Heat Transfer Correlations

From Thermal-FluidsPedia

Table 2 summarizes the existing correlations in literature for various heat transfer modes for both single-phase and two-phase systems in different geometric configurations. It can be seen that the heat transfer coefficient depends on surface geometry, the driving force of the fluid motion, thermal properties of the fluid, and flow properties.

Table 2 Correlations for convective heat transfer for various modes and geometries

Heat transfer mode	Geometry	Nusselt number	Comments and restrictions	Dimensionless numbers
Forced convection	Flow parallel to a flat plate	$Nu_x = 0.332Re_x^{1/2}Pr^{1/3}(Pr>0.6)$ $\overline {Nu} = 0.664Re_L^{1/2}Pr^{1/3}(Pr>0.6)$ $Nu_x = 0.565Re_x^{1/2}Pr^{1/2}(Pr \le 0.05)$	Isothermal surface $Re_x <5 \times 10^5$ (laminar)	$\begin{array}{l} {\rm{N}}{{\rm{u}}_x} = \frac{{hx}}{k} \\ {\rm{R}}{{\rm{e}}_x} = \frac{{{u_\infty }x}}{\nu } \\ \end{array}$ $\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hL}}{k} \\ {\rm{R}}{{\rm{e}}_L} = \frac{{{u_\infty }L}}{\nu } \\ \end{array}$
	Flow parallel to a flat plate	$\overline {Nu} = 0.037({Re_L^{0.8}}-871)Pr^{0.33}$ $(0.6 \le Pr \le 60)$	$5 \times 10^5 < {Re_L} < 10^5$ (Turbulent)
	Flow in a pipe (conventional size)	$\begin{array}{l} \overline {{\rm{Nu}}} = 3.66 \\ + \frac{{0.0668(D/L){\mathop{\rm Re}} \Pr }}{{1 + 0.04{{[(D/L){\mathop{\rm Re}} \Pr ]}^{2/3}}}} \\ \end{array}$	Isothermal surface ${\mathop{\rm Re}} \le 2300$ Thermal entry region	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hD}}{k} \\ {\mathop{\rm Re}} = \frac{{\bar uD}}{\nu } \\ \end{array}$ $\bar u$ is mean velocity
	Flow in a pipe (conventional size)	$\begin{array}{l} {\rm{Nu}} = 0.027{{\mathop{\rm Re}} ^{0.8}} \\ {\rm{ }} \times {\rm{P}}{{\rm{r}}^{0.33}}{\left( {\mu /{\mu _w}} \right)^{0.14}} \\ {\rm{ (0}}{\rm{.7}} \le \Pr \le 16700) \\ \end{array}$	$\begin{array}{l} L/D \ge 10 \\ {\mathop{\rm Re}} > 10,000 \\ \end{array}$ (Fully developed turbulent) $μ w$ is viscosity evaluated at $T w$
	Flow in a pipe (miniature)	$\begin{array}{l} \overline {{\rm{Nu}}} = (1 + F) \\ \times \frac{{(f/8)({\mathop{\rm Re}} - 1000)\Pr }}{{1 + 12.7{{(f/8)}^{0.5}}({{\Pr }^{2/3}} - 1)}} \\ f = {[1.82\log ({\mathop{\rm Re}} ) - 1.64]^{ - 2}} \\ F = 7.6 \times {10^{ - 5}}{\mathop{\rm Re}} \\ {\rm{ }} \times [1 - {(D/{D_0})^2}] \\ \end{array}$	$D 0$ =1.164 mm is reference diameter. Correlation was obtained for water at D=0.102, 0.76 and 1.09 mm.	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hD}}{k} \\ {\mathop{\rm Re}} = \frac{{\bar uD}}{\nu } \\ \end{array}$
	Flow between parallel plates	$\begin{array}{l} \overline {{\rm{Nu}}} = 7.54 \\ + \frac{{0.03({D_h}/L){\mathop{\rm Re}} \Pr }}{{1 + 0.016{{[({D_h}/L){\mathop{\rm Re}} \Pr ]}^{2/3}}}} \\ \end{array}$	Isothermal surface ${\mathop{\rm Re}} \le 2800$ (Laminar)	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar h{D_h}}}{k} \\ {\mathop{\rm Re}} = \frac{{\bar u{D_h}}}{\nu } \\ \end{array}$
