CHAPTER13:Gas Mixtures

Ex13.3:Pg-533

In [2]:
#ques3
#calculating humidity ratio, dew point,mass of air, mass of vapor

r=0.70;#relative humidity
Pg=5.628;#saturation pressure in kPa
Pv=r*Pg;#vapour pressure in kPa
P=100;#net pressure kPa 
Pa=P-Pv;#Partial pressure of air
w=0.622*Pv/Pa;#humidity ratio formula
V=100;#volume in m^3
Ra=0.287;#gas constant for water vapour
T=308.2;#Temperature in K
ma=Pa*V/(Ra*T);#mass in kg
mv=w*ma;#mass of vapour
print" Mass of vapour is",round(mv,3),"Kg "
 Mass of vapour is 2.77 Kg 

Ex13.3E:Pg-533

In [11]:
#ques3
#calculating humidity ratio, dew point,mass of air, mass of vapor

r=0.70;#relative humidity
Pg=0.6988 ;#saturation pressure in lbf/in^2
Pv=r*Pg;#vapour pressure in lbf/in^2
P=14.7;#net pressure in lbf/in^2
Pa=P-Pv;#Partial pressure of air
w=0.622*Pv/Pa;#humidity ratio formula
V=2000;#volume in ft^3
Ra=53.34;#gas constant for water vapour
T=70;#Temperature in F
ma=Pa*V*144/(Ra*(T+480));#mass in lbm
mv=w*ma;#mass of vapour
print" Mass of vapour is",round(mv,2),"lbm"
 Mass of vapour is 2.99 lbm

Ex13.4:Pg-534

In [6]:
#ques4 
#calculating amount of water vapour condensed on cooling

#from example 3
w1=0.0255;#w1=w, humidity ratio at initial temperature
ma=108.6;#mass of air in kg
P=100;#kPa net pressure
#at 5 C mixture is saturated so Pv2=Pg2
Pg2=0.8721;
Pv2=Pg2;
w2=0.622*Pv2/(P-Pg2);
mc=ma*(w1-w2);
print"Mass of vapour condense is ",round(mc,3),"kg "
Mass of vapour condense is  2.175 kg 

Ex13.4E:Pg-535

In [13]:
#ques4 
#calculating amount of water vapour condensed on cooling

#from example 3 
w1=0.02135;#w1=w, humidity ratio at initial temperature
ma=139.6;#mass of air in lbm
P=14.7;#net pressure in lbf/in^2
#at 40 degree F mixture is saturated so Pv2=Pg2
Pg2=0.1217 # lbf/in^2
Pv2=Pg2;
w2=0.622*Pv2/(P-Pg2);# humidity ratio at final temperature
mc=ma*(w1-w2); #Mass of vapour condense in lbm
print"Mass of vapour condense is ",round(mc,3),"lbm "
Mass of vapour condense is  2.256 lbm 

Ex13.5:Pg-536

In [11]:
#ques5
#calculating heat transfer per kilogram of dry air

#1-inlet state
#2-Exit state
r1=0.80;#realtive humidity at state 1
Pg1=4.246;#saturation pressure of vapour in kPa
P1=105.0;#net pressure at state 1 in kPa
P2=100.0;#net pressure at state 2 in kPa
Pv1=r1*Pg1;#partial pressure of vapour in kPa
w1=0.622*Pv1/(P1-Pv1);#humidity ratio at state 1
r2=0.95;#relative humidity at state 2
Pg2=1.7051;#saturation pressure of vapour in kPa
Pv2=r2*Pg2;#partial pressure of vapour in kPa
w2=0.622*Pv2/(P2-Pv2);#humidity ratio at state 2
T1=30.0;#C
T2=15.0;#C
Cp=1.004;#specific heat of water vapour in kJ/kg
hv2=2528.9;#enthalpy of vapourisation of vapour in kJ/kg
hv1=2556.3;#enthalpy of vapourisation of vapour in kJ/kg
hl2=62.99;#enthalpy of 
q=Cp*(T2-T1)+w2*hv2-w1*hv1+hl2*(w1-w2);#kJ/kg
print" Heat transferred per unit mass =",round(q,1),"kJ/kg of dry air"
 Heat transferred per unit mass = -41.7 kJ/kg of dry air

