Electric heat tracing calculation (is it correct?)

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Calculation problem: Given that the pipe diameter is Φ711×10, the density of crude oil is 0.98, the steel grade used is 20#, the insulation thickness is δ=60 mm, and the lowest ambient temperature is -16.9°, calculate the power in watts required per meter of pipeline to raise the temperature from 0° to 35° within 36 hours Is the following calculation correct? 1. Theoretical heat loss of the φ711 pipeline in the tank area: 1.15×2πK(Tm – Ta). 1.15×2π×0.035×Q1 = 84.2 W/m. The actual heat loss, Q2, is calculated as Q2 = C1×C2×Q1 = 1.05×1×84.2 = 88.4 W/m.
Q1: Theoretical heat loss per meter.
Q2: Actual heat loss per meter.
K: Thermal conductivity of the insulation material; the thermal conductivity of rock wool is 0.035 W/m·°C.
Tm: Temperature that needs to be maintained, 35°C.
Ta: Lowest ambient temperature, -16.9°C.
δ: Thickness of the insulation layer, 60 mm.
C1: Environmental correction factor, taken as 1.05.
C2: Pipeline material correction factor, taken as 1.

2. Calculation of the thawing power for the φ711 pipeline in the tank area:
1) Calculation of the heating power for the φ711 steel pipe:
P1 = 1.1×C1×m1×△t = 1.1×0.47×149.7×(35–0) = 752.5 W/m. This value is calculated over a time period of 3.6×h; since h = 1, P1 = 752.5 W/m.
2) Calculation of the thawing power for the φ711 crude oil:
P2 = 1.1×C2×m2×△t = 1.1×2.2×309.9×(35–0)×2/3 = 4861 W/m. Again, this is calculated over a time period of 3.6×h; with h = 1, P2 = 4861 W/m.
P_total = P1 + P2 + 2/3×heat loss due to the pipeline = 752.5 + 4861 + 56.1 = 5669.6 W/m.
The calculations are based on principles from engineering thermodynamics, verified through years of practical experience, as well as methods used by leading international companies. 3. Thawing time required for the φ711 crude oil pipeline: For each meter of this pipeline, 6 units of 15DFI-BS elements are used in series with heating elements; thus, the heating power per meter of pipeline is: 15 W/m × 6 = 90 W. The time required to thaw one meter of pipeline is: 5669.6 ÷ 90 = 63 hours. 4. If the thawing time is considered over a 24-hour period, the required power is: 5669.6 ÷ 24 = 236.2 W/m. Therefore, the heating power needed per meter of pipeline for a 24-hour thawing process is 236.2 W/m. 5. The power required for thawing over 36 hours is calculated as follows: 5669.6 ÷ 36 = 157.5 W/m. Thus, the heating power needed to thaw each meter of pipeline over 36 hours is 157.5 W/m.