Air Conditioning Engineering Lecture 1 PDF

University of Omar Al-Mukhtar Engineering College Mechanical Engineering Department Air Conditioning Engineering Presented by Lecturer Fadeel Al-Areeda 1 𝑠𝑡 Lecture Outline  Introduction  Ideal Gases Mixture Model Dalton Model Amaga...

University of Omar Al-Mukhtar Engineering College Mechanical Engineering Department Air Conditioning Engineering Presented by Lecturer Fadeel Al-Areeda 1 𝑠𝑡 Lecture Outline  Introduction  Ideal Gases Mixture Model Dalton Model Amagat Model  Psychrometry of Atmospheric Air -Properties of Psychrometry Dry-bulb Temperature Wet-bulb Temperature Relative Humidity Humidity Ratio / Specific Humidity Absolute Humidity Degree of Saturation Dew point Temperature Specific Volume Enthalpy Pressure - Example.1 - Example.2 - Example.3 - Home Work - Question.1 - Question.2 - Question.3 - Question.4 7/23/2024 2 Introduction: Refrigeration and air conditioning have been a necessity for living. Nowadays, vaccines, medicines, and meat…etc, can’t be kept preserved except in air-conditioned or refrigerated spaces. Also, closed work areas want to be well ventilated and air conditioned in order to increasing productivity, for example. Air conditioning and refrigeration, all in all, are important for: preserving food, human comfort, keeping industrial equipment in a functional situation such as in some processes in which machines or computers deal with huge amount of data. Humans had found their ways to preserving food before inventing modern refrigerators. Ages ago, In summer, they used to keep their food deep inside water wells or inside caves. They also used to use ice blocks after transferring them from high cold mountains’ tops to lower wormer areas for preserving food. Ancient Egyptians had their strategy in which they kept their food in pottery jars filled of water, so water absorbs heat from the content of the jars and evaporates, as shown in the figure on the right. 7/23/2024 3  Humans found out that when storing food at temperature lower than 10𝐶 ° and higher than 0𝐶 ° , food is preserved and protected of both microbial growth and freezing. For fish and other types of meats need to be kept at temperature lower than 0𝐶 ° ; this temperature range is called ‘food safety zone’. By the end of the first quarter of the 19𝑡ℎ century, a machine of ice production was invented. By the beginning of the 20𝑡ℎ century, exactly in 1912, J. M. Larsen invented his first mechanical refrigeration machine, as shown in the figure on the right. In terms of ventilation and air conditioning, humans have shape their living conditions by using fire for heating in winter, and by digging holes as if they were windows in caves’ walls for ventilation in summer. Romans, for example, used to heat their baths using enclosed cannels that carry hot water. Since the 19𝑡ℎ century the refrigeration and air conditioning have been witnessing accelerating development, starting with ventilation fans, boilers…etc up to what we use of cutting-edge refrigeration and air conditioning equipment. 