1. EGR 334 Thermodynamics Chapter 12:
Lecture 40:
Pyschrometic Chart
Quiz Today?

2. 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.
Final Exam:
1:00 p.m. on Tuesday, May 15
Reading Assignment:
no assignment
Homework Assignment:
No Assignment

3. Heat from “hot side” is used for evaporation of the coolant.
Sec 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. Evaporation of water provides cooling.
Sec 12.8 : Analyzing Air-Conditioning Processes
“Swamp” Coolers
(evaporative cooler)
Evaporation of water provides cooling.

5. Procedure for analysis of air conditioning systems:
1) Identify State Properties of as many individual mixture components
Use ideal gas law: or Table A20 and A22
Steam tables: Tables A2, A3, A4, etc.
Humidity definitions:
Constant process data: isobaric, isothermal, isentropic, polytropic, etc.
Psychrometric Chart: Figures A9 and A9E.
2) Apply mass balance to each individual component of the mixture.
3) Apply energy balance to each separate stream of the mixture.
4) Solve equations

6. Sec 12.7 : Psychrometric Charts
Figure A-9

7. To open the windows or not?
Sec 12.7 : Psychrometric Charts
To open the windows or not?
Inside: T = 85°F,  = 60%
Outside: T = 80°F,  = 75%
Open only if ωi< ωo.
Figure A-9 (pages 920,1)

8. To open the windows or not?
Sec 12.7 : Psychrometric Charts
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.
Figure A-9 (pages 920,1)

9. Problems can be solved using 1) tabulated data 2) Psychrometric chart.
Sec 12.8 : Analyzing Air-Conditioning Processes
Problems can be solved using
1) tabulated data
2) Psychrometric chart.
Mass balance:
Often can neglect
with
and
Energy balance:

10. But, we can re-write in a more convenient form. with and and
Sec 12.8 : Analyzing Air-Conditioning Processes
Air-Conditioning
Energy balance:
But, we can re-write in a more convenient form.
with
and
and
Evaluate hv1, hv2 & hv3 at steam tables
for hv1 & hv3 use saturation point
Evaluate ha1 & ha3 using Table A-22 or
Evaluate moist air specific enthalpy using the pyschrometric chart

11. b) the temperature in oC.
Example (12.67): Moist air at 22°C, and a wet bulb temperature of of 9°C
enters 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 exit
a) the humidity ratio
b) the temperature in oC.
Moist air
T2=? ω2=?
T1=22°C, Twb=9°C 1 2 3
Saturated vapor:
T3=110 oC

12. b) the temperature in oC.
Example (12.67):
a) the humidity ratio
b) the temperature in oC. 1 2 3
Mass Balance Equations:
for air:
for H20:
with
Energy Balance Equations:

13. b) the temperature in oC.
Example (12.67):
a) the humidity ratio
b) the temperature in oC. 1 2 3
State 1:
Use the psychrometric chart to find properties.
State 3:
Saturated vapor at 110oC
Use Steam table, A2:

14. b) the temperature in oC.
Example (12.67):
a) the humidity ratio
b) the temperature in oC. 1 2 3
Mass Balance Equations:
and
therefore:
From the energy balance:

15. b) the temperature in oC.
Example (12.67):
a) the humidity ratio
b) the temperature in oC. 1 2 3
therefore at State 2:
Using the pyschrometric chart:
At the intersection of the humidity ratio and the specific enthalpy of moist
air, the dry bulb temperature can
be directly looked up: 24 oC.
24oC

16. End of slides for lecture 40