Shape Factor Example 2

1. Shape Factor Example 2 The shape factor can be modeled as a thermal resistance where Rt=1/kS. Therefore, it can be integrated into the electrical circuit analogy discussed earlier. As an example, let us place a thick insulation layer with a thickness of 10 cm around the pipe line. Now, determine the heat loss. (Note: the following calculation will be an approximation, since the addition of insulation the will cause the outer surface temperature of the insulation to be non-constant. Accordingly, the assumption of isothermal surfaces used in shape factors is not strictly valid. However, it is still a reasonable assumption if the temperature variation is not very large. ) If we accept this, we can model the heat transfer as two-step process. First, from the pipe through the insulator, followed by the second stage: from the outer surface of the insulator to the ground.

2. Shape Factor Example (cont.) Rsoil=1/kS Insulator through the soil to the ground D2: outer diameter Pipe through insulator D1: pipe diameter Tground Tpipe

3. Shape Factor Example (cont.) • The heat loss is significantly lower than that without the insulator (q=181.2 W) • Although the shape factor assumption is not exactly valid, but the approximation should be good enough for most applications. Especially in cases where only a first-order estimation is needed.