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23-Chem-B1 Transport Phenomena · May 2015

Question 2 of 6: A2 — Pressure drop, benzene vs. kerosene

Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)

Notes on this paper

Paper format: National Exams, May 2015 — 04-CHEM-B1 Transport Phenomena; 3 hours, open book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer), each worth 25 marks; candidates attempt one problem from each section plus a fourth from any section (four of six marked). The worked solutions below cover all six problems. A summary of the conservation equations (continuity, Navier–Stokes, energy, species) is supplied as Appendix A in the paper and is used throughout.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change and the differential momentum / energy / species balances used across every problem; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe-friction and boundary-layer mass-transfer correlations; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — conduction with variable conductivity, exchanger analysis; D. Q. Kern, Process Heat Transfer (McGraw-Hill) — the Kern casing-side method and the LMTD–correction-factor chart for multiple casing passes; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — physical properties and the Moody chart.

Question 2: A2 — Pressure drop, benzene vs. kerosene (25 marks)

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.

QuantityBenzeneKerosene
Density $\rho$$899\ \text{kg/m}^3$$820\ \text{kg/m}^3$
Viscosity $\mu$$8\times10^{-4}\ \text{Pa}\cdot\text{s}$$2.5\times10^{-3}\ \text{Pa}\cdot\text{s}$
Flow rate $Q$$15\ \text{L/min}=2.5\times10^{-4}\ \text{m}^3/\text{s}$
Pipe $D$, $L$$0.055\ \text{m}$, $150\ \text{m}$
Roughness $\varepsilon$$8.5\times10^{-4}\ \text{ft}=2.59\times10^{-4}\ \text{m}$

Find. (a) pressure drop for benzene; (b) pressure drop for kerosene at the same volumetric flow; (c) why they differ.

Approach. Both fluids share the same velocity (same $Q$, same pipe); compute the Reynolds number for each, pick the correct friction-factor law for its flow regime, then apply the Darcy–Weisbach equation $\Delta P=f\,\tfrac{L}{D}\,\tfrac{\rho u^2}{2}$.

  1. Common mean velocity. $u=\dfrac{Q}{\tfrac{\pi}{4}D^2}=\dfrac{2.5\times10^{-4}}{\tfrac{\pi}{4}(0.055)^2}=0.1052\ \text{m/s}.$ This value is identical for both fluids.
  2. Benzene Reynolds number and regime. $Re=\dfrac{\rho u D}{\mu}=\dfrac{(899)(0.1052)(0.055)}{8\times10^{-4}}=6.50\times10^{3}.$ Since $Re>4000$ the flow is turbulent, so the friction factor comes from the Colebrook/Moody relation with relative roughness $\varepsilon/D=2.59\times10^{-4}/0.055=4.71\times10^{-3}.$
  3. Benzene friction factor (Colebrook). Solving $\dfrac{1}{\sqrt f}=-2\log_{10}\!\Big(\dfrac{\varepsilon/D}{3.7}+\dfrac{2.51}{Re\sqrt f}\Big)$ iteratively gives $f=0.0402$ (Darcy), consistent with reading the Moody chart at $Re=6.5\times10^3,\ \varepsilon/D\approx0.0047$.
  4. Benzene pressure drop. $$\Delta P_b=f\frac{L}{D}\frac{\rho u^2}{2}=0.0402\left(\frac{150}{0.055}\right)\frac{(899)(0.1052)^2}{2}=\boxed{545\ \text{Pa}}.$$
  5. Kerosene Reynolds number and regime. $Re=\dfrac{(820)(0.1052)(0.055)}{2.5\times10^{-3}}=1.90\times10^{3}.$ Now $Re<2100$, so the flow is laminar and roughness is irrelevant: $f=\dfrac{64}{Re}=\dfrac{64}{1898}=0.0337.$
  6. Kerosene pressure drop. $$\Delta P_k=f\frac{L}{D}\frac{\rho u^2}{2}=0.0337\left(\frac{150}{0.055}\right)\frac{(820)(0.1052)^2}{2}=\boxed{417\ \text{Pa}}.$$ As a check, the Hagen–Poiseuille form gives the same value: $\Delta P_k=\dfrac{32\mu L u}{D^2}=\dfrac{32(2.5\times10^{-3})(150)(0.1052)}{(0.055)^2}=417\ \text{Pa}.$
  7. Explain the difference (part c). Despite kerosene being roughly three times more viscous, its pressure drop is lower. The higher viscosity pushes the kerosene Reynolds number below the transition value, so it flows laminarly with a smaller friction factor, while benzene is turbulent with a larger, roughness-augmented factor and a higher density. The change of flow regime dominates the direct effect of viscosity.
QuantityValue
Mean velocity $u$$0.105\ \text{m/s}$
Benzene $Re$ / regime / $f$$6.50\times10^3$ / turbulent / $0.0402$
(a) $\Delta P$ benzene$545\ \text{Pa}$
Kerosene $Re$ / regime / $f$$1.90\times10^3$ / laminar / $0.0337$
(b) $\Delta P$ kerosene$417\ \text{Pa}$
(c) ReasonRegime flip: kerosene is laminar → lower $f$ outweighs its higher $\mu$