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BRE2031 – Environmental Science:
Lecture 5 – Psychrometrics
Dr. Meng Ni (倪萌)
Associate Professor
Department of Building and Real Estate, The Hong
Kong Polytechnic University
Office: ZN713
Tel: 2766 4152
Email: [email protected]
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Content
1. Moist air and their properties
2. Psychrometric chart
3. Psychrometric process analysis
Objectives
After studying this lecture, you will be able to:
1. Understand the physical meaning of moist air properties
2. Read moist air properties from the psychrometric chart.
3. Determine heat addition for heating/cooling of moist air.
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Basic concepts
• The surrounding air is a mixture of dry air and water
vapor – moist air, plus some pollutants;
• Psychrometrics is a subject of studying how moist air
behaves when it is cooled or heated;
• It is a tool to analyze typical air-conditioning processes.
Atmospheric
air
Dry air
Water vapor
Various pollutants
Moist air
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Illustration of Dalton’s law – Concept of partial pressure
• For the mixture of gases A and B, pmix = pA + pB
• pA is the pressure gas
A would have if gas A
is alone in the tank.
For a mixture of gases occupying a given volume at a certain temperature, the total pressure
of the mixture is equal to the sum of the partial pressures of the constituents of the mixture.
From ideal gas law, the
pressure of gas depends on
its number of moles (at a
given temperature and
volume)RT
P nV
=
mixture A Bn n n= +
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Atmospheric pressure
Standard atmospheric pressure at the sea level:
P=101.325 (kPa) = 14.696 (psi) = 30 (In. Hg)
For the elevation above the sea level H<1220 m (or <4000 ft):
P=101.325-0.01153H (kPa)
P=29.92-0.001025H (In. Hg)
For H>1220 m (or >4000 ft):
P=99.436-0.010H (kPa)
P=29.42-0.0009H (In. Hg)
psi: pound force per square inch
Pressure can be measured by height of water or other liquid.
An elevation increase by
1m, how many Pascal
decrease?
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Composition of Dry and Moist Air
• Atmospheric air contains many gaseous components as well as
water vapor and miscellaneous contaminants (e.g., smoke, pollen, and
gaseous pollutants not normally present in free air far from pollution
sources).
• Dry air is atmospheric air with all water vapor and contaminants
removed. Its composition is relatively constant, but small variations in the
amounts of individual components occur with time, geographic location, and
altitude.
• Moist air is a binary (two-component) mixture of dry air and water
vapor.
• Saturation is a state of neutral equilibrium between moist air and the
condensed water phase (liquid or solid); Air sample in a saturation state
– contains the maximum amount of vapor possible at a given
Temperature.
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A Microscopic Comparison of Gases, Liquids and Solids
In a liquid
• Molecules are in random
motion;
• There are appreciable
intermolecular forces holding
molecules close together
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LIQUID VAPOR
break IM bonds
make IM bonds
Add energy
Remove energy
<---condensation
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On the physics of saturation – in a closed system
Evaporation process
• Molecules in liquid water
attract each other
• In motion
(courtesy T.S. Zhao)
• Collisions
• Molecules near surface
gain velocity by collisions
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Evaporation
• Fast moving molecules leave
the surface (Only those with
enough KE escape)
• Evaporation
Twater
• Rate of evaporation – Function of water temperature
• Evaporation is a cooling
process.
• It requires heat.
• Endothermic.
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Evaporation
• Soon, there are many water
molecules in the air
• Slower molecules return
to water surface
• Condensation
Change from gas to liquid
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Evaporation
• Net Evaporation
– Number leaving water surface is greater than the number returning
– Evaporation greater than condensation
– Evaporation continues to pump moisture into air
– Water vapor increases with time
Net Evaporation
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Equilibrium
• Eventually, equal rates of
condensation and evaporation
• “Air is saturated”
• Equilibrium
At Equilibrium – Rate of evaporation is a
function of temperature;
Rate of condensation depends on water
vapor mass; Also a function of temperature
Rate of
evaporationRate of
condensation=
Evaporation = f(T)
Condensation = f(T)
Tair = Twater
Achieves a dynamic equilibrium
with vaporization in a closed
system.
A closed system means matter can’t
go in or out.
Any difference between evaporation and boiling?
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• Vaporization is an endothermic process - it requires heat.
• Energy is required to overcome intermolecular forces
• Responsible for cool earth.
• Why we sweat ?
