Question 4 of 6: B2 — Allowable fouling resistance of a tubular exchanger
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format: National Exams, May 2015 — 04-CHEM-B1 Transport Phenomena; 3 hours, open book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer), each worth 25 marks; candidates attempt one problem from each section plus a fourth from any section (four of six marked). The worked solutions below cover all six problems. A summary of the conservation equations (continuity, Navier–Stokes, energy, species) is supplied as Appendix A in the paper and is used throughout.
Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances used across every problem; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe-friction and boundary-layer mass-transfer correlations; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — conduction with variable conductivity, exchanger analysis; D. Q. Kern, Process Heat Transfer (McGraw-Hill) — the Kern casing-side method and the LMTD–correction-factor chart for multiple casing passes; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — physical properties and the Moody chart.
Question 4: B2 — Allowable fouling resistance of a tubular exchanger (25 marks)
$19/15$ mm, $5$ m, $166$ per unit ($\times2$ units)
Tube-side coeff. $h_i$
$1000\ \text{W/m}^2\text{K}$
Find. The maximum scale (fouling) resistance $R_{scale}$ (referred to the outside area) that still lets the two available units deliver the duty.
Fig. B2: Countercurrent temperature paths. The terminal differences $\Delta T_1=55$ K and $\Delta T_2=10$ K set the LMTD; because the cold outlet (345 K) exceeds the hot outlet (315 K) there is a temperature cross, which controls how many casing passes are feasible.
Approach. This is a rating problem: compute the required overall coefficient $U_{req}=Q/(A\,F\,\Delta T_{lm})$, compute the clean coefficient $U_{clean}$ from the two film coefficients and the wall, and the allowable fouling resistance is their reciprocal difference.
Duty from the hot-stream energy balance. $Q=\dot m\,c_p\,\Delta T=(4.0)(2380)(400-315)=8.09\times10^{5}\ \text{W}.$
Log-mean temperature difference (countercurrent). With $\Delta T_1=400-345=55\ \text{K}$ and $\Delta T_2=315-305=10\ \text{K}$, $$\Delta T_{lm}=\frac{55-10}{\ln(55/10)}=26.4\ \text{K}.$$
Correction factor from the supplied double-casing-pass chart. $P=\dfrac{t_o-t_i}{T_i-t_i}=\dfrac{345-305}{400-305}=0.421$ and $Z=\dfrac{T_i-T_o}{t_o-t_i}=\dfrac{400-315}{345-305}=2.125.$ The supplied chart is drawn for two exchangers connected in series (two casing passes, tube passes in multiples of two) — exactly the two available units working together. Reading at $P=0.42$ just to the left of the $Z=2.0$ curve gives $F\approx0.73$ (the analytic two-casing-pass expression gives $F=0.730$). This sits just below the usual $F\ge0.75$ guideline because of the temperature cross (benzene leaves at 345 K, above the 315 K nitrobenzene outlet), so the operating point is on the steep part of the curve.
Casing-side coefficient (true cross-flow over the tube bank). The problem tells us to treat the casing flow as true cross-flow, so the tube-bank correlation based on the tube outside diameter is used (Coulson & Richardson, Vol. 1): $\dfrac{h_od_o}{k}=0.33\,C_h\,Re^{0.6}Pr^{0.3}$, with $C_h=1$. Flow area between baffles $A_s=\dfrac{D_s(p_t-d_o)l_B}{p_t}=\dfrac{0.44(0.025-0.019)(0.15)}{0.025}=0.01584\ \text{m}^2$; mass flux $G_s=\dot m/A_s=252.5\ \text{kg/m}^2\text{s}$. Then $Re=\dfrac{d_oG_s}{\mu}=\dfrac{(0.019)(252.5)}{7\times10^{-4}}=6.85\times10^{3}$ and $Pr=\dfrac{c_p\mu}{k}=\dfrac{(2380)(7\times10^{-4})}{0.15}=11.1$, so $$h_o=\frac{0.15}{0.019}\,(0.33)\,(6854)^{0.6}(11.1)^{0.3}=1074\ \text{W/m}^2\text{K}.$$
Clean overall coefficient (outside basis). Combining the casing film, the steel tube wall ($k_w\approx45\ \text{W/m K}$, assumed) and the tube film $h_i=1000$ referred to the outside area: $$\frac{1}{U_{clean}}=\frac{1}{1074}+\frac{0.019\ln(19/15)}{2(45)}+\frac{19/15}{1000}=2.25\times10^{-3}\ \Rightarrow\ U_{clean}=445\ \text{W/m}^2\text{K}.$$
Required coefficient from the available area. Outside heat-transfer area of both units, $A=2\,n_t\,\pi d_o L=2(166)\pi(0.019)(5)=99.1\ \text{m}^2.$ Hence $$U_{req}=\frac{Q}{A\,F\,\Delta T_{lm}}=\frac{8.09\times10^{5}}{(99.1)(0.73)(26.4)}=424\ \text{W/m}^2\text{K}.$$
Allowable scale resistance. $$\boxed{\;R_{scale}=\frac{1}{U_{req}}-\frac{1}{U_{clean}}=\frac{1}{424}-\frac{1}{445}=1.1\times10^{-4}\ \text{m}^2\text{K/W}\;}$$ The value is positive, so the two units can deliver the duty, but the margin is thin: about $1.1\times10^{-4}\ \text{m}^2\text{K/W}$, below the roughly $1.8\times10^{-4}\ \text{m}^2\text{K/W}$ commonly allowed for organic liquids. These exchangers are therefore marginal and would need to be kept fairly clean.
Quantity
Value
Duty $Q$
$8.09\times10^5\ \text{W}$
$\Delta T_{lm}$ (countercurrent)
$26.4\ \text{K}$
$P$, $Z$, $F$ (two casing passes)
$0.421$, $2.125$, $0.73$
Casing-side $h_o$ (true cross-flow)
$1074\ \text{W/m}^2\text{K}$
$U_{clean}$ / $U_{req}$
$445$ / $424\ \text{W/m}^2\text{K}$
Allowable scale resistance $R_{scale}$
$1.1\times10^{-4}\ \text{m}^2\text{K/W}$
Check — assumptions and sensitivity
The tube-wall conductivity ($k_w\approx45\ \text{W/m K}$, carbon steel) is not given; leaving the wall term out raises the allowance to $1.6\times10^{-4}\ \text{m}^2\text{K/W}$. $C_h=1$ is taken in the cross-flow correlation. The result depends strongly on the stated true-cross-flow assumption: Kern’s baffled-casing correlation on the equivalent diameter gives only $h_o\approx751\ \text{W/m}^2\text{K}$, so $U_{clean}\approx378<U_{req}$ and there would be no scale allowance at all. $F=0.73$ is read on the steep part of the supplied chart, so small errors in the terminal temperatures move $F$ noticeably.