	Flow between parallel plates	$\begin{array}{l} \overline {Nu} = 0.023{{\mathop{\rm Re}} ^{0.8}}{\Pr ^{0.33}} \\ {\rm{ (}}\Pr > 0.5) \\ \end{array}$	${\mathop{\rm Re}} > 10,000$ (Turbulent)
	Flow across a circular cylinder	$\begin{array}{l} \overline {{\rm{Nu}}} = 0.3 \\ + \frac{{0.62{{{\mathop{\rm Re}} }^{1/2}}{{\Pr }^{1/3}}}}{{{{[1 + {{(0.4/\Pr /)}^{2/3}}]}^{1/4}}}} \\ \times {\left[ {1 + {{\left( {\frac{{{\mathop{\rm Re}} }}{{282000}}} \right)}^{5/8}}} \right]^{4/5}} \\ \end{array}$	${\mathop{\rm Re}} \Pr > 0.2$ (Both laminar and turbulent)	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hD}}{k} \\ Re = \frac{{{u_\infty }D}}{\nu } \\ \end{array}$
	Flow across a sphere	$\begin{array}{l} \overline {{\rm{Nu}}} = 2 + (0.4{{\mathop{\rm Re}} ^{0.5}} \\ + 0.06{{\mathop{\rm Re}} ^{2/3}}){\Pr ^{0.4}}{(\mu /{\mu _w})^{\frac{1}{4}}} \\ \end{array}$	$\begin{array}{l} 3.5 < {\mathop{\rm Re}} \\ < 76000 \\ \end{array}$ $0.71 \le \Pr \le 380$ $μ w$ is viscosity evaluated at $T w$
	Flow through a packed bed of spheres	$\overline {{\rm{Nu}}} = 1.625{{\mathop{\rm Re}} ^{1/2}}{\Pr ^{1/3}}$	$15 \le {\mathop{\rm Re}} \le 120$ D – diameter of sphere A – bed cross- sectional area	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hD}}{k} \\ {\mathop{\rm Re}} = \frac{{\dot mD}}{{A\mu }} \\ \end{array}$
Free convection	On a vertical surface	$\begin{array}{l} {\overline {{\rm{Nu}}} ^{1/2}} = 0.825 + \\ \frac{{0.387{\rm{R}}{{\rm{a}}^{1/6}}}}{{{{[1 + {{(0.492/\Pr )}^{9/16}}]}^{8/27}}}} \\ \end{array}$	$\Delta T = \left\| {{T_w} - {T_\infty }} \right\|$ Applicable to both laminar and turbulent	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hL}}{k} \\ {\rm{Ra}} = \frac{{g\beta \Delta T{L^3}}}{{\nu \alpha }} \\ \end{array}$
	On a horizontal heated square facing up	$\overline {{\rm{Nu}}} = 0.54{({\rm{Gr}}\Pr )^{1/4}}$	Isothermal surface ${10^5} \le {\rm{Gr}}$ $\le 7 \times {10^7}$ For rectangle, use shorter side of $L$	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hL}}{k} \\ {\rm{Gr}} = \frac{{g\beta \Delta T{L^3}}}{{{\nu ^2}}} \\ \end{array}$
	On a horizontal heated square facing down	$\overline {{\rm{Nu}}} = 0.27{({\rm{Gr}}\Pr )^{1/4}}$	Isothermal surface $\begin{array}{l} 3 \times {10^5} \le {\rm{Gr}} \\ \le 3 \times {10^{10}} \\ \end{array}$ For rectangle, use shorter side of $L$	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hL}}{k} \\ {\rm{Gr}} = \frac{{g\beta \Delta T{L^3}}}{{{\nu ^2}}} \\ \end{array}$
	On a horizontal cylinder	$\begin{array}{l} {\overline {{\rm{Nu}}} ^{1/2}} = 0.60 + \\ + \frac{{0.387{\rm{R}}{{\rm{a}}^{1/6}}}}{{{{[1 + {{(0.559/\Pr )}^{9/16}}]}^{8/27}}}} \\ \end{array}$	$Ra < 10 12$	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hD}}{k} \\ {\rm{Ra}} = \frac{{g\beta \Delta T{D^3}}}{{\nu \alpha }} \\ \end{array}$
