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Chapter 7 PSYCHROMETRY 1 PSYCHROMETRY The study of the properties of air and vapour pertaining to air conditioning problems is called psychrometry. Dry air: it is the mixture of nitrogen, oxygen and small percentage of other gases. Air contains 79% nitrogen, 21% oxygen by volume and has a molecular weight of 29. Moist air: it is the mixture of dry air and water vapour. The amount of water vapour varies according to the temperature of air and reaches saturation at one point. Saturated air mixture: saturated air mixture is a mixture of dry air and water vapour in which the partial pressure of the vapour is equal to the saturation pressure of water at the temperature of the mixture. 2 Unsaturated air mixture: it is a mixture of dry air and super heated water vapour. The partial pressure of vapour being less than the saturation pressure of water at the temperature of the mixture. Super saturated air mixture: it is a mixture of dry air and water vapour in which the partial pressure of the water vapour is greater than the saturation pressure of water at the temperature of the mixture. Dew point temperature: when the unsaturated air is cooled at constant pressure, the mixture reaches saturation temperature corresponding to the partial pressure of water vapour. This temperature at which condensation of the vapour begins resulting in formation of liquid droplets or dew when the mixture is cooled is called dew point temperature. 3 Absolute humidity or specific humidity or humidity ratio: The ratio of mass of vapour to the mass of dry air is called absolute humidity. Denoted by ‘ω’. 4 Dalton’s law of partial pressures: the total pressure of the mixture of gases is the sum of the partial pressures exerted by each gas when it occupies the same volume of the mixture at the same temperature of the mixture. Dry bulb temperature: it is the equilibrium temperature of the mixture indicated by an ordinary thermometer denoted by Tdb. Wet bulb temperature: it is the temperature indicated by a wet bulb thermometer which has its temperature sensitive element (bulb) covered with a wick soaked in water. It is denoted by Twb. 5 6 Relative humidity: it is defined as the ratio of partial pressure of water vapour in a mixture to the saturation pressure of water at dry bulb temperature. using perfect gas relation, pv.vv = pg.vg Hence Hence relative humidity is also defined as the ratio of the mass of water vapour in a certain volume of moist air at a given temperature to the mass of water vapour in the same volume of saturated air at the same temperature. 7 Degree of saturation or saturation ratio: it is defined as the ratio of specific humidity of actual air (ω) to the specific humidity of saturated air (ωg) at the same temperature. 8 Enthalpy of moist air: Enthalpy of moist air = enthalpy of dry air + enthalpy of water vapour associated. h = hair + ω hvapour = Cp tdb + ω hvapour But, hair = Cpa tdb = 1.005 tdb Where Cpa = is the specific heat of air = 1.005 kJ/kg/K hvapour = hg + Cps (tdb – tdp) = 2500 + 1.88tdb Cps = Specific heat of water vapour = 1.88kJ/kg/K & tdp = 0 Therefore, h = 1.005 tdb + ω(2500 + 1.88tdb) kJ/kg of dry air 9 Carrier equation: When DBT and WBT are given, for calculating the partial pressure of water vapour in air many equations have been proposed of which Dr. Carriers equation is most widely used. Where, (pg)wb = saturation pressure at wet bulb temperature. pv = partial pressure of water vapour pg = partial pressure of saturated vapour p = total pressure of moist air tdb = dry bulb temperature, 0C twb = wet bulb temperature, 0C 10 Psychrometric chart: As the calculations of various properties of moist air are tedious and time consuming, all the essential data for complete thermodynamic and psychrometric analysis of air conditioning processes can be summarized and can be presented in the form of a chart. Such a chart which makes it possible to obtain the necessary information readily for engineering calculations related to moist air known as a psychrometric chart. A psychrometric chart is constructed for a given mixture pressure, using dry bulb temperature and the specific humidity as co-ordinates. 