For Heat Tracing Design - Sample Heat Loss Calculation for 450 MM NB Pipe

Formula:

q = 2∏K x (Tm-Ta) / ln(D2/D1)

q = Heat loss per unit length of pipeline (W/mtr.)

Tm = Maintenance Temperature (deg. C)

Ta = Minimum Ambient Temperature (deg. C)

D1 = Diameter of pipe (mtr)

D2 = Diameter of pipe including Insulation thickness (mtr)

K1 = Thermal Conductivity of Insulation material evaluated at its mean temperature (W/m .deg.K)

Sample Calculation:

Pipe Size (OD) = 457.2 mm

Maintenance Temperature (Tm) = 40 deg. C

Min. Ambient Temperature (Ta) = -0.7 deg. C

Insulation Material = Mineral Wool

Thermal Conductivity of Insulation at mean temp. 65 deg. C (K) = 0.038 W/m. deg. C

Insulation Thickness (t) = 45 mm

Diameter of Pipe (D1) = OD of Pipe = 457.2 mm

Diameter of Pipe including Insulation thickness (D2) = D1+2t = 547.2 mm

Putting the above values in heat loss formula, we have

Heat loss/mtr of pipeline (q) = 54.08 Watts

Heat loss/mtr of pipeline with 10% design Margin = 59.5 Watts

Hence to compensate the required heat loss, one need to select the correct heat Tracing cable and design the system accordingly.