7/23/2024 4 Modern air conditioning systems consist of devices that receive air, and then handle it to be suitable for applications, then distributing and delivering it to areas that wanted to be air-conditioned. Air play a role within air conditioning systems. In general, air is consisted of Nitrogen 78% Oxygen 21% Pollutants and Other gases 1% (such as Argon, water vapor, and some other gases). The concentration of water vapor and pollutants decreases with altitude. At 10 kilometers or higher, atmospheric air consists of only dry air. 7/23/2024 5 NOTE! Air is described as ‘dry air’ if there is no water content in it. Air is described as ‘ wet air’ if there is water content in it. Air is described as ‘ saturated air’ if the moisture in it is at its maximum. Within an air conditioning system, air is treated as if it’s a mixture of ideal gases that consists, basically, of dry air and water vapor. Atmospheric Air  Molecular weight/mass of air, 𝑴𝒂 = 28.966 kg/kmol  Gas constant for air, 𝑹𝒂 = 287.035 J/kg.K  Molecular weight/mass of water vapor, 𝑴𝒗 = 18.015 kg/kmol  Gas constant for water vapor 𝑹𝒗 = 461.52 J/kg.K 7/23/2024 6  Ideal Gases Mixtures: Suppose that a container, as shown in the figure, contains moist air at pressure P, temperature T, and Volume V. Moles number of air is 𝑁𝑎 , and moles number of vapor is 𝑁𝑣 ; therefore, the total number of moles of the moist atmospheric air inside the container is 𝑁 = 𝑁𝑎 + 𝑁𝑣 (1) By dividing the eq,(1) on the total number of moles 𝑁𝑎 𝑁𝑣 1= + (2) 𝑁 𝑁 This equation can be expressed as 1 = 𝑦𝑎 + 𝑦𝑣 (3) Where 𝑦𝑎 : is the partial fraction of dry air 𝑦𝑣 : is the partial fraction of water/ vapor 7/23/2024 7  Dalton and Amagat Models: When dealing with atmospheric air as an ideal gas, it’s important to know that there are two separate models that have been sat by the scientists Dalton, and Amagat to understand and explain ideal gases’ behavior. - Dalton Model: Dalton Assumptions: 1- the properties of each gas within the mixture are not affected by other gases of the mixture. 2- the temperature and volume of each gas are equal to the temperature and volume of the whole mixture. From the ideal law of gases 𝑃 𝑉 = 𝑁𝑅𝑇 (4) 𝑃𝑉 ∴𝑁= (5) 𝑅𝑇 Where R is the constant of ideal gases (R=8.3144kJ/Kmol.K) 7/23/2024 8 From equation (1), and by considering water vapor as an ideal gas too, the equation (4) for both air and vapor becomes 𝑃𝑎 𝑉 = 𝑁𝑎 𝑅𝑇 (6) 𝑃𝑣 𝑉 = 𝑁𝑣 𝑅𝑇 (7) So equation (5) can be rewritten for both air and water vapor to be 𝑃𝑎 𝑉 𝑁𝑎 = (8) 𝑅𝑇 𝑃𝑣 𝑉 𝑁𝑣 = (9) 𝑅𝑇 From equations (5), (8), (9) into equation (1) 𝑃𝑉 𝑃𝑎 𝑉 𝑃𝑣 𝑉 = + (10) 𝑅𝑇 𝑅𝑇 𝑅𝑇 ∴ 𝑃 = 𝑃𝑎 + 𝑃𝑣 (11) This leads to that the total pressure of a mixture of ideal gases equals the algebraic sum of its consists’ partial pressures. 