23-Chem-A2 Unit Operations and Separation Processes · December 2016
Question 1 of 6: Pump Power for a Crude-Oil Transfer Line
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, December 2016. 3 hours, open book. Six problems (Section A Mechanical Operations: A1–A3; Section B Thermal Operations: B1–B3), each 25 marks; candidates attempt at least two from each section (only the first two per section are marked). All six are worked below for completeness.
Reference texts. Coulson & Richardson, Chemical Engineering Vol. 2 (particle technology, sedimentation, fluidization, filtration, crystallization, evaporation, drying) and Vol. 1 (fluid flow, heat transfer); McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Perry's Chemical Engineers' Handbook (9th ed.); Kern, Process Heat Transfer.
Note on chart and property data. A1 uses a Fanning friction-factor chart, B1 a KNO3 solubility curve (mole-fraction basis), and B2/B3 standard steam-table / transport-property values. Engineering choices are flagged in Check callouts.
Question A1: Pump Power for a Crude-Oil Transfer Line (25 marks)
Find. The brake power the pump must deliver to move the oil at 4000 L/min.
Figure A1 — Flow path. The mechanical-energy balance from the free surface of the feed tank to that of the discharge tank must supply the 10 m static lift plus all friction (straight pipe + fittings + entrance/exit).
Approach. Write the steady mechanical-energy (Bernoulli-with-friction) balance between the two free surfaces; the pump head equals the static lift plus the total friction head, which we build from a Fanning friction factor (Colebrook) over an equivalent length that folds in the valve and bends, plus the stated 1.5 velocity heads of entrance/exit loss.
Mean velocity in the pipe. $$v=\frac{Q}{\tfrac{\pi}{4}D^2}=\frac{0.0667}{\tfrac{\pi}{4}(0.2027)^2}=\frac{0.0667}{0.03227}=2.066\ \mathrm{m/s}.$$
Reynolds number and flow regime. $$Re=\frac{\rho v D}{\mu}=\frac{(880)(2.066)(0.2027)}{0.085}=4335.$$ Just into the turbulent range, so a friction chart / Colebrook applies (not the laminar $16/Re$).
Fanning friction factor. With $\varepsilon/D=0.046/202.7=2.3\times10^{-4}$, the Colebrook equation gives $$f=0.0098.$$ (A Fanning chart read at $Re=4335$ on the $\varepsilon/D\approx0.0002$ curve returns the same value.)
Equivalent length of the fittings. The valve and bends are converted to added pipe length: $$L_\text{fit}=\big(340+2\times30\big)D=(400)(0.2027)=81.1\ \mathrm{m}.$$ Total equivalent length $L=195+81.1=276.1\ \mathrm{m}$.
Straight-and-fitting friction head. Using the Fanning form $h_f=4f\dfrac{L}{D}\dfrac{v^2}{2g}$ with one velocity head $v^2/2g=0.2175$ m: $$h_f=4(0.0098)\frac{276.1}{0.2027}(0.2175)=11.6\ \mathrm{m}.$$
Entrance and exit loss (given as 1.5 velocity heads). $$h_\text{ee}=1.5\frac{v^2}{2g}=1.5(0.2175)=0.33\ \mathrm{m}.$$ Prose connector: adding the 10 m static lift closes the head balance.
Total pump head. $$H=\Delta z+h_f+h_\text{ee}=10+11.6+0.33=\;\boxed{21.9\ \mathrm{m}}.$$
Fluid power and brake power. Fluid power $=\rho g Q H=(880)(9.81)(0.0667)(21.9)=12.63\ \mathrm{kW}$; dividing by the 70% efficiency, $$P=\frac{12.63}{0.70}=\;\boxed{18.0\ \mathrm{kW}}.$$
