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

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

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

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

Mixture Analysis (non-reacting)

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)

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

Dehumidification

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

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

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

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.

Humidification

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?

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

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

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

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