Vaporization (evaporation)
• A liquid boils when its temperature reaches its boiling point
• Normal Boiling point is the temperature a substance boils at
1 atm pressure.
• The temperature of a liquid can never rise above its boiling
point.
Boiling
Changing the Boiling Point
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• Lower the pressure (going up into the mountains).
• Lower vapor pressure means lower boiling point.
• Food cooks slower.
• Raise the external pressure (Use a pressure cooker).
• Raises the boiling point.
• Food cooks faster.
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Moist Air
Dry air + Water vapor = Moist air
V
Moist Air: Mda+Mw=m
T Pda
da w
nRTp p p
V= = +
w daM M<<
w dap p<<
Typically
thus
Pw is called partial pressure
of water vapor.
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Humidity Parameters
( )
( )
mass of water vapor
mass of dry air
18.015268 /( )
( ) 28.966 /
18.0152680.622
28.966
w
da
total w
total da
w w
da da
MW
M
g moleN X mole
N X mole g mole
X X
X X
= =
= ×
= × =
0.622 0.622 0.622( )w w w
da da w
X p pW
X p p p= = =
−
Humidity ratio W (alternatively, the moisture content
or mixing ratio) of a given moist air sample is defined
as the ratio of the mass of water vapor to the mass of
dry air in the sample:
is the molar fraction ratio of water vapor and dry air.w
da
X
X
ww
total
NX
N=
Molar fraction of
water vapor.
Number of moles of water vapor
over the total number of mole of
the gas mixture.
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Humidity ParametersSpecific humidity γ is the ratio of the mass of water vapor to total mass
of the moist air sample:
1
w
w da
w daw da
da da
M
M M W
M MM M W
M M
γ = = =+ +
+
Absolute humidity (alternatively, water vapor density) dv is the ratio of
the mass of water vapor to total volume (V) of the sample:
/v wd M V=
Density ρ of a moist air mixture is the ratio of total mass to total volume:
( ) /da wM M Vρ = +
Saturation humidity ratio Ws(t, p) is the humidity ratio of moist air saturated
with respect to water (or ice) at the same temperature t and pressure p.
Degree of saturation µ is the ratio of air humidity ratio W to humidity
ratio Ws of saturated moist air at the same temperature and pressure:
,s t p
W
Wµ = 0.622 ws
s
ws
pW
p p=
−
Very close to W
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Relative HumidityRelative humidity (RH) ϕ is the ratio of the mole fraction of water
vapor xw in a given moist air sample to the mole fraction xws in an air
sample saturated at the same temperature and pressure:
, ,
w w
ws wst p t p
x p
x pφ = =
Change in RH can come from:
–Change in air temperature or pressure
–For a constant air temperature, RH increases with addition of water vapor
–For a constant water vapor pressure, increase in air temperature lowers the RH
–Highest RH occurs in the early morning (coolest time corresponds to the highest
RH); lowest RH occurs during the warmest part of the afternoon.
xws increases with T;
warmer air can hold more
moisture than cold air.
The ratio of actual water vapor content to the maximum possible moisture content at a
given temperature and pressure.
pws is the saturated vapor pressure (SVP) – the vapor pressure of the water vapor in an
air sample that contains the maximum amount of vapor possible at that temperature.
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RH highest in the cool morning; lowest in warmest time
Why droplets are usually
observed in the morning time?
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Saturated Vapor Pressure (SVP) of water vapor
Temperature (oC) SVP (Pa) Temperature (oC) SVP (Pa)
0 610 13 1497
1 657 14 1598
2 705 15 1704
3 758 16 1818
4 813 17 1937
5 872 18 2063
6 935 19 2197
7 1001 20 2337
8 1072 25 3166
9 1148 30 4242
10 1227 40 7375
11 1312 50 12351
12 1402 100 101325
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An exampleA sample of air has an RH of 40% at a temperature of 20oC. Calculate
the vapor pressure of the air.
What we know?
RH = 40% T = 20oC → SVP = 2337 Pa
, ,
40%2337
w w w
ws wst p t p
x p p
x pφ = = = =
2337 40% 934.8wp Pa= × =
1 millibars (mb) = 100 Pa
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Humidity Parameters
The mixture also obeys the perfect gas equation:
pV nRT=
( ) ( )da w da wp p V n n RT+ = +
where p = pda + pw is the total mixture pressure and n = nda + nw is the total
number of moles in the mixture.
