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EGR 334 Thermodynamics Chapter 12: Sections 5-7

EGR 334 Thermodynamics Chapter 12: Sections 5-7. Lecture 39: Humidity and Psychrometric Applications. Quiz Today?. Today’s main concepts:. Demonstrate understanding of psychrometric terminology, including humidity ratio, relative humidity, mixture enthalpy, and dew point temperature.

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EGR 334 Thermodynamics Chapter 12: Sections 5-7

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  1. EGR 334 ThermodynamicsChapter 12: Sections 5-7 Lecture 39: Humidity and Psychrometric Applications Quiz Today?

  2. Today’s main concepts: • Demonstrate understanding of psychrometric terminology, including humidity ratio, relative humidity, mixture enthalpy, and dew point temperature. • Apply mass, energy, and entropy balances to analyze air-conditioning processes. Reading Assignment: Read Chapter 13, Sections 1-5 Homework Assignment: Problems from Chap 12: 46,51, 55, 67

  3. Sec 12.5 : Psychrometric applications Greek: psuchra = cold Metron = measure Psychrometric: Study of systems containing “dry air” and water vapor May also include condensed water. Humidity: a measure of the amount of water in the air or “moist air” Terms to understand: • Absolute humidity • % humidity • Wet bulb temperature • Relative humidity • Dew Point Temperature • Humidity Ratio • Mixture Enthalpy

  4. for Moist Air 1. The overall mixture and each component, dry air and water vapor, obey the ideal gas equation of state. 2. Dry air and water vapor within the mixture are considered as if they each exist alone in volume V at the mixture temperature T while each exerts part of the mixture pressure. 3.The partial pressures pa and pv of dry air and water vapor are, respectively and whereyaand yvare the mole fractions of the dry air and water vapor. 4.Humidity Ratio is the ratio of the mass of the vapor to the mass of the “dry air”. Humidity Ratio:

  5. for Moist Air Mixture pressure, p , T 5.The mixture pressure is the sum of the partial pressures of the dry air and the water vapor: 6. A typical state of water vapor in moist air is fixed using partial pressure pv and the mixture temperature T. The water vapor is superheated at this state. Typical state of the water vapor in moist air 7.When pv corresponds to pg at temperature T, the mixture is said to be saturated. Relative humidity: 8.The ratio of pv and pg is called the relative humidity, f:

  6. Sec 12.5 : Psychrometric applications Humidity relates to temperature

  7. Sec 12.5 : Psychrometric applications Humidity can be measured using a hygrometer: Hair/Fiber hygrometer Paper-disk hygrometer Whirling hygrometer (sling psychrometer) Capacitive hygrometer

  8. Sec 12.5.4 : Evaluating the Dew Point Temperature Humidity can be measured using a “wet bulb” The temperature of the “wet bulb” is lower than the dry thermometer. The evaporation of water is an endothermic process (requires heat) which is obtained from the environment (bulk water). The “wet bulb” temperature = dew point temperature. This is the lowest temperature that can hold the current water vapor content of the air.

  9. Mixture pressure, p , T Summarize: Relative Humidity: Temperature at which pv=pg: Typical state of the water vapor in moist air Dew point Humidity Ratio: Relative humidity:

  10. Sec 12.5.3 : Evaluating U, H, and S For mixtures, recall the internal energy, enthalpy, and entropy are equal to the parts added together Using the definition of the humidity ratio, : and

  11. Sec 12.5.3 : Evaluating U, H, and S To evaluate enthalpy: For air: a) use Table A-22 • or b) use Δh=cp_air (Δ T) For vapor: use hv hg(T) notice how h ≈ constant for low pressure superheated vapor on Molier diagram. To evaluate entropy: Here is the relative humidity Note: hg and sg are taken from the steam table.

  12. Example (12.47): A lecture hall having a volume of 106 ft3 contains air at 80 oF, 1 atm, and a humidity ratio of 0.01 lbm of water per lbm of dry air. Determine a) relative humidity b) the dew point temperature in degrees F. c) the mass of water vapor contained in the room

  13. Example (12.47): A lecture hall having a volume of 106 ft3 contains air at 80 oF, 1 atm, and a humidity ratio of 0.01 lbm of water per lbm of dry air. Determine a) relative humidity b) the dew point temperature in degrees F. c) the mass of water vapor contained in the room  At T = 80 F: Look up pg using Table A2E

