egr 334 thermodynamics chapter 12: lecture 40: pyschrometic chart quiz today?
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EGR 334 ThermodynamicsChapter 12:
Lecture 40: Pyschrometic Chart
Quiz Today?
Today’s main concepts:• Understand the structure of the pyschrometric chart
and identify air/vapor properties from it.• Be able to solve air conditioning problems using the
chart.
Reading Assignment:
Homework Assignment:
no assignment
No Assignment
Final Exam: 1:00 p.m. on Tuesday, May 15
3Sec 12.8 : Analyzing Air-Conditioning Processes
Air-Conditioning
Heat from “hot side” is used for evaporation of the coolant.
Heat is rejected to the outside by condensation.
4
“Swamp” Coolers (evaporative cooler)
Evaporation of water provides cooling.
Sec 12.8 : Analyzing Air-Conditioning Processes
5
Procedure for analysis of air conditioning systems:
1) Identify State Properties of as many individual mixture componentsUse ideal gas law: or Table A20
and A22Steam tables: Tables A2, A3, A4, etc.Humidity definitions:Constant process data: isobaric, isothermal, isentropic, polytropic, etc. Psychrometric Chart: Figures A9 and A9E.
1 1 1, ,pV mRT pv RT pV n RT
/V gp p )/ 0.622 /(V a V Vm m p p p
2) Apply mass balance to each individual component of the mixture.
3) Apply energy balance to each separate stream of the mixture.
in outair airm m 2 2in outH O H Om m
2 2 2 20 0 0 00in in in in out out out outair air H H air air H HQ W m h m h m h m h
4) Solve equations
6Sec 12.7 : Psychrometric Charts
Figure A-9
Psychrometric Chart
7Sec 12.7 : Psychrometric Charts
Figure A-9 (pages 920,1)
To open the windows or not?Inside: T = 85°F, = 60%
Outside: T = 80°F, = 75%
Open only if ωi< ωo.
8Sec 12.7 : Psychrometric Charts
Figure A-9 (pages 920,1)
To open the windows or not?Inside: T = 85°F, = 60%
Outside: T = 80°F, = 80%
Open only if ωi> ωo.
Since ωi< ωo, don’t open, house will cool. Don’t let extra moisture in.
9Sec 12.8 : Analyzing Air-Conditioning Processes
Problems can be solved using 1) tabulated data 2) Psychrometric chart.
Mass balance:
132 vvv mmm
Energy balance:
3322110 hmhmhmQCV
aaa mmm 12
Often can neglect
withav mm 11 and
av mm 22
amm 123
e
ee
eei
ii
iiCVCV gz
vhmgz
vhmWQ
220
22
000000 ee
eii
iCV hmhmQ
10Sec 12.8 : Analyzing Air-Conditioning Processes
Air-Conditioning Energy balance:
3322110 hmhmhmQCV
332221110 vvvvaavvaaCV hmhmhmhmhmQ
for hv1 & hv3 use saturation point
But, we can re-write in a more convenient form.with
av mm 11 andav mm 22
3122221110 vavaavaaCV hmhhmhhmQ and av mm 123
3122211210 vvvaaaCV hhhhhmQ
Evaluate hv1, hv2 & hv3 at steam tables
Evaluate ha1 & ha3 using Table A-22
or
a gh hEvaluate moist air specific enthalpy using the pyschrometric chart
11
Example (12.67): Moist air at 22°C, and a wet bulb temperature of of 9°Centers a steam spray humidifier. The mass flow of the dry air is 90 kg/min. Saturated water vapor at 110 C is injected into the mixture at a rat of 52 kg/hr. There is no heat transfer with the surroundings, and the pressure is constant throughout at 1 bar. Determine at the exita) the humidity ratiob) the temperature in oC.
T1=22°C, Twb=9°C
Saturated vapor:T3=110 oC
Moist airT2=? ω2=?1 2
3
1 1 1a vm m m 2 2 2a vm m m
3 3vm m
3 52 /vm kg hr
1 90 / minam kg
12
Example (12.67): a) the humidity ratiob) the temperature in oC.
Mass Balance Equations:
1 2
3
1 1 1a vm m m 2 2 2a vm m m
3 3vm m
1 1 1 1 2 2 2 2 3 30 ( ) ( ) ( )a a v v a a v v vQ W m h m h m h m h m h
1 2 90 / mina am m kg for air:
for H20:
1 3 2v v vm m m 3 52 / 0.867 / minvm kg hr kg with
Energy Balance Equations:
1 1 1 2 2 2 3 30 ( ) ( ) ( )a a a v a a a v vm h m h m h m h m h
1 1 2 2 3 30 ( ) ( )a a g a a g gm h h m h h m h
13
Example (12.67): a) the humidity ratiob) the temperature in oC.
Use the psychrometric chart to find properties.
1 2
3
1 1 1a vm m m 2 2 2a vm m m
3 3vm m1 122 9o o
dpT C T C State 1:
State 3: Saturated vapor at 110oC
1 0.002 /vapor airkg kg
1( ) 27.2 /a g airh h kJ kg
Use Steam table, A2:
3 2691.5 /g vaporh kJ kg
14
Example (12.67): a) the humidity ratiob) the temperature in oC.
Mass Balance Equations:
1 2
3
1 1 1a vm m m 2 2 2a vm m m
3 3vm m
2 3 1 0.867 0.180 1.047 / minv v v vaporm m m kg
From the energy balance:
1 1 1 (0.002 / )(90 / min) 0.18 / minv a vapor air air vaporm m kg kg kg kg
and
therefore:
32 2 1 1 3( ) ( )a g a g g
a
mh h h h h
m
22
2
1.0470.0116 /
90v
vapor aira
mkg kg
m
0.86727.2 / (2691.5 / ) 53.13 /
90vapor
air vapor airair
kgkJ kg kJ kg kJ kg
kg
15
Example (12.67): a) the humidity ratiob) the temperature in oC.
therefore at State 2:
1 2
3
1 1 1a vm m m 2 2 2a vm m m
3 3vm m
Using the pyschrometric chart:
2 0.0116 /vapor airkg kg
2 2( ) 53.13 /a g airh h kJ kg
24oC
At the intersection of the humidity ratio and the specific enthalpy of moistair, the dry bulb temperature canbe directly looked up: 24 oC.
16
End of slides for lecture 40
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