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a) first calculate Reynold number (Re)R e=\frac{d v \rho}{\mu} \text { where } \begin{aligned}d & =0.015 \mathrm{~m} \\v & =0.3 \mathrm{~m} / \mathrm{s} \\\rho & =1,240 \mathrm{~g} / \mathrm{m}^{3}\end{aligned}\mu of glycerol at 60^{\circ} \mathrm{C} is 0.082226 \mathrm{~Pa}-\mathrm{s}\text { So } \begin{aligned}R e & =\frac{0.015(\mathrm{~m}) \times 0.3(\mathrm{~m} / \mathrm{s}) \times 1240\left(\mathrm{~kg} / \mathrm{m}^{3}\right)}{0.082226(\mathrm{Pa-s})} \\R e & =67.8617\end{aligned}\operatorname{Re}<2300 that mean laminar JlowSo transition length, L_{H}=0.05 \operatorname{Re}(d)\begin{array}{l}L_{H}=0.05 \times 67.8617 \times 0.015 \\L_{H}=0.0509 \mathrm{~m}\end{array}b) find Re, \quad R e=\frac{d v \rho}{\mu}\begin{array}{l}d=3 \mathrm{in}=0.0762 \mathrm{~m}, \quad v=7 \mathrm{ft} / \mathrm{s}=2.1336 \mathrm{~m} / \mathrm{s} \\\rho=50 \mathrm{~b} / \mathrm{ff}^{3}=800.923 \mathrm{~kg} / \mathrm{m}^{3}\end{array}\mu of n-propyl alcohol at 30^{\circ} \mathrm{C}=1.8 \times 10^{-3} \mathrm{~Pa}-\mathrm{s}\text { So } \begin{aligned}R e & =\frac{0.0762(\mathrm{~m}) \times 2.1336(\mathrm{~m} / \mathrm{s}) \times 800.923\left(\mathrm{~kg} / \mathrm{m}^{3}\right)}{1.8 \times 10^{-3}(\mathrm{~Pa}-\mathrm{s})} \\R e & =72,341.2878\end{aligned}R e .72900, so that mean turbulent flow for turbulent flow, transition length, H_{1}=10 D\begin{array}{l}L_{H}=10 \mathrm{~d}, \quad L_{H}=10 \times 0.0762=0.762 \mathrm{~m} \\L_{H}=0.762 \mathrm{~m} \text { or } L_{H}=30 \mathrm{in}\end{array} ...