Psychometric Properties and Processes

Dry Air Composition

Gas % by Volume Molecular Weight

Nitrogen 78.084 28.0134

Oxygen 20.9476 31.9988

Argon 0.934 39.943

Carbon Dioxide 0.0314 44.00995

Hydrogen 0.000001 2.02

99.997001

Molecular Weight of air = 28.9645 ≈ 28.97Molecular Weight of water = 18.02Density of air at 1.01325 Bar & 0°C = 1.293 kg/m3 Density of water at 4°C = 1000.0 kg/m3

Molar Volume of air = 22.4145 m3/kgmolUniversal Gas Constant = 8.314 J/gmolKDry Air Gas Constant = 287 J/kgKVapour (Steam) Gas Constant = 461 J/kgK

Mol. Wt. of H2O 18.02

Mol. Wt. of Air 28.9645 622.0622140.0

1. Humidity Ratio / Specific Humidityd.a.

vap

vap

d.a.

vap

d.a.

vv

V/vV/v

mmw

))((

)./).(/1()./).(/1(

vap

air

vap

air

airoair

vapovap

pp

MM

TMRpTMRp

vap

vaptotal

ppp X622.0

))()((

)(. 1 Tv

Tvpo

o

RMp

RM

pvap = Partial Pressure of Vappair = Partial Pressure of Airp = Total / Atmospheric Pressure

If Ptotal is constant say 1.01325 Bar then w = f (Pvap)

2. Degree of Saturation μ

wwsaturatedμ

)/(622.0 )/(622.0

vv

vsvs

pppppp

}{ vs

v

v

vs

pppp

pp

T-S Diag. for Steam

2

1

- - - - - - - - - - - - - - - - - - - - T

S

p vs

p v

3. Relative Humidity ϕ

ϕ = v

vs

vVvV

//

vs

v

vv

When Boyle’s Law holds good

i.e. when pv . vv = pvs . vvs

then ϕ =

v

vs

pp

- - - - - - - - - - - - - - - - - - - - T

S

p vs

p v

4. Dew Point Temperature td

- - - - - - - - - - - - - - - - - - -

td

T

S

2

1

Dew Point Temp.

Formation of FogCondensation starts on nuclei of dust or other floating particles in air

5. Enthalpy of Moist Airh = ha + w.hv 3

cpair = 1.005 kJ/kgK

cpwater(liq.) = 4.1868 kJ/kgK

cpvap = 1.88 kJ/kgK(from 0 to 60°C)

ha = hair = cp . t = (1.005)t kJ/kg 1

hv at A = cpw . td +(hfg)td + cpv(t-td)

= (4.1868)td +(hfg)td +1.88(t-td)

hv at A = hv at B

= (4.1868)X(0.0) + (hfg)0°C + 1.88(t-0.0)

= 2500 + (1.88)t kJ/kg 2

- - - - - - - - - - - - - - - - - -

- - - - - - - - - -

p v

t

td

0°C

A B

S

T

5. Enthalpy of Moist Air (contd.)

hmoist air = ha + w.hv at B

= 1.005t + w(2500+1.88t) kJ/kg of d.a.

HUMID SPECIFIC HEAT hmoist air = ha + w.hv

= (cpa + w.cpv).t + w.hfg (at 0°C)

= (cphumid).t + w.hfg (at 0°C)

where cphumid = (1.005+1.88w)

cphumid 1.0216 kJ/kg of d.a. (approximately)

6. Wet Bulb Temperature

• S

• T

- - - - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - -

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - t

td

p v

p vs p v1

p s1 = p v1

p s1

Wet WickConsidering Mass Transfermv = kd.A.(Pv’ – Pv ) 1Considering Heat Transfermv . h’fg = fg.A.(t-t’) 2

kd = mass tr. coeff. Based on partial pr. diff.fg = heat transfer coeff.

6. Wet Bulb Temperature (contd.)

Substituting for mv from 1 into 2

kd.A.(Pv’ – Pv).h’fg = fg.A.(t-t’)

t’ = t – (kd/fg).(Pv’ – Pv).h’fg 3

Here – kd is based on partial pr. diff.

Also t’ = t – (kw/fg).(w’-w).(h’fg) 4

where kw = mass tr. coeff. based on hum. ratios

Effect of Velocity of Air fg = hconvective + hradiation

At high velocity (hconv. + hrad.)/hconv. 1

7. Thermodynamic Wet Bulb Temp.

Air Alongwith Moisture Get Cooled from t1 to t2

• Sensible cooling of dry air from t1 to t2 = Cpa (t1-t2)

• Moisture w1 cooling from t1 to t2 = w1.Cpv.(t1-t2)

Supplied water gets heated from tw to t2 and evaporates

• Sensible heating of water = (w2 – w1).(t2-tw).Cpw

• Latent Heat for evaporation = (w2-w1).hfg at t2

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - WATER

t2, w2, h2t1, w1, h1

AIR

tw (w2-w1), hw (tw<t2 is assumed)

