Question 1 of 4: Laminar Pressure Drop in Small-Bore Tubing
Nivaar worked solution (AI-drafted; not reviewed by a licensed engineer)
Notes on this paper
04-Agric-A4, Fluid Flow — National Exams, December 2019. Open-book, 3-hour exam; four questions of equal value, all requiring calculation.
Reference texts. White, Fluid Mechanics (pipe friction/laminar duct flow, cavitation, open-channel flow, pumps & the energy equation) — the standard undergraduate text for this subject.
Problem 1: Laminar Pressure Drop in Small-Bore Tubing (25 points)
Check: the paper prints $\mu=1.79\times10^{-3}\ \text{N}\cdot\text{s/m}^2$, but that value is inconsistent with the question's own statement that the flow "would normally be turbulent" — at $\mu=1.79\times10^{-3}$ the Reynolds number at these conditions is only $\approx137$ (already laminar even without special precautions). Standard air viscosity at these "standard conditions" ($\rho=1.23\ \text{kg/m}^3$, i.e. $\approx15^{\circ}\text{C}$, 1 atm) is $\mu=1.79\times10^{-5}\ \text{N}\cdot\text{s/m}^2$ (White's air-property table); with that value $\text{Re}\approx13{,}700$, which is indeed normally turbulent, matching the narrative. Treated as a misprinted exponent ($-3$ for $-5$) and solved with the standard $1.79\times10^{-5}$ value.
Given. Air at standard conditions flows through a short length of very smooth, vibration-free drawn tubing, small enough in diameter that special precautions can suppress the usual turbulence.
Quantity
Value
Air density, $\rho$
1.23 kg/m³
Air viscosity, $\mu$
$1.79\times10^{-5}\ \text{N}\cdot\text{s/m}^2$
Tube diameter, $D$
4.0 mm $=0.004$ m
Average velocity, $V$
50 m/s
Section length, $L$
0.1 m
Find. The pressure drop $\Delta p$ over the 0.1-m section, assuming the flow has been kept laminar.
Figure 1 — Air at $V=50$ m/s enters a 4.0-mm drawn tube; the pressure drop is required over a 0.1-m section, assuming disturbances have been suppressed enough to keep the flow laminar.
Approach. First confirm the flow is normally turbulent at these conditions (justifying the problem's premise), then, since we are told the flow has in fact been kept laminar, apply the Hagen–Poiseuille laminar friction factor $f=64/\text{Re}$ in the Darcy–Weisbach relation (equivalently, the direct Hagen–Poiseuille pressure-drop formula) to get $\Delta p$.
Reynolds number under normal (undisturbed) conditions.
$$\text{Re} = \frac{\rho V D}{\mu} = \frac{(1.23)(50)(0.004)}{1.79\times10^{-5}} = \boxed{\text{Re}\approx13{,}740}.$$
Since $\text{Re}\gg4000$, the flow would indeed normally be turbulent, exactly as the problem states — the special precautions are what let laminar flow persist instead.
Laminar friction factor and Darcy–Weisbach pressure drop. With the flow held laminar, $f=64/\text{Re}=64/13{,}740=4.658\times10^{-3}$, so
$$\Delta p = f\,\frac{L}{D}\,\frac{\rho V^2}{2} = (4.658\times10^{-3})\left(\frac{0.1}{0.004}\right)\frac{(1.23)(50)^2}{2} = \boxed{\Delta p\approx179\ \text{Pa}}.$$
Cross-check via Hagen–Poiseuille. The same result follows directly from
$$\Delta p = \frac{32\mu L V}{D^2} = \frac{32(1.79\times10^{-5})(0.1)(50)}{(0.004)^2} = \boxed{\Delta p\approx179\ \text{Pa}},$$
confirming the laminar-$f$ and closed-form routes agree.
