Question 12.1: Air flows through a smooth circular duct of internal diamete......
Air flows through a smooth circular duct of internal diameter 250 mm at an average velocity of 15 m/s. Calculate the fluid velocity at points 50 mm and 5 mm from the wall. What will be the thickness of the laminar sub-layer if this extends to u^{+} = y^{+} = 5? The density and viscosity of air may be taken as 1.10 kg/m³ and 20 x 10^{-6} N s/m² respectively.
Reynolds number: Re = {\frac{(0.250\times15\times1.10)}{(20\times10^{-6})}}=2.06\times10^{5}
Hence, from Figure 3.7: {\frac{R}{\rho u^{2}}}=0.0018
u_{s}={\frac{u}{0.817}}=\left({\frac{15}{0.817}}\right)=18.4~{\mathrm{m/s}}u^{*}=u{\sqrt{\frac{R}{\rho u^{2}}}}=15{\sqrt{0.0018}}=0.636~\mathrm{m/s}
At 50 mm from the wall: {\frac{y}{r}}=\left({\frac{0.050}{0.125}}\right)=0.40
Hence, from equation 12.34:
\frac{u_{s}\ -u_{x}}{u^{*}}=2.5\ln\frac{r}{y} (12.34)
u_{x}=u_{s}+2.5u^{*}\ln\left({\frac{y}{r}}\right)= 18.4 + (2.5 x 0.636 ln 0.4)
= \underline{\underline{16.9\ m/s}}
At 5 mm from the wall: y/r = 0.005/0.125 = 0.04
Hence: u_{x} = 18.4 + 2.5 x 0.636 ln 0.04
= \underline{\underline{13.3\ m/s}}
The thickness of the laminar sub-layer is given by equation 12.54:
\frac{\delta_{b}}{d}=5R e^{-1}\phi^{-1/2} (12.54)
\delta_{b}=\frac{5d}{R e\sqrt{(R/\rho u^{2})}}= {\frac{(5\times0.250)}{(2.06\times10^{5}\sqrt{(0.0018)})}}
= 1.43 x 10^{-4} m
or: \underline{\underline{0.143\ mm}}
