Measuring radiant thermal heat loss.

Hello,

I can read the temperature with an Arduino, but I would also like to know the radiant heat loss of my body for a room.
A temperature sensor measures only the air temperature.

To measure the radiant heat loss, I was thinking of a piece of metal inside a block of expanded polystyrene. The metal has a temperature sensor and is kept at 37 degrees Celsius (99 degrees Fahrenheit) with a heating wire. I measure the energy to keep it at 37 degrees, and that tells something about the radiant heat loss.

The block polystyrene is to avoid the influence of draft. The metal will be inside completely surrounded by polystyrene without air. Perhaps the metal has to be black.

Does this make any sense ?
Does something like this exist ?
How accurate would the temperature sensor has to be ?

Dosrd: Does this make any sense ?

No.

Perhaps radiant heat loss means something different to you. To me it means the heat energy being emitted by radiation i.e. infra-red, visible light, ultra-voilet. The amount of radiated power is proportional to the fourth power of temperature and for things at human body temperature the power density is very low indeed. If you're trying to measure the proportion of heat you're losing from radiation and conduction, I think a good first approximation is that 100% of it is lost by conduction.

If you want to measure the radiated heat at body temperature you'd need the infra-red equivalent of a light intensity meter. I've never seen one, but I'd imagine they're rather specialised bits of kit and not cheap.

I’m trying to measure a room.
This depends greatly on the temperature outside, but if I can have some numbers for different rooms and different temperatures outside, perhaps I can make sense of it.

The loss of heat by conduction is certainly not 100%.
Infrared heat radiation loss could be 10% - 30%. I read that in some special cases it could be up to 50%.

Some jackets have a radiant heat reflective layer. Those jackets do actually prevent some radiant heat loss.
Try searching for example for : “columbia omni-heat reflective”.

Dosrd: I'm trying to measure a room. This depends greatly on the temperature outside, but if I can have some numbers for different rooms and different temperatures outside, perhaps I can make sense of it.

Measure what about the room? How much it is being heated by heat radiated from hot bodies inside it?

Dosrd: Some jackets have a radiant heat reflective layer. Those jackets do actually prevent some radiant heat loss. Try searching for example for : "columbia omni-heat reflective".

I don't think that a sportswear supplier is a good source of impartial scientific information.

It's possible to calculate the radiated heat loss, knowing the temperature of the object and the temperature of it's surroundings, and the surface area, and the thermal emissivity of the surface material. Here's an explanation:

http://www.ehow.com/how_7168415_calculate-radiant-heat-loss.html

Here is a reference to a table of emissivity values for various materials.

http://www.ib.cnea.gov.ar/~experim2/Cosas/omega/emisivity.htm

It suggests figures around 0.7 for various cloths.

Assuming your human has a surface area of two square meters and is at 38C in a room at 20C, my back-of-the-fag-packet arithmetic calculated the radiated power as 0.2W. Maybe I fumbled the sums, but that's the sort of number I was expecting.

PeterH:

Dosrd: Does this make any sense ?

No.

Perhaps radiant heat loss means something different to you. To me it means the heat energy being emitted by radiation i.e. infra-red, visible light, ultra-voilet. The amount of radiated power is proportional to the fourth power of temperature and for things at human body temperature the power density is very low indeed. If you're trying to measure the proportion of heat you're losing from radiation and conduction, I think a good first approximation is that 100% of it is lost by conduction.

If you want to measure the radiated heat at body temperature you'd need the infra-red equivalent of a light intensity meter. I've never seen one, but I'd imagine they're rather specialised bits of kit and not cheap.

Most organic chemicals have strong lines in the IR spectrum so the human body could be approximated as a black-body radiator. For a 35C object in a room of ambient temp 25C the heat loss would be 57 W/m^2. Outside on a cloudless night the loss would be 450 W/m^2 if facing skywards. So the ambient surroundings temperature really dominates radiative heat loss.

MarkT: Most organic chemicals have strong lines in the IR spectrum so the human body could be approximated as a black-body radiator. For a 35C object in a room of ambient temp 25C the heat loss would be 57 W/m^2.

How did you calculate that?

PeterH: Measure what about the room? How much it is being heated by heat radiated from hot bodies inside it?

No, I want to know about the warmth my body has to produce in that room.