	On a sphere	$\begin{array}{l} \overline {{\rm{Nu}}} = 2 + \\ + \frac{{0.589{\rm{R}}{{\rm{a}}^{1/4}}}}{{{{[1 + {{(0.469/\Pr )}^{9/16}}]}^{4/9}}}} \\ \end{array}$	$\begin{array}{l}\Delta T = {T_w} - {T_\infty } \\ {\rm{Ra}} < {10^{11}} \\ \Pr \ge 0.7 \\ \end{array}$	$\begin{array}{l} \overline {{\rm{Nu}}} = \frac{{\bar hD}}{k} \\ {\rm{Ra}} = \frac{{g\beta \Delta T{D^3}}}{{\nu \alpha }} \\ \end{array}$
Evaporation	Falling film evaporation	Laminar $\begin{array}{l} {\rm{Nu}} = 1.10{\mathop{\rm Re}} _\delta ^{ - 1/3} \\ ({{\mathop{\rm Re}} _\delta } \le 30) \\ \end{array}$	Nu - local Nusselt number $Γ$ - mass flow rate per unit width of the vertical surface	$\begin{array}{l} {\rm{Nu}} = \frac{{h{{(\nu _\ell ^2/g)}^{\frac{1}{3}}}}}{k} \\ {{\mathop{\rm Re}} _\delta } = \frac{{4\Gamma }}{\mu } \\ \end{array}$
		Wavy laminar $\begin{array}{l} {\rm{Nu}} = 0.828{\mathop{\rm Re}} _\delta ^{ - 0.22} \\ (30 \le {{\mathop{\rm Re}} _\delta } \le 1800) \\ \end{array}$
		Turbulent $\begin{array}{l}{\rm{Nu}} = 0.0038{\mathop{\rm Re}} _\delta ^{0.4}{\Pr ^{0.65}} \\ {\rm{ }}({{\mathop{\rm Re}} _\delta } > 1800) \\ \end{array}$
Condensation	On a vertical surface	Laminar (Nusselt) $\begin{array}{l}{\rm{Nu}} = 1.10{\mathop{\rm Re}} _\delta ^{ - 1/3} \\ ({{\mathop{\rm Re}} _\delta } \le 30) \\ \end{array}$	Nu - local Nusselt number $Γ$ - mass flow rate per unit width of the vertical surface	$\begin{array}{l} {\rm{Nu}} = \frac{{h{{(\nu _\ell ^2/g)}^{\frac{1}{3}}}}}{k} \\ {{\mathop{\rm Re}} _\delta } = \frac{{4\Gamma }}{\mu } \\ \end{array}$
		Wavy laminar $\begin{array}{l} {\rm{Nu}} = \frac{{{{{\mathop{\rm Re}} }_\delta }}}{{{\mathop{\rm Re}} _\delta ^{1.22} - 5.22}} \\ (30 \le {{\mathop{\rm Re}} _\delta } \le 1800) \\ \end{array}$
		Turbulent ${\rm{Nu}} = 0.023{\mathop{\rm Re}} _\delta ^{0.25}{\Pr ^{ - 0.5}}$
	On tubes	$\begin{array}{l} \overline {Nu} = 0.729 \\ \times {\left[ {\frac{{{D^3}{h_{\ell v}}g\left( {{\rho _\ell } - {\rho _v}} \right)}}{{n{k_\ell }{\nu _\ell }\Delta T}}} \right]^{\frac{1}{4}}} \\ \end{array}$	$Δ T = T s a t - T w$ n - number of tubes	$\overline {Nu} = \frac{{\bar hD}}{{{k_\ell }}}$
	In microscale channel ( $D h < 1.5mm$ )	${\rm{Nu}} = {\rm{W}}{{\rm{e}}^{ - {\rm{Ja}}}}{\mathop{\rm Re}} {\Pr ^Y}$	$\begin{array}{l} Y = 1.3{\rm{ }} \\ {\rm{for }}{\mathop{\rm Re}} \le 65 \\ Y = (0.5{D_h} - 1) \\ {\rm{ }}/(2{D_h}) \\ {\rm{for }}{\mathop{\rm Re}} > 65 \\ \end{array}$	${\rm{We}} = \frac{{{\rho _\ell }{V^2}L}}{\sigma }$ ${\rm{Ja}} = \frac{{{c_{p\ell }}({T_{sat}} - {T_w})}}{{{h_{\ell v}}}}$ ${\mathop{\rm Re}} = \frac{{\dot m''{D_h}}}{{{\mu _\ell }}}$ $\dot m''$ –mass flux (kg/s-m²)
Boiling	Nucleate, saturated pool boiling	$\overline {{\rm{Nu}}} = \frac{{{\rm{Ja}}_\ell ^2}}{{{C^3}\Pr _\ell ^m}}$	m=2 for water m=4.1 for other fluids C=0.013 water- copper or stainless steel C=0.006 for water- nickel or brass	$\begin{array}{l} \overline {Nu} = \frac{{\bar h{L_c}}}{{{k_\ell }}} \\ {L_c} = \sqrt {\frac{{{\sigma _\ell }}}{{g({\rho _\ell } - {\rho _v})}}} \\ {\rm{J}}{{\rm{a}}_\ell } = \frac{{{c_{p,\ell }}\Delta T}}{{{h_{\ell v}}}} \\ \end{array} \Delta T = {T_w} - {T_{sat}}$