11 At a given mixture pressure, the vapour pressure pv is a function of specific humidity only and hence there is only one value of pv for each value of ω. 12 Saturation curve: the saturation line represents the states of saturated air at different temperatures. Wet bulb and dew point temperatures fall on this curve. Relative humidity on the curve is 100%. Relative humidity lines: these are the curves with humidity ranging from maximum 100% to minimum 0%. Constant specific volume lines: these are inclined lines and are equally spaced. The WBT lines are much flatter than constant enthalpy lines. DBT lines: these are vertical lines and are equally spaced. DBT increases as we move from left to right. Specific humidity lines: these are horizontal lines and value of ω increases from bottom towards top of chart. 13 Sensible heating: Heating of air without addition or subtraction of moisture is called sensible heating. This can be achieved by passing the air over a heating coil. The heat added increases the DBT of air. This is useful in winter air conditioning. The heat added is given by, Qs = ma (h2 – h1) 14 Sensible cooling: Cooling of air without addition or subtraction of moisture is called sensible cooling. This can be achieved by passing the air over a cooling coil. This is useful in summer air conditioning. The heat removed is given by, Qr = ma (h1 – h2) 15 Cooling and dehumidification: Water vapour may be removed from air by cooling it below its dew point temperature. As a result of the cooling process a portion of the vapour in the air is condensed. Dehumidification will take place along with cooling when the saturation temperature of the cooling coil is below the dew point temperature of the cooling air. 16 In the above process warm air t1 enters the cooling coil maintained at temperature t2. The surface temperature of the cooling coil t2 is lesser than the dew point temperature of the incoming air, t4. Under ideal conditions air leaves at t2, but due to inefficient cooling it leaves at a higher temperature t3. 17 The temperature t2 corresponding to point 2 on the saturation curve is known as apparatus dew point, ADP. The ratio of actual heating/cooling to the ideal heating/cooling is known as by-pass factor. In this case it is given by 18 Adiabatic humidification: if humidification is carried out adiabatically, the energy required for the evaporation of the added moisture must come from the entering air. 19 As the dry bulb temperature of air decreases during adiabatic humidification process, it is also known as evaporative cooling process or cooling with adiabatic humidification of air. When warm air is passed through a spray chamber, part of water is vaporized and carried away with air. This results in humidification of air. 20 Mixing process: In this process two or more streams are mixed to produce a stream with desirable state of temperature and relative humidity. 21 Making energy balance, ma1(ha1 + ω1hv1) + ma2(ha2 + ωhv2) = ma3(ha3 + ω3hv3) Making mass balance ma1 + ma2 = ma3 Making moisture balance ma1 ω1 + ma2 ω2 = ma3 ω3 Combining we get, 22 Requirement of Comfort air conditioning: The following 5 factors determine the comfort feeling of the people in an air conditioned space. 1. Supply of O2 and removal of CO2. 2. Removal of body moisture dissipated by the occupants. 3. To provide sufficient air movement and air distribution in the occupied space. 