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9/18/2012
1
PsychrometricsTheory and Applications
Dr. Murat Kacira, Associate ProfessorA g r i c u l t u r a l a n d B i o s y s t e m s E n g i n e e r i n g
C o n t r o l l e d E n v i r o n m e n t A g r i c u l t u r e C e n t e r
Theory and Applications
C o n t r o l l e d E n v i r o n m e n t A g r i c u l t u r e C e n t e r
U n i v e r s i t y o f A r i z o n a
m k a c i r a @ c a l s . a r i z o n a . e d u
ObjectivesLearn to answer these question at the end of Psychrometrics lecture session!!
What is Psychrometrics?
What are thermodynamic properties of moist air? What are thermodynamic properties of moist air?
What is the Psychrometric Chart?
How do we use the Psychrometric Chart?
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Applications:
· HVAC systems, animal, plant and human comfort· Air-conditioning devices· Cooling towers· Industrial processes requiring close control of the vapor content in airF d i d i i· Food sciences and engineering
· Animal housing· Plant systems (Greenhouses, growth chambers, plant based bioregenerative
life support systems etc.)· Evaporative coolers· Mold problems & health related issues (Dew & Condensation!)· Water harvesting in arid lands
What is Psychrometrics ?
The term derives from the Greek psuchron (ψυχρόν) (cold) and metron (μέτρον) (means of measurement)(cold) and metron (μέτρον) (means of measurement)
Simple definition:
the study of moist air properties!
in Engineering terms:in Engineering terms:
the science involving thermodynamic properties of moist air.
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What are thermodynamic properties of moist air?
Properties that describe the state of a Psychrometricprocess process .
What is Moist Air?
A binary mixture of dry
air and water vaporair and water vapor.
- Each mixture component behave as ideal gases at the states under present consideration.
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Dalton model (Partial pressures)
Each component is considered to act as if it existed alone in the volume V at the mixture temperature T while exerting a part of the pressure
(John Dalton, 1766 – 1844)E li h h i h i i )part of the pressure.
P = Pa + Pv
P = mixture pressurePa = partial pressure of dry airPv = partial pressure of water vapor
English chemist, physicist)
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Psychrometric definitionsA ) Temperature
Dry Bulb Temperature (TDB; °C or F)
The temperature of air measured by a thermometerThe temperature of air measured by a thermometerfreely exposed to the air but shielded from radiationand moisture.
It is the true air
temperature we “feel ”temperature we feel.
Wet Bulb Temperature (TWB; °C or °F)The temperature at which water, by evaporating into air,
b i h i i di b i ll h
A ) Temperature
can bring the air to saturation adiabatically at the same temperature.In other words, TWB is the minimum temperature that the moist air could achieve if enough water was added to achieve saturation (RH = 100%).T is often used to indicate how TWB is often used to indicate how much water can be added to the air through evaporation.Also called the adiabatic saturation temperature.
Dry-bulb temperature (oC, F)
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Sling Psychrometer
One device that uses the wet/dry bulb method is the slingpsychrometer where the thermometers are attached to apsychrometer, where the thermometers are attached to ahandle or length of rope and spun
around in the air for a few minutesWet bulb thermometer
Wet wick
What does the reading tell you?
Dry bulb thermometer
A. If the wet bulb temperature is lowerthan the dry bulb the air-vapor mixture is unsaturated.B. If the wet bulb temperature is the sameas the dry bulb, the air-vapor is saturated.
Assmann Psychrometer
Aspirated type portable hygrometerAspirated type portable hygrometer
In the Assman psychrometer, the wet-bulb thermometer is installed in a duct where the air is flowing at reasonable velocity (2.5 m/s wind into 2
lower pipes)
Richard Assmann (German name Richard Aßmann); (1845 -1918), meteorologist and physician
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A ) Temperature
Dew Point Temperature (TDP; °C or F)The temperature at which saturation is reached (RH = 100%) when the moisture content of the air (W) stays constant.In other words, TDP is the temperature at which water will begin to condense out of moist air.C d i hCondensation occurs when:
TDP > TairDew point temperature is typically achieved by sensible cooling.
