Radiators and Fans

[Roscal]

Ok Thanks BillA for trying , I'm curious to see results.

[Since87]

Originally posted by unregistered

Since87 - did you adjust for such with the rad data ?

I didn't make any attempt to account for variations in tank geometry or inlet/outlet tubing size.

I just lumped all that into the 'overhead' factor mentioned above. For the simulated results, I added the same 'overhead' flow resistance as a single 'component' in series with the tubing flow resistance, for all three heatercores.

I'm interested in seeing the results for the reversed flow test. I'm not really expecting to see much difference, but if there is a clear difference it will be an indication that tank geometry, etc. is a significant factor. That would put a serious kink in our ability to generate models based on a few easily obtainable measurements.

More later.

Edit: Corrected "four heatercores" to three heatercores".

[Les]

Originally posted by Since87
I didn't make any attempt to account for variations in tank geometry or inlet/outlet tubing size.
.............

Done some manipulation for inlet/outlet tube size.
Original P/Q by inpection of gif
Connector correction used "SF Pressure Drop5.0"



Too much like hard work.
Don't think I like this game.

[Since87]

Originally posted by Les

Too much like hard work.
Don't think I like this game.

LOL

Thanks for the info Les. It gives me hope that there is an equation that works for all three heatercores.

I'll work on it some more. I downloaded "SF Pressure Drop 5.0". It looks very handy.

This is going to take a while. Obviously the inlet and outlet sizes need to become inputs to the model. They are much bigger factors than I had guessed.

I'm probably going to need to write some code to search for the right combination of 4 or more 'weighting factors' to minimize the model error for all three heatercores. (The only way I know to do this is very much brute force.)

Yuck.

[bigben2k]

Originally posted by unregistered
deeppow's '86 Chevette rad (# not known)


2-pass, 5/8 OD tubing connections (one cut back from 3/4" end)
effective dimensions: 6 1/8" long, 6 3/8" wide, 1 15/16" thick
14 corrugated tubes, with folded and louvered fins @ 10/in.

Sorry for taking so long...

The heatercore in question is Fedco# 2-261.


I just ordered 2-304: here's hoping it's a double pass! (Also a Chevy core)

[Blackeagle]

I'm curious Ben, Bill's data shows the single pass is lower restriction, which offers some advantages.

I'm wondering what advantage you are seeing in a double pass type?

What special set up do you have in mind for it?

[BillA]

swirling water will draw out more heat

diz-zy, I'm so diz-zy . . . .

[Since87]

Bill,

I was going to start beating my head against this data tonight, and realized I'm lacking some info.

Would you tell me the ID's for all the inlets and outlets?

Or, if they all use the same wall thickness, what is the wall thickness of the inlet/outlet tubing?

Also, is there any significant ID transition between where your pressure sense ports are and the ends of the inlet/outlet tubes?

Anything else I'm missing to be able to be able to calculate a reasonable approximation of the flow resistance attributable to the inlet/outlet tubes?

[BillA]

the Chevette:
0.540 and 0.565

yours:
0.565 - both

Craigs:
0.565 and 0.680

taps:
0.570 - both

I think you're chasing your tail, treat 'em as black boxes

[Since87]

Originally posted by unregistered

I think you're chasing your tail, treat 'em as black boxes

May well be, but I did come up with a better match. (For these three heatercores at least.)

First off, I played around with, "SF Pressure Drop5.0".

The results didn't make sense to me. The software would sometimes calculate negative pressure drops for some features. Quite likely a user problem, but I did come across the following in the FAQ:

"I have calculated a standard orifice acc. DIN 1952 or EN ISO 5167. The program calculates other pressure drops as I. Why?

The DIN 1952 or EN ISO 5167 calculate the pressure drop direct behind the orifice. These values are used to calculate the flow rate. The program SF Pressure Drop calculates the remain pressure drop which you will find 6 D behind the orifice."

Anyway, I modified my treatment of the flat tubes. My previous go at this, just had a factor based on the length of the tubes.

This time I split that up into a per tube factor and a per inch factor. The basis for this, was my dilletante belief, that there should be a pressure drop associated with entering and exiting the flat tube regardless of length, and that there should be a second pressure drop that was dependent on the length of the tubing.

Using these two factors and ignoring the main inlet and outlet altogether gave me the following:



And zoomed in more on the area of interest to most...



The only inputs required to generate these curves is the length of the flat tubes and the number of tubes in each pass.

Reasonably useful general equation for this 'family' of heatercores, or mental masturbation?

I don't know.

[BillA]

I'll not quibble with the rationale,
the fit is close enough for our purposes
(and likely more consistent than the data you're comparing it to !)

I'm putting: together a bunch of data for ThermoChill which should be more than sufficient to extract flow and dissipation parameters for oval tube/folded fin rads

right now I'm in Excel hell with a compressed axis title box (not autosizing)
any Excel gods out there please help

[bigben2k]

Nice work, Since87.

