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23-Chem-A2 Unit Operations and Separation Processes · December 2016

Question 6 of 6: Overall Coefficient of a Tubular Heat Exchanger

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

Notes on this paper

Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, December 2016. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt at least two from each section (only the first two per section are marked). All six are worked below for completeness.

Reference texts. Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, sedimentation, fluidization, filtration, crystallization, evaporation, drying) and Vol. 1 (fluid flow, heat transfer); McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Perry's Chemical Engineers' Handbook (9th ed.); Kern, Process Heat Transfer.

Note on chart and property data. A1 uses a Fanning friction-factor chart, B1 a KNO3 solubility curve (mole-fraction basis), and B2/B3 standard steam-table / transport-property values. Engineering choices are flagged in Check callouts.

Question B3: Overall Coefficient of a Tubular Heat Exchanger (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.

Geometry / propertyValue
Shell dia. / tubes / length0.635 m / 532 / 4.8 m
Tube $d_o/d_i$ / pitch (△) / baffle0.019 / 0.016 m / 0.024 m / 0.241 m
Wall $k$ / fouling coeff.17 W/m·K / 5670 W/m²·K
Tube fluid (organic)63.0 kg/s, $c_p$2100, $\mu$5e−4, $\rho$800, $k$0.13
Shell fluid (water)25.2 kg/s, $c_p$4190, $\mu$8.3e−4, $\rho$1000, $k$0.62

Find. The overall heat-transfer coefficient $U_o$ (outside-area basis).

Resistances in series (outside-area basis) organic tube d_i=16, d_o=19 mm 1/h_i wall fouling 1/h_o → U_o
Figure B3 — The overall coefficient combines the tube-side organic film (controlling), the stainless wall, the fouling layer, and the water film in the shell, all referred to the tube outside area.

Approach. Compute the tube-side film coefficient with the Dittus–Boelter correlation and the shell-side coefficient with Kern's method (equivalent diameter and cross-flow area), then add the film, wall-conduction and fouling resistances on the outside-area basis to obtain $U_o$.

  1. Tube-side velocity and Reynolds number (organic). Flow area $=532(\tfrac{\pi}{4}d_i^2)=0.107$ m²; $u=63.0/(800\times0.107)=0.736$ m/s, so $$Re_i=\frac{\rho u d_i}{\mu}=\frac{800(0.736)(0.016)}{5\times10^{-4}}=18{,}850,\quad Pr_i=\frac{c_p\mu}{k}=8.08.$$
  2. Tube-side film coefficient (Dittus–Boelter). $$h_i=\frac{k}{d_i}\big(0.023\,Re_i^{0.8}Pr_i^{1/3}\big)=\frac{0.13}{0.016}(0.023)(18850)^{0.8}(8.08)^{1/3}=987\ \mathrm{W/m^2K}.$$ This organic film is the smallest coefficient and will control.
  3. Shell geometry (Kern). Equivalent diameter for triangular pitch $d_e=0.0144$ m; cross-flow area $A_s=\dfrac{D_s\,(p_t-d_o)\,B}{p_t}=\dfrac{0.635(0.005)(0.241)}{0.024}=0.0319$ m²; mass velocity $G_s=25.2/0.0319=790$ kg/m²s.
  4. Shell-side film coefficient (water). $$Re_s=\frac{d_eG_s}{\mu}=13{,}740,\quad Pr_s=5.6,\quad h_o=\frac{k}{d_e}\big(0.36\,Re_s^{0.55}Pr_s^{1/3}\big)=5190\ \mathrm{W/m^2K}.$$
  5. Combine resistances on the outside area. $$\frac{1}{U_o}=\frac{d_o}{d_i h_i}+\frac{d_o\ln(d_o/d_i)}{2k_w}+\frac{1}{h_o}+\frac{1}{h_d}.$$ Numerically $\tfrac{1}{U_o}=1.203\times10^{-3}+9.6\times10^{-5}+1.93\times10^{-4}+1.76\times10^{-4}=1.67\times10^{-3}$.
  6. Overall coefficient. $$U_o=\frac{1}{1.67\times10^{-3}}=\;\boxed{599\ \mathrm{W/m^2K}}.$$ The tube-side organic film alone contributes about 72% of the total resistance.
QuantityValue
Tube-side $Re_i$ / $h_i$18,850 / 987 W/m²K
Shell $Re_s$ / $h_o$13,740 / 5190 W/m²K
Wall + fouling resistance$2.7\times10^{-4}$ m²K/W
Overall coefficient $U_o$599 W/m²K
Check (fouling placement): the exam gives a single fouling coefficient (5670 W/m²K); it is applied here once, as a combined dirt resistance on the outside-area basis. If the same allowance were instead credited to each surface, the inside fouling referred to the outside area would add $d_o/(d_ih_d)=2.09\times10^{-4}$. $U_o$ would then fall to about 532 W/m²K, roughly 11% lower. The single-allowance reading matches the wording (“the fouling heat-transfer coefficient”), so 599 W/m²K stands.
Check (tube-side correlation): the tube film uses the Colburn form of the Dittus–Boelter equation, $Nu=0.023Re^{0.8}Pr^{1/3}$, because the problem does not say which stream is heated. Kern's own tube-side method is Sieder–Tate, $Nu=0.027Re^{0.8}Pr^{1/3}(\mu/\mu_w)^{0.14}$. The statement that the viscosity change is negligible sets $(\mu/\mu_w)^{0.14}=1$, and on that basis $h_i\approx1160$ W/m²K and $U_o\approx670$ W/m²K. Either correlation is defensible if you name it, so expect $U_o$ between 600 and 670 W/m²K.
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