Feser's New Lineup

[Martinm210]

so what sort of solder do you suspect is used on a common radiator?

I know they are rohs now, so lead is out...is it mostly tin?

I searched wikki and see there are other options...would be interesting if there was one soldered with a high silver concentration.

http://en.m.wikipedia.org/wiki/Solde...edirected=true

I think I read somewhere that the typical folded fins is not ideal either. If the fins could be folded such that the fins formed rectangles instead of triangles, that you'd get some gain out of it...

I'm just happy to see some new ideas being tried, look forward to skinnees testing..

[Monkey Puzzled]

so what sort of solder do you suspect is used on a common radiator?

I know they are rohs now, so lead is out...is it mostly tin?

I searched wikki and see there are other options...would be interesting if there was one soldered with a high silver concentration.

http://en.m.wikipedia.org/wiki/Solde...edirected=true

I think I read somewhere that the typical folded fins is not ideal either. If the fins could be folded such that the fins formed rectangles instead of triangles, that you'd get some gain out of it...

I'm just happy to see some new ideas being tried, look forward to skinnees testing..

Yeah solder is mostly tin (usually around 95% Sn, 4% Ag, 1% Cu IIRC), with some silver and copper for wetting properties. The really high silver content stuff is very expensive;- it's used by jewellers to hide the solder joints on silver. Can't remember what they use for higher temperature brazing rods - I think it's mostly copper.

[asura]

Not to hide solder joints but to make a structural alloy.

When we (jewellers) solder, the two pieces of metal need to be as clean and flat and close to each other as possible, if you can see a line of light between the two pieces they're (ideally) too far away from each other, there is a little bit of leway, but hard soldering is not for joining gaps. The metal (both pieces) must be heated to an appropriate temperature for the solder (usually in the red-heat range, silver solders go from about 650-800c) when the metal reaches the appropriate temperature the solder flows into the joint and penetrates the metal on either side, the heating anneals the metal and creates freedom in the crystalline matrix into which the solder is drawn by capillary action. This makes for a very, very strong connection.

Soft soldering (tin solder) doesn't make such a strong connection, it simply joins to things together using a different material; it's working like a (not hugely) conductive hot glue gun. However because of the much lower melting point of the solder it is a much more versatile material than hard solder e.g. you cant form a hard solder joint between steel and silver but you can with soft solder; by the time the steel is hot enough to alloy with the solder the silver has melted too...

This is a layman's explanation of hard soldering, that being what I am for a detailed explanation ask a metallurgist.

[Kaldskryke]

It's in the equation, but it most likely has less to do with distance. It's probably because the thermal resistance due to the convective boundary, water to copper and brass to air, is much greater than that of the copper to brass. but I'm just speculating... Waterlogged was speculating earlier that the brass to air term that dominates... Maybe feser knows

Don't you mean water to brass and copper to air?

Than homogeneous Al? Why would you average?

...

XS needs a resident thermodynamicist...

I don't mean to take the mean of the two values. I'm suggesting you approximate the system as a (very) simplified one-dimensional geometry that goes: free-stream water -> boundary layer water -> brass tube wall -> solder material -> copper fin -> air boundary layer -> free-stream air. To determine the overall heat transfer coefficient for the system you'll need to work out the convective heat transfer terms separately (Nusselt number differs, etc) but you can join the conductive terms if you weight their conductivities according to their length. It's just a convenient simplification that doesn't change the answer in this kind of geometry, and I was just using it to illustrate that I think that a thin tube wall and a thin layer of solder are less dominant terms than the much longer length of copper fin, even if the solder is particularly resistant.

In reality, the fact that convection occurs along the entire length of the copper fin makes the one-dimensional simplification all but useless for actually doing the math, but I was only intending to invoke it for illustration.

I should maybe dig out my heat transfer textbook from college and see if I can take the simple fin example and change the boundary conditions to fit louvered fins and come up with a more realistic answer, but I don't feel it's worth the effort. Afterall you may be right that the convective terms dominate

[Monkey Puzzled]

Not to hide solder joints but to make a structural alloy.

When we (jewellers) solder, the two pieces of metal need to be as clean and flat and close to each other as possible, if you can see a line of light between the two pieces they're (ideally) too far away from each other, there is a little bit of leway, but hard soldering is not for joining gaps. The metal (both pieces) must be heated to an appropriate temperature for the solder (usually in the red-heat range, silver solders go from about 650-800c) when the metal reaches the appropriate temperature the solder flows into the joint and penetrates the metal on either side, the heating anneals the metal and creates freedom in the crystalline matrix into which the solder is drawn by capillary action. This makes for a very, very strong connection.

