مخلوط هوا – بخار و تهویه مطبوع

مخلوط هوا – بخار و تهویه مطبوع

اسلاید 1: Chapter 14 مخلوط هوا - بخار و تهویه مطبوع Thermodynamics: An Engineering Approach, 7th edition by Yunus A. Çengel and Michael A. Boles

اسلاید 2: 2 در بحث مخلوط گازها در فصل گذشته با هیچ گونه فرایند تبدیل فاز روبرو نبودیم. در این فصل مخلوط هوای خشک و بخار آب مورد بررسی قرار می گیرد که ممکن است با چگالش بخار و تبدیل آن به مایع روبرو باشیم. به مخلوط هوا و بخار آب، هوای اتمسفری می گویند. تغییرات دمای هوای اتمسفری در مسایل تهویه مطبوع معمولا از -10oC تا 50oC است. در این شرایط هوا را می توان گاز آرمانی با گرمای ویژه ثابت در نظر گرفت. (Cpa=1.005 kJ/kgK) در این صورت انتالپی هوای خشک با فرض مقدار انتالپی صفر در دمای صفر درجه سلسیوس به عنوان نقطه مرجع چنین خواهد بود:

اسلاید 3: 3Note: For the dry air-water vapor mixture, the partial pressure of the water vapor in the mixture is less that its saturation pressure at the temperature. فشار اشباع متناظر با دمای 50oC برابر با 12.3 kPaمی باشد. در این محدوده فشار، همان طور که از شکل روبرو پیداست، در دمای ثابت، انتالپی ثابت می ماند. بنابراین، انتالپی بخار آب را می توان معادل انتالپی بخار اشباع در دمای مخلوط هوا و بخار گرفت. انتالپی بخار اشباع بنابراین به صورت تابعی از دما به شکل زیر قابل بیان خواهد بود.

اسلاید 4: 4Consider increasing the total pressure of an air-water vapor mixture while the temperature of the mixture is held constant. See if you can sketch the process on the P-v diagram relative to the saturation lines for the water alone given below. Assume that the water vapor is initially superheated.PvWhen the mixture pressure is increased while keeping the mixture temperature constant, the vapor partial pressure increases up to the vapor saturation pressure at the mixture temperature and condensation begins. Therefore, the partial pressure of the water vapor can never be greater than its saturation pressure corresponding to the temperature of the mixture.

اسلاید 5: 5DefinitionsDew Point, TdpThe dew point is the temperature at which vapor condenses or solidifies when cooled at constant pressure. Consider cooling an air-water vapor mixture while the mixture total pressure is held constant. When the mixture is cooled to a temperature equal to the saturation temperature for the water-vapor partial pressure, condensation begins.When an atmospheric air-vapor mixture is cooled at constant pressure such that the partial pressure of the water vapor is 1.491 kPa, then the dew point temperature of that mixture is 12.95oC.

اسلاید 6: 6Relative Humidity, ϕ Pv and Pg are shown on the following T-s diagram for the water-vapor alone.Since

اسلاید 7: 7Absolute humidity or specific humidity (sometimes called humidity ratio),  Using the definition of the specific humidity, the relative humidity may be expressed as Volume of mixture per mass of dry air, vAfter several steps, we can show (you should try this)

اسلاید 8: 8So the volume of the mixture per unit mass of dry air is the specific volume of the dry air calculated at the mixture temperature and the partial pressure of the dry air.Mass of mixtureMass flow rate of dry air, Based on the volume flow rate of mixture at a given state, the mass flow rate of dry air isEnthalpy of mixture per mass dry air, h

اسلاید 9: 9Example 14-1 Atmospheric air at 30oC, 100 kPa, has a dew point of 21.3oC. Find the relative humidity, humidity ratio, and h of the mixture per mass of dry air.

اسلاید 10: 10

اسلاید 11: 11Example 14-2If the atmospheric air in the last example is conditioned to 20oC, 40 percent relative humidity, what mass of water is added or removed per unit mass of dry air?At 20oC, Pg = 2.339 kPa.The change in mass of water per mass of dry air is

اسلاید 12: 12Or, as the mixture changes from state 1 to state 2, 0.01038 kg of water vapor is condensed for each kg of dry air. Example 14-3 Atmospheric air is at 25oC, 0.1 MPa, 50 percent relative humidity. If the mixture is cooled at constant pressure to 10oC, find the amount of water removed per mass of dry air.Sketch the water-vapor states relative to the saturation lines on the following T-s diagram. TsAt 25oC, Psat = 3.170 kPa, and with = 50%

اسلاید 13: 13Therefore, when the mixture gets cooled to T2 = 10oC < Tdp,1, the mixture is saturated, and = 100%. Then Pv,2 = Pg,2 = 1.228 kPa. The change in mass of water per mass of dry air is

اسلاید 14: 14Or as the mixture changes from state 1 to state 2, 0.00228 kg of water vapor is condensed for each kg of dry air.Steady-Flow Analysis Applied to Gas-Vapor MixturesWe will review the conservation of mass and conservation of energy principles as they apply to gas-vapor mixtures in the following example.Example 14-3 Given the inlet and exit conditions to an air conditioner shown below. What is the heat transfer to be removed per kg dry air flowing through the device? If the volume flow rate of the inlet atmospheric air is 17 m3/min, determine the required rate of heat transfer.Cooling fluidInOutAtmospheric air T1 = 30oC P1 =100 kPa 1 = 80% 1 = 17m3/minCondensateat 20oCInsulated flow ductT2 = 20oCP2 = 98 kPa2 = 95%

اسلاید 15: 15Before we apply the steady-flow conservation of mass and energy, we need to decide if any water is condensed in the process. Is the mixture cooled below the dew point for state 1?So for T2 = 20oC < Tdp, 1, some water-vapor will condense. Lets assume that the condensed water leaves the air conditioner at 20oC. Some say the water leaves at the average of 26 and 20oC; however, 20oC is adequate for our use here.Apply the conservation of energy to the steady-flow control volumeNeglecting the kinetic and potential energies and noting that the work is zero, we getConservation of mass for the steady-flow control volume is

اسلاید 16: 16For the dry air: For the water vapor:The mass of water that is condensed and leaves the control volume isDivide the conservation of energy equation by , then

اسلاید 17: 17Now to find the s and hs.

