Question 9 of 13: Pressure Drop in a Corrugated Aluminum Air Duct

Question 9: Pressure Drop in a Corrugated Aluminum Air Duct (5 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.

Find. Pressure drop per 100 m of straight pipe.

Approach. Compute the mean velocity and Reynolds number, read (or solve via Colebrook) the friction factor at the given relative roughness, then apply the Darcy–Weisbach head-loss formula.

1. Mean velocity and Reynolds number. $$V = \frac{Q}{A} = \frac{0.08}{\frac{\pi}{4}(0.15)^2} = 4.53\ \text{m/s}, \qquad Re = \frac{\rho VD}{\mu} = \frac{1.19(4.53)(0.15)}{1.8\times10^{-5}} = 4.49\times10^4$$
2. Relative roughness and friction factor. $$\frac{e}{D} = \frac{0.003}{0.150} = 0.02$$ Reading the Moody chart at Re = 4.5×10⁴ along the e/D=0.02 curve (confirmed by the Colebrook equation): $$f \approx \boxed{0.0495}$$
3. Darcy–Weisbach head loss per 100 m, converted to a pressure drop. $$\Delta p = f\left(\frac{L}{D}\right)\left(\frac{1}{2}\rho V^2\right) = 0.0495\left(\frac{100}{0.15}\right)\left(\frac{1}{2}(1.19)(4.53)^2\right)$$ $$\Delta p = 0.0495(666.7)(12.19) = \boxed{402\ \text{Pa} = 0.402\ \text{kPa per 100 m}}$$