Dew-point temperature td is the temperature of moist air saturated at
pressure p, with the same humidity ratio W as that of the given sample of
moist air. It is defined as the solution td( p, W) of the following equation:
( ),s dW p t W=
When moist air is considered a mixture of independent ideal gases (i.e., dry
air and water vapor), each is assumed to obey the ideal gas equation of state
as follows:Dry air:
da dap V n RT=
Water vapor:w wp V n RT=
V: total volume of the mixture; pda and pw are partial pressures of dry air and water vapor;
nda and nw are the number of moles of dry air and water vapor.
Dew or condensationAt which a fixed sample of air becomes
saturated (condensation occurs).
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How to measure the energy of moist air ? –
Enthalpy
The unit of enthalpy (H) is J or kJ.
Specific enthalpy (enthalpy per unit mass): h = H/m (J/kg, or kJ/kg)
Sensible energy (sensible heat)
Latent energy (latent heat)
The specific volume v of a moist air mixture is expressed in
terms of a unit mass v = V/Mda (m3/kg)
Mass of dry air1
da
vρ
= Density of dry
air (kg/m3)
How to measure the amount of moist air?
Why is v based on mass of dry air?
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Hygrometry (Psychrometry) – measurement of humidity
Hygrometers (Psychrometers) – instruments that measure the humidity of air
Hair/paper hygrometers: make use of the fact that hair or paper
change their dimensions with changing moisture content.
Dew-point hygrometer: measuring the dew-point temperature by
cooling a surface until water vapor condenses on it (then use the dew-point
T and room T to obtain an RH from tables/charts).
Wet-and-dry-bulb hygrometer: measuring the difference between the
wet and dry thermometers, as it indicates the relative humidity (from tables).
Electronic hygrometer: electronic sensors whose resistance changes with
changing humidity (this affect the amount of water absorbed by the sensors
from the air).
Psychrometric Chart
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Content
1. Moist air and their properties
2. Psychrometrics chart
3. Psychrometric process analysis
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Psychrometric Chart• Dry-bulb temperature t: ordinary air temperature
• Wet-bulb temperature twb: temperature read by a thermometer
wrapped in a wet fabric.
• Humidity ratio W
• Relative humidity φ
• Partial pressure of water vapor pw
• enthalpy of moist air: h = hda + Whw
Once two of the above properties are
known, the others can be readily
found out from a two-dimensional
chart – Psychrometric Chart.
hda is the specific enthalpy of dry air (kJ/kgda)
hw is the specific enthalpy of water vapor (kJ/kgda)
Approximately, 1.006dah t≈
t is in oC 2501 1.86wh t≈ +
At 0oC, the specific enthalpy of dry air is set as 0.
The specific enthalpy of water vapor at
0oC is 2501 kJ/kg.
Are they the same?
Which one is larger?
Why is the specific enthalpy based
on kg dry air?
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Psychrometric Chart
Dry-bulb temperature
Wpw
twb
twb
twb
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Psychrometric Chart
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Psychrometric chart
Linear
interpolation
should be done
when
necessary.
There are 2 horizontal lines:1. Solid lines are moisture content
(humidity ratio)
2. Dashed lines are vapor pressure
Search the interception of two
curves to find the state of
moisture air.
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Examples
Moist air sample has a dry
bulb temperature of 10oC,
humidity ratio (moisture
content) of 0.003 kg/kg
dry air, find (1) the wet-
bulb temperature; (2)
vapor pressure (millibars);
And (3) relative humidity
Answer:
(1) Wet-bulb T: 4.5oC
(2) Vapor pressure:
About 4.85millibars
(3) Relative humidity:
About 40%.
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Examples
Moist air sample has a dry
bulb temperature of 15oC,
wet-bulb temperature of
11oC, find (1) humidity
ratio (moisture content);
(2) vapor pressure
(millibars);
And (3) relative humidity
Answer:
(1) Humidity ratio:
0.0066 kg/kg dry air
(2) Vapor pressure:
About 10.5millibars
(3) Relative humidity:
About 60%.
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Another example - Textbook
(1) The RH of the heated air;
(2) The RH of the heated air
if 0.005kg/kg of moisture
is added;
(3) The temperature at which
this moistened air would
first condense.
External air at 0oC and 80% RH is
heated to 18oC. Use the
Psychrometric chart to determine
the following information:
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Another example - Textbook
(1) When the moist air is heated, the
mass of dry air and vapor does not
change. So the moisture content
remain unchanged!
