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

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

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

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

12 57.5 kj/kg d.a C 63% 17.6 C 12.6 g/kg d.a. 25 C 0.86 m 3 /kg d.a.

13 PSYCHROMETRIC PROCESSES

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

16 Sensible heating: Example 5

17 T 2 =60ºC T w2 =50ºC RH 2 =58.8% T 3 =70ºC T w3 =? RH 3 =?

18 287.0 kj/kg d.a kj/kg d.a. 51 C 50 C 58.8% % 60 C 70 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 2 1 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 52 C 80% g/kg d.a. 18 g/kg d.a. 27 C 60 C 0.87 m 3 /kg d.a.

23 Heating and humidifying: Example 7 HR = ( ) g/kg dry air = 74.0 g/kg dry air w a1 = drying rate/ HR = (12.5 kg/hour)/(0.074 kg/kg dry air) = kg dry air/hour VF 1 =(w a1 )(v 1 ) =(168.9 kg dry air/hour)(0.87 m 3 /kg dry air) = 147 m 3 /hour = 2.45 m 3 /minute

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

25 2 1

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 m 3 of red oak lumber is 4 kg/hour, determine the kw of refrigeration required.

27 115.7 kj/kg d.a kj/kg d.a C 2 32% g/kg d.a g/kg d.a. 18 C 50 C

28 Cooling and dehumidifying: Example 9 HR = ( ) g water/kg dry air = 12.3 g water/kg dry air w a = = drying rate HR 4 kg water h kg water kg dry air = kg dry air h

29 Cooling and dehumidifying: Example 9 h = ( ) kj/kg dry air = 64.9 kj/kg dry air q = h w a ( )( ) kj kg dry air = kg dry air h kj = = 5.9 kw h

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 2 1

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 m 3. 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.

33 T=65ºC T w =57ºC T=62ºC

34 Evaporative cooling: Example 10 Given: T 1 = 65 C; T w1 = 57 C Adiabatic cooling to T 2 = 62 C Air flow rate = 2 m/s Volume of lumber = 2.5 m 3 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 57 C g/kg d.a g/kg d.a. 62 C 65 C 1.14 m 3 /kg d.a m 3 /kg d.a.

36 Evaporative cooling: Example 10 (a) At state point 1: T 1 = 65 C T w1 = 57 C T dp1 = 56.3 C RH 1 = 66.9% HR 1 = g/kg of dry air v 1 = 1.15 m 3 /kg of dry air h 1 = kj/kg of dry air At state point 2: T 2 = 62 C T w2 = 57 C T dp2 = 56.5 C RH 2 = 77.3% HR 2 = g/kg of dry air v 2 = 1.14 m 3 /kg of dry air h 2 = kj/kg of dry air

37 Evaporative cooling: Example 10 (b) Drying rate = ( HR)( w ) a VF w= a v 2 VF = ( A)( air flow rate)

38 Evaporative cooling: Example 10 V Pl + S s A= Pl St St Sw PP l w Bt Ss A = 3.6* * *1.2* A= 1.47m

39 Evaporative cooling: Example 10 2 A= 1.47m VF = ( A)( air flow rate) ( 3 ) m m VF = 1.47 m 2 = 2.9 s s 3

40 Evaporative cooling: Example 10 3 m VF =2.9 s VF w= a v 2 3 m 2.9 w = s a m = 1.14 kg dry air kg dry air s

41 Evaporative cooling: Example 10 w = 2.6 a kg dry air s Drying rate = w ( )( ) a HR kg dry air g Drying rate = s kg dry air g kg = 3.6 = 13.0 s h

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 3 2 1

44 Adiabatic mixing: Example 11

45 T 2 =26.7ºC RH 2 =80% VF 2 =28 m 3 /min T 3 =43.3ºC T w3 =37.8ºC T 1 =43.3ºC T w1 =37.8ºC VF 1 =112 m 3 /min

46 37.8 C % 26.7 C 43.3 C 0.87 m 3 /kg d.a m 3 /kg d.a.

47 Adiabatic mixing: Example 11 a VF w= v 3 m 112 w = minute a m = 0.95 kg dry air kg dry air minute 3 m 28 w = minute a m = 0.87 kg dry air kg dry air minute

48 Adiabatic mixing: Example 11 line 1-3 w 32.2 = = line 1-2 w +w a2 = 0.21 a2 a1 Therefore, length of line segment 1-3 is 0.21 times the length of line 1-2

49 37.8 C 35.6 C % 26.7 C 40 C 43.3 C 0.87 m 3 /kg d.a m 3 /kg d.a.

50 Adiabatic mixing: Example 11 T 3 = 40.0 C T w3 = 35.6 C

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