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

Question 6 of 7: Cooling-Tower Sizing and Air/Make-up Balances

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

Notes on this paper

National Exams — May 2013 — 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 constitute a complete paper (Parts A/B carry 20% each, Part C 30% each). All seven are solved below for completeness. Property data are stated in each Given block. Two chart appendices (SI humidity–temperature charts) accompany Q6.

Reference texts: Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular diffusion, mass-transfer coefficients, absorption, humidification; Welty, Wicks, Wilson & Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (5th/6th ed., Wiley) — transient diffusion, boundary-layer mass transfer; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — packed-tower and cooling-tower design; supporting data from Perry's Chemical Engineers' Handbook (9th ed.) and Incropera & DeWitt, Fundamentals of Heat and Mass Transfer.

Question 6: Cooling-Tower Sizing and Air/Make-up Balances (Part C — 30%)

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). $L=5000$ kg/s water, $310\to296$ K (range 14 K, approach 13 K to the 283 K wet bulb); dry bulb 288 K; performance coefficient $C_t=5.2$, read as the required $K_aV/L$.
Given (b). $L=1000$ kg/s water $40\to30\,°\text{C}$; air in $40\,°\text{C}$, 1.0 bar, 30% RH; air out $35\,°\text{C}$, 70% RH.

Find. (a) the tower base diameter and height; (b)(i) the air mass flow rate and (ii) the make-up water rate.

natural-drafthyperbolic towerhot water 40°C, 1000 kg/s(310 K → 296 K in 6a)moist air out 35°C, 70% RHair in40°C 30% RHcold water 30°C + make-up
Figure 6 — Natural-draft counter-flow cooling tower. Warm water falls against rising air; the enthalpy driving force (Merkel) sets the required transfer, and mass/energy balances close the air and make-up rates.

Approach. Part (a): treat $C_t=5.2$ as the required tower characteristic $K_aV/L$, find the operating $L/G$ that makes the Merkel enthalpy integral equal to it, hence the air rate, then size the plan area and height from a design water loading and aspect ratio. Part (b): humid-air mass and energy balances between the two measured air states.

  1. (a) Merkel demand vs. tower characteristic. The required transfer units are $$\frac{K_aV}{L}=\int_{T_c}^{T_h}\frac{c_{pw}\,dT}{h^{*}(T)-h_{\text{air}}(T)},\qquad h_{\text{air}}=h_{\text{in}}+\frac{L}{G}c_{pw}(T-T_c).$$ Setting this equal to $C_t=5.2$ (inlet air enthalpy $\approx$ saturated air at the 283 K wet bulb, $28.8$ kJ/kg) and integrating numerically gives the operating $\boxed{L/G=1.85}$ — just inside the pinch at $L/G\approx1.92$.
  2. Air mass flow. $$G=\frac{L}{L/G}=\frac{5000}{1.85}=\boxed{2.70\times10^{3}\ \text{kg/s of air}}.$$
  3. Base diameter and height (estimate). At a design plan-area water loading of $1.8\ \text{kg}/(\text{m}^2\cdot\text{s})$, the plan area is $A=5000/1.8=2778\ \text{m}^2$, so $$D_\text{base}=\sqrt{4A/\pi}=\boxed{\approx60\ \text{m}},\qquad H\approx1.5\,D_\text{base}=\boxed{\approx90\ \text{m}},$$ with a superficial air velocity $G/(\rho A)\approx0.8$ m/s — typical of large natural-draft towers.
  4. (b) Humid-air states. With $H=0.622\,p_w/(P-p_w)$ and $h=(1.005+1.88H)T+2500.9H$ (kJ/kg dry air): $$H_1=0.0140,\ h_{a1}=76.4;\qquad H_2=0.0254,\ h_{a2}=100.4\ \text{kJ/kg}.$$
  5. (b)(i) Air rate from the tower energy balance. With water $40\to30\,°\text{C}$ ($c_{pw}=4.187$) and evaporation $E=G(H_2-H_1)$, $$L\,c_{pw}(40)-\big(L-E\big)c_{pw}(30)=G(h_{a2}-h_{a1})\ \Rightarrow\ \boxed{G\approx1.85\times10^{3}\ \text{kg dry air/s}}.$$
  6. (b)(ii) Make-up water. The make-up replaces the evaporated water, $$E=G(H_2-H_1)=1851(0.0254-0.0140)=\boxed{\approx21\ \text{kg/s}}.$$
QuantityResult
(a) Operating $L/G$ (for $K_aV/L=5.2$)1.85
(a) Air rate / base diameter / height$\approx2.7\times10^{3}$ kg/s · $\approx60$ m · $\approx90$ m
(b)(i) Air flow (dry)$\approx1.85\times10^{3}$ kg/s
(b)(ii) Make-up water$\approx21$ kg/s

Check: the natural-draft dimensions in (a) are engineering estimates — the base diameter follows from an assumed $1.8\ \text{kg}/(\text{m}^2\cdot\text{s})$ plan-area water loading and the height from a 1.5 aspect ratio (both typical of large hyperbolic towers). The air rate is fixed once $C_t=5.2$ is read as the required $K_aV/L$. Coulson & Richardson Vol. 1 (the source of this paper's humidity-chart appendices) sizes natural-draft towers from a performance-coefficient chart for water loading and height; that chart is not reproduced with the exam, so the loading and aspect ratio above are stated assumptions and the two dimensions should be read as order-of-magnitude estimates. Part (b) is a closed balance and is not assumption-sensitive.