Dew Point Temperature
State 1
State 2
T1
Tdp
SODAT < TDP
When the temperature of cold drink is below the dew-point temperature of the surrounding air, it “sweats.”
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Psychrometric definitionsB ) Water Content
Humidity Ratio/Absolute Humidity/Specific Humidity (W/AH; kg H2O/kg DA or lb H2O/lbDA)
- The actual water content of the air.- Expressed as a ratio of water
vapor content to total amount of dry air.
mv: mass of water vapor (kg H2O)ma: mass of dry air (kg dry air)P : Total pressure (kPa)
When the water vapor is added to the dry air, specific humidityor humidity ratio (W) increases. As we keep adding water vapor, the air becomessaturated and can not hold water vapor anymore.
Any more moisture addition to saturated air will condense.The amount of water vapor in saturated air can be determined using:
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B ) Water Content
Relative Humidity (φ/RH; %)
- The ratio of actual vapor pressure to saturation vapor pressure at the same temperaturesame temperature
- Simply, it is a measure of how much water is
in the air versus how much water the air can
hold at the same temperature.
Important: Relative humidity is relative
Because it depends on the air temperature.
100g
v
g
v
PP
mm
==φ
p p
The air can hold more moisture at higher
Temperatures (TDB) than lower temperatures
(TDB). Therefore, air at 20°C and 40% RH
will have a lower water content (ω) than at
40°C and 40% RH.
g
v
PP
=φCombining and
( ) gPPω
ωφ+
=622.0 g
g
PPP
φφ
ω−
=622.0
( ) g gφ
9/18/2012
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B ) Water Content
Vapor Pressure Deficit (VPD; kPa or mm Hg or in. H2O)
- The difference between actual vapor pressure and saturation vapor pressure at a given temperaturevapor pressure at a given temperature.
- Like relative humidity, VPD is a measure of how much water is in the air versus how much water the air can hold at the same temperature.
- Many plant scientists use VPD rather than RH because transpiration by plants depends largely on the gradient between water at the plant (PSat,plant) and water in the air (PAct, air).
VPD = Psat. vapor – Pactual vapor
or VPD = Pg – Pv
or VPD = e* – e
C) Energy Content
Enthalpy (h; kJ/kg dry air or BTU/lb air)
- This is the amount of energy contained in the moist air.This is the amount of energy contained in the moist air.
- Enthalpy represents the
amount of sensible and
latent energy contained
in the moist air.
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Dry air1 kg
ha
Moistureω kg
hgva
mhHHHH
=+=
Atmospheric air is mixture of dry and moist air, so enthalpy of air is expressed in terms of enthalpies of dry and moist air as:
va
vava
va
a
a
vvaa
hhh
hhhmmh
mHh
mbydividinghmhmH
ω
ω
+=
+=+==
+=:
(kJ kg dry air-1)
(1+ ω) kg of moist air
hv ≅ hg (T) ga hhh ω+=
Reference to steam table data or a Mollier diagram for water shows that theenthalpy of superheated vapor at low vapor pressures is very closely given by thesaturated vapor value corresponding to the given temperature. Hence, the enthalpyof the water vapor hv , can be taken as hg at the mixture temperature. That is:
Example 1
5 m x 5 m x 5 m
GivenFind:a) The partial pressure of dry airb) The specific humidity (humidity ratio) of the air
Containing air @T = 25 oCP = 100 kPaφ = 75%
b) The specific humidity (humidity ratio) of the airc) The enthalpy per unit mass of dry air (specific enthalpy)d) The masses of dry air and water vapor in the room
Solution:a) Pa = P – Pv
Pv = φ Pg = φ Psat@25o
C = 0.75 x 3.169 kPa = 2.38 kPaTherefore;
Pa = P – Pv = (100 – 2.38) kPa = 97.62 kPa
b) 0152.038.2100
38.2622.0622.0 =−
=−
=v
v
PPPω kg H2O kg dry air-1
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c) h = ha + ωhv
h C T h h h k k f blh ≅ CpT +ωhg ; hg=hsat@25 oC = 2547.2 kJ kg-1 from Table
= [1.005 (kJ kg-1 oC-1) x 25 oC] + [0.152 x 2547.2 (kJ kg-1)]
h = 63.8 kJ kg-1
d) Both dry air and water vapor fill the entire room.
