Chapter 14: Gas–Vapor Mixtures and Air-Conditioning¶

Example 14-1 ,Page No.720¶

In [2]:
#given data
V=5*5*3.0;#volume of the room in m^3
RH=0.75;#relative humidity
P=100.0;#pressure of air in kPa
T=25.0;#temperature of air in C

#constants used
Ra=0.287;#in kPa.m^3 / kg.k
Rv=0.4615;#in kPa.m^3 / kg.k

#from Table A-2a and A-4
cp=1.005;
Psat=3.1698;
hg=2564.6;

#calculation
Pv=RH*Psat;
Pa=P-Pv;
w=0.622*Pv/(P-Pv);
h=cp*T+w*hg;
ma=V*Pa/(Ra*(T+273));
mv=V*Pv/(Rv*(T+273));
print'the partial pressure of dry air %f kPa'%round(Pa,2);
print'the specific humidity %f kg water/kg of dry air'%round(w,4);
print'the enthalpy per unit mass of the dry air %f kJ'%round(h,1);
print'mass of air %f kg'%round(ma,2);
print'mass of water vapour %f kg'%round(mv,1);

the partial pressure of dry air 97.620000 kPa
the specific humidity 0.015100 kg water/kg of dry air
the enthalpy per unit mass of the dry air 64.000000 kJ
mass of air 85.610000 kg
mass of water vapour 1.300000 kg


Example 14-2 ,Page No.722¶

In [3]:
#given data
T=20;#temperature of air in C
RH=0.75;#relative humidity

#from Table A-4
Psat=2.3392;
Pv=RH*Psat;
#thus at this from Eq 14-13
Tdp=15.4;
print'window temperature %f C'%Tdp

window temperature 15.400000 C


Example 14-3 ,Page No.725¶

In [5]:
#given data
T1=25.0;#dry bulb temperature in K
T2=15.0;#wet bulb temperature in K
P2=101.325;#pressure of air in kPa

#from Table A-2a & A-4
#at T1
Psat1=3.1698;
hg1=2546.5;
#at T2
Psat2=1.7057;
hfg2=2465.4;
hf2=62.982;
cp=1.005;

#calculations
w2=0.622*Psat2/(P2-Psat2);
w1=(cp*(T2-T1)+w2*hfg2)/(hg1-hf2);
print'the specific humidity %f kg water/kg of dry air'%round(w1,5);
RH1=w1*P2/((0.622+w1)*Psat1);
print'the relative humidity is %f'%round(RH1,3);
h=cp*T1+w1*hg1;
print'the enthalpy of the air %f kJ/kg of dry air'%round(h,1)

the specific humidity 0.006530 kg water/kg of dry air
the relative humidity is 0.332000
the enthalpy of the air 41.700000 kJ/kg of dry air


Example 14-5 ,Page No.731¶

In [17]:
#given data
RH1=0.3;#outdoor relative humidity
P1=100.0;#pressure of air in kPa
T1=10.0;#outdoor temperature in C
T2=22.0;#first heated temperature in C
RH3=0.6;#final relative humidity
T3=25.0;#final temperature in C

#from Table A-2a & A-4
cp=1.005;
Ra=0.287;
Pg1=1.2281;
hg1=2519.2;
hg2=2541.0;
Pg3=3.1698;

#calculations
Pv1=RH1*Pg1;
Pa1=P1-Pv1;
v1=Ra*(T1+273)/Pa1;
ma=V1/v1;
w1=0.622*Pv1/(P1-Pv1);
h1=cp*T1+w1*hg1;
h1=round(h1,1);
w2=w1;
h2=cp*T2+w2*hg2;
h2=round(h2)
Q=ma*(h2-h1);
# ma2*w2 + mw = ma3*w3
#which reduces to mw = ma * (w3 - w2)
w3=0.622*RH3*Pg3/(P1-(RH3*Pg3));
mw=ma*(w3-w2);
print'the rate of heat supply in the heating section %f kJ/min'%round(Q);
print'the mass flow rate of the steam required in the humidifying section %f kg/min'%round(mw,3)

the rate of heat supply in the heating section 673.000000 kJ/min
the mass flow rate of the steam required in the humidifying section 0.539000 kg/min


Example 14-6 ,Page No.733¶

In [18]:
#given data
V1=10;#volumertric rate in m^3/min
T1=30;#intial temperature in C
RH1=0.8;#intial relative humidity
T2=14;#final temperature in C
RH2=1;#final relative humidity

#from Table A-4
hw=58.8;
h1=85.4;
h2=39.3;
w1=0.0216;
w2=0.0100;
v1=0.889;

#calculations
#mw= ma*(w1-w2)
#Qout=ma*(h1-h2) - mw*hw
ma=V1/v1;
mw= ma*(w1-w2);
Qout=ma*(h1-h2) - mw*hw;
print'rates of moisture removal from the air %f kg/min'%round(mw,3);
print'rate of moisture removal from the air %i kJ/min'%round(Qout);

rates of moisture removal from the air 0.130000 kg/min
rate of moisture removal from the air 511 kJ/min


Example 14-8 ,Page No.736¶

In [20]:
#given values
V1=50;#saturated air volumertric rate in m^3/min
T1=14;#saturated air temperature in C
V2=20;#outside air volumertric rate in m^3/min
T2=32;#outside air temperature in C
RH2=60;#outside air relative humidity

#from psychrometric chart
h1=39.4;
w1=0.010;
v1=0.826;
h2=79;
w2=0.0182;
v2=0.889;

#calculations
ma1=V1/v1;
ma2=V2/v2;
ma3=ma1+ma2;
#from Eq 14-24
w3=(w2*ma2+w1*ma1)/(ma1+ma2);
h3=(h2*ma2+h1*ma1)/(ma1+ma2);
print'the specific humidity %f kg of water/kg of dry air'%round(w3,3);
#from psychrometric chart
T3=19;
RH3=0.89;
v3=0.844;
V3=ma3*v3;
print'the dry-bulb temperature %f C'%T3;
print'the relative humidity is %f'%RH3
print'the volume flow rate of the mixture %f m^3/min'%round(V3,1)

the specific humidity 0.012000 kg of water/kg of dry air
the dry-bulb temperature 19.000000 C
the relative humidity is 0.890000
the volume flow rate of the mixture 70.100000 m^3/min


Example 14-9 ,Page No.738¶

In [22]:
#given data
m=100;#mass flow rate in kg/s
T1=20;#entry temperature in C
P1=1;#entry pressure im kPa
RH1=60;#entry relative humidity
T2=30;#exit temperature in C
RH2=1;#exit relative humidity
T3=35;#entry tempearture of wet cooling tower in C
T4=22;#water temperature in cooling tower in C

#from Table A-4
h1=42.2;
w1=0.0087;
v1=0.842;
h2=100;
w2=0.0273;
h3=146.64;
h4=92.28;

#calculations
#Dry air balane = ma1 = ma2 = ma
#Water balance = m3 - m4 = ma*(w2 - w1)
#Energy balance = ma1*h1 + m3*h3 = ma2*h2 + m4*h4
ma= m*(h3-h4)/(h2-h1-(w2-w1)*h4);
V1=ma*v1;
mmakeup=ma*(w2-w1);
print'the volume flow rate of air into the cooling tower %f m^3/s'%round(V1,1);
print'the mass flow rate of the required makeup water %f kg/s'%round(mmakeup,2)

the volume flow rate of air into the cooling tower 81.600000 m^3/s
the mass flow rate of the required makeup water 1.800000 kg/s