Soft soldering (tin solder) doesn't make such a strong connection, it simply joins to things together using a different material; it's working like a (not hugely) conductive hot glue gun. However because of the much lower melting point of the solder it is a much more versatile material than hard solder e.g. you cant form a hard solder joint between steel and silver but you can with soft solder; by the time the steel is hot enough to alloy with the solder the silver has melted too...

This is a layman's explanation of hard soldering, that being what I am for a detailed explanation ask a metallurgist.

Thanks for the explanation - very interesting. I was looking into hard-soldering/brazing copper heatfins to copper pipe at one stage, but the costs and difficulty getting it to such high temperature put an end to that idea.

[meanmoe]

Don't you mean water to brass and copper to air?

sure... whatever

Are you a thermodynamicist?

In you previous post, it sounded like you were implying that the thermal resistances were averaged and that result was close to solid aluminium. This isn't the case of course.

Also, I did break out an old book, and I don't think that there is a simple example which this follows - that I could find anyway. Assumptions have to be made about the flow on both sides of the solid material. but I ran from thermo after school, so if you got one then throw it out...

[mlwood37]

PMSL just thought of some thing

If there nano plating is any thing like there nano fluid then there lol ....

[Martinm210]

Yeah solder is mostly tin (usually around 95% Sn, 4% Ag, 1% Cu IIRC), with some silver and copper for wetting properties. The really high silver content stuff is very expensive;- it's used by jewellers to hide the solder joints on silver. Can't remember what they use for higher temperature brazing rods - I think it's mostly copper.

Cool thanks!

Soo...according to this reference:

http://www.engineeringtoolbox.com/th...ity-d_429.html

Conductivities (k - (W/mK)) are as follows:
Aluminum = 250
Brass = 109
Copper = 401
Tin (Solder) = 67

Most rads are sort of a Brass Tube>Solder Interface>Copper Fin type construction.

I'm guessing this conductivity is some sort of functional of thickness (distance), but I don't know the details.

[meanmoe]

it's power per unit of length per unit temperature.... where the length is the distance through the material. IIRC unless you assume 1D like Kaldskryke was talking about it becomes a differential. The heat equation you'd end up with a dA and a dT or dx and dT. If you want to look at 1D conduction, you can follow the link on the bottom of that page... http://www.engineeringtoolbox.com/co...fer-d_428.html That'll get you 1D conduction to the surfaces of the boundary materials. Then it becomes a different problem though because of the convection from the surface to the fluid boundary layer and then to the fluid freestream. As Kaldskryke said. The freestream isn't as tough, but the boundary layer part is. You could assume no boundary layer, but I really don't know how good of an assumption that is. I'm hoping that Kaldskryke tosses us an example.

the equation caller Fourier's Law in that link assumes a single material... for the conduction problem the conductive heat flow is equal to the total delta T / the sum of the resistences in the case of multiple materials... where the resistances are each dx/(k * A) ... The numbers you threw out are 'k's and A is area. Assume dx to be linear distance.

[Kaldskryke]

sure... whatever

Are you a thermodynamicist?

I graduated from university with a BSc in Chemical Eng. about a year ago. Thermo isn't my specialty, but I've taken several courses that focus on heat transfer within metals and through fluid boundary layers on flat plates, fins, shell and tube heat exchangers etc. I don't have any practical experience with the material, my day job is more related to process controls

Also, I did break out an old book, and I don't think that there is a simple example which this follows. Assumptions have to be made about the flow on both sides of the solid material. but I ran from thermo after school, so if you got one then throw it out...

For the louvered fins I think it should be pretty straightforward to adapt a simple model for flow over a generic fin and change the boundary conditions so that both "ends" of the fin are heated. If I can assume that the water and air streams remain at constant temperatures throughout the rad the rest shouldn't be too hard. I think I would use the "hydraulic radius" as the characteristic length in determining reynold's number, which is just a function of the flatness of the brass tubes or the FPI of the fins. When I get home today I'll see if I can't take a closer look.

With respect to the argument over aluminum versus brass & copper, I still think that the solder joint would have to be particularly large or sloppy for it to be worse than solid aluminum.

