# Chapter 18 : Two Phase Gas Liquid Systems Partial Saturation and Humidity¶

### Example 18.1 Page no.539¶

In [5]:
# Variables
V = 1. ;			# Volume of water vapour-[cubic metre]
rel_h =  43. ;			# relative humidity -[%]
vp_H2O = 1.61 ;			# vapour pressure of water at 94 F-[in. of Hg]
P_H2O = vp_H2O*(rel_h/100) ;			# Pressure of water vapour in air-[in. of Hg]
P =  29.92 ;			# [in of Hg]
T = 94+460. ;			# Temperature -[Rankine]
Ts =  492. ;			#Temperature std. -[Rankine]
mw_H2O = 18. ;			# molecular mass of water -[lb]

# Calculations
H2O = (5280**3*Ts*P_H2O*mw_H2O)/(T*P*359) ;			#mass of H2O-[lb]
# The dew point is temperature at which water vapour in air first condense ,i.e at realative humidity 100 %, therefore
psat_H2O = P_H2O ;			# Saturation pressure of H2O -[in. of Hg]

# Results
print 'Saturation pressure of H2O %.3f in. of Hg'%psat_H2O
print 'Use saturation  pressure of H2O to get dew point temperature T from steam table: T is about 68-69 F.'

Saturation pressure of H2O 0.692 in. of Hg
Use saturation  pressure of H2O to get dew point temperature T from steam table: T is about 68-69 F.


### Example 18.2 Page no. 541¶

In [6]:
# Variables
# Data from steam table
psat_H2O = 31.8 ;			# Saturation pressure -[mm of Hg]

#(c)
H =  .0055 ;			# Humidity
mw_H2O =  18. ;			# Molecular wt.  of water-[lb]
mw_air =  29. ;			# Molecular wt. of air -[lb]
P = 750. ;			# Pressure total -[mm of Hg]

# Calculations
p_H2O = ((H*mw_air*P)/mw_H2O)/(1+(H*mw_air/mw_H2O)) ;			# Partial pressure of water vapour in air-[mm of Hg]
#(a)
rel_H = (p_H2O/psat_H2O)*100  ;			# relative humidity -[%]
#(b)
mol_H = (p_H2O)/(P-p_H2O) ;			# Molal humidity

# Results
print '(a)Relative humidity is %.0f%% .'%rel_H
print '(b)Molal humidity is %.1e '%mol_H
print '(c)Partial pressure of water vapour in air is %.1f  mm of Hg.'%p_H2O

(a)Relative humidity is 21% .
(b)Molal humidity is 8.9e-03
(c)Partial pressure of water vapour in air is 6.6  mm of Hg.


### Example 18.3 Page No. 544¶

In [7]:
# Variables
V_BDA  = 1000. ;			# Volume of bone dry air(BDA) at 20 C & 108.0 kPa
mol_V =  22.4 ;			# Molar volume of gas at standard condition-[m**3]
T = 20+273. ;			# Temperature of BDA-[K]
P = 108.0 ;			#Pressure of BDA-[kPa]
Ts = 273. ;			# Standard temperature-[K]
Ps = 101.3 ;			#Standard pressure-[kPa]
W = 0.93 ;			# [kg]
mw_W =  18. ;			# mol. wt. of 1kmol water -[kg]

# Calculations
mol_W = W/mw_W ;			# amount of water vapour(W)-[kg mol]
mol_BDA = (V_BDA*Ts*P)/(T*Ps*mol_V) ;			#  amount of BDA-[kg mol]
p_H2O =  (mol_W/(mol_W+mol_BDA))*P ;			# Partial pressure of H2O-[kPa]

# Get vapour pressure for water at 15 C , namely 1.70 kPa
psat_H2O  =  1.70 ;			#vapour pressure for water at 15 C-[kPa]
rel_H = (p_H2O/psat_H2O) ;			#Fractional relative humidity-[]

# Results
print '(a)Fractional relative humidity of original air was %.3f .'%rel_H

(a)Fractional relative humidity of original air was 0.074 .


