23-Chem-A2 Unit Operations and Separation Processes · December 2019
Question 1 of 6: Pump Power for Liquid Transfer Between Tanks
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 — December 2019, 3-hour open-book exam. Six 25-point problems: Part A (Unit Operations) — A1 pump-power from a full mechanical-energy balance, A2 terminal velocity & matched fluidized-bed densities, A3 constant-pressure plate-and-frame filtration; Part B (Separation Processes) — B1 countercurrent rotary-drier air rate, B2 Swenson–Walker crystallizer sizing, B3 double-effect forward-feed evaporator capacity. 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.); Coulson & Richardson, Chemical Engineering Vol. 2 (5th 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, §11 heat transfer, §12 drying, §18 solid–liquid separation).
Question A1: Pump Power for Liquid Transfer Between Tanks (25 points)
2 × 90° standard elbows, 1 angle valve (fully open)
Static lift
3 m (supply level) → 12 m (discharge) ⇒ $\Delta z=9$ m
Pump efficiency $\eta$
0.60
Find. The shaft power (W) the pump must deliver to move 1 kg/s against the 9 m lift plus all frictional losses.
Figure A1 — Mechanical-energy balance from the supply free surface to the discharge. Both surfaces are at atmospheric pressure, so only the static lift and the pipe/fitting friction (and the exit kinetic loss) load the pump.
Approach. Write the steady-flow mechanical-energy (Bernoulli-with-friction) balance between the two open surfaces; the pump work equals the static lift plus the total friction head, and the shaft power is that work divided by the efficiency.
Velocity in the pipe. With $A=\tfrac{\pi}{4}D^2=\tfrac{\pi}{4}(0.02291)^2=4.123\times10^{-4}\ \text{m}^2$ and volumetric flow $Q=\dot m/\rho=1/997.1=1.003\times10^{-3}\ \text{m}^3/\text{s}$, $$u=\frac{Q}{A}=\frac{1.003\times10^{-3}}{4.123\times10^{-4}}=\boxed{2.433\ \text{m/s}}.$$
Reynolds number and friction factor. $$Re=\frac{\rho u D}{\mu}=\frac{997.1(2.433)(0.02291)}{2.1\times10^{-3}}=26{,}465\ (\text{turbulent}).$$ For a smooth pipe the Blasius correlation gives the Fanning factor $f=0.079\,Re^{-0.25}=0.079(26{,}465)^{-0.25}=\boxed{0.00619}$.
Total friction loss. Combining straight-pipe friction ($4fL/D$) with the fitting loss coefficients (entrance 0.5, two elbows 2×0.75, angle valve 2.0, exit 1.0 ⇒ $\sum K=5.0$), $$h_f=\left(\frac{4fL}{D}+\sum K\right)\frac{u^2}{2}=\left(\frac{4(0.00619)(30)}{0.02291}+5.0\right)\frac{2.433^2}{2}=\boxed{110.8\ \text{J/kg}}.$$ The straight-pipe term ($4fL/D=32.4$) dominates the fittings ($\sum K=5.0$).
Static-lift work. Raising the liquid 9 m costs $g\,\Delta z=9.81(12-3)=88.3\ \text{J/kg}$. Since both tanks are open ($\Delta P=0$) and the supply surface is essentially stationary, the specific pump work is $$W_p=g\,\Delta z+h_f=88.3+110.8=\boxed{199.1\ \text{J/kg}}.$$
Shaft power. Multiplying by the mass rate and dividing by efficiency, $$\dot W_{\text{shaft}}=\frac{\dot m\,W_p}{\eta}=\frac{1(199.1)}{0.60}=331.8\ \text{W}\approx\boxed{332\ \text{W}}.$$
Check — fitting loss coefficients
The problem does not tabulate $K$ values, so standard Crane/Perry figures are assumed: entrance 0.5, 90° standard elbow 0.75 each, fully-open angle valve 2.0, exit into the tank 1.0. These fittings add only $\sum K\,u^2/2\approx15$ J/kg (about 7% of the total), so the pump duty is dominated by the straight-pipe friction and lift; a different but reasonable set of $K$ values shifts the power by only a few watts.
