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23-Chem-B1 Transport Phenomena · December 2017

Question 3 of 6: B1 — Tube-side exit temperature by the transfer analogies

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

Notes on this paper

Paper format. EGBC 16-CHEM-B1 Transport Phenomena, December 2017, 3 hours, open-book (one textbook permitted, no loose notes). 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, and only the first four in the answer book are marked. All six problems are solved below as a complete study resource. Appendix A of the paper supplies the equations of change (continuity, Navier–Stokes, energy and species-continuity tables) in rectangular, cylindrical and spherical coordinates; these are quoted rather than re-derived.

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; J. R. Welty, C. E. Wicks, R. E. Wilson & G. L. Rorrer, Fundamentals of Momentum, Heat and Mass Transfer (Wiley) — pipe friction, the transfer analogies, boundary-layer and film coefficients; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — internal-flow convection and the sphere/flat-plate correlations; F. M. White, Fluid Mechanics (McGraw-Hill) — the three-reservoir branching-pipe problem; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook — transport properties.

Question 3: B1 — Tube-side exit temperature by the transfer analogies (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.

QuantitySymbolValue
Inside diameter$D$$1.25\ \text{in}=0.1042\ \text{ft}$
Tube length$L$$9\ \text{ft}$
Volumetric flow$Q$$18\ \text{gpm}=0.0401\ \text{ft}^3/\text{s}$
Inlet / wall temperature$T_{in},T_w$$45\ \text{°F},\ 200\ \text{°F}$
Prandtl number$Pr$$6.78$
Kinematic viscosity$\nu$$1.06\times10^{-5}\ \text{ft}^2/\text{s}$

Find. The water outlet temperature $T_{out}$ predicted by each of the four momentum–heat transfer analogies.

tube wall at T_w = 200 °F (constant) in: 45 °F, 18 gpm out: T_out = ? D = 1.25 in, L = 9 ft, smooth
Fig. B1: Constant-wall-temperature internal heating. Each analogy converts the tube’s friction factor into a Stanton number, which sets the exponential approach of the bulk temperature to the wall value.

Approach. Evaluate the Reynolds number and the smooth-tube Fanning friction factor, form the Stanton number from each analogy (which differ only in how they account for $Pr\neq1$), then apply the constant-wall-temperature energy balance $\tfrac{T_w-T_{out}}{T_w-T_{in}}=e^{-4\,St\,L/D}$ to get each outlet temperature.

  1. Velocity and Reynolds number. $A_c=\tfrac{\pi}{4}D^2=8.52\times10^{-3}\ \text{ft}^2$, so $V=Q/A_c=0.0401/8.52\times10^{-3}=4.71\ \text{ft/s}$ and $$Re=\frac{VD}{\nu}=\frac{(4.71)(0.1042)}{1.06\times10^{-5}}\approx4.62\times10^{4}\quad(\text{turbulent}).$$
  2. Smooth-tube friction factor. From the Fanning chart (or Blasius $f=0.0791\,Re^{-1/4}$ for $Re\lesssim10^5$), $$f=\frac{0.0791}{(4.62\times10^{4})^{1/4}}\ \Longrightarrow\ \boxed{f\approx5.4\times10^{-3},\qquad \tfrac{f}{2}=2.69\times10^{-3}.}$$
  3. Stanton number from each analogy. All four start from $St=f/2$ and correct for the water’s thermal sublayer ($Pr=6.78$):
    • Reynolds ($Pr=1$): $St=\tfrac{f}{2}=2.69\times10^{-3}$.
    • Colburn: $St=\tfrac{f}{2}Pr^{-2/3}=2.69\times10^{-3}/3.58=7.52\times10^{-4}$.
    • Prandtl (two-layer): $St=\dfrac{f/2}{1+5\sqrt{f/2}\,(Pr-1)}=1.08\times10^{-3}$.
    • von Kármán (three-layer): $St=\dfrac{f/2}{1+5\sqrt{f/2}\,[(Pr-1)+\ln(\frac{5Pr+1}{6})]}=9.11\times10^{-4}$.
  4. Energy balance to the exit. For a constant wall temperature, integrating $\dot m c_p\,dT=h(T_w-T)\,\pi D\,dx$ gives $\tfrac{T_w-T_{out}}{T_w-T_{in}}=\exp\!\big(-\tfrac{4L}{D}St\big)$, with $\tfrac{4L}{D}=\tfrac{4(9)}{0.1042}=345.6$. Thus $T_{out}=200-155\,e^{-345.6\,St}$.
  5. Outlet temperatures. Substituting each $St$: $$\boxed{T_{out}^{\text{Rey}}\approx139\ \text{°F},\quad T_{out}^{\text{Prandtl}}\approx93\ \text{°F},\quad T_{out}^{\text{vK}}\approx87\ \text{°F},\quad T_{out}^{\text{Colburn}}\approx80\ \text{°F}.}$$ The pure Reynolds analogy assumes $Pr=1$ and so grossly over-predicts the heat pickup for water; the $Pr$-corrected analogies (Prandtl, von Kármán, Colburn) cluster near $80$–$93\ \text{°F}$ and are the physically reliable estimates.
AnalogyStanton number $St$Exit temperature $T_{out}$
Reynolds ($Pr=1$)$2.69\times10^{-3}$$\approx139\ \text{°F}$ (over-predicts)
Prandtl (two-layer)$1.08\times10^{-3}$$\approx93\ \text{°F}$
von Kármán (three-layer)$9.11\times10^{-4}$$\approx87\ \text{°F}$
Colburn ($j$-factor)$7.52\times10^{-4}$$\approx80\ \text{°F}$
Check — friction-factor read and analogy scope

The exit temperatures scale with the friction factor read from the smooth-tube chart; a $\pm5\%$ change in $f$ moves each $T_{out}$ by a degree or two. The Reynolds analogy is exact only at $Pr=1$ and is included for comparison — for water ($Pr\approx7$) the von Kármán and Colburn results are the trustworthy ones. Entry-length effects are neglected ($L/D\approx86$, so the flow is essentially fully developed).