Main Page | Recent changes | View source | Page history

Printable version | Disclaimers | Privacy policy

Not logged in
Log in | Help

Temperature changes of coolant within the loop

From OCAU Wiki



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.


dT = Q' / ( m' * c )

This is simply a re-arrangement of a 1st law rule for relating a change in energy to a change in temperature:

Q' = m' * c * dT


For the most part we are not specifying SI units here, as they become impractical to work with for such small applications as PC watercooling loops. The only important thing is that you use the same standard of units throughout (E.g. if you specify Q' as kJ/s, then you damn better specify c as kJ/(kg.K) too :P).

dT = Temperature rise (In K (kelvin) or °C)

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 (given as J/(g.K)).


We have a loop with 5 L/min 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 reached 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 due to pump: T = 10W / ( 83.33g/s * 4.18J/(g.K) ) T = 0.028°C

Second, temperature rise due to CPU block: T = 90 / ( 83.33g/s * 4.18J/(g.k) ) T = 0.258°C

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


Water has a specific heat of 4.18 kJ/(kg.K). This means that every kilogram of water takes 4.18 kilojoules of energy to change its temperature by 1°C (1K).

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 min/60 sec * 1.00 g/cm³, which is 80 grams or 0.08kg per second.

If each kilogram requires 4.18kJ to raise its temperature by 1°C, then the 0.08kg will require 1°C x 4.18 kJ/(kg.K) x 0.08kg = 0.334 kJ.

Now 80 grams come and go each second, so to heat them all by 1°C requires 0.334kJ/s or 334.88 joules per second.

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

So, for the water to rise in temperature by 1°C on one trip around the cooling loop, we'd need to be adding 335W of heat to the water.

If only 50W is being applied, there is only a 50 / 335 = 0.149°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.

[Main Page]
OCAU Forums
PC Database

Main Page
Recent changes
Random page
All pages

View source
Discuss this page
Page history
What links here
Related changes

Special pages