How Do You Correctly Calculate the Heat Transfer Load?

It's analogous to an electrical circuit, where heat flow (q) is like current (I), temperature differential (To-Ti) is like voltage (V).

Ohm's law for electrical current: I = V / R, current = voltage / resistance.

Equivalent law for flow of heat: (q = (To - Ti ) ÷ [(1/ho ) + (1/hi ) + R] x 5.67

To pass from inside to outside the heat has three "obstacles", internal convection (1/hi), conduction through the cabinet wall (R), and external convection (1/ho). It's like three resistors in series for an electrical circuit. Note that the convection "resistance" is the inverse of the convective "conductance", (hi or ho).

I hope this helps...

Hi Rixter, thanks for your answer. The part that is not clear to me is the calculation of the ∆T, for me the value will be negative... I've considered To as 30°c while the Ti which is the rated temperature of the electronic components running inside my enclosure is 60°c.

So To-Ti is -30...

PWSlack i thought about that but I wasn't sure!

Thank you so much!

The guy that derived the formula picked the opposite direction. Maybe he was figuring for an ice chest.

Presumably you need to ensure the calculated total heat loss (q*cabinet surface area) at least = the heat generated by the electronic components. If so, for safety you would most likely use the lowest figures for h, but have no insulation, as insulation would reduce heat flow.

What is the blower doing? If it draws outside air through the cabinet and then vents it, it might keep the inside temperature within limits without worrying about the convective heat losses.

Funny about your analogically example,... i had explained to green engineer about an electrical circuit by using a pneumatic circuit about 2 months ago.

And one can even see the light bulb lighting up in their eyes... so to speak...

Not quite that simple,... hence your question. I’ve written excel sheets with vbasic that did that, but it’s been quite a while since, so I did a quick internet search ... and Concluded, I was born about 25 years too late, could have saved me some time because there’re all over the internet.

there are a number of variables you need to solve concurrently, and to do this without going through details, you would need to enter an approximate starting point (number) and then solve using iterations. And the more iterations that are used, the more accurate it is... theoretically accurate that is.

the other issue of course is units

seeing that ugly 5.67 suggests to me that this is a calculation in Imperial / US Customary units. Its also not clear to me if the 5.67 should be in the numerator or denominator

This is a rearrangement of Q = U A ΔT

Where U - overall heat transfer coeff 1/U = 1/hi + 1/ho + R

so you have Q/A = (To - Ti) / (1/hi + 1/ho + R)

I would normally go with

Q - W

A - m2

q - W/m2

T - K (or °C) they are the same for temp changes

h - W/m2K

R - m2K/W if using Thermal Insulance (the R value you quote)

or if using Specific thermal resistance mK/W and you will multiply by the insulation thickness (symbol t measured in m)

In this case you end up not needing a conversion factor and

q = (To - Ti) / (1/hi + 1/ho + R)

or q = (To - Ti) / (1/hi + 1/ho + Rt)

Your data is throwing around various Imperial units with different length and time dimensions and so you should carefully work through the calculation to determine the unit correction factor.

The insulation term (R or Rt) would be the first place I would look.

You seem to be overly worried about the sign, without focussing on the numbers, Mildred. Those who give a damn about the sign will be few and far between.

Hi again, can anyone explain to me this table?

I don't understand why the bigger is the ΔT the less air flow you need...
I would have thought that if you have a bigger number it means you need more air to cool the enclosure. Sorry if I'm being stupid but I can't read it.

You can't cool below ambient with just air....the greater the differential the more air you need...

I know that, that's the thing if the ambiente temperature is 35°c and inside the enclosure the temperature reaches 75°c the ΔT is 40. If you cross reference 40 with lets say 1500w load you get a lower air flow than if your ΔT was 20 for instance.
That's not clear to me...

I would imagine that the bigger is the ΔT the bigger is the volume of air necessary to lower the internal temperature close to the ambient one.

You are calculating the airflow needed to dissipate the heat being produced according to maximum temperature allowed...If your components have a max operating temp of 120°, and your max ambient is 95°, and your equipment is generating 200 watts of heat, you need enough airflow to dissipate 200 watts of heat with a 25° operating range above the max ambient of 95°...The amount of air flow will depend on the enclosure area, let's say you have a 1 cu ft enclosure...then you need to determine the static pressure of the enclosure...

The volume of air will always allow for some heat rise in the enclosure, that's what you are looking at...You have to keep the level of heat rise below the max operating temp minus the ambient....

https://blog.exair.com/2017/03/22/not-a-fan-of-fans-because-rising-air-temp-will-kill-your-electronics/

http://cyntech.co.uk/cooling-power-supplies/

I still don't get the last table...

It's not clear to me why the bigger is the difference between the ambient temperature and the internal temperature and the least air flow you need.

Sorry if being stupid but I don't get how to read that table.

