1. Ideal Gas Mixture and Psychrometric Applications
Chapter 12:
Ideal Gas Mixture and Psychrometric Applications

2. Cooling Towers, 2000MW coal-fired Cottam Power Station, Nottinghamshire. Photo courtesy of FreeFoto.com

3. Mixture Composition For Gases: Dalton’s Model Amagat’s Model
Describing mixture composition
For Gases:
Dalton’s Model
Amagat’s Model

4. Ideal Gas Properties for Mixtures
Evaluated at the mixture temperature
Evaluated at the mixture temperature, and appropriate partial pressure, pi

5. Mixture Analysis (non-reacting)

6. mw_ar=molarmass(argon) mw_et=molarmass(ethane)
Example Problem-1
Two insulated tanks A and B are connected by a valve. Tank A has a volume of 1
m3 and initially contains argon at 300 kPa, 10°C. Tank B has a volume of 2 m3
and initially contains ethane at 200 kPa, 50°C. The valve is opened and remains
open until the resulting gas mixture comes to a uniform state. Determine the final
pressure and temperature.
v1_ar=volume(ar,p=300,t=10)
m1_ar=1/v1_ar
mw_ar=molarmass(argon)
mw_et=molarmass(ethane)
n=m1_ar/mw_ar+m1_et/mw_et
v1_et=volume(c2h6,p=200,t=50)
m1_et=2/v1_et
u1=intenergy(ar,t=30)*m1_ar+intenergy(c2h6,t=50)*m1_et
u2=intenergy(ar,t=t2)*m1_ar+intenergy(c2h6,t=t2)*m1_et
u1=u2
pf*3=n*r#*(t2+273)

7. Example problem-2: A mixture of 2 kg of oxygen and 2 kg of argon is in an insulated piston cylinder arrangement at 100kpa , 300K. The piston now compresses the mixture to half its Initial volume in a reversible adiabatic process. Find the final pressure, temp and the work done by the piston.
no2=2/32; nar=2/39.94
yo2=no2/(no2+nar); yar=nar/(no2+nar)
u1=intenergy(o2,t=300)*2+intenergy(ar,t=300)*2
u2=intenergy(o2,t=t2)*2+intenergy(ar,t=t2)*2
s1=entropy(o2,t=300,p=100*yo2)*2+entropy(ar,t=300,p=100*yar)*2
s2=entropy(o2,t=t2,p=p2*yo2)*2+entropy(ar,t=t2,p=p2*yar)*2
s1=s2
v1=volume(o2,t=300,p=100*yo2)*2+volume(ar,t=300,p=100*yar)*2
v2=volume(o2,t=t2,p=p2*yo2)*2+volume(ar,t=t2,p=p2*yar)*2
v2=v1/2
w+u2-u1=0

8. Dehumidification

9. Ideal Gas Cases For Ideal Gases using Tables A-22, A-23:
Assuming Constant Specific Heats:

10. Psychrometric Principles
Dry air + Water vapor = Moist Air
Humidity Ratio
Relative Humidity

11. Psychrometric Properties
Mixture Enthalpy
Water Vapor Enthalpy
Water Vapor Entropy
Dew Point Temperature
Sling Psychrometer picture courtesy of USA Today

12. Relative Humidity f Mixture Enthalpy Wet Bulb Temp. Humidity Ratio w
ASHRAE “Comfort Zone”
HUMID
HOT
COLD
DRY
Dry Bulb Temp.
Copyright ©
John Wiley & Sons Ltd.

13. Humidification

14. Use of psychrometric chart
{find humidity ratio and wet bulb temp; given dry bulb temp =25C, r=.8}
{w=humrat(airh2o,p=100,t=25,r=.8)
b=wetbulb(airh2o,p=100,t=25,r=.8)}
{ given dry bulb temp=25C, wet bulb temp=23 find r and w}
{w=humrat(airh2o,p=100,t=25,b=23)
r=relhum(airh2o,p=100, t=25, b=23)}
{ given w=.008, wet bulb temp=17C find r and dry bulb temp}
{r=relhum(airh2o,p=100,w=.008, b=17)
t=temperature(airh2o,p=100, w=.008, b=17)}
{ given wet bulb temp=15c, r=.6 find w and dry bulb temp}
w=humrat(airh2o,p=100, b=15, r=.6)
t=temperature(airh2o, p =100, b=15, r=.6)
Example problem-1
A class room of IIT kgp is of size 10m x 8m x 3.5m. In a summer day of room temp 34C and relative humidity of
85% what is the amount of water vapor present in the class room, amount of dry air? Find the dew point and
humidity ratio. If the room is cooled to 22C (p =const) then how much of water has condensed?

