A tube with a bell-mouth inlet configuration is subjected to uniform wall heat flux of \(3 \mathrm{~kW} / \mathrm{m}^{2}\). The tube has an inside diameter of \(0.0158 \mathrm{~m}(0.622 \mathrm{in})\) and a flow rate of \(1.43 \times\) \(10^{-4} \mathrm{~m}^{3} / \mathrm{s}(2.27 \mathrm{gpm})\). The liquid flowing inside the tube is ethylene glycol-distilled water mixture with a mass fraction of \(2.27\). Determine the fully developed friction coefficient at a location along the tube where the Grashof number is \(\mathrm{Gr}=\) 16,600 . The physical properties of the ethylene glycol-distilled water mixture at the location of interest are \(\operatorname{Pr}=14.85, \nu=\) \(1.93 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\), and \(\mu_{b} / \mu_{s}=1.07\).

Given in the exercise is the Grashof number: \(\mathrm{Gr} =\) 16,600. We will use this value later in the analysis.

We can find the Reynolds number using the flow rate \((Q)\), the kinematic viscosity \((\nu)\), and the diameter of the tube \((D)\), using the relation: $$\textit{Re} = \frac{4Q}{\pi D \nu}.$$ For our problem, we have \(Q = 1.43 \times 10^{-4} \mathrm{~m}^{3} / \mathrm{s}\), \(D = 0.0158 \mathrm{~m}\), and \(\nu = 1.93 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\). Plugging these values, we get: $$\textit{Re} = \frac{4(1.43 \times 10^{-4})}{\pi(0.0158)(1.93 \times 10^{-6})} \approx 1026.$$

Use the given ratio of \(\mu_b/\mu_s = 1.07\) (bulk viscosity to surface viscosity) to calculate the buoyancy adjusted Reynolds number by the following equation: $$\textit{Re}_b = \textit{Re}(\frac{\mu_b}{\mu_s}) = 1026(1.07) \approx 1097.81.$$

Now, with the given Prandtl number of the fluid, \(\operatorname{Pr} = 14.85\), we can use the Grashof number \((\mathrm{Gr})\) and the buoyancy adjusted Reynolds number \((\textit{Re}_b)\) to calculate the fully developed friction factor \((f)\), given by the Petukhov-Kirillov equation: $$f = \frac{0.79 \operatorname{ln}\left(\textit{Re}_b\right) - 1.64}{\textit{Re}_b} \left[1+\textstyle\frac{185}{\textit{Re}_b^{1.5}} \frac{\mathrm{Gr}}{\textit{Re}_b \operatorname{Pr}}\right]^{\frac{1}{0.3}}.$$ Substituting the given values and simplifying, we get: $$f = \frac{0.79 \operatorname{ln}\left(1097.81\right) - 1.64}{1097.81} \left[1+\frac{185}{1097.81^{1.5}} \frac{16600}{1097.81 \times 14.85}\right]^{\frac{1}{0.3}} \approx 0.01910.$$ The fully developed friction coefficient at the location along the tube with Grashof number 16,600 is 0.01910.