassed — the humid bypass re-adds just enough moisture to hit 0.03.
Dry-air mass in the 100 m³ feed. The feed is saturated at 320 K, so $y_w=p_w^{\text{sat}}/P=10.6/101.3=0.1046$. Total moles and dry-air mass: $$n_{\text{tot}}=\frac{PV}{RT}=\frac{101.3(100{,}000)}{8.314(320)}=3808\ \text{mol},\quad m_{\text{dry}}=n_{\text{tot}}(1-y_w)\frac{M_{\text{air}}}{1000}=98.3\ \text{kg}.$$
Water condensed (part b). The stream through the dehumidifier loses $Y_1-Y_2$ per kg dry air: $$m_{\text{cond}}=m_{\text{dry}}\,(Y_1-Y_2)=98.3\,(0.0727-0.00872)=\boxed{6.29\ \text{kg per 100 m}^3}.$$
Final-air volume (part c). Total dry air after adding the bypass is $m_{\text{dry}}(1+x)=98.3(1.498)=147.3$ kg, carrying water at $Y_f$. In moles, $$n_{\text{final}}=\frac{m_{\text{dry}}(1+x)\,10^3}{M_{\text{air}}}+\frac{m_{\text{dry}}(1+x)\,Y_f\,10^3}{M_w}=5107+245=5352\ \text{mol},$$ $$V_{\text{final}}=\frac{n_{\text{final}}RT}{P}=\frac{5352(8.314)(320)}{101.3(1000)}=\boxed{140.6\ \text{m}^3\ \text{at 320 K}}.$$

Check — target humidity taken as the design value

The problem gives the product humidity as a maximum ("not exceeding 0.03"). We design at exactly $Y_f=0.03$ (the least bypass, most dehumidification margin); a lower target would simply bypass less air.

Quantity	Result
(a) Dry air bypassed	0.498 kg per kg through the dehumidifier
(b) Water condensed	6.29 kg per 100 m³ fed
(c) Final air volume	140.6 m³ per 100 m³ fed (at 320 K)

Concept & Theory

Topic. Psychrometric (humidity) balances on a dehumidifier–bypass–reheat process.

Key relations. Absolute humidity $Y=\dfrac{M_w}{M_{\text{air}}}\dfrac{p_w}{P-p_w}$ (saturated when $p_w=p_w^{\text{sat}}$); mixer moisture balance $Y_2+xY_1=(1+x)Y_f$; ideal-gas law $PV=nRT$ for the volume conversions.

Why this works. Dry air is the tie (inert) component, conserved through cooling, mixing and reheating, so all balances are written per kg dry air. Reheating does not change the humidity ratio — it only lowers the relative humidity — so the product humidity is fixed entirely at the mixing junction. The bypass supplies the moisture that a fully-dehumidified stream would lack.

Common pitfall. Using a wet-air basis, or forgetting the 0.622 mass ratio and writing $Y=p_w/(P-p_w)$ (that gives the molar humidity, off by a factor of 0.622). Also, condensate is set by the stream through the unit, not the whole feed.

Source. Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (humidity & saturation); Felder & Rousseau §8.4 (psychrometry) and §4.5–4.7 (bypass balances).

Question 2 of 6: Air Dehumidification with a Bypass

Question 2: Air Dehumidification with a Bypass (Part A — equal value)