Number of parallel tubes in a steam-heated exchanger: 22-Mec-A1 Applied Thermodynamics and Heat Transfer, December 2017, Q8

Question 8 — Number of parallel tubes in a steam-heated exchanger (Part B, equal value)

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. Duty 200 kW; solution heated 65 → 93 °C; steam condensing at 250 kPa (constant $T_\text{sat}$) on the outside of tubes $D_o=0.04$ m, $D_i=0.03$ m, $L=3$ m, $k=111\ \text{W/m}$·°C, $h_i=3400$, $h_o=7300\ \text{W/m}^2$·°C. Find. the number of parallel tubes.

Quantity	Value
Total duty, $\dot Q$	200 kW
Solution inlet / outlet	65 / 93 °C
Steam saturation temp (250 kPa)	127.4 °C
Tube OD / ID / length	0.04 / 0.03 / 3 m
$k$ / $h_i$ / $h_o$	111 / 3400 / 7300

Figure 8 — Steam condenses on the outside of $N$ parallel tubes; the chemical solution flows inside and is heated from 65 to 93 °C. Each tube contributes an equal share of the 200 kW.

Approach. Build the per-tube overall $UA$ from the inside-film, wall and outside-film resistances, evaluate the LMTD against the constant condensing temperature, get the duty of one tube, and divide the total duty by it (rounding up).

Per-tube resistances. With $A_i=\pi D_i L = 0.2827\ \text{m}^2$ and $A_o=\pi D_o L = 0.3770\ \text{m}^2$, $$R_i=\frac{1}{h_iA_i}=1.040\times10^{-3},\quad R_\text{wall}=\frac{\ln(D_o/D_i)}{2\pi kL}=1.38\times10^{-4},\quad R_o=\frac{1}{h_oA_o}=3.63\times10^{-4}\ \text{K/W}$$
Per-tube conductance. $$\left(\tfrac{1}{UA}\right)=R_i+R_\text{wall}+R_o=1.541\times10^{-3}\ \text{K/W}\ \Rightarrow\ UA = 649\ \text{W/K}$$
Log-mean temperature difference (condensing steam ≈ constant). $$\Delta T_\text{lm}=\frac{(127.4-65)-(127.4-93)}{\ln\frac{127.4-65}{127.4-93}}=\frac{62.4-34.4}{\ln(62.4/34.4)}=47.0\ \text{°C}$$
Duty of one tube. $$\dot q_1 = UA\,\Delta T_\text{lm} = (649)(47.0) = 3.05\times10^{4}\ \text{W} = 30.5\ \text{kW}$$
Number of tubes. $$N = \frac{\dot Q}{\dot q_1} = \frac{200}{30.5} = 6.55\ \Rightarrow\ \text{round up}$$ $N = 7$ parallel tubes

Quantity	Result
Per-tube conductance $UA$	649 W/K
$\Delta T_\text{lm}$	47.0 °C
Duty per tube	30.5 kW
Tubes required	6.55 → 7 tubes

Concept & Theory

Topic. Sizing (tube count) of a condensing-steam exchanger by the overall-$UA$ / LMTD method with series conduction–convection resistances.

Key relations. $1/(UA)=1/(h_iA_i)+\ln(D_o/D_i)/(2\pi kL)+1/(h_oA_o)$; condenser LMTD with constant hot temperature; $\dot q_1=UA\,\Delta T_\text{lm}$; $N=\lceil \dot Q/\dot q_1\rceil$.

Why this works. Condensing steam holds the outside surface at essentially one temperature, so the driving difference is captured by the two end differences of the solution. Each identical tube presents the same series resistance and therefore carries the same duty, so the required count is simply the total load divided by a single tube’s capability, rounded up to guarantee the full 200 kW is delivered.

Common pitfall. Adding the resistances on mismatched areas (they must each use their own surface area), using an arithmetic-mean instead of log-mean temperature difference, or rounding the tube count down.

Source. Çengel & Ghajar, Heat and Mass Transfer §11 (heat exchangers; overall $U$, LMTD method, condensers).

Question 8 of 8: Number of parallel tubes in a steam-heated exchanger

Question 8 — Number of parallel tubes in a steam-heated exchanger (Part B, equal value)