# Fully-developed flow and heat transfer

(Difference between revisions)
 Revision as of 03:47, 31 May 2010 (view source) (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…')← Older edit Revision as of 03:48, 31 May 2010 (view source)Newer edit → 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:${{T}_{m}}=\frac{\int_{A}^{{}}{uTdA}}{\int_{A}^{{}}{udA}}=\frac{2\int_{0}^{{{r}_{o}}}{\pi ruTdr}}{\pi r_{o}^{2}{{u}_{m}}}$ + Using the definition of mean temperature presented in the last section with the above profile for temperature, and assuming constant properties: + +
[itex]{{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: 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)