How to calculate the heat loss of insulated pipes

The heat loss of insulated pipes can be estimated using the following formula: $$ Q = 2 \pi L (T_{inside\ pipe} - T_{environment}) / \ln(\frac{r_2}{r_1}) / k $$ Where: - \(Q\) represents the heat loss per hour, with units typically being watts (W) or kilowatts (kW). - \(L\) is the length of the pipe, in meters (m). - \(T_{in-tube} is the temperature of the fluid inside the tube, measured in degrees Celsius (°C). - \(T_{environment} is the environmental temperature, measured in degrees Celsius (°C). - \\(r_1\) is half of the outer diameter of the pipe, in meters (m). - \\(r_2\) is half of the outer diameter of the insulation material, that is, the outer diameter of the pipe plus half of the thickness of the insulation material, measured in meters (m). - \(k\) is the thermal conductivity of the insulation material, with the unit of watts per meter per degree Celsius (W/m·°C). Regarding your specific question, assuming that the thermal conductivity of rock wool is approximately 0.035 W/m·°C, the outer diameter of a DN80 pipe is about 89 mm or 0.089 m, and the thickness of the insulation material is 50 mm or 0.05 m, we can calculate as follows: 1. \(r_1 = \frac{0.089}{2} = 0.0445\) m 2. \(r_2 = \frac{0.089}{2} + 0.05 = 0.0945\) m 3. \(T_{inside\ pipe} = 95\)°C 4. \(T_{environment} = 25\)°C 5. \(L = 35\) m 6. \(k = 0.035\) W/m·°C Substituting these values into the formula gives: $ Q = 2 \pi \times 35 \times (95 - 25) / \ln(\frac{0.0945}{0.0445}) / 0.035 $ Let’s calculate the actual value of this expression. from math import pi, log # Define variables L = 35 # Pipe length, in meters T_pipe = 95 # Temperature inside the pipe, in °C T_env = 25 # Ambient temperature, in °C r1 = 0.0445 # Half of the outer diameter of the pipe, in meters r2 = 0.0945 # Half of the outer diameter of the insulation material, in meters k = 0.035 # Thermal conductivity of rock wool, in W/m·°C # Calculate the heat dissipation amount Q Q = 2 * pi * L * (T_pipe - T_env) / log(r2 / r1) / k The result is: Was the calculation successful? 584008.4378455528 The heat dissipation rate is approximately 584008 watts (W), or about 584 kilowatts (kW) per hour. This calculation is an estimate; the actual heat dissipation may vary depending on factors such as the installation method and the actual performance of the insulation material. .

This post was last edited by fmch6605 on 2024-4-17 08:46. Thank you very much for your reply, but I didn’t quite understand it. Could you please list the calculation steps in detail? How did I not come up with this result? Q=2*35*(95-25)/log(r2/r1)/k=70*70/log(0.0945/0.0445)/0.035=428134.56; I’m not sure if this is correct. Also, based on the context, should it be ln or log in the formula? I earnestly request a reply. Thank you!

This requires a design institute to perform calculations in order to produce a calculation report

I don’t need a calculation book; I just need to know the formulas and how to use them. Am I doing this correctly? Q=2*35*(95-25)/log(r2/r1)/k=70*70/log(0.0945/0.0445)/0.035=428134.56, thank you!