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EGR 334 Thermodynamics Chapter 12: Sections 5-7 Lecture 39: Humidity and Psychrometric Applications Quiz Today?

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Presentation on theme: "EGR 334 Thermodynamics Chapter 12: Sections 5-7 Lecture 39: Humidity and Psychrometric Applications Quiz Today?"— Presentation transcript:

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

2 Todays 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: Homework Assignment: Read Chapter 13, Sections 1-5 Problems from Chap 12: 46,51, 55, 67

3 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 Absolute humidity % humidity Wet bulb temperature Relative humidity Dew Point Temperature Humidity Ratio Mixture Enthalpy Terms to understand:

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 p a and p v of dry air and water vapor are, respectively where y a and y v are the mole fractions of the dry air and water vapor. and Humidity Ratio: 4. Humidity Ratio is the ratio of the mass of the vapor to the mass of the dry air.

5 Mixture pressure, p, T for Moist Air 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 p v and the mixture temperature T. The water vapor is superheated at this state. Typical state of the water vapor in moist air 7. When p v corresponds to p g at temperature T, the mixture is said to be saturated. 8. The ratio of p v and p g is called the relative humidity, : Relative humidity:

6 6 Sec 12.5 : Psychrometric applications Humidity relates to temperature

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

8 8 Sec : 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 Typical state of the water vapor in moist air Summarize: Relative humidity: Relative Humidity: Dew point Temperature at which p v =p g : Humidity Ratio:

10 10 Sec : 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 11 Sec : Evaluating U, H, and S To evaluate enthalpy: For air: a) use Table A-22 or b) use Δh=c p_air (Δ T) Note: h g and s g are taken from the steam table. For vapor: use h v h g (T ) notice how h constant for low pressure superheated vapor on Molier diagram. To evaluate entropy: Here is the relative humidity

12 12 Example (12.47): A lecture hall having a volume of 10 6 ft 3 contains air at 80 o F, 1 atm, and a humidity ratio of 0.01 lb m of water per lb m 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 13 Example (12.47): A lecture hall having a volume of 10 6 ft 3 contains air at 80 o F, 1 atm, and a humidity ratio of 0.01 lb m of water per lb m 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 p g using Table A2E

14 14 Example (12.47): A lecture hall having a volume of 10 6 ft 3 contains air at 80 o F, 1 atm, and a humidity ratio of 0.01 lb m of water per lb m 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: From Table A3E: at p v =p g To find the mass, use the ideal gas law: p V (T) p g (T) T dp

15 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 kg air /kg wet_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 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 kg air /kg wet_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. Dry air 90°C,1 atm Wet grain, 20°C 40% moisture Dry grain ?% moisture Wet air 38°C,1 atm Mass balance on water: Mass balance on air:Mass balance of grain:

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

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

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

20 Procedure for analysis of air conditioning systems: 20 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. 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 21 Example (12.67): Moist air at 90°F, 1 atm, 60% relative humidity and a volumetric flow rate of 2000 ft 3 /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 Moist Air at 90°F and 1 atm RH = 60%, (AV) 1 =2000 ft 3 /min Condensate at 54°F Q Saturated air at 54°F (a)The rate of heat transfer, in kW (b) The rate of enthalpy change, in KW/K.

22 22 Example (12.67): Set property states: State 1: T 1 =90 °F, p 1 = 1 atm, RH = 60%, (AV) 1 =2000 ft 3 /min Q State 2: saturated air at T 2 = 54 °F State 3: condensate at T 3 =54 °F refers to saturated fluid at 54 deg. F. Using from Table A2E: p g at 90 F = psi then for saturated air: p v = p g From Table A2E: p g at 54 F = psi = atm

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

24 24 Example (12.67): Determine (a)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 Q For air in: (ideal gas) For water vapor in: Air Mass Balance: For water/vapor out:

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

26 end of slides for lecture 39 26


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