# Chapter 12: Properties of Gaseous Mixtures¶

## Exa 12.1¶

In [ ]:
#In a dew point apparatus a metal beaer is cooled by gradually adding ice water
#to the water initally in the beaker at room temp. The moisture from the room
#circulating around the beaker beins to condense on the beaker when its temp
#is 70 F and the pressure is 15 psia, find (1) the partial pressure of water
#vapor in the room air and (2) the parts by mass of the same.
#initialisation of variables
P= 15.0 				#psia
T= 55 					#F
P1= 0.2141 				#psia
ma= 29. 				#lb
mb= 18. 				#lb
P2= 0.2141 				#psia
P3= 0.3631 				#psia
#CALCULATIONS
dp= P-P1 				#Change in pressure
r= (dp*ma)/(P1*mb) 		#parts by mass
r1= r/(r+1) 			#ratio
r2= 1/(r+1) 			#ratio
r4= r2/r1 				#specific humidity
P= P2/P3 				#relative humidity
#RESULTS
print '%s %.2f' %('relative humidity= ',P)
print '%s %.4f' %(' \n specific humidity (lb vapour/lb air) = ',r4)
raw_input('press enter key to exit')

relative humidity=  0.59

specific humidity (lb vapour/lb air) =  0.0090


## Exa 12.2¶

In :
#Air supplied to a furnacce has RH 75%, T=80 F and P=10 in of water. The
#barometer reads 29.5 in of mercury. How many pounds of water vapor enter
#the furnace per pound of dry air?
#initialisation of variables
h= 29.5 				#in
n= 75.
T= 80 					#F
h1= 10 					#in
mw= 0.380*18
ma= 14.47*29
d= 13.6 				#kg/m^3
P= 0.5069 				#psi
#CALCULATIONS
Pw= (n/100.)*P 			#partial pressure
P= (h+(h1/d))*(0.491) 	#Total pressure
pa= P-Pw 				#partial pressure of air
r= mw/ma 				# pounds of water
#RESULTS
print '%s %.4f' %('Pounds of water vapour enter the furnance per pound of dry air (lb vapour/lb air) = ',r)
raw_input('press enter key to exit')

Pounds of water vapour enter the furnance per pound of dry air (lb vapour/lb air) =  0.0163
press enter key to exit

Out:
''

## Exa 12.3¶

In :
#A mixture of air and water vapor at 75 F and 14.7 psia has RH 0.5; find its
#specific humidity and its dew-point temp.?
#initialisation of variables
n= 0.5
T= 75 			#F
P= 14.7 		#psia
pg= 0.4298 		#psia
pw= 0.2149 		#psia
#CALCULATIONS
pw1= n*pg 		#partial pressure of water
pa= P-pw1 		#partial pressure of air
r= 0.622*pw/pa 	#RH
#RESULTS
print '%s %.5f' %('relative humidity (lb water/lb dry air) = ',r)

relative humidity (lb water/lb dry air) =  0.00923


## Exa 12.4¶

In :
#In the process shown, the air is received at 1 atm, 40 F and Rh 60% and it is
#desired to discharge it at 70 F, RH 50%. How much water at 45F must be supplied
#per pound of dry air passing through the apparatus?
#initialisation of variables
r2= 0.0078 						#lb water /lb dry air
r1= 0.0032 						#lb water /lb dry air
h2= 25.33 						#Btu/lb
h1= 12.9 						#Btu/lb
pg= 0.1217 						#psia
p= 14.7 						#psia
h3= 13 							#Btu/lb
n= 60.
t2=70.
t1=40.
cpa=0.240
R2= 0.00788 					#lb/lb of dry air
w1= 0.00477 					#lb/lb of dry air
#CALCULATIONS
print '%s' %('Method 1')
w= r2-r1 						#water to be supplied
q= h2-h1-w*h3 					#energy supplied
print '%s %.2f' %('\n In method 1, Enthalpy (Btu/lb of dry air) = ',q)
print('\n Method 2')
R1= 0.622*(n/100.)*(pg/(p-pg))	#Gamma 1
R2=0.00788						#Gamma 2
w2=R2-R1 						#weight 2
#All constants are obtained from steam tables
Q=cpa*(t2-t1)+R2*(1092.6)-R1*(1079.6) -w2*h3
print '%s %.2f' %('In method 2, Enthalpy (btu/lb of dry air) = ',Q)
raw_input('press enter key to exit')

Method 1

In method 1, Enthalpy (Btu/lb of dry air) =  12.37

Method 2
In method 2, Enthalpy (btu/lb of dry air) =  12.38
press enter key to exit

Out:
''

## Exa 12.5¶

In :
#Air at 1 am,75F,70% relative humidity is to be brought to 70 F and 60 % RH
#. To what temp. must the mixture be cooled?How much heat must be removed by
#the cooling coil and how much be supplied by the heatin coil per pound of dry
#air?what fraction of the heat removed in the cooling coil is required to cool
#and condense the water removed?
#initialisation of variables
P= 1. 			 	#atm
n= 70.
T= 75 				#F
T1= 70 				#F
Td=55.3 			#F
r1= 0.0131 			#lb water/lb dry air
r2= 0.0093 			#lb water/lb dry air
h1= 32.36 			#Btu/lb dry air
h2= 27.03 			#Btu/lb dry air
hd2= 23.40 			#Btu/lb dry air
hf= 23.4 			#Btu/lb dry liquid
hg= 1094.5 			#Btu/lb dry liquid
#CALCULATIONS
R1= r1-r2			#Gamma 1
Qc= hd2-h1+R1*hf 	#Cooling
Qh= h2-hd2 			#Heating
x= R1*(hg-hf)  		#Mole fraction
y= x/(-Qc) 			#fraction of heat removed
#RESULTS
print '%s %.2f' %('The mixture must be cooled to the dew point temperature (F) = ',Td)
print '%s %.2f' %('Heat removed by the cooling coil (Btu/lb dry air)= ',Qc)
print '%s %.2f' %('Heat supplied by the heating coil (Btu/lb dry air)= ',Qh)
print '%s %.2f' %('Fraction of heat removed in the coil = ',y)
raw_input('press enter key to exit')

The mixture must be cooled to the dew point temperature (F) =  55.30
Heat removed by the cooling coil (Btu/lb dry air)=  -8.87
Heat supplied by the heating coil (Btu/lb dry air)=  3.63
Fraction of heat removed in the coil =  0.46
press enter key to exit

Out:
''