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**EGR 334 Thermodynamics Chapter 12: Sections 5-7**

Lecture 39: Humidity and Psychrometric Applications Quiz Today?

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**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

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**Psychrometric: Study of systems containing “dry air” and water vapor**

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

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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 where ya and yv are 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:

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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 pv and the mixture temperature T. The water vapor is superheated at this state. Mixture pressure, p , T 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:

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**Humidity relates to temperature**

Sec 12.5 : Psychrometric applications Humidity relates to temperature

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**Humidity can be measured using a hygrometer:**

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

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**Humidity can be measured using a “wet bulb”**

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.

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**Temperature at which pv=pg:**

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

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**Using the definition of the humidity ratio, :**

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

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**or b) use Δh=cp_air (Δ T)**

Sec : 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.

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**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

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**b) the dew point temperature in degrees F. **

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

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**b) the dew point temperature in degrees F. **

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: pV(T) pg(T) Tdp From Table A3E: at pv=pg To find the mass, use the ideal gas law:

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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.

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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:

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**15 kgair / 1 kg wet grain comes in:**

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

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**Unknowns so far: (10 unknowns)**

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

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**which means that can be set as our basis or let =1 kg/s**

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

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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. 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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**The rate of heat transfer, in kW **

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. Moist Air at 90°F and 1 atm RH = 60%, (AV)1=2000 ft3/min Q Saturated air at 54°F 1 2 3 Condensate at 54°F

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**Example (12.67): Set property states:**

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

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**Example (12.67): Set property states:**

Q 1 2 3 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: (from Table A22E) (from Table A2E) State 3: (from Table A2E)

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**The rate of heat transfer, in kW **

Example (12.67): Determine Q 1 2 3 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: Air Mass Balance: For water/vapor out:

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**The rate of heat transfer, in kW **

Example (12.67): Determine Q 1 2 3 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:

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**end of slides for lecture 39**

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