23-Chem-A2 Unit Operations and Separation Processes · May 2018
Question 2 of 6: Pump Power for Pipe Flow with Fittings
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
Paper format. 16-CHEM-A2, Unit Operations and Separation Processes — May 2018, 3-hour open-book exam. Six 25-point problems: Part A (Unit Operations) — A1 filter-cake washing, A2 pipe flow & pump power, A3 free-settling of a particle; Part B (Separation Processes) — B1 packed-tower flooding/diameter, B2 adsorption isotherms, B3 multistage liquid–liquid extraction. The rubric asks for two problems per part; all six are worked in full below.
Reference texts (subject). McCabe, Smith & Harriott, Unit Operations of Chemical Engineering (7th ed.); Geankoplis, Transport Processes and Separation Process Principles (4th ed.); Treybal, Mass-Transfer Operations (3rd ed.); Perry's Chemical Engineers' Handbook (§6 fluid & particle mechanics, §14 packed towers, §15 extraction, §16 adsorption).
Question A2: Pump Power for Pipe Flow with Fittings (25 points)
Find. The shaft power the pump must draw (HP and kW).
Figure A2 — Mechanical-energy balance runs surface-to-surface; the 200-ft static lift and all pipe/fitting friction are supplied by the pump. The 10-ft inlet submergence affects only suction/NPSH.
Approach. Size the velocity from $Q$ and $D$, get the Fanning friction factor from Reynolds number and relative roughness (Colebrook), then apply the engineering-Bernoulli balance between the two quiescent free surfaces and divide the fluid power by the efficiency.
Velocity and Reynolds number. $A=\tfrac{\pi}{4}D^2=0.08727\ \text{ft}^2$, so $u=Q/A=1.359/0.08727=15.57\ \text{ft/s}$ and $$Re=\frac{\rho uD}{\mu}=\frac{62.37(15.57)(0.3333)}{7.587\times10^{-4}}=\boxed{4.27\times10^{5}}\quad(\text{fully turbulent}),\qquad \tfrac{\epsilon}{D}=4.5\times10^{-4}.$$
Fanning friction factor. The Colebrook correlation (equivalently the supplied Fanning chart) at this $Re$ and roughness gives $$\tfrac{1}{\sqrt{f}}=-4\log_{10}\!\Big(\tfrac{\epsilon/D}{3.7}+\tfrac{1.256}{Re\sqrt{f}}\Big)\ \Rightarrow\ f=4.38\times10^{-3}.$$
Mechanical-energy balance, surface to surface. Both surfaces are at atmospheric pressure with velocity $\approx0$, and the reservoir sits 200 ft below the tank, so with $\sum K_f=2(0.2)+5(0.9)=4.9$: $$W_p=\frac{g}{g_c}\Delta z+\Big(4f\tfrac{L}{D}+\sum K_f\Big)\frac{u^2}{2g_c}.$$ The velocity head is $u^2/2g_c=15.57^2/(2\cdot32.174)=3.769\ \text{ft}\cdot\text{lbf/lbm}$ and $4f(L/D)=4(0.00438)(525/0.3333)=27.6$.
Pump head. $$W_p=200+(27.6+4.9)(3.769)=200+122.4=\boxed{322.4\ \text{ft}\cdot\text{lbf/lbm}}.$$ The fittings alone ($\sum K_f=4.9$) add only $\approx18$ ft·lbf/lbm — small next to the straight-pipe friction at $\approx15.6$ ft/s. The supplied $K_f$ table lists no entrance or exit loss, so only the named fittings are counted. Adding a sharp entrance (0.5) and the exit (1.0) would raise $W_p$ by $1.5(3.769)=5.7$ to $328$ ft·lbf/lbm and the power to 84.3 HP, under 2% more.
Shaft power. With $\dot m=\rho Q=62.37(1.359)=84.77\ \text{lbm/s}$, the fluid power is $\dot m W_p=27{,}325\ \text{ft}\cdot\text{lbf/s}$; dividing by $\eta=0.60$: $$P_\text{shaft}=\frac{\dot m W_p}{\eta}=\frac{27{,}325}{0.60}=45{,}540\ \text{ft}\cdot\text{lbf/s}=\boxed{82.8\ \text{HP}\ (61.7\ \text{kW})}.$$
