Investigations on thermal characteristics in a double pipe fitted with circular finned and frequently spaced helical twisted inserts and Graphene oxide nanofluid

Abstract

In this experimental study, the influence of combined passive technique on heat transfer, friction factor, and thermal performance factor are investigated in a double pipe heat exchanger fitted with different turbulators and GO-Water nanofluid. Experiments were conducted for a fixed flow rate of (Re = 2500) in the inner tube (hot fluid); for varying flow rates (500 ≤ Re ≤ 5000) and volume concentrations (0.05%–0.15%) of GO-Water nanofluid in the outer tube (cold fluid), under uniform heat flux condition, using i) circular finned twisted tape inserts with TR = 20, 13.3 and 9.8 ii) frequently spaced helically twisted inserts with the number of helices = 5, 7, 9). The experimental results showed that the Nusselt number and Thermal Performance Factor (TPF) were increased with decreased twist ratio and increased helices and volume concentration of nanofluid. The enhancement of heat transfer was 21.35% and 22.21%, respectively, whereas the enhancement in the Thermal Performance Factor (TPF) was 7.16% and 8.06%, respectively, for the above two test conditions. The augmentation was maximum for configurations corresponding to the circular finned twisted tape insert with TR = 9.8 and frequently spaced helical screw-tape with 9 helices in 0.15% volume concentrations of GO-water nanofluid as compared to other combinations, with a slight penalty in pressure drop.

Appendix: Uncertainty Calculations

1.1 Reynolds number

$$ {\displaystyle \begin{array}{c}\frac{{\mathrm{U}}_{\mathrm{Re}}}{\operatorname{Re}}={\left[{\left(\frac{{\mathrm{U}}_{\uprho}}{\uprho}\right)}^2+{\left(\frac{{\mathrm{U}}_{\mathrm{m}}}{\mathrm{m}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\mathrm{d}\mathrm{i}}}{{\mathrm{d}}_{\mathrm{i}}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\upmu}}{\upmu}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {}={\left[{\left(1\times {10}^{-5}\right)}^2+{\left(2.635\times {10}^{-3}\right)}^2+{(0.01388)}^2+{(0.01)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0.0173=1.73\%\end{array}} $$

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1.2 Heat transfer coefficient

$$ {\displaystyle \begin{array}{c}\frac{{\mathrm{U}}_{\mathrm{h}\mathrm{o}}}{{\mathrm{h}}_{\mathrm{o}}}={\left[{\left(\frac{{\mathrm{U}}_{\mathrm{Q}\mathrm{avg}}}{{\mathrm{Q}}_{\mathrm{avg}}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\mathrm{A}}}{\mathrm{A}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\Delta \mathrm{T}}}{\Delta \mathrm{T}}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {}={\left[{\left(2.058\times {10}^{-6}\right)}^2+{\left({0.01}^2\right)}^2+{\left(1.449\times {10}^{\hbox{-} 3}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0.0101=1.01\%\end{array}} $$

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1.3 Nusselt Number

$$ {\displaystyle \begin{array}{c}\frac{{\mathrm{U}}_{\mathrm{Nu}}}{\mathrm{Nu}}={\left[{\left(\frac{{\mathrm{U}}_{\mathrm{h}}}{\mathrm{h}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\mathrm{d}\mathrm{i}}}{{\mathrm{d}}_{\mathrm{i}}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\mathrm{K}}}{\mathrm{K}}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {}={\left[{\left(7.2811\times {10}^{-5}\right)}^2+{\left({0.01}^2\right)}^2+{(0.0159)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0.0188=1.88\%\end{array}} $$

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1.4 Friction factor

$$ {\displaystyle \begin{array}{c}\frac{{\mathrm{U}}_{\mathrm{f}\mathrm{c}}}{{\mathrm{f}}_{\mathrm{c}}}={\left[{\left(\frac{{\mathrm{U}}_{\mathrm{R}\mathrm{e}}}{{\mathrm{R}}_{\mathrm{e}}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\mathrm{d}\mathrm{i}}}{{\mathrm{d}}_{\mathrm{i}}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\Delta \mathrm{P}}}{\Delta \mathrm{P}}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {}={\left[{\left(3.46\times {10}^{-5}\right)}^2+{(0.01)}^2+{\left(1.1904\times {10}^{-2}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0.0101=1.01\%\end{array}} $$

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1.5 Overall uncertainty

$$ {\displaystyle \begin{array}{c}{\mathrm{U}}_{\mathrm{Overall}}={\left[{\left(\frac{{\mathrm{U}}_{\Delta \mathrm{T}}}{\Delta \mathrm{T}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\mathrm{m}}}{\mathrm{m}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\Delta \mathrm{P}}}{\Delta \mathrm{P}}\right)}^2+{\left(\frac{{\mathrm{U}}_{\mathrm{W}}}{\mathrm{W}}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}\\ {}={\left[{\left(1.449\times {10}^{\hbox{-} 3}\right)}^2+{\left(2.635\times {10}^{-3}\right)}^2+{\left(1.1904\times {10}^{-2}\right)}^2+{\left(1\times {10}^{\hbox{-} 5}\right)}^2\right]}^{\raisebox{1ex}{$1$}\!\left/ \!\raisebox{-1ex}{$2$}\right.}=0.0289=2.89\%\end{array}} $$

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