RH: 23%
(2) When 0.005kg/kg of moisture is
added, the moisture content become:
0.003 + 0.005 = 0.008 kg/kg
RH: 62%
(2) For condensation
(dew), RH = 100%,
moisture content of
0.008kg/kg
Dew point: 10.8 oC
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An exercise
Moist air sample has a dry
bulb temperature of 15oC,
relative humidity of 90%,
find (1) wet-bulb
temperature;(2) humidity
ratio (moisture content);
(3) vapor pressure
(millibars); and (4) dew
point.
Answer:
(1) wet-bulb T:
14.2oC;
(2) Humidity ratio:
0.0097 kg/kg dry air;
(3) Vapor pressure:
About 15.3millibars
(4) Dew point:
14.3oC.
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Psychrometrics
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An example - ASHRAEMoist air exists at 40°C dry-bulb temperature, 20°C
thermodynamic wet-bulb temperature, and 101.325 kPa
pressure. Determine the humidity ratio, enthalpy, dew-point
temperature, relative humidity, and specific volume.
Why is the specific volume based on dry air?
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The example
Humidity ratio: 0.0065 kg/kg dry air
Enthalpy: about 56.7 kJ/kgda
Dew point temperature: 7oC
Relative humidity: 14%
Specific volume: 0.896 m3/kgda
Moist air exists at 40°C dry-bulb temperature, 20°C
thermodynamic wet-bulb temperature, and 101.325 kPa
pressure. Determine the humidity ratio, enthalpy, dew-point
temperature, relative humidity, and specific volume.
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Content
1. Moist air and their properties
2. Psychrometrics chart
3. Psychrometric process analysis
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Psychrometric processes
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Psychrometric
Processes
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Analysis of Psychrometric Systems
1 da1 w1m m m= +ɺ ɺ ɺ
Moist air
wmɺLiquid or vapor
QɺHeat
Moist air
2 da2 w2m m m= +ɺ ɺ ɺ
The property with a dot above it means the rate form.
m :i
:m mass (kg)
rate of mass flow (kg/s) E :i
:E Energy (kJ)
Rate of energy flow (kJ/s, or kW)
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Mass & Energy Conservation Equations
Mass
C.V. INTO
RATE
FLOWMASS RATE OF
MASS Generation/Consumption
in THE C.V.
=
−
C.V. OUT
RATE
FLOWMASS
Energy
C.V.
INTO ENERGY
OFRATE RATE OF ENERGY
Gen/Consumption
in the C.V.
=
−
C.V.OF
OUT ENERGY
OFRATE
Typical Assumptions• Steady state, steady flow
• No work
• Kinetic energy change is zero - usually
• Potential energy change is zero - usually
C.V. – Control volume, in
thermodynamics, it is also
called “system”
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Mass Equation
w da 2 1m m ( )W W= −ɺ ɺ
w1 w w2m m m+ =ɺ ɺ ɺWater(vapor):
dam= ɺda1 da2m m=ɺ ɺDry air:
w1 da 1m m W=i iw
da
MW
M=
w2 da 2m m W=i i
and
Then we
can get:
No mass gen/con in the C.V.
Mass flowing into the CV = Mass flowing out of the CV
da1 w1 w da2 w2m m m m m+ + = +i i i i i
W2>W1: humidifying
W2<W1: dehumidifying
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Energy Equation
=
OUT RATE
ENERGY
IN RATE
ENERGY
QIN RATE
ENERGYɺ=
w wm h+ ɺ da da1 da 1 w1m m W hh+ +ɺ ɺ
=
OUT RATE
ENERGY
da da2 da 2 w2m m W hh +ɺ ɺ
Thusw wmQ h+
i
ɺ da da1 da 1 w1m m W hh+ +ɺ ɺ
da da2 da 2 w2= m m W hh +ɺ ɺ
Energy (specific heat) added
when adding water vapor
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Energy Equation
w da 2 1m m ( )W W= −ɺ ɺ
w wmQ h+i
ɺ da da1 da 1 w1m m W hh+ +ɺ ɺ
da da2 da 2 w2= m m W hh +ɺ ɺ
( ) ( ) ( )da 2 1 w da da1 1 w1 da da2 2 w2 m W W m W h m W h 0Q h h h+ − + + − + =i
ɺ ɺ ɺ
Read from ASHRAE Psychrometric chart
Why is the specific enthalpy based on the mass of dry air?
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Classic Moist Air Processes
� Heating of moist air;
� Cooling of moist air;
� Cooling and dehumidification of moist air;
� Heating and humidification of moist air;
� Mixing process of moist air streams.
