Question 2 of 6: Tube Length to Heat Water with Condensing Steam
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) — conduction resistance networks, critical radius, convection correlations, radiation exchange between surfaces; Coulson & Richardson, Chemical Engineering Vol. 1 (Butterworth-Heinemann) — heat-exchanger sizing and diffusion through stagnant films; Treybal, Mass-Transfer Operations (3rd ed., McGraw-Hill) — differential (Rayleigh) distillation and penetration-theory absorption; Geankoplis, Transport Processes and Separation Process Principles (4th ed., Prentice Hall) — molecular-diffusion flux relations. Property data from Perry's Chemical Engineers' Handbook (9th ed.).
Exam format: six problems worth 25 points each — Part A (Heat Transfer, Q1–Q3) and Part B (Mass Transfer, Q4–Q6). The rubric asks for at least two problems from each part; every problem is solved here for completeness. Open-book, three-hour paper.
Part A — Heat Transfer
Question 2: Tube Length to Heat Water with Condensing Steam (25 points)
The DATA block prints both conductivities in $\text{W/m}^2\cdot\text{K}$; a conductivity has units $\text{W/m}\cdot\text{K}$, which is how they are used here. The water $c_p$ is not given and is taken as $4180\ \text{J/kg}\cdot\text{K}$; both films are comparable, so the tube length is insensitive to a few percent in $c_p$.
Given. Water $30\to70$ °C, $\dot V=1200$ L/hr $=3.333\times10^{-4}$ m³/s; $d_i=0.025$ m, $d_o=0.028$ m; condensing steam at $T_s=120$ °C, outside coefficient $h_o=5800\ \text{W/m}^2\text{K}$.
Quantity
Value
Water inlet / outlet
30 °C → 70 °C
Volumetric flow $\dot V$
1200 L/hr = $3.333\times10^{-4}$ m³/s
Tube $d_i$ / $d_o$
0.025 m / 0.028 m
Steam temperature $T_s$
120 °C (isothermal, condensing)
Outside coefficient $h_o$
5800 W/m²·K
$\rho,\ \mu,\ k_w,\ k_{\text{wall}}$
980 kg/m³, $6\times10^{-4}$ Pa·s, 0.63, 950 W/m·K
Find. The heated length $L$ of tube.
Figure 2 — Double-pipe arrangement: water flows inside the tube and gains sensible heat; saturated steam condenses isothermally on the outside surface, so the driving temperature difference falls from $90$ K at inlet to $50$ K at outlet (a log-mean of $68.1$ K).
Approach. Compute the duty from the water energy balance, the inside coefficient from Dittus–Boelter, combine the three resistances into $U_o$, then divide the duty by $U_o$, the LMTD and the outer perimeter to get the length.
Duty and velocity. Mass flow $\dot m=\rho\dot V=980\times3.333\times10^{-4}=0.327\ \text{kg/s}$, so$$q=\dot m c_p\Delta T=0.327\times4180\times(70-30)=\boxed{54.6\ \text{kW}}.$$The tube velocity is $v=\dot V/(\tfrac{\pi}{4}d_i^2)=3.333\times10^{-4}/4.909\times10^{-4}=0.679$ m/s.
Inside coefficient (Dittus–Boelter). $Re=\rho v d_i/\mu=980(0.679)(0.025)/6\times10^{-4}=2.77\times10^{4}$ (turbulent) and $Pr=c_p\mu/k_w=4180(6\times10^{-4})/0.63=3.98$. With heating ($n=0.4$),$$Nu=0.023\,Re^{0.8}Pr^{0.4}=143,\qquad h_i=\frac{Nu\,k_w}{d_i}=3610\ \text{W/m}^2\text{K}.$$
Overall coefficient on the outside area.$$\frac{1}{U_o}=\frac{d_o}{d_i h_i}+\frac{d_o\ln(d_o/d_i)}{2k_{\text{wall}}}+\frac{1}{h_o}=3.10\times10^{-4}+1.7\times10^{-6}+1.72\times10^{-4}=4.84\times10^{-4},$$so $U_o=\boxed{2064\ \text{W/m}^2\text{K}}$. The thin metal wall is negligible; the water film and the steam film are comparable.
LMTD and length. The steam is isothermal, so with $\Delta T_1=90$ K and $\Delta T_2=50$ K, $\text{LMTD}=(90-50)/\ln(90/50)=68.1$ K. Then$$A_o=\frac{q}{U_o\,\text{LMTD}}=\frac{54{,}600}{2064\times68.1}=0.389\ \text{m}^2,\qquad L=\frac{A_o}{\pi d_o}=\frac{0.389}{\pi(0.028)}=\boxed{4.42\ \text{m}}.$$
