Fully-developed flow and heat transfer

From Thermal-FluidsPedia

(Difference between revisions)
Jump to: navigation, search
(Created page with 'In this section, we consider the case of fully developed laminar flow and constant properties in a circular tube with a fully developed temperature and concentration profiles. We…')
Line 35: Line 35:
|{{EquationRef|(4)}}
|{{EquationRef|(4)}}
|}
|}
-
Using the definition of mean temperature presented in the last section with the above profile for temperature, and assuming constant properties:<math>{{T}_{m}}=\frac{\int_{A}^{{}}{uTdA}}{\int_{A}^{{}}{udA}}=\frac{2\int_{0}^{{{r}_{o}}}{\pi ruTdr}}{\pi r_{o}^{2}{{u}_{m}}}</math>
+
Using the definition of mean temperature presented in the last section with the above profile for temperature, and assuming constant properties:
 +
 
 +
<center><math>{{T}_{m}}=\frac{\int_{A}^{{}}{uTdA}}{\int_{A}^{{}}{udA}}=\frac{2\int_{0}^{{{r}_{o}}}{\pi ruTdr}}{\pi r_{o}^{2}{{u}_{m}}}</math></center>
Substituting eq. (5.39) into the above expression yields:
Substituting eq. (5.39) into the above expression yields:

Revision as of 03:48, 31 May 2010

In this section, we consider the case of fully developed laminar flow and constant properties in a circular tube with a fully developed temperature and concentration profiles. We first consider the case of constant heat rate per unit surface area for steady, laminar, fully developed flow. The energy equation in a circular tube, by neglecting axial heat conduction and viscous dissipation terms, is:

u\frac{\partial T}{\partial x}=\alpha \left[ \frac{1}{r}\frac{\partial }{\partial r}\left( r\frac{\partial T}{\partial r} \right) \right]

(1)

For a fully developed flow with constant wall heat flux, eq. (5.20) can be substituted into eq. (5.36) to obtain

u\frac{d{{T}_{m}}}{dx}=\alpha \left[ \frac{1}{r}\frac{\partial }{\partial r}\left( r\frac{\partial T}{\partial r} \right) \right]

(2)

The boundary conditions are

\begin{align}& -k\frac{\partial T}{\partial r}={{{{q}''}}_{w}}\begin{matrix}{} & {}  \\\end{matrix}\text{at}\ r={{r}_{o}} \\  & \frac{\partial T}{\partial r}=0\begin{matrix}   {} & {} & {}  \\\end{matrix}\text{at}\ r=0 \\ \end{align}

(3)

Integrating eq. (5.37) twice and applying the boundary conditions in eq. (5.38) to get the temperature distribution gives us

T={{T}_{w}}-\frac{2{{u}_{m}}}{\alpha }\frac{d{{T}_{m}}}{dx}\left( \frac{3r_{o}^{2}}{16}-\frac{{{r}^{2}}}{4}+\frac{{{r}^{4}}}{16r_{o}^{2}} \right)

(4)

Using the definition of mean temperature presented in the last section with the above profile for temperature, and assuming constant properties:

{{T}_{m}}=\frac{\int_{A}^{{}}{uTdA}}{\int_{A}^{{}}{udA}}=\frac{2\int_{0}^{{{r}_{o}}}{\pi ruTdr}}{\pi r_{o}^{2}{{u}_{m}}}

Substituting eq. (5.39) into the above expression yields:

{{T}_{m}}={{T}_{w}}-\frac{11}{96}\left( \frac{2{{u}_{m}}}{\alpha } \right)\frac{d{{T}_{m}}}{dx}r_{o}^{2}

(5)

The heat flux at the wall can be obtained using the above relation for Tm

{{q}_{w}}^{\prime \prime }=h\left( {{T}_{w}}-{{T}_{m}} \right)=h\left( \frac{11}{96} \right)\left( \frac{2{{u}_{m}}}{\alpha } \right)\left( \frac{d{{T}_{m}}}{dx} \right)r_{o}^{2}

(6)

The heat flux at the wall can also be calculated using eq. (5.39) for the temperature profile and Fourier’s law of heat conduction

{{{q}''}_{w}}={{\left. -k\frac{\partial T}{\partial r} \right|}_{r={{r}_{o}}}}=\rho {{c}_{p}}{{r}_{o}}\left( \frac{{{u}_{m}}}{2} \right)\left( \frac{d{{T}_{m}}}{dx} \right)

(7)

Combining eqs. (5.41) and (5.42) and solving for the heat transfer coefficient, h, yields h = 4.364 k / D or in terms of the Nusselt number,

Nu = 4.364

(8)