Ex13.6:Pg-537

In [13]:
#ques6
#calculating heat transferred in gas vapour mixture

#n-Nitrogen
#v-water vapour
Pn2=1995;#Pressure of nitrogen in kPa
V=0.5;#Volume in m^3
Rn2=0.2968;#Gas constant for nitrogen in kJ/kg.K
Rv=0.4615;#gas constant for vapour
T1=323.2;#Temperature in K
T2=283.2;#Temperature in K
Pv1=5;#Pressure of water vapour in kPa at state 1
Pv2=1.2276;#Pressure of water vapour in kPa at state 2
mn2=Pn2*V/(Rn2*T1);#mass of nitrogen
mv1=Pv1*V/(Rv*T1);#mass of vapour in kg
mv2=Pv2*V/(Rv*T2);#mass of vapour in kg
ml2=mv1-mv2;#mass of liquid condensed n kg
uv1=2443.1;#specific internal energy of vapour in kJ/kg at state 1
uv2=2389.2;#specific internal energy of vapour in kJ/kg at state 2
ul2=42.0;#specific internal energy of liquid water in kJ/kg
Cv=0.745;#specific heat at constant volume in kJ/kg.K
Q=mn2*Cv*(T2-T1)+mv2*uv2+ml2*ul2-mv1*uv1;
print"Heat transferred =",round(Q,2),"kJ "

#the answer is different in book due to intermediate approximization 
Heat transferred = -339.1 kJ 

Ex13.7:Pg-539

In [16]:
#ques7
#calculating humidity ratio and relative humidity 

#1-Inlet state
#2-Exit state
P=100;#net pressure in kPa 
#it is steady state adiabatic process
#water vapour leaving is saturated so Pv2=Pg2
Pg2=2.339;#saturation pressure of vapour in kPa
Pv2=Pg2;#partial pressure of vapour
w2=0.622*Pv2/(P-Pg2);
Cpa=1.004;#specific heat n kJ/kg/K
T2=20;# final temp in C
T1=30;# initial temp in C
Hfg2=2454.1;#specific heat difference at state 2 in kJ/kg
hv1=2556.3;#enthalpy of water vapour at state 1 in kJ/kg
hl2=83.96;#enthalpy of liquid water in kJ/kg
w1=(Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2);
print " Relative humidity =",round(w1,4)
#also w1=0.622*Pv1/(100-Pv2)
Pv1=100*w1/(0.622+w1);
Pg1=4.246;#saturation pressure at state 1 in kPa
r=Pv1/Pg1;#humidity ratio
print" Humidity ratio =",round(r,4)
 Relative humidity = 0.0107
 Humidity ratio = 0.3993

Ex13.7E:Pg-540

In [18]:
#ques7
#calculating humidity ratio and relative humidity 

#1-Inlet state
#2-Exit state
P=14.7;#net pressure in lbf/in^2
#it is steady state adiabatic process
#water vapour leaving is saturated so Pv2=Pg2
Pg2=0.3632;#saturation pressure of vapour in lbf/in^2
Pv2=Pg2;#partial pressure of vapour
w2=0.622*Pv2/(P-Pg2);
Cpa=0.24;#specific heat n Btu/lbm/F
T2=70;# final temp in F
T1=84;# initial temp in F
Hfg2=1054.0;#specific heat difference at state 2 in Btu/lbm
hv1=1098.1;#enthalpy of water vapour at state 1 in Btu/lbm
hl2=38.1;#enthalpy of liquid water in Btu/lbm
w1=(Cpa*(T2-T1)+w2*Hfg2)/(hv1-hl2);
print " Relative humidity =",round(w1,4)
#also w1=0.622*Pv1/(100-Pv2)
Pv1=14.7*w1/(0.622+w1);
Pg1=0.584;#saturation pressure at state 1 in lbf/in^2
r=Pv1/Pg1;#humidity ratio
print" Humidity ratio =",round(r,4)
 Relative humidity = 0.0125
 Humidity ratio = 0.4958