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**EGR 334 Thermodynamics Chapter 12:**

Lecture 40: Pyschrometic Chart Quiz Today?

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**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

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**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.

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**Evaporation of water provides cooling.**

Sec 12.8 : Analyzing Air-Conditioning Processes “Swamp” Coolers (evaporative cooler) Evaporation of water provides cooling.

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**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

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**Sec 12.7 : Psychrometric Charts**

Figure A-9

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**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)

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**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)

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**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:

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**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

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**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

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**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:

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**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:

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**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:

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**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

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**End of slides for lecture 40**

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