Suppose my body is 37 degrees, and the room is 22 degrees. I can sense clearly that when it is outside 20 degrees I have less problems staying warm than with an outside temperature of -10 degrees. But in both cases I am in the room with an air temperature of 22 degrees.

I don't understand the calculations, but if I can sense it clearly, perhaps I can measure it. I know that's not very scientific, but that's why I want to do some measurements.

Dosrd: No, I want to know about the warmth my body has to produce in that room.

Suppose my body is 37 degrees, and the room is 22 degrees. I can sense clearly that when it is outside 20 degrees I have less problems staying warm than with an outside temperature of -10 degrees. But in both cases I am in the room with an air temperature of 22 degrees.

I don't understand the calculations, but if I can sense it clearly, perhaps I can measure it. I know that's not very scientific, but that's why I want to do some measurements.

I doubt that radiant heat is the cause. I'd be more inclined to suspect it's a draft, or cold spots in the room. I find that the floor temperature drops substantially in winter and this seems to pull down the air temperature near the floor. For example, where I am now the outside temperature is just about freezing; the internal and external walls are at 20C and so is the air temperature in most of the room. But the floor is down at 17C and the air temperature at floor level is also around 17C. Having cold feet makes the room feel much colder. These temperature variations are also likely to set up convection currents which will make the room feel significantly colder. I've got a friend with underfloor heating and at the same room temperature that house feels warm and cosy because there's no convection and no cold spots.

Does this make any sense ?

Yes.

How accurate would the temperature sensor has to be ?

Not much.

You can incorporate this with a pwm controller for the heating element: whenver the heater is turned on, you accumulate a value, which will be proportional to the amount of energy radiated out.

PeterH, I live on the second floor, and the room below me is also 22 degrees. For a healthy active person, the radiation heat loss would perhaps be unnoticeable. For me it makes a difference. I think it makes even a big difference.

dhenry, Thank you. But I'm not sure about the accuracy of the temperature sensor. For the PWM signal I could use a 16-bit timer, or do it in software to get a higher accuracy than 8 bits. But the PWM value is only accurate if the temperature is very accurate. The temperature doesn't have to be precise 37 degrees, but perhaps I need to be able to measure a variation of 0.1 degrees Celsius. Or 0.01 degrees, I don't know. I need the measure the room temperature with the same accuracy.

I started to think about a metal sphere inside a block of expanded polystyrene. But reading your replies I think it is about surface area, so I better use a flat metal surface (aluminum) inside a block of polystyrene. Perhaps I need to place the heating wires inside the metal, so all radiation heat is from the metal and not directly from the heating wires. Instead of heating wires, I could use a resistor of 10 Ohm 10 Watt.

There is a whole lot of issues here.

If you have ever seen "infra-red" cameras for hunting or surveillance, heat radiates at a different rate from different parts of the body. And then there is the effect of clothing. And heat can transfer by radiation from your skin to your clothes and then by convection and conduction from the clothes to the air. It's not simple, and I get the impression you don't know what you are doing.

A better idea, might be to go back to square one and rethink what you are actually trying to achieve ?

Maybe this post would be of interest to you for measuring heat http://arduino.cc/forum/index.php/topic,54655.0.html

Dosrd: PeterH, I live on the second floor, and the room below me is also 22 degrees. For a healthy active person, the radiation heat loss would perhaps be unnoticeable. For me it makes a difference. I think it makes even a big difference.

I completely accept that there is a perceptible difference between the two situations.

The bad news is that measuring heat radiation will be very hard to do, probably impossible to get any meaningful results with a DIY solution. The solution you're proposing is a good start towards measuring total heat loss but gives you no information at all about how much of that heat was lost by radiation, conduction, convection, evaporative cooling etc.

The good news is that what you really want to measure, it seems to me, IS the total heat loss. I think a sensible first step to understand why you feel less comfortable in the winter is to find whether your total heat loss is changing measurably. If you find it is then you can try to find what mechanism is causing it. I think you're barking up the wrong tree with radiative cooling. My guess is that there are temperature variations and cold spots making you feel colder, and drafts/convection currents causing wind chill, and the air is drier so evaporative cooling makes you even more susceptible to wind chill.

What you're proposing so far measures heat less from all mechanisms except evaporative cooling. It won't measure convection at all accurately because that depends on the size of the object driving the convection, but it will measure some convection.

