×
INTELLIGENT WORK FORUMS
FOR ENGINEERING PROFESSIONALS

Are you an
Engineering professional?
Join Eng-Tips Forums!
• Talk With Other Members
• Be Notified Of Responses
• Keyword Search
Favorite Forums
• Automated Signatures
• Best Of All, It's Free!

*Eng-Tips's functionality depends on members receiving e-mail. By joining you are opting in to receive e-mail.

#### Posting Guidelines

Promoting, selling, recruiting, coursework and thesis posting is forbidden.

# Heating of air in enclosure - calculation and experiment discrepancy4

## Heating of air in enclosure - calculation and experiment discrepancy

(OP)
Hello everybody,

I would like feedback please on a large discrepancy between a theoretical calculation and an experimental measurement I have taken.

I ran a 470 Ohm 5 Watt ceramic (at 50 V and 0.1 A, both confirmed using a multimeter) resistor inside of an 8 litre insulated cooler box and measured the air temperature (two thermocouples used, both in close agreement) over time. The resistor was suspended inside of the box and was not in contact with the sides or bottom. I also had a DC fan blowing air across the resistor in an effort to make sure heat is being transferred out of the ceramic element and into the air.

If I calculate the amount of energy needed to raise the temperature of air from 26 °C to 31 °C:

Q = m∙cp∙ΔT

Where:

Q is energy provided by the resistor (Joules),

m is the mass of the air in the box, 0.0095 (kg),

cp is the specific heat capacity of air, 1006 (J / (kg∙K))

ΔT is the change in temperature, 5 (°C)

After converting 8 litres to 0.008 m^3 and multiplying by air density of 1.184 (kg / m^3) I get around 0.0095 kg.

So to change the temperature of air by 5 °C (from 26 °C to 31 °C):

Q = (0.0095) * (1006) * (31 - 26) = 47.8 J

In the experiment it took 15 minutes (900 seconds) for the air to heat from 26 °C to 31 °C. 5 Watts multiplied by 900 seconds is 4500 Joules. So I calculated that it would take roughly 50 J to heat the air by a certain amount. But the experiment required nearly 95 times the amount of energy.

The only two explanations I have are that:

1. The box is not a very effective insulator and leaks quickly. Though the sides of the box felt cool to the touch during the experiment. If energy were conducting through the material, I should be able to feel the box warming up.

2. Some of the energy produced by the resistor goes into heating a) the ceramic body of the resistor itself, b) the interior material of the box. b) does not seem very likely to me as the resistor is hanging in the air.

I’m not sure what I am doing wrong, but would appreciate feedback and advice.

My reason for doing this test is because I would like to be able to put some electronics inside the box, power them and calculate how much heat they are dissipating based on the temperature that I measure. But I am now unsure if that will be reliable based on these results.

### RE: Heating of air in enclosure - calculation and experiment discrepancy

Your air needs to be in thermal equilibrium with its container; that implies both 1 and 2. Moreover, the thermal mass of the interior wall of your container is easily 100 times the mass of the air you are trying to heat to the same temperature; this is why people usually use water as the medium for this sort of experiment.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

### RE: Heating of air in enclosure - calculation and experiment discrepancy

Your objective seems misguided to me. If you have the ability to measure the air temperature, you should just measure the temperature of the electronics; in most datasheets for components, there are thermal resistance equations and parameters that allow you to calculate the limits of the components relative to ambient or surface temperature of the component.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

### RE: Heating of air in enclosure - calculation and experiment discrepancy

I am with IR here.
Even if you only heated the inner 2-3mm of the cooler that mass will dwarf that of the air.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

### RE: Heating of air in enclosure - calculation and experiment discrepancy

Your calcs are correct. Would agree with your observation at 2a - how much ceramic is there in this heater ? It is likely that this heater, in 900sec, would have reached temperatures exceeding 100degC ??
Are thermocouples installed with tips exposed to the inside air, or are they sitting in a metal thermowell ?
Otherwise, use thermistors for reduced thermal inertia in the sensing device.
It is likely that the metal wall of this enclosure also will be heated up. What if the mean temp of the wall goes up to say 28degC( since you've got forced convection with this internal fan) ? Can you install a thermistor on the outside face of the metal wall so you get a reading of this too ?

### RE: Heating of air in enclosure - calculation and experiment discrepancy

#### Quote (Miran)

I would like to be able to put some electronics inside the box, power them and calculate how much heat they are dissipating based on the temperature that I measure.

If you put some electronics in a box and power them then 100% of the electrical power that you provide will be dissipated as heat. All of it.

Just measure the power input, much easier.

### RE: Heating of air in enclosure - calculation and experiment discrepancy

(OP)

#### Quote (IRStuff)

Moreover, the thermal mass of the interior wall of your container is easily 100 times the mass of the air you are trying to heat to the same temperature; this is why people usually use water as the medium for this sort of experiment.

#### Quote (IRStuff)

Your objective seems misguided to me. If you have the ability to measure the air temperature, you should just measure the temperature of the electronics; in most datasheets for components, there are thermal resistance equations and parameters that allow you to calculate the limits of the components relative to ambient or surface temperature of the component.

