Conjoint effect of turbulator and Al₂O₃ nanofluids on DPHEs thermal performance: Experimental study

Abstract

The energy crisis we are currently experiencing is merely the start of a very challenging and wide transformation. The sectors of power, coal, and natural gas encountered the biggest vibrations. To help with energy conservation, a compact and effective heat exchanger was made available that may be utilized to collect waste heat from power plants and industries. This study investigates the effects of combining passive techniques on the performance of a double-pipe heat exchanger equipped with a metal turbulator on the hot side and Al₂O₃ nanofluid on the cold side. The experiments used different volume fractions of Al₂O₃ nanofluid (Vol.%: 0.05, 0.1, and 0.15) as cold fluid with varying flow rates (500 ≤ Re ≤ 5000) in the annulus, as well as variously configured twisted tapes (Twist ratio: 20, 13.3, and 9.8) and frequently spaced helical screw tapes (Number of helices: 5, 7 and 9). The results show that the Nusselt number increases by 11.11% and the thermal performance factor increases by 1.116 times in case of twisted tapes with twist ratio 20 and 0.05% nanofluid combination, and by 24.93% and 1.269 times in case of frequently spaced helical screw tape with 9 number of helices and 0.15% nanofluid combination, respectively. Therefore, even at the expense of a small amount of pressure loss, 9 helices with 0.15% of Al₂O₃ nanofluid offered better performance in the combinations evaluated.

Appendix: Uncertainty calculations

1.1 Reynolds number

$$\begin{aligned}\frac{{\text{U}}_{\text{Re}}}{\text{Re}}&={\left[{\left(\frac{{\mathrm{U}_\rho}}{{\rho}}\right)}^{2}+{\left(\frac{{\text{U}}_{\text{m}}}{\text{m}}\right)}^{2}+{\left(\frac{{\text{U}}_{\text{di}}}{{\text{d}}_{\text{i}}}\right)}^{2}+{\left(\frac{{\text{U}_\mu}}{\mu}\right)}^{2}\right]}^{^{1}/_{2}} \\&={\left[{\left({1}\times {10}^{-{5}}\right)}^{2}+{\left(\text{2.635}\times {10}^{-{3}}\right)}^{2}+{\left(0.01388\right)}^{2}+{\left(\text{0.01}\right)}^{2}\right]}^{^{1}/_{2}}\\&=\text{0.0173 }={ 1.73 \%}\end{aligned}$$

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1.2 Heat transfer coefficient

$$\begin{aligned}\frac{{\text{U}}_{\text{ho}}}{{\text{h}}_{\text{o}}}&={\left[{\left(\frac{{\text{U}}_{\text{Qavg}}}{{\text{Q}}_{\text{avg}}}\right)}^{2}+{\left(\frac{{\text{U}}_{\text{A}}}{\text{A}}\right)}^{2}+{\left(\frac{{\text{U}}_{\Delta {\text{T}}}}{\Delta {\text{T}}}\right)}^{2}\right]}^{^1/_{2}}\\&={\left[{\left(\text{2.058}\times {10}^{-{6}}\right)}^{2}+{\left({0.01}^{2}\right)}^{2}+{\left(\text{1.449}\times {10}^{-3}\right)}^{2}\right]}^{^1/_{2}}\\&=\text{0.0101 }= 1.01 \%\end{aligned}$$

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1.3 Nusselt number

$$\begin{aligned}\frac{{\text{U}}_{\text{Nu}}}{\text{Nu}}&={\left[{\left(\frac{{\text{U}}_{\text{h}}}{\text{h}}\right)}^{2}+{\left(\frac{{\text{U}}_{\text{di}}}{{\text{d}}_{\text{i}}}\right)}^{2}+{\left(\frac{{\text{U}}_{\text{K}}}{\text{K}}\right)}^{2}\right]}^{^1/_{2}}\\&={\left[{\left(7.2811\times {10}^{-5}\right)}^{2}+{\left({0.01}^{2}\right)}^{2}+{\left(0.0159\right)}^{2}\right]}^{^1/_{2}}\\&=\text{0.0188 }=1.88 \%\end{aligned}$$

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1.4 Friction factor

$$\begin{aligned}\frac{{\text{U}}_{\text{fc}}}{{\text{f}}_{\text{c}}}&={\left[{\left(\frac{{\text{U}}_{\text{Re}}}{{\text{R}}_{\text{e}}}\right)}^{2}+{\left(\frac{{\text{U}}_{\text{di}}}{{\text{d}}_{\text{i}}}\right)}^{2}+{\left(\frac{{\text{U}}_{\Delta {\text{P}}}}{\Delta {\text{P}}}\right)}^{2}\right]}^{^1/_{2}}\\&={\left[{\left(3.46\times {10}^{-5}\right)}^{2}+{\left(0.01\right)}^{2}+{\left(1.1904\times {10}^{-2}\right)}^{2}\right]}^{^1/_{2}}\\&=0.0101=1.01\%\end{aligned}$$

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1.5 Overall uncertainty

$$\begin{aligned}{\text{U}}_{\text{Overall}}&={\left[{\left(\frac{{\text{U}}_{\Delta\text{T}}}{\Delta \text{T}}\right)}^{2}+{\left(\frac{{\text{U}}_{\text{m}}}{\text{m}}\right)}^{2}+{\left(\frac{{\text{U}}_{\Delta\text{P}}}{\Delta\text{P}}\right)}^{2}+{\left(\frac{{\text{U}}_{\text{W}}}{\text{W}}\right)}^{2}\right]}^{1/_{2}}\\&=\Big[{\left(\text{1.449}\times {10}^{-3}\right)}^{2}+{\left(\text{2.635}\times {10}^{-{3}}\right)}^{2}\\&\quad\,+{\left(1.1904\times {10}^{-2}\right)}^{2}+{\left({1}\times {10}^{-5}\right)}^{2}\Big]^{1/_{2}}\\&=\text{0.}{0289} \, = 2.89 \%\end{aligned}$$

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