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23-Chem-A3 Heat and Mass Transfer · December 2019

Question 3 of 6: Restoring Duty on a Fouled Tube Condenser

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

Notes on this paper

National Exams — December 2019 — 16-Chem-A3 Heat and Mass Transfer. Three-hour, open-book exam (one textbook of the candidate’s choice; any non-communicating calculator). Format: two parts — Part A (Q1–Q3) Heat Transfer and Part B (Q1–Q3) Mass Transfer; at least two questions must be attempted from each part and only the first two in each part are marked, so four questions (each 25 points) constitute a complete paper. All six questions are solved below for completeness. Property values not printed on the paper (molar masses, water latent heat, the dimensionless free-convection peak velocity $f'_{max}$, benzene/toluene physical properties) are stated explicitly in each Given block as open-book look-ups.

Reference texts: Coulson & Richardson (Backhurst, Harker & Richardson), Chemical Engineering, Vol. 1 — Fluid Flow, Heat Transfer and Mass Transfer (6th ed., Butterworth-Heinemann) — the source family for the crystalliser, tube-condenser, Stefan-tube and distillation problems; Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (free- and forced-convection correlations, the Ostrach similarity solution); Treybal, Mass-Transfer Operations (3rd ed.) and McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.) — Stefan diffusion, Chilton–Colburn analogy, McCabe–Thiele; supporting property data from Perry’s Chemical Engineers’ Handbook (9th ed.) and the NIST Chemistry WebBook.

The obscured verb in Part B Q1 is taken as “calculate”, and the Part B Q2 water viscosity at 27 °C is $8.76\times10^{-4}$ Pa·s. One column of the Part A Q2 air-property table, the buoyancy group $g\beta/\nu^{2}$, is internally inconsistent: its printed values are about 40% below what $\rho$, $\mu$ and $\beta=1/T_f$ give. That column is not used; the Grashof number is built from the $\rho$ and $\mu$ columns and $\beta=1/T_f$.

Question 3: Restoring Duty on a Fouled Tube Condenser (Part A — 25 pts)

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. $N=120$ tubes, $d_i=22$ mm, $L=2.5$ m; benzene condenses isothermally at 350 K, water in at 290 K; clean duty 4 kg/s benzene at $v_0=0.70$ m/s. After fouling, $R_s=2\times10^{-4}$ m²K/W is added inside; $h_i\propto v^{0.8}$, $h_o=2250$ W/m²K, $\lambda_{benzene}=400$ kJ/kg. Water $c_p=3.98$ kJ/kg·K (from the paper), $\rho_w=1000$ kg/m³.

QuantityValue
Condensation duty $Q=\dot m\lambda$$4\times400=1600$ kW (fixed)
Inside area $A_i=N\pi d_i L$20.74 m²
Flow area $A_f=N(\pi/4)d_i^2$0.04562 m²
Condensing coefficient $h_o$2250 W/m²K
Scale resistance $R_s$2×10⁻⁴ m²K/W

Find. The new water velocity that keeps the condensation rate at 4 kg/s after the scale forms.

Tube-bundle condenser120 tubes, di=22 mm, L=2.5 mBenzene vapour 350 KBenzene condensate 4 kg/sCooling water 290 Kv = 0.70 m/s (clean)Warm water out
Figure 3 — Single-pass surface condenser: benzene condenses on the outer (bundle) side at 350 K while cooling water is heated as it passes through the tubes. Scale on the tube inside adds a resistance that must be offset by a faster water velocity.

Approach. The duty and area are fixed, so back-calculate the clean water coefficient from the clean operating point, express the fouled overall coefficient as a function of the unknown velocity (both through $h_i\propto v^{0.8}$ and through the velocity-dependent LMTD), and solve $U'A_i\,\Delta T_{lm}'=Q$.

  1. Clean operating point. Water flow $\dot m_w=\rho_w v_0 A_f=1000(0.70)(0.04562)=31.9$ kg/s. Its temperature rise from the fixed duty is $\Delta T_w=Q/(\dot m_w c_p)=1600/(31.9\times3.98)=12.6$ K, so $T_{w,out}=302.6$ K and$$\Delta T_{lm}=\frac{(350-290)-(350-302.6)}{\ln[(350-290)/(350-302.6)]}=\boxed{53.5\ \text{K}}.$$
  2. Clean coefficients. $U_{clean}=Q/(A_i\Delta T_{lm})=1.6\times10^{6}/(20.74\times53.5)=1444$ W/m²K. Removing the condensing film gives the clean water coefficient:$$\frac1{h_{i,clean}}=\frac1{U_{clean}}-\frac1{h_o}=\frac1{1444}-\frac1{2250}\;\Rightarrow\; \boxed{h_{i,clean}=4030\ \text{W/m}^2\text{K}}\ \text{at }0.70\text{ m/s}.$$
  3. Fouled resistance network. With velocity $v$ the water film scales as $h_i=4030\,(v/0.70)^{0.8}$ and the overall coefficient (inside basis) becomes$$\frac1{U'}=\frac1{h_i}+R_s+\frac1{h_o}=\frac1{4030(v/0.70)^{0.8}}+2\times10^{-4}+\frac1{2250}.$$At the same time faster water heats less, so $\Delta T_w=1600/(1000\,v\,0.04562\times3.98)$ and the LMTD rises toward 60 K as $v$ grows.
  4. Solve the duty constraint. Requiring $U'A_i\Delta T_{lm}'=1600$ kW and solving (bracketed root-find) gives$$\boxed{v'\approx2.05\ \text{m/s}},\qquad \Delta T_w'=4.3\ \text{K},\ T_{w,out}'=294.3\ \text{K}.$$The velocity must almost triple (0.70 → 2.05 m/s) to push enough extra water-side coefficient through the new scale layer.

Check (which data are load-bearing): the DATA block also lists two steam latent heats (2202 and 2383 kJ/kg) — these belong to a different problem and are ignored; a benzene condenser is governed only by $\lambda_{benzene}=400$ kJ/kg. Pumping power scales roughly as $v^{2.8}$, so tripling the velocity multiplies pump duty ~20-fold — the real cost of fouling. $c_p=3.98$ kJ/kg·K is used for water as printed on the paper.

QuantityResult
Fixed condensation duty1600 kW
Clean LMTD / $U$ / $h_i$53.5 K / 1444 / 4030 W/m²K
Clean water velocity0.70 m/s
Required velocity after fouling≈ 2.05 m/s