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Psychrometry & Air-
Conditioning
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Introduction
Moist air (atmospheric air) is a mixture of twogases:
� dry air – mixture of a number of gases(nitrogen, oxygen and traces of argon, carbondioxide, neon etc.)
� water vapour (which may exist in a saturatedor superheated state – can be treated as anideal gas).
Although the amount of water vapour is small, itplays a major role in human comfort.
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� In psychrometry analysis, the water vapour is
treated as the variable component since the
amount of water vapour changes as a result of
condensation and evaporation from oceans,
lakes etc.
� The dry air is treated as the fixed component.
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AssumptionsAssumptions:
� Water vapour contains no dissolved gases.
� Both dry air and water vapour can be
considered as an ideal gas mixture since both
exist in the atmosphere at low pressures.
� No interaction between components and all
components are at the mixture temperature.
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Psychrometric Properties
(Properties of Moist Air)
� Dry air and water vapour form a binarymixture.
� A mixture of two substances requires 3properties to completely define itsthermodynamic state, unlike a pure substancewhich requires only 2.
� One of the three properties can be thecompositions.
� The properties of moist air (atmospheric air)are called psychrometric properties.
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Psychrometric Properties
(Properties of Moist Air)
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Volume V
Mass m
Pressure P (barometric pressure)
Temperature T
(T is the dry bulb temperature (DBT))
For defining and calculating the relevant
psychrometric properties, we may consider a
certain volume V of moist air at pressure P and
temperature T, containing ma kg of dry air and mv
kg of water vapour.
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Specific (or absolute) humidity of air
� The amount of water vapour can be specified
by the mass of water vapour present in a
unit mass of dry air.
� This is called humidity ratio, moisture
content, specific or absolute humidity and is
denoted by ω:
(kg water vapor/kg dry air)v
a
m
mω =
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Specific (or absolute) humidity of air
It can also be expressed as:
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air)dry our/kg water vap(kg 622.0
or
ly.respectiveour water vapandair dry toreferring and Subscript
622.0/
/
/
/
v
v
a
v
aa
vv
aa
vv
a
v
PP
P
va
P
P
RP
RP
TRVP
TRVP
m
m
−=
====
ω
ω
Since:
P = Pa + Pv (kPa) (Where P = atmospheric pressure)
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Relative humidity
� The amount of moisture in the air has a
definite effect on how we feel in an
environment.
� This comfort level depends more on the
amount of moisture the air holds (mv)
relative to the maximum amount of moisture
the air can hold at the same temperature
(mg).
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Relative humidity
The ratio of these two quantities is called the
relative humidity, φ.
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g
v
vg
vv
g
v
P
P
TRVP
TRVP
m
m===
/
/φ
Note:
Pg = Psat @T
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Relative humidity
� The relative humidity and specific humidity can also be
expressed as,
� The relative humidity ranges from 0 for dry air to 1 for
saturated air.
� The amount of moisture air can hold depends on its
temperature.
� Therefore, the relative humidity of air changes with
temperature even when its specific humidity remains constant.
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( )
g
g
g
PP
P
P
P
φ
φω
ω
ωφ
−=
+=
622.0
and
622.0
Specific enthalpy
� The enthalpy of moist air is expressed in terms of the
enthalpies of the dry air and the water vapour.
� The total enthalpy (an extensive property) of
atmospheric air (moist air) is,
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( )airdry kJ/kg
gives, by Dividing
va
v
a
va
a
a
vvaava
hhh
hm
mh
m
Hh
m
hmhmHHH
ω+=
+==
+=+=
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Specific enthalpy
The enthalpy of dry air can be determined from,
where, T is the dry bulb temperature.
The enthalpy of water vapour can be expressed by,
where hg is the enthalpy of saturated vapour at the
same temperature.
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( ) (kJ/kg) CkJ/kg.005.1 o TTch pa ==
( ) ( ) (kJ/kg) low, ThPTh gv ≅
Specific enthalpy
hg, can also be determine approximately from,
in the temperature range of -10oC to 50oC with negligible error.
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( ) (kJ/kg) 82.19.2500 TThg +≅
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Example 1Example 1
A tank contains 21 kg of dry air and 0.3 kg of
water vapour at 30oC and 100 kPa total
pressure. Determine,
a) the specific humidity;
b) the relative humidity;
c) the volume of the tank.
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Example 2Example 2
A room contains air at 20oC and 98 kPa at a relative
humidity of 85 percent. Determine,
a) the partial pressure of dry air;
b) the specific humidity of the air;
c) the enthalpy per unit mass of dry air.