	Film boiling on a horizontal plate	$\begin{array}{l} \overline {{\rm{Nu}}} = 0.425 \\ \times {\left[ {Gr{{\Pr }_v}\left( {\frac{{1 + 0.4J{a_v}}}{{J{a_v}}}} \right)} \right]^{\frac{1}{4}}} \\ \end{array}$	Term in parentheses accounts for sensible heating effect in vapor film	$\begin{array}{l} \overline {Nu} = \frac{{\bar h{L_c}}}{{{k_v}}} \\ {L_c} = \sqrt {\frac{{{\sigma _\ell }}}{{g({\rho _\ell } - {\rho _v})}}} \\ Gr = \frac{{g[({\rho _\ell } - {\rho _v})/{\rho _v}]L_c^3}}{{\nu _v^2}} \\ {\rm{J}}{{\rm{a}}_v} = \frac{{{c_{p,v}}\Delta T}}{{{h_{\ell v}}}} \\ \end{array}$
	Film boiling on a horizontal cylinder	$\begin{array}{l} \overline {{\rm{Nu}}} = 0.62 \\ \times {\left[ {{\rm{Gr}}{{\Pr }_v}\left( {\frac{{1 + 0.4J{a_v}}}{{J{a_v}}}} \right)} \right]^{\frac{1}{4}}} \\ \end{array}$	$D \gg$ film thickness	$\begin{array}{l} \overline {Nu} = \frac{{\bar hD}}{{{k_v}}} \\ {\rm{Gr}} = \\ \frac{{g[({\rho _\ell } - {\rho _v})/{\rho _v}]{D^3}}}{{\nu_v^2}} \\ {\rm{J}}{{\rm{a}}_v} = \frac{{{c_{p,v}}\Delta T}}{{{h_{\ell v}}}} \\ \end{array}$
	Film boiling on a sphere	$\begin{array}{l} \overline {{\rm{Nu}}} = 0.4 \\ \times {\left[ {{\rm{Gr}}{{\Pr }_v}\left( {\frac{{1 + 0.4{\rm{J}}{{\rm{a}}_v}}}{{{\rm{J}}{{\rm{a}}_v}}}} \right)} \right]^{\frac{1}{3}}} \\ \end{array}$	$D\gg$ film thickness
	Boiling in microchannel (D=1.39 – 1.69 mm)	$\begin{array}{l} {\rm{Nu}} = 30{{\mathop{\rm Re}} ^{0.857}} \\ \times {\rm{B}}{{\rm{o}}^{0.714}}{(1 - x)^{ - 0.143}} \\ \end{array}$	Correlation obtained by using Freon ® 141 x is quality	$\overline {Nu} = \frac{{\bar hD}}{{{k_\ell }}} {\rm{Bo}} = \frac{{q''}}{{{h_{\ell v}}\dot m''}}$ $\dot m''$ – mass flux (kg/s-m2)
Melting	Melting in a rectangular cavity	$\begin{array}{l} {\rm{Nu}} = {(2\tau )^{ - 1/2}} \\ + [{c_1}{\rm{R}}{{\rm{a}}^{1/4}} - {(2\tau )^{ - 1/2}}] \\ \times {[1 + {({c_2}{\rm{R}}{{\rm{a}}^{3/4}}{\tau ^{3/2}})^n}]^{1/n}} \\ \end{array}$ $c 1 = 0.35, c 2 = 0.175$ $n = - 2$	Nusselt number is function of time	$\begin{array}{l} \overline {Nu} = \frac{{\bar hH}}{k} \\ {\rm{Ra}} = \frac{{g\beta \Delta T{H^3}}}{{\nu \alpha }} \\ \tau = {\rm{SteFo}} \\ \end{array}$ ${\rm{Fo}} = \frac{{{\alpha _\ell }t}}{{{H^2}}}$
Solidification	Solidification around a horizontal tube	$\overline {{\rm{Nu}}} = 0.52{\rm{R}}{{\rm{a}}^{1/4}}$	$D$ is transient equivalent outer diameter of the solid $Ra \le {10^9}$	$\begin{array}{l} \overline {Nu} = \frac{{\bar hD}}{k} \\ {\rm{Ra}} = \frac{{g\beta \Delta T{D^3}}}{{\nu \alpha }} \\ \end{array}$
Sublimation		$\begin{array}{l} {\rm{N}}{{\rm{u}}_x} = 0.458{\mathop{\rm Re}} _x^{1/2}{\Pr ^{1/3}} \\ {\rm{S}}{{\rm{h}}_x} = 0.459{\mathop{\rm Re}} _x^{1/2}{{\mathop{\rm Sc}} ^{1/3}} \\ \end{array}$	Uniform heat flux surface ${{\mathop{\rm Re}} _x} < 5 \times {10^5}$	$\begin{array}{l} {{\mathop{\rm Nu}} _x} = \frac{{hx}}{k} \\ {{\mathop{\rm Sh}} _x} = \frac{{{h_m}x}}{D} \\ \end{array}$