Sarthit Toolthaisong

1 / 65

# Sarthit Toolthaisong - PowerPoint PPT Presentation

Heat Transfer In Channels Flow. Sarthit Toolthaisong. 6.5 Channels with Uniform Surface Temperature. Sarthit Toolthaisong. We wish to determine the following:. 6.5 Channels with Uniform Surface Temperature. Sarthit Toolthaisong. Applying conservation of energy to the element dx.

I am the owner, or an agent authorized to act on behalf of the owner, of the copyrighted work described.

## PowerPoint Slideshow about ' Sarthit Toolthaisong' - benedict-davis

Download Policy: Content on the Website is provided to you AS IS for your information and personal use and may not be sold / licensed / shared on other websites without getting consent from its author.While downloading, if for some reason you are not able to download a presentation, the publisher may have deleted the file from their server.

- - - - - - - - - - - - - - - - - - - - - - - - - - E N D - - - - - - - - - - - - - - - - - - - - - - - - - -
Presentation Transcript

Heat Transfer In Channels Flow

Sarthit Toolthaisong

6.5 Channels with Uniform Surface Temperature

Sarthit Toolthaisong

We wish to determine the following:

6.5 Channels with Uniform Surface Temperature

Sarthit Toolthaisong

Applying conservation of energy to the element dx

Eq. (a) = Eq. (b), we get

6.5 Channels with Uniform Surface Temperature

Sarthit Toolthaisong

From the average heat transfer coefficient over the length x

We get

(d)

6.5 Channels with Uniform Surface Temperature

Sarthit Toolthaisong

Introducing (d) into (6.11) and solving the resulting equation for Tm(x)

Application of conservation of energy between the inlet of the channel and a section x gives

Application of Newton’s law of cooling gives the heat flux q”s(x) at location x gives

6.5 Channels with Uniform Surface Temperature

Sarthit Toolthaisong

Solution

For flow through a tube at uniform surface temperature, applying Eq.(6.13)

At the outlet of the heat section (x=L) and solving for L

Where

6.5 Channels with Uniform Surface Temperature

Sarthit Toolthaisong

The properties of air using at the mean temperature Tm(x)

Check the flow is laminar or turbulent

6.5 Channels with Uniform Surface Temperature

Sarthit Toolthaisong

Since the Reynolds number is smaller than 2300, the flow is laminar. Thus

The mass flow rate.

6.5 Channels with Uniform Surface Temperature

Sarthit Toolthaisong

The perimeter.

Finally, the length of tube

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

6.6.1 Scale Analysis

Equating Fourier’s law with Newton’s law

A scale for r is

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

From Eq. (6.18) applying thermal thickness of external flow

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

6.6.2 Basic Considerations for the Analytical Determination of Heat Flux, Heat Transfer Coefficient and Nusselt Number

(1) Fourier’s law and Newton’s law.

(6.21)

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

Substituting into (a)

(6.22)

We define h using Newton’s law of cooling

(6.23)

Combining (6.22) and (6.23)

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

Where

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

(2) The Energy Equation

The last term in Eq.(6.28) can be neglected for

where

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

Thus, under such conditions, Eq.(6.28) becomes

3) Mean (Bulk) Temperature, Tm

Where

6.6 Determination of Heat Transfer Coefficient h(x) and Nusselt Number NuD

Sarthit Toolthaisong

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

This section focuses on the fully developed region.

6.7.1 Definition of Fully Developed Temperature Profile

Far away from the entrance of a channel

We introduce a dimensionless temperature defined as

For fully developed is independent of x. That is

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

Thus.

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

6.7.2 Heat Transfer Coefficient and Nusselt Number

Equating Fourier’s with Newton’s law

Using Eq.(6.37) in the definition of the Nusselt number, give

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

For scale analysis of temperature gradient

Compared Eq.(6.19)

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

6.7.3 Fully Developed Region for Tubes at Uniform Surface flux

Application of Newton’s law of cooling gives

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

Using energy balance on element dx for detemine eq.(6.41)

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

Assume Cp and m constant

Substituting eq.(6.42) into (6.41)

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

For determine fluid temperature distribution T(r,x) and surface temperature Ts(x), from energy equation

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

The axial velocity for fully developed flow is

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

Substituting eq.(6.46) and (6.49) into (6.32a)

gives

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

Substituting T(r,x), Tm(x) and Ts(x) into eq.(6.33) gives

Differentiating (6.54) and substituting into (6.38) gives the Nusselt number

From scaling analysis

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

From eq.(6.44) and (6.50), we get

Substituting (6.51) into (6.49)

Surface temperature, by setting r=ro in (6.52)

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

6.7.4 Fully Developed Region for Tubes at Uniform Surface Temperature

By energy equation

- Neglecting axial conduction and dissipation

- vr = 0

Simplifies to

Boundary conditions

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

Using equation (6.36a) to eliminate

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

Applied boundary condition to Eq.(6.58)

6.7 Heat Transfer Coefficient in the Fully Developed Temperature Region

Sarthit Toolthaisong

6.7.5 Nusselt Number for Laminar Fully Developed Velocity and Temperature in Channels of Various Cross-Sections

Example 6.4: Maximum Surface Temperature in an Air Duct

Sarthit Toolthaisong

Solution

Temperature distribution for uniform heat flux, given by eq.(6.10)

Example 6.4: Maximum Surface Temperature in an Air Duct

Sarthit Toolthaisong

Using Energy conservation to determine L

Example 6.4: Maximum Surface Temperature in an Air Duct

Sarthit Toolthaisong

Laminar flow

From Table.6.2 for uniform heat flux

6.8 Thermal Entrance Region: Laminar Flow through Tubes

Sarthit Toolthaisong

6.8.1 Uniform Surface Temperature: Graetz Solution

Consider laminar flow in Fig. 6.8 Fluid enters a heated or cooled section with a fully developed velocity

We neglect axial conduction (Pe >100)

6.8 Thermal Entrance Region: Laminar Flow through Tubes

Sarthit Toolthaisong

Assume product solution as the form

6.8 Thermal Entrance Region: Laminar Flow through Tubes

Sarthit Toolthaisong

Substitution the solution of (b) and (c) into (a)

Where Cn is constant

6.8 Thermal Entrance Region: Laminar Flow through Tubes

Sarthit Toolthaisong

The surface heat flux is given by

Example 6.5 hot water heater

Sarthit Toolthaisong

Solution

For flow through the tube at uniform surface temperature, from Eq.(6.13)

Example 6.5 hot water heater

Sarthit Toolthaisong

Example 6.5 hot water heater

Sarthit Toolthaisong

Example 6.5 hot water heater

Sarthit Toolthaisong

Compute the thermal entrance length, from Eq.(6.6)

Example 6.5 hot water heater

Sarthit Toolthaisong

Compute the heat transfer coefficient

Example 6.5 hot water heater

Sarthit Toolthaisong

6.8 Thermal Entrance Region: Laminar Flow through Tubes

Sarthit Toolthaisong

6.8.2 Uniform Surface Heat Flux

6.8 Thermal Entrance Region: Laminar Flow through Tubes

Sarthit Toolthaisong

The solution for Nusselt number is

6.8 Thermal Entrance Region: Laminar Flow through Tubes

Sarthit Toolthaisong

The average Nusselt number is given by.