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23-Chem-A2 Unit Operations and Separation Processes · December 2018

Question 2 of 6: Frictional Pressure Drop in Series Pipes with a Reducer

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 2018, 3-hour open-book exam. Six 25-point problems: Part A (Unit Operations) — A1 continuous-thickener sizing, A2 series-pipe friction pressure drop, A3 packed/fluidized-bed pressure drop & power; Part B (Separation Processes) — B1 flash vs. differential distillation, B2 batch-drying time, B3 adsorption isotherm fitting & surface area. 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, §12 drying, §16 adsorption, §18 solid–liquid separation).

Question A2: Frictional Pressure Drop in Series Pipes with a Reducer (25 points)

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.

QuantityValue
Mass flow $\dot m$12,000 kg/hr $=3.333$ kg/s
Pipe 1 (ID / length)50 mm / 100 m (smooth)
Pipe 2 (ID / length)38 mm / 60 m (smooth)
Reducer loss$0.2$ velocity heads (downstream $u$)
Viscosity $\mu$ / density $\rho$$3\times10^{-3}$ N·s/m² / 850 kg/m³

Find. The total frictional pressure drop $\Delta P$ across the horizontal series system (both straight runs plus the reducer).

Two smooth pipes in series with a 50→38 mm reducer D₁ = 50 mm, L = 100 m reducer 0.2 vel. heads D₂ = 38 mm, L = 60 m flow ṁ = 3.333 kg/s → u₁ = 2.00 m/s, u₂ = 3.46 m/s
Figure A2 — Same mass flow in both runs; the velocity rises in the narrower pipe. Total loss = friction in pipe 1 + reducer minor loss + friction in pipe 2. Horizontal, so no elevation term.

Approach. Get the constant volumetric flow, then in each pipe compute velocity, Reynolds number, and a smooth-pipe (Blasius) Fanning factor; sum the two straight-pipe losses and the reducer velocity-head loss.

  1. Volumetric flow (constant through the series). $Q=\dot m/\rho=3.333/850=\boxed{3.92\times10^{-3}\ \text{m}^3/\text{s}}$.
  2. Pipe 1 — velocity and Reynolds number. $A_1=\tfrac{\pi}{4}(0.050)^2=1.963\times10^{-3}\ \text{m}^2$, so $u_1=Q/A_1=2.00\ \text{m/s}$ and $$Re_1=\frac{\rho u_1 D_1}{\mu}=\frac{850(2.00)(0.050)}{3\times10^{-3}}=2.83\times10^{4}\ (\text{turbulent}).$$ Smooth-pipe Blasius Fanning factor: $f_1=0.079\,Re_1^{-1/4}=6.09\times10^{-3}$.
  3. Pipe 1 pressure drop. $$\Delta P_1=4f_1\frac{L_1}{D_1}\cdot\frac{\rho u_1^2}{2}=4(0.00609)\frac{100}{0.050}\cdot\frac{850(2.00)^2}{2}=\boxed{82.6\ \text{kPa}}.$$
  4. Pipe 2 — velocity, Reynolds, factor. $A_2=\tfrac{\pi}{4}(0.038)^2=1.134\times10^{-3}\ \text{m}^2$, so $u_2=3.46\ \text{m/s}$, $$Re_2=\frac{850(3.46)(0.038)}{3\times10^{-3}}=3.72\times10^{4},\qquad f_2=0.079\,Re_2^{-1/4}=5.69\times10^{-3}.$$ Then $\Delta P_2=4f_2\frac{L_2}{D_2}\frac{\rho u_2^2}{2}=4(0.00569)\frac{60}{0.038}\cdot\frac{850(3.46)^2}{2}=\boxed{182.5\ \text{kPa}}.$
  5. Reducer minor loss. Charged on the higher downstream velocity head: $$\Delta P_r=0.2\cdot\frac{\rho u_2^2}{2}=0.2\cdot\frac{850(3.46)^2}{2}=\boxed{1.0\ \text{kPa}}.$$
  6. Total frictional pressure drop. $$\Delta P=\Delta P_1+\Delta P_2+\Delta P_r=82.6+182.5+1.0=\boxed{266\ \text{kPa}}.$$ The 38-mm run dominates: reducing the bore from 50 to 38 mm roughly triples the velocity head, since $(50/38)^4\approx3.0$, and raises $L/D$, so the shorter narrow pipe carries about two-thirds of the total loss.
QuantityPipe 1 (50 mm)Pipe 2 (38 mm)
Velocity $u$2.00 m/s3.46 m/s
Reynolds number$2.83\times10^{4}$$3.72\times10^{4}$
Fanning $f$ (Blasius)$6.09\times10^{-3}$$5.69\times10^{-3}$
$\Delta P$82.6 kPa182.5 kPa
Reducer loss1.0 kPa
Total $\Delta P$266 kPa