"You have to keep the level of heat rise below the max operating temp minus the ambient"

That's what I want to do but I'm not getti how I can determin the airflow needed to lower the internal temperature at least 30° [outside is 35°c and inside is 76°c]

I think you're getting 2 different thiongs mixed up.

To calculate heat flow through a system of known conductivity, heat flow q is proportional to ΔT, as in your formula q = (To - Ti ) ÷ [(1/ho ) + (1/hi ) + R] x 5.67.

But if you want to calculate air flow q_air (m³/s) to remove a given heat load q,

q_air = q/ΔT/C where q is in kW, ΔT = max temp acceptable in the enclosure minus incoming air temp, and C = air specific heat, in the region of 1.2 kJ/m³/K

Yeah I need the second formula then.

But how does it work with the numbers I have?

Incoming air temp. 35°c

Max acceptable internal temp 35°c

Internal temperature during normal working cycle 76°c [when outside is 35°c]

If your max operating temperature for internal components is 76°C and your ambient temperature is 35°C, then the differential is 41°C....So if you have 1500 watts total load just look at the chart and determine how much air flow you need to keep the temperature within operating parameters...I would allow a safety margin, you always design cooling to overcome the temperature on the hottest day, that is your design temperature...If your hottest day is going to be 48°C, then that is your design ambient...So your max temperature for your components should target below 76°C to say 60°C...So you need to make a determination what duty cycle you want for the fan, over a certain temperature it runs all the time or it cycles on temperature with a thermostat...So when the temperature hits 60°C the fan kicks on and stays on until the temperature hits 45°C, say as an example, so you have a 15°C differential...

Now the amount of air flow you need can be found with the chart... the difference between 76°C and 60°C these are your max operating conditions, is 16°C, 1500 watts is equal to 5100 btu's/hr, so we look at the chart and see we need about 350 cfm...So to overcome static pressure and recovery time we would probably go with about 450 cfm...

Much more clear, one thing though, the operating temperature is not 76°c. That is the temperature that the totem reaches inside while operating in a sunny day. The max operating temperature for the electric components inside should not be bigger than 35°c

The design temperature here in Florida is 95°F - 75°F , that means we size the A/C units at the tonnage that will pull the temperature down to 75°F on a 95°F day....The design temperature of the electrical components is the maximum temperature that they will continue to function properly...

..."Conventional electronic components are designed to operate over a specified temperature range with upper limits generally set at 70°C for commercial applications, 85°C for industrial applications, and 125°C for military applications."...

Ref...

So when you are designing a system for cooling, there is a reason, for dwellings it's comfort and humidity control, but for electrical components, it's to keep them operating properly...In some cases that requires forced ventilation...in some cases it requires A/C... In any case you use the design temperature as a guide for sizing cooling components...Now whether it's cooler outside than the design temperature only means that the unit will have to run less, but on the days that it reaches the design temperature, it will still do it's job....

Most of the stuff inside there will work with no problems around 70° but I have also quite big LCD display in there and the company specified that they have to be around maximum 50°C to avoid quick pixels deterioration and invalidate the warranty.

To be on the safe side I wanted to keep the temperature around the ambient one.

The 2 statements

Max acceptable internal temp 35°c

Internal temperature during normal working cycle 76°c [when outside is 35°c]

appear to be contradictory. Is the 76°C the temperature reached in the sun? (which your design is presumably trying to prevent). But that wouldn't be Internal temperature during normal working cycle. Please explain.

If possible, a sunshade could be provided. Otherwise, you would need to estimate the heat input from the sun, add this to the heat generated by the internal kit, and design the ventilation flow for that.

But if the outside temperature (hence cooling air input temperature) is 35°C you clearly cannot have inside temperature the same or air flow is ∞. There has to be a compromise. When you have decided on the ΔT available you can find the air flow either from the chart posted or my formula.

So "appear to be contradictory. Is the 76°C the temperature reached in the sun?" yes that is the temperature that the volume inside my enclosure will reach in a summer sunny day,
during normal working cycle. That's why I must reduce that temperature to avoid deterioration of the components inside. [critical temperature for some of those electrical components is 50°c so I would want to try to lower the temperature to 35°-40° max]

"If possible, a sunshade could be provided. Otherwise, you would need to estimate the heat input from the sun, add this to the heat generated by the internal kit, and design the ventilation flow for that."
No shading is possible and yes I did all the calculations considering internal heat load, heat dissipation through the enclosure walls, solar gain etc. and I found out that the temperature hit 76°c.
"design the ventilation flow for that"
that's where I'm stuck, I can't get around the determination of the air flow volume necessary to lower that temperature.

"yes that is the temperature that the volume inside my enclosure will reach in a summer sunny day," In that case it's not the "Internal temperature during normal working cycle."

What heat load have you arrived at?

With outside air temperature 35° and maximum inside 40° that gives ΔT 5°. From that it's easy to calculate air flow using my formula. Give me your heat load and I'll do it for you.