4. To maintain the purity of air by removing odour and dust. 23 1. Oxygen supply: human body takes in oxygen through the system and gives out CO2. Normally each person requires nearly 0.65m3 of O2 per hour and produces 0.2m3 of CO2. The percentage of CO2 in atmosphere is about 0.6% and it is necessary to maintain this to ensure comfort for easy breathing. Thus the quantity of air supply to an air conditioned space should be regulated properly to see that percentage of CO2 should not be exceeded. 2. Heat removal: it is a well known fact that human beings dissipate good amount of heat to the atmosphere during breathing etc. 24 The atmosphere should be capable of absorbing the heat dissipated by persons otherwise discomfort exist. Thus sufficient circulation of air should be provided through proper ventilation system to avoid rise in temperature of air in the air conditioned space. 3. Moisture removal: A moisture loss of up to 50% from human body is commonly observed. This should be properly taken into account while designing a air conditioning unit. The ventilation system must be capable of maintaining an RH of below 70%. 25 4. Air movement: In addition to providing air motion, proper air distribution is very important. Air distribution is defined as a uniform supply of air to an air conditioned system. Air movement without proper air-distribution is permissible for local cooling sensation known as draft. A velocity of about 8m/min associated with temperature differential of 10C do not result in noticeable draft. Velocities greater than this produces uncomfortable drafting conditions. 26 5. Purity of air: It is important to maintain quality of air in any airconditioned space. Odour, dust, toxic gases and bacteria are considered for defining the purity of air. The various factors which makes air impure are: i. The evaporation on the surface of the body adds odour to the air. ii. The smoke from the surroundings which has a bad effect on nose, eyes and heart. iii. The toxic gases are objectionable as they cause irritation. 27 Summer air conditioning systems (for Hot and dry outdoor conditions): this system is used when outdoor conditions are hot and dry. That means atmospheric temperature is higher than the comfort temperature with less moisture content (lesser RH). 28 Atmospheric air at condition ‘1’ (higher DBT and lower RH) enters the air dampers and passes over the cooling coil via an air filter. Temperature of the air is reduced to condition ‘2’ in the cooling coil. Air now enters an adiabatic humidifier and is passed over water eliminators. 29 Air enters the conditioned space at condition ‘3’. Point ‘4’ represents the condition of air after passing over the cooling coil if efficiency of the coil were 100%. Condition line 1-2-4 represents the changes in DBT of air when passes over the cooling coil and the condition line 2-3-5 represents the changes in DBT of cooled air in the adiabatic humidifier. 30 Summer air conditioning systems: Hot and humid outdoor conditions: this system is used when the out door condition is hot and humid, like in coastal areas. That means the atmospheric temperature is higher but at the same time contains large moisture content. This condition eliminates the need for an adiabatic humidifier as shown in fig. 31 Atmospheric air at condition ‘1’ enter the air filter via air dampers. Filtered air is cooled when it passes over the cooling coil at condition ‘3’. Condition line 1-2-3 represents the changes in DBT of air. While passing over the cooling coil with an efficient coil the air would have been cooled to 2. 