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23-Chem-B1 Transport Phenomena · Undated paper

Question 1 of 6: A1 — Pressure drop for benzene and kerosene in a steel pipe

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

Notes on this paper

Paper. National Examination (Engineers Canada) — 16-Chem-B1 Transport Phenomena, May 2019. Open book, 3 hours. Six 25-point problems in three sections — A Fluid Mechanics (A1, A2), B Heat Transfer (B1, B2), C Mass Transfer (C1, C2); one problem from each section must be attempted, plus a fourth from any section. All six problems are solved and fully worked below.

Reference texts: R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — the equations of change, the power-law slit-flow momentum balance and the film/annular diffusion balances (A2, B2, C2); J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, the Moody chart, and the external-flow convection/mass-transfer correlations (A1, B1, C1); F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (8th ed., Wiley) — Churchill–Chu free convection, horizontal-plate correlations, slug-flow internal convection and air properties (B1, B2); C. J. Geankoplis, Transport Processes and Separation Process Principles (Prentice Hall) — boundary-layer mass transfer and the film model (C1); R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.

Question 1: A1 — Pressure drop for benzene and kerosene in a steel pipe (25 marks: 13 + 10 + 2)

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.

QuantitySymbolValue
Pipe length, inside diameter$L,\ D$$150\ \text{m},\ 0.055\ \text{m}$
Volumetric flow rate$Q$$15\ \text{L/min}=2.50\times10^{-4}\ \text{m}^3/\text{s}$
Equivalent roughness$\varepsilon$$8.5\times10^{-4}\ \text{ft}=2.59\times10^{-4}\ \text{m}$
Benzene $\rho,\ \mu$—$899\ \text{kg/m}^3,\ 8\times10^{-4}\ \text{Pa}\cdot\text{s}$
Kerosene $\rho,\ \mu$—$820\ \text{kg/m}^3,\ 2.5\times10^{-3}\ \text{Pa}\cdot\text{s}$

Find. (a) the frictional pressure drop for benzene; (b) the pressure drop for the same volumetric flow (15 L/min) of kerosene in the same pipe; (c) a physical explanation of why they differ.

Q = 15 L/min D = 5.5 cm steel pipe (ε = 8.5×10⁻⁴ ft) P₁ P₂ L = 150 m (horizontal ⇒ no elevation term) ΔP = P₁ − P₂ = f (L/D)(ρV²/2)
Fig. A1: Fully-developed, horizontal, single-phase flow. With no elevation change and constant area, the mechanical-energy balance reduces to the Darcy–Weisbach friction term, so $\Delta P = f\,(L/D)\,\tfrac12\rho V^2$.

Approach. Convert the flow rate to a mean velocity, form the Reynolds number and relative roughness, obtain the Darcy friction factor (Colebrook, the algebraic form of the turbulent Moody curves, or $64/Re$ if laminar), then apply Darcy–Weisbach; repeat for kerosene at the same flow rate.

  1. Mean velocity. $A=\pi D^2/4=\pi(0.055)^2/4=2.376\times10^{-3}\ \text{m}^2$, so $$V=\frac{Q}{A}=\frac{2.50\times10^{-4}}{2.376\times10^{-3}}=0.1052\ \text{m/s}.$$ With $\varepsilon=8.5\times10^{-4}(0.3048)=2.591\times10^{-4}$ m, the relative roughness is $\varepsilon/D=4.71\times10^{-3}$ and $L/D=2727$.
  2. (a) Reynolds number, benzene. $$Re=\frac{\rho V D}{\mu}=\frac{899(0.1052)(0.055)}{8\times10^{-4}}=6.50\times10^{3},$$ above about 4000, so the flow is turbulent (at the low-$Re$ end of the turbulent curves on the chart).
  3. (a) Friction factor. The Colebrook equation $$\frac{1}{\sqrt f}=-2\log_{10}\!\left(\frac{\varepsilon/D}{3.7}+\frac{2.51}{Re\sqrt f}\right)$$ converges to $f=0.0402$ (Darcy). Reading the supplied Moody chart at $Re\approx6.5\times10^3$ between the $\varepsilon/D=0.004$ and $0.006$ curves gives $f\approx0.040$ — consistent.
  4. (a) Pressure drop. The dynamic pressure is $\rho V^2/2=899(0.1052)^2/2=4.98$ Pa, so $$\boxed{\;\Delta P_{\text{benzene}}=f\frac{L}{D}\frac{\rho V^2}{2}=0.0402\,(2727)\,(4.98)\approx545\ \text{Pa}.\;}$$
  5. (b) Reynolds number, kerosene. “The same amount” through the same pipe is taken as the same volumetric flow, so $V=0.1052$ m/s again: $$Re=\frac{820(0.1052)(0.055)}{2.5\times10^{-3}}=1.90\times10^{3}<2100,$$ so the kerosene flow is laminar.
  6. (b) Friction factor and pressure drop. In laminar flow $f=64/Re=64/1898=0.0337$, independent of wall roughness. Then $$\boxed{\;\Delta P_{\text{kerosene}}=0.0337\,(2727)\,\frac{820(0.1052)^2}{2}\approx417\ \text{Pa},\;}$$ identical to the Hagen–Poiseuille form $\Delta P=32\mu LV/D^2=32(2.5\times10^{-3})(150)(0.1052)/(0.055)^2=417$ Pa.
  7. (c) Why kerosene gives the smaller drop. Kerosene is about three times more viscous, which drops $Re$ from $6.5\times10^3$ to $1.9\times10^3$ and moves the flow from the turbulent regime into the laminar one. Laminar flow has no turbulent eddy (Reynolds-stress) dissipation and does not feel the wall roughness, so its friction factor ($0.034$) is lower than benzene’s turbulent value ($0.040$); combined with kerosene’s lower density ($820$ vs $899$ kg/m³) the pressure drop falls by about 23 %. Had kerosene stayed turbulent, its higher viscosity would have raised the drop instead.
Fluid$Re$$f$ (Darcy)$\Delta P$
Benzene (turbulent)$6.50\times10^3$$0.0402$$\approx545\ \text{Pa}$
Kerosene (laminar)$1.90\times10^3$$0.0337$$\approx417\ \text{Pa}$
Check — interpretation of “same amount” and the near-critical kerosene flow

The Darcy friction factor is used throughout ($\Delta P=f\tfrac{L}{D}\tfrac12\rho V^2$); a Fanning value $f_F=f/4$ needs $\Delta P=4f_F\tfrac{L}{D}\tfrac12\rho V^2$. If “same amount” is read as the same mass flow, $V=0.1154$ m/s, $Re=2.08\times10^3$ (still laminar) and $\Delta P\approx458$ Pa — the conclusion is unchanged. Kerosene’s $Re=1.9\times10^3$ is close to the critical value; if the flow were disturbed into turbulence, Colebrook would give $f=0.054$ and $\Delta P\approx666$ Pa, so the laminar answer assumes an undisturbed, fully-developed line. Minor and entrance losses are neglected.

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