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