اسلاید 18: 18Using the steam tables, the hs for the water areThe required heat transfer per unit mass of dry air becomes

اسلاید 19: 19The heat transfer from the atmospheric air is The mass flow rate of dry air is given by

اسلاید 20: 20The Adiabatic Saturation Process Air having a relative humidity less than 100 percent flows over water contained in a well-insulated duct. Since the air has < 100 percent, some of the water will evaporate and the temperature of the air-vapor mixture will decrease.

اسلاید 21: 21If the mixture leaving the duct is saturated and if the process is adiabatic, the temperature of the mixture on leaving the device is known as the adiabatic saturation temperature. For this to be a steady-flow process, makeup water at the adiabatic saturation temperature is added at the same rate at which water is evaporated.We assume that the total pressure is constant during the process.Apply the conservation of energy to the steady-flow control volume Neglecting the kinetic and potential energies and noting that the heat transfer and work are zero, we getConservation of mass for the steady-flow control volume is

اسلاید 22: 22For the dry air:For the water vapor:The mass flow rate water that must be supplied to maintain steady-flow is,Divide the conservation of energy equation by , thenWhat are the knowns and unknowns in this equation?

اسلاید 23: 23Since 1 is also defined byWe can solve for Pv1. Then, the relative humidity at state 1 isSolving for 1

اسلاید 24: 24Example 14-4 For the adiabatic saturation process shown below, determine the relative humidity, humidity ratio (specific humidity), and enthalpy of the atmospheric air per mass of dry air at state 1.

اسلاید 25: 25Using the steam tables:From the above analysis

اسلاید 26: 26We can solve for Pv1. Then the relative humidity at state 1 isThe enthalpy of the mixture at state 1 is

اسلاید 27: 27Wet-Bulb and Dry-Bulb TemperaturesIn normal practice, the state of atmospheric air is specified by determining the wet-bulb and dry-bulb temperatures. These temperatures are measured by using a device called a psychrometer. The psychrometer is composed of two thermometers mounted on a sling. One thermometer is fitted with a wet gauze and reads the wet-bulb temperature. The other thermometer reads the dry-bulb, or ordinary, temperature. As the psychrometer is slung through the air, water vaporizes from the wet gauze, resulting in a lower temperature to be registered by the thermometer. The dryer the atmospheric air, the lower the wet-bulb temperature will be. When the relative humidity of the air is near 100 percent, there will be little difference between the wet-bulb and dry-bulb temperatures. The wet-bulb temperature is approximately equal to the adiabatic saturation temperature. The wet-bulb and dry-bulb temperatures and the atmospheric pressure uniquely determine the state of the atmospheric air.

اسلاید 28: 28The Psychrometric ChartFor a given, fixed, total air-vapor pressure, the properties of the mixture are given in graphical form on a psychrometric chart. The air-conditioning processes:

اسلاید 29: 29

اسلاید 30: 30Example 14-5 Determine the relative humidity, humidity ratio (specific humidity), enthalpy of the atmospheric air per mass of dry air, and the specific volume of the mixture per mass of dry air at a state where the dry-bulb temperature is 24oC, the wet-bulb temperature is 16oC, and atmospheric pressure is 100 kPa. From the psychrometric chart readNOTE: THE ENTHALPY READ FROM THE PSYCHROMETRIC CHART IS THE TOTAL ENTHALPY OF THE AIR-VAPOR MIXTURE PER UNIT MASS OF DRY AIR. h= H/ma = ha + ωhv

اسلاید 31: 31Example 14-6 For the air-conditioning system shown below in which atmospheric air is first heated and then humidified with a steam spray, determine the required heat transfer rate in the heating section and the required steam temperature in the humidification section when the steam pressure is 1 MPa.

اسلاید 32: 32The psychrometric diagram isApply conservation of mass and conservation of energy for steady-flow to process 1-2.Conservation of mass for the steady-flow control volume is

اسلاید 33: 33For the dry airFor the water vapor (note: no water is added or condensed during simple heating)Thus,Neglecting the kinetic and potential energies and noting that the work is zero, and letting the enthalpy of the mixture per unit mass of air h be defined as we obtain

اسلاید 34: 34Now to find the and hs using the psychrometric chart.At T1 = 5oC, 1 = 90%, and T2 = 24oC:The mass flow rate of dry air is given by

اسلاید 35: 35The required heat transfer rate for the heating section isThis is the required heat transfer to the atmospheric air. List some ways in which this amount of heat can be supplied.At the exit, state 3, T3 = 25oC and 3 = 45%. The psychrometric chart gives

اسلاید 36: 36Apply conservation of mass and conservation of energy to process 2-3.Conservation of mass for the steady-flow control volume isFor the dry airFor the water vapor (note: water is added during the humidification process)

اسلاید 37: 37Neglecting the kinetic and potential energies and noting that the heat transfer and work are zero, the conservation of energy yieldsSolving for the enthalpy of the steam,

اسلاید 38: 38At Ps = 1 MPa and hs = 2750 kJ/kgv, Ts = 179.88oC and the quality xs = 0.985.See the text for applications involving cooling with dehumidification, evaporative cooling, adiabatic mixing of airstreams, and wet cooling towers.