Air at a temperature of 20^{\circ} C flows through a 19mm ID tube of copper which may be assumed to be smooth. The mass flow rate through the tube is 0.03 kg/s. Is the flow laminar or turbulent? Calculate the pressure loss across a meter length of tube if the mean pressure is 2 bar.
The properties of air are taken at p=2 \text { bar }=2 \times 10^{5} Pa, T=20^{\circ} C = 290 K. Assuming air to behave as an ideal gas with R = 287 J/kg K, we have
\rho=\frac{p}{R T}=\frac{2 \times 10^{5}}{287 \times 290}=2.791 kg / m ^{3}From air tables (pressure equal to one atmosphere and temperature of 20° C),we have the dynamic viscosity of air as \mu=18.09 \times 10^{-6} kg / ms. The kinematic viscosity of air is then obtained as
\nu=\frac{\mu}{\rho}=\frac{18.09 \times 10^{-6}}{2.791}=6.482 \times 10^{-6} m ^{2} / sThe diameter of the tube is given as D = 19 mm = 0.019 m. The mass flow of air is specified as \dot{m}=0.03 kg / s. Hence, the air velocity may be calculated as
U=\frac{\dot{m}}{\rho \pi \frac{D^{2}}{4}}=\frac{0.03}{2.791 \times \pi \times \frac{0.019^{2}}{4}}=37.91 m / sThe tube Reynolds number is then calculated as
R e_{D}=\frac{U D}{\nu}=\frac{37.91 \times 0.019}{6.482 \times 10^{-6}}=111124The flow is turbulent. Hence, the friction factor may be calculated using Eq. 14.24(b) as
f=\frac{0.184}{R e_{D}^{0.2}} \text { for } R e_{D}>2 \times 10^{4} (14.24)b
f=\frac{0.184}{111124^{0.2}}=0.018The reader may verify that this value of friction factor agrees with that read off Moody chart. The pressure drop over a meter length of tube is then calculated as
-\frac{\Delta p}{L}=\frac{f \rho U^{2}}{2 D}=\frac{0.018 \times 2.791 \times 37.91^{2}}{2 \times 0.019}=1900 Pa / m