Sizing a Tubular (Water-to-Water) Heat Exchanger: 23-Chem-A2 Unit Operations and Separation Processes, December 2015, Q6

Question B3: Sizing a Tubular (Water-to-Water) Heat Exchanger (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. A 1-outer-pass, multi-tube-pass water-to-water exchanger sized inside-out from the tube-side duty, the tube-side design velocity, and a length limit.

Quantity	Value
Tube-side water: $\dot m$, $30\to45$ °C	14,400 kg/h = 4 kg/s
Outer-vessel water: $\dot m$, in 90 °C	120 kg/min = 2 kg/s
$U_i$ (inside-area basis)	1390 W/m²·°C
Tube inside diameter $d_i$	1.875 cm
Design tube velocity	37.7 cm/s = 0.377 m/s
Maximum tube length	1.75 m
Water $\rho$ / $c_p$	993 kg/m³ / 4174 J/kg·K

Find. (a) total number of tubes, (b) tubes per pass, (c) length of each tube.

Figure B3 — 1-outer-pass / 2-tube-pass exchanger: cold tube water heated $30\to45$ °C by outer water cooling $90\to60$ °C; the design velocity sets tubes-per-pass, the area sets total tubes and length.

Approach. Two nearly independent constraints size the bundle: the thermal duty fixes the total inside area (via $UF\Delta T_{\text{lm}}$), while the tube-side velocity fixes tubes-per-pass; the length limit then sets the number of passes.

Duty and outer-stream outlet. $\dot m_t=14400/3600=4$ kg/s: $$Q=\dot m_t c_p\Delta T=4(4174)(15)=2.504\times10^{5}\ \text{W},\quad T_{\text{out}}=90-\frac{Q}{\dot m_s c_p}=90-\frac{2.504\times10^{5}}{2(4174)}=60\,{}^\circ\text{C}.$$
LMTD and correction factor. Counter-current terminal differences $90\to45$ and $60\to30$: $$\Delta T_{\text{lm}}=\frac{45-30}{\ln(45/30)}=37.0\ \text{K}.$$ With $P=(45-30)/(90-30)=0.25$ and $R=(90-60)/(45-30)=2.0$, the 1-outer / 2-tube-pass factor (chart, confirmed in closed form) is $F=0.94$, so $$A_i=\frac{Q}{U_iF\Delta T_{\text{lm}}}=\frac{2.504\times10^{5}}{1390(0.94)(37.0)}=5.17\ \text{m}^2.$$
Part (b): tubes per pass from the velocity. Tube bore area $\tfrac{\pi}{4}(0.01875)^2=2.761\times10^{-4}$ m², so flow per tube $=0.377\times2.761\times10^{-4}=1.041\times10^{-4}$ m³/s. Total flow $\dot m_t/\rho=4/993=4.028\times10^{-3}$ m³/s: $$n_{\text{pass}}=\frac{4.028\times10^{-3}}{1.041\times10^{-4}}=38.7.$$ $n_{\text{pass}} = 39\ \text{tubes per pass}$
Parts (a) and (c): passes, total tubes, length. Spreading $A_i$ over $39\,n_p$ tubes, $A_i=(39n_p)\pi d_iL$ gives $n_pL=A_i/(39\pi d_i)=2.25$ m. One pass would need $L=2.25>1.75$ m, so use two tube passes: $$n_p=2\;\Rightarrow\;L=\frac{2.25}{2}=1.13\ \text{m}\ (\le1.75),\quad N=39\times2.$$ $N = 78\ \text{tubes},\quad L = 1.13\ \text{m}$ Check: $78\pi(0.01875)(1.13)=5.17$ m², matching $A_i$; two tube passes is exactly the geometry the $F$-chart was drawn for.

Quantity	Result
Duty $Q$ / outer outlet	$2.50\times10^{5}$ W / 60 °C
$\Delta T_{\text{lm}}$ / $F$	37.0 K / 0.94
Required inside area $A_i$	5.17 m²
(b) Tubes per pass	39
(a) Total tubes / number of passes	78 / 2
(c) Length of each tube	1.13 m

Concept & Theory

Topic. Sizing a multi-pass tubular exchanger from independent thermal and hydraulic constraints.

Key relations. Duty $Q=\dot m c_p\Delta T$; $\Delta T_{\text{lm}}=(\Delta T_1-\Delta T_2)/\ln(\Delta T_1/\Delta T_2)$; multi-pass factor $F(P,R)$; area $A_i=Q/(U_iF\Delta T_{\text{lm}})$; tubes-per-pass $=(\dot m/\rho)/(v\,\tfrac{\pi}{4}d_i^2)$; $A_i=N\pi d_iL$.

Why this works. The thermal requirement fixes total area; the hydraulic requirement fixes how many tubes run in parallel per pass so the design velocity (good tube-side coefficient without erosion) is met. Dividing area by per-tube area and tube count gives the length, and whenever that overflows the length limit you add passes — each pass reusing the same parallel tube count but stacking more length into the same footprint. Velocity alone gives tubes-per-pass; area plus the length limit gives the pass count and total tubes.

Common pitfall. Forgetting the correction factor $F$ (using bare LMTD over-predicts $\Delta T$ and under-sizes the unit), or mixing inside- and outside-area bases for $U$. Also, the $F$ value read must match the pass arrangement finally chosen — here 1-outer / 2-tube-pass is self-consistent with $F=0.94$.

Source. Incropera & DeWitt, Fundamentals of Heat and Mass Transfer ch. 11 (heat exchangers; LMTD & $F$-charts); Coulson & Richardson Vol. 1, ch. 9; Perry's §11.

Question 6 of 6: Sizing a Tubular (Water-to-Water) Heat Exchanger

Question B3: Sizing a Tubular (Water-to-Water) Heat Exchanger (25 marks)