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23-Chem-A2 Unit Operations and Separation Processes · May 2015

Question 6 of 6: Steam Condenser Sizing and Tube Count

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, May 2015. 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, evaporation); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Perry's Chemical Engineers' Handbook (9th ed.).

Note on the charts and tables. Page 3 (Fanning friction-factor chart), page 5 (LMTD correction-factor plot) and page 6 (Table B.2, saturated water by pressure) were read from the printed figure. The smooth-pipe friction factor is computed from the Colebrook equation, which the chart plots; the B3 condenser is isothermal on the steam side, so the correction factor is exactly $F=1$; and the B2 saturation temperatures are read from the supplied Table B.2.

Question B3: Steam Condenser Sizing and Tube Count (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. Steam condenses isothermally at 50 °C; cooling water is heated 15→25 °C inside the tubes. The 35,000 kg/h is taken as the rate of dry saturated steam condensed (see the note below).

Check (interpretation): the paper prints “handles 35,000 kg of dry air per hour and saturated steam at 50 °C”. Read literally, no steam flow is given, so the duty in (a) cannot be found. The only reading that uses every datum supplied (the steam latent heat and the steam-side coefficient) is 35,000 kg/h of dry saturated steam, so “air” is treated as a misprint for “steam”. Any non-condensable air would lower $h_o$ and is not modelled.
QuantityValue
Steam rate35,000 kg/h = 9.722 kg/s
Steam / water temperatures50 °C (isothermal) / 15→25 °C
Tube $d_i$ / $d_o$ / $k$22.5 mm / 25 mm / 45 W/m·K
Water velocity $V$2 m/s
Steam-side $h_o$5000 W/m²K
Water $\rho,c_p,\nu,k$998.8, 4180, $1.0006\times10^{-6}$, 0.59859
Latent heat $\lambda$2374 kJ/kg

Find. (a) water mass flow; (b) outside heat-transfer area; (c) tubes required for the water flow; (d) number of passes for tube length ≤ 2.5 m.

vessel: steam condensing at 50 °C (h_o=5000) water 15 °C, 2 m/s water 25 °C steam in condensate (sat. water)
Figure B3 — Surface condenser (steam in the vessel, water in the tubes): steam condenses isothermally; cooling water passes through the tubes. Isothermal steam ⇒ LMTD correction $F=1$.

Approach. The condensing duty fixes the water flow (a); the overall coefficient (tube-wall + Dittus–Boelter water film + steam film) with the isothermal LMTD gives the area (b); the per-tube water hydraulics give the tube count per pass (c); and the total tube length divided by the 2.5 m limit gives the passes (d).

  1. (a) Water mass flow. Condensing duty $Q=\dot m_s\lambda = 9.722(2374\times10^3)=2.31\times10^{7}\ \mathrm{W}$; from the water side, $$\dot m_w=\frac{Q}{c_p\Delta T_w}=\frac{2.31\times10^{7}}{4180(25-15)}=552\ \mathrm{kg/s}.$$ $\dot m_w=552\ \mathrm{kg/s}$
  2. (b) Overall coefficient and area. Water-side $\mathrm{Re}=Vd_i/\nu=4.50\times10^4$, $\Pr=\nu\rho c_p/k=6.98$; heating Dittus–Boelter $\mathrm{Nu}=0.023\mathrm{Re}^{0.8}\Pr^{0.4}=264$, so $h_i=\mathrm{Nu}\,k/d_i=7.02\times10^3\ \mathrm{W/m^2K}$. On the outside area, $$\frac1{U_o}=\frac1{h_o}+\frac{d_o\ln(d_o/d_i)}{2k_w}+\frac{d_o}{d_i h_i}\Rightarrow U_o=2.58\times10^3\ \mathrm{W/m^2K}.$$ Isothermal steam gives $\Delta T_{lm}=[(50-15)-(50-25)]/\ln(35/25)=29.7\ \mathrm{°C}$, hence $$A_o=\frac{Q}{U_o\Delta T_{lm}}=\frac{2.31\times10^{7}}{2580(29.7)}=301\ \mathrm{m^2}.$$ $A_o=301\ \mathrm{m^2}$ (The printed correction-factor chart is not needed: $F=1$ for an isothermal side.)
  3. (c) Number of tubes for the water flow. Each tube carries $q_t=V\cdot\tfrac{\pi}{4}d_i^2=2(\tfrac{\pi}{4})(0.0225)^2=7.95\times10^{-4}\ \mathrm{m^3/s}$. The total volumetric water flow is $\dot m_w/\rho=552/998.8=0.5527\ \mathrm{m^3/s}$, so $$N=\frac{\dot m_w/\rho}{q_t}=\frac{0.5527}{7.95\times10^{-4}}=695.2\Rightarrow 696\ \text{tubes (per pass).}$$ $N=696\ \text{tubes per pass}$
  4. (d) Number of passes. The total tube length that provides $A_o$ over $N$ tubes is $L_{tot}=A_o/(\pi d_o N)=301/[\pi(0.025)(696)]=5.51\ \mathrm{m}$. Limiting each tube to 2.5 m per pass, $$n_{pass}=\left\lceil\frac{5.51}{2.5}\right\rceil=3\ \text{passes}\ (1.84\ \mathrm{m\ per\ pass}).$$ 3 passes ⇒ ≈2088 tubes total
QuantityResult
Condensing duty $Q$$2.31\times10^7$ W
(a) Water mass flow552 kg/s
Overall $U_o$ / LMTD2.58 kW/m²K / 29.7 °C
(b) Heat-transfer area $A_o$301 m²
(c) Tubes per pass696
(d) Passes (≤2.5 m each)3 (≈2088 tubes total)
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