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

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Presentation on theme: "EGR 334 Thermodynamics Chapter 12: Lecture 40: Pyschrometic Chart Quiz Today?"— Presentation transcript:

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

2 Todays 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

3 3 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 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: 5 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 6 Sec 12.7 : Psychrometric Charts Figure A-9 Psychrometric Chart

7 7 Sec 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 ω i < ω o.

8 8 Sec 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 ω i > ω o. Since ω i < ω o, dont open, house will cool. Dont let extra moisture in.

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

10 10 Sec 12.8 : Analyzing Air-Conditioning Processes Air-ConditioningEnergy balance: for h v1 & h v3 use saturation point But, we can re-write in a more convenient form. withand Evaluate h v1, h v2 & h v3 at steam tables Evaluate h a1 & h a3 using Table A-22 or Evaluate moist air specific enthalpy using the pyschrometric chart

11 11 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 o C. T 1 =22°C, T wb =9°C Saturated vapor: T 3 =110 o C Moist air T 2 =? ω 2 =? 12 3

12 12 Example (12.67): a) the humidity ratio b) the temperature in o C. Mass Balance Equations: 12 3 for air: for H 2 0: with Energy Balance Equations:

13 13 Example (12.67): a) the humidity ratio b) the temperature in o C. Use the psychrometric chart to find properties. 12 3 State 1: State 3: Saturated vapor at 110 o C Use Steam table, A2:

14 14 Example (12.67): a) the humidity ratio b) the temperature in o C. Mass Balance Equations: 12 3 From the energy balance: and therefore:

15 15 Example (12.67): a) the humidity ratio b) the temperature in o C. therefore at State 2: 12 3 Using the pyschrometric chart: 24 o C At the intersection of the humidity ratio and the specific enthalpy of moist air, the dry bulb temperature can be directly looked up: 24 o C.

16 End of slides for lecture 40 16


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