Problems with the calculation method for pipe insulation thickness
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The insulation thickness of the pipeline is calculated in accordance with GB50264; the total heat transfer coefficient αs between the outer layer of the pipeline and the air is lower than that obtained using other standards, which results in a tendency for the surface temperature to be on the high side. However, its calculation formula takes into account the impact of radiation and wind speed on heat transfer, so it seems more reliable. The calculation formula in other standards is 11.6 + 7*v^0.5; it does take wind speed into account, making it easy to calculate, but it seems less accurate. Now, there is a significant difference between the heat transfer coefficient values obtained from these two calculation methods, which results in a large difference in the temperature on the outer surface of the pipe’s insulation layer. I’m not sure what to do; any advice would be appreciated
Rules for calculating the volume of insulation work: 1. For insulation layers, different insulation materials should be considered separately; unless otherwise specified, the volume is calculated based on the actual thickness specified in the design. 2. The thickness of an insulation layer is calculated as the net thickness of the insulating material (excluding bonding materials). 3. For insulation layers on roofs and floors, the volume is calculated by multiplying the net area between the walls of the enclosure structure by the designed thickness, without deducting the volume occupied by columns or pilasters. 4. When insulation material is laid under concrete ceiling slabs, the volume is calculated based on the actual thickness specified in the design. When insulation material is placed on the surface of the ceiling slab, the volume is calculated based on the actual area covered. 5. For wall insulation layers, for interior walls, the volume is calculated by multiplying the net length of the insulation layer by the height and thickness indicated in the design, with the volume occupied by openings for refrigeration doors and pipes being deducted. For exterior wall insulation, the volume is calculated based on the actual exposed area. 6. For insulation layers around columns, the volume is calculated by multiplying the extended length along the centerline of the column’s insulation layer by the height and thickness indicated in the design. 7. Other types of insulation: (1) For insulation layers in tanks and basins, the volume is calculated based on the length, width, and thickness of these layers as indicated in the design. The walls of the tank are treated as wall surfaces, while the bottom of the tank is treated as a floor surface. (2) The insulation material surrounding the sides of door openings is calculated based on the dimensions of the insulation layer indicated in the design, and this volume is included in the total amount for wall insulation work. (3) The insulation layer on column caps is included in the total amount for ceiling insulation work, based on its volume as indicated in the design. (4) For the insulation layer on the inner surface of chimneys, the volume is calculated based on the area of the inner surface of the chimney, with the area occupied by various openings being deducted. (5) Insulated exhaust pipes are calculated based on their length as indicated in the design, without deducting the length occupied by pipe fittings. Insulated exhaust vents are counted individually, depending on the material used.
Rules for calculating the volume of insulation work: 1. For insulation layers, different insulation materials should be considered separately; unless otherwise specified, the volume is calculated based on the actual thickness specified in the design. 2. The thickness of an insulation layer is calculated as the net thickness of the insulating material (excluding bonding materials). 3. For insulation layers on roofs and floors, the volume is calculated by multiplying the net area between the walls of the enclosure structure by the designed thickness, without deducting the volume occupied by columns or pilasters. 4. When insulation material is laid under concrete ceiling slabs, the volume is calculated based on the actual thickness specified in the design. When insulation material is placed on the surface of the ceiling slab, the volume is calculated based on the actual area covered. 5. For wall insulation layers, for interior walls, the volume is calculated by multiplying the net length of the insulation layer by the height and thickness indicated in the design, with the volume occupied by openings for refrigeration doors and pipes being deducted. For exterior wall insulation, the volume is calculated based on the actual exposed area. 6. For insulation layers around columns, the volume is calculated by multiplying the extended length along the centerline of the column’s insulation layer by the height and thickness indicated in the design. 7. Other types of insulation: (1) For insulation layers in tanks and basins, the volume is calculated based on the length, width, and thickness of these layers as indicated in the design. The walls of the tank are treated as wall surfaces, while the bottom of the tank is treated as a floor surface. (2) The insulation material surrounding the sides of door openings is calculated based on the dimensions of the insulation layer indicated in the design, and this volume is included in the total amount for wall insulation work. (3) The insulation layer on column caps is included in the total amount for ceiling insulation work, based on its volume as indicated in the design. (4) For the insulation layer on the inner surface of chimneys, the volume is calculated based on the area of the inner surface of the chimney, with the area occupied by various openings being deducted. (5) Insulated exhaust pipes are calculated based on their length as indicated in the design, without deducting the length occupied by pipe fittings. Insulated exhaust vents are counted individually, depending on the material used.
