Psychrometricsoperation 2 Unit Example 1. Air at 60 C dry bulb temperature and 27.5 C wet bulb temperature, and a humidity ratio of 0.01 kg water/kg dry air is mixed with water adiabatically and is cooled and humidified to a humidity ratio of 0.02 kg water/kg dry air. What is the final temperature of the conditioned air? Given Inlet: dry bulb temperature = 60C wet bulb temperature = 27.5 C Initial humidity ratio W 1 = 0.01 kg water/kg dry air Final humidity ratio W 2 =0.02 kg water/kg dry air Solution From Table A.4.2, latent heat of vaporization at 27.5 C = 2436.37 kJ/kg Example 2. Calculate the rate of thermal energy required to heat 10 m 3 /s of outside air at 30 C dry bulb temperature and 80% relative humidity to a dry bulb temperature of 80 C. Solution 1. Using the psychrometric chart, we find at 30 C dry bulb temperature and 80% relative humidity, the enthalpy H1 = 85.2 kJ/kg dry air, humidity ratio W 1 = 0.0215 kg water/kg dry air, and specific volume V1= 0.89 m3 /kg dry air. At the end of the heating process, the dry bulb temperature is 80 C with a humidity ratio of 0.0215 kg water/kg dry air. The remaining values are read from the chart as follows: enthalpy H 2 =140 kJ/kg dry air; relative humidity φ2 =7%. 2. 3. The rate of heat required to accomplish the given process is 615.7 kW 4. In these calculations, it is assumed that during the heating process there is no gain of moisture. This will not be true if a directly fired gas or oil combustion system is used, since in such processes small amounts of water are produced as part of the combustion reaction. Example 3. In efforts to conserve energy, a food dryer is being modified to reuse part of the exhaust air along with ambient air. The exhaust air flow of 10 m 3/s at 70 C and 30% relative humidity is mixed with 20 m3/s of ambient air at 30 C and 60% relative humidity. Using the psychrometric chart, determine the dry bulb temperature and humidity ratio of the mixed air. Solution 1. From the given data, locate the state points A and B, identifying the exit and ambient air as shown on the skeleton chart ( Fig. E9.3 ). read the humidity ratio 0. Determine the amount of water removed per kg of dry air. point B. The amount of moisture removed from rice 0. 4. 2.019 – 0. the shorter length of line AC corresponds to larger air mass. 3.2. Locate point A on the psychrometric chart. The mixed air.0078 kg water/kg dry air. Follow the constant enthalpy line to the saturation curve. Read humidity ratio 0.0112 kg water/kg dry air. Solution 1. Example 4. Join points A and B with a straight line. At point B.019 kg water/kg dry air. Thus. will have a dry bulb temperature of 44 C and a humidity ratio of 0. 3.032 kg water/kg dry air. 4. Heated air at 50 C and 10% relative humidity is used to dry rice in a bin dryer. The air exits the bin under saturated conditions. . represented by point C. The division of line AB is done according to the relative influence of the particular air mass. Since the mixed air contains 2 parts ambient air and 1 part exhaust air. as shown in Figure. line AB is divided in 1:2 proportion to locate point C.0078 = 0. EXAMPLE 6 . and the relative humidity of the heated air. to the condition at the higher (dry bulb) temperature of 40°C.906 = 1104kg. Mass of 1000 m3 is 1000/0. So rate of heating required = 1104 x 16 kJ h -1 = (1104 x 16)/3600 kJ s -1 = 5 kW . Similar examination of the enthalpy lines gives an estimated enthalpy of 57 kJ kg-1. specific volume is 0. On heating. calculate the rate of heat supply needed for a flow of 1000 m3 h-1 of this hot air for a dryer. the air condition moves. Examining the location of this point of intersection with reference to the lines of constant relative humidity.EXAMPLE 5. the enthalpy is 73 kJ kg-1.906 m3 kg-1 and RH 27%.57) = 16 kJ kg-1. Relative humidity of heated air If the air in Example 5 is then to be heated to a dry-bulb temperature of 40°C. enthalpy and specific volume of air If the wet-bulb temperature in a particular room is measured and found to be 20°C in air whose dry-bulb temperature is 25°C (that is the wet-bulb depression is 5°C) estimate the relative humidity. and from the volume lines a specific volume of 0. Once the properties of the air have been determined other