The following typical processes of moist air are
frequently encountered in air conditioning applications:
In each of the following examples, the process takes
place at a constant total pressure of 101.325 kPa.
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Moist Air Heating or CoolingAdding heat alone to or removing heat alone from moist air is represented by
a horizontal line on the Psychrometric chart, because the humidity ratio
remains unchanged.
Below figure shows a device that adds heat to a stream of moist air. For
steady-flow conditions, the required rate of heat addition is
( )2 1daQ m h h= −i i
( ) ( ) ( )da 2 1 w da da1 1 w1 da da2 2 w2 m W W m W h m W h 0Q h h h+ − + + − + =i
ɺ ɺ ɺ
1 da1 1 w1W hh h= + 2 da2 2 w2W hh h= +
1 2
Heating coil
Qɺ
damɺ
h1
W1
damɺ
h2
W2
w da 2 1m m ( )W W= −ɺ ɺ
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Heating of moist air
tdb
W
t1
W1
t2
h1
h2
2 1( )daQ m h h= −ɺ ɺ
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An example – heating/coolingMoist air, saturated at 2°C, enters a heating coil at a rate of 10
m3/s. Air leaves the coil at 40°C. Find the required rate of heat
addition.
State 1 is located on the saturation curve at 2°C. Thus, h1 =
13.0 kJ/kg, W1 = 4.38 gw/kg, and v1 = 0.785 m3/kgda. State 2 is
located at the intersection of t = 40°C and W2 = W1 = 4.38
gw/kg. Thus, h2 = 51.5 kJ/kg. The mass flow is:
3
3
10 // 12.74 /
0.785 /da
da
m sm Q v kg s
m kg= = =
i
( ) ( )2 1 12.74 / 51.5 13.0 / 490kWda da daq m h h kg s kJ kg= − = × − =i
Why is the specific volume based on mass of dry air?
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Cooling of Moist Air
1 2
Cooling
coilQi
damɺ
h1
W1
damɺ
h2
W2
da1 da2 dam m m= =ɺ ɺ ɺMass balance: w1 w2m m=ɺ ɺ
1 2( )daQ m h h= −ɺ ɺEnergy balance:
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Cooling of Moist Air
tdb
W
t2
W1
t1
h2
h1
1 2( )daQ m h h= −ɺ ɺ
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An exerciseOutside air at dry bulb temperature of 5oC, 80% RH enters a preheater
coil and leaves at 24oC (dry bulb). The air volume flow rate is 6.5
m3/s. Find (a) the outdoor air wet-bulb temperature and specific
volume; (b) the heated air moisture content and RH; and (c) the
heating coil power (rate of heating addition to the coil)
Find the enthalpy of the air sample before the coil and after the
coil, the change in enthalpy is caused by the preheater.
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The exercise
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Cooling and Dehumidification
1 2
3
damɺ
h1
W1
damɺ
h2
W2
Cooling
coilQi
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Cooling and Dehumidification
What if the supply air temperature is higher than 14oC?
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Heating and Humidifying
1 2Qi
Heating coil
damɺ
h1
W1
damɺ
h2
W2
wmɺ hwa
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Heating and Humidifying
tdb
W
t1
W1
t2
h1
h2
W2
2
1a
ha
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Adiabatic mixing of two moist air streams
,1 1 ,2 2 ,3 3da da dam h m h m h+ =ɺ ɺ ɺ
,1 ,2 ,3da da dam m m+ =ɺ ɺ ɺ
,1 1 ,2 2 ,3 3da da dam W m W m W+ =ɺ ɺ ɺ
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( )
( ) ( )
( )
( )
,1 1 ,2 2 ,1 ,2 3
,1 1 3 ,2 3 2
,13 2
1 3 ,2
da da da da
da da
da
da
m h m h m m h
m h h m h h
mh h
h h m
+ = +
⇒ − = −
−⇒ =
−
ɺ ɺ ɺ ɺ
ɺ ɺ
ɺ
ɺ
( )
( )
( )
( )
( )
,1 1 ,2 2 ,1 ,2 3
,13 2 3 2
1 3 ,2 1 3
da da da da
da
da
m W m W m m W
mW W h h
W W m h h
+ = +
− −⇒ = =
− −
ɺ ɺ ɺ ɺ
ɺ
ɺ
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Thank you very much for your attention!
Please feel free to contact me if you have any
questions.
Meng Ni
ZN713
Tel: 2766 4152 (office)
Email: [email protected]