Conductive Heat Transfer Case study

Conductive Heat Transfer Case study

Conduction involves the transfer of heat by the interaction between adjacent molecules of a material. Heat transfer by conduction is dependent upon the driving "force" of temperature difference and the resistance to heat transfer. The resistance to heat transfer is dependent upon the nature and dimensions of the heat transfer medium. All heat transfer problems involve the temperature difference,  

In conduction heat transfer, the most common means of correlation is through Fourier’s Law of Conduction. The law, in its equation form, is used most often in its rectangular or cylindrical form (pipes and cylinders), both of which are presented below.

Rectangular     Q = K  A  ( ∆T / ∆X)

FIG 1

Cylindrical     Q = K A  ( ∆T / ∆r )

where:
Q= rate of heat transfer (Btu/hr) 
A = cross-sectional area of heat transfer (ft²)
∆x = thickness of slab (ft)
∆r = thickness of cylindrical wall (ft)
∆T = temperature difference (°F)

k = thermal conductivity of slab (Btu/ft-hr-°F)

Equivalent Resistance

It is possible to compare heat transfer to current flow in electrical circuits. The heat transfer rate may be considered as a current flow and the combination of thermal conductivity, thickness of material, and area as a resistance to this flow. The temperature difference is the potential or driving function for the heat flow,

FIG 2

As shown in figure 2 resulting in the Fourier equation being written in a form similar to Ohm’s Law of Electrical Circuit Theory. If the thermal resistance term ∆x/k is written as a resistance term where the resistance is the reciprocal of the thermal conductivity divided by the thickness of the material

R =  ∆x/k  

WHERE

R = Thermal Resistance (∆x/k) (hr-ft²-°F/Btu)

∆x = thickness of slab (ft)
k = thermal conductivity of slab (Btu/ft-hr-°F)

Equivalent Resistance

Rth = ∆xA/kA  + ∆xB/kB + ∆xC/kC 

Heat Flux

Q" = Q/A = ∆T/ Rth

WHERE

Q"= Heat Flux (Q /A) (Btu/hr-ft²)

∆T = Temperature Difference (°F)

Rth = Thermal Resistance (∆x/k) (hr-ft²-°F/Btu)

Conduction-Cylindrical

Across a cylindrical wall, the heat transfer surface area is continually increasing or decreasing. Figure 3 is a cross-sectional view of a pipe

FIG 3

The surface area (A) for transferring heat through the pipe (neglecting the pipe ends) is directly proportional to the radius (r) of the pipe and the length (L) of the pipe.

A = 2𝝿 r L

As the radius increases from the inner wall to the outer wall, the heat transfer area increases.The development of an equation evaluating heat transfer through an object with cylindrical geometry begins with Fourier’s law Equation

Q = K A  ( ∆T / ∆r )

so that  the heat transfer rate for cylindrical geometries  

where:
L = length of pipe (ft)
ri = inside pipe radius (ft)

ro = outside pipe radius (ft)



Composite Cylindrical Layers

Conduction Heat Transfer Summary
• Conduction heat transfer is the transfer of thermal energy by interactions between adjacent molecules of a material.
• Fourier’s Law of Conduction can be used to solve for rectangular and cylindrical coordinate problems.
• Heat flux (Q" ) is the heat transfer rate (Q ) divided by the area (A). ˙
• Heat conductance problems can be solved using equivalent resistance formulas analogous to electrical circuit problems.