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Dew-Point Temperature
� The dew-point temperature Tdp
is defined as the temperature at
which condensation begins
when the air is cooled at
constant pressure.
� Or, the saturation temperature of
water corresponding to the
vapour pressure.
vPsatdp TT @=
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Constant-pressure cooling of
moist air and the dew-point
temperature on the T-s
diagram of water.
Dew-Point Temperature
� As the air cools at constant
pressure, the vapor
pressure Pv remains
constant.
� Therefore, the vapor in the
air (state 1) undergoes a
constant-pressure cooling
process until it strikes the
saturated vapor line (state
2).
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Constant-pressure cooling of
moist air and the dew-point
temperature on the T-s
diagram of water.
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Dew-Point Temperature
� The temperature at this
point is Tdp, and if the
temperature drops any
further, some vapor
condenses out.
� As a result, the amount of
vapor in the air decreases,
which results in a decrease
in Pv.
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Constant-pressure cooling of
moist air and the dew-point
temperature on the T-s
diagram of water.
Dew-Point Temperature
� The air remains saturated
during the condensation
process and thus follows a
path of 100 % relative
humidity (the saturated
vapor line).
� The ordinary temperature
and the dew-point
temperature of saturated air
are identical.
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Constant-pressure cooling of
moist air and the dew-point
temperature on the T-s
diagram of water.
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Example 3Example 3
In cold weather, condensation frequently occurs on
the inner surfaces of the windows due to the lower air
temperatures near the window surface. Consider a
house that contains air at 20oC and 75% relative
humidity. At what window temperature will the
moisture in the air start condensing on the inner
surfaces of the windows.
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Adiabatic Saturation And Wet-Bulb
Temperatures
� One way to determine relative humidity and specific humidity is by determining the dew-point temperature of air, viz.,
1. knowing Tdp,
2. determine the Pv,
3. able to determine the ω,
4. finally, able to determine the φ.
� This approach is simple, but not quite practical.
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Constant-pressure cooling of
moist air and the dew-point
temperature on the T-s
diagram of water.
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Adiabatic Saturation And Wet-Bulb
Temperatures
� A more practical approach is to use a thermometer whose bulb is covered with a cotton wick saturated water and to blow air over the wick.
� The temperature measured in this manner is called the wet-bulb temperature, Twb
and it is commonly used in air-conditioning applications.
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A simple arrangement to
measure the wet-bulb
temperature
Adiabatic Saturation And Wet-Bulb
Temperatures
� This method is related to an
adiabatic saturation process,
shown schematically and on
a T-s.
� Where it can be analyzed as
a steady-flow process, which
involves no heat and work
interactions, and kinetic and
potential energy changes
are neglected.
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The adiabatic saturation
process & its representation on
a T-s diagram of water
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Adiabatic Saturation And Wet-Bulb
Temperatures
Thus, the ω and φ of air can be determined from
equations below,
( )
2
2
21
2
2
2
212
1
622.0
where,
g
g
fg
fgp
PP
P
hh
hTTc
−=
−
+−=
ω
ωω
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Adiabatic Saturation And Wet-Bulb
Temperatures
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22
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@
@
2
@
@
22
1
1
2
o
pressure catmospheri
at humidity specific
air ofhumidity specific
e)temperatur(ordinary re temperatubulb-dry
re temperatubulb-wet
C)kJ/kg. (1.005heat specific pressure-constant
Where,
Tfgfg
Tsatg
Tff
Tgg
p
hh
PP
P
hh
hh
T
T
T
c
=
=
=
=
=
=
=
=
=
=
ω
ω
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Adiabatic Saturation And Wet-Bulb
Temperatures
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Sling psychrometer
Example Example 44
The air in a room has a dry-bulb temperature of
22oC and a wet-bulb temperature of 16oC.
Assuming a pressure of 100 kPa, determine,
a) the specific humidity;
b) the relative humidity;
c) the dew-point temperature.
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The Psycrometric Chart
� Psychrometric charts
present the moist air data
in the form of easily
readable charts,
calculated from the
previous relations.
� They are used extensively
in air-conditioning
applications.
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Schematic for a psyhcrometricchart
Example 5Example 5
A room contains air at 1 atm, 26oC, and 70 %
relative humidity. Using the psychrometric
chart, determine,
a) the specific humidity;
b) the enthalpy;
c) the wet-bulb temperature;
d) the dew-point temperature;
e) the specific volume of the air.
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Air-Conditioning Processes
� Air-conditioning processes are processes to
maintain a living space or an industrial facility
at the desired temperature and humidity.