Care to take a guess at my core? 2-304 (not received yet).

[Les]

Originally posted by Since87
......, or mental masturbation?

I don't know.

Neither do I, but some more :-
Simulated using "SF Pressure Drop5.0".
No connectors or pleni(?).
Sean Simulated as 6.5(500x50x0.79mm) Channels
Ralph Simulated as 7(320x50x0.79mm) Channels
Craig Simulated as 13(250x50x0.79mm) Channels
Simulation for tubes only . The 0.79mm Dimension chosen( can't remember ,will edit when do)
All flow is clrssified as Laminar. Transition Laminar/Turbulent ~ 2.7LPM per Channel.


EDIT
Sean
Have only got negative Pressure Drop from "SF Pressure Drop5.0" for enlargement of pipe(so have not used until have thought about ita bit).

[BillA]

Les
the channels are quite corrugated (and your sims proportionally low)
- just add a 'fudge factor' to shift to match
??

and the heat dissipation ?
unfortunatly we have only one datum
just convert to "dissipated W/in²/CFM" (or metric) ?

[Since87]

Originally posted by bigben2k

Care to take a guess at my core? 2-304 (not received yet).

You might want to wait till you have the heatercore in hand. Blackeagle's did not meet the specs shown on the website.

Keep in mind that the numbers my spreadsheet generates may be total BS. I realized I'd made a fundamental mistake in how the 'per tube' fudge factor was applied. I've since redone it with that factor applied 'correctly' and have redone the spreadsheet.

Correcting the 'per tube' fudge factor, required me to add an 'overhead' fudge factor. The overhead fudge factor accounts for inlet/outlet tubing and tanks and is assumed to be the same for all heatercores. (Obviously an oversimpification which begs the question, 'Why does the inlet and outlet tubing size vary between heatercores?")

My updated spreadsheet generates nearly identical curves to the previous one with substantially different equations and fudge factors. Don't believe that because it gets a reasonable match with test data, that it must be 'right'.

That said, I do believe that what I have now is much more plausible. (It bothered me that there was no 'overhead factor' in my last pass at it.)

I'll create a cleaned up, user friendly version of the spreadsheet tonight, and make it available for download here. Anyone with Excel can play around with it, and possibly refine it.

[BillA]

heater core connectors are sized and oriented simply to fit up with the rest of the heating system;
the engine water pump has more than enough head to push whatever,
and normally the heater control valve is not fully open

we are the ones with pissant pumps

[Since87]

Originally posted by unregistered
heater core connectors are sized and oriented simply to fit up with the rest of the heating system;

It seems to me, like it would be easier on the manufacturers to use consistent tubing sizes, but what do I know? Maybe where the tubing has to be routed is the major factor?

Originally posted by unregistered

and normally the heater control valve is not fully open

You clearly live well south of me.

[Les]

Originally posted by unregistered

and the heat dissipation ?
unfortunatly we have only one datum
just convert to "dissipated W/in²/CFM" (or metric) ?

Metric but ugh.
Still a bit fuzzy what this quantity "dissipated W/in²/CFM" is called.
Current favoured flavour for Air Flow is "m^3/min"
This would m^3/min per radiator(?)


Area cm^2 and probably referring to frontal area i guess

Am only slow learning the ways of radiators.Using this series of pdfs by Wolverine
http://www.wlv.com/products/databook/ . Finding some interesting snippets (e.g suggested Erosion Velocity Velocity limit 6ft/s for Water impinging Cu) in addition apparantly sound theory.

[BillA]

seriously good link Les

the metric units I'm using are:
Pa for air pressure, m³/min for air volume
mH2O for pump related (liquid) pressures, lpm for flow rate

for the dissipation 'constant',
how about: W/cm²/m³
but note that this will be a very small # (the air unit is 'too' big)

think you need to get into 3D graphs
air flow and backpressure,
coolant flow and head loss;
for a given rad type - will then yield the dissipation
-> at the 'design' air/coolant temp differential of course

there is nothing that cannot be made more complex

[Since87]

Here is a 'user' spreadsheet for calculating PQ curves for heatercores. No guarantees as to the accuracy of the results.

[BillA]

very professional, a worthwhile addition
I'm impressed

somewhere a note needs to be made that this applies to 1 15/16" thick heater cores

[Since87]

Originally posted by unregistered

somewhere a note needs to be made that this applies to 1 15/16" thick heater cores

True. Probably other notes that should be attached as well.

I'll leave this one up for a bit and collect suggestions then put a revised one up.