calculations can easily be made. reading from the chart at 40°C and humidity 0. Relative humidity.862 m3 kg-1. it lies between 60% and 70% RH and about 4/10 of the way between them but nearer to the 60% line.0125 kg kg-1. On the humidity chart follow down the wet-bulb line for a temperature of 20°C until it meets the dry-bulb temperature line for 25°C. DH = (73 . at constant absolute humidity as no water vapour is added or subtracted. At this condition. the enthalpy and the specific volume of the air in the room. Therefore the RH is estimated to be 64%. It is found that the first traces of moisture appear on this surface when it is at 40°C. 0. Through the dryer. with the heat for evaporation being supplied by the hot air passing over a wet solid surface. so at 35°C its condition is a humidity of 0.96 = 1794 m3 h-1 If air is cooled. It is then reheated to 140°C and passed over another set of trays .0107kg water.010) = 0. EXAMPLE 7 . passes over a surface which is gradually cooled. the saturation temperature is 40°C and proceeding at constant humidity from this. the inlet air condition shows the humidity of the drying air to be 0.e. the system behaves like the adiabatic saturation system.If the air is used for drying. Reheating of air in a dryer A flow of 1800 m3 h-1 of air initially at a temperature of 18°C and 50% RH is to be used in an air dryer. this is a drying process. with an exit temperature of 45°C. from the viewpoint of the air it is humidification. i.0107 kg kg-1 of air. So each kg. until it reaches the saturation curve at its dew point. EXAMPLE 8.96 m3. with water condensing to adjust the humidity as the saturation humidity cannot be exceeded. Volume of air to remove 20 kg h-1 = (20/0.0207 . Water removed in air drying Air at 60°C and 8% RH is blown through a continuous dryer from which it emerges at a temperature of 35°C. and the volume of drying air required to remove 20 kg water per hour.0.01 kg kg-1 and its specific volume to be 0. On the psychrometric chart. Estimate the relative humidity of the air leaving the dryer.96 m3 kg-1. Estimate the quantity of water removed per kg of air passing. It is heated to 140°C and passed over a set of trays in a shelf dryer. Further cooling then proceeds down the saturation line to the final temperature. of air passing will remove 0. Relative humidity of air leaving a dryer The air emerging from a dryer.0107) x 0. the 45°C line is intersected at a point indicating: relative humidity = 76% EXAMPLE 9. which it leaves at 60 % RH. It is adiabatic because no heat is obtained from any source external to the air and the wet solid. horizontally on a psychrometric chart. and the latent heat of evaporation must be obtained by cooling the hot air. Water removed = (0. the condition of the air follows a constant wet-bulb line of about 27°C . Looked at from the viewpoint of the solid.0207kg kg-1. then initially its condition moves along a line of constant humidity. Using the psychrometric chart. 0062 = 0.0. the enthalpy is found to be 160 kJ kg-1. EXAMPLE 10.0062 kg kg-1.082 kg kg-1. Reheating to 140°C keeps humidity constant (0. it is sometimes useful to reheat the air so as to reduce its relative humidity and thus to give it an additional capacity to evaporate more water from the material being dried.6 kg s-1 Energy taken in by air = 233 x 0.834 = 2158 kg h-1 = 0.6 x 0.045 kg kg-1 )and enthalpy goes to 268 kJ kg-1.045 kg kg-1.0758 = 0. the quantity of water removed and the amount of reheating necessary. Estimate the energy necessary to heat the air and the quantity of water removed per hour.0266 kg kg-1. The processing room has a volume of 1650 m3 and it is estimated to require six air changes per hour. determine the temperature to which the air should be cooled. and what has been said about them. will show that they can be used for calculations focused on the air. for the purposes of air conditioning as well as for drying. The ambient air is at 31.35 = 233 kJ kg-1 Total water removed = DY = 0. .045 kg s-1 = 163kgh -1 Exit temperature of air (from chart) = 60°C.0758 kg kg-1 1800 m3 of air per hour = 1800/0. Total energy supplied = DH in heating and reheating = 268 . specific volume is 0. then to reheat if necessary. If the chosen method is to cool the