The psychrometric chart theory and application
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The document discusses the psychrometric chart and various psychrometric processes involving moist air. It begins by identifying parts of the psychrometric chart and explaining how it can be used to determine moist air properties and analyze processes involving moist air. Several examples are then provided to illustrate key psychrometric processes including sensible heating/cooling, heating and humidifying, cooling and dehumidifying, adiabatic or evaporative cooling, and adiabatic mixing of moist air streams.
The psychrometric chart theory and application
- 1. THE PSYCHROMETRIC CHART: Theory and Application Perry PeraltaPerry Peralta NC State UniversityNC State University
- 2. PSYCHROMETRIC CHART Identify parts of the chartIdentify parts of the chart Determine moist air propertiesDetermine moist air properties Use chart to analyze processesUse chart to analyze processes involving moist airinvolving moist air
- 11. Psychrometric chart: Example 1 Given: T = 25°C Tw =20°C Required: (a) RH, (b) Tdp, (c) HR, (d) v, (e) h
- 12. 25°C 20.0°C 57.5 kJ/kg d.a. 63% 0.86 m3/kg d.a. 17.6°C 12.6 g/kg d.a.
- 14. Sensible Heating or Cooling a psychrometric process that involves the increase or decrease in the temperature of air without changing its humidity ratio Example: passing moist air over a room space heater and of kiln air over the heating coils
- 15. 1 2
- 18. 60°C 70°C 50°C 58.8%37.6% 2 3 275.5 kJ/kg d.a. 287.0 kJ/kg d.a. 51°C
- 19. Heating and Humidifying a psychrometric process that involves the simultaneous increase in both the dry bulb temperature and humidity ratio of the air
- 20. 1 2 0
- 21. Heating and humidifying: Example 7 Two and a half cubic meters of lumber is being dried at 60°C dry bulb temperature and 52°C wet bulb temperature. The drying rate of the lumber is 12.5 kg of water per hour. If outside air is at 27°C dry bulb temperature and 80% relative humidity, how much outside air is needed per minute to carry away the evaporated moisture?
- 22. 27°C 60°C 80% 1 2 92 g/kg d.a. 18 g/kg d.a. 0.87 m3/kg d.a. 52°C
- 23. Heating and humidifying: Example 7 ∆HR = (92.0 – 18.0) g/kg dry air = 74.0 g/kg dry air wa1 = drying rate/∆HR = (12.5 kg/hour)/(0.074 kg/kg dry air) = 168.9 kg dry air/hour VF1=(wa1)(v1) =(168.9 kg dry air/hour)(0.87 m3/kg dry air) = 147 m3/hour = 2.45 m3/minute
- 24. Cooling and Dehumidifying a psychrometric process that involves the removal of water from the air as the air temperature falls below the dew- point temperature
- 25. 1 2
- 26. Cooling and dehumidifying: Example 9 Moist air at 50°C dry bulb temperature and 32% relative humidity enters the cooling coil of a dehumidification kiln heat pump system and is cooled to a temperature of 18°C. If the drying rate of 6 m3 of red oak lumber is 4 kg/hour, determine the kW of refrigeration required.
- 27. 18°C 50°C 32% 2 1 25.2 g/kg d.a. 12.9 g/kg d.a. 28.8°C 115.7 kJ/kg d.a. 50.8 kJ/kg d.a.
- 28. Cooling and dehumidifying: Example 9 a drying rate w HR kg water4 h kg water0.0123 kg dry air kg dry air 325.2 h = ∆ = = ∆HR = (25.2 – 12.9) g water/kg dry air = 12.3 g water/kg dry air
- 29. Cooling and dehumidifying: Example 9 ( )( )aq h w kJ kg dry air 64.9 325.2 kg dry air h kJ 21105.7 5.9 kW h = ∆ = = = ∆h = (115.7 – 50.8) kJ/kg dry air = 64.9 kJ/kg dry air
- 30. Adiabatic or Evaporative Cooling a psychrometric process that involves the cooling of air without heat loss or gain. Sensible heat lost by the air is converted to latent heat in the added water vapor
- 31. 1 2
- 32. Evaporative cooling: Example 10 Referring to Figure 21, air at state point 1 (65°C dry bulb temperature and 57°C wet bulb temperature) experiences a temperature drop of 3°C as it passes through the 1.2-m wide stack of lumber. Determine the properties of the air at state point 2 and compare them with those at state point 1. If the air is flowing at a rate of 2 meters per second, determine the drying rate assuming that the volume of the stack of 2.5-cm-thick lumber is 2.5 m3. The stack is 1.2 m wide x 3.6 m long, and the boards are separated by stickers 3.8 cm wide x 1.9 cm thick that are spaced 0.6 m apart.
