Temperature changes of coolant within the loop

Revision as of 16:35, 11 November 2008

Overview

Within a typical rig flowing 4.8 litres per minute through a CPU waterblock where it is cooling a 50W of components (including the pump), the water will be heated up by a mere 0.15°C during the heating phase of each cycle.

Hence, any consideration to the order of components due to heating/cooling of the coolant should only come into effect if there are multiple CPU blocks, or extremely low flow rates.

Rule

dT = Q' / ( m' * c )

Properties

dT = Temperature rise

Q' = Rate at which heat is being applied (E.g. 60W), this is the rate at which the waterblocks are rejecting heat into the coolant.

m' = Mass flow rate of coolant (g/s), this is the rate at which the coolant is flowing.

c = Specific Heat of coolant (AKA Specific Heat Capacity), this is the energy required to institute a temperature change of 1°C for 1g of coolant.

Example

We have a loop with 5 LPM of coolant flow resulting from a pump that dumps 10W of heat into the loop. The loop also consists of a waterblock where 90W of heat enters via a CPU, and a radiator, where all 100W of heat leave. The system has reach equilibrium, and it is assumed all coolant is heated equally by the blocks and cooled equally by the radiator.

Question: What is the impact of running pump -> cpu -> radiator versus pump -> radiator -> cpu?

Answer First, temperature rise caused by pump: T = 10 / ( 5 * 69.77 ) T = 0.028°C Second, temperature rise caused by CPU block: T = 90 / ( 5 * 69.77 ) T = 0.258°C

Hence, the second layout will deliver coolant that is 0.286°C cooler than the first.

Proof

Water has a specific heat of 4186 J/kg-°C. This means that every kilogram of water takes 4186 joules of energy to change its temperature by 1°C.

Let's assume an example loop with a flow rate of 4.8 litres per minute. 4.8 litres per minute, is 4800 cm³/min. Water has a density of 1.00 g/cm³. Therefore water flow of 4.8 Litres per minute is 4800 cm³/min * 1.00 g/cm³, or 4800 grams per minute, which is 80 grams or 0.08kg per second.

If each kilogram requires 4186 joules to raise its temperature by 1°C, then the 0.08 kilograms will require 1°C x 4186 J/kg-°C x 0.08kg = 334.88 Joules.

Now 80 grams come and go each second, so to heat them all by 1°C requires 334.88 joules per second.

"Joules per second", or J/s, has a familar name: Watts.

So, for the for the water to rise in temperature by 1 degree celsius on one trip around the cooling loop, we'd need to add 335 Watts of heat energy to the water.

If only 50W is being applied, there is only a 50 / 335 = 0.149307 °C temperature rise.

Assumptions: The above assumes that all 80 grams of liquid per second are being heated equally; ie, the waterblocks introduce sufficient turbulence or other means to stop portions of coolant remaining unheated. Important: The above example only calculates the temperature rise of the water after one complete trip through the system. The actual final equilibrium temperature of the water after many loops through the system is a separate value, and is a function of the heat-exchanger (typically a radiator) and its subsquent ability to cool the applied heat load to hold the water temperature at some point consistent with the heat exchanger's capacity.