Effect Of Wind Speed On Heat Loss From A Heated Building

Hi All.
Back again with another query to drive you all daft.
I can easily work out the heat loss through a building fabric if I know the internal temperature, the external temperature and the overall thermal resistance of the fabric and if the external air is static.
Does anyone know how to account for wind velocity?
I know this could become very complicated - Is the flow laminar or turbulent? What effect does wind direction have (as most buildings are rectangular in shape)? And so on. But there might be some mechanism to approximate the effect. And anything that could make some sort of allowance for this would have to be more accurate than ignoring the effect altogether.
Take care.
Mike

AC Me wrote:

I can easily work out the heat loss through a building fabric if I know the internal temperature, the external temperature and the overall thermal resistance of the fabric and if the external air is static.
Does anyone know how to account for wind velocity?

You might try a web search for "wind chill" effect...
-- Morris Dovey DeSoto Solar DeSoto, Iowa USA http://www.iedu.com/DeSoto/

On Dec 1, 7:35�pm, Tim Jackson wrote:

AC Me wrote: Hi All.
Back again with another query to drive you all daft.
I can easily work out the heat loss through a building fabric if I know the internal temperature, the external temperature and the overall thermal resistance of the fabric and if the external air is static.
Does anyone know how to account for wind velocity?
I know this could become very complicated - Is the flow laminar or turbulent? What effect does wind direction have (as most buildings are rectangular in shape)? And so on. But there might be some mechanism to approximate the effect. And anything that could make some sort of allowance for this would have to be more accurate than ignoring the effect altogether.
Take care.
Mike
Well you can have my theoretical view, which is untested.
In dry conditions, this is simply forced convection. �I believe that the convection effect accounts for about 0.1 �C.m�/W of wall's R-value. I know this bit is true (and have measured it) for heat flows around 10-20 w/m� through a horizontal surface in the lab. �For higher heat flows it reduces. �For vertical surfaces I would also expect it to be less. �The effect of forced convection can only be to reduce this. �So at worst any amount of wind can only reduce your insulance by 0.1 �C.m�/W.

Hi Tim.
I see where you're coming from, but I hadn't thought of it this way before.

For wet surfaces the wind-chill effect which depends on among other things wind speed and relative humidity, will lower the outer surface temperature, and so will produce increased losses in proportion to the increased temperature drop from inside to outside. �So if for example we have 20�C inside and 5�C outside then we have 15� drop, but if there is a 2� wind-chill then your will have 17� drop which will increase the losses by 13%.

Do you mean wind chill as we normally refer to it? Isn't this the effect humans feel when exposed to a cooling wind? That is, isn't this predicated on the cooling effect on a body at a temperature of 36 degrees Celcius? I don't think we can use the standard wind chill measuring technique when attempting to find a wind chill effect on a structure that has an internal temp. of, say, 20 degrees Celcius. Or am I wrong?

For detached houses on level ground I doubt that wind direction and turbulence have much effect, but for terraced (row) blocks and well sheltered houses it might make a little difference.
Not that the dry effect is an subtractive change in insulance, therefore becomes a smaller proportion if the insulation is better, whereas the wet effect is a simple percentage of your losses (although at least it will dry quicker if the losses are worse).
Of course for most older buildings, the biggest effect of wind comes in the form of draughts. which are hard to predict even if you know the size of orifices, as it depends critically of the positioning of the orifices, the wind direction and any labyrinth effect e.g. due to closed internal doors.
Tim

Take care.
Mike

AC Me wrote:

Hi All.
Back again with another query to drive you all daft.
I can easily work out the heat loss through a building fabric if I know the internal temperature, the external temperature and the overall thermal resistance of the fabric and if the external air is static.
Does anyone know how to account for wind velocity?
I know this could become very complicated - Is the flow laminar or turbulent? What effect does wind direction have (as most buildings are rectangular in shape)? And so on. But there might be some mechanism to approximate the effect. And anything that could make some sort of allowance for this would have to be more accurate than ignoring the effect altogether.
Take care.
Mike