7/23/2024 9 By dividing equation (8) on equation (5) the partial fraction of air can be a function of both the partial pressure of air and the total pressure. 𝑁𝑎 𝑃𝑎 = = 𝑦𝑎 (12) 𝑁 𝑃 Now, by dividing equation (9) on equation (5) the partial fraction of water vapor can be a function of both the partial pressure of water vapor and the total pressure. 𝑁𝑣 𝑃𝑣 = = 𝑦𝑣 (13) 𝑁 𝑃 -Amagat Model: Amagat Assumptions: Unlike Dalton, Amagat assumed that the temperature and pressure of each gas are equal to the temperature and pressure of the whole mixture; this means that the volume of each gas is lower than the volume of the whole mixture. From equation (1), and by considering water vapor as an ideal gas too, the equation (4) for both air and vapor becomes 𝑃 𝑉𝑎 = 𝑁𝑎 𝑅𝑇 (14) 𝑃𝑉𝑣 = 𝑁𝑣 𝑅𝑇 7/23/2024 (15) 10  𝑃𝑉𝑎 ∴ 𝑁𝑎 = (16) 𝑅𝑇 𝑃𝑉 𝑁𝑣 = 𝑣 (17) 𝑅𝑇 From equations (5), (16), (17) into equation (1) 𝑃𝑉 𝑃𝑉𝑎 𝑃𝑉𝑣 = + (18) 𝑅𝑇 𝑅𝑇 𝑅𝑇 ∴ 𝑉 = 𝑉𝑎 + 𝑉𝑣 This leads to that the total volume of a mixture of ideal gases equals the algebraic sum of its consists’ partial volumes. By dividing equation (16) on equation (5) the partial fraction of air can be a function of both the partial volume of air and the total volume. 𝑁𝑎 𝑉 = 𝑎 = 𝑦𝑎 (19) 𝑁 𝑉 In the same way, by dividing equation (17) on equation (5) the partial fraction of water vapor can be a function of both the partial volume of water vapor and the total volume. 𝑁𝑣 𝑉 = 𝑣 = 𝑦𝑣 (20) 𝑁 𝑉 Note!: According to both Dalton’s model and Amagat’s model, For any gas , 𝑖, in any ideal gas mixture 𝑁𝑖 𝑉 𝑃 = 𝑖 = 𝑖 = 𝑦𝑖 (21) 𝑁 𝑉 𝑃 7/23/2024 11 Psychrometry of Atmospheric Air: Psychrometry is the science that deal with the physical laws of air-water vapor mixtures. When designing an air conditioning system, the temperature and moisture content of the air, that is wanted to be conditioned, should be controlled. In other words, we can say that Psychrometry is the study of moist air or mixture of dry air and water vapor. - Properties of Psychrometry: Dry-bulb Temperature (DBT), 𝑻𝒅𝒃 : Dry-bulb temperature is the temperature that is indicated by a thermometer exposed to the air in a place sheltered from direct solar radiation. The term ‘dry-bulb’ is customarily added to temperature to distinguish it from wet-bulb or dew point temperature. Wet-bulb Temperature (WBT), 𝑻𝒘𝒃 : Wet bulb temperature is the temperature recorded by a thermometer when its bulb is enveloped by a cotton wick saturated with water. Wet Bulb temperature is measured by passing air over a wet bulb tip. Dry and wet bulb temperature are measured by using a kit that includes two thermometers. It is called ‘psychrometer’ or ‘hygrometer’. 