Heat Transfer in Fully-Developed Internal Turbulent Flow

Heat transfer in fully-developed turbulent flow in a circular tube subject to constant heat flux (q''w = const) will be considered in this article [1]. When the turbulent flow in the tube is fully developed, we have $\bar{v}=0$ and the energy equation becomes

 $\bar{u}\frac{\partial \bar{T}}{\partial x}=\frac{1}{r}\frac{\partial }{\partial r}\left[ r(\alpha +{{\varepsilon }_{H}})\frac{\partial \bar{T}}{\partial r} \right]$ (1)

After the turbulent flow is hydrodynamically and thermally fully developed, the time-averaged temperature profile is no longer a function of axial distance from the inlet, i.e.,

 $\frac{\partial }{\partial x}\left( \frac{{{T}_{w}}-\bar{T}}{{{T}_{w}}-{{{\bar{T}}}_{c}}} \right)=0$ (2)

where ${{\bar{T}}_{c}}$ is the time-averaged temperature at the centerline of the tube, and Tw is the wall temperature. Thus, $({{T}_{w}}-\bar{T})/({{T}_{w}}-{{\bar{T}}_{c}})$ is a function of r only, i.e.,

 $\frac{{{T}_{w}}-\bar{T}}{{{T}_{w}}-{{{\bar{T}}}_{c}}}=f\left( r \right)$ (3)

where f is independent from x. Differentiating eq. (2) yields

 $\frac{\partial \bar{T}}{\partial x}=\frac{d{{T}_{w}}}{dx}-\left( \frac{{{T}_{w}}-\bar{T}}{{{T}_{w}}-{{{\bar{T}}}_{c}}} \right)\left( \frac{d{{T}_{w}}}{dx}-\frac{d{{{\bar{T}}}_{c}}}{dx} \right)$ (4)

At the wall, the contribution of eddy diffusivity on the heat transfer is negligible, and the heat flux at the wall becomes

 ${{{q}''}_{w}}=k{{\left. \frac{\partial \bar{T}}{\partial r} \right|}_{r={{r}_{0}}}}$ (5)

Substituting eq. (3) into eq. (5), one obtains:

 ${{{q}''}_{w}}=-k({{T}_{w}}-{{\bar{T}}_{c}}){f}'({{r}_{0}})$ (6)

Since the heat flux is constant, q''w = const, it follows that $({{T}_{w}}-{{\bar{T}}_{c}})= Const$, i.e.,

 $\frac{d{{T}_{w}}}{dx}=\frac{d{{{\bar{T}}}_{c}}}{dx}$ (7)

Therefore, eq. (4) becomes:

 $\frac{\partial \bar{T}}{\partial x}=\frac{d{{T}_{w}}}{dx}$ (8)

For fully developed flow, the local heat transfer coefficient is:

 ${{h}_{x}}=\frac{{{{{q}''}}_{w}}}{{{T}_{w}}-{{{\bar{T}}}_{m}}}=\text{const}$ (9)

where ${{\bar{T}}_{m}}$ is the time-averaged mean temperature defined as:

 ${{\bar{T}}_{m}}=\frac{2}{r_{0}^{2}}\int_{0}^{{{r}_{0}}}{\bar{u}\bar{T}rdr}$ (10)

Since q''w = const, it follows from eq. (9) that $({{T}_{w}}-{{\bar{T}}_{m}})= Const$, i.e.,

 $\frac{d{{T}_{w}}}{dx}=\frac{d{{{\bar{T}}}_{m}}}{dx}$ (11)

Combining eqs. (7), (8) and (11), the following relationships are obtained:

 $\frac{\partial \bar{T}}{\partial x}=\frac{d{{T}_{w}}}{dx}=\frac{d{{{\bar{T}}}_{c}}}{dx}=\frac{d{{{\bar{T}}}_{m}}}{dx}$ (12)

The time-averaged mean temperature, ${{\bar{T}}_{m}}$, changes with x as the result of heat transfer from the tube wall. The rate of mean temperature change can be obtained as follows:

 $\frac{d{{{\bar{T}}}_{m}}}{dx}=\frac{4\pi {{{{q}''}}_{w}}}{\rho {{c}_{p}}D{{{\bar{u}}}_{m}}}$ (13)

Substituting eq. (12) into eq. (1), the energy equation becomes:

 $\bar{u}\frac{d{{{\bar{T}}}_{m}}}{dx}=\frac{1}{{{r}_{0}}-y}\frac{\partial }{\partial y}\left[ ({{r}_{0}}-y)(\alpha +{{\varepsilon }_{H}})\frac{\partial \bar{T}}{\partial y} \right]$ (14)

where y = r0r is the distance measured from the tube wall. Equation (14) is subject to the following two boundary conditions:

 $\frac{\partial \bar{T}}{\partial y}=0,\text{ }y={{r}_{0}}$ (axisymmetric condition) (15)
 $\bar{T}={{T}_{w}}\text{ (unknown)},\text{ }y=0$ (16)

Integrating eq. (14) in the interval of (r0, r) and considering eq. (15), we have:

 $({{r}_{0}}-y)(\alpha +{{\varepsilon }_{H}})\frac{\partial \bar{T}}{\partial y}=\frac{d{{{\bar{T}}}_{m}}}{dx}\int_{{{r}_{0}}}^{y}{({{r}_{0}}-y)\bar{u}dy}$ (17)

which can be rearranged to

 $\frac{\partial \bar{T}}{\partial y}=\frac{I(y)}{({{r}_{0}}-y)(\alpha +{{\varepsilon }_{H}})}\frac{d{{{\bar{T}}}_{m}}}{dx}$ (18)

where

 $I(y)=\int_{{{r}_{o}}}^{y}{({{r}_{o}}-y)\bar{u}dy}$ (19)

Integrating eq. (18) in the interval of (0, y) and considering eq. (16), one obtains:

 $\bar{T}-{{T}_{w}}=\frac{d{{{\bar{T}}}_{m}}}{dx}\int_{0}^{y}{\frac{I(y)}{({{r}_{o}}-y)(\alpha +{{\varepsilon }_{H}})}dy}$ (20)

If the profiles of axial velocity and the thermal eddy diffusivity are known, eq. (20) can be used to obtain the correlation for internal forced convection heat transfer. With the exception of the very thin viscous sublayer, the velocity profile in the most part of the tube is fairly flat. Therefore, it is assumed that the time-averaged velocity, $\bar{u}$, in eq. (19) can be replaced by ${{\bar{u}}_{m}}$, and I(y) becomes:

 $I(y)\simeq -\frac{\bar{u}_{m}^{2}}{2}{{({{r}_{o}}-y)}^{2}}$ (21)

Substituting eqs. (21) and (13) into eq. (20) yields:

 $\bar{T}-{{T}_{w}}=-\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}\int_{0}^{y}{\frac{(1-y/{{r}_{o}})}{(\alpha +{{\varepsilon }_{H}})}dy}$ (22)

which can be rewritten in terms of wall coordinate

 $\bar{T}-{{T}_{w}}=-\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}\sqrt{\frac{\rho }{{{\tau }_{w}}}}\int_{0}^{{{y}^{+}}}{\frac{(1-y/{{r}_{o}})}{[1/\Pr +({{\varepsilon }_{M}}/\nu )/{{\Pr }^{t}}]}d{{y}^{+}}}$ (23)

where y+ is defined as ${{y}^{+}}=\frac{y{{u}_{\tau }}}{\nu }$.

To consider heat transfer in an internal turbulent flow, the entire turbulent boundary layer is divided into three regions: (1) inner region (y + < 5), (2) buffer region ($5\le {{y}^{+}}\le 30$), and (3) outer region (y + > 30). In the inner region ${{\varepsilon }_{M}}={{\varepsilon }_{H}}=0$ and eq. (23) becomes

 $\bar{T}-{{T}_{w}}=-\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}\sqrt{\frac{\rho }{{{\tau }_{w}}}}\Pr \int_{0}^{{{y}^{+}}}{(1-y/{{r}_{o}})d{{y}^{+}}}$ (24)

Since the inner region is very thin, $y/{{r}_{0}}\ll 1$ and 1 − y / r0 is effectively equal to 1. Therefore, the temperature profile in the inner region becomes:

 $\bar{T}-{{T}_{w}}=-\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \right)\sqrt{\frac{\rho }{{{\tau }_{w}}}}\Pr {{y}^{+}}$ (25)

The temperature at the boundary between the inner and buffer regions (y + = 5), ${{\bar{T}}_{s}}$, can be obtained from eq. (25) as

 ${{\bar{T}}_{s}}-{{T}_{w}}=-5\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \right)\sqrt{\frac{\rho }{{{\tau }_{w}}}}\Pr$ (26)

In the buffer region where $5\le {{y}^{+}}\le 30$, the eddy diffusivity in the buffer region is:

