Chapter 3: Steady-State Conduction
3.1 Steady State and Transient State

If you heat a pan on a stove, it takes a while for the pan to heat up to cooking temperature, after which the temperature of the pan remains relatively constant. The latter state is called the steady state, where there is no temporal change in temperatures. When the system is still changing with time, it is in transient state. The rate of conduction through an object at steady-state is given by:

 
(Eq. 3.1)

where k is the conductivity of the material, A is the cross-sectional area through which the heat is conducting, and ΔT is the temperature difference between two surfaces separated by a distance Δx.

3.2 One-Dimensional Conduction

One-dimensional heat transfer refers to special cases where there is only one spatial variable – the temperature varies in one direction only. A model used often to calculate the heat transfer through a 1-D system is called the thermal circuit model. This model simplifies the analysis of heat conduction through composite materials.

In this model, each layer is replaced by an equivalent resistor called the thermal resistance. An analysis much like a circuit analysis follows. For conduction, the thermal resistance is expressed as:

 
(Eq. 3.2)

where L is the thickness of the layer, k is the thermal conductivity of the layer, and A is the cross-sectional area. When there is more than one layer in the composite, the total resistance of the circuit must be calculated. The total resistance for layers in series is simply the sum of the resistances:

 
(Eq. 3.3)

For resistors in parallel, the total resistance is given by:

 
(Eq. 3.4)

The convection at the surface must also be expressed as a resistor:

 
(Eq. 3.5)

Once the total resistance of a structure is found, the heat flow through the layers can be found by:

 
(Eq. 3.6)

where Tinitial and Tfinal refers to the temperatures at the two ends of the thermal circuit (analogous to voltage difference in an electrical circuit) and q is the heat flow through the circuit (current).

Example Problem

Consider a composite structure shown on below. Conductivities of the layer are: k1 = k3 = 10 W/mK, k2 = 16 W/mK, and k4 = 46 W/mK. The convection coefficient on the right side of the composite is 30 W/m2K. Calculate the total resistance and the heat flow through the composite.

First, draw the thermal circuit for the composite. The circuit must span between the two known temperatures; that is, T1 and T∞.

Next, the thermal resistances corresponding to each layer are calculated:

Similarly, R2 = 0.09, R3 = 0.15, and R4 = 0.36

To find the total resistance, an equivalent resistance for layers 1, 2, and 3 is found first. These three layers are combined in series:

The equivalent resistor R1,2,3 is in parallel with R4:

Finally, R1,2,3,4 is in series with R5. The total resistance of the circuit is:
  Rtotal = R1,2,3,4 + R5 = 0.46 ← total thermal resistance

The heat transfer through the composite is:

  = 173.9 W. ← heat flow through the composite

Click here for References

Go to Chapter 3 Problems

TOC     << | 1 | 2 | 3 | 4 | 5 | 6 | 7 | 8 | >>

HOME | GENERAL INFO | LECTURE NOTES | SYLLABUS | HOMEWORK | LAB INFO
Send questions and comments to Haruna Tada
©2002 Mechanical Engineering Department*Tufts University*Medford, MA 02155