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Conduction One Dimensional - 3 - Heat Transfer - Mechanical Engineering

Lecture 5 - Conduction: One Dimensional, Heat Transfer

Illustration 2.3
Consider a composite wall containing 5-different materials as shown in the fig. 2.7. Calculate the rate of heat flow through the composite from the following data?

Conduction: One Dimensional - 3

Solution 2.3

Assumptions

  • Steady-state one-dimensional conduction.
  • Thermal conductivity of each material is constant over the temperature range of interest.
  • Heat loss from the edges (side surfaces) is negligible compared with heat flow in the primary direction.
  • Layers are in perfect thermal contact (no contact resistance included in this calculation).
  • Area normal to heat flow is 1 m2.

The geometry indicates that the height of the first layer is 4 m (h1 = h2 + h3), and the layers are arranged in series in the direction of heat flow. The composite may therefore be represented by an equivalent thermal circuit where each layer is modelled by a conduction resistance.

Solution 2.3

Use the following relation for conduction resistance of a plane layer:

R = L / (k A), where L is layer thickness, k the thermal conductivity and A the area (here A = 1 m2).

The total resistance for series layers is the sum of individual resistances:

Rtotal = Σ Ri = Σ (Li / (ki A)).

On calculating the equivalent resistance with the given data (note: thickness of layer 2 = thickness of layer 3, and thickness of layer 4 = thickness of layer 5, in the heat flow direction), we obtain the component resistances and their sum as shown below.

Solution 2.3
Solution 2.3
Solution 2.3

Fig. 2.7: Composite of illustration 2.3; (a) composite, (b) corresponding electrical circuit

Once the total thermal resistance is known, the steady heat flow rate through the composite is given by:

Q̇ = (Thot - Tcold) / Rtotal

Solution 2.3

The calculation sequence followed (each line is a logical step of the solution):

Write down the temperatures of the two external faces (Thot and Tcold) as given.

Write the conduction resistance for each layer using R = L/(kA) with A = 1 m2.

Use symmetry/relationships given in the problem (for example L2 = L3, L4 = L5) to simplify expressions and combine equal resistances where applicable.

Sum all resistances in series to obtain Rtotal.

Compute Q̇ using Q̇ = (Thot - Tcold)/Rtotal.

Report Q̇ in appropriate units (W) since A = 1 m2 and lengths and conductivities are in SI units.

2.2 Thermal contact resistance

In the preceding analysis the interfaces between layers were assumed to be perfect: that is, temperatures at the interface were taken identical on both sides. In practical assemblies this assumption often fails because real solid surfaces are rough. When two solids are pressed together the actual contact occurs only at discrete asperity points and interstitial voids remain. These voids are usually filled by air or another fluid. The heat transfer across such an interface therefore occurs by conduction through the microscopic solid contact spots and by conduction/convection (or radiation at high temperatures) through the trapped fluid.

2.2 Thermal contact resistance

Because of these effects an apparent temperature drop may appear across the interface even though the macroscopic surfaces appear to touch. If TI and TII denote the theoretical temperatures of the nominal interface on the two bodies, the thermal contact resistance Rc is defined by the relation

Rc = (TI - TII) / Q̇, where Q̇ is the heat flow across the joint.

The value of Rc depends on the materials in contact, surface roughness, contact pressure, the medium filling the voids, temperature, and any interstitial films (e.g., oxides, lubricants). Typical smooth metallic surfaces have roughness in the order of micrometres and measured Rc values are usually obtained from experiments. Several theoretical models exist that predict trends in Rc with changing parameters, but experimental data are commonly used for design.

Two main heat-transfer paths exist across a joint:

  • Conduction through the entrapped gas or fluid in the void spaces.
  • Solid-to-solid conduction at the microscopic contact spots.

Because the thermal conductivity of gases is low compared with solids, the primary contribution to the interfacial thermal resistance often comes from the fluid-filled voids rather than the contact spots, especially at low contact pressures.

2.2 Thermal contact resistance

If Av is the total void area in the joint and Ac the total actual contact area, the heat flow across the joint can be written as the sum of the contributions through voids and through contacts. Using lg for the characteristic thickness of the void region and kf for the thermal conductivity of the filling fluid, and assuming for evenly rough surfaces that lg/2 is an effective half-thickness allocated to each solid surface, the two contributions may be expressed schematically as follows.

2.2 Thermal contact resistance
2.2 Thermal contact resistance

In symbolic form, the total heat flow Q̇ across the interface may be represented by a sum of two conduction terms (one through the voids and one through the contact spots):

Q̇ = Q̇void + Q̇contact

void ≈ (kf Av / lg) × (TI - TII)

contact ≈ (ks Ac / (lg/2)) × (TI - TII), where ks is the solid thermal conductivity and an effective thickness of contact conduction is taken as lg/2 for each body when surfaces are evenly rough.

Combining these terms gives an effective interfacial conductance and therefore the contact resistance Rc. The proportions of heat carried by each path change with contact pressure, surface finish, and the thermal conductivity of the filling medium.

Practical implications and remedies:

  • Increasing contact pressure enlarges the real contact area Ac, thereby reducing Rc.
  • Using a thermally conductive filler (e.g., thermal grease, metallic foils, or solder) within voids reduces the resistance associated with Av.
  • Improving surface finish (reducing roughness) increases actual contact and lowers Rc.
  • At high temperatures, radiative transfer across gaps may become significant and must be included in the interface heat-transfer model.

Understanding and accounting for thermal contact resistance is essential in accurate prediction of heat flow in multi-layer systems, heat exchanger joints, electronic device interfaces, and many engineering assemblies where interfaces cannot be assumed ideal.

The document Conduction: One Dimensional - 3 is a part of the Mechanical Engineering Course Heat Transfer.
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FAQs on Conduction: One Dimensional - 3

1. What is one-dimensional conduction in chemical engineering?
Ans. One-dimensional conduction in chemical engineering refers to the transfer of heat within a material in a single direction. It occurs when there is a temperature gradient along a straight line, and the heat flows parallel to that line. This concept is commonly used to analyze heat transfer in pipes, rods, and other cylindrical or planar geometries.
2. How is one-dimensional conduction different from multi-dimensional conduction?
Ans. One-dimensional conduction is characterized by heat transfer occurring only in one direction, whereas multi-dimensional conduction involves heat transfer in multiple directions. In one-dimensional conduction, temperature gradients exist only along a single line, while in multi-dimensional conduction, temperature gradients can exist in multiple directions, such as radial or axial directions.
3. What factors affect the rate of one-dimensional conduction?
Ans. The rate of one-dimensional conduction is influenced by several factors, including the thermal conductivity of the material, the temperature difference across the material, the thickness of the material, and the surface area through which heat is being transferred. Additionally, the presence of any insulation or boundary conditions can also affect the rate of conduction.
4. How is one-dimensional conduction analyzed in chemical engineering?
Ans. In chemical engineering, one-dimensional conduction is often analyzed using Fourier's law of heat conduction. This law relates the rate of heat transfer through a material to the temperature gradient and the material's thermal conductivity. By solving the one-dimensional heat conduction equation, which is a partial differential equation, engineers can determine the temperature distribution and the heat transfer rate within the material.
5. What are some practical applications of one-dimensional conduction in chemical engineering?
Ans. One-dimensional conduction has several practical applications in chemical engineering. It is commonly used to analyze heat transfer in pipes, heat exchangers, and reactors. For example, understanding one-dimensional conduction is crucial for designing efficient heat exchangers that maximize heat transfer between two fluids. It is also important in the design of insulation materials to minimize heat loss in industrial processes.
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