I should have been more clear in that the 'electronics' is actually a small off the shelf apparatus that generates, according to the datasheet, ~ 60 W of heat. I wasn't too sure how accurate that was and wanted to measure it, but as you say my setup isn't really appropriate. Unfortunately, I obviously can't submerge it in water as it will be damaged/destroyed. It's not something I would feel confident taking voltage/current measurements on.

#### Quote (EdStainless)

Even if you only heated the inner 2-3mm of the cooler that mass will dwarf that of the air.

Yes I was becoming suspicious that was the case, but I managed to talk myself out of it. But thank you both for pointing this out, I will trust your judgment on that.

#### Quote (georgeverghese)

Would agree with your observation at 2a - how much ceramic is there in this heater ? It is likely that this heater, in 900sec, would have reached temperatures exceeding 100degC ??
Are thermocouples installed with tips exposed to the inside air, or are they sitting in a metal thermowell ?

The ceramic element is quite small ~ 10 mm x 10 mm x 20 mm. I placed a thermocouple on it after powering on the power supply and it reached about 125 °C quite quickly and sat there (5 W dissipated). The thermocouples have the tips exposed to the air, they are not in contact with anything.

#### Quote (georgeverghese)

It is likely that the metal wall of this enclosure also will be heated up. What if the mean temp of the wall goes up to say 28degC( since you've got forced convection with this internal fan) ? Can you install a thermistor on the outside face of the metal wall so you get a reading of this too ?

The enclosure is all plastic, it is literally a small drinks cooler. I may rerun the test with the thermocouples in different locations i.e. inner wall, outer wall, surface of resistor just for completeness. But as others have pointed out, this experiment seems like the wrong track for me to be on.

#### Quote (MintJulep)

If you put some electronics in a box and power them then 100% of the electrical power that you provide will be dissipated as heat. All of it. Just measure the power input, much easier.

It is an off the shelf apparatus that I can't really go pulling apart, or messing with the power leads. Otherwise I agree with you.

Thank you all for your feedback and advice. As IRStuff has said this isn't really the right way to go about testing something like this. I could trust the datasheet but I think what is listed is possibly its peak power output and I don't trust that it is putting out 60 W at all times. Is there some other way I could approach this? Perhaps characterise the leak rate of the enclosure? If I know how much it is leaking, the next time I test a different load inside I can take that into account to get a more accurate measurement?

### RE: Heating of air in enclosure - calculation and experiment discrepancy

#### Quote:

according to the datasheet, ~ 60 W of heat.

So, just measure its input power; conservation of energy is always applicable.

TTFN (ta ta for now)
I can do absolutely anything. I'm an expert! https://www.youtube.com/watch?v=BKorP55Aqvg
FAQ731-376: Eng-Tips.com Forum Policies forum1529: Translation Assistance for Engineers Entire Forum list http://www.eng-tips.com/forumlist.cfm

### RE: Heating of air in enclosure - calculation and experiment discrepancy

Feed the power through a metered source and then you will know the exact consumption.
Often the spec sheet value is for maximum power consumption.
Even when typical operation or standby may draw a lot less.

= = = = = = = = = = = = = = = = = = = =
P.E. Metallurgy, consulting work welcomed

### RE: Heating of air in enclosure - calculation and experiment discrepancy

(OP)

#### Quote (IRStuff)

So, just measure its input power; conservation of energy is always applicable.

#### Quote (EdStainless)

Feed the power through a metered source and then you will know the exact consumption.
Often the spec sheet value is for maximum power consumption.
Even when typical operation or standby may draw a lot less.

I'll see if I can buy a power meter and measure the input as you both suggest. I will update when I have measured it.

Thank you both for your help, I really appreciate it.

### RE: Heating of air in enclosure - calculation and experiment discrepancy

My meter is a simple plug-in design.
It displays voltage, current, power, freq, and power factor all at once.
I think that I paid about $30 for it. They are very handy for trouble shooting appliances and such. = = = = = = = = = = = = = = = = = = = = P.E. Metallurgy, consulting work welcomed ### RE: Heating of air in enclosure - calculation and experiment discrepancy (OP) #### Quote (EdStainless) My meter is a simple plug-in design. It displays voltage, current, power, freq, and power factor all at once. I think that I paid about$30 for it.
They are very handy for trouble shooting appliances and such.

I bought one of these today with the same functions as the one you described. I've only done some quick tests (I haven't ran the apparatus at full-load, which will eventually operate at, but close to) and I can see that it will dissipate close to what the datasheet listed.

Thanks for making the power meter suggestion EdStainless, IRStuff and MintJulep, very quick and easy method to do what I wanted.

Thanks again for the help.

#### Red Flag This Post

Please let us know here why this post is inappropriate. Reasons such as off-topic, duplicates, flames, illegal, vulgar, or students posting their homework.

#### Red Flag Submitted

Thank you for helping keep Eng-Tips Forums free from inappropriate posts.
The Eng-Tips staff will check this out and take appropriate action.

Close Box

# Join Eng-Tips® Today!

Join your peers on the Internet's largest technical engineering professional community.
It's easy to join and it's free.

Here's Why Members Love Eng-Tips Forums:

• Talk To Other Members
• Notification Of Responses To Questions
• Favorite Forums One Click Access
• Keyword Search Of All Posts, And More...

Register now while it's still free!