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23-Chem-B1 Transport Phenomena · December 2014

Question 4 of 6: B2 — Length of a blood-cooling coil

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

Notes on this paper

Paper format: National Exams, December 2014 — 04-CHEM-B1 Transport Phenomena; 3 hours, open book. Six problems in three sections (A Fluid Mechanics, B Heat Transfer, C Mass Transfer); candidates attempt one from each section plus a fourth from any section (four of six marked, 25 marks each). The worked solutions below cover all six problems. A summary of the conservation equations (continuity, Navier–Stokes, energy, species) is provided as Appendix A in the paper and is quoted throughout rather than re-derived.

Reference texts: R. S. Brodkey & H. C. Hershey, Transport Phenomena — A Unified Approach (McGraw-Hill) — the paper’s own reference and source of the appended conservation-equation tables; R. B. Bird, W. E. Stewart & E. N. Lightfoot, Transport Phenomena (2nd ed., Wiley) — 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) — annular flow, variable-conductivity conduction and diffusion; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — internal-flow convection ($Nu=3.66$) and transient conduction/diffusion; C. J. Geankoplis, Transport Processes and Separation Process Principles (Prentice Hall) — Stefan-tube / evaporating-drop mass transfer.

Question 4: B2 — Length of a blood-cooling coil

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. Blood cooled 40 → 30 °C in a tube of $d=2.5\ \text{mm}$ immersed in a $0\ \text{°C}$ ice bath ($h_o=500\ \text{W/m}^2\text{K}$ — printed as “W/m2”, but a film coefficient carries W/m$^2\cdot$K; wall resistance neglected).

SymbolValue
Volume flow $Q$$6\ \text{L/hr}=1.667\times10^{-6}\ \text{m}^3/\text{s}$
Inside diameter $d$$2.5\times10^{-3}\ \text{m}$
$\rho,\ c_p,\ k$$1000\ \text{kg/m}^3,\ 4000\ \text{J/kg}\cdot\text{K},\ 0.5\ \text{W/m}\cdot\text{K}$
$\nu$; outside coeff. $h_o$$7\times10^{-7}\ \text{m}^2/\text{s}$; $500\ \text{W/m}^2\text{K}$

Find. The coil length $L$ that achieves the 10 °C drop.

ice bath, 0 °C blood in 40 °C out 30 °C 6 L/hr, d=2.5 mm; heat leaves through the wall to the bath (ho=500)
Fig. B2: Blood flows laminarly ($Re\approx1213$) through a coil in a $0\ \text{°C}$ bath. With a constant-temperature sink the fully developed Nusselt number is $Nu=3.66$, fixing the inside coefficient and hence the length needed for the 10 °C drop.

Approach. Get the velocity and Reynolds number to confirm laminar flow; take $Nu=3.66$ (fully developed, constant wall temperature) for the inside coefficient; combine with $h_o$ into $U$; then size the area from the duty and the log-mean temperature difference.

  1. Velocity and regime. $v=\dfrac{Q}{\pi d^2/4}=\dfrac{1.667\times10^{-6}}{4.909\times10^{-6}}=0.340\ \text{m/s}$; $Re=\dfrac{vd}{\nu}=\dfrac{(0.340)(0.0025)}{7\times10^{-7}}=1213$ — laminar.
  2. Inside coefficient. For laminar, fully developed flow with a constant wall temperature, $Nu=3.66$, so $$h_i=\frac{Nu\,k}{d}=\frac{3.66(0.5)}{0.0025}=732\ \text{W/m}^2\text{K}.$$
  3. Overall coefficient. With negligible wall resistance and $h_o=500$, $$U=\left(\frac{1}{h_i}+\frac{1}{h_o}\right)^{-1}=\left(\frac{1}{732}+\frac{1}{500}\right)^{-1}=297\ \text{W/m}^2\text{K}.$$
  4. Duty. $\dot m=\rho Q=1.667\times10^{-3}\ \text{kg/s}$, so $q=\dot m\,c_p\,\Delta T=(1.667\times10^{-3})(4000)(10)=66.7\ \text{W}.$
  5. Driving force. Bath at $0\ \text{°C}$: $\Delta T_1=40,\ \Delta T_2=30$, so $$\Delta T_{lm}=\frac{40-30}{\ln(40/30)}=34.8\ \text{K}.$$
  6. Required length. $A=\dfrac{q}{U\,\Delta T_{lm}}=\dfrac{66.7}{(297)(34.8)}=6.46\times10^{-3}\ \text{m}^2$, and $A=\pi d L$ gives $$L=\frac{A}{\pi d}=\frac{6.46\times10^{-3}}{\pi(0.0025)}=\boxed{0.82\ \text{m}} .$$
Check — developing-flow caveat

The thermal entry length is $L_t\approx0.05\,Re\,Pr\,d$ with $Pr=\mu c_p/k=\nu\rho c_p/k=5.6$, giving $L_t\approx0.05(1213)(5.6)(0.0025)\approx0.85\ \text{m}$ — comparable to the coil itself. Flow is therefore thermally developing, where the local Nusselt number exceeds 3.66; using $Nu=3.66$ is conservative and slightly over-sizes the length. Curvature of the coil enhances $h_i$ further, reinforcing the margin.

QuantityValue
Velocity / Reynolds$0.340\ \text{m/s}$ / $Re=1213$ (laminar)
$h_i$ ($Nu=3.66$)$732\ \text{W/m}^2\text{K}$
Overall $U$$297\ \text{W/m}^2\text{K}$
Duty / $\Delta T_{lm}$$66.7\ \text{W}$ / $34.8\ \text{K}$
Required length $L$$0.82\ \text{m}$