Insulation Thickness Calculation for Pipe

 Sample Problem Statement

Calculate insulation thickness (minimum value) required for a pipe carrying temperature at 2000C. The pipe size is 10" and the maximum allowable temperature of outer wall of insulation is 600C. Thermal conductivity of the insulation material for the temperature range of the pipe can be taken as 0.046 W/m·C. The heat loss from steam per meter of pipe length has to be limited to 100 W/m.

Solution:-

For radial heat transfer by conduction across a cylindrical wall, the heat transfer rate is expressed by following equation,

For the given sample problem,

T1 = 600C
T2 = 2000C
r1 = 10" = 10 × 0.0254 m = 0.254 m
k = 0.046 W/m·C
N = length of the cylinder

Q = Heat loss

Q/N = Heat loss per unit length of pipe

Q/N = 100 W/m

Hence, inserting the given numbers in the radial heat transfer rate equation from above,

100 = 2π × 0.046 × (200-60) ÷ ln(r2/0.254)

ln(r2/0.254) = 2π × 0.046 × (200-60) / 100 = 0.4046

Hence, r2= r1 × e0.4046
r2= 0.254 × 1.4987 = 0.3806 m

Hence, insulation thickness = r2 - r1
thickness = 0.3806 - 0.254 = 0.1266 m or 126.6mm

Some margin should be taken on the insulation thickness because if the conductive heat transfer rate happens to be higher than the convective heat transfer rate outside the insulation wall, the outer insulation wall temperature will shoot up to higher values than 600C. Hence conductive heat transfer rate should be limited to lower values than estimates used in this sample problem. The purpose of this sample problem is to demonstrate radial heat conduction calculations and practical calculations of insulation thickness also require consideration of convective heat transfer on the outside of insulation wall.