CPD Programme
v=0.84m�/kg
Specific volume
P
e
rcentage saturation
W
θ’
e
=13.9°C
t-bulb
μ =50%
Moisture content
g = 7.4g/kg
Specific enthalp
h =39kJ/kg
Dry-bulb
θ =20.0°C
y
Figure 1 - Plotting psychrometric properties
> moist air, m
a
is the mass of the dry air (that And so to determine v and ρ devices (such as orifice plates)
[3]
is combined with the mass of water vapour, v = 287·T / (p
at
– p
v
) = Considering the formula for air density
m , to give [m 287 (273+20)/(101325 - 1179) above, moist air is less dense than dry air.
v a
+ m
v
] kg moist air), p
a
is dry
air partial pressure, p
v
is water vapour partial = 0.839 m
3
·kg
-1
This may seem counter intuitive however the
pressure and p
at
is atmospheric pressure. ρ = (p
at
/287·T)[1-(0.378·p
v
/p
at
)] ideal gas law states that a fixed volume of air
Density, ρ (kg·m
-3
), is the mass of moist air = [101325/(287 x (273+20))][1-(0.378 x at a certain pressure has a fixed number of
that occupies a volume of 1 m
3
. To evaluate the 1179/101325)] = 1.200 kg·m
-3
molecules and each of those molecules has a
mass of 1 m
3
of moist air the masses of the dry And to see what the practical difference is particular mass. Looking back at the previous
air and the water vapour are added together. between the inverse of specific volume and article
[4]
dry air is made up principally of
So as m
a
/V=p
a
/R
a
T and m
v
/V=p
v
/R
v
T the density (at this particular condition) consider nitrogen (N2) and oxygen (O2) molecules with
overall density of the single unit volume of that the reciprocal of the specific volume molar masses of 28 and 32 respectively. Water
moist air will be (p
a
/R
a
T + p
v
/R
v
T). calculated above is 1.192 and this compares vapour has a molar mass of 18. Hence, when
Rearranging this and putting in values for with a density of 1.200 - a difference of less a given volume of air has water molecules
R and R gives than 1 per cent. The difference will rise as displacing other heavier molecules its overall
a v
ρ = (p
at
/287·T)[1-(0.378·p
v
/p
at
)] kg·m
-3
moisture content increases and in conditions mass and hence density will be less.
(with all pressures in Pa) that are likely to prevail in heating, ventilation
As an example consider air at 20.0 deg and air conditioning applications there is likely Plotting the air point
C dry bulb and 13.9 deg C wet bulb at to be maximum of about 4 per cent difference Using the relationships described in this,
an atmospheric pressure of 101.325 kPa at high temperatures and high moisture and the previous, article and using nothing
= 101325Pa (ie standard atmospheric contents between the simple reciprocal value more than a sling psychrometer and a
pressure). of specific volume and the true density. psychrometric chart or a hand held calculator
To find the vapour pressure Specific volume is particularly useful when (mobile phone, palmtop etc) the properties of
p
v
= p
s
’ - p
at
A (θ – θ’) undertaking calculations where there is a air may readily be obtained.
a value of p
s
’ is required and so using …. change in moisture content during a process As an example the air point of 20 deg C
log p
s
’ = 30.5905 – 8.2 log (13.9 + 273.16) (humidification and dehumidification), as dry-bulb and 13.9 deg C wet-bulb (sling) is
+ 2.4804 x 10
-3
(13.9 + 273.16) – [3142.31 / the amount of dry air will remain constant shown in Figure 1. The dry-bulb and wet-
(13.9 + 273.16)] and so p
s
’ = 1.591 kPa. while the amount of water vapour changes. bulb temperatures may well have been
Hence p
v
= 1.591 – [101.325 x 6.66 x 10
-4
x However density is useful when measuring obtained using a sling psychrometer in a
(20 – 13.9)] = 1.179kPa the flowrate of air through pressure drop room - the standard chart is drawn up for an
50 CIBSE Journal August 2009 www.cibsejournal.com
CIBSEaug09 pp49-52 cpd.indd 50 7/23/09 4:06:33 PM
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