So, the volume of each element is equal to the room volume:
Va = Vv = Vroom = 5 X 5 X 5 = 125 m3
P V R TPaVa=maRaT
PvVv=mvRvT
kgKkgKkPam
mkPaTvRvVvP
vm
kgKkgKkPam
mkPaTaRaVaP
am
3.1298/34619.0
312538.2
61.85298/3287.0
312562.97
=×⎥⎦⎤
⎢⎣⎡
×==
=×⎥⎦⎤
⎢⎣⎡
×==
kgk amv 98/6 9.0 ⎥⎦⎢⎣
The mass of water vapor could also be computed as :
mv = ω ma = 0.0152 x 85.61 kg = 1.3 kg water vapor
9/18/2012
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Specific Volume (ν; m3/kg dry air or ft3/lb dry air)
• This is the volume of the moist air mixture (volume
D) Volume
• This is the volume of the moist air mixture (volume occupied by both dry air and water vapor) versus the unit mass of dry air.
• At higher temperatures, the air
molecules are more energetic
causing the volume of the moist causing the volume of the moist
air mixture to expand and the
density to decrease.
mVv =
v = specific volume (m3 / kg dry air)
• The specific volume of air is the inverse of density
(ν = 1/ρ)
am
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The Psychrometric Chart
What is a Psychrometric Chart?What is a Psychrometric Chart?A graphical representation of the relations between the thermodynamic properties of moist air
Charts are constructed for a single barometric pressure
If you know only two properties of the moist air you If you know only two properties of the moist air, you can determine all other thermodynamic properties of the moist air!!
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Example 2
Given: TDB = 25°C, TWB =20°C
Required: Required:
(a) Relative humidity (RH) (%),
(b) Dew-point temperature (TDP) (oC)
(c) Humidity ratio (Absolute Humidity) (g per kg dry air)
(d) S ifi l ( ) ( /k d i )(d) Specific volume (v) (m3/kg dry air)
(e) Specific enthalpy (h) (kJ/kg dry air)
RH = 63%
TDT = 17.6 oC
TWT = 20 oCh= 57.5 kJ/kg DA
v= 0.862 m3/kg DA
W = 12.6 g H2O/kg DA
TDB = 25 oC
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Example 2 (solutions)
Given: TDB = 25°C, TWB =20°C
Required: Required:
(a) Relative humidity = 63 (%)
(b) Dew-point temperature = 17.6 (oC)
(c) Humidity ratio = 12.6 (g/kg dry air)
(d) Specific volume = 0.862 (m3/kg dry air)
(e) Specific enthalpy (h) = 57.5 (kJ/kg dry air)
Example 3Based on given Find the thermodynamic properties of moist air at the two states. Draw the process line.
State Point 1 State Point 2TDB 1 = 40°C TDB 2= 25°CDB,1 DB,2
ω1= 14 g/kg
ν2= 0.85m3/kg
TWB,1= 25CTDP,1= 19.1CRH1= 30 %
h1= 76 kJ/kgv1= 0.907 m3/kg
TWB,2= 12.5 CTDP,2= 0.67 CRH2= 20.7%ω2= 3.9 g/kgh2= 35 kJ/kg
P1= 16.5 mmHg(2.23 kPa)
P2= 4.8 mmHg(0.64 kPa)
to convert from mmHg, multiply by 0.133 kPa/mm Hg
9/18/2012
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Example
State Point 1 State Point 2TDB1 = 40°C
1 ) Find the properties.2) Draw the process line.
State Point 1 State Point 2TDB1 = 40°C TDB,2=22.1°CDB1
T2WB,= 18.5°C
RH1= 10% RH2= 70%
) p3) What is the process called?
DB1 DB,2TWB,1= 18.5 °C T2WB,= 18.5°CTDP,1= 2.66 °C TDP,2= 17.1 °CRH1= 10% RH2= 70%ω1= 4.6 gr/kg ω2= 12.1 gr/kgh1= 51.9 kJ/kg h2= 52 kJ/kgν1= 0.89 m3/kg ν2= 0.85 m3/kgν1 0.89 m /kg ν2 0.85 m /kgP1= 5.5 mmHg P2= 13.98 mmHg
How is the Psychometric Chart used?
