Psychrometrics 1) Ideal Mixing 2) Ideal Gas Air 3) Ideal Gas Water .

Psychrometrics 1) 2) 3) 4) Ideal Mixing Ideal Gas Air Ideal Gas Water Vapor Adiabatic Saturation 1

Ideal Gas Law AVOGADRO'S LAW One (1) mole of any gas 22.4 liters. 6.023 1023 molecules /mole of gas at STP (1 atm and 0o C ) BOLYES LAW p1 v1 p 2 v 2 CHARLES LAW v1 T1 v 2 T2 p1 T1 p 2 T2 IDEAL (PERFECT) GAS LAW pv RT pV mRT p - absolutepressure, psia, kPa T - absolute temperature, o R,o K R* R molecularweight lbf/lbmo R R 1545.15 lbmole kJ kPa m3 * R 8.314 or o kmole K kmole o K * mass moles MolecularWeight m n MolecularWeight pv R *T pV nR*T 2

The specific volume of air at 75o F and 14.7 psia ( RT 53.35 ft lbf/lbm R 459.69 o R 75o F v p 14.7 lbf/in 2 144 in 2 /ft 2 ) v 13.476 ft 3 /lb The specific volume of air at 24 o C and 101.325 kPa ( RT .287 kPa m 3 /kg 273.15o K 24o C v p 101.325 kPa 2 ) v .8417 m 3 /kg 3

æ ¶T ö çç JT Coefficient è ¶p ø h h constant 4

Specific and Relative Humidity Basis of calculation - 1 lb m dry air, 1 kgm dry air Specific Humidity, Mass Fraction, ω mass water vap or mass dry air æç pV ö mass water vap or è RT ø water p water MW Twater ω mass dry air p air MW Tair æç pV ö è RT ø air 18 p w ω (12.43) 29(p ambient - p w ) Relative humidity f actual mass water vapor mass water vap or at saturation æç pV ö actual mass water vap or è RT ø f mass water vap or at saturation æ pV ö ç RT è øsaturation p w vg f (12.44) pg v w H H air H w BTU/lb dryair , kJ/kg dryair h h a ω h v c p T ω h v c p T ω c pv T 5

Metric Psychrometric Chart Moran Table A-9, A-9E 6

Given, 25o C, f 50 %, barometer 100 kPa. Find : a) p air c) specific humidity, ω b) specific volume a) f pw pg p w f p g .5 3.1698 ω p w 1.5849 kPa p a p atm - p w 98.415 kPa b) ω 18 p v 1.5849 .622 29 (p atm - p v ) (100 - 1.5849) v c) v v air R a T/p a kPa 289.15 K 3 m K v .843 kg/m 3 (100 - 1.5849) v v vapor ωR v T/p v .287 v .001 .4615 kPa 289.15 K 3 mK .842 kg/m 3 1.5849 7

Given: lb moist air, 75o F, f 50%, barometer14.69 psia Find : pw, w, va , ma , mv f pw pgat 75 pw .5 .43016 .21508psia ω 18 pw .21508 .622 29 patm - pw 14.69- .21508 ω .00924lb water /lb dry air V Va Vw pv RT Va m RT 53.35 (459.69 75) 1 13.685ft3 p (14.69- .21508) 144 pa Va (14.69- .21508) 144 1ft3 ma .0731lb RT 53.35 (459.69 75) pv Vv .21508 144 1ft3 mv .000675lb RT 85.766 (459.69 75) mv .00924lb water /lb dry air (check) ma Tdp T at pv .21508 55.11 F o Psychrometric Chart ω .093 lb water/lb dry air h 28.5 BTU/lb dry air T dp 55o F v 13.68 ft 3 / lb 8

Adiabatic Saturation h g1 1 db 2 wb Tdb h f2 basis of calculation 1 mass unit dryair Steady Flow Energy Equation E air vapor in E water added E air vapor out 2 Twb T 3 T 0 h v1 1 4 h g2 s h a1 ω1h v1 (ω2 - ω1 )h fg2 h 2a ω 2 h v2 1) h v1 saturation vapor enthalpy @(T Tdb , p p v ) substituti ng, (h a1 - h a2 ) c p (T1 - T2 ) 3) h v1 h g@ Twb .45 (Tdb - Twb ) (h v2 - h f2 ) h fg c p (T2 - T1 ) ω1 (h v1 h f2 ) ω2 h fg2 ω h - c (T - T ) ω1 2 fg2 p 2 1 (h v1 -h f2 ) ω wb h fg2 - c p (Tdb - Twb ) ω (h v db -h f wb ) 2) h v1 h g@ Tdb 4) h v1 h g@ 0 F .44 (Tdb - Twb ) 1061.8 .44 (Tdb ) for 75F db,70 F wb 1) h v1 1094.07 BTU/ lb dry air 2) h v1 1093.9 BTU/lb dry air Exact easiest to use 3) h v1 1094.05BTU/lb dry air most accurate 4) h v1 1094.80BTU/lb dry air 9