[Kaldskryke]

it's power per unit of length per unit temperature.... where the length is the distance through the material. IIRC unless you assume 1D like Kaldskryke was talking about it becomes a differential. The heat equation you'd end up with a dA and a dT or dx and dT. If you want to look at 1D conduction, you can follow the link on the bottom of that page... http://www.engineeringtoolbox.com/co...fer-d_428.html That'll get you 1D conduction to the surfaces of the boundary materials. ... the equation caller Fourier's Law in that link assumes a single material... for the conduction problem the conductive heat flow is equal to the total delta T / the sum of the resistences in the case of multiple materials... where the resistances are each dx/(k * A) ... The numbers you threw out are 'k's and A is area. Assume dx to be linear distance.

As long as the temperature/heat flux doesn't vary within the cross-sectional area of your fin/rod/whatever then you don't need to make the math complicated. Once you use Fourier's Law to get the heat flux through the conductive element you can just multiply by the cross-sectional area to get the total heat flow. Generally any "hot spots" that might be in the middle of your element are insignificant enough to be ignored because usually conduction within metals is much more rapid than convection losses on the outside of your element. As long as you know that the heat flow in your element is primarily in one direction, analyzing in one dimension is quite valid. This generalization breaks down for fins because they are so thin. They have lots of surface area for convection relative to their little cross-sectional area for conduction...

Then it becomes a different problem though because of the convection from the surface to the fluid boundary layer and then to the fluid freestream. As Kaldskryke said. The freestream isn't as tough, but the boundary layer part is. You could assume no boundary layer, but I really don't know how good of an assumption that is. I'm hoping that Kaldskryke tosses us an example.

I'm not even sure how I would go about ignoring the boundary layer, as all of my handy equations for characterizing flow over plates/tubes used the no-slip condition as the basis of their derivation. All I need is the reynolds number of the free stream and some geometry info and I can characterize the convection from the fins and tube.
An example is forthcoming. You'll have to give me some time on that.

[meanmoe]

As long as the temperature/heat flux doesn't vary within the cross-sectional area of your fin/rod/whatever then you don't need to make the math complicated.

...

no slip

Do you know what the theory is for the handy equations? Is it Couette flow? or Poiseulle? They both assume a no slip condition for velocity as an approximation and make the other assumptions plus some more like laminar flow...

[SpuTnicK]

Here is a pic how these water-channels cut can look like(sorry for small pic size):

Spiral-wound finned tubes

[Metroid]

http://www.tfc-admiral.com/

So today is 1st of October 2010 and where are the products?

I gotta love marketing

[MpG]

http://www.tfc-admiral.com/

So today is 1st of October 2010 and where are the products?

I gotta love marketing

Nah, still Sept 30th here, mate. I'm sure they'll be released in the next 3 1/2 hours, so don't you worry.

[Factotum]

feser site updated^ http://www.feser-one.com/admiral/ADMIRAL.swf

[skinnee]

takes 2-3 minutes to load...

It looks great, but please improve the load times.

[affiliate13]

Anyone know when these will get reviewed at all.
I would like to see some more photos and info that isnt marketing.

[SpuTnicK]

[Martinm210]

takes 2-3 minutes to load...

It looks great, but please improve the load times.

No kidding!...most people will think the site is down...took forever on 10MB cable..

I thought I was still on dial up..

There web guy is pretty good with flash though...feel like I'm playing BF2142 or something..

I like the Holy Moses..makes me laugh every time.

Oh..and 6L of fluid to fill that Monster 700...wow...

[skinnee]

Anyone know when these will get reviewed at all.
I would like to see some more photos and info that isnt marketing.

I've got the first test and review, but I am still waiting to get one in my hands. Should be this month from what I've been told, time will tell.

[bundymania]

No kidding!...most people will think the site is down...took forever on 10MB cable..

I thought I was still on dial up..

There web guy is pretty good with flash though...feel like I'm playing BF2142 or something..

You know what, "Webguy" & one of the 2 CEO´s = same Person in the 2 Man Team

[rheydt]

I´d rather believe Skinnee´s test review than any marketing hypes, good to know they have developed this product for one year, but they should be able to send a test sample at least NOW to Skinnee!

[affiliate13]

I've got the first test and review, but I am still waiting to get one in my hands. Should be this month from what I've been told, time will tell.

Ah splendid. For good for good or for bad at least that will put an end to the speculation on these.

[skinnee]

Yeah, I lean back and forth on perceived performance... and in the end I always end up at "I guess we'll know for sure when it is finally on the bench."