Max efficiency of heatercore at 1.5 gpm?

[bigben2k]

Ok, I had to read that twice to get it...

The delta t involved between the block inlet and outlet is, as Bill reported, so small, that even he would be hard pressed to measure it, and does not have the required instrumentation to do it (right?).

So if that's true, then you'd also be hard pressed to measure a temp difference under a higher flow rate.

Keep in mind that a system will typically have a capacity that is less than 1/2 gallon, and typical flow is ~ 1 gpm (60 gph). That translates in the water passing by the same point twice per minute.

Hummm...

Let's try another angle: if the same flow carries the same heat, and you double the flow rate, the higher flow is absorbing the same energy, as you stated, but the same flow is also releasing the same energy through the rad.

I'm either rambling, or going in circles...

It's easier for me to see it as a series of thermal resistances, where the higher flow rate decreases the thermal resistance, and where the rad will behave in an odd way at different flow rates, for reasons that only a handful of us understand clearly, and that does NOT include me!

Sorry, someone else take this one.

[Alchemy]

Each line on that chart is at a constant air flow rate at constant input temperature, or as near as possible. The variables are the water flow rate, the inlet and outlet water temperatures, and the heat load. As freeloadingbum notes, the heat load is a function of the difference between inlet and outlet water temperatures and the flow rate.

Is this an open system, that is, one without a waterblock imparting a constant, known amount of heat into the system? If it is, then the question is whether or not BillA took that into account in his calculations.

The answer to that question is yes.

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

Originally posted by bigben2k
Ok, I had to read that twice to get it...

The delta t involved between the block inlet and outlet is, as Bill reported, so small, that even he would be hard pressed to measure it, and does not have the required instrumentation to do it (right?).

So if that's true, then you'd also be hard pressed to measure a temp difference under a higher flow rate.

There's no block, only the rad. If he didn't know the temp difference, then the entire test is pointless. If you don't know the difference between the inlet and outlet fluid temperatures, *you don't know how many watts are being dissipated*. That means no C/W values. Nothing.

Keep in mind that a system will typically have a capacity that is less than 1/2 gallon, and typical flow is ~ 1 gpm (60 gph). That translates in the water passing by the same point twice per minute.

True but irrelevant.

Let's try another angle: if the same flow carries the same heat, and you double the flow rate, the higher flow is absorbing the same energy, as you stated, but the same flow is also releasing the same energy through the rad.

Invalid. Nonsensical. Untrue. "Flow" does not absorb heat. Water does.

I'm either rambling, or going in circles...

Not so much a circle as a Mobius strip.

It's easier for me to see it as a series of thermal resistances, where the higher flow rate decreases the thermal resistance, and where the rad will behave in an odd way at different flow rates, for reasons that only a handful of us understand clearly, and that does NOT include me!

I think the whole "odd way at different flow rates" was never well-settled among the engineers on this board. I've always thought it to be mostly a function of experimental error, but since I'm not quite willing to repeat BillA's tests, I would not make any assumptions.

Alchemy