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GB8175-87 Guidelines for Equipment and Thermal Insulation Design 5. Calculation and Determination of Thermal Insulation Thickness 5.1 For pipes and cylindrical equipment with an outer diameter greater than 1020 mm, the thickness of the thermal insulation layer can be calculated on a planar basis; The thickness of the insulation layer for the rest is calculated based on the cylindrical surface. 5.2 To reduce heat loss, the thickness of the insulation layer should be calculated using the economic thickness method. 5.2.1 When the heat price is low, the cost of insulation materials or construction is high, and the economically optimal thickness calculated using the formula turns out to be too small, resulting in heat loss exceeding the maximum allowable heat loss values specified in Tables 1 and 2 of GB4272, the insulation thickness should be recalculated as 80%-90% of the maximum allowable heat loss value stated in those tables. 5.2.2 For pipes with high heat transfer costs, insulation materials, or low construction costs, and those laid side by side, the support structure and overall economic benefits such as floor space should also be taken into consideration; in such cases, their thickness can be less than the economically optimal thickness. 5.3 Calculation of insulation layer thickness and heat loss 5.3.1 Formula for calculating the economic thickness of the insulation layer a. Planar formula (1): б=A1λ*t(T-Ta)/Pi/S]1/2-λ/a (1) Where б is the thickness of the insulation layer, in meters ; A1 —— a constant; calculated using the legal units of measurement in the People’s Republic of China, A1=1.8975*10-3 (when using metric units, A1=1.0-3) ; fn——Heat price, yuan/106kj (yuan/106kcal) ; λ —— Thermal conductivity of the thermal insulation material product; for soft materials, the thermal conductivity at the installation density should be used, in W/(m·K) ; t —— operating time in years, h ; T — Temperature of the outer surface of equipment and pipelines, K (°C) ; Ta——Ambient temperature, K (°C) ; Pi — Unit cost of the thermal insulation structure, yuan/m³ ; S — Annual loan amortization rate for investment in insulation projects, calculated on a compound interest basis: S = i(1+i)^n / n, % ; i——annual interest rate (compounded), % ; n — number of years for interest calculation, in years ; a — Heat transfer coefficient from the outer surface of the insulation layer to the atmosphere, W/(㎡*k). In the aforementioned national standard formula, (1) fn — how should the heat value be determined? (2) Pi — How is the unit cost of the thermal insulation structure calculated? (3) S — Annual loan amortization rate for insulation project investment: How is it calculated? How should it be valued if there is no loan?
According to Article 5.1 of the revision notes in the latest **standard GB50264-2013, ‘Design Code for Thermal Insulation of Industrial Equipment and Piping’, the new standard has removed the requirement that the calculation boundary line for equipment and piping be set at an outer diameter of 1000 mm.
Okay, thank you very much. Regarding the thermal insulation calculation for that cylinder, if there is no value for S – the annual loan amortization rate for investment in thermal insulation projects – how should it be calculated?