In the case of liquid, the total heat transfer coefficient $($h$)$ for a long tube is expressed by the following empirical equation:

h$=0\mathrm{.}023\backslash$frac$\{$G$^\{0\mathrm{.}8\}$k$^\{0\mathrm{.}67\}\{$C$\_\{$p$\}\}^\{0\mathrm{.}33\}\}\{$D$^\{0\mathrm{.}2\}\backslash$mu$^\{0\mathrm{.}47\}\}$

where

G$=$ Mass velocity $($kg$/$m$^\{2\}$s$)$

k$=$ Thermal conductivity $($W$/$m$^\{\backslash$circ$\}$C$)$

C$\_\{$p$\}=$ Heat capacity $($J$/$kg$^\{\backslash$circ$\}$C$)$

D$=$ Diameter of the tube $($m$)$

$\backslash$mu$=$ Viscosity of the liquid $($kg$/$m~s$)$

The above equation is dimensionally consistent. Determine the unit of the total heat transfer coefficient $($h$).$ Note that I W$=1$~J$/$s$.($Mark: $7)$

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