HCBBS Forum (English)
Submit Chemical Projects / Find Solutions
Amplify Your Requirements on a Broader Chemical Platform *Engineering · Technology · Equipment · Solutions*
Submit Request

Seeking help from experts; need the algorithm for equivalent diameter

2021-07-20View Original

Thread Content

A tube ring has a circular cross-section; the inner diameter of the larger tube is 100 mm, and the outer diameter of the smaller tube is 60 mm. What is the equivalent diameter? Is there a formula?
Reply #22021-07-20
The equivalent diameter should be used for converting non-circular shapes, right?
Reply #32021-07-20
I’m not quite sure about the usefulness of this equivalent diameter, but it might be possible to convert it into an equivalent flow diameter using the total flow area… If it’s used for calculating reinforcement through openings, then this method cannot be applied
Reply #42021-07-20
This post was last edited by zjq1962 on 2021-7-20 at 10:12. The equivalent diameter of the annular channel: 4 * annular area / (wetting perimeter) = 40 mm. Using this method, the equivalent diameter of cross-sections with other shapes can be calculated.
Reply #52021-07-20
It’s mentioned in any book on heat exchangers; you can just pick any one. . . .
Reply #62021-07-21
Calculated using this formula, the equivalent diameter of the casing annulus is larger than the diameter of the outer tube, which doesn’t make sense.
Reply #72021-07-21
This post was last edited by wanlirn on 2021-7-21 11:27: (100^2-60^2)/60; I calculated the equivalent diameter to be 106.66. Is there anything wrong with that? That’s just how it’s specified. You also wrote up there 4 times the area of the ring divided by the perimeter length of the heat exchange surface; simplify it, and that’s the formula ; /(60*PI)
Reply #82021-07-21
The wetting perimeter of the annular gap is (Do+Di)π. My calculation method yields the same result as that obtained using the formula provided at the end of your illustration.
Reply #92021-07-22
I don’t know what you want to say. The formulas in the manual are very clear. If you have doubts about the manual, I don’t either; I just use it as it is

Submit a Project

**Looking for Chemical Technology, Equipment & Solutions?** No Registration Required Broader Platform Exposure | Global Chemical Service Provider Connections

Submit Request — Free Consultation

Disclaimer

This is an automated machine translation of the original thread. Some technical terms may have inaccuracies; the original text shall prevail. Click "View Original" at the top right to access the source page, which supports IP-based automatic real-time language translation. Please watch out for contact details and sales inducements to prevent fraud. All content and translations are for reference only, representing solely the poster's personal views. For enquiries, email service@hcbbs.com.