7. Thermodynamic Wet Bulb Temp.(contd.)Considering Energy BalanceCpa(t1-t2) + Cpv.w.(t1-t2) = Cpw.(w2-w1)(t2-tw)+(w2-w1).hfg at t2

t2 = t1 – (w2-w1){(t2-tw).cpw+hfg at t2}

cpa + cpv.w1

If t2 and tw are equal

t2 = t1 – (w2-w1).hfg at t2

cpa +cpv.w1

OR t* = t – (w*-w).h*fg

cpa+cpv.w

8. Wet Bulb & Thermodynamic Wet Bulb Temperature

t’ = t – (kw/fg).(w’-w).h’fg

t* = t – (h*fg/Cp).(w*-w)

Le = fg/kw.Cp

If Le=1 then t1=t*

For Air-Water Vap. Mixture Le 0.945

Various Psychometric Properties• Humidity Ratio or Sp.Humidity, (w)

w = 0.622 X = 0.622.ϕ.(pvs/pair)

• Degree of Saturation, μμ

• Relative Humidity ϕ

• Dew Point Temp.• Enthalpy of Moist Air

h = 1.005t + (2500 + 1.88t).w• Wet Bulb Temperature• Thermodynamic Wet Bulb Temperature

vap

vaptotal

ppp

}{ vs

v

v

vs

pppp

pp

/). 1(1

622.0 . pp

ppvs

wpp vs

airv

s

Psychometric Chart

Enthalpy

R.H., ϕ

Dry BulbTemp.

Dew Point Temp.

Moisture Content (w)

Wet Bulb Temp. t*

HumidSp. Vol.

Various Processes Using Chart1. Sensible Heating and Cooling (Constant w)

td t2 t1

3 2 1

Standard Air20°C and 50% R.H.ρ = 1.2 kg/m3 of dry airCp = 1.0216 kJ/kg pf d.a.ma = cmm. ρ/60 kg of d.a./s

Qs (cooling) = ma.cp.(t1-t2)

= ma.cpa.(t1-t2) + ma.w.cpv.(t1-t2)

= ma.(h1-h2)

= (cmm. ρ/60).cp.∆t

= (cmmx12x1.0216/60). ∆t kW

= {0.0204.cmm. ∆t} kW

Lowest temp. possible is dew point temp. td

2. Humidification & Dehumidification (constant t)

h2

h1

w2

w1

dbt

2

1

Heat Transfer to air (Latent Heat)= ma.(h2-h1)= ma.{(Cp.t2+hfg(0°C).w2) –

(Cp.t1+hfg(0°C).w1)}= cmm/60. ρ.(w2-w1). hfg0°C

= 50.cmm.(w2-w1) kW

3. Total Heat Process

Q = Qsensible + Qlatent

= 0.0204.cmm. ∆t + 50. ∆w.cmm

= (0.0204. ∆t + 50. ∆w).cmm kW

h1 h2

h3

1 2

3 w3

w1

t1 t2=t3

Sensible Heat Factors (S.H.F.)SHF =

=

=

s

l s

Q Q Q

12

13

hhhh

0.0204. ) 50. 0204.0(

Mixing of two Streams

w1

w3

w2

1

3

2

h1

h2

h3

2mm1

mm

.hm.hmm 3

..m 3

3a3 2a2 1a1

a3a2 1

h. h.

h also

.wm .wm .wm

m m m

a2

a3

a1

a3

2a21a1

a3

2211

a3

wmwm

a

aaw

Now, h=1.005t + w(2500+1.88t) = Cp.t + hfg0°C.w

h3 = cp.t3 + hfg0°C.w3

= (ma1/ma3).(Cp.t1+ hfg0°C.w1) + (ma2/ma3)(Cp.t2+ hfg0°C.w2)

}w{ t 3.wm

m .wm

m hC

.tm.tmm 3

2a2

a3

1a1

a3

fg0C

p

2a21a1

a3

The second term will be 0 only if Cp is considered as constant which is not

.tm.tm

mm 32a21a1

a2a1t

Mixing With Condensation

w4 = w3 – wc

? w, ? t

.hw h

w w

44

f4c.hm .hm

mm 4

c.wm .wm

mm 4

2211

21

2211

21

1

2

34

w3

w4

t3

13

2

wc

4

1

2

ITERATIVE PROCEDURE

1. Assume value of t4 slightly more than t3

2. Find w4 and h4 for saturated air at t4

3. Obtain hf4 from steam tables

4. Find wc using equation 1

5. Use this wc and hf4 in equation 2 (equation 2 should be valid)

6. Otherwise assume another t4 and repeat

- - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - - -

Inlet Stream

Pump

Heating or

CoolingMake-up water

~~~~~~~~~

~~~~~~~~~

Outlet Stream

EliminatorPlates

Spray Chamber Process

Means Of Humidification

• Spray Chamber with large no. of droplets and cross flow of moist air

• Flow of air over large wetted surface• Injection of aerosol size droplets directly