Does this make sense?

https://drive.google.com/file/d/1QmHRDlzNZZUMBrME2DPNE3EBie3XJfHA/view?usp=sharing

Maximum temperature reached sitting the Sun has no meaning in calculating heat gain, which is the btuh gain....

..."Heat Load Transfer Heat load transfer is the heat lost (negative heat load transfer) or gained (positive heat load transfer) through the enclosure walls with the surrounding ambient air.

This can be calculated by the following formula:

Heat load transfer (BTU/H) = 1.25 x surface area (sq. ft.) x (max. outside ambient air (°F) – max. allowable internal enclosure temperature air (°F)) Surface Area (sq. ft.) = 2 [(H x W) + (H x D) + (W x D)] / 144 sq. inches Note: 1.25 is an industry standard constant for metal enclosures;

...0.62 should be used for plastic enclosures. Once you have determined your Internal Heat Load and the Heat Load Transfer, you can choose the proper size unit by calculating the needed cooling capacity. Cooling capacity (BTU/H) = Internal Heat Load ± Heat Load Transfer"...

https://www.stego-usa.com/nc/support/calculation-tools/cooling-calculation/

https://cdn.automationdirect.com/static/specs/enccoolingselection.pdf

In #23 OP said he's already done the calcs.

There's no mention of wall conductivity in your formula, so I assume it just applies to sheet metal enclosure. I can't see anything about heat from the sun.

Does this make sense?

https://drive.google.com/file/d/1QmHRDlzNZZUMBrME2DPNE3EBie3XJfHA/view?usp=sharing

What are the measurements of the cabinet? Your heat gain seems reeally high...I think you are figuring heat gain in BTU's per hour and CFM air flow in CF per minute...Is this cabinet to be in the shade or direct sun exposure...You might just need louvered ventilation...

In #23 OP said it's in the sun and can't be shaded.

If the inputs are correct I don't see a problem. Assuming the heat transfer load calc with the 5.67 factor is correct (I don't know the source).

No calc is given for the solar heat gain.

Possibly insulation would help, this would reduce the solar gain and also the -ve heat transfer load, and there might be a net reduction.

The 76.6° figure isn't used as far as I can see. Using ΔT 5° and 1156 watt I make air flow just over 400 cfm, so 500 cfm for some safety is reasonable.

The point of the graph is that the less airflow you have, the greater your temperature rise will be for a particular value of thermal power to be dissipated.

The opposite is true. The closer the approach desired to the selected temperature, the more fluid at that temperature will be needed to make the approach from the calculated temperature.

Because of the heat dissipation, the selected temperature can never be equal to the ambient temperature, as that would require an infinitely large fluid flow.

Guys, I have a question, when I calculate the internal heat load I take into account everything that absorb power.

I have a doubt...if I have for instance 15w led module obviously goes in the calculation but what about its transformer [which is indeed a 15W driver]

How does it work, do I take in to account both like 15W + 15W?

Obviously both produce heat...

Thanks

You seem to be looking at the nominal size rating of these components instead of the actual wattage load...If you look at a circuit breaker at 15 amp rating and 120v line that = 1800 watts, that's the max safe load, not the actual wattage being used...The efficiency of the driver is likely to be around 80%, so you would add 20% of the load as heat to the 15 watt LED, or 3 watts...

You might have added that an 1800 watt breaker only dissipates a few watts (maybe, if it has a coil) whether it's at max load or lower actual load.

For the LEDs the driver data might include input power at design 15 watt output, hence heat loss, and the LEDs may not consume 15 watt, reducing the internal heat load.

Friends,

Lots of good information and questions in this topic. A few I will add:

• Always include the units as you work the formulas--a mixture of imperial and metric will result in errors that are easy to catch when the units are included.
• For the display, the manufacturer seems to be stating the maximum ambient temperature for it, not the actual maximum surface temperature the components are safely operating at internally. So the 35ºC needs to be heeded.
• A few posts have mentioned radiative heating/cooling, but the various formulas I have seen don't include this. This is directly related to the surface's reflectivity (the inverse of emissivity), a unitless number that ranges from nearly 0 to 1. To minimize heat gain due to direct solar exposure, the reflectivity needs to be as high as possible, so a black or average metal surface is a poor choice (white is much better, as is a polished silvery surface). To increase heat transfer OUT of the enclosure, its interior walls need a low reflectivity, but if the walls are warmer than the desired interior temperature a high reflectivity is desired.
• Wally 89, you are a newer member of the community. Welcome. Keep asking questions and keep studying. Mistakes are made by all of us, so lean from them also.
• The lifespan of electronic components is significantly reduced (or increased) by changes in their operating temperature. An approximate rule-of-thumb is a 10ºC increase above the specified operating temperature cuts the life span by 50%; a similar decrease results in a much greater life span. So yes, you can operate devices at too high a temperature if you are willing to sacrifice their life. Of course there are limits on how high, when the component fries instantly.

--JMM