Question 2 of 6: Length of a Double-Pipe Cooler — Countercurrent and Cocurrent
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Reference texts: Incropera, DeWitt, Bergman & Lavine, Fundamentals of Heat and Mass Transfer (Wiley) — variable-conductivity conduction, resistance networks, convection correlations; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — double-pipe exchanger sizing, evaporator heat load, diffusion through stagnant films; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations and two-film interphase transfer; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — gas absorption. Property data from Perry's Chemical Engineers' Handbook (9th ed.).
Exam format: six problems worth 25 points each — three in Part A (Heat Transfer) and three in Part B (Mass Transfer, printed as Problems 1–3 of Part B and numbered Questions 4–6 here). At least two problems from each part must be attempted and only the first two from each section are marked. Open-book, three-hour paper. All six problems are solved here for completeness.
Paper note
This is the 16-CHEM-A3 Heat and Mass Transfer paper of May 2019. The printed paper is five pages (cover notes, Part A Problems 1–3, Part B Problems 1–3), followed by two sheets of graph paper and a periodic table.
Part A — Heat Transfer
Question 2: Length of a Double-Pipe Cooler — Countercurrent and Cocurrent (25 points)
The DATA list also prints “Thermal conductivity of water = 0.63 W/m²·K”. No water is present in a glycol/toluene exchanger, and the units given are those of a heat-transfer coefficient, so this line is not used. Because both pipes are 3 mm thick, the inner-pipe ID is $43-2(3)=37$ mm and the outer-pipe ID, the outer boundary of the annulus, is $70-2(3)=64$ mm. The table gives one set of properties per fluid, so they are taken as constant.
Given.
Quantity
Value
Ethylene glycol (hot, inner pipe)
$\dot m=5500$ kg/hr $=1.528$ kg/s, $85\to68$ °C
Toluene (cold, annulus)
$30\to62$ °C
Inner pipe ID / OD
0.037 m / 0.043 m (3 mm wall)
Outer pipe OD / ID (annulus outer $D_2$)
0.070 m / 0.064 m (3 mm wall)
Pipe-wall conductivity $k_{\text{wall}}$
46.52 W/m·K
Glycol $\rho,c_p,k,\mu$
1080, 2680 J/kg·K, 0.248, $3.4\times10^{-3}$
Toluene $\rho,c_p,k,\mu$
840, 1800 J/kg·K, 0.146, $4.4\times10^{-4}$
Find. The total tube length $L$ for (a) countercurrent and (b) cocurrent operation.
Figure 2 — Countercurrent double-pipe cooler. Glycol enters the tube at 85 °C and leaves at 68 °C; toluene flows the opposite way in the annulus, warming 30→62 °C. The approach is $23$ K where the glycol is hottest and $38$ K where it is coolest. In cocurrent operation (b) both streams enter at the same end, and the approaches become $55$ K and $6$ K.
Approach. Fix the duty and the toluene flow from energy balances. Get the two film coefficients from Dittus–Boelter (tube side, and the annulus on its hydraulic diameter) and combine the three series resistances into $U_o$, referred to the inner-pipe outer surface. Then $L=q/(U_o\,\pi d_o\,\text{LMTD})$ for each flow arrangement. The flows, properties and therefore $U_o$ are the same in (a) and (b); only the LMTD changes.
Duty and toluene flow. $\dot m_h=5500/3600=1.528$ kg/s, so$$q=\dot m_h c_{p,h}\Delta T_h=1.528(2680)(85-68)=\boxed{69.6\ \text{kW}}.$$An energy balance on the coolant gives $\dot m_c=q/[c_{p,c}\Delta T_c]=69{,}606/[1800(62-30)]=1.208$ kg/s $=4350$ kg/hr.
(a) LMTD, countercurrent. The end approaches are $\Delta T_1=85-62=23$ K and $\Delta T_2=68-30=38$ K, so$$\text{LMTD}=\frac{38-23}{\ln(38/23)}=29.9\ \text{K}.$$
Tube-side film (glycol, cooled). $Re=\dfrac{4\dot m_h}{\pi d_i\mu}=\dfrac{4(1.528)}{\pi(0.037)(3.4\times10^{-3})}=1.55\times10^{4}$ (turbulent), $Pr=c_p\mu/k=2680(3.4\times10^{-3})/0.248=36.7$. With $n=0.3$ (cooling),$$Nu=0.023\,Re^{0.8}Pr^{0.3}=152,\qquad h_i=\frac{Nu\,k}{d_i}=1021\ \text{W/m}^2\text{K}.$$
Annulus film (toluene, heated). The annulus runs from $d_o=0.043$ m to the outer-pipe ID $D_2=0.064$ m, so the hydraulic diameter is $D_h=D_2-d_o=0.021$ m and the flow area is $\tfrac\pi4(0.064^2-0.043^2)=1.765\times10^{-3}$ m². Then $u=1.208/(840\times1.765\times10^{-3})=0.815$ m/s, $Re=\rho uD_h/\mu=840(0.815)(0.021)/(4.4\times10^{-4})=3.27\times10^{4}$ (turbulent) and $Pr=1800(4.4\times10^{-4})/0.146=5.42$. With $n=0.4$ (heating),$$Nu=0.023\,Re^{0.8}Pr^{0.4}=185,\qquad h_o=\frac{Nu\,k}{D_h}=\frac{185(0.146)}{0.021}=1285\ \text{W/m}^2\text{K}.$$
Overall coefficient (on outer surface of inner pipe).$$\frac{1}{U_o}=\frac{d_o}{d_i h_i}+\frac{d_o\ln(d_o/d_i)}{2k_{\text{wall}}}+\frac{1}{h_o}=1.14\times10^{-3}+6.9\times10^{-5}+7.78\times10^{-4}=1.986\times10^{-3},$$so $U_o=\boxed{504\ \text{W/m}^2\text{K}}$. The metal wall is negligible (3.5%). The viscous glycol film carries the larger share of the resistance (57%).
(a) Area and length, countercurrent.$$A_o=\frac{q}{U_o\,\text{LMTD}}=\frac{69{,}606}{504(29.9)}=4.63\ \text{m}^2,\qquad L=\frac{A_o}{\pi d_o}=\frac{4.63}{\pi(0.043)}=\boxed{34.3\ \text{m}}.$$
(b) Cocurrent operation. Both fluids now enter at the same end. The end differences are $\Delta T_1=85-30=55$ K at the inlet and $\Delta T_2=68-62=6$ K at the outlet, so$$\text{LMTD}=\frac{55-6}{\ln(55/6)}=22.1\ \text{K}.$$The flows and properties are unchanged, so $h_i$, $h_o$ and $U_o$ are the same, and$$A_o=\frac{69{,}606}{504(22.1)}=6.25\ \text{m}^2,\qquad L=\frac{6.25}{\pi(0.043)}=\boxed{46.3\ \text{m}}.$$Cocurrent flow needs $29.9/22.1=1.35$ times the countercurrent length. It is feasible only because the toluene leaves (62 °C) below the glycol outlet (68 °C), and that 6 K exit pinch is what depresses the log-mean.