Adiabatic Saturation basis of calculation 1 mass unit dryair 1 db 2 wb Steady Flow Energy Equation E air vapor in E water added E air vapor out h a1 ω1h v1 (ω 2 - ω1 )h fg2 h 2a ω 2 h v2 substituti ng, (h a1 - h a2 ) c p (T1 - T2 ) (h v2 - h f2 ) h fg c p (T1 - T2 ) ω1 (h v1 h f2 ) ω 2 h fg2 ω 2 h fg2 - c p (T1 - T2 ) ω1 (h v1 -h f2 ) ω wb h fg2 - c p (Tdb - Twb ) ω (h v db - h f wb ) 10

95 F db 65 F wb 14.7 atm Specific Humidity ? T p Tdb 95 hf h fg Twb 65 F w(Twb ) hg Tdb 95 F 1102.6 Twb 65 .30578 33.08 1056.5 ω(Twb ) .622 ω(Tdb ) ω(Tdb ) p saturation (Twb ) .30578 .622 .0132 lb water /l b dry air p atm - p saturation (Twb ) 14.7 - .30578 ω(Twb ) h fg (Twb ) c p (Twb - Tdb ) h g (Tdb ) - h f (Twb ) .0132 1056.5 .24 (65 - 96) 13.946 - 7.2 .0063 lb water /l b dry air 1102.6 - 33.08 1069.52 11

h g1 Tdb 2 Twb T h f2 3 T 0 h v1 1 4 h g2 s 1) h v1 saturation vapor enthalpy @(T Tdb , p p v ) 2) h v1 h g@ Tdb 3) h v1 h g@ Twb .45 (Tdb - Twb ) 4) h v1 h g@ 0 F .44 (Tdb - Twb ) 1061.8 .44 (Tdb ) for 75F db,70 F wb 1) h v1 1094.07 BTU/ lb dry air 2) h v1 1093.9 BTU/lb dry air 3) h v1 1094.05BTU/lb dry air Exact easiest to use most accurate 4) h v1 1094.80BTU/lb dry air 12

Air at 40 F db, 35F wb is heated and humidified to 70 F db, 40 % relative humidity at 14.7 psia. Find the mass of water added. 35 F 1 40 F 1g 35 F 2 .09998 .00426 14.7 - .09998 c p (T1 - T2 ) ωg1 h fg at 35 1 ω1g .622 ω1 h v at 40 - h l at 35 .24(35 - 40) .00426 1073.5 ω1 1078.7 - 3.004 ω1 .00314 lb water/ lb dry air ω 2 .622 pv p atm - p v p v2 f p g2 .4 .36335 .1453 in Hg .1453 .00621 14.7 - .1453 water added ω 2 .622 ω 2 - ω1 .00649 - .00326 .00323 lb water lb dry air 13 3-10

USCU Psychrometric Chart 14

ENTHALPY h h air ωh vapor To Build a Psychrometric Chart h vapor h g @Tdb select p atm h .24 Tdb ω (1061.8 .44 Tdb ) (1) repeatedly select h and ω Twb h 1.005 Tdb ω (1061.8 1.8 Tdb ) with h find Tdb from (1) p atm SPECIFIC HUMIDITY p atm p vapor p air with ω find p vapor from (2) p vapor ω Tdb p air h v ω f ω p vapor 18 p vapor .622 29 p air p atm - p vapor (ω h fg ) wb c pair (Twb - Tdb ) h vapor - h liquid db (2) SPECIFIC VOLUME RT v v air 1 db p air R T ω air db p vapor with p vapor find φ from (4) with Tdb, p vapor find v from (5) (3) Common HVAC State Point Specifications Twb andTdb wb RELATIVE HUMIDITY p vapor f p saturation @ Tdb v v vapor with ω find Twb from (3) by itteration (4) (5) f andTdb Metric English c p air 1.005 .24 c p liquid 4.18 1.0 water (6) c p water 1.8 vapor .44 15

