Countercurrent Rotary Dryer Air Rate: 23-Chem-A2 Unit Operations and Separation Processes, December 2016, Q5

Question B2: Countercurrent Rotary Dryer Air Rate (25 marks)

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.

Stream / property	Value
Wet feed	0.125 kg/s, 40% moisture, 295 K
Product	5% moisture, 305 K
Air in / out	385 K, $H=0.006$ / 310 K
Radiation loss	20 kJ/kg dry air
$c_p$ solid / air / vapor	0.88 / 1.00 / 2.01 kJ/kg·K
Latent heat @ 295 K	2449 kJ/kg

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

Figure B2 — Hot air enters at the dry (product) end and leaves humid and cool at the feed end; the enthalpy the air gives up supplies evaporation, product heating and radiation loss.

Approach. Close the moisture balance to get the water evaporated, then an enthalpy balance (sensible heat released by the cooling air stream = latent + superheat of the evaporated water + product sensible heat + radiation) for the dry-air rate; back-substitute for the exit humidity.

Solids and moisture balance. Dry solids $=0.125(0.60)=0.075$ kg/s. Product mass $=0.075/0.95=0.0789$ kg/s, carrying $0.00395$ kg/s water. Water evaporated $$\dot m_w=0.125(0.40)-0.00395=0.0461\ \mathrm{kg/s}.$$
Sensible heat released by the air (per kg dry air). Cooling the dry air and its entering moisture from 385 K to 310 K: $$q_\text{air}=\big[c_{p,a}+H_1c_{p,v}\big](385-310)=[1.00+0.006(2.01)](75)=75.9\ \mathrm{kJ/kg}.$$
Heat to evaporate and superheat the moisture (reference 295 K). $$q_\text{evap}=\dot m_w\big[\lambda+c_{p,v}(310-295)\big]=0.0461[2449+2.01(15)]=114.2\ \mathrm{kW}.$$
Product sensible heat. Warming solids and residual water from 295 K to 305 K: $q_\text{prod}=0.075(0.88)(10)+0.00395(4.19)(10)=0.83\ \mathrm{kW}$ — small next to evaporation.
Energy balance for the dry-air rate. The air supplies evaporation, product heating and radiation ($20G$ kW): $$G\,q_\text{air}=q_\text{evap}+q_\text{prod}+20G\;\Rightarrow\;G=\frac{114.2+0.83}{75.9-20}=\;\boxed{2.06\ \mathrm{kg/s}}.$$
Exit humidity. All evaporated water enters the air: $$H_2=H_1+\frac{\dot m_w}{G}=0.006+\frac{0.0461}{2.06}=\;\boxed{0.0284\ \mathrm{kg/kg}}.$$

Quantity	Value
Dry solids / water evaporated	0.075 / 0.046 kg/s
Air sensible heat per kg	75.9 kJ/kg
Evaporation + superheat duty	114.2 kW
(a) Dry-air mass flow $G$	2.06 kg/s
(b) Exit humidity $H_2$	0.0284 kg/kg

Concept & Theory

Topic. Continuous convective drying — coupled mass and enthalpy balances.

Key relations. Moisture balance $\dot m_w=F(x_1-x_2')$; enthalpy balance $G[c_{p,a}+H_1c_{p,v}]\Delta T_\text{air}=\dot m_w[\lambda+c_{p,v}\Delta T]+q_\text{prod}+q_\text{rad}$; $H_2=H_1+\dot m_w/G$.

Why this works. The dryer is adiabatic apart from the stated radiation loss, so the enthalpy the air loses as it cools must appear as evaporation, product heating and that loss. Referencing water to its liquid state at the feed temperature (295 K) lets the tabulated latent heat enter directly.

Common pitfall. Using the wet-feed mass instead of the dry solids as the conserved stream, or omitting the vapor already carried by the inlet air when computing the air's heat capacity. Also forgetting to superheat the evaporated water to the exit-air temperature.

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

Question 5 of 6: Countercurrent Rotary Dryer Air Rate

Question B2: Countercurrent Rotary Dryer Air Rate (25 marks)