# Sample Problem – Insulation thickness calculation for a pipe

### Sample Problem Statement

Calculate insulation thickness (minimum value) required for a pipe carrying steam at 1800C. The pipe size is 8" and the maximum allowable temperature of outer wall of insulation is 500C. Thermal conductivity of the insulation material for the temperature range of the pipe can be taken as 0.04 W/m·K. The heat loss from steam per meter of pipe length has to be limited to 80 W/m.

### Solution Solution to this sample problem is quite straightforward as demonstrated below.

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 = 500C
T2 = 1800C
r1 = 8" = 8 × 0.0254 m = 0.2032 m
k = 0.04 W/m·K
N = length of the cylinder

Q/N = Heat loss per unit length of pipe
Q/N = 80 W/m

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

80 = 2π × 0.04 × (180-50) ÷ ln(r2/0.2032)

ln(r2/0.2032) = 2π × 0.04 × (180-50) / 80 = 0.4084

Hence, r2/= r1 × e0.4084
r2/= 0.2032 × 1.5044 = 0.3057 m

Hence, insulation thickness = r2 - r1
thickness = 305.7 - 203.2 = 102.5 mm

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 500C. 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.