To be representative, you would need to address the fact that you are (presumably) wearing clothing so the surface temperature exposed to the cold air is not at body temperature. However, when you're wrapped up against the cold I think that a large proportion of heat loss is from the head. All you need to do is insulate the block sufficiently so that the outer surface exposed to the air is at the same temperature as your face when the core is at nominal body temperature.

But I'm not sure about the accuracy of the temperature sensor.

You don't need the sensor to be (super) accurate. You just need it to be consistent.

Thanks for all your replies.

I know about thermal camera's and the MLX90614 IR sensor. I could go outside to see if a room radiates more heat then another room, but that's not what I want to know.

It is indeed almost impossible to measure only the heat radiation for an object of 37 degrees. There is too much going on. A lot of heat is lost by breathing, by evaporation, and so on. That's all true.

I don't want to measure the total heat loss, but I want to know if it is possible to measure the radiation heat loss for a certain room (with a certain outside temperature, humidity, etc.).

I guess that I also have to measure the humidity of the room. That's another variable.

Meanwhile I build a small version (a cube of about 10x10x10 cm) with a resistor of 27 Ohm 3W. I have start somewhere.

Dosrd:
I don’t want to measure the total heat loss, but I want to know if it is possible to measure the radiation heat loss for a certain room (with a certain outside temperature, humidity, etc.).

If your goal is to understand why the room feels less comfortable, it seems to me that the total heat loss would be the right thing to measure. If you have decided not to try to measure that, that’s your business.

When you talk about measuring the heat loss ‘for a certain room’, I assume you are still talking about the heat transfer between a person and the room they’re in.

I can only see a couple of approaches. One is to find and beg/steal/borrow/buy some sort of infrared intensity meter to measure how much heat is being transmitted. I suspect this route would be difficult and expensive. If you managed it you’d need to plot out the intensity over the whole surface of the subject and then integrate that over the area to give you the total heat power being radiated.

The other, much simpler and easier approach is to use an IR thermometer or similar to measure the surface temperature at representative points over the whole area, measure or look up the approximate emissivity of the materials at the surface, measure the temperature over the room’s inner surface in a similar way, and plug those temperature differences, emissivity and areas into the Radiant Heat Loss equation I linked you to yesterday. You won’t be able to do this in one go - you’ll need to conceptually divide the person’s surface up into pieces with similar emissivity and temperature and calculate the heat loss for each piece, then add them up to get the total.

To avoid wasting too much time on this, as the first step I suggest you make worst-case approximations for all these values (round the surface area up, assume the whole surface is at the temperature of the hottest point, assume the emissivity of the whole surface is the emissivity of the highest point) and calculate an upper bound for the radiated heat. You should already have a rough idea how much heat a resting person puts out in total so you can decide whether the radiant heat loss is actually of any relevance to your problem before you try to estimate it more accurately. I already explained that my ‘back of a fag packet’ calc suggest that it would account for less than a tenth of a percent of the total heat loss.

PeterH,

Thanks for you thoughts.

I already measured a few things with an IR thermometer. But it will give some indication, and it is hard to get a single number. I can't set the outside temperature, so I have to think about how make good measurements that I can repeat a few times a year. Luckily I have a nearby weather station with a website with good data. I could add that to my measurements.

About the less than a tenth of a percent. I disagree with that. If I use a aluminized foil (emergency blanket) I certainly feel a lot warmer after about 20 minutes. Without the aluminum I don't feel that. As I wrote before, in some special conditions, the radiation heat could be up to 50% of the heat loss for a human. If you could block the radiation heat of your house, your house would be a lot warmer in the winter and a lot cooler in the summer.

I have the first data of my small test. But due to overshoot of the temperature control, the data has too much variation. So I can't compare it with other conditions.

Dosrd: If you could block the radiation heat of your house, your house would be a lot warmer in the winter and a lot cooler in the summer.

That part I don't have any trouble believing, because there's about a 6000C temperature difference between the house and the sun, and about 280C difference between the house and the cold night sky.

It's the radiated heat when there is only a 20C differential that I have trouble believing is as big as you expect. I've pointed you to an explanation of how radiated heat varies, and suggested ways to get a rough guesstimate and a more accurate reading. I believe you have all the information you need to prove or disprove your theory.

Good point.

I need to collect a lot of data, and in the end I probably won't be able to a number of the radiant heat. So I will collect as many data as I can, and will try to see if I can pinpoint the effects that actually do occur.