The psychrometric chart

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THE PSYCHROMETRIC CHART: Theory and Application
Perry Peralta NC State University

Identify parts of the chart Determine moist air properties Use chart to analyze processes involving moist air

Psychrometric chart: Example 1
Given: Required: T = 25°C Tw =20°C (a) RH, (b) Tdp, (c) HR, (d) v, (e) h

57.5 kJ/kg d.a.

20.0°C 17.6°C

63% 12.6 g/kg d.a.25°C 0.86 m3/kg d.a.


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



Sensible heating: Example 5

T3=70ºC T2=60ºC Tw2=50ºC Tw3=? RH2=58.8% RH3=?

287.0kJ/kg d.a. 275.5 kJ/kg d.a. 51°C 50°C



37 .


58 .8%

Heating and Humidifying
a psychrometric process that involves the simultaneous increase in both the dry bulb temperature and humidity ratio of the air

2 1 0

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 bulbtemperature. 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?

80 %

2 1

92 g/kg d.a. 18 g/kg d.a.


60°C 0.87 m3/kg d.a.

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

Cooling and Dehumidifying
a psychrometric process that involves the removal of water from the air as the air temperature falls below the dewpoint temperature



Cooling and dehumidifying: Example 9
Moist air at50°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.

115.7 kJ/kg d.a.

50.8 kJ/kg d.a. 28.8°C

32 %

25.2 g/kg d.a. 12.9 g/kg d.a.



Cooling and dehumidifying:Example 9
∆HR = (25.2 – 12.9) g water/kg dry air = 12.3 g water/kg dry air
drying rate wa = ∆HR 4 kg water h = kg water 0.0123 kg dry air kg dry air = 325.2 h

Cooling and dehumidifying: Example 9
∆h = (115.7 – 50.8) kJ/kg dry air = 64.9 kJ/kg dry air
q = ( ∆h ) ( w a )   kJ kg dry air  = 64.9  325.2  kg dry air   h   kJ = 21105.7 = 5.9 kW h

Adiabatic or Evaporative Coolinga 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

2 1

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 oflumber. 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. T=62ºC T=65ºC Tw=57ºC

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)...
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