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?
Assumptions
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.
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.
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
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.
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.
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:
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.
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.
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
Q̇void ≈ (kf Av / lg) × (TI - TII)
Q̇contact ≈ (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:
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.
| 1. What is one-dimensional conduction in chemical engineering? | ![]() |
| 2. How is one-dimensional conduction different from multi-dimensional conduction? | ![]() |
| 3. What factors affect the rate of one-dimensional conduction? | ![]() |
| 4. How is one-dimensional conduction analyzed in chemical engineering? | ![]() |
| 5. What are some practical applications of one-dimensional conduction in chemical engineering? | ![]() |