[Les]

Neat
Big Momma* no problem
Works with Alien Fudge Factor(0.0159722222222222222222222222222222)
Use Bill's http://forums.overclockers.com.au/sh...threadid=58005 to correct for 3/8" barb to 0.488"ID(Works with my sums).
Experimental Big Momma* ~ 1.1mH2O at 7.5LPM


Tried a Wider HC( re Bills comment)
Radiator E* 30 Tubes(9/8"wide) 2 pass (0.4" ID entry) , (2 Pass ?)
Dunno but perhaps not so clever
Used an Alien Fudge(0.008151) and Quadrupled(4x) "per tube resistance coeffs"
Experimental* ~ 1.75mH2O at 7.5LPM



* {url]http://thermal-management-testing.com/radiator%20testing%201.htm[/url]

EDIT Corrected some stupidity.

[Since87]

Originally posted by Les
Neat
Big Momma* no problem
Works with Alien Fudge Factor(0.0159722222222222222222222222222222)
Use Bill's http://forums.overclockers.com.au/sh...threadid=58005 to correct for 3/8" barb to 0.488"ID(Works with my sums).
Experimental Big Momma* ~ 1.1mH2O at 7.5LPM

Hmm. I need to add something explaining 'one pass' vs 'two pass'. It looks like you setup the spreadsheet for Big Momma with only one pass. Because the water flows down one set of seven parallel tubes and up another set of six parallel tubes, the pressure drops for the two sets of tubes will be in series rather than parallel.

I think the flow resistance of the 3/8" barbs, swamps the resistance of the heatercore itself so much, that whether the heatercore were one pass or two would make a few percent difference in the total pressure drop of the two systems. (No time for a more in depth look right now.)

It does look like it would be reasonable to add a factor to the spreadsheet accounting for the ID's of barbs that have been added to the heatercore.

Originally posted by Les

Tried a Wider HC( re Bills comment)
Radiator E* 30 Tubes(9/8"wide) 2 pass (0.4" ID entry) , (2 Pass ?)

Radiator E is a single pass, but regardless, I would expect a substantially different set of fudge factors to be required to approximate the PQ behavior of Rad E. Too big a difference in the flat tubes. I think it would take data from a group of three rads, with flat tubes identical to Rad E's tubes, but different overall lengths and/or configurations. That gives you three equations using the three unknowns. (The fudge factors.)

[Les]

Originally posted by Since87
. It looks like you setup the spreadsheet for Big Momma with only one pass.

Yes
For the Big Mamma all is well
The only reputable report* marks no mention of 2-Pass for the Big Mamma.


"Rad F - OCWC PN: Big Momma
Tube: 13 full thickness corrugated brass “plates” "

. I am onlly suggesting it is very difficult.

* http://thermal-management-testing.co...esting%201.htm

[Les]

[i]

Radiator E is a single pass, but regardless, I would expect a substantially different set of fudge factors to be required to approximate the PQ behavior of Rad E. Too big a difference in the flat tubes. I think it would take data from a group of three rads, with flat tubes identical to Rad E's tubes, but different overall lengths and/or configurations. That gives you three equations using the three unknowns. (The fudge factors.) [/b]

Single pass worsens the agreement.
Fudge factors were changed in favour of a correlation.

Think it is difficult

[Since87]

Originally posted by Les
Single pass worsens the agreement.
Fudge factors were changed in favour of a correlation.

Think it is difficult

I'm not sure if you mean 'Coming up with a set of fudge factors to get a curve match is difficult.', or 'Coming up with a plausible set of fudge factors is difficult.'

I wouldn't even attempt the latter with only data from a single HC from a given 'family'. I'm grossly unqualified to even offer a guess.

The former is simple algebra though. There is an infinite set of bogus combinations of fudge factors, that will give a curve that appears to be a reasonable match to Bill's data for RAD E. (See attached image.) All output curves are of the equation:

y = k * x^2

User inputs and fudge factors only affect k.

I hope I'm not stating the obvious. Wondering if you're giving me too much credit.

[Les]

Originally posted by Since87
I'm not sure if you mean 'Coming up with a set of fudge factors to get a curve match is difficult.', or 'Coming up with a plausible set of fudge factors is difficult.'

I wouldn't even attempt the latter with only data from a single HC from a given 'family'. I'm grossly unqualified to even offer a guess.

By "It is difficult"
I was meaning "It is difficult to assess pressure drop fom a set of dimensions".
Which I think is akin to the "latter".

[Les]

Would have helped if I had looked at the photos of the radiators.
Yes the Big Momma cannot be 1 Pass, and Rad E cannot be 2 Pass(could be 3,5,7 etc)
. For the record the difference between 1Pass and 2Pass on Big Momma .:-


Apologies for confusing the issue.
To quote Bill "there is nothing that cannot be made more complex"
I am certainly capable of that.

[Blackeagle]

It would seem that in the case of the Big Momma rad that those highly restrictive 1/4" ID barbs are acting as a leveler when comparing single vs dual pass.

Be interesting to see the differance if somebody moded the inlet/oulets to 1/2" ID.

BE