23-Chem-A2 Unit Operations and Separation Processes · December 2014
Question 6 of 6: Tubular Condenser — Restoring Duty After Fouling
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 2014. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt two from each section. All six are worked below for completeness.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, filtration, crystallisation); Perry's Chemical Engineers' Handbook (9th ed.); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.).
Given. A fixed condensation duty must be held after fouling adds a series resistance; raising the water velocity boosts the tube-side coefficient (as $v^{0.8}$) and steepens the temperature profile.
Quantity
Value
Tubes $N$ / $D_i$ / $L$
120 / 0.022 m / 2.5 m
Benzene $T_{\mathrm{sat}}$ / $\lambda$
350 K / 400 kJ/kg
Water inlet / clean velocity
290 K / 0.70 m/s
Condensation rate (held)
4 kg/s
Vapour coeff. $h_o$ (inside basis)
2.25 kW/m$^2$K
Added scale resistance $R_s$
$2\times10^{-4}\ \mathrm{m^2K/W}$
Find. the water velocity that restores 4 kg/s of condensate after fouling.
Figure B3 — Single-pass tubular condenser: series resistances $1/h_i + R_s + 1/h_o$ on the inside area; the duty is held by raising $h_i$ through the water velocity.
Approach. Establish the clean tube-side coefficient from the observed clean duty (which fixes $U_i$ and hence $h_i$ by subtracting the vapour film), then require the same duty $Q/A_i$ after adding scale, letting $h_i\propto v^{0.8}$ and the outlet temperature respond to the new water rate; solve the resulting balance for the velocity.
Duty and clean water outlet. The condenser duty is $Q=\dot m\lambda=4(400)=1.60\ \mathrm{MW}$. The clean water rate is $\dot m_w=\rho v \,N\tfrac{\pi}{4}D_i^2=1000(0.70)(120)\tfrac{\pi}{4}(0.022)^2=31.9\ \mathrm{kg/s}$, giving an outlet temperature $$T_{\mathrm{out}}=290+\frac{Q}{\dot m_w c_p}=290+\frac{1.60\times10^{6}}{31.9(4180)}=302.0\ \mathrm{K}.$$
Clean overall and tube-side coefficient. Inside area $A_i=N\pi D_i L=120\pi(0.022)(2.5)=20.7\ \mathrm{m^2}$. With $\mathrm{LMTD}=\dfrac{60-48.0}{\ln(60/48.0)}=53.8\ \mathrm{K}$, $$U_i=\frac{Q}{A_i\,\mathrm{LMTD}}=\frac{1.60\times10^{6}}{20.7(53.8)}=1435\ \mathrm{W/m^2K}.$$ Removing the vapour film ($h_o=2250$) leaves the clean water-side coefficient $$\frac1{h_i}=\frac1{U_i}-\frac1{h_o}\;\Rightarrow\; h_i=3960\ \mathrm{W/m^2K}\ \text{at }0.70\ \mathrm{m/s}.$$
Fouled balance at fixed duty. Holding $Q$ (hence $Q/A_i=7.72\times10^{4}\ \mathrm{W/m^2}$) with the velocity ratio $r=v'/0.70$: the water-side coefficient becomes $h_i'=3960\,r^{0.8}$, the outlet temperature $T_{\mathrm{out}}'=290+12.0/r$ (higher rate, smaller rise), and $$U_i'=\left(\frac1{3960\,r^{0.8}}+\frac1{2250}+2\times10^{-4}\right)^{-1},\qquad U_i'\,\mathrm{LMTD}'(r)=7.72\times10^{4}.$$
Solve for the velocity. Solving this single equation (bisection) gives $r=2.945$, so $$v' = 0.70(2.945)=2.06\ \mathrm{m/s},$$ with $h_i'=3960(2.945)^{0.8}=9.40\times10^{3}\ \mathrm{W/m^2K}$. The water must run about three times faster to overcome the scale and restore full condensation. $v' \approx 2.06\ \mathrm{m/s}$ (from 0.70 m/s)