We can determine property changes caused by a system (process)system (process).
We can design HVAC systems based on heating/cooling needs to achieve the final (desired) state point.
We can determine system efficiencies.
We can determine energy changes related to moist We can determine energy changes related to moist air properties and the systems that change them (ventilation, evaporative cooling, heating, etc)
9/18/2012
19
AIR CONDITIONING PROCESSES
Hu
mid
ifyi
ng
g Sensible heating
Sensible cooling
De
hu
mid
ifyi
ng Sensible heating
Modeling air conditioning (AC) processes
Most AC processes can be modeled as steady-state.Apply:
Conservation of mass (both dry air and water vapor)Conservation of energy
∑ ∑ ∑ ∑∑ ∑
==
=
eeaiiaewiw
eaia
mmormm
mm
ωω ,,,,
,,Dry air mass
Water mass
∑∑ −=− iiee hmhmWQEnergy
i and e denotes inlet and exit states, respectively.
9/18/2012
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Sensible heating
This process involves heat transfer of only sensibleheat (remember that’s the heat we feel). We canheat (remember that s the heat we feel). We candetermine the total energy added to the air byknowing the change of enthalpy fromState 1 to State 2, as well as the total amount of airthat is receiving the energy.
Sensible Heating and Cooling (ω = constant)
Heating coils
T1, ω1, φ1
Air T2
ω2= ω1
φ2 < φ1
21 mmm aaa ==
( )12
12
21
hhqhhmQ a
−=−=
= ωω
21
ω = constant
φ1 φ2
9/18/2012
21
Example 4
Moist air (TDB = 45F, RH = 80%) enters a heating coil at 10 ft3 min-1 and exits the coil at TDB = 84°F, at 10 ft min and exits the coil at TDB 84 F, RH=70%.
1. Find the psychrometric properties of moist air entering and leaving the coil.
2. What is the rate of heat transfer to the air?
3 If the coil is operated for 10 minutes what is the 3. If the coil is operated for 10 minutes, what is the total heat added to the air?
State Point 1 State Point 22) How much heat is added?
Sensible Heat Energy Balance
1) Properties
Solution 4
TDB1 = 45°F TDB,2= 84°FTWB,1= T2WB,= 18.5°CTDP,1= TDP,2= RH1= 80 % RH2= 70%ω1= ω2= h1= h2=
)( 1212 hhmQ a −= &
ina
FlowRatemν
=&
i/780min/10 3
lbft
gy
Mass flow rate of air
42 F39 F
0.005 gr/lb 0.005 gr/lb
26 BTU/lb 16.5 BTU/lb
39 F
1 2vin=
min/78.0/8.12 3 lblbft
f==
min/41.7)5.1626(78.012 BtuQ =−=BtuBtuQ 1.74min10
min41.712 =×=
Rate of Heat Added to Air
3) Total heat added to air after 10 min
26 BTU/lb 5 / b12.8 ft3/lb
9/18/2012
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Heating with humidification
Heating coils
Air
ω3> ω21 2 3ω1= ω2
Heating section Humidifying section
(ω3- ω2)
(ω3- ω2)
9/18/2012
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where the left side of the equation represents the slope of the humidificationprocess on a psychrometric chart. Thus the direction of the process can bedetermined from the enthalpy of the steam added to the air stream and the
(ω3- ω2)
pyenthalpy – moisture protractor on a psychrometric chart.
9/18/2012
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Example 5
An air conditioning system is taking in outside air at10 oC and 30% relative humidity at a steady rate of 453 y y 45m3 min-1 and is conditioning it to 25 oC and 60%relative humidity. The outdoor air is first heated to22 oC in the heating section and then humidified byinjection of hot steam in the humidifying section.Assume that the processes take place at 100 kPapressure. Determine:a) The rate of heat supply in the heating section?b) The mass flow rate of the steam required in thehumidifying section?
