Calculating power dissipated in a heat exchanger

[SgtSixpack]

TL;DR Summary

Is this calculation right?

I want some help in figuring out any flaws in these calculations. I was given this picture but I was looking into it and found some things which are either wrong or don't apply to my watercooling loop (for my PC).

First thing I notice is that litres per hour divided by 3600 (hours to minutes) gives units in kg.

I don't know if it maters (since I can't find specific numbers on heat capacity of the fluid) but I'm not using distilled water but this stuff (it was £20 when I bought it):
https://mayhems.store/mayhems-x1-clear-premixed-watercooling-fluid-5-litres.html

For background information here is the thread where I got the picture from:
https://forum.aquacomputer.de/weite...t/?s=d38a467e13100c26b8c0293df1023b395dbbd769

Also appologies if this is the wrong section of the forums.

 

[renormalize]

TL;DR Summary: Is this calculation right?

The "flowchart" shown is a correct (but cumbersome) way to calculate the cooling capacity based on the simple formula in the gray box, which I enlarge here for clarity:

Note though that the specific figures used there (kg/sec = liters-per-hour/3600 and Cp = 4.178) apply only to distilled water. To use the formula with your cooling fluid, you'll need to contact the manufacturer to obtain the proper values of mass-density and heat-capacity for that particular product.

 

[SgtSixpack]

Thanks for the response. I'm still not happy with the first part of the calculation.

LPH/(a constant)
can't end up as kg from my limited knowledge of mathematics.

To find the mass in kg, one would need something like density of water. Guess I'm right/wrong according to google:

"One litre of water has a mass of almost exactly one kilogram when measured at its maximal density, which occurs at 3.984 °C. It follows, therefore, that ⁠1/1000⁠ of a litre, known as one millilitre (1 mL), of water has a mass of about 1 g, while 1000 litres of water has a mass of about 1000 kg (1 tonne or megagram)."

 

[renormalize]

I'm still not happy with the first part of the calculation.
LPH/(a constant)
can't end up as kg from my limited knowledge of mathematics.
To find the mass in kg, one would need something like density of water.

That's exactly right. Use the formula:$$\text{mass-rate}\left(\text{kg/s}\right)=\text{mass-density}\left(\text{kg/m}^{3}\right)\times\text{volume-rate}\left(\text{m}^{3}\text{/s}\right)$$For water, the mass-density is $\text{997}\approx\text{1000}\,\text{kg/m}^3$, but as I said, you'll need to contact your supplier to find the mass-density of your cooling fluid.

 

[SgtSixpack]

Thanks for explaining that. I can't like your post unfortunately.