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- Angela, so we've been given a set of averages
for a particular climate zone in Australia.
Could you should me how you could plot
that particular climate zone on our psychrometric chart?
- Sure, to start off with,
we take the T min temperature
which is 13.6 degrees which is around about here
and then we also take the relative humidity
at 9.am. which is 66%
which if we follow down, it comes to about here,
so we put our first point there
and then we take our T max term
which is 28.5 degrees,
which is approximately here
and then we also take the relative humidity at 3 p.m.
which is 35%, so if that's 50%,
40%, 30% and then we go up to about here,
so that's our second point
and then we wanna join these two
with a straight line like that.
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- Thank you, Angela.
So, one thing with this particular diagram
is that each of these lines described are monthly,
so these temperatures and these averages
are obviously mean maximum temperatures
for each month averaged
and of course the same for the relative humidities.
So, when you go through and you actually plot this
on your diagram, it's not gonna be exactly
what we have here
but you will have something of the same form.
So, we've now plotted our climate zone
on our psychrometric chart from our average values.
We now wanna look at how we can plot the comfort zone,
the thermal comfort zone on this psychrometric chart.
Angela, could you show me
how you begin plotting the comfort zone?
- Sure, so to start off with,
we need to find Tn which is the temperature of neutrality
and from that we find the lower and upper temperatures
of the thermal comfort box
and we plot that on the 50% relative humidity line
but to start off with, we need to find Tn,
so to find Tn, we use an empirical formula
which is given by Tn is equal to 17.6
plus 0.31 times the DBT average
and the DBT average that comes from these two values here
which is T max minus T min divided by two
which is the average,
so to get this,
we do 20 plus 7.1 over two
and you can just put that
into your calculator straight away,
subbing that in and you can get Tn
is equal to 21.92 degrees Celsius
and using that, you can find your upper and lower limits.
So, if the question asks for an annual thermal comfort box,
you use plus or minus two
which is what we're gonna use
but if it asks you for monthly, you'd do plus or minus 1.75
from this value here.
So, to find the upper,
we do 21.92
plus two degrees
which is going to give us 23.92 degrees Celsius
and to find the lower one,
we need to do 21.92 minus two degrees
which is gonna give us 19.92 degrees Celsius,
so now that we have these values,
we can plot them onto the 50% relative humidity line.
So, to start off with,
we have 23.92 which is approximately around here
and we wanna put that onto the 50% relative humidity line,
so that is our upper
and then for our lower,
we have 19.92 which is approximately here
and we're gonna put that on the 50% relative humidity line,
so that's about here.
- So, we've now found the upper and lower bounds
of our thermal comfort box,
so what we now have to do is find out
where we're actually intersecting on the diagram
to actually draw our box,
so we're looking to draw our set lines.
Those are our standard affected temperature lines,
so these describe the change in effective temperature
with relative humidity.
So, these lines are gonna go through our lower points
and upper points and intersect
on the X axis somewhere.
These are arbitrary at the moment,
so I'm not drawing them in exact
because we haven't calculated them
but just to illustrate,
somewhere they'll intersect on the diagram,
so Angela, can you show me
how we calculate the intersection
for these set lines?
- Sure, so to do that,
you're going to need to use an empirical formula
which is given by delta DBT
over delta omega equal to minus 0.025
multiplied by DBT minus 14.
Just to explain what these terms actually are,
so the delta DBT is the difference
between your dry-bulb temperature
which will either be from your L or your U,
for the moment we're just focusing on L,
so we'll use L
and where this set line intersects
with the X axis,
and your delta omega
will come from the psychrometric chart,
so for L you come across here
and you find what this absolute humidity value is.
It so happens to be that this value
is 0.0071 but for this formula,
you need it in grams, not kilograms,
so we'll be using 7.1, so to start off with,
we need to rearrange this,
so we will start by doing delta DBT
is equal to negative 0.025
but we know that this value here has to be negative,
so that negative would turn into a positive,
multiplied by DBT minus 14
times by delta omega.
Now that we have this equation,
we can simplify it even further,
so we know that our delta DBT is the difference
between here, so we will do XL minus L
which is what we found before
is equal to 0.025
multiplied by DBT which is our L again
minus 14 by our delta omega
which we found before was 7.1
remembering that we need grams, not kilograms.
So, to get XL, we move L over the other side
and we get plus 0.025
multiplied by L minus 14 times by 7.1
and from before we know that our L value is 19.92,
so we sub that into this equation
to find what our X intercept for this line is.
And we get that our value for XL is 20.9 degrees Celsius.
So, once we have that we can plot it onto here,
so 20.9 is approximately here,
this line was arbitrary before
but now we're gonna draw in the correct one,
so we go through this point
and through this point here
to get something that looks like that.
And then for U, we have a very, very similar calculation
except wherever we've had an L,
we now put a U, so for instance,
here we would have XU minus U times this
and our omega W will comes from the U,
so we come across here
and it will turn out to be something like 0.0095
but again we want grams, not kilograms,
so if you put it through the calculation,
you'll get that your XU value
is equal to 26.27 degrees Celsius,
so you'll come over here,
plot it on here, it's around about here
and then draw that line,
so once we have those, we can complete our box
by putting in the set lines,
so we have a line coming through here from 0.012
which will be the top of our box
and we also have a line that comes from the 0.004
which will be the bottom of our box.
- Excellent.
So, that is our thermal comfort region here.
Thank you.
So, we're looking at the solar temperature,
so we're looking to calculate the solar temperature
for this particular opaque surface on the building
for this particular setup that we have.
So, if we look back, we know that the solar temperature,
if we look back at our diagram in our lectures,
the solar temperature
is our instantaneous excess temperature
on top of the outdoor temperature,
so that is caused by radiation on the surface of your wall.
So, Angela, can I get you to calculate for me
the solar air temperature
and the additional heat gained
as a result of that solar temperature
to this opaque element of a building?
- Sure, so to find the solar air temperature,
we use this equation,
so solar air temperature is equal
to the outside air temperature
plus the excess solar air temperature
which is given by G which is irradiance,
alpha which is absorptivity,
RSO which is the outside air film.
We know what the outside air temperature is,
that's given to us in the question,
that's 18 degrees C
and we also know what irradiance is,
that's also given to us,
that's 800 watts per metre squared.
We're given absorptivity which is 0.14,
that's dimensionless and we're also given
the outside air resistance, film resistance
which is 0.04 metres squared kelvin per watt.
So the interesting part
is is that the two watts cancel out
and that should be made a squared,
the metres squared cancel out,
so we end up with 18 degrees Celsius
which is with Kelvin and as you know,
per degree Celsius and per Kelvin are the same,
so 18 plus the solar air excess temperature
is calculated to be 12.8 degrees,
so that ends up being 30.8 degrees C.
And then now that we have this,
we can find the additional heat transfer
through the wall, so to do the additional heat transfer
through the wall, we use this value here
and as always, the heat transfer through a wall
is given by Q is equal to UA delta T.
U is given to us in the equation,
U wall, A is the area of the wall
which is also given to us
and the delta T is just this value here
which we calculated before.
So, if we put that into the equation,
we get 1.47 watts per metres squared Kelvin
multiplied by 12 metres squared
times by our difference in temperature
which as before is from here, 12.8 degrees C
and we have some cancelling of units going on,
so the metres squared and the metres squared
and the degree Kelvin and the degree C go
and we need up with a value that is 226 watts,
so the additional heat transfer to the wall
because of the excess solar air temperature is 226 watts.
- Excellent.
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