Solution (Example 5)
Dry air mass 21,, mmmmm aaaeaia ==→=
∑ ∑∑ ∑
Water massEnergy
φ3=60%
3φ 30%
Heating coils
( )12
2211,,
hhmQmmmm
a
aaewiw
−=
=→=∑ ∑ ωω
3
1φ1=30%
10 C
ω = constant
2
22 C 25 C
ω3> ω2Air
1 2 3ω1= ω2
T1 =10 Cφ1 =30 %V1=45 m3/min
T3 =25 Cφ3 =60 %T2 =22 C
9/18/2012
25
kPaPPP
kPakPaPPP Csatgv
632993680100
368.02276.13.020@1111
=−=−=
=×=== φφa)
[ ] kgdryairmkPa
KkgKkPamP
TRv
kPaPPP
a
a
va
/815.0632.99
283/287.0
632.99368.0100
33
1
11
111
=×
==
=−=−=
/0023.03680100368.0622.0622.0
min/2.55/815.0
min/45
11
3
3
1
11
kgdryairkgwaterkPakPa
PPP
kgkgm
mvV
m
v
a
=×
=×
=
===
ω
( ) ( )min/4.673
/8.1528min]/2.55[
/0.28/2.25410023.0)22/005.1(
/8.15/8.25190023.0)10/005.1(368.0100
12
21
2222
1111
11
kJQkgkJkghhmQ
kgkJkgkJCkgCkJhTCh
kgkJkgkJCkgCkJhTChkPaPP
a
gp
gp
v
=−×=−=
=
=×+×=+=
=×+×=+=−−
ωω
ω
ω
3322 mmm awa =+ ωω
b)
( )
[ ]/01206.0
169.360.0100169.360.0622.0622.0
3
333
333
23
3322
kgdryairkgwater
kPakPa
PPP
mm
g
g
aw
awa
=
×−××
=−
=
−=
ω
φφ
ω
ωω
( )min/539.0
0023.001206.0)/(2.55,
kgmkgdryairkgwaterm
So
w
w
=−×=
9/18/2012
26
Cooling with Dehumidification
This process involves both the removal ofwater vapor and heat from the moist air.Therefore, the energy balance equation willinclude a change in enthalpy as well as aterm for the removal of moisture from thesystem, both in the forms of vapor andliquid condensate that will likely form onthe cooling coil.
Example 6
Air enters a window air conditioner at 1 atm, 30 C and80% RH at a rate of 10m3/min and leaves at saturated80% RH at a rate of 10m3/min and leaves at saturatedair at 14 C. Part of the moisture in the air whichcondenses during the process is also removed at 14 C.Determine:
a) The rates of heat and moisture removal from the air
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Solution 6
Cooling coils
AirAir
T1 =30 Cφ1 =80%
V1=10 m3/min
T2 =14 Cφ3 =100 %
aaa mmm == 21
Condensate removalat 14 C
( )
wwaaaa
iiee
aw
waa
hmhmhmQhmhmWQ
mmmmm
+−=
−=−
−=+=
∑ ∑1122
21
2211
ωωωω
St t 1 St t 1’ St t 2
In let and exit states are completely specified and the total pressure is at 1 atm, so we can use psychrometric chart to determine the properties of the air
Solution 6 continued
2ω = constant
φ2=100%φ1=80%
1’1
State 1 State 1’ State 2
TDB1 = 30 C TDB,2= 14 C TDB,2= 14 C
φ1= 80 % φ1’= 100 % φ2= 100 %ω1= 0.0216kgH2O/kg dry air
ω1’= 0.0216kgH2O/kg dry air
ω2= 0.0100kgH2O/kg dry air
h1= 85.4 kJ/kg hw=hf@14C
hw=58.8 kJ/kgh2= 39.3 kJ/kg
v1=0.889 m3/kg dry air 2
14 C 30 C
g y
( )( )
min/513)]/8.58min)(/131.0[(/4.853.39min/3.11
min/0131.00100.00216.0min/3.11
min/3.11/889.0
min/103
3
1
11
kJQkgkJkgkgkJkgQ
kgkgm
kgkgdryairm
mvVm
w
a
−=+−==−=
===
9/18/2012
28
Evaporative Cooling
Evaporative cooling is an adiabatic process, meaningthere is no energy transfer to or from the system.there is no energy transfer to or from the system.However, the energy is transformed from sensible tolatent.
Conventional cooling systems operate on arefrigeration cycle, and they can be used in any part ofthe world. But they have high initial and operationcosts!!
