Heat transfer coefficient calculation standards (solved)
Thread Content
This post was last edited by burtgood on 2010-3-8 21:11. When performing heat transfer calculations, the selection of reference surfaces is important; generally, the outer surface is used as the reference. So, is it the case that the surface with the greater heat transfer resistance is taken as the reference?
Does anyone know? I’ve been waiting for days with no reply.
It seems not; on the side with high heat transfer resistance, it is possible to find ways to improve it, such as by increasing the flow rate. Or fins, etc., can be used on the side with greater resistance.
First, analyze the heat transfer mechanism, calculate the thermal resistance for each section, and determine the overall heat transfer coefficient
The calculation of the heat transfer coefficient requires differentiation based on different conditions, such as fluids, gases, liquids, or gas-liquid mixtures. The general principle is that the side with weaker heat transfer capacity should have a larger heat exchange area; the nominal heat exchange area is usually indicated by the outer diameter of the tube involved in heat exchange.
1# burtgood No, it’s not. In your heat transfer calculation formula, isn’t there an item called “heat transfer area A”? If this A represents the external surface area, then when calculating the \"heat transfer coefficient K\", the external surface area is used as the basis; the heat transfer coefficient inside the tube, the fouling thermal resistance inside the tube, and the wall thermal resistance all need to be adjusted accordingly ； Similarly, if the heat transfer area refers to the internal surface area, then the external heat transfer coefficient, the external fouling thermal resistance, and the wall thermal resistance need to be adjusted accordingly ； If the heat transfer area is calculated based on the diameter of the heat exchange tube, then the remaining four terms, apart from the tube wall thermal resistance, need to be adjusted accordingly. However, when high calculation accuracy is not required, it can be ignored. Taking a tube with an outer diameter of 25 and a wall thickness of 2 as an example, its external surface area is only 19% larger than its internal surface area.