PSYCHROMETRIC PROCESSES CoolingDehumidification Heating-Cooling w w Tdb Tdb Mixing Humidification w Tdb w Tdb 16

Heating Cooling ω heating heating Tdry bulb Q cooling ω constant p w constant φ, Tdry bulb , v varying Q ΔH ΔH air ΔH water Q m dry air Δh air m dry air ω Δh water vapor Q m dry air c p ΔT m dry air ω c p water ΔT, kJ, BTU Q m dry air .24 ΔT m dry air ω .45 ΔT, kJ, BTU 17

Cooling Dehumidification 1 Q 1 2 2 Tdp 1 apparatus 2 Δω kg water kg dryair Q ΔH dryair ΔH water vapor ΔH condensate condensate ω1 - ω 2 , q m dryair æ ö Δh air ω 2Δh water vapor (ω1 - ω 2 )çç (h1 - h1 ) (h1 - h 2 ) liquid liquid ø è vapor liquid q sensible Δh m dry air air ω 2 (h v1 - h v2 ) (ω1 - ω 2 )(h1 - h2 liquid ) liquid Q sensible c p ΔTdry air ω 2 c p water ΔTdry air (ω1 - ω2 )c p liquidΔTdry air air vapor Q latent (ω1 - ω 2 )(h1 vapor - h1 ) (ω1 - ω 2 )h fg liquid Sensible Heat Factor, SHF water Q sensible Q sensible Q latent 1 18

Conditioned Space R - return, space conditions mass water mass dryair Q ΔH dry air ΔH water vapor ΔH condensate water gain Dω, S - supply q sensible Δh air ωSΔh water vapor (ω R h R vapor - ωS h S vapor ) m dry air q sensible Δh air ωSΔh water c p ΔTdry air ωSc p ΔTdry air (ω R - ωS )c p ΔTdry air m dry air vapor air water liquid vapor water q latent (ω R - ωS )(h R - h R ) (ω R - ωS )h fg m dry air vapor liquid R Sensible Heat Factor, SHF Q sensible Q sensible Q latent 19

Adiabatic Humidification 1 2 2 H1 H w H 2 m a h1 m a h w m a h 2 1 2 h1 (ω2 - ω1 )h w h 2 h 2 - h1 hw ω2 - ω1 Qsensible 32o F water h w 0 BTU/lb 1 30 psi steam h w 1164 BTU/lb Δh hw Δω Q sensible m dry air Δh air m dry air ω1Δh water vapor Q latent (ω 2 - ω1 ) (h 2 vapor - h1 vapor ) steam humidification Q latent (ω 2 - ω1 ) (h 2 vapor - h1 liquid ) water wash 20

q l Δh Δω h fg q s Δh s Δh - Δh l SHF SHF qs ql qs Δh - Δω h fg Δω h fg Δh - Δω h fg ql q qs Δh - h fg SHF Δω Δh h fg - h fg Δω h fg Δh Δω SHF - 1 21

Mixing 2 3 2 m m h m m1h 1 m 2 h 2 m1 m2 hm h1 h2 (m1 m 2 )3 (m1 m 2 ) m1 m2 ωm ω1 ω2 (m1 m 2 ) (m1 m 2 ) ω 1 1 H m H1 H 2 2 3 Tdb m 1 Δq1 m Δq1 m m1c p (Tm - T1 ) m 2 c p (T2 - Tm ) m1 (T2 - Tm ) m 2 (Tm - T1 ) 22