32 As the air temperature is still below the required comfort condition, it is now passed over a heating coil. The air coming out of the coil is at condition ‘5’ which is delivered to the space. Line 3-5-4 represents changes in DBT of air while it passes over the heating coil. Point ‘4’ represents the maximum temperature that could have been attained by using an efficient heating coil. 33 Winter air conditioning system: During winter air in the atmosphere is at lower temperature than the required conditions. Also the relative humidity may be more or lesser than the actual relative humidity required for human comfort. Fig shows an arrangement for such a system where dry conditions prevail in the atmosphere. 34 Cold air from the atmosphere is filtered in an air filter before it passes over the heating coil. Air at condition ‘1’ is heated to ‘3’ in the heating coil. However due to losses, condition of air leaving the heat coil is ‘2’. Condition line 1-2-3 represents the changes in the DBT of air while it passes over the heating coil. Hot air at ‘2’ now enters the humidifier and it is cooled to ‘4’. 35 Any water particles suspended in air is removed by the water eliminator. As the humidity and temperature of air have still not reached comfort conditions air is heated again up to point ‘7’ by another heating coil. Due to heat losses the final temperature of air entering the conditioned space is at point ‘6’. 36 Problem 1. Moist air at 30°C,1.01325 bar has a relative humidity of 80%. Determine without using the psychrometry chart Partial pressures of water vapour and air , Specific humidity , Specific Volume and Dew point temperature (V.T.U. July2004) Solution: At 30C from t able ps 4.2461kPa p p 0.8 x 4.2461 3.397kPa ps 0.622p 3.397 0.622x p p 101.325 3.397 0.213kg/kg of dry air. Corresponding to Pv =3.397 kPa from tables, we get dew point temperature = 28.9°C 37 Problem 2: Atmospheric air at 101.325 kPa ha 30°C DBT and 15°C DPT. Without using the psychometric chart, using the property values from the table, Calculate Partial pressure of air and water vapour, Specific humidity , Relative humidity,Vapour density and Enthalpy of moist air Solution: p 101.325kpa 1.01325bar DBT 30C , DP T 15C From table Corresponding t o DBT 30C, we haveps 0.042461bar Corresponding t o DP T 15C, we havep 0.017051bar Partial pressure of air p - p 1.01325 0.017051 0.984274bar 38 p 0.622x0.017051 Specifichum idity 0.622 pa 0.984274 0.01077kJ/ kg of dry air p 0.017051 Re lative hum idity 0.4015 ps 0.042461 Ent halphy 1.005t db 40.15% ( 2500 1088t db ) 1.005x30 0.010775(2 500 1.88x30) 57.69kJ/kgof dry air RT Specific volumeof dry air,a P 0.2872x303 0.874m3 / kg 0.98425x100 0.010775 Vapourdensity w 0.12kg / m3 a 0.847 39 Problem 3: Air at 30°C DBT and 25°C WBT is heated to 40°C. if the air is 300 m3/min, find the amount of heat added/min and RH and WBT of air. Take air pressure to be 1 bar Solution: At 25°C WBT from tables page no 14 Pvs(wbt)=0.03166 bar 40 p ( PVS ) wbt ( p pswbt )(tdb t wb ) 1547 1.44t wb (1 - 0.03166)(3 0 - 25) 0.031661547 1.44x 25 0.0284bar p 1 0.622 p p 0.0284 0.622 1 0.0284 0.0179kJ / kg of dry air 41 At 40C DBT PVS 0.07375bar During sensible heating and p rem ainconst ant p 0.0284bar p 0.0284 RH ps 0.07375 0.385 38.5% H 2 1.005x 40 0.0179( 2500 1.88x 40) 86.29kJ/kgof dry air ( p p )V 3 Weight of 300m / min ofair RT (1- 0.0284)x30 0x102 335.18kg / min 0.287x303 Heat added/min 335.18(86. 29 - 76) 3449kJ/min From chart WBT 27.2C 42 4) One stream of air at 5.5m3/min at 15°C and 60% RH flows into another stream of air at 35m3/min at 25°C and 70%RH, calculate for the mixture 1) Dry bulb temperature, 2) Wet bulb temperature 3) Specific Humidity and 4) Enthalpy Solution: For air at 15°C and 60%RH, V=5.5m3/min ps 0.017051bar p RH p 0.6 x0.017051 0.01023bar ps (p - p )V (1.01325 0.01023) x102 x5.5 Mass of air RT 0.287x 288 m1 6.672kg / min 0.622p 0.622x0.01023 1 (p - p ) (1.01325 0.01023) 0.006343kg / kg of dry air 43 H1 1.005t db 1 (2500 1.88t db ) 1.008x18 0.006343(2 500 1.88x15) 34.12J/kgof dry air For air at 25C and 70% RH, V 35m3 / min,Ps 0.03169bar p RH p 0.03169x0.7 0.02218bar ps 2 (1.01325- 0.02218x10 x35) Mass of air m 2 40.55kg. min 0.287x 298 0.622x0.02218 2 0.01392kg / kg of dry air (1.01325 0.02218) H 2 (1.005x 25) 0.01392(2500 1.88x 25) H 2 60.59kJ / kg of dry air Mass of dry air/Unit mass of moist air m a1 m1 6.672 6.6299 1 1 1 0.006343 44 m2 40.55 Since m a2 39.993 1 2 1 0.01392 Then enthalpyof themixedair, ma1 ( H1 ) ma 2 ( H 2 ) 6.6299(34. 