Countercurrent Rotary-Drier Air Rate: 23-Chem-A2 Unit Operations and Separation Processes, December 2019, Q4

Question B1: Countercurrent Rotary-Drier Air Rate (25 points: 18 + 7)

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
Wet stock feed	0.125 kg/s, 40% moisture in → 5% out
Solid temperatures	in 22 °C, out 32 °C
Air in / out	112 °C / 37 °C, inlet humidity $H_1=0.006$
Radiation loss	20,000 J/kg dry air
$c_p$ (solid / air / vapour)	880 / 1000 / 2010 J/kg·K
$\lambda$ (water at 22 °C)	2.449 × 10⁶ J/kg

Find. (a) The dry-air mass flow $G$; (b) the outlet humidity $H_2$.

Figure B1 — Overall mass balance fixes the water evaporated; an overall energy balance (reference 22 °C) equates the enthalpy dropped by the hot air to the sensible heating of the product, the evaporation duty, and the radiation loss.

Approach. First close the moisture balance to get the evaporation rate; then write an overall energy balance about the whole drier (reference 22 °C) and solve the single equation for the dry-air rate $G$, from which the outlet humidity follows.

Solids and moisture balance. Dry solids $=0.125(1-0.40)=0.075$ kg/s. Water in $=0.05$; product $=0.075/(1-0.05)=0.0789$ kg/s carrying $0.00395$ kg/s water out. Evaporated water $$W=0.05-0.00395=\boxed{0.0461\ \text{kg/s}}.$$
Inlet-air enthalpy (per kg dry air, above 22 °C). $$h_1=c_{pa}(112-22)+H_1[\lambda+c_{pv}(112-22)]=1000(90)+0.006[2.449\times10^6+2010(90)]=105{,}779\ \text{J/kg}.$$
Product sensible duty. Heating dry solids and their residual water from 22 → 32 °C (liquid-water $c_p=4190$ J/kg·K assumed, as the paper does not list it), $$Q_s=0.075(880)(10)+0.00395(4190)(10)=825.4\ \text{J/s}.$$
Overall energy balance → dry-air rate. The hot air gives up enthalpy to (i) heat the product, (ii) evaporate $W$ and superheat that vapour to 37 °C, and (iii) supply the radiation loss. Writing the balance per unit $G$ with $a=\lambda+c_{pv}(37-22)$, $$G=\frac{Q_s+W\,a}{h_1-c_{pa}(37-22)-H_1a-20{,}000}=\boxed{2.06\ \text{kg dry air/s}}.$$
(b) Outlet humidity. All evaporated water joins the air, so $$H_2=H_1+\frac{W}{G}=0.006+\frac{0.0461}{2.057}=\boxed{0.0284\ \text{kg/kg dry air}}.$$

Quantity	Value
Water evaporated $W$	0.0461 kg/s
(a) Dry-air rate $G$	≈ 2.06 kg/s
(b) Outlet humidity $H_2$	0.0284 kg/kg dry air

Concept & Theory

Topic. Coupled mass and energy balance on a continuous convective drier.

Key relations. Moisture balance $W=$ (water in stock) − (water in product); humid enthalpy $h=c_{pa}\Delta T+H[\lambda+c_{pv}\Delta T]$; overall energy balance air-side = product sensible + evaporation + radiation; $H_2=H_1+W/G$.

Why this works. An overall balance about the whole drier sidesteps the internal temperature and humidity profiles: only the terminal states of the two streams and the fixed radiation loss enter. Choosing 22 °C as the datum makes the incoming solid contribute zero sensible enthalpy and lets the tabulated latent heat (given at 22 °C) be used directly.

Common pitfall. Omitting the sensible superheat of the evaporated vapour to the exit-air temperature, or forgetting that the radiation loss is quoted per kg of dry air (so it multiplies $G$ and belongs on the denominator side of the $G$ equation).

Source. Coulson & Richardson Vol. 2 ch. 16 (drying); Geankoplis ch. 9; McCabe ch. 24.

Question 4 of 6: Countercurrent Rotary-Drier Air Rate

Question B1: Countercurrent Rotary-Drier Air Rate (25 points: 18 + 7)