NivaarExam PrepOfficial exam papers ↗

23-Chem-A3 Heat and Mass Transfer · December 2013

Question 5 of 6: Height of a Natural-Draft Cooling Tower

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

Notes on this paper

National Exams — December 2013 — 04-Chem-A3 Mass Transfer Operations. Three-hour, open-book exam; any non-communicating calculator permitted. Format: six questions in two parts — answer two of Q1–Q3 (Part A) and two of Q4–Q6 (Part B); four questions of equal value constitute a complete paper. All six are solved below for completeness. Property data (diffusivities, vapour pressures, psychrometric enthalpies) are stated in each Given block; the psychrometric appendix supplied with the exam furnishes the saturated-air enthalpies used in Q5.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, unsteady-state diffusion, absorption and humidification; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — Stefan diffusion, gas absorption, cooling-tower design; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (boundary-layer mass transfer); supporting property/psychrometric data from Perry's Chemical Engineers' Handbook (9th ed.).

Question 5: Height of a Natural-Draft Cooling Tower (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 counter-current water-cooling tower analysed by the enthalpy (Merkel) method. Saturated-air enthalpies are read from the exam's psychrometric appendix (datum: dry air and liquid water at 0 °C).

QuantityValue
Water rate $L$2000 kg/h, cooled 50 → 30 °C
Air inDB 30 °C, WB 24 °C ⇒ $H_{G1}\approx72$ kJ/kg
$k_ya$2400 kg/m³·h
Excess air20% above minimum
Cross-section $A$1.0 m²
$H^*$: 30/40/50 °C99.5 / 165.7 / 273.4 kJ/kg

Find. The packed height $Z$ of the tower.

water temperature T (°C)enthalpy H (kJ/kg dry air)saturation H*(T)air in H₁=72air out H₂=226operating line3035404550
Figure 5 — Enthalpy–temperature diagram: the operating line runs from the air-inlet state ($H_1\approx72$ kJ/kg at 30 °C) to the air-outlet state at 50 °C, staying below the saturated-air enthalpy curve $H^*(T)$. The vertical gap is the enthalpy driving force.

Approach. Fix the air-inlet enthalpy from its wet-bulb temperature, find the minimum air rate from the point where the operating line is tangent to the saturation curve, apply the 20% excess to set the operating-line slope and outlet enthalpy, integrate the number of gas-enthalpy transfer units, and multiply by the height of a transfer unit $H_{tOG}=G_s/k_ya$.

  1. Air-inlet enthalpy. For the air-water system the wet-bulb line is the adiabatic-saturation line, so the entering-air enthalpy equals the saturated enthalpy at the wet bulb: $H_{G1}=H^*(24\,{}^\circ\text{C})\approx72$ kJ/kg.
  2. Minimum air rate (tangent pinch). The operating line is anchored at the bottom $(30\,{}^\circ\text{C},\,72)$. Rotating it up until it just touches the convex saturation curve gives the steepest allowable slope; the pinch occurs near 42 °C, where $\big(L\,c_w/G_s\big)_{max}=9.26$ kJ/kg·°C, so $G_{s,min}=\dfrac{L\,c_w}{9.26}=\dfrac{2000\times4.187}{9.26}=904\ \text{kg/h}.$
  3. Design air rate and outlet enthalpy. With 20% excess, $G_s=1.2\times904=\boxed{1085\ \text{kg/h}}$, giving an operating slope $L\,c_w/G_s=7.72$ kJ/kg·°C and an outlet enthalpy $H_{G2}=72+7.72(50-30)=226\ \text{kJ/kg}.$ The driving force stays positive throughout (minimum ≈ 16 kJ/kg near 40 °C).
  4. Number of transfer units. Integrating the gas-enthalpy balance up the tower, $$N_{tOG}=\int_{H_{G1}}^{H_{G2}}\frac{dH_G}{H^*-H_G}=7.22$$ (evaluated by stepping the water temperature 30 → 50 °C, reading $H^*(T)$ and the operating $H_G(T)$).
  5. Tower height. The height of a gas-enthalpy transfer unit is $H_{tOG}=\dfrac{G_s}{k_ya\,A}=\dfrac{1085}{2400\times1.0}=0.452\ \text{m}$, so $$Z=H_{tOG}\,N_{tOG}=0.452\times7.22=\boxed{3.3\ \text{m}}.$$
QuantityResult
Minimum air $G_{s,min}$≈ 904 kg/h
Design air $G_s$ (20% excess)≈ 1085 kg/h
Outlet air enthalpy $H_{G2}$≈ 226 kJ/kg
$N_{tOG}$ / $H_{tOG}$7.22 / 0.452 m
Tower height $Z$≈ 3.3 m
Check
Enthalpies were taken from the standard saturated-air relation $H^*=(1.005+1.88\mathcal{H})T+2501\,\mathcal{H}$ (datum 0 °C), which reproduces the supplied psychrometric chart to within chart-reading accuracy. Because $N_{tOG}$ depends only on enthalpy differences, the answer is insensitive to the choice of datum.