# How to calculate required air mass flow to keep CPU below certain temperature?

I have been assigned to work out how much air flow is needed through each partition of our servers in order to keep our processors at (or below) 60C (333K).

I have worked out the current mass flow rate going through each partition, and the heat flow rate, and I know the temperature of each processor from the system logs.

However I don't know how to calculate the required mass flow rate to bring down the temperature of the CPU's to a certain value.

For example, let's assume on one partition I have found:

air mass flow rate = 0.01 kg/s
heat flow rate = 150 J/s
temperature of inlet air = 298K
temperature of outlet air = 308K
temperature of CPU = 353K
Goal CPU temperature = 333K.


So I need to reduce the CPU temperature by approximately 6% (1 - 333/353).

Does this mean I simply need to increase the airflow by 6%? I originally figured this could be true because a decrease in temperature of the CPU would be a linear decrease in the heat flow. So according to the heat capacity formula (q=mc$\Delta$T), where q is the heat flow, m is the mass flow, c is the heat capacity at constant pressure, and $\Delta$T is the change in temperature:

1.06 * 150 J/s = 1.06 * 0.01kg/s * 1000J/kgK * (308K - 298K)
Therefore the mass flow rate needs to be 0.0106m/s.


The problem with this is I am assuming the outlet temperature remains the same, which I don't expect would happen. And the mass flow increases by such a marginal amount. I just don't feel like an increase in mass flow by 6% would decrease the CPU temperature by 20C.

Could someone please show me how I should be doing this? (Please note the goal is simply to find how much I need to increase the airflow)

• Are you familiar with H/X calculations? Dec 4 '17 at 8:27
• This should cover it : springer.com/in/book/9780333459997 Dec 4 '17 at 10:10
• Mike - I have a similar book but short of reading the entire thing, I am not sure on what sections focuses specifically on my problem. Selter - I'm not aware of those calculations, but will google it now Dec 4 '17 at 10:12
• Sorry Setler, yes I am familiar with enthalpy and quality diagrams. But not sure how that applies here? Dec 4 '17 at 10:15
• One of the critical things that will affect the mass flow rate required into each partition will be the temperature of the air used for cooling. If you're using ambient air, with its daily fluctuations in temperature, then the required mass flow rate will vary daily. If your using air delivered at a specific temperature (possibly pre-cooled), the required mass flow rate of delivered air will most likely be constant - heat generated by the computers is constant.
– Fred
Dec 4 '17 at 11:13

For anyone interested in how to solve this, I have reached out to my thermo professor who has come back with a very helpful way of working this out using the formula:

$$rate.of.cooling [W] = rate.of.heat.production = h A (T_{surface} - T_{air})$$

Where, $h$ is the convection coefficient, $A$ is the surface area of the processor, $T_{surface}$ is the surface temperature of the processor and $T_{air}$ is the average temperature of the inlet air.

The rate of cooling will remain the same no matter the airflow. The convection coefficient is proportional to the velocity of the air, so any changes to it's value will require an equivalent change in velocity.

The initial $T_{surface}$ value is $353K$, so the temperature difference that is driving the heat exchange is $353K - 298K = 50K$. The goal $T_{surface}$ value is $333K$, which has a temperature difference of only $30K$ driving the cooling process.

Since the rate of cooling remains the same at $150J/s$, the other side of the equation must also remain the same. So:

$$hA(50) = x\cdot h A (30)$$ $$x = 50/30 = 1.7$$

So if the convection coefficient is increased by a factor of 1.7, and velocity is proportional to the convection coefficient, the velocity must be increased a factor of 1.7 in order to maintain the same rate of cooling, but decrease the surface temperature of the processor to $333K$. This is assuming the vented area remains the same.