Well you can have my theoretical view, which is untested.
In dry conditions, this is simply forced convection. I believe that the convection effect accounts for about 0.1 �C.m�/W of wall's R-value. I know this bit is true (and have measured it) for heat flows around 10-20 w/m� through a horizontal surface in the lab. For higher heat flows it reduces. For vertical surfaces I would also expect it to be less. The effect of forced convection can only be to reduce this. So at worst any amount of wind can only reduce your insulance by 0.1 �C.m�/W.
For wet surfaces the wind-chill effect which depends on among other things wind speed and relative humidity, will lower the outer surface temperature, and so will produce increased losses in proportion to the increased temperature drop from inside to outside. So if for example we have 20�C inside and 5�C outside then we have 15� drop, but if there is a 2� wind-chill then your will have 17� drop which will increase the losses by 13%.
For detached houses on level ground I doubt that wind direction and turbulence have much effect, but for terraced (row) blocks and well sheltered houses it might make a little difference.
Not that the dry effect is an subtractive change in insulance, therefore becomes a smaller proportion if the insulation is better, whereas the wet effect is a simple percentage of your losses (although at least it will dry quicker if the losses are worse).
Of course for most older buildings, the biggest effect of wind comes in the form of draughts. which are hard to predict even if you know the size of orifices, as it depends critically of the positioning of the orifices, the wind direction and any labyrinth effect e.g. due to closed internal doors.
Tim

AC Me wrote:

Hi Tim.
I see where you're coming from, but I hadn't thought of it this way before. For wet surfaces the wind-chill effect which depends on among other things wind speed and relative humidity, will lower the outer surface temperature, and so will produce increased losses in proportion to the increased temperature drop from inside to outside. So if for example we have 20�C inside and 5�C outside then we have 15� drop, but if there is a 2� wind-chill then your will have 17� drop which will increase the losses by 13%.
Do you mean wind chill as we normally refer to it? Isn't this the effect humans feel when exposed to a cooling wind? That is, isn't this predicated on the cooling effect on a body at a temperature of 36 degrees Celcius? I don't think we can use the standard wind chill measuring technique when attempting to find a wind chill effect on a structure that has an internal temp. of, say, 20 degrees Celcius. Or am I wrong?

Internal temperature is immaterial, it is *surface* temperature and heat-flow that matter.
Think of it as an electrical circuit - in terms of the effect we are looking at the internal temperature is a big voltage, the wall (or the flesh) a big resistor so we have an approximate constant current source and the moving atmosphere is a large heat sink representing a reference voltage point. So the calculations are simplest if referenced to ambient.
I guess it works like the viscous air layer, with a saturated layer of water vapour clinging to the wall. And like the viscous layer I don't know how to calculate its dependence on wind speed, but I can likewise put bounds on it.
Obviously the more heat-flow there is the more water flow there will be across the layer and the less chill. The surface temperature will always be somewhere between the dry-bulb temperature (wind speed zero) and the wet-bulb temperature (wind speed infinite).
The "dry bulb" temperature is the one we get in dry conditions (including viscous layer conduction rise) - at zero air speed the wall becomes covered with saturated water vapour and there is no further evaporation so no evaporative cooling.
The "wet bulb" temperature is where the SVP of water matches the ambient humidity. It corresponds to infinite air speed or zero heat flow, where there is also no conduction drop. (In a wet bulb humidity meter readings are taken at equilibrium, when there is zero heat flow.)
You can measure it by a modification of a wet-bulb humidity meter, with wet and dry temperature sensors in contact with the wall surface, using a Stephenson's screen type of apparatus to keep rain off (the dry bulb) but allow the wind to pass.
As the heat flow density through most house walls is hopefully rather less than that through bare skin, I'd expect the wind chill temperature drop to be somewhat greater, within these bounds.
I suppose there is also the possibility of cold rain to consider, thunderstorm rain can I suppose be a lot colder than the surface air. But I doubt that is a frequent issue in Ireland any more than it is here in the Pennines (if it's hot enough for thunder then heat loss isn't a worry).
Tim

On Mon, 01 Dec 2008 09:01:06 -0600, Morris Dovey wrote:

AC Me wrote:
I can easily work out the heat loss through a building fabric if I know the internal temperature, the external temperature and the overall thermal resistance of the fabric and if the external air is static.
Does anyone know how to account for wind velocity?
You might try a web search for "wind chill" effect... Wind chill only applies to warm damp "objects" like bodies

On Mon, 01 Dec 2008 19:35:39 +0000, Tim Jackson wrote:

AC Me wrote: Hi All.
Back again with another query to drive you all daft.
I can easily work out the heat loss through a building fabric if I know the internal temperature, the external temperature and the overall thermal resistance of the fabric and if the external air is static.
Does anyone know how to account for wind velocity?
I know this could become very complicated - Is the flow laminar or turbulent? What effect does wind direction have (as most buildings are rectangular in shape)? And so on. But there might be some mechanism to approximate the effect. And anything that could make some sort of allowance for this would have to be more accurate than ignoring the effect altogether.
Take care.
Mike
Well you can have my theoretical view, which is untested.
In dry conditions, this is simply forced convection. I believe that the convection effect accounts for about 0.1 °C.m²/W of wall's R-value. I know this bit is true (and have measured it) for heat flows around 10-20 w/m² through a horizontal surface in the lab. For higher heat flows it reduces. For vertical surfaces I would also expect it to be less. The effect of forced convection can only be to reduce this. So at worst any amount of wind can only reduce your insulance by 0.1 °C.m²/W.
For wet surfaces the wind-chill effect which depends on among other things wind speed and relative humidity, will lower the outer surface temperature, and so will produce increased losses in proportion to the increased temperature drop from inside to outside. So if for example we have 20°C inside and 5°C outside then we have 15° drop, but if there is a 2° wind-chill then your will have 17° drop which will increase the losses by 13%.
For detached houses on level ground I doubt that wind direction and turbulence have much effect, but for terraced (row) blocks and well sheltered houses it might make a little difference.
Not that the dry effect is an subtractive change in insulance, therefore becomes a smaller proportion if the insulation is better, whereas the wet effect is a simple percentage of your losses (although at least it will dry quicker if the losses are worse).
Of course for most older buildings, the biggest effect of wind comes in the form of draughts. which are hard to predict even if you know the size of orifices, as it depends critically of the positioning of the orifices, the wind direction and any labyrinth effect e.g. due to closed internal doors.
Tim It has been widely theorised that heat loss through leakage is the

only appreciable change with wind speed.

On Tue, 02 Dec 2008 00:01:29 +0000, Tim Jackson wrote:

AC Me wrote:
Hi Tim.
I see where you're coming from, but I hadn't thought of it this way before. For wet surfaces the wind-chill effect which depends on among other things wind speed and relative humidity, will lower the outer surface temperature, and so will produce increased losses in proportion to the increased temperature drop from inside to outside. So if for example we have 20°C inside and 5°C outside then we have 15° drop, but if there is a 2° wind-chill then your will have 17° drop which will increase the losses by 13%.
Do you mean wind chill as we normally refer to it? Isn't this the effect humans feel when exposed to a cooling wind? That is, isn't this predicated on the cooling effect on a body at a temperature of 36 degrees Celcius? I don't think we can use the standard wind chill measuring technique when attempting to find a wind chill effect on a structure that has an internal temp. of, say, 20 degrees Celcius. Or am I wrong?
Internal temperature is immaterial, it is *surface* temperature and heat-flow that matter.
Think of it as an electrical circuit - in terms of the effect we are looking at the internal temperature is a big voltage, the wall (or the flesh) a big resistor so we have an approximate constant current source and the moving atmosphere is a large heat sink representing a reference voltage point. So the calculations are simplest if referenced to ambient.
I guess it works like the viscous air layer, with a saturated layer of water vapour clinging to the wall. And like the viscous layer I don't know how to calculate its dependence on wind speed, but I can likewise put bounds on it.
Obviously the more heat-flow there is the more water flow there will be across the layer and the less chill. The surface temperature will always be somewhere between the dry-bulb temperature (wind speed zero) and the wet-bulb temperature (wind speed infinite).
The "dry bulb" temperature is the one we get in dry conditions (including viscous layer conduction rise) - at zero air speed the wall becomes covered with saturated water vapour and there is no further evaporation so no evaporative cooling.
The "wet bulb" temperature is where the SVP of water matches the ambient humidity. It corresponds to infinite air speed or zero heat flow, where there is also no conduction drop. (In a wet bulb humidity meter readings are taken at equilibrium, when there is zero heat flow.)
You can measure it by a modification of a wet-bulb humidity meter, with wet and dry temperature sensors in contact with the wall surface, using a Stephenson's screen type of apparatus to keep rain off (the dry bulb) but allow the wind to pass.
As the heat flow density through most house walls is hopefully rather less than that through bare skin, I'd expect the wind chill temperature drop to be somewhat greater, within these bounds.
I suppose there is also the possibility of cold rain to consider, thunderstorm rain can I suppose be a lot colder than the surface air. But I doubt that is a frequent issue in Ireland any more than it is here in the Pennines (if it's hot enough for thunder then heat loss isn't a worry).

