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23-Chem-A1 Process Balances and Chemical Thermodynamics · May 2018

Question 3 of 6: Countercurrent Rotary Dryer — Air Requirement

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

Notes on this paper

National Exams — May 2018 — 16-Chem-A1 Process Balances and Chemical Thermodynamics. Three-hour, open-book exam; any non-communicating calculator permitted. Format: two parts — Part A (Q1–Q3) Process Mass and Energy Balances, Part B (Q4–Q6) Chemical Thermodynamics; the candidate answers two questions from each part (four constitute a complete paper, equal value). All six questions are solved below for completeness. Property data not printed on the paper (molar volume of an ideal gas, the gas constant, air molar mass, the psychrometric ratio 0.622, and the gas-phase heat-capacity polynomials in the attached Table C-4) are stated explicitly in each Given block as open-book look-ups.

Reference texts: Felder, Rousseau & Bullard, Elementary Principles of Chemical Processes (4th ed., Wiley) — combustion/excess-air, humidity and drying energy balances; Himmelblau & Riggs, Basic Principles and Calculations in Chemical Engineering (8th ed.) — psychrometric and dryer balances; Smith, Van Ness, Abbott & Swihart, Introduction to Chemical Engineering Thermodynamics (8th ed., McGraw-Hill) — generalized virial fugacity, reaction-equilibrium ΔG°(T) from heat-capacity data, and residual/real-gas property changes; supporting property data from the attached Perry’s / Poling “Properties of Gases and Liquids” Table C-4.

Question 3: Countercurrent Rotary Dryer — Air Requirement (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. Product: 1000 kg/hr at 0.2 wt% moisture. Feed: same dry solid at 4.0 wt% moisture. Air in: 363 K, humidity $Y_{\text{in}}=0.01$; air out: 305 K. Solid: 298 K in, 333 K out. Heat loss $Q_L=4\times10^4$ kJ/hr. Reference state 273 K (liquid water, dry air, dry solid). Property data as listed.

QuantityValue (kg/hr)
Dry solid (tie component)$1000\times0.998=998$
Water in product (0.2%)2.0
Wet feed $=998/0.96$1039.6
Water in feed (4.0%)41.58
Water evaporated$41.58-2.0=39.58$

Find. The mass flow of drying air, $G$ (kg dry air/hr and kg humid air/hr).

Countercurrentrotary drierWet (NH4)2SO44.0% H2O, 298 KDry product 1000 kg/hr0.2% H2O, 333 KHot air in: 363 KY=0.01, G = ?Moist air out305 KHeat loss 4 x 10^4 kJ/hr
Figure 3 — Solids move left→right (298→333 K) while hot air moves countercurrently right→left (363→305 K), picking up the evaporated water. An overall energy balance about the whole drum yields the air rate.

Approach. A water balance (via the bone-dry-solid tie) fixes the evaporation; an overall enthalpy balance about the whole drum — moist air in/out, wet/dry solid in/out, heat loss — then gives the dry-air rate $G$, with the exit humidity written as $Y_{\text{out}}=Y_{\text{in}}+w_{\text{evap}}/G$.

  1. Water evaporated (mass balance). The 998 kg/hr of bone-dry solid is conserved; feed water is $998(0.04/0.96)=41.58$ and product water $998(0.002/0.998)=2.0$: $$w_{\text{evap}}=41.58-2.0=\boxed{39.58\ \text{kg water/hr}}.$$
  2. Moist-air enthalpy per kg dry air (ref 273 K). $H_{\text{air}}=C_{p,a}(T-273)+Y\big[\lambda_0+C_{p,v}(T-273)\big]$. At inlet (363 K, $Y=0.01$): $$H_{\text{in}}=1.005(90)+0.01\big[2502.3+1.884(90)\big]=90.45+26.72=117.2\ \tfrac{\text{kJ}}{\text{kg dry air}}.$$ The outlet (305 K) carries $Y_{\text{out}}=0.01+39.58/G$, so $G\,H_{\text{out}}=32.16\,G+(0.01G+39.58)(2562.6)$.
  3. Solid-stream enthalpies (ref 273 K). Dry solid plus its liquid water, in and out: $$H_{\text{sol,in}}=998(1.507)(25)+41.58(4.2)(25)=41{,}966\ \text{kJ/hr},$$ $$H_{\text{sol,out}}=998(1.507)(60)+2.0(4.2)(60)=90{,}743\ \text{kJ/hr}.$$
  4. Overall energy balance → air rate. With $G\,H_{\text{in}}+H_{\text{sol,in}}=G\,H_{\text{out}}+H_{\text{sol,out}}+Q_L$, the $G$-terms collect to $$G\,(117.2-57.79)=H_{\text{sol,out}}+Q_L+101{,}428-H_{\text{sol,in}}=190{,}205,$$ $$\boxed{G=3203\ \text{kg dry air/hr}\;\;\Rightarrow\;\;G_{\text{humid}}=G(1+Y_{\text{in}})=3235\ \text{kg/hr}.}$$

The exit humidity is $Y_{\text{out}}=0.01+39.58/3203=0.0224$ — comfortably below saturation at 305 K, confirming the air can hold the evaporated water.

QuantityResult
Water evaporated39.58 kg/hr
Dry-air requirement $G$3203 kg dry air/hr
Humid-air requirement≈ 3235 kg/hr (at $Y=0.01$)
Exit-air humidity0.0224 kg/kg dry air