air to condense out enough water to reduce the water content of the air sufficiently. Proceeding along a wetbulb line to an RH of 60% gives the corresponding temperature as 48°C and humidity as 0. From the psychrometric chart the humidity of the initial air is 0.5°C and 90% RH. Consideration of psychrometric charts.082 .0085 kg kg-1. Air conditioning In a tropical country. Final humidity is 0.which it leaves at 60 % RH again. This process can easily be followed on a psychrometric chart. Proceeding at constant humidity to a temperature of 140°C.834 m3 kg-1. and enthalpy 35 kJ kg-1. Saturation temperature for this humidity is 13°C. Thence along a wet-bulb line to 60 % RH gives humidity of 0. In dryers. it is desired to provide processing air conditions of 15°C and 80% RH. Using the psychrometric chart : Initial humidity is 0.6 kJ s-1 = 140 kW Water removed in dryer = 0. and the humidifying efficiency of the recirculated spray water.970 x 0. The following options are available for this objective: (a) by passing air through a heated water-spray air washer. From the psychrometric chart. b.6 kJ s-1 Problem 1. The moist air enters at 25C and wet bulb temperature of 20 C. c. Mass of air to be conditioned = (1650 x 6)/0. and (c) dry bulb temperature of the two streams mixed together.827 m3 kg-1.827 = 11. The exit air leaves at 30 C and relative humidity of 80%. Dew-point temperature. determine the (a) humidity ratio. determine how much heat is added per m3 initial moist air and what the final dew-point temperature is. d. 4 Moist air at 35 C and 55% relative humidity is heated using a common furnace to 70C.970 x 3. and then passing through a water spray washer with recirculated water until relative humidity rises to 95% and then again heating sensibly to the final required state. Moist air flowing at 2 kg/s and a dry bulb temperature of 46 C and wet bulb temperature of 20 C mixes with another stream of moist air fl owing at 3 kg/s at 25 C and relative humidity of 60%. Problem 2.5 = 41.33.0.018 kg kg-1 Mass of water removed per hour = 11. Assuming that the air changes are calculated at the conditions in the working space. the enthalpy is 37 kJ kg-1 and the specific volume of air is 0. Enthalpy of air per kg dry air.0085 = 0. e. Problem 5. 15°C and 80 % RH. Determine the flow rate of water. Humidity ratio.5 kJ kg-1 Total reheat power required = 11. Using a psychrometric chart. (b) by preheating sensibly. Volume of moist air/kg dry air. in kg/s.0266 .6 kW. Problem. Relative humidity. (b) enthalpy.5)= 3. Problem 3. A water-cooling tower is to be designed with a blower capacity of 75 m 3/s. Atmospheric air at 760 mm Hg is at 22 C dry bulb temperature and 20 C wet bulb temperature. the enthalpy is 33. that can be . Using the psychrometric chart. Determine for (a) and (b) the total heating required.018 = 215 kg h-1 Reheat required DH = (37 .Therefore the air should be cooled to 13°C At the saturation temperature of 13°C. determine: a.970 kg h-1 Water removed per kg of dry air DY = 0.5 kJ kg-1 At the final conditions. Air at a dry bulb temperature of 20C and relative humidity of 80% is to be heated and humidified to 40C and 40% relative humidity.895 kJ h-1 = 11. = 11. the makeup water required in water-spray air washer. Specific volume Problem 7. . Relative humidity c. Dew point e. The water enters the tower at 40C and leaves the tower at 25C. Enthalpy d. Air at a dbt (dry bulb temprature) of 30 C and a relative humidity of 30% is conveyed through a heated dryer where it is heated to a dbt of 80 C. Problem 6. Clearly show the paths of air. Then it is conveyed through a bed of granular pet food to dry it. The air exits the dryer at a dbt of 60 C.cooled if the cooled water is not recycled. Air is at a dry bulb temperature of 20C and a wet bulb temperature of 15C. a. starting from the ambient air to the saturated air exiting the second dryer on a copy of a psychrometric chart. Moisture content b. The exit air from the second dryer leave at saturation. The exit air is again heated to 80 C and conveyed through another dryer containing another batch of pet food. Determine the following properties from a psychrometric chart. Determine the amount of water removed in the first and second dryer per kg of dry air.