12) 39.993(60. 56) H mix m1 m2 6.672 40.55) 55.96kJ/kg of dry air Specific Humidityof themixedair, ma1 (1 ) ma 2 (2 ) (6.6299x0. 006343) (39.993x0. 01932) mix m1 m2 6.672 40.55 0.01268kg/kg of dry air But H mix 1.005t db mix (2500 1.88t db ) 55.96 1.005xtdb 0.01234(2500 1.88t db ) t db 24.42C DBT of t hemixt ure 24.42C From chart WBT 19C , RH 67% 45 Problem 5: An air conditioning system is designed under the following conditions Outdoor conditions: 30°C DBT, 75% RH Required indoor conditions: 22°C DBT,70% RH Amount of Free air circulated 3.33 m3/s Coil dew point temperature DPT=14° The required condition is achieved first by cooling and dehumidification and then by heating. Estimate The capacity of the cooling coil in tons of refrigeration Capacity of the heating coil in kW The amount of water vapour removed in kg/hr 46 Solution: Locate point' a' 30C DBT ,75%RH out door condit ion Locate point' d' 22C DBT ,70%RH required condit ion Locate point' b' 14C DP T ,coil surface t emperat re u Join ab at d, draw a horizont alline t o cut t he line ab at pointc. ac coolingand dehumidificat ion cd heat ing 47 From chart H a 83kJ / kg of air , H b 40kJ / kg of air H d 53kJ / kg of air , H c 48kJ / kg of air , Wa 0.0202kg / kg of dry air Wc Wd 0.0118kg / kg of dry air , Vsa 0.88m / kg 3 V 3.33 Mass of air 3.78kg / s Va 0.88 ma (H a H c ) Capacit yof coolingcoil 3. 5 3.78(83 48) 37.84tons of refrigeration 3. 5 Capacit yof heat ingcoil m a ( H d H c ) 3.78(53- 48) 18.92kW 48 Am ountof water vapour removed ma (a d )3600 3.78(0.020 2 - 0.0118)360 0 114.3kg/hr 49 Problem 6: A summer air conditioning system for hot and humid weather (DBT=32°Cand 70% RH) Consists in passing the atmosphere air over a cooling coil where the air is cooled and dehumidified. The air leaving the cooling coil is saturated at the coil temperature. It is then sensibly heated to the required comfort condition of 24°C and 50%RH by passing it over an electric heater then delivered to the room. Sketch the flow diagram of the arrangement and represent the process undergone by the air on a skeleton psychometric chart and determine 1. The temperature of the cooling coil 2. The amount of moisture removed per kg of dry air in the cooling coil. 3. The heat removed per kg of dry air in the cooling coil and 4. The heat added per kg of dry air in the heating coil 50 From chart H a 86kJ / kg of air H b 38kJ / kg of air H c 48.5kJ / kg of air a 0.021kg / kg of dry air b 0.0092kg / kg of dry air 51 The t emperat re u of t hecoolingcoil Tb 13C Am ountof moist ureremoved a b 0.021- 0.0092 0.0108kg/kg of dry air Heat removed H a H b 86 38 48 kJ/kg of dry air Heat added H c H b 48.5 38 10.5kJ/kg of dry air Locatepoint' a' 32C , 70%RH out door condit ion Locatepoint' c' 24C DBT ,50%RH required condit ion At c draw a horizont alline t o cut t hesat urat ion line at point' b' Join ab ab coolingand dehumidificat ion bc heat ing 52 Problem 7 It is required to design an air conditioning plant for an office room with the following conditions. Outdoor conditions: 14°CDBT, 10°CWBT Required conditions: 20°CDBT,60% RH Amount of air circulated 0.3m3/min/person Starting capacity of the office= 60 The required condition is achieved first by heating and then by adiabatic humidifying. Determine the following. Heating capacity of the coil in kW and the surface temperature required, if the by pass factor of the coil is 0.4 Capacity of the humidifier. 