  14. Example (12.47): A lecture hall having a volume of 106 ft3 contains air at 80 oF, 1 atm, and a humidity ratio of 0.01 lbm of water per lbm of dry air. Determine a) relative humidity b) the dew point temperature in degrees F. c) the mass of water vapor contained in the room To find the Dew Point: pg(T) pV(T) From Table A3E: at pv=pg Tdp To find the mass, use the ideal gas law:

  15. Example (12.54): Wet grain at 20°C containing 40% moisture by mass enters a dryer operating at steady state. Dry air enters the dryer at 90°C, 1 atm at a rate of 15 kgair/kgwet_grain entering. Moist air exits the dryer at 38°C, 1 atm, 53% relative humidity. For the grain exiting the dryer, determine the percent moisture by mass.

  16. Example (12.54): Wet grain at 20°C containing 40% moisture by mass enters a dryer operating at steady state. Dry air enters the dryer at 90°C, 1 atm at a rate of 15 kgair/kgwet_grain entering. Moist air exits the dryer at 38°C, 1 atm, 53% relative humidity. For the grain exiting the dryer, determine the percent moisture by mass. Wet air 38°C,1 atm Dry air 90°C,1 atm 4 3 1 2 Dry grain ?% moisture Wet grain, 20°C • 40% moisture • Mass balance on water: • Mass balance of grain: • Mass balance on air:

  17. Example (12.54): • Mass balance on water: • 40% of wet grain is water: Dry air 90°C,1 atm • 15kgair/kggrain Wet air 38°C,1 atm 4 3 • mass balance of air: • 15 kgair / 1 kg wet grain comes in: 1 2 Wet grain 20°C • 40% moisture Dry grain • ?% moisture • Relative Humidity of State 3: • (from Table A2 at T=38C, pg=0.06632 bar)

  18. Example (12.54): • Humidity ratio: Dry air 90°C,1 atm • 15kgair/kggrain Wet air 38°C,1 atm 4 3 1 2 Wet grain 20°C • 40% moisture Dry grain • ?% moisture • Unknowns so far: (10 unknowns) • Equations so far: (9 equations)

  19. Example (12.54): • But wait: • Recall the answer is asked for as • % moisture per grain out Dry air 90°C,1 atm • 15kgair/kggrain Wet air 38°C,1 atm 4 3 • which means that can be set as our basis or let =1 kg/s 1 2 • Solving using IT for the other mass flow rates: Wet grain 20°C • 40% moisture Dry grain • ?% moisture • Therefore:

  20. 1) Identify State Properties of as many individual mixture components Procedure for analysis of air conditioning systems: 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. 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

  21. Example (12.67): Moist air at 90°F, 1 atm, 60% relative humidity and a volumetric flow rate of 2000 ft3/min enters a control volume at steady state and flows along a surface maintained at 40°F through which heat transfer occurs. Saturated moist air and condensate, each at 54°F, exit the control volume For the control volume W = KE = PE = 0. • Determine • The rate of heat transfer, in kW • (b) The rate of enthalpy change, in KW/K. Q • Moist Air at 90°F and 1 atm RH = 60%, (AV)1=2000 ft3/min Saturated air at 54°F 1 2 3 Condensate at 54°F

  22. Example (12.67): Set property states: State 1: T1=90 °F, p1 = 1 atm , RH = 60%, (AV)1=2000 ft3/min Using from Table A2E: pg at 90 F = 0.6988 psi then  Q State 2: saturated air at T2= 54 °F for saturated air: pv = pg From Table A2E: pg at 54 F = 0.2064 psi = 0.01404 atm 1 2 3  State 3: condensate at T3=54 °F refers to saturated fluid at 54 deg. F.

  23. Example (12.67): Set property states: State 1: T1=90 °F, p1 = 1 atm , RH = 60%, (AV)1=2000 ft3/min air: water: (from Table A22E) (from Table A2E) State 2: air: water: Q 1 2 (from Table A22E) (from Table A2E) 3 State 3: (from Table A2E)

  24. Example (12.67): Determine • The rate of heat transfer, in kW • (b) The rate of enthalpy change, in KW/K. Use ideal gas law to establish mass flow rate For air in: (ideal gas) For water vapor in: Q 1 2 Air Mass Balance:  3 For water/vapor out:

  25. Example (12.67): Determine • The rate of heat transfer, in kW • (b) The rate of enthalpy change, in KW/K. Set up Mass balances: For air: For water/vapor: Set up Energy balance: Q 1 2 3

  26. end of slides for lecture 39

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