 $\frac{{{\varepsilon }_{M}}}{\nu }=\frac{{{y}^{+}}}{5}-1$ (27)

Substituting eq. (27) into eq. (23) and assuming the turbulent Prandtl number ${{\Pr }^{t}}=1$, the following expression is obtained:

 $\bar{T}-{{\bar{T}}_{s}}=-\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}\sqrt{\frac{\rho }{{{\tau }_{w}}}}\int_{5}^{{{y}^{+}}}{\frac{(1-y/{{r}_{o}})}{[1/\Pr +{{y}^{+}}/5-1]}d{{y}^{+}}}$ (28)

Since the buffer region is also very thin, 1 − y / r0 in eq. (28) is effectively equal to 1. Defining ${{T}^{+}}={(\bar{T}-{{T}_{w}})}/{\left( -\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}\sqrt{\frac{\rho }{{{\tau }_{w}}}} \right)}\;$ and Integrating eq. (28) yields

 $\int_{5\Pr }^{{{T}^{+}}}{d{{T}^{+}}}=\int_{5}^{{{y}^{+}}}{\frac{d{{y}^{+}}}{1/\Pr +({{y}^{+}}-5)/(5{{\Pr }^{t}})}}$ (29)

i.e.,

 $\bar{T}-{{\bar{T}}_{s}}=-5\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \right)\sqrt{\frac{\rho }{{{\tau }_{w}}}}\ln \left( \frac{{{y}^{+}}}{5}\Pr -\Pr +1 \right),\text{ }5<{{y}^{+}}<30$ (30)

The temperature at the top of the buffer region where y + = 30, ${{\bar{T}}_{b}}$, becomes

 ${{\bar{T}}_{b}}-{{\bar{T}}_{s}}=-5\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \right)\sqrt{\frac{\rho }{{{\tau }_{w}}}}\ln (5\Pr +1)$ (31)

For the outer region where ${{\varepsilon }_{M}}\gg \nu \text{ and }{{\varepsilon }_{H}}\gg \alpha$, eq. (23) becomes

 $\bar{T}-{{\bar{T}}_{b}}=-\frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}}\sqrt{\frac{\rho }{{{\tau }_{w}}}}\int_{30}^{{{y}^{+}}}{\frac{(1-y/{{r}_{o}})}{{{\varepsilon }_{M}}/\nu }d{{y}^{+}}}$ (32)

where the turbulent Prandtl number is assumed to be equal to 1.

It is assumed that the Nikuradse equation is valid in the outer region and the velocity gradient in this region becomes:

 $\frac{\partial {{u}^{+}}}{\partial {{y}^{+}}}=\frac{2.5}{{{y}^{+}}}$ (33)

The expression of apparent shear stress in this region, can be non-dimensionalized as:

 $\frac{{{\tau }_{app}}}{{{\tau }_{w}}}=\frac{{{\varepsilon }_{M}}}{\nu }\frac{\partial {{u}^{+}}}{\partial {{y}^{+}}}$ (34)

Substituting

${{\tau }_{app}}={{\tau }_{w}}\cdot \left( \frac{r}{{{r}_{o}}} \right)={{\tau }_{w}}\cdot \left( 1-\frac{y}{{{r}_{o}}} \right)$

and eq. (33) into eq. (34), the eddy diffusivity in the outer region is obtained as:

 $\frac{{{\varepsilon }_{M}}}{\nu }=\left( 1-\frac{y}{{{r}_{o}}} \right)\frac{{{y}^{+}}}{2.5}$ (35)

Substituting eq. (35) into eq. (32), the temperature distribution in this region becomes:

 $\bar{T}-{{\bar{T}}_{b}}=-2.5\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \right)\sqrt{\frac{\rho }{{{\tau }_{w}}}}\int_{30}^{{{y}^{+}}}{\frac{1}{{{y}^{+}}}d{{y}^{+}}}=-2.5\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \right)\sqrt{\frac{\rho }{{{\tau }_{w}}}}\ln \left( \frac{{{y}^{+}}}{30} \right)$ (36)

which is valid from y + = 30 to the center of the tube where yc = r0 or

 $y_{c}^{+}=\frac{{{r}_{0}}}{\nu }\sqrt{\frac{{{\tau }_{w}}}{\rho }}$ (37)

The temperature at the center of the tube, ${{\bar{T}}_{c}}$, can be obtained by letting ${{y}^{+}}=y_{c}^{+}$ in eq. (36), i.e.