7/23/2024 12 As mentioned earlier the moisture of atmospheric air can be determined by measuring both dry and wet temperature that can be read on thermometers when passing a stream of moist air on both wet and dry thermometers. There is a variety of psychrometers. The psychrometer at the lower right corner is provided with a fan that blows a stream of air that passes over both wet and dry thermometers. If the air is not moist, some of the water that socked by the cotton wick evaporates, causing a decrease of temperature; this happens due to the transformation of heat from the surrounding air and thermometer to the water. There is another type of psychrometers that called a ‘sling’ or ‘ rotary’ psycrometer since it has a handle that can be rotated by so air is forced to pass over both dry and wet thermometers. Sling psychrometer is one of those types that are common in almost all refrigeration and air-conditioning application for its easy usage, as shown in the figure at the lower left corner. 7/23/2024 13  Relative Humidity (RH), ϕ: Is a ratio of the mole fraction of water vapor in moist air to the mole fraction of water vapor in saturated air at the same temperature. 𝑦𝑣 ϕ = (22) 𝑦𝑠 - Relative humidity is therefore defined as the ratio of vapor pressure in a sample of air to vapor pressure of saturated air at the same temperature. 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑢𝑟 𝑃𝑣 ϕ= 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑢𝑟 = (23) 𝑃𝑠 Relative humidity is measured in percentage. It has a great influence on evaporation of water in the air and therefore on the comfort of human beings. Note!: Relative humidity equals zero when moist air is totally dry, i.e. when it does not contain water vapor. 7/23/2024 14  Humidity Ratio / Specific Humidity (SH)/ Moisture Content, ω: Specific Humidity is the ratio of the mass of water in the atmospheric moist air (𝑚𝑣 ) to the mass of dry air (𝑚𝑎 ) 𝑃𝑣 𝑉 𝑚𝑣 ൗ𝑅𝑣 𝑇 𝑃𝑣 𝑅𝑎 𝑃𝑣 0.287 𝜔= = 𝑃𝑎 𝑉 = × = × (24) 𝑚𝑎 ൗ𝑅𝑎 𝑇 𝑃𝑎 𝑅𝑣 𝑃𝑎 0.46152 𝑃𝑣 𝑃𝑣 ∴ 𝜔 = 0.622 = 0.622 (25) 𝑃𝑎 𝑃−𝑃𝑣 since 𝑝𝑎𝑟𝑡𝑖𝑎𝑙 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑢𝑟 𝑃𝑣 ϕ= 𝑠𝑎𝑡𝑢𝑟𝑎𝑡𝑖𝑜𝑛 𝑝𝑟𝑒𝑠𝑠𝑢𝑟𝑒 𝑜𝑓 𝑤𝑎𝑡𝑒𝑟 𝑣𝑎𝑝𝑜𝑢𝑟 = 𝑃𝑠 and 𝑃𝑣 =𝑃𝑠 × ϕ 𝜔×𝑃𝑎 ∴ ϕ= (26) 0.622𝑃𝑠 7/23/2024 15  Absolute Humidity : It is the mass of water vapor present in one cubic meter of dry air. It is expressed in terms of grains per cubic meter of dry air (𝑔ൗ𝑚3 of dry air). 1 kg of water vapor is equal to 15,430 grains. 