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23-Chem-B1 Transport Phenomena · May 2016

Question 1 of 6: A1 — Gravity-fed IV bag flow rate and emptying time

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

Notes on this paper

Paper format. EGBC 04-CHEM-B1 Transport Phenomena, May 2016, 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 (four of six at 25 marks each). 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 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, internal-flow heat transfer and boundary-layer mass transfer; F. P. Incropera & D. P. DeWitt, Fundamentals of Heat and Mass Transfer (Wiley) — the Dittus–Boelter correlation and constant-heat-flux internal flow; R. H. Perry & D. W. Green, Perry’s Chemical Engineers’ Handbook (9th ed.) — transport properties; T. B. Reddy & standard metallurgical mass-transfer literature (Eisenberg, Tobias & Wilke, J. Electrochem. Soc. 1954) — the rotating-cylinder correlation.

Question 1: A1 — Gravity-fed IV bag flow rate and emptying time (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
Bag volume$V_{bag}$500 mL $=5.0\times10^{-4}\ \text{m}^3$
Tube internal diameter (18-gage)$D$$0.953\ \text{mm}=9.53\times10^{-4}\ \text{m}$
Tube length$L$2 m
Tube entrance height above vein$z_e$1.0 m
Bag (fluid column) length—0.30 m
Venous gage pressure$p_{vein}$0 (gage)
Fluid (aqueous $\approx$ water)$\rho,\ \mu$$1000\ \text{kg/m}^3,\ 1.0\times10^{-3}\ \text{Pa}\!\cdot\!\text{s}$

Find. (i) the volumetric flow rate $Q$ of IV fluid, and (ii) the time to empty the 500 mL bag.

500 mL bag column 0.30 m tube L = 2 m, D = 0.953 mm vein (0 mm Hg gage) H ≈ 1.0–1.3 m
Fig. A1: Gravity feed from the IV bag. The driving head $H$ is the height of the fluid surface above the vein; it falls from 1.30 m (full bag) to 1.00 m (empty), so a representative $H\approx1.15$ m is used. The narrow bore makes viscous (Hagen–Poiseuille) friction dominate.

Approach. Write a mechanical-energy (Bernoulli-with-friction) balance from the fluid surface to the vein, recognise that the tiny bore makes laminar wall friction dominate the loss, and solve for velocity, flow rate and emptying time; check the Reynolds number to confirm laminar flow.

  1. Energy balance, surface → vein. With both ends at gage pressure zero (open surface; venous $p=0$) and negligible surface velocity, the mechanical-energy balance is $gH=\tfrac12 v^2+h_f$, where $H$ is the surface height above the vein and $h_f$ the friction loss in the tube.
  2. Friction is laminar Hagen–Poiseuille. For fully-developed laminar tube flow the loss is $h_f=\dfrac{32\,\mu L v}{\rho D^2}$. Anticipating a small velocity, the kinetic term $v^2/2$ is negligible next to $gH$ (verified below), so $gH\simeq\dfrac{32\,\mu L v}{\rho D^2}.$
  3. Solve for the velocity. $$v=\frac{\rho g H D^2}{32\,\mu L}=\frac{1000(9.81)(1.15)(9.53\times10^{-4})^2}{32(1.0\times10^{-3})(2.0)}\;\Longrightarrow\;\boxed{v\approx0.160\ \text{m/s}}$$ using the representative head $H=1.15$ m.
  4. Confirm laminar & negligible kinetic term. $Re=\dfrac{\rho v D}{\mu}=\dfrac{1000(0.160)(9.53\times10^{-4})}{1.0\times10^{-3}}\approx152\ (<2100$, laminar ✓$)$. The kinetic head $v^2/2=0.0128\ \text{J/kg}$ is $\sim10^{-3}$ of $gH=11.3\ \text{J/kg}$ — negligible, as assumed.
  5. Flow rate. $Q=vA=v\,\dfrac{\pi D^2}{4}=0.160\cdot\dfrac{\pi(9.53\times10^{-4})^2}{4}=1.14\times10^{-7}\ \text{m}^3/\text{s}.$ That is $$\boxed{Q\approx0.114\ \text{mL/s}=6.85\ \text{mL/min}}.$$
  6. Emptying time. Treating $Q$ at the representative head as roughly constant, $$t=\frac{V_{bag}}{Q}=\frac{5.0\times10^{-4}}{1.14\times10^{-7}}\approx4.4\times10^{3}\ \text{s}\;\Longrightarrow\;\boxed{t\approx73\ \text{min}}.$$ (The head falls only 1.30→1.00 m over the drain, so $Q$ varies by $\pm13\%$; the mid-head estimate is well within clinical tolerance.)
QuantityResult
Fluid velocity in tube$v\approx0.160\ \text{m/s}$
Reynolds number$Re\approx152$ (laminar)
Volumetric flow rate$Q\approx6.85\ \text{mL/min}\ (0.114\ \text{mL/s})$
Time to empty 500 mL bag$t\approx73\ \text{min}$
Check — the driving head

The problem gives the tube entrance at 1.0 m and a 0.30 m bag; the fluid surface therefore sits between 1.00 m (empty) and 1.30 m (full) above the vein. A mid-value $H=1.15$ m is used for a single representative flow. Taking $H=1.0$ m (empty-bag, conservative) gives $Q\approx5.95$ mL/min and $t\approx84$ min; either bracket is acceptable if the assumption is stated.

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