# 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.'
# 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
# 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
# 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
# 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)