𝑃𝑣 𝑉 = 𝑚𝑣 𝑅𝑣 𝑇 (27) ∴ Vapor’s density equals 𝜌 = 𝑚𝑣 /𝑉 = 𝑃𝑣 /𝑅𝑣 𝑇 (28) Degree of Saturation, µ: It is the mass of water vapor in a sample of air to the mass of water vapor in the same sample when it is saturated at the same temperature. 𝜔 𝜇= (29) 𝜔𝑠 where 𝜔 and 𝜔𝑠 are specific humidity of air and saturated air, respectively 𝑃𝑣 0.622 𝑃𝑣 𝑃−𝑃𝑠 𝑃−𝑃𝑣 ∴𝜇= 𝑃𝑠 = (30) 0.622𝑃−𝑃 𝑃𝑠 𝑃−𝑃𝑣 𝑠 7/23/2024 16  Dew Point Temperature (DPT), Tdp : It is the temperature of air recorded by a thermometer when the moisture present in it starts condensing. Consider that a certain sample of unsaturated moist air shown by state (A) in the figure, on the right, is cooled slowly at constant pressure by passing it over a cooling coil. Its temperature decreases until it reaches (DPT), at which the first drop of dew is formed at point (B). It means that the water vapor in the air starts condensing. NOTE! In the case of dehumidification of air, it is required to maintain the temperature of cooling coil well below7/23/2024 DPT. 17 During a cooling process, the partial pressure of water vapor and the specific humidity (𝜔) remain constant until the vapor starts condensing. NOTE! - The saturation temperature can be found from the steam tables, corresponding to the partial pressure of water vapor 𝑷𝒗. - Dew point temperature can be also a function of the dry bulb temperature 4030(𝑇𝑑𝑏 +235) 𝑇𝑑𝑝 = − 235 (31) 4030− 𝑇𝑑𝑏 +235 𝑙𝑛𝜙 Where both 𝑇𝑑𝑝 and 𝑇𝑑𝑏 are in Celsius Specific Volume, 𝝑 : Specific Volume indicates the space occupied by air. 𝑅𝑎 𝑇 𝜗= (32) 𝑃𝑎 Or 𝑅𝑎 𝑇 𝜗= (33) 𝑃−𝑃𝑣 7/23/2024 18  Enthalpy, h: Air is a homogeneous mixture of dry air and water vapor. Therefore, the enthalpy of air is found by taking the sum of enthalpy of dry air and the enthalpy of water vapor in the moist air. This means that air’s enthalpy=ℎ𝑎 + 𝜔ℎ𝑣 (34) Considering the change in enthalpy of perfect gas as a function of temperature only, the enthalpy of dry air part, above a datum of 0°C, can be found as 𝑘𝐽 ℎ𝑎 = 𝑐𝑝𝑎 × 𝑇𝑑𝑏 = 1.005 × 𝑇𝑑𝑏 (𝑘𝑔) (35) Assuming the enthalpy of saturated liquid at 0°C equals to zero, the enthalpy of water vapor at point A in the figure below, is expressed as ℎ𝑣= 𝑐𝑝𝑤 × 𝑇𝑑𝑝 + ℎ𝑓𝑔 𝑑𝑝 + 𝑐𝑝𝑣 (𝑇𝑑𝑏 − 𝑇𝑑𝑝 ) (36) 7/23/2024 19 Where: 𝑐𝑝𝑤 = specific heat of water (kJ/kg.C) ℎ𝑓𝑔 𝑑𝑝= latent heat of vaporization at dew point temperature 𝑐𝑝𝑣 = specific heat of water vapor (kJ/kg.C) For reference state at 0°C, ℎ = ℎ𝑎 + 𝜔ℎ𝑣 ∴ ℎ = 𝑐𝑝𝑎 × 𝑇𝑑𝑏 + 𝜔 𝑐𝑝𝑤 × 𝑇𝑑𝑝 + ℎ𝑓𝑔 𝑑𝑝 + 𝑐𝑝𝑣 (𝑇𝑑𝑏 − 𝑇𝑑𝑝 ) (37) As the reference state of water vapor is at 0°C , 𝑇𝑑𝑝 = 0, instead of ℎ𝑓𝑔 𝑑𝑝 one can take latent heat of vaporization at 0°C as 2501 kJ/kg ∴ ℎ = 𝑐𝑝𝑎 × 𝑇𝑑𝑏 + 𝜔(2501 + 𝑐𝑝𝑣 𝑇𝑑𝑏 − 0 ) (38) ∴ ℎ = 1.005 × 𝑇𝑑𝑏 + 𝜔(2501 + 1.88 × 𝑇𝑑𝑏 ) 7/23/2024 20 Also, Equation (37) can be simplified as ℎ = (𝑐𝑝𝑎 + 𝜔𝑐𝑝𝑣 ) × 𝑇𝑑𝑏 + 𝜔(ℎ𝑔 − 𝑐𝑝𝑣 × 𝑇𝑑𝑝 ) (39) ⟹ ℎ = (𝑐𝑝𝑚 × 𝑇𝑑𝑏 ) + 𝜔(ℎ𝑔 − 𝑐𝑝𝑣 × 𝑇𝑑𝑝 ) (40) where 𝜔 = specific humidity in kg/kg of dry air ℎ𝑔 = enthalpy of saturated water vapor at dew point temperature in kJ/kg 𝑇𝑑𝑏 = dry-bulb temperature in °C 𝑇𝑑𝑝 = dew point temperature in °C 𝑐𝑝𝑎 = specific heat of dry air = 1.005 kJ/kg-C 𝑐𝑝𝑣 = specific heat of water vapor = 1.88 kJ/kg-C 𝑐𝑝𝑚 = specific heat of moist air in kJ/kg-C 7/23/2024 21  Pressure, P: Air means a mixture of water vapor and some other gases. The partial pressure of water vapor in the mixture can be found by some empirical relations of vapor pressure of water in moist air; theses empirical equations are a- Carrier’s Equation: (𝑃−𝑃𝑤 )×(𝑇𝑑𝑏 −𝑇𝑤𝑏 )×1.8 𝑃𝑣 = 𝑃𝑤 − (41) 2800−(1.3× 1.8×𝑇𝑑𝑏 +32 ) b- Apjohn’s Equation: 1.8×𝑃×(𝑇𝑑𝑏 −𝑇𝑤𝑏 ) 𝑃𝑣 = 𝑃𝑤 − (42) 2700 c- Modified Ferrell’s Equation 1.8×𝑇𝑑𝑏 𝑃𝑣 = 𝑃𝑤 − 0.00066 × 𝑃 × (𝑇𝑑𝑏 − 𝑇𝑤𝑏 ) × 1 + 1571 (43) 𝑃𝑤 : Is the saturation pressure of vapor, corresponding to the wet bulb temperature of vapor. P : is the barometric pressure. Note! All pressure units in above equations should be consistent. 7/23/2024 22 Example.1: A container with a volume of 100𝑚3 contains atmospheric air at pressure, dry-bulb temperature, and relative humidity of 0.1MPa, 35℃, 70% , respectively. Find its specific humidity, dew point temperature, and both the mass of dry air and the mass of water vapor within the mixture. Solution: Given: P=0.1MPa, 𝑇𝑑𝑏= 35℃, 𝜙 = 70% 1- To calculate the specific humidity It’s required to find the pressure of vapor by using the equation of relative humidity and water vapor tables. Since 𝑃𝑣 𝜔 = 0.622 𝑃𝑎 𝑃𝑣 ϕ= ⇒ 𝑃𝑣 = ϕ × 𝑃𝑠 𝑃𝑠 The saturation pressure at 35℃ can be found from steam tables ⇒ 𝑃𝑠 = 5.6291𝑘𝑃𝑎 𝑃𝑣 = 0.7 × 5.6291 = 3.94𝑘𝑃𝑎 7/23/2024 23 To find the partial pressure of air 𝑃𝑎 = 𝑃 − 𝑃𝑣 ⇒ 𝑃𝑎 = 100 − 3.94 = 96.06𝑘𝑃𝑎 3.94 ∴ 𝜔 = 0.622 = 0.0255 96.06 2-The mass of dry air can be found from the ideal gas law 𝑃𝑎 𝑉 𝑃𝑎 𝑉 = 𝑚𝑎 𝑅𝑎 𝑇 ⇒ 𝑚𝑎 = 𝑅𝑎 𝑇 96.06 × 100 ∴ 𝑚𝑎 = = 108.6𝑘𝑔 0.287 × 308.2 3-The mass of water vapor can be found from the ideal gas law 𝑃𝑣 𝑉 𝑃𝑣 𝑉 = 𝑚𝑣 𝑅𝑣 𝑇 ⇒ 𝑚𝑣 = 𝑅𝑣 𝑇 3.94 × 100 ∴ 𝑚𝑣 = = 2.77𝑘𝑔 0.46152 × 308.2 4- The dew point temperature can be calculated from the equation (31) 4030(35 + 235) 𝑇𝑑𝑝 = − 235 = 28.7℃ 4030 − 35 + 235 ln(0.7) 7/23/2024 24 Example.2: Find the absolute humidity of an air sample which has a dew point temperature of 16°C. Solution: Given: 𝑇𝑑𝑝 = 16°C Assumptions: Assume that the volume of the sample is 𝑉 = 1𝑚3. From steam tables, the vapor pressure for the saturation temperature of 16°C is 0.018168bar. 