10 cubic meters per second of 10 C db, 5 C wb air is mixed with 6 cubic meters per second of air at 25 C db and 18 C wb. Compute the mixture conditions and compare them to results from the psychrometric chart. Pt 1 - 10 c db, 5 c wb .622p v@ Twb ω g1 p atm - p v 6 7 kg/sec .858 m3 /kg h 2 50.9 kJ/kg m2 m m ω m m1ω1 m 2 ω 2 m m m1 m 2 ωm 1.005 kJ/kgK (5 - 10) (.0054 kg/kg 2489.1 kJ/kg ) ω1 2510.1 kJ/kg - 21.02 kJ/kg ω1 .00338 kg/kg p a V (101.325 kPa - .8725 kPa ) 10 m /sec RT .287 kPa m 3 /kg (273.15 10) m a1 12.36 kg/sec 3 pa V ( .8725 kPa ) 10 m 3 /sec m v1 .0668 kg/sec RT .4615 kPa m 3 /kg (273.15 10) m m 12.36 .0668 12.427 kg/sec h1 c pa (Tdb ) ω1 h v1 h1 1.005 kJ/kg K 10 .00338 kg/kg 2519.2 kJ/kg h1 18.56 kJ/kg v 2 .858 m 3 /kg Calculated Mixing .622 .8725 kPa ω g1 .0054 kg/kg 101.325 kPa - .8725 kPa c (T - Tdb ) (ω g1 h fg )@Twb ω1 pa wb h v@Tdb - h w @Twb m a1 Chart, Pt 2 25 C db, 18 C wb ω 2 10 g/kg m1 m ω1 2 ω 2 mm mm 12.36 7 .00338 .010 19.36 19.36 ω m .00577 kg/kg ωm 12.36 7 18.56 kJ/kg 50.9 kJ/kg 19.36 19.36 h m 30.25 kJ/kg hm Chart Mixing distance from point 2 to point m 12.361 2.65 in 1.686 in 19.428 read ω m 6g/kg and h m 30.5 kJ/kg 23 3-26

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51.8 BTU/lbm 24 BTU/lbm h h dry air ωh vapor at 100O , f 0% h .24 100 0. 1104.7 24 BTU/lb dry air at 100O , f 60% h .24 100 .0252 1104.7 51.84 BTU/ lb dry air 100 F 25

Combined Processes - mixing, cooling and dehumidification, reheating, conditioned space. 1 1 Outside air R Return air from conditionded space 2 w 2 After mixing 3 After dehumidifucation R 3 S S Supply Tdb 1 2 3 S R Q water Q 26

Combined Air Processes O 2 S 4 R 2 S 3 1 O R Winter - Heating Summer - Cooling cool 3 O o S 2 1 4 Pre-heat R Heat o Humidify 27

Natural Draft Cooling Tower Air water vapor out Hot water Air water vapor in Cold water makeup blowdown Mass Balance Energy Balance m air in m airout net energy lost by liquid water mass water in - mass water out m hot water m makeup m air ωin energy gained by air water vapor m cold m blow down m air ωout water m makeup m air (ωout - ωin ) 28

Cooling Tower Example An evaporative counterflow air cooling tower ω1 removes1 10 BTU/hrfrom a waterflow. 6 18 .6 p g90 o F ( ) 18 .6 .69904 29 (14.3 - .6 .69904 ) The temperature of the wateris reducedfrom120o F 29 14.3 - .6 p g90 o F to110o F. Air enters the cooling towerat 90o F and ω 1 .01876 lb water/lb dry air 60% relative humidity, and the air leaves at 100o and 82% relative humidity. Calculatea) the air flow rate and b) the quantityof makeup water. 2 100 F 82% 1 18 .82 .95052 .03578lb water/lb dry air 29 (14.3 - .82 .95052 ) Mass and Energy Balance ω2 ΔQ air ΔQ water 1,000,000B TU/hr Assume makeup and blowdown at 110 o F Q dryair m a c p ΔT m a .24 (100 - 90 ) 2.4 m a 120 F Q ω1 m a ω 1 c p ΔT m a .01876 .45 (100 - 90) 90 F Q ω1 .0844 m a Q Δω m a (ω 2 - ω 1 ) (h v 100 o F - h f 120 o F ) 60% Q Δω m a (.03578 - .01876 ) (1104.7 - 88. ) 2 Q Δω 17.3 m a 1,000,000 Q drtair Q ω1 Q Δω 19.78 m a 110 F m a 50,556 lb dry air/hr 1 mass air m a ω1 m a (1 ω 1 ) m a 51,504lb/h r make up m a (ω 2 - ω 1 ) 51,504 (.03578 - .01876 ) make up 876.6 lb/hr 29

To Build a Psychrometric Chart db, vapor vapor wb vapor db atm db wb db andT T andT Common HVAC State Point Specifications f f É v h p p p T T air vapor atm wb db c 1.8 .44 c 4.18 1.0 c air 1.005 .24 Metric English vapor p water water p liquid p