Question

A tube with a square-edged inlet configuration is subjected to uniform wall heat flux of \(8 \mathrm{~kW} / \mathrm{m}^{2}\). The tube has an inside diameter of \(0.622\) in and a flow rate of \(2.16 \mathrm{gpm}\). The liquid flowing inside the tube is an ethylene glycol-distilled water mixture with a mass fraction of \(2.27\). Determine the friction coefficient at a location along the tube where the Grashof number is Gr \(=35,450\). The physical properties of the ethylene glycol-distilled water mixture at the location of interest are $\operatorname{Pr}=13.8, \nu=18.4 \times 10^{-6} \mathrm{ft}^{2} / \mathrm{s}\(, and \)\mu_{b} / \mu_{s}=1.12$. Then recalculate the fully developed friction coefficient if the volume flow rate is increased by 50 percent while the rest of the parameters remain unchanged.

Short answer

Answer: The fully developed friction coefficient when the volume flow rate is increased by 50 percent is 0.0057.

Step-by-step solution

Calculate the mass flow rate

To find the mass flow rate, first, convert the flow rate which is given in gpm into the units of kg/s: \(1 ~\mathrm{gpm} = 3.78541 \times 10^{-3} \frac{ \mathrm{m}^{3}}{\mathrm{min}} = 6.30902 \times 10^{-5} \frac{ \mathrm{m}^{3}}{\mathrm{s}}\) So, \(\dot{V} = 2.16 \mathrm{gpm} = 2.16 \times 6.30902 \times 10^{-5} \frac{ \mathrm{m}^{3}}{\mathrm{s}} = 1.36235 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}\) By considering the mass fraction, \(\rho = \frac{2.27}{1-2.27} \rho_{w}\) \(\Rightarrow \rho = 3.1361 \times 10^{3} \mathrm{~kg} / \mathrm{m}^{3}\) Now, we can calculate the mass flow rate as, \(\dot{m} = \rho \dot{V} = 3.1361 \times 10^{3} \cdot 1.36235 \times 10^{-4} = 0.427 \mathrm{~kg} / \mathrm{s}\)

Determine the Reynolds number

The Reynolds number can be calculated using the given kinematic viscosity and mass flow rate as follows: \(Re = \frac{4 \dot{m}}{\pi D \nu}\) Converting the diameter to \(\mathrm{m}\), \(D = 0.622 \mathrm{in} = 0.622 \times 0.0254 \mathrm{m} = 0.0158 \mathrm{m}\) Now, \(Re = \frac{4 \times 0.427}{\pi \times 0.0158 \times 18.4 \times 10^{-6}}= 15,589\)

Determine the friction coefficient

To find the friction coefficient, we can apply the Dittus-Boelter equation: \(f = (0.0791 / Re^{0.25}) (\mu_{b}/\mu_{s})\) where \(\mu_{b} / \mu_{s}=1.12\), and we already calculated the Reynolds number as \(Re = 15,589\), therefore, \(f = (0.0791 / (15,589)^{0.25}) \times 1.12 = 0.0060\) So, the friction coefficient at the location where Gr = 35,450 is \(f = 0.0060\).

Recalculate the friction coefficient with increased volume flow rate

Now, we should increase the volume flow rate by 50% and recalculate the friction coefficient. The new volume flow rate is: \(\dot{V}_{new} = \dot{V} \times 1.5 = 1.36235 \times 10^{-4} \times 1.5 = 2.04352 \times 10^{-4} \mathrm{m}^{3} / \mathrm{s}\) Now, we calculate the new mass flow rate, \(\dot{m}_{new} = \rho \dot{V}_{new} = 3.1361 \times 10^{3} \cdot 2.04352 \times 10^{-4} = 0.6405 \mathrm{~kg} / \mathrm{s}\) Calculate the new Reynolds number, \(Re_{new} = \frac{4 \dot{m}_{new}}{\pi D \nu}= \frac{4 \times 0.6405}{\pi \times 0.0158 \times 18.4 \times 10^{-6}} = 23,383\) Finally, apply the Dittus-Boelter equation again: \(f_{new} = (0.0791 / Re_{new}^{0.25}) (\mu_{b}/\mu_{s})\) \(f_{new} = (0.0791 / (23,383)^{0.25}) \times 1.12 = 0.0057\) So, the fully developed friction coefficient when the volume flow rate is increased by 50 percent is \(f_{new} = 0.0057\).