15. v1=volume(airh2o,p=100, t=34, r=.85)
w1=humrat(airh2o,p=100, t=34, r=.85)
m1=vol/v1
m_vap=m1*w1 { water vapor present }
t_dew=dewpoint(airh2o,p=100, t=34, r=.85)
w2=humrat(airh2o,p=100, t=22, r=1)
m_wat=m1*(w1-w2) {water that condensed}
Example problem-2 : To an air conditioner air enters at 105kPa, t=30C, r=80% and comes
out at 100 kPa, t=15C, r=95%. Compute the heat transfer per kg of dry air
ha1=enthalpy(airh2o,p=105,t=30,r=.8)
ha2=enthalpy(airh2o,p=100,t=15,r=.95)
w1=humrat(airh2o,p=105,t=30,r=.8)
w2=humrat(airh2o,p=100,t=15,r=.95)
hl2=(w1-w2)*enthalpy(water,t=15,x=0) { enthalpy of condensed water}
ha1+q_cv=ha2+hl2

16. Example prob-3: A 0.5m^3 tank contains nitrogen and water vapor at 50C , 2MPa. The partial
Pressure of water vapor is 5kpa, If the tank is cooled to 10C find the heat transfer from the tank.
vw1=volume(water,p=5,t=50)
mv1=.5/vw1 { mass of wat vap}
vn1=volume(n2,p=1995,t=50)
mn1=.5/vn1 { mass of n2}
pv1=pressure(water,t=10,x=1) { assume saturated wat vap and see if final wat is less than the initial}
vw2=volume(water,x=1,t=10)
mv2=.5/vw2 { final mass of wat vap}
ml2=mv1-mv2 {liquid that has condensed}
u1=intenergy(water,t=50,p=5)*mv1+intenergy(n2,t=50)*mn1
u2=intenergy(water,t=10,x=1)*mv2+intenergy(n2,t=10)*mn1+ml2*ul2
ul2=intenergy(water,t=10,x=0)
q=u2-u1

17. w2=humrat(airh2o,p=100,t=20,r=1) w1=humrat(airh2o,p=100,t=30,r=r1)
Example problem-4
A mixture of air and wat vapor enters an adiabatic saturator at 0.1MPa, 30C and leaves the
at 20C, which is the adiabatic saturation temp. Compute the humidity ratio and relative
Humidity of the mixture at the entrance.
w2=humrat(airh2o,p=100,t=20,r=1)
w1=humrat(airh2o,p=100,t=30,r=r1)
h1=enthalpy(airh2o,p=100,t=30,r=r1)
h2=enthalpy(airh2o,p=100,t=20,r=1)
hl2=enthalpy(water,p=100, t=20) {hl2 to be computed at exit condition}
h1+(w2-w1)*hl2=h2
Air+vap
Air + saturated vapor
water
Adiabatic saturation process

18. w1=humrat(airh2o,p=100,t=5,r=1) mv1/(100-mv1)=w1 ma1=100-mv1
Example Prob-5: A piston cylinder has 100kg of saturated moist air at 5C, 100kPa. If it is
heated to 45C in an isobaric process, find Q12 and the final relative humidity. If it is compressed
from the initial state to 200 kPa, in an isothermal process, find the mass of water that
has condensed.
w1=humrat(airh2o,p=100,t=5,r=1)
mv1/(100-mv1)=w1
ma1=100-mv1
r2=relhum(airh2o,p=100,t=45,w=w1)
v1=volume(airh2o,p=100,t=5,r=1)
vol1=ma1*v1
v2=volume(airh2o,p=100,t=45,w=w1)
vol2=ma1*v2
w12=100*(vol2-vol1)
q12=w12+ma1*(u2-u1)
u1=intenergy(airh2o,p=100,t=5,r=1)
u2=intenergy(airh2o,p=100,t=45,w=w1)
w2=humrat(airh2o,p=200,t=5,r=1)
m_wat=(w1-w2)*ma1