OK last thing first - you haven't seen some of the thunderstorms we've had here in Central Ontario, Canada. Thunder and lightning and SNOW!!!!!
As for the "wind chill" the temperature difference between the outside ambient air and the temperature of the outside of a decently insulated wall will be EXTREMELY small - almost negligible when it comes to figuring wind chill.
If the outside wall was say 20C and the wind was -10, I could concurr you MIGHT experience a significant chilling effect. Thebiggest heat loss problem is leakage on the lea side of the building (the low pressure side) allowing the warm air from inside to escape. Even with a solid west wall (our prevailings are westerlies) the weatherstripping on east side windows and doors needs to be VERY good. >Tim

On Dec 2, 9:00�am, Tim Jackson wrote:

cl...@snyder.on.ca wrote: � Wind chill only applies to warm damp "objects" like bodies
They don't have to be warm, although wet houses in cold weather fit that description. �Published wind-chill values make assumptions about the perception of temperature which are not valid here. �For example it subtracts the natural speed of movement of people from the wind speed, and for houses that number is very small. :)
OK last thing first - you haven't seen some of the thunderstorms we've had here in Central Ontario, Canada. Thunder and lightning and SNOW!!!!!
I have. That's the sort of thing I was thinking of when I mentioned it, but such continental weather is pretty unheard of in IE. �The tendency there is toward overcasts and drizzle.
As for the "wind chill" the temperature difference between the outside ambient air and the temperature of the outside of a decently insulated wall will be EXTREMELY small - almost negligible when it comes to figuring wind chill.
I'm aware we are talking small numbers, the object of my exercise was to set numerical limits on just how small. �It is dangerous to disregard a mechanism before you have quantified it's effect. �Many scientists have had ignored effects come and bite them on the ass.
But what you say is not quite true. The better the insulation the deeper the chill, although of course the smaller the loss it is affecting, as it is primarily a temperature effect. Evaporation causes the surface (wet bulb) temperature to go *below* ambient (dry bulb) - the heat of evaporation is provided by heat flow *from* the air to the cold surface.
Theoretically the temperature depression could be as great as 14�C, (en.wikipedia.org/wiki/Image:PsychrometricChart-SeaLevel-SI.jpg) although that would require improbable weather conditions, snow-melt in a warm dry wind is the probably worst case; the practical sustainable limit is probably around 5�C.
I'm in total agreement that in most houses, leakage is the major effect.
If the outside wall was say 20C and the wind was -10, I could concurr you MIGHT experience a significant chilling effect. Thebiggest heat loss problem is leakage on the lea side of the building (the low pressure side) allowing the warm air from inside to escape. �Even with a solid west wall (our prevailings are westerlies) the weatherstripping on east side windows and doors needs to be VERY good.
� It doesn't really matter whether it is the weather or lee side that is sealed, air has to come in AND go out, it requires two holes to make a draught. �(A fact ignored by leakage standards which refer to pressurising the house - OK for checking new build but meaningless as a measure of draughtiness.) My own house has deliberate ventilation to the lee side (for evaporation, it's built over a spring), and I work to maintain a pressure seal to windward.
Tim

Hi all, again.
Me being me, I'm easily confused :), or should that be :(?
As I understand the normal use of the term 'wind chill' this is the cooling effect felt by an individual when that individual is exposed to a wind that carries away some of that individual's heat. The person has a certain level of heat, a temperature, that is, say, 36 degrees Celcius (or whatever). The ambient temperature is, say, 10 degrees Celcius. Without any wind there would be a certain heat loss from the body. With a wind of 10km/h that heat loss would increase. With a wind speed of 20km/h the heat loss would be greater still.
Would the same principle not apply to a heated structure?
There are formulae/techniques and information available on the web to calculate the wind chill effect on a human. These allow the adjustment of the ambient temperature to take account of the effect of wind speed. In simple terms, the ambient temperature is reduced to what it would be if the wind speed was zero but with the same heat loss.
If a wind chill effect exists for a heated structure, can a formula be applied to adjust the ambient temperature so that it would now be the temperature at which the same heat loss would occur? If the answer to both these questions is yes - and if the answer to the first is yes, then the second should also be yes - it would be possible to make some account for the effect of wind velocity on a heated structure.
Or am I missing something?
Take care.
Mike

clare@snyder.on.ca wrote:

Wind chill only applies to warm damp "objects" like bodies

They don't have to be warm, although wet houses in cold weather fit that description. Published wind-chill values make assumptions about the perception of temperature which are not valid here. For example it subtracts the natural speed of movement of people from the wind speed, and for houses that number is very small. :)

OK last thing first - you haven't seen some of the thunderstorms we've had here in Central Ontario, Canada. Thunder and lightning and SNOW!!!!!
I have. That's the sort of thing I was thinking of when I mentioned it,

but such continental weather is pretty unheard of in IE. The tendency there is toward overcasts and drizzle.

