23-Chem-A2 Unit Operations and Separation Processes · May 2015
Question 3 of 6: Which Cooling Pipe Gives the Least Pressure Drop?
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. National Exams — 04-CHEM-A2 Mechanical and Thermal Operations, May 2015. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt two from each section. All six are worked below for completeness.
Reference texts. McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Coulson & Richardson, Chemical Engineering Vol. 1 (fluid flow, heat transfer) and Vol. 2 (particle technology, evaporation); Incropera & DeWitt, Fundamentals of Heat and Mass Transfer (8th ed.); Perry's Chemical Engineers' Handbook (9th ed.).
Note on the charts and tables. Page 3 (Fanning friction-factor chart), page 5 (LMTD correction-factor plot) and page 6 (Table B.2, saturated water by pressure) were read from the printed figure. The smooth-pipe friction factor is computed from the Colebrook equation, which the chart plots; the B3 condenser is isothermal on the steam side, so the correction factor is exactly $F=1$; and the B2 saturation temperatures are read from the supplied Table B.2.
Question A3: Which Cooling Pipe Gives the Least Pressure Drop? (25 marks)
Given. The same water duty (cool 0.454 kg/s from 65.56 to 26.67 °C) is achieved in two constant-wall-temperature pipes of different bore; the required length differs, and length drives the pressure drop.
Quantity
Value
Mass flow $\dot m$
0.454 kg/s
Water in / out
65.56 °C → 26.67 °C
System 1
$D$ = 1 in (0.0254 m), $T_w$ = 21.11 °C
System 2
$D$ = 0.5 in (0.0127 m), $T_w$ = 4.44 °C
$\rho,\ c_p$
988.6 kg/m³, 4186.8 J/kg·°C
$\mu,\ k$
$5.829\times10^{-4}$ kg/m·s, 0.6542 W/m·°C
Find. Which pipe produces the smaller pressure drop for the fixed cooling duty.
Figure A3 — Same cooling duty in two bores: the length needed to reach the outlet temperature and the resulting friction pressure drop trade off against each other.
Approach. Length is the hidden variable: size each pipe's length from the constant-wall-temperature heat-transfer requirement $Q=hA\,\Delta T_{lm}$ (Dittus–Boelter for cooling), then compute the friction pressure drop $\Delta P=4f(L/D)(\rho v^2/2)$ and compare.
Common duty and Prandtl number. $Q=\dot m c_p\Delta T = 0.454(4186.8)(65.56-26.67)=7.39\times10^{4}\ \mathrm{W}$; $\Pr = c_p\mu/k = 4186.8(5.829\times10^{-4})/0.6542 = 3.73$.
System 1 (1-inch) — velocity, film coefficient, length. $v=\dot m/(\rho A)=0.906\ \mathrm{m/s}$, $\mathrm{Re}=3.90\times10^4$; cooling Dittus–Boelter $\mathrm{Nu}=0.023\mathrm{Re}^{0.8}\Pr^{0.3}=161$, so $h=\mathrm{Nu}\,k/D=4.14\times10^3\ \mathrm{W/m^2K}$. With $\Delta T_{lm}=[(65.56-21.11)-(26.67-21.11)]/\ln(\cdots)=18.71$ °C, $$L_1=\frac{Q}{h\,\pi D\,\Delta T_{lm}}=11.95\ \mathrm{m}.$$ $L_1=11.95\ \mathrm{m}$
System 1 — pressure drop. Smooth-pipe Colebrook $f=0.00552$, so $$\Delta P_1=4f\frac{L_1}{D}\frac{\rho v^2}{2}=4(0.00552)\frac{11.95}{0.0254}\frac{988.6(0.906)^2}{2}=4.22\ \mathrm{kPa}.$$ $\Delta P_1 \approx 4.2\ \mathrm{kPa}$
System 2 (0.5-inch) — velocity, film coefficient, length. Halving the diameter quadruples the velocity: $v=3.63\ \mathrm{m/s}$, $\mathrm{Re}=7.81\times10^4$; $\mathrm{Nu}=280$, $h=1.44\times10^4\ \mathrm{W/m^2K}$. The colder wall gives $\Delta T_{lm}=38.45$ °C, so $$L_2=\frac{Q}{h\,\pi D\,\Delta T_{lm}}=3.34\ \mathrm{m}.$$ $L_2=3.34\ \mathrm{m}$ The narrow, colder pipe needs only about a quarter of the length.
System 2 — pressure drop and comparison. With $f=0.00474$, $$\Delta P_2=4(0.00474)\frac{3.34}{0.0127}\frac{988.6(3.63)^2}{2}=32.4\ \mathrm{kPa}.$$ Although $L_2\ll L_1$, the fourfold-higher velocity ($\Delta P\propto v^2/D$) dominates. $\Delta P_2 \approx 32.4\ \mathrm{kPa}$ » $\Delta P_1$
