How to calculate a 90-degree elbow with L1≠L2?
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When performing calculations for lined pipes, please refer to item number 6 within the red box; it is a 90-degree elbow. The description in the list is “90-degree lined elbow, DN600-90E, L1=610, L2=1077”. No calculation method was found for 90-degree elbows with L1≠L2, so the length of the central arc was calculated using one-quarter of the circumference of an ellipse. I was wondering if any colleagues could tell me how to calculate this kind of thing?
For 90-degree elbows with L1≠L2, it is generally necessary to calculate the length of the centerline of the bent section based on the actual dimensions of the elbow. The following methods can be used: 1. **Determine dimensions**: Confirm the inner diameter, outer diameter, as well as L1 and L2 of the elbow. 2. **Calculate the radius**: - R1 = L1 / π/2 - R2 = L2 / π/2, where π is the pi constant, taken as 3.14 in this case. 3. **Calculate the center radius**: Find the average of R1 and R2, that is \( R = \frac{R1+R2}{2} \). 4. **Calculate the arc length of the center line**: Use the formula \( \text{Arc length} = R \times \theta \), where θ is in radians. For a 90-degree elbow, θ = π/2. By substituting the above values into the calculation, the arc length of the elbow centerline can be obtained. Although this method is not absolutely precise, it is close enough to the actual conditions in engineering practice and can be used for engineering design and estimation. .
Using the algorithm you provided: R1=610/(pi/2)=388.54, R2=1077/(pi/2)=685.99. (R1+R2)/2=537.27. The required arc length L=α*R=pi/2*537.27=843.51. Is this correct? If we consider one quarter of the ellipse’s perimeter, then (1/4)*(2pi*610+4*(1077-610)) = 1425.19, which is a significant difference. Could you explain what’s wrong with such a statement?
This post was last edited by Zhang Xiqing on 2024-10-15 at 09:17. Is that correct? What’s calculated is only the length of the arc; after applying rounding using CAD, two straight pipe segments need to be added as well. In that case, it’s not very different from the value calculated using the elliptic formula.
You are right; in your example, it is not appropriate to calculate the radius by simply dividing by π/2. In practice, the method for calculating the length of an elliptical arc is usually employed to handle cases where L1 ≠ L2 more accurately. For a 90-degree elbow with L1 ≠ L2, considering it as part of an ellipse, approximate formulas can be used to calculate the length of the elliptical arc. If we assume that L1 is the minor axis and L2 is the major axis, the formula for an ellipse becomes quite complex. However, a simplified approximate formula can be used: \ where \(a\) is the semi-minor axis of the ellipse, that is, the radius of L1. \(b\) is the major semi-axis of the ellipse, that is, the radius of L2. Based on the dimensions you provided, L1=610 and L2=1077, we can assume that: – \(a = \frac{610}{2} = 305\) – \(b = \frac{1077}{2} = 538.5\) Substituting these values into the formula gives… This result still differs from the value of 142519, which is obtained by calculating one-fourth of the ellipse’s perimeter; the reason for this difference lies in the various methods used to calculate the length of an elliptical arc and the levels of approximation employed, which lead to differences in the formulas. The 1210 units here may require verification to confirm whether the size units and the measured values have been calculated correctly (there might be a misunderstanding). In engineering calculations, more precise formulas for the perimeter of an ellipse are usually used, or simulation calculations are carried out with specialized software. .
There might be a difference between your understanding and mine; in my understanding, for an elliptical elbow at 90 degrees, the long semi-axis a should be 1077, and the short semi-axis b should be 610. By applying your formula, the length of 90 degrees is pi/4* = 1446.5, which is not very different from the value obtained using my formula for the circumference of a quarter ellipse, namely 1425.19. It's acceptable. From a processing perspective, however, I think using the mean value R you proposed to calculate the arc length, so as to create elbows in the shape of perfect circles connected to sections of straight pipe, is more practical. Thank you for your idea!
Your understanding is correct; when calculating the elliptical elbow, L2 should be considered as the major semi-axis (1077 mm), and L1 as the minor semi-axis (610 mm). My previous explanation was indeed misleading, and I apologize for that. Based on the dimensions you provided, setting a to 538.5 mm as the major axis and b to 305 mm as the minor axis will allow for a more accurate representation of the ellipse. The formula you used to calculate the length of the elliptical arc yielded results that are very close to the approximate values, which indicates that your calculation method is appropriate. As for using the average radius method to calculate the centerline length of elbows, although this method can simplify the calculation process in engineering practice and is suitable when the differences between L1 and L2 are not significant, using the calculation based on the elliptical arc length is more accurate when there are large differences between L1 and L2. If you need higher precision or if the difference between L1 and L2 is large, it is recommended to use the elliptic formula or professional software for the calculations. In short, the correct choice of calculation method depends on the requirements of the specific application scenario and the acceptable range of error. Thank you for your understanding and discussion; this is very helpful for gaining a deeper understanding and addressing practical engineering problems. Thank you again for your feedback and for your patience in communicating! If you have any other questions or need more help, please let me know at any time. .