- 34. Evaporative cooling: Example 10 Given: T1 = 65°C; Tw1 = 57°C Adiabatic cooling to T2 = 62°C Air flow rate = 2 m/s Volume of lumber = 2.5 m3 Board thickness = 2.5 cm Stack dimensions: 1.2 m wide x 3.6 m long Sticker dimensions: 3.8 cm wide x 1.9 cm thick Sticker spacing = 0.6 m Required: (a) Properties of the air at state point 2 relative to that at state point 1 (b) Drying rate Solution:
- 35. 1 2 65°C62°C 1.14 m3/kg d.a. 1.15 m3/kg d.a. 123.1 g/kg d.a. 124.5 g/kg d.a. 57°C
- 36. Evaporative cooling: Example 10 (a) At state point 1: T1 = 65°C Tw1 = 57°C Tdp1 = 56.3°C RH1 = 66.9% HR1 = 123.1 g/kg of dry air v1 = 1.15 m3/kg of dry air h1 = 387.7 kJ/kg of dry air At state point 2: T2 = 62°C Tw2 = 57°C Tdp2 = 56.5°C RH2 = 77.3% HR2 = 124.5 g/kg of dry air v2 = 1.14 m3/kg of dry air h2 = 387.7 kJ/kg of dry air
- 37. Evaporative cooling: Example 10 (b) ( )( )aDrying rate = HR w∆ a 2 VF w = v ( )( )VF = A air flow rate
- 38. Evaporative cooling: Example 10 l s l t t w l w t s V P S A P S S S P P B S + = − 2.5 3.6 0.6 A 3.6*0.019 0.019*0.038 3.6*1.2*0.025 0.6 + = − 2 A 1.47 m=
- 39. Evaporative cooling: Example 10 2 A 1.47 m= ( ) 3 3 m m VF = 1.47m 2 2.9 s s = ( )( )VF = A air flow rate
- 40. Evaporative cooling: Example 10 3 m VF =2.9 s 3 a 3 m 2.9 kg dry airsw 2.6 m s 1.14 kg dry air = = a 2 VF w = v
- 41. Evaporative cooling: Example 10 kg dry air g Drying rate 2.6 1.4 s kg dry air g kg 3.6 13.0 s h = = = a kg dry air w 2.6 s = ( )( )aDrying rate = w HR∆
- 42. Adiabatic Mixing of Moist Air Stream A psychrometric process that involves no net heat loss or gain during the mixing of two air streams
- 43. 1 3 2
- 44. Adiabatic mixing: Example 11
- 46. 2 3 1 0.95 m3/kg d.a.0.87 m3/kg d.a. 43.3°C 37.8°C 26.7°C 80%
- 47. Adiabatic mixing: Example 11 a VF w = v 3 a1 3 m 112 kg dry airminutew 117.9 m minute 0.95 kg dry air = = 3 a2 3 m 28 kg dry airminutew 32.2 m minute 0.87 kg dry air = =
- 48. Adiabatic mixing: Example 11 a2 a2 a1 line 1-3 w 32.2 = 0.21 line 1-2 w +w 32.2 117.9 = = + Therefore, length of line segment 1-3 is 0.21 times the length of line 1-2
- 49. 2 3 1 0.95 m3/kg d.a.0.87 m3/kg d.a. 43.3°C 37.8°C 26.7°C 80% 40°C 35.6°C
- 50. Adiabatic mixing: Example 11 T3 = 40.0°C Tw3 = 35.6°C