### Example 18.4 Page no.545¶

In [8]:
# Variables
F  = 1000. ;			# Volume of entering moist air at 22 C & 101.0 kPa
mol_V =  22.4 ;			# Molar volume of gas at standard condition-[m**3]
T_in = 22.+273 ;			# Temperature of entering moist air-[K]
P_in = 101.0 ;			#Pressure of entering moist air -[kPa]
dp_in = 11.+273 ;			# Dew point of entering air-[K]
Ts = 273. ;			# Standard temperature-[K]
Ps = 101.3 ;			#Standard pressure-[kPa]
T_out = 58.+273 ;			# Temperature of exiting moist air-[K]
P_out = 98. ;			#Pressure of exiting moist air -[kPa]

# Additional vapour pressure data
psat_in = 1.31 ;			#Vapour pressure of entering moist air -[kPa]
psat_out = 18.14 ;			# Vapour pressure of exiting moist air -[kPa]
pBDA_in = P_in-psat_in ;			# Pressure of entering dry air - [kPa]
pBDA_out  =  P_out - psat_out ;			# Pressure of exiting dry air - [kPa]

# Calculations
mol_F = (F*P_in*Ts)/(Ps*T_in*mol_V) ;			# Moles of moist air entering-[kg mol]
mol_P = (mol_F*(pBDA_in/P_in))/(pBDA_out/P_out); 			#BDA balance- [kg mol]
mol_W = mol_P-mol_F ;			# Total balance -[kg mol]
mw_BDA = 29. ;			# Mol. wt. of dry air
mw_H2O = 18. ;			# Mol. wt. of water vapour
m_BDA = (mol_F*pBDA_in/P_in)*mw_BDA ;			# Mass of dry air entering-[kg]
m_H2O = (mol_F*psat_in/P_in)*mw_H2O ;			# Mass of water vapour entering-[kg]
wa_in = m_BDA+m_H2O ;			#Total wet air entering -[kg]
H2O_ad = mol_W*mw_H2O/wa_in ;			#Water added to each kg of wet air entering the process-[kg]

# Results
print 'Water added to each kg of wet air entering the process is %.3f kg.'%H2O_ad

Water added to each kg of wet air entering the process is 0.132 kg.


### Example 18.5 Page No.547¶

In [1]:
# Variables
# Given data
#Basis: F = 29.76 lb mol
F  = 29.76 ;			# amount of entering moist air -[lb mol]
F_rh = 90/100. ;			# Relative humidity
T_in = 100 + 460. ;			# Temperature of entering moist air-[Rankine]
P_in = 29.76 ;			#Pressure of entering moist air -[in. of Hg]
psat_in =  1.93 ;			# Saturation pressure from steam table-[in. of Hg]
T_out = 120 + 460. ;			# Temperature of exiting dry air-[Rankine]
P_out = 131.7 ;			#Pressure of exiting dry air -[in. of Hg]
psat_out =  3.45 ;			# Saturation pressure from steam table-[in. of Hg]
mol_V =  22.4 ;			# Molar volume of gas at standard condition-[m**3]
mw_H2O  =  18.02 ;			# Mol. wt. of water -[lb]
mw_air  =  29. ;			# Mol. wt. of air -[lb]
p_H2O_in = F_rh*psat_in ;			# Partial pressure of water vapour at inlet--[in. of Hg]
p_air_in = P_in-p_H2O_in ;			# Partial pressure of air at inlet--[in. of Hg]

# Calculations
# Assume condensation takes place , therefore output gas P is saturated,
P_rh =  1;			# Relative humidity of output gas
p_H2O_out = P_rh*psat_out ;			# Partial pressure of water vapour at outlet--[in. of Hg]
p_air_out = P_out-p_H2O_out ;			# Partial pressure of air at outlet--[in. of Hg]

# Get W and P from  balance of air and water
P = (p_air_in*F/P_in)/(p_air_out/P_out) ;			# From air balance-[ lb mol]
W = (p_H2O_in*F/P_in)-(P*p_H2O_out/P_out);			# From water balance -[lb mol]
W_ton = (W*mw_H2O*2000)/(p_air_in*mw_air) ;			# Moles of water condenses per ton dry air-[lb mol]
W_m = mw_H2O*W_ton ;			# Mass of water condenses per ton dry air-[lb]

# Results
#  Since W is positive our assumption(condensation takes place ) is right .
print '(a) Yes water condense out during compression ,since W(%.3f  lb mol) is positive our assumption(condensation takes place ) \nis right .'%W
print '(b) Amount of water condenses per ton dry air is %.1f lb mol i.e %.0f lb water.'%(W_ton,W_m)

(a) Yes water condense out during compression ,since W(0.983  lb mol) is positive our assumption(condensation takes place )
is right .
(b) Amount of water condenses per ton dry air is 43.6 lb mol i.e 786 lb water.