53 Locate point' a'14C , and10CWBT (out door condition) Locate point' c' 20C DBT ,60%RH required condition At a draw a horizontalline At ' c' draw a constantenthalpyline to cut thehorozontalline at point' b' Join ab ab heating bc adiabatichumidification 54 From chart H a 30kJ / kg of air , H b H c 43kJ / kg of air a b 0.006kg / kg of dry air c 0.00875kg/kg of dry air Specific volomeVsa 0.8175m 3 / kg 0.3x60 Volum eof air supplied V 0.3m 3 / sec 60 V 0.3 Weight of air supplied m a Va 0.8175 0.3669kg/sec Capacityof theheatingcoil m a ( H b H a ) 0.3669(43- 30) 4.77kW From chart Tb 26.5C Let coil surface temperatu re be Td 55 Td Tb By passing factor 0.4Td 5.6 Td 26kJ Td Ta Td 26.5 0 .4 Td 34.8C Td 1.4 Capacityof thehumidifier m a (c b ) x3600 0.3669(0.00875- 0.006)3600 3.63kg/hour 56 Problem 8 An air conditioned system is to be designed for a hall of 200 seating capacity when the following conditions are given: Atmospheric condition = 300C DBT and 50% RH Indoor condition = 220C DBT and 60% RH Volume of air required = 0.4m3/min/person The required condition is achieved first by chemical dehumidification and after that by sensible cooling. Find the following . 1. DBT of the air leaving the dehumidifier. 2. The quantity of water vapour removed in the dehumidifier per hour. 3. The capacity of cooling coil in tons of refrigeration. 4. Surface temperature of the coil if the by pass factor of the coil is 0.25. 57 Solution: Locate point ‘a’, 300C DBT, 50% RH, the atmospheric condition. Locate point ‘c’, 220C DBT, 60% RH, the required indoor condition. “Since chemical dehumidification process follows constant enthalpy line” at a draw a line parallel to constant enthalpy line. At ‘c’ draw a constant line to cut the previous line at point b. DBT of air leaving the dehumidifier Tb = 40.50C From chart Hb = Ha = 65kJ/kg, a = 0.013 kg/kg of dry air Hc = 45 kJ/kg, b = 0.009 kg/kg of dry air, Vsa = 0.875 m3/min Volume of air = 200 X 0.4 = 80 m3/min Wa = Weight of air = V/Vsa = 80/0.875 = 91.42 kg/min 58 Quantity of water vapour removed/hour = Wa(a-b)60 = 91.42(0.13-0.009)60 = 21.94 kg/hr Capacity of cooling coil = Wa(Ha-Hb)/ (60 X 3.5) = 91.42(65-45)/(60 X 3.5) = 8.7 tons By pass factor = (Tc-Td)/( Tb-Td) = 0.25 Td = Temperature of cooling coil = 15.830C 59 Problem 9 An air conditioned system is to be designed for a cinema hall of 1000 seating capacity when the following conditions are given: Outdoor condition = 110C DBT and 70% RH Required indoor condition = 200C DBT and 60% RH Amount of air required = 0.3m3/min/person The required condition is achieved first by heating, then by humidifying and finally by heating. The condition of air coming out of the humidifier is 75% RH. Find the following . Heating capacity of the first heater in kW and condition of the air coming out of the first heater in kW and condition of the air 60 Locate point ‘a’, 300C DBT, 50% RH, the atmospheric condition. Locate point ‘c’, 220C DBT, 60% RH, the required indoor condition. “Since chemical dehumidification process follows constant enthalpy line” at a draw a line parallel to constant enthalpy line. At ‘c’ draw a constant line to cut the previous line at point b. DBT of air leaving the dehumidifier Tb = 40.50C From chart Hb = Ha = 65kJ/kg, a = 0.013 kg/kg of dry air Hc = 45 kJ/kg, b = 0.009 kg/kg of dry air Vsa = 0.875 m3/min Volume of air = 200 X 0.4 = 80 m3/min Wa = Weight of air = V/Vsa = 80/0.875 = 91.42 kg/min 61 Quantity of water vapour removed/hour = Wa(a-b)60 = 91.42(0.13-0.009)60 = 21.94 kg/hr Capacity of cooling coil = Wa(Ha-Hb)/ (60 X 3.5) = 91.42(65-45)/(60 X 3.5) = 8.7 tons By pass factor = (Tc-Td)/( Tb-Td) = 0.25 Td = Temperature of cooling coil = 15.830C 62