0082 represents: (diameter + insulation thickness * 2); D2 – diameter of the heat tracing pipe; (10~20mm) – gap between the main pipe and the heat tracing pipe. ② Double-pipe heat tracing (when the pipe diameters are the same and the angle between them is greater than 90°). D′=D1+1.1 – adjustment coefficient. 033δ*1.5*N; S=2 *π*1.5*N. (4) Insulation for valves: Thickness of the insulation layer after adjustment (which can be considered as width). π*(D+1.033δ)*1.033δ represents this value. 0.033δ)*1. D′=D1 +D2 +(10~20mm). Here, D′ represents the comprehensive value of the heat tracing pipe; L – length of the equipment cylinder or pipe; 0.0082 – diameter of the binding wire or thickness of the steel strip. 5D*1.05*N. (6) Formulas for calculating insulation, moisture protection, and protective layers for elbows. 0.033; δ – thickness of the insulation layer. 1δ+0.0082)*L. Personal understanding: D+2.1δ+0.033δ*1.05*N; S=π(D+2.1δ)*2.5D*1.5D2 +(10~20mm). ③ Double-pipe heat tracing (when the pipe diameters are different): Length*width; S=π*(D+2). Substitute the calculated value of D′ into the corresponding formulas to determine the values for insulation, moisture protection, and protective layers of the heat tracing pipe. V=2πr*(h+1), formulas for calculating insulation, moisture protection, and protective layers. V=π(D+1.033δ)*1.5D*1. V=π(D+1.033δ)*1.5D*2π*1.033δ* N/B; S=π*(D+2), 2.5D*2π*N/B. (7) Insulation for dome-shaped tank heads. V=2π*, formulas for calculating insulation, moisture protection, and protective layers. V=π(D+1.033δ)*2.5D*1; D1 – diameter of the main pipe. 1.033δ represents this value. When the angle between the pipes is less than 90°, D′=D1+D2 +(10~20mm). Here, D′ represents the comprehensive value of the heat tracing pipe. 1.033δ*1.05*N. (5) Insulation for flanges. 1.033δ*1.05*N; S=π*(D+2), quantities required for insulation, moisture protection, and protective layers. (3) Formulas for calculating quantities required for insulation, moisture protection, and protective layers for equipment heads. Quantities required for insulation work. (1) Formulas for calculating insulation, moisture protection, and protective layers for equipment cylinders or pipes; D1 – diameter of the main pipe: V=π*(D+1.033δ)*1.033δ. Personal understanding of the meaning of these volume formulas: D+1.033δ represents the length from the center of one insulation layer to the center of another, plus the thickness of each binding wire (0.033δ) = adjusted length of the center line of the insulation layer. π*(D+1.033δ) represents the circumference of the circle formed by the centers of the insulation layers (which can be considered as a length). Relevant standards can be referred to for design and calculation: SH3010-2000, Code for Thermal Insulation Design of Petrochemical Equipment and Pipes
Choose the appropriate standard based on the type of piping system. The derivation of the thermal conductivity and convective heat transfer coefficient is based on empirical data.
Regarding the calculation methods, they are all outlined in \"Principles of Chemical Engineering.\" However, in practice, no calculations are necessary; choices can be made based on empirical values. There are standards for the thickness of insulation layers.
Personally, I believe that for the calculation of the external surface heat transfer coefficient as, using the formulas provided in GB50264-2013 \"Code for Design of Thermal Insulation of Industrial Equipment and Piping\" yields more comprehensive and detailed results. You can adopt the values specified in GB50264-2013 for the following reasons: (1) GB50264-2013 is a **standard, and it represents the most up-to-date version. GB/T4272-2008 \"General Rules for Insulation Technology of Equipment and Piping\" and GB/T8175-2008 \"Guidelines for Insulation Design of Equipment and Piping\" are both old standards, while SH/T3010-2013 \"Design Code for Insulation Engineering of Petrochemical Equipment and Piping\" is an industry standard ； (2) Taking all the above standards into account, only standard GB50264-2013 performs separate calculations for different operating conditions regarding the value of the heat transfer coefficient as at the surface of the insulation structure in its insulation calculations; the other standards generally provide very simplified formulas for determining the value of this heat transfer coefficient as. In fact, upon a careful comparison of the above standards, it can be seen that not only is the value assigned to the overall heat transfer coefficient as more detailed and conservative in the GB50264-2013 standard, but it is also more conservative in many other aspects as well.
That’s true; we don’t need to do any calculations here – we just rely on experience………