Calculating power dissipated in a heat exchanger

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Discussion Overview

The discussion focuses on calculating the power dissipated in a heat exchanger, specifically in the context of a watercooling loop for a PC. Participants explore the validity of certain calculations and the implications of using a non-distilled water cooling fluid.

Discussion Character

  • Technical explanation
  • Debate/contested
  • Mathematical reasoning

Main Points Raised

  • One participant questions the calculations provided, noting that litres per hour divided by 3600 results in units of kg, which may not be appropriate without knowing the specific heat capacity of the fluid being used.
  • Another participant confirms that the flowchart used for calculating cooling capacity is correct but emphasizes that the specific figures apply only to distilled water, suggesting that the manufacturer should be contacted for the cooling fluid's mass-density and heat-capacity values.
  • A participant expresses confusion regarding the calculation, stating that LPH divided by a constant cannot yield kg without knowing the density of the fluid, referencing the relationship between volume and mass for water.
  • Another participant reiterates the need for density to convert volume to mass, providing the formula for mass-rate and noting the density of water while again suggesting contacting the supplier for the cooling fluid's density.

Areas of Agreement / Disagreement

Participants generally agree on the necessity of knowing the density of the cooling fluid to perform accurate calculations. However, there is no consensus on the correctness of the initial calculations, as some participants express uncertainty and challenge the assumptions made.

Contextual Notes

Participants highlight limitations in the calculations due to the lack of specific information about the cooling fluid's properties, which are necessary for accurate mass and heat capacity calculations.

SgtSixpack
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TL;DR
Is this calculation right?
2024-0804 Rad Heat Calc-C.png

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.
 
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SgtSixpack said:
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:
1744986155606.png

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.
 
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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)."
 
SgtSixpack said:
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.
 
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Likes   Reactions: SgtSixpack
Thanks for explaining that. I can't like your post unfortunately.
 
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