In desert (hot and dry) climates, evaporative coolerscan be used to avoid high costs of cooling.
Swamp coolers
9/18/2012
29
PAD-FAN system in Greenhouses
Example 7
Outside air enters a greenhouse pad at 95 F and 20% relative humidity and enter the greenhouse interior at relative humidity and enter the greenhouse interior at 80% relative humidity. Determine:
a) The exit temperature of the air from the pad.
b) The lowest temperature to which the air can be cooled by pad-fan system.
9/18/2012
30
T 95 F φ %T1 = 95 Fφ1 = 20%
P = 14.7 psiφ2 = 80%
Twb and h = constant
φ2’=100%
φ1=20%
1
2
2’ φ2=80%
Twb1@95 F and 20% = 66 F
T = T = 66 F95 FT2Tmin
Twb1= Twb2 = 66 F
T2@Twb=66 F and 80% = 70.4 F
Tmin@Twb=66 F and 100% = T2’= 66 F
100,,
,, ×−−
=OutWBOutDB
InDBOutDBPad TT
TTη
%8.841006695
4.7095=×
−−
=Padη
9/18/2012
31
Adiabatic mixing of airstreams
ω1
h1
ω2
h2
h1
ω3
h3
Mixing section
2
3
1
2
3
ω1
ω3
ω2h1
h3
h2
h3-h1
h2-h3
Dry air mass 321 mmm aaa =+
95 FT2Tmin
1y
Water mass
Energy
13
32
13
32
2
1
332211
332211
321
hhhh
mm
hmhmhmmmm
a
a
aaa
aaa
aaa
−−
=−−
=
=+=+
ωωωω
ωωω
Eliminating ma3
Example 8
A stream of 3 m3/s of outdoor air at 4°C dry-bulbtemperature and 2°C thermodynamic wet-bulb temperatureis adiabatically mixed with 7.25 m3/s of recirculated air at25°C dry-bulb temperature and 60% rh. Find the dry-bulbtemperature and thermodynamic wet-bulb temperature ofthe resulting mixture.
9/18/2012
32
The above figure shows the schematic solution.
v1 = 0.789 m3/kg (dry air), and v2 = 0.86 m3/kg (dry air). ma1= 3 / 0.789 = 3.8 kg/s (dry air)ma2= 7.25 / 0.86 = 8.43 kg/s (dry air)ma3=ma1+ma2= 12.23 kg/s (dry air)
23 mLine
φ2 = 60%
69.023.1243.8
21313123
3
2
2
1
===−−
=−−
a
a
a
a
mm
LineLine
ormm
LineLine
Consequently, the length of line segment 1–3 is 0.69 times the length of entire line 1–2. Using a ruler State 3 is located
Tdb1=4 oC
Tw1= 2 oC
Tdb2=25 oC
Using a ruler, State 3 is located,Line(1-2)=6 cmLine (1-3)=6*0.69=4.14 cm
And, Tdb3 = 18.5°C Twb3 = 15.0°C
Example 9
Moist air at 25°C dry-bulb and 10°C thermodynamicwet-bulb temperature is to be processed to a final dew-wet bulb temperature is to be processed to a final dewpoint temperature of 15°C by adiabatic injection ofsaturated steam at 110°C. The rate of dry airflow is 3kg/s (dry air). Find the final dry-bulb temperature ofthe moist air and the rate of steam flow.
9/18/2012
33
From a sarurated water-vapor properties table, the enthalpy of the steam hg@110 C = 2691.5 kJ/kg ( or hw). g@110 C g ( w)Therefore, according to the psychrometricequation, the condition line on the psychrometric chart connecting States 1 and 2 must have a direction:Δh/ΔW = 2.691 kJ/g (water)
State 2 is established at the intersection of the condition linewith the horizontal line extended from the
10 C
15C W2= 10.5
with the horizontal line extended from the saturation curve at 15°C (td2 = 15°C). And, tdb2 = 27.5Values of W2 and W1 can be read from the chart. The required steam flow is,mw= ma (W2 – W1 ) = 3 kg dry air/s× (10.5 – 2)g/kg dry air = 25.5 g/s (steam)
25 C
W1= 2