I’m facing a problem today; I hope someone experienced can give me some advice.
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Everyone is aware of the issue of pipe welding equivalence, right? Welding workload = Welding equivalent × Length of weld joint. I would like to know how to calculate the welding workload for thick-walled pipes This post was last edited by lxzzl on 2008-7-24 15:00]
Whether from a mathematical perspective or in terms of time required, the welding workload = welding equivalent × length of the weld joint × number of passes required to complete the welding (for example, if the entire weld is completed after 5 passes, then 5 should be multiplied by)
How is the total number of trips mentioned above determined? Is it related to the wall thickness? Is there also a conversion rule between it and the diameter?
When welding thick walls, it is not possible to complete the welding in just one pass; the number of passes required is determined by the level of stress involved. For example, when welding ship hulls, multiple passes are needed until the cross-section of the weld joint is filled and the welding requirements are met. The greater the force, the larger and deeper the groove becomes.
This is how the volume of pipeline work is calculated for many companies in Taiwan and abroad these days. If thick-walled pipes need to be taken into consideration when working on a project, some of them will be mentioned separately while others will be considered together; one has to assess the quantities on their own. After all, welding thick-walled pipelines also requires consideration of issues such as heat treatment, welding volume, and inspection. . . .
For areas that are not equipment areas, namely ductwork areas, the calculation can be done using the formula mentioned by LIANGZONGHUI; however, for equipment areas, using this method results in an overestimate! It can be calculated using an empirical formula: Welding workload in the equipment area = total length of pipelines × 0.127 (correction factor) × pipeline diameter + (number of elbows × pipeline diameter × 2) + (number of tees × pipeline diameter × 3) + (number of flanges × pipeline diameter) + (number of reducers × pipeline diameter × 2)
DIN (Dia inch), also known as welding equivalent, is referred to as dynes abroad. It means that a weld with a diameter of 1 inch counts as 1 welding equivalent (1 dyne); 10 welds with a 1-inch diameter total 10 dynes, and 2 welds with a 5-inch diameter also total 10 dynes. This method of calculation takes into account only the diameter of the welds and does not consider the influence of wall thickness, so it is only applicable to welds with a wall thickness of 8 millimeters or less. For values over 8 mm, a coefficient of 0.1 is added for every additional 2 mm.
Did everyone find this information in books? It’s quite a lot of knowledge. Are there any books that can introduce this topic?
The unit used to measure welding workload, also known as welding equivalent, is called dyne abroad. One weld with a diameter of 1 inch is considered 1 welding equivalent (1 dyne); 10 welds of 1 inch each equal 10 dynes, and 2 welds of 5 inches each also equal 10 dynes. This method of calculation takes into account only the diameter of the welds and does not consider the effect of wall thickness, so it is only applicable to welds with a wall thickness of 8 millimeters or less. During operation, a coefficient of 0.1 can be added for every additional 2 millimeters beyond 8 millimeters. The specific coefficient can also be determined precisely. 9# ailiy
It’s excellent and much more detailed; it really helped me understand the current method for calculating pipeline work quantities.
How exactly is addition and multiplication done in this case? For a 4-inch pipe with a wall thickness of 12 millimeters, should we first calculate the value for the portion below 8 millimeters? 4 inches is equivalent to 4 dynes, and the additional 4 millimeters would then be considered as 0.4 dynes. So in total, it’s 4.4 dynes? Thank you!