If Mw is mass of moisture addedm1.w1 + Mw = m2.w2

m1.h1 + Mw.hw = m2.h2

hair1=1.005t1 + w1(2500+1.88t2)hair2 = hair1 + w2(2500+1.88t2)

21

S

Water Injected at 0°C

1. h1=h2’ (Process is isenthalpic)

2. tw=wBT

Le=1 (Process is constant wBT)

3. tw = 100°C

(In this case max. heating possible)

Max. Angle Between Processes 1 & 3 is just 7°

1

7°

tw = 100°C

tw = wBT

tw = 0°C

Various Processes on Psychometric Chart

2A

2B

2C

2D

2E

2F2G

tsc

t1 ’=wBT

w1

td t1

1-2A Sensible Cooling at Constant wTemp. of cooling water >= td

1-2B Humidification at Constant DBTWater is externally heated

1-2C Cooling with humidificationWater is externally heated

WBT < tsc <t1 Enthalpy Increases though DBT Falls

1-2D Adiabatic Saturationtsupply = t1’ and water continuously re-circulated

1-2E Cooling and HumidificationWater externally cooled, enthalpy decreases

1-2F Cooling and Dehumidification (tsupply<td)

1-2G Lowest Cooling Coil Temp. that can be used

Heating with Dehumidification with absorbent

1-2 Ideal Process (without heat of adsorption)1-3 Actual Process

1

2 3

t1’=t2’

Adsorption process occurs in following steps• Process of dehumidification along constant wBT• Moisture condenses on surface• Condensate enters pores because of vap. pr. diff. by

capillary action• Slowly vap. pr. diff. decreases• When vap. pr. diff. equals zero adsorbent is saturated

Ideal process

Actual Process

• As condensation occurs, latent heat is released and air DBT increases (1-2)

• Additionally heat of adsorption is also released, thus further increasing DBT along (1-3)

1

2 3

t1’=t2’

Chilled Hygroscopic Solution for Dehumidifying

Brines and glycol solutions exert quite low vapour pressure.

1. Thus, initially the partial pr. diff. is quite large and hence rate of moisture removal is high.

2. As the solution/brine becomes weaker the partial pr. diff reduces and the process slows down.

3. The process thus cannot be continued over long time period.

brine

12

3

ts

ts = temp. of cold surface or cold brine

General A/C Process

RSH = Qs = 0.0204 cmm (ti-ts) 1

RLH = QL = 50 cmm (wi-ws) 2

ts, ws, cmm are three unknowns

Therefore, normally ts is fixed & ws & cmm are obtained

ts, ws

cmm ROOM

ti, wi

A/C Apparatus

Qs

QL

Summer A/C Process

Dehumidified rise = (ti-ts1)

Dehumidified air quantity = cmms1

Bypass factor = X = hs1-hs

hi-hs

tADD = Apparatus Dew PointRSHF = RSH .

RSH + RLH

RSH = Qs = 0.0204 cmm (ti-ts)cmmmin = RSH .

0.0204(ti-tADP) = RLH . 50 (wi-wADP) = RTH .

0.02 (hi-hADP)

tADP

S S1

i

Summer A/C With Ventilation

Room Load RSH RLH RTH

Ventilation Air(Outside Air) Load

OASH OALH OATH

A/C Apparatus(Load = Q)

(RSH + OASH) = TSH (RLH + OALH) = TLH (RTH + OATH)

RSHF2 i

GSHF 1

0

ms

ROOMti, wi

mi

mo

ms - mi

1 2

Q = ms (h1-h2)

= (mi.hi + mo.ho) – ms.h2

= {ms-mo}hi + mo.ho – ms.h2

= ms(hi-h2) + mo{ho-hi}

= Room Load + Ventilation Air Load

Grand Sensible Heat FactorGSHF = TSH .

TSH + TLH Even though ventilation heat load is an additional load on apparatus, for the room its effect is not felt and hence supply air condition stilllies on the RSHF line as well as on GSHF line.

RSHF = Qs/QL

Minimum dehumidified air quantity = cmm at 2*When ventilation is considered without B.P.F., the mixed air (mo+mi) is given by state 1* for minimum dehumified quantity.

Summer A/C with Ventilation & B.P.F

ROOMti, wi

mi

mo

mi

1 2

Qs

QL

2*

S

2 1

i

0

1*

RSHF

ESHF

GSHF

When mo is decided on other considerations and with BPF for A/C apparatus

with trials we fix state 2 such that

After fixing state 2, dehumidified air quantity cmmd is obtained

Thus, knowing cmmd – cmmi = cmmo

the actual state 1 of mixture is obtained.

Extending line 1-2 we get required cold surface temperature.

Line joining i and s gives Effective SHF.

BPF BPF)(1

S-22-1