As for the "wind chill" the temperature difference between the outside ambient air and the temperature of the outside of a decently insulated wall will be EXTREMELY small - almost negligible when it comes to figuring wind chill.
I'm aware we are talking small numbers, the object of my exercise was to

set numerical limits on just how small. It is dangerous to disregard a mechanism before you have quantified it's effect. Many scientists have had ignored effects come and bite them on the ass.
But what you say is not quite true. The better the insulation the deeper the chill, although of course the smaller the loss it is affecting, as it is primarily a temperature effect. Evaporation causes the surface (wet bulb) temperature to go *below* ambient (dry bulb) - the heat of evaporation is provided by heat flow *from* the air to the cold surface.
Theoretically the temperature depression could be as great as 14°C, (en.wikipedia.org/wiki/Image:PsychrometricChart-SeaLevel-SI.jpg) although that would require improbable weather conditions, snow-melt in a warm dry wind is the probably worst case; the practical sustainable limit is probably around 5°C.
I'm in total agreement that in most houses, leakage is the major effect.

If the outside wall was say 20C and the wind was -10, I could concurr you MIGHT experience a significant chilling effect. Thebiggest heat loss problem is leakage on the lea side of the building (the low pressure side) allowing the warm air from inside to escape. Even with a solid west wall (our prevailings are westerlies) the weatherstripping on east side windows and doors needs to be VERY good.
It doesn't really matter whether it is the weather or lee side that is

sealed, air has to come in AND go out, it requires two holes to make a draught. (A fact ignored by leakage standards which refer to pressurising the house - OK for checking new build but meaningless as a measure of draughtiness.) My own house has deliberate ventilation to the lee side (for evaporation, it's built over a spring), and I work to maintain a pressure seal to windward.
Tim

AC Me wrote:

Hi all, again.
Me being me, I'm easily confused :), or should that be :(?
As I understand the normal use of the term 'wind chill' this is the cooling effect felt by an individual when that individual is exposed to a wind that carries away some of that individual's heat. The person has a certain level of heat, a temperature, that is, say, 36 degrees Celcius (or whatever). The ambient temperature is, say, 10 degrees Celcius. Without any wind there would be a certain heat loss from the body. With a wind of 10km/h that heat loss would increase. With a wind speed of 20km/h the heat loss would be greater still.
Would the same principle not apply to a heated structure?
There are formulae/techniques and information available on the web to calculate the wind chill effect on a human. These allow the adjustment of the ambient temperature to take account of the effect of wind speed. In simple terms, the ambient temperature is reduced to what it would be if the wind speed was zero but with the same heat loss.
If a wind chill effect exists for a heated structure, can a formula be applied to adjust the ambient temperature so that it would now be the temperature at which the same heat loss would occur? If the answer to both these questions is yes - and if the answer to the first is yes, then the second should also be yes - it would be possible to make some account for the effect of wind velocity on a heated structure.
Or am I missing something?
Take care.
Mike

The same *principle* applies to heated, or even unheated structures, but not the same *numbers*.
The calculation of human perceived wind chill took a lot of messing with until the numbers came out reflecting typical human perception, and they are still controversial: a single temperature cannot be applied to all situations, the perceived temperature change partly depends for example on what clothing you are wearing. (e.g. you are much more sensitive to wind when naked because stripping increases your evaporative losses more than your conductive losses.)
The formulae and tables that exist for that are not really based on theory, so much as empirical observation, figuring out a mathematical equation that fits the observed profile.
I am sure someone could produce a table of wind chill factors for buildings, although they also would be very approximate, as the cooling at any particular surface would depend on such things as the wind direction across it, and the wetness and texture of the wall, as well as temperature, wind speed and relative humidity.
What is the point of being able to predict the outside surface temperature accurately under specific conditions? After all you can't predict the conditions themselves. What are you trying to achieve here?
If you want to use it in a return-on-investment calculation, you need the local statistics for wind and humidity, which you'd have to determine empirically anyway.
Tim