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23-Chem-B1 Transport Phenomena · December 2016

Question 3 of 6: B1 — Counterflow double-pipe exchanger sizing and rating

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

Notes on this paper

Paper format. EGBC 04-CHEM-B1 Transport Phenomena, December 2016, 3 hours, open-book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer); candidates attempt one from each section plus a fourth (four of six at 25 marks each, only the first four in the answer book are marked). All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species tables) in rectangular, cylindrical and spherical coordinates — these are quoted rather than re-derived.

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; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, internal-flow heat transfer and film mass transfer; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — the log-mean-temperature-difference and effectiveness–NTU methods; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — transport properties and the Colebrook friction correlation.

Question 3: B1 — Counterflow double-pipe exchanger sizing and rating (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.

QuantitySymbolValue
Water flow (base case)$\dot m_w$$70\ \text{kg/min}=1.167\ \text{kg/s}$
Water temperatures$T_{w,i}\to T_{w,o}$$40\to85\ \degree\text{C}$
Oil temperatures$T_{o,i}\to T_{o,o}$$120\to85\ \degree\text{C}$
Overall coefficient$U$$400\ \text{W/m}^2\!\cdot\!\text{K}$
Heat capacities$c_{p,o},\,c_{p,w}$$1700,\ 4183\ \text{J/kg}\!\cdot\!\text{K}$

The printed phrase “inlet and temperatures” is read as the oil inlet and outlet temperatures (120 and 85 °C), and the printed unit of $U$, “W/m²”, as W/m²·K.

Find. (a) area $A$; (b) water outlet temperature at 50 kg/min in the same exchanger; (c) area needed to still reach 85 °C at 50 kg/min.

T length oil 120°C oil out 85°C water out 85°C water in 40°C ΔT₁=35 ΔT₂=45
Fig. B1: Counterflow temperature profile. The hot oil (120→85 °C) and cold water (40→85 °C) move in opposite directions; the terminal differences $\Delta T_1=35$ K and $\Delta T_2=45$ K set the log-mean driving force.

Approach. Size the base case with $Q=UA\,\Delta T_{\text{lm}}$ (water duty and counterflow LMTD). For the reduced-flow rating (part b) the area is fixed, so switch to the effectiveness–NTU method; for part c revert to an LMTD sizing at the new duty.

  1. Heat duty from the water side. $$Q=\dot m_w c_{p,w}\Delta T_w=1.167(4183)(85-40)=2.196\times10^{5}\ \text{W}=219.6\ \text{kW}.$$
  2. Oil flow (needed for the rating parts). From the oil energy balance, $\dot m_o=\dfrac{Q}{c_{p,o}(120-85)}=\dfrac{2.196\times10^5}{1700(35)}=3.69\ \text{kg/s}$, giving capacity rates $C_w=\dot m_w c_{p,w}=4880\ \text{W/K}$ and $C_o=\dot m_o c_{p,o}=6275\ \text{W/K}$ (water is $C_{\min}$).
  3. Counterflow LMTD. $\Delta T_1=T_{o,i}-T_{w,o}=120-85=35$ K and $\Delta T_2=T_{o,o}-T_{w,i}=85-40=45$ K, so $$\Delta T_{\text{lm}}=\frac{\Delta T_2-\Delta T_1}{\ln(\Delta T_2/\Delta T_1)}=\frac{45-35}{\ln(45/35)}=39.8\ \text{K}.$$
  4. Required area (part a). $$A=\frac{Q}{U\,\Delta T_{\text{lm}}}=\frac{2.196\times10^5}{400(39.8)}\;\Longrightarrow\;\boxed{A\approx13.8\ \text{m}^2.}$$
  5. Reduced flow — switch to effectiveness–NTU (part b). With the same exchanger ($A=13.8\ \text{m}^2$, same $U$ and oil flow) but $\dot m_w=50\ \text{kg/min}=0.833\ \text{kg/s}$: $C_w=3486\ \text{W/K}$ (still $C_{\min}$), $C_o=6275\ \text{W/K}$, so $C_r=C_{\min}/C_{\max}=0.556$ and $$\text{NTU}=\frac{UA}{C_{\min}}=\frac{400(13.8)}{3486}=1.58.$$
  6. Effectiveness and outlet temperature. The counterflow relation $\varepsilon=\dfrac{1-e^{-\text{NTU}(1-C_r)}}{1-C_r e^{-\text{NTU}(1-C_r)}}=0.697$ gives $$Q'=\varepsilon C_{\min}(T_{o,i}-T_{w,i})=0.697(3486)(120-40)=1.94\times10^5\ \text{W},$$ so the water outlet is $$T_{w,o}=40+\frac{Q'}{C_w}=40+\frac{1.94\times10^5}{3486}\;\Longrightarrow\;\boxed{T_{w,o}\approx95.7\ \degree\text{C}.}$$ (The oil now leaves at $120-Q'/C_o\approx89\ \degree\text{C}$.)
  7. Area to reach 85 °C at 50 kg/min (part c). Target duty $Q''=C_w(85-40)=3486(45)=1.569\times10^5\ \text{W}$; oil now leaves at $120-Q''/C_o=95.0\ \degree\text{C}$, so $\Delta T_1=120-85=35$, $\Delta T_2=95-40=55$, $\Delta T_{\text{lm}}=\dfrac{55-35}{\ln(55/35)}=44.2\ \text{K}$ and $$A=\frac{Q''}{U\,\Delta T_{\text{lm}}}=\frac{1.569\times10^5}{400(44.2)}\;\Longrightarrow\;\boxed{A\approx8.86\ \text{m}^2.}$$
QuantityResult
Heat duty (base case)$Q\approx219.6\ \text{kW}$
(a) Required area$A\approx13.8\ \text{m}^2$ ($\Delta T_{\text{lm}}=39.8$ K)
(b) Water outlet at 50 kg/min (same exchanger)$T_{w,o}\approx95.7\ \degree\text{C}$ ($\varepsilon=0.70$)
(c) Area for 85 °C at 50 kg/min$A\approx8.86\ \text{m}^2$
Check — oil-side assumptions in parts (b) and (c)

The statement changes only the water flow, so the oil mass flow (3.69 kg/s), inlet temperature (120 °C) and $U$ are held fixed. Part (b) is a rating of the fixed part-(a) exchanger — hence effectiveness–NTU — while part (c) is a fresh sizing to the original 85 °C target. If instead the oil flow were re-optimised, the numbers would shift; the assumption is standard for this "what-if" pairing and matches the supplied $\varepsilon$–NTU chart.