1
\$\begingroup\$

Before anything: I am not an electrical engineer: I don't know very much coaxial cables (just the very basics). I need to do some orders of magnitude calculation of heat transport of typical coaxial cable designed for cryogenic systems, over a range of temperatures. I could use the heat transport properties given by the documentations but the problem is that they are usually given for a fixed couple of temperatures at the both ends of the wire. I need to have arbitrary ending temperatures for my purpose.

What I need is then to use the properties of the materials of inner and outter conductor, the property of the dielectric in between, properties of the jacket. In addition to that I need the diameter of those components. Then I will simply use Fourier Heat law to find what I need.

But I have some troubles to understand aspects of documentation.

For example consider the following:

https://www.tek-stock.com/ut-085-ss-ss/

What are the inner and outter diameter of the jacket ? I don't find it. Is it because the jacket is not provided with the rest ?

Consider also this one:

https://www.lakeshore.com/products/categories/specification/temperature-products/cryogenic-accessories/cryogenic-cable

In "Jacket outer dimension", for type C, they give 0.7874 mm × 1.016 mm: what does that mean ? I would expect a diameter (the outer diameter of the jacket). Here they give me a surface as if it was a square.

For type SR the shield diameter and Jacket outer dimension are the same. What does that even mean ? The jacket must have a bigger diameter than the shield.

You see my point: it is probably very basic questions to understand how to read those documentations.

\$\endgroup\$
  • \$\begingroup\$ You should post a screengrab of the details into your question so we don't all have to follow two links to understand it and so that the question still makes sense when the links die. \$\endgroup\$ – Transistor May 6 at 19:26
  • \$\begingroup\$ Does this answer your question? Coaxial cable terminology : how to read? \$\endgroup\$ – Elliot Alderson May 6 at 20:59
1
\$\begingroup\$

Case 1

enter image description here

Figure 1. The jacket is stainless steel. There is no outer insulation. The inner diameter of the outer jacket would be the dielectric diameter.

Case 2

enter image description here

Figure 2. The drain wire makes the cable non-circular.

| improve this answer | |
\$\endgroup\$
  • \$\begingroup\$ I see, thank you. But in practice we should always put a jacket around right ? It is just that it has to be bought "separately" ? I guess the insulation would not be great without it ? Is it common to put coaxial cable without insulation around ? \$\endgroup\$ – StarBucK May 6 at 20:13
  • \$\begingroup\$ I don't do cryogenics but I'm familiar with thermocouples used in heating circuits but we would never bother to insulate the braided sheath because the heat loss would be insignificant in relation to the power we're putting in. \$\endgroup\$ – Transistor May 6 at 20:33
  • \$\begingroup\$ Thanks. Actually the power that would be put in the cable are quite low. Also, I mainly need to know the passive heating, i.e without any signal. But in the end, people always put jacket around right ? I need to have access to order of magnitude passive heating load (not a very precise calculation). As I am not experimentalist I am not aware of all the details of those cable which is the reason of my questions. It is in the context of quantum computing: coaxial cable that are put in cryostat in order to control the quantum chip. \$\endgroup\$ – StarBucK May 6 at 21:45
  • \$\begingroup\$ Oh actually the insulator will way less conduct heat than conductor. The jacket being insulator indeed I think I can neglect it for an order of magnitude calculation for heat transport. \$\endgroup\$ – StarBucK May 7 at 14:36
1
\$\begingroup\$
  1. There's no jacket on those, just bare outer conductor. Those are straightforward because they are solids.

  2. The outer dimension is not circular because of the drain wire that's sitting alongside the spiral wound aluminized polyester. The wire is about 0.2mm diameter while the shield is only 1 or 2 layers about 1/10 as thick. So you would expect the dimension across the outside to vary by about 0.2mm, as it does.

  3. You didn't ask this, but be careful with heat loss calculations in cryogenics, you really have to integrate the heat loss from one temperature to the next rather than just assuming a fixed thermal resistance. Suppliers of cryogenic-qualified cables such as Lakeshore should have that data available.
| improve this answer | |
\$\endgroup\$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.