� These processes include,
i. Simple heating (raising the temperature)
ii. Simple cooling (lowering the temperature)
iii. Humidifying (adding moisture)
iv. Dehumidifying (removing moisture)
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Air-Conditioning Processes
� Sometimes two or more of
these processes are needed to
bring the air to a desired
temperature and humidity
level.
� Various air-conditioning
processes can be illustrated
on the psychrometric chart.
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Various air-conditioning
processes
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Air-Conditioning Processes
Simple heating and cooling
processes appear as horizontal
lines on this chart since the
moisture content of the air remains
constant (ω = constant) during
these processes.
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Various air-conditioning
processes
Air-Conditioning Processes
Air commonly heated and
humidified in winter.
And it commonly cooled
and dehumidified in
summer.
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Various air-conditioning
processes
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Air-Conditioning Processes
Most air-conditioning processes can be
modelled as steady-flow processes, and thus
the mass balance relation can be
expressed for dry air and water as,outin mm
••
=
∑∑••
==out
a
in
a mmairdryforbalanceMass
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Air-Conditioning Processes
∑∑
∑∑
••
••
=
==
out
a
in
a
out
w
in
w
mm
mmwaterforbalanceMass
ωω
or
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Air-Conditioning Processes
Disregarding the kinetic and potential energy changes, the
steady-flow energy balance relation can be expressed in
this case as,
The work term usually consists of fan work input, which is
small relative to other terms in energy balance relation.
∑ ∑••••••
++=++in out
outoutininhmWQhmWQ
outin EE••
=
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Simple Heating and Cooling (ω = constant)
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During simple heating, specific
humidity remains constant, but
relative humidity decreases.
During simple cooling, specific
humidity remains constant, but
relative humidity increases.
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Simple Heating and Cooling (ω = constant)
During simple heating, specific humidity remains
constant, but relative humidity decreases.
Many residential heating systems consist of a stove, a heat pump, or an electric
resistance heater. The air in these systems is heated by circulating it through a duct
that contains the tubing for the hot gases or the electric resistance wires.
Cooling can be accomplished by passing the air over some coils through which a
refrigerant or chilled water flows.
Heating and cooling appear as a horizontal line since no moisture is added to or
removed from the air.
Dry air mass balance
Water mass balance
Energy balance
During simple cooling, specific
humidity remains constant, but
relative humidity increases.
Example Example 66
A heating section consists of a 38 cm-diameter
duct that houses a 4-kW electric resistance
heater. Air enters the heating section at 1 atm,
10oC, and 40%rh at a velocity of 8 m/s.
Determine,
a) the exit temperature,
b) the relative humidity of the air,
c) the exit velocity.
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Heating with Humidification
Problems with the low relative humidity resulting from simple heating can be
eliminated by humidifying the heated air. This is accomplished by passing the air
first through a heating section and then through a humidifying section.
Example Example 77
Air at 1 atm, 15oC, and 60%rh is first heated to 20oC in a
heating section and then humidified by introducing water
vapor. The air leaves the humidifying section at 25oC
and 65%rh. Determine,
a) the amount of steam added to the air [kg w.v/kg d.a],
b) the amount of heat transfer to the air in the heating
section [kJ/kg d.a].
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15oC 20oC
Cooling with Dehumidification
� The specific humidity of air remains constant
during a simple cooling process, but its relative
humidity increases.
� If the φ reaches undesirably high levels, it may be
necessary to remove some moisture from the air,
that is, to dehumidify it.
� This requires cooling the air below its dew point
temperature.
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Cooling with Humidification
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Example of schematic and psychrometric chart
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Cooling with DehumidificationThe specific humidity of air remains constant during a simple cooling process, but
its relative humidity increases. If the relative humidity reaches undesirably high
levels, it may be necessary to remove some moisture from the air, that is, to
dehumidify it. This requires cooling the air below its dew-point temperature.
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Example 8Example 8
Air enters a window air conditioner at 1 atm, 30oC,
and 80% relative humidity at a rate of 10 m3/min,
and it leaves at saturated air at 14oC. Part of the
moisture in the air that condenses during the
process is also removed at 14oC. Determine the rates
of heat and moisture removal from the air.
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Example Example 99
2000 m3/hr of atmospheric air at 28oC with a dew point of
25oC flows into an air conditioner that uses chilled water as
the working fluid. The atmospheric air is to be cooled to
18oC. Determine mass flow rate of the condensate water, if
any, leaving the air conditioner, in kg/hr. If the cooling
water has a 10oC temperature rise while flowing through
the air conditioner, determine the volume flow rate of
chilled water supplied to the air conditioner heat
exchanger, m3/min. The air conditioning process takes
place at 100 kPa.
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