Question

A 4-m-long tube is subjected to uniform wall heat flux. The tube has an inside diameter of \(0.0149 \mathrm{~m}\) and a flow rate of $7.8 \times 10^{-5} \mathrm{~m}^{3} / \mathrm{s}$. The liquid flowing inside the tube is an ethylene glycol-distilled water mixture with a mass fraction of \(0.5\). Determine the apparent (developing) friction factor at the location \(x / D=20\) if the inlet configuration of the tube is: \((a)\) re-entrant and \((b)\) square- edged. At this location, the local Grashof number is Gr \(=24,000\), and the properties of the mixture of distilled water and ethylene glycol are: \(\operatorname{Pr}=20.9\), $\nu=2.33 \times 10^{-6} \mathrm{~m}^{2} / \mathrm{s}\(, and \)\mu_{y} / \mu_{s}=1.25$.

Short answer

The apparent friction factors for the given flow conditions and inlet configurations are: a) Re-entrant inlet: 0.02125 b) Square-edged inlet: 0.02325

Step-by-step solution

Calculate average fluid velocity

To calculate the average fluid velocity, use the equation for volumetric flow rate: $$ Q = \frac{\pi D^{2}}{4} V $$ Rearrange to solve for the average fluid velocity, \(V\): $$ V = \frac{4Q}{\pi D^{2}} $$ Plug in the given values for the flow rate, \(Q = 7.8 x 10^{-5} m^3/s\), and the inside diameter of the tube, \(D = 0.0149 m\): $$ V = \frac{4(7.8 \times 10^{-5})}{\pi (0.0149)^{2}} = 2.236 m/s $$ So the average fluid velocity is \(2.236 m/s\).

Calculate the Reynolds number

To calculate the Reynolds number (\(Re\)) for the flow, use the formula: $$ Re = \frac{VD}{\nu} $$ Plug in the values for the average fluid velocity calculated in Step 1, \(V = 2.236 m/s\), the inside diameter of the tube, \(D = 0.0149 m\), and the kinematic viscosity, \(\nu = 2.33 \times 10^{-6} m^2/s\): $$ Re = \frac{(2.236)(0.0149)}{2.33 \times 10^{-6}} = 14335 $$ The Reynolds number for this flow is \(14335\).

Calculate the apparent developing friction factors

Now we determine the developing friction factors for both inlet configurations from their corresponding empirical correlations. For re-entrant inlets - Gnielinski correlation: $$ f_{aR} = \frac{1.14\left[\frac{2.42}{(x/D)^{1/6}}\left(\frac{\mu_{m}}{\mu_{w}}\right)^{0.14}\sqrt{\frac{Gr}{Re}}\right]}{1 + \frac{1.14\sqrt{Gr}}{(x/D)^{1/6}\left[\frac{14.9}{1.07+19.9 Pr^{0.630-2.34 \times 10^{-4} Gr^{1/3}}}\right]}} $$ For square-edged inlets - Bergles-Rohsenow correlation: $$ f_{aS} = \frac{f_{c}\left[\frac{1}{(x/D)^{2/9}}\sqrt{\frac{Gr}{f_c}}\right]}{1 + \frac{2.86 \sqrt{Gr}}{(x/D)^{1/6}\left[\frac{14.9}{1.07+19.9 Pr^{0.630-2.34 \times 10^{-4} Gr^{1/3}}}\right]}} $$ To apply these formulas, we need to calculate the fully-developed friction factor, \(f_c\). We can use the Blasius correlation for smooth tubes since \(Re<10^5\): $$ f_c = 0.0791 Re^{-0.25} $$ Plugging in the Reynolds number calculated in Step 2: $$ f_c = 0.0791 (14335)^{-0.25} = 0.0206 $$ We have all necessary values to plug into the formulas and determine the developing friction factors for both inlet configurations: Re-entrant inlet friction, \(f_{aR}\): $$ f_{aR} = \frac{1.14\left[\frac{2.42}{(20)^{1/6}}\left(\frac{1.25}{1}\right)^{0.14}\sqrt{\frac{24000}{14335}}\right]}{1 + \frac{1.14\sqrt{24000}}{(20)^{1/6}\left[\frac{14.9}{1.07+19.9 (20.9)^{0.630-2.34 \times 10^{-4} 24000^{1/3}}}\right]}} = 0.02125 $$ Square-edged inlet friction, \(f_{aS}\): $$ f_{aS} = \frac{0.0206\left[\frac{1}{(20)^{2/9}}\sqrt{\frac{24000}{0.0206}}\right]}{1 + \frac{2.86 \sqrt{24000}}{(20)^{1/6}\left[\frac{14.9}{1.07+19.9 (20.9)^{0.630-2.34 \times 10^{-4} 24000^{1/3}}}\right]}} = 0.02325 $$ The apparent (developing) friction factor at the specified location (\(x/D=20\)) for the re-entrant inlet configuration is \(0.02125\), and for the square-edged inlet configuration, it is \(0.02325\).