# How to calculate heat transfer?

1. Jan 4, 2018

### yahastu

I have a practical engineering problem for which I really have no background on how to solve. I am looking for help, but don't even know what field of engineering would be likely to know the answer, so not quite sure where to ask. Any formulas that I might use to calculate the answer would also be useful.

My problem is this:

I have a closed space with known volume, inside which some machinery is computing X Btu/hr. I want to get rid of this heat by putting an exhaust fan in the ceiling. Obviously I would also need an intake vent that is rated for the same airflow as the exhaust fan. My question is, how do I calculate the required Cubic Feet per Minute (CFM) of the fan that would be necessary to dissipate the generated heat?

I presume this would also depend somewhat on the ambient air temperature, pressure, and humidity.

Any help would be greatly appreciated. Thanks!

2. Jan 4, 2018

### Staff: Mentor

It would also depend on the temperature that you are willing to allow the temperature in the room to rise to.

3. Jan 4, 2018

### yahastu

Can you please explain why that matters? If I'm generating X Btu/hr, and I remove anything less than X Btu/h, then as time goes to infinity, wouldn't any arbitrary temperature threshold eventually be exceeded?

Last edited: Jan 4, 2018
4. Jan 4, 2018

### Staff: Mentor

No. If the air in the room is "well mixed," then the temperature of the air leaving the room in the exit stream is equal to the room temperature. The equation that describes your situation is $$WC_p(T-T_{in})=Q$$ where W is the mass flow rate of air flowing through the room, Cp is the mass heat capacity of the air, $T_{in}$ is the air inlet temperature, Q is the heat generation rate by the equipment, and T is both the temperature within the room and the temperature of the exit stream. So, $$T=T_{in}+\frac{Q}{WC_p}$$
This assumes that there are no heat losses through the walls. So the faster the flow rate of air through the room (purge rate), the less of a temperature rise you get in the room.

5. Jan 4, 2018

### yahastu

I think I actually solved the problem. I'll show me work here, and anyone can correct me.

1) Suppose I have a 60 kW heater in a 1360 ft^3 space, and the maximum allowable temperature rise is 10 degC.

I found an equation from here that calculates how much time it will take to achieve this temperature rise:
https://www.quora.com/How-much-ener...-temperature-of-an-average-room-by-10-degrees

E = Sh * M * dT
E = energy (kJ) = power (kw) * time (sec)
Sh = specific heat of air (1.006 kJ/kgC)
M = mass of air = density (0.036 kg/ft^3) * volume(ft^3) = 50 kg

Rearranging to solve for time,

time(sec) = Sh * M * dT / power(kw) = (1.006 kJ/kgC) * (50 kg) *( 10 degC ) / (60 kW) = 8.4 sec = 0.14 min

Now in order to maintain this temperature, it implies I would need a fan capable of bringing back to ambient temperature within this same amount of time. This is simply done by replacing all the air in the volume. Thus,

Required airflow (ft^3/min) = volume (ft^3) / time(min) = (1360 ft^3) / (0.14 min) = 9,714 ft^3/min

Does that seem right?

6. Jan 4, 2018

### yahastu

Thanks for explaining and the formula! I want to check and see if your formula is equivalent to mine.

"Cp is the mass heat capacity of the air"
Same as specific heat of air, right?

"Q is the heat generation rate by the equipment"
Is W measured in Watts?

7. Jan 4, 2018

### Staff: Mentor

Yes.
Q is measured in Watts and W is measured in kg/min (or whatever units are appropriate)

So $$W=\frac{Q}{C_p\Delta T}=\frac{60}{(1.006)(10)}=5.96\ kg/sec=358\ kg/min$$
Dividing by the density of air gives 9940 CFM. So it agrees with your answer.

Note that use of this equation is independent of the volume of the room. This means that, in your calculation, if you had doubled the volume of the room, you would have gotten the same result. So the method you devised is a round about way of getting the same result as the simple equation.

8. Jan 4, 2018

### yahastu

Awesome! It's pretty cool that my roundabout way of thinking is equivalent, helps me to better understand it. I was actually just realizing that volume might be able to cancel out of the equation, but hadn't gotten around to checking if that worked.

There's just one more aspect of this problem I am trying to figure out: what happens when the fan is replaced with a hole in the ceiling (again, assuming no heat loss through the walls)?

There would still be a significant airflow simply due to the natural convection of the hot air rising out of the hole. If I knew what this rate was, then I could calculate how long it takes to empty the box, and hence what the steady state temperature increase would converge to...but I am at a complete loss, as the rate of convection seems to be dependent on the temperature, making it a self-referencing problem. Without resorting to simulation, can you think of a way to get a reasonable answer?

9. Jan 4, 2018

### Staff: Mentor

The natural convection part of it will not be important. The important part will be forcing air into the room at the required inlet rate with the inlet fan. This will make the same rate of air exiting the hole in the ceiling (provided the hole isn't too small, thus causing a pressure buildup in the room). The actual flow rate from the fan will be determined by the pressure buildup and the fan operating characteristic (provided by the fan manufacturer). The operating characteristic is defined as the volume flow rate as a function of the fan rotational speed and the pressure difference.

10. Jan 4, 2018

### yahastu

I was actually thinking of an inlet vent and an outlet fan, so that removing the outlet fan means the only pressure differential is due to natural convection through the open air hole that remains, which is a function of the exit hole size.

Your answer raises another interesting point, though -- does it make a difference if the fan is placed at the intake vs the exit? My intuition tells me that it would make a difference, because an exit fan pushes air into open air with no resistance, whereas an intake fan would push air directly into an interior wall, causing turbulence and increased resistance, hence slowing down the fan blades and reducing the airflow at both the intake and the outtake.

11. Jan 4, 2018

### Staff: Mentor

In my judgment, there would not be much difference between using an inlet fan or using an outlet fan, if there were a corresponding outlet hole or inlet hole in each case.

12. Jan 10, 2018

### Staff: Mentor

Convection is difficult. As upper limit on the convection rate: Find the pressure difference between inlet and outlet with the outside temperature, find the same with the room’s temperature. Both will depend on the absolute temperature, the height difference and the gravitational acceleration. The difference us the energy per volume released by convection, and as upper limit all this energy could accelerate the gas (guve it kinetic energy) twice, at the inlet and outlet. Setting these energies equal and solving for the mass flow will give a rough estimate of the convection rate.