 ${{\bar{T}}_{c}}-{{\bar{T}}_{b}}=-2.5\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \right)\sqrt{\frac{\rho }{{{\tau }_{w}}}}\ln \left( \frac{{{r}_{0}}}{30\nu }\sqrt{\frac{{{\tau }_{w}}}{\rho }} \right)$ (38)

The overall temperature change from the wall to the center of the tube can be obtained by adding eqs. (26), (31) and (38):

 ${{T}_{w}}-{{\bar{T}}_{c}}=\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}} \right)\sqrt{\frac{\rho }{{{\tau }_{w}}}}\left[ 2.5\ln \left( \frac{{{r}_{0}}}{30\nu }\sqrt{\frac{{{\tau }_{w}}}{\rho }} \right)+5\ln (5\Pr +1)+5\Pr \right]$ (39)

It follows from the definition of friction factor, ${{c}_{f}}=\frac{{{\tau }_{w}}}{\rho \bar{u}_{m}^{2}/2}$, that

 ${{\tau }_{w}}=\frac{1}{2}{{c}_{f}}\rho \bar{u}_{m}^{2}$ (40)

Substituting eq. (40) into eq. (39) and considering the definition of Reynolds number, ${{\operatorname{Re}}_{D}}={{\bar{u}}_{m}}D/\nu$, eq. (39) becomes:

 ${{T}_{w}}-{{\bar{T}}_{c}}=\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}{{{\bar{u}}}_{m}}} \right)\sqrt{\frac{2}{{{c}_{f}}}}\left[ 2.5\ln \left( \frac{{{\operatorname{Re}}_{D}}}{60}\sqrt{\frac{{{c}_{f}}}{2}} \right)+5\ln (5\Pr +1)+5\Pr \right]$ (41)

In order to obtain the heat transfer coefficient, $h={{{q}''}_{w}}/({{T}_{w}}-{{\bar{T}}_{m}})$, the temperature difference ${{T}_{w}}-{{\bar{T}}_{m}}$ must be obtained. If the velocity profile can be approximated by $\frac{{\bar{u}}}{{{{\bar{u}}}_{c}}}={{\left( \frac{y}{{{r}_{o}}} \right)}^{1/7}}$, and the temperature and velocity can also be approximated by the one-seventh law, i.e.,

 $\frac{{{T}_{w}}-\bar{T}}{{{T}_{w}}-{{{\bar{T}}}_{c}}}={{\left( \frac{y}{{{r}_{o}}} \right)}^{1/7}}{{,}_{_{\text{ }}}}\frac{{\bar{u}}}{{{{\bar{u}}}_{c}}}={{\left( \frac{y}{{{r}_{o}}} \right)}^{1/7}}$ (42)

it follows that

 ${{T}_{w}}-{{\bar{T}}_{m}}=\frac{\int_{0}^{{{r}_{o}}}{\bar{u}({{T}_{w}}-\bar{T})2\pi rdr}}{\int_{0}^{{{r}_{o}}}{\bar{u}2\pi rdr}}=\frac{5}{6}({{T}_{w}}-{{\bar{T}}_{c}})$ (43)

Substituting eq. (41) into eq. (43) results in:

 ${{T}_{w}}-{{\bar{T}}_{m}}=\frac{5}{6}\left( \frac{{{{{q}''}}_{w}}}{\rho {{c}_{p}}{{{\bar{u}}}_{m}}} \right)\sqrt{\frac{2}{{{c}_{f}}}}\left[ 2.5\ln \left( \frac{{{\operatorname{Re}}_{D}}}{60}\sqrt{\frac{{{c}_{f}}}{2}} \right)+5\ln (5\Pr +1)+5\Pr \right]$ (44)

which can be rearranged to the following empirical correlation

 $\text{N}{{\text{u}}_{D}}=\frac{{{\operatorname{Re}}_{D}}\Pr \sqrt{\frac{{{c}_{f}}}{2}}}{\frac{5}{6}\left[ 2.5\ln \left( \frac{{{\operatorname{Re}}_{D}}}{60}\sqrt{\frac{{{c}_{f}}}{2}} \right)+5\ln (5\Pr +1)+5\Pr \right]}$ (45)

which can be used together with appropriate friction coefficient to obtain the Nusselt number.

References

1. Oosthuizen, P.H., and Naylor, D., 1999, Introduction to Convective Heat Transfer Analysis, WCB/McGraw-Hill, New York.