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23-Chem-A3 Heat and Mass Transfer · May 2015

Question 7 of 7: Rotary drier — air rate and exit humidity

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

National Exams — May 2015 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: seven questions in three parts — answer one of Q1–Q2 (Part A), one of Q3–Q4 (Part B) and two of Q5–Q7 (Part C); four questions of equal value constitute a complete paper. All seven are solved below for completeness. Every property datum (diffusivities, vapour pressures, solubilities, humid-air heat capacities) is stated in the question, so each answer is self-contained.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, convective mass transfer, gas absorption and drying; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (6th ed., Wiley) — boundary-layer and transient diffusion; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer — transient conduction/diffusion charts and the sphere in a stagnant medium; Treybal, Mass-Transfer Operations (3rd ed.) — packed-tower design; Crank, The Mathematics of Diffusion (2nd ed.) — series solutions for the sphere; supporting data from Perry's Chemical Engineers' Handbook (9th ed.).

Question 7: Rotary drier — air rate and exit humidity (Part C — 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. A rotary drier (sketched countercurrent; the overall balances do not depend on the flow arrangement): hot air evaporates moisture from wet sand. Two balances close the problem — a water/solids mass balance for the evaporation rate, and an enthalpy balance (datum = feed temperature 294 K) for the dry-air rate.

QuantityValue
Sand feed / moisture1.0 kg/s, 50% water
Discharge moisture3%
Air in: $T$, $\mathcal H_1$380 K, 0.007 kg/kg
Air out $T$310 K
Solids in / out $T$294 K / 309 K
$\lambda_{294}$; $c_{pa},c_{pv},c_{ps}$2450; 0.99, 2.01, 0.88 kJ/kg·K
Radiation loss25 kJ/kg dry air

Find. The dry-air mass flow rate $G$ and the exit humidity $\mathcal H_2$.

rotary drier loss = 25 kJ/kg dry air wet sand 1.0 kg/s, 50% H₂O, 294 K dry sand, 3% H₂O, 309 K air in 380 K, H₁ = 0.007 air out 310 K, H₂ = ?
Figure 7 — Countercurrent rotary drier. A water/solids balance fixes the evaporation rate; an enthalpy balance about the whole unit (datum 294 K) then fixes the dry-air rate, from which the exit humidity follows.

Approach. First close the solids/water balance for the evaporation rate; then write an overall enthalpy balance with the feed temperature (294 K) as datum, substitute $\mathcal H_2=\mathcal H_1+\dot W/G$, and solve the single linear equation for $G$.

  1. Solids and water balance. Bone-dry sand $=1.0(1-0.50)=0.5$ kg/s. Discharged wet solids carry 3% water: discharge $=0.5/(1-0.03)=0.515$ kg/s, of which water $=0.0155$ kg/s. Evaporated water: $$\dot W=(0.5)-(0.0155)=0.485\ \text{kg/s}.$$
  2. Inlet moist-air enthalpy (datum 294 K). $$h_1=c_{pa}(380-294)+\mathcal H_1\!\left[\lambda+c_{pv}(380-294)\right]=0.99(86)+0.007[2450+2.01(86)]=103.5\ \text{kJ/kg}.$$
  3. Group the outlet air terms. Per kg dry air leaving at 310 K: dry-air sensible $b=c_{pa}(310-294)=15.84$; vapour enthalpy coefficient $a=\lambda+c_{pv}(310-294)=2450+2.01(16)=2482$.
  4. Product-solids enthalpy (datum 294 K). $$h_\text{prod}=0.5\,c_{ps}(309-294)+0.0155\,c_{pw}(309-294)=0.5(0.88)(15)+0.0155(4.187)(15)=7.57\ \text{kJ/s}.$$
  5. Overall enthalpy balance → dry-air rate. With $G\mathcal H_2=G\mathcal H_1+\dot W$, the balance $G h_1=G b+(G\mathcal H_1+\dot W)a+h_\text{prod}+25G$ rearranges to $$G=\frac{a\,\dot W+h_\text{prod}}{h_1-b-a\,\mathcal H_1-25}=\frac{2482(0.485)+7.57}{103.5-15.84-2482(0.007)-25}=\boxed{26.7\ \text{kg/s dry air}}.$$
  6. Exit humidity. $$\mathcal H_2=\mathcal H_1+\frac{\dot W}{G}=0.007+\frac{0.485}{26.7}=\boxed{0.0251\ \text{kg/kg}}.$$
QuantityResult
Evaporation rate $\dot W$0.485 kg/s
Inlet air enthalpy $h_1$103.5 kJ/kg dry air
Dry-air mass flow $G$26.7 kg/s
Exit humidity $\mathcal H_2$0.0251 kg/kg
Check
The latent load dominates: evaporating 0.485 kg/s of water needs $\approx1200$ kJ/s, which is why the air rate is large (26.7 kg/s) despite the modest 70 K air temperature drop. The exit humidity rises only from 0.007 to 0.025 kg/kg, confirming the air is far from saturation and the drier is not humidity-limited.
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