𝑅 8314.14 ∵ 𝑅𝑣 = 𝑀 = = 461.52 J/kg.K 𝑣 18.015 ∵ 𝜌 = 𝑚𝑣 /𝑉 = 𝑃𝑣 /𝑅𝑣 𝑇 1816.8 ∴𝜌= = 0.01362 𝑘𝑔/𝑚3 461.52 × 289 The absolute humidity = 0.01362 × 15,430= 210.1566 grains/𝑚3 7/23/2024 25 Example.3: A sample of air is at dry-bulb temperature of 25°C and a relative humidity of 50%, if the barometric pressure is 740 mm Hg, calculate, without using the psychrometric chart, the partial pressure of water vapor and dry air, dew point temperature, specific humidity, Specific volume, and enthalpy. Solution: Given: 𝑇𝑑𝑏 = 25°C, ϕ=50% 1- Calculating partial pressures of both vapor and dry air: ∵ 𝑃𝑏 = 740𝑚𝑚𝐻𝑔 98420𝑁 ∴ 𝑃𝑎𝑡𝑚 = 740 × 133 = = 98.420𝑘𝑃𝑎 𝑚2 From steam tables, the corresponding saturation pressure of water vapor at 25°C is 3.17𝑘𝑃𝑎 𝑃𝑣 ∵ ϕ= ⇒ 𝑃𝑣 = ϕ × 𝑃𝑠 𝑃𝑠 The partial pressure of water vapor is 𝑃𝑣 = 0.5 × 3.17𝑘𝑃𝑎 = 1.585𝑘𝑃𝑎 ∵ 𝑃 =7/23/2024 𝑃𝑎 + 𝑃𝑣 ⇒ 𝑃𝑎 = 𝑃 − 𝑃𝑣 26 The partial pressure of air is ∴ 𝑃𝑎 = 98.420 − 1.585 = 96.835𝑘𝑃𝑎 2- Dew-point temperature: From steam tables at 𝑃𝑣 = 1.585𝑘𝑃𝑎, the corresponding dew-point temperature is 14°C 3- Specific humidity: 𝑃𝑣 1.585 𝜔 = 0.622 = 0.622 × = 0.01018 𝑃𝑎 96.835 4- Specific volume: 𝑅𝑎 𝑇 287.3 × (25 + 273) 0.884 𝑚3 ∵𝜗= ⇒𝜗= = 𝑃𝑎 96.835 × 103 𝑘𝑔 5- The enthalpy of moist air: ∵ ℎ = ℎ𝑎 + 𝜔ℎ𝑣 ∴ ℎ = 1.005𝑇𝑑𝑏 + 𝜔 (2501 + 1.88 × 𝑇𝑑𝑏 ) 51.0636𝑘𝐽 ∴ ℎ = 1.005 × 25 + 0.01018 × (2501 + 1.88 × 25) = 7/23/2024 𝑘𝑔 27 Home Work Q.1. A sample of air at the atmospheric pressure has a dry-bulb temperature of 43°C and a wet-bulb temperature of 29°C. Without using the psychrometric chart, calculate: 1- The partial pressure of water vapor, using Carriers' equation 𝑃 − 𝑃𝑤 × 𝑇𝑑𝑏 − 𝑇𝑤𝑏 𝑃𝑣 = 𝑃𝑤 − × 1.8 2800 − (1.3 × (1.8 × 𝑇𝑑𝑏 + 32) Where 𝑃𝑤 : Is the saturation pressure of vapor, corresponding to the wet bulb temperature of vapor. 2- The specific humidity 3- The relative humidity 4- The dew point temperature 5- The enthalpy 6- The Degree of saturation 7/23/2024 28 Q.2. Moist air at 1 atm. pressure has a dry bulb temperature of 32°C and a wet bulb temperature of 26°C. Using the law of perfect gases model and psychrometric equations calculate: 1- The partial pressure of water vapor, 2- Humidity ratio, 3- Relative humidity, 4- Dew point temperature, 5- Density of dry air in the mixture, 6- Density of water vapor in the mixture and 7- Enthalpy of moist air 7/23/2024 29 Q.3. On a particular day the weather forecast states that the dry bulb temperature is 37°C, while the relative humidity is 50% and the barometric pressure is 101.325 kPa. Find 1- The humidity ratio 2- Dew point temperature 3- Enthalpy of moist air on this day Q.4. A sample of air its both dray and wet temperature were measured using a sling psychrometer that reads 40°C for the dry- bulb temperature, and 28°C for the wet-bulb temperature. If the atmospheric pressure was 75cm of Hg, calculate: 1- The specific humidity 2- Relative humidity 3- Dew point temperature 4- Enthalpy 5- Vapor density 7/23/2024 30