Chapter 15:Psychrometrics

Ex15.1:pg-631

In [1]:
import math
Ps = 0.033363 #Saturation pressure in bar
P = 1.0132 # Atmospheric pressure in bar
W2 = (0.622*Ps)/(P-Ps) # mass fraction of moisture
hfg2 = 2439.9 # Latent heat of vaporization in kJ/kg
hf2 = 109.1 # Enthalpy of liquid moisture in kJ/kg
cpa = 1.005 # Constant pressure heat capacity in kJ/kg
hg = 2559.9 # Enthalpy of gas moisture in kJ/kg
hw1 = hg # constant enthalpy
T2 = 26 # wbt in degree Celsius 
T1 = 32 # dbt in degree Celsius 
W1 = (cpa*(T2-T1)+(W2*hfg2))/(hw1-hf2)
Pw = ((W1/0.622)*P)/(1+(W1/0.622))

Psat = 0.048 # Saturation pressure in bar at 32 degree
fi = Pw/Psat # Relative humidity

mu = (Pw/Psat)*((P-Psat)/(P-Pw)) # Degree of Saturation
Pa = P-Pw # Air pressure
Ra = 0.287 # Gase constant
Tdb = T1+273 #  dbt in K
rho_a = (Pa*100)/(Ra*Tdb) # Density of air 
rho_w = W1*rho_a # Water vapor density
ta = 32 # air temperature in degree Celsius  
tdb = 32 # dbt in degree Celsius 
tdp = 24.1# Dew point temperature in degree Celsius 
h = cpa*ta + W1*(hg+1.88*(tdb-tdp))
print "\n Example 15.1\n"
print "\n Specific humidity is ",W1 ," kg vap./kg dry air"
print "\n Partial pressure of water vapour is ",Pw ," bar"
print "\n Dew point temperature is ",tdp ," degree celcius"
print "\n Relative humidity is ",fi*100 ," percent "
print "\n Degree of saturation is ",mu
print "\n Density of dry air is ",rho_a ," kg/m**3"
print "\n Density of water vapor is ",rho_w ," kg/m**3"
print "\n Enthalpy of the mixture is ",h ," kJ/kg"
#The answers vary due to round off error
 Example 15.1


 Specific humidity is  0.0186241999923  kg vap./kg dry air

 Partial pressure of water vapour is  0.0294557080928  bar

 Dew point temperature is  24.1  degree celcius

 Relative humidity is  61.3660585267  percent 

 Degree of saturation is  0.602092639086

 Density of dry air is  1.12382965889  kg/m**3

 Density of water vapor is  0.0209304283244  kg/m**3

 Enthalpy of the mixture is  80.1126961785  kJ/kg

Ex15.2:pg-632

In [2]:
import math
Ps = 2.339 # Satutation pressure in kPa
P = 100.0 # Atmospheric pressure in kPa
W2 = (0.622*Ps)/(P-Ps) # Specific humidity
hfg2 = 2454.1 # Latent heat of vaporization in kJ/kg
hf2 = 83.96 # Enthalpy of fluid in kJ/kg
cpa = 1.005 # COnstant pressure heat capacity of air
hw1 = 2556.3# ENthalpy of water
T2 = 20.0  # Exit tempeature of mixture in degree Celsius
T1 = 30.0 # Inlet tempeature of mixture in degree Celsius
W1 = (cpa*(T2-T1)+(W2*hfg2))/(hw1-hf2) # Specific humidity at inlet
Pw1 = ((W1/0.622)*P)/(1+(W1/0.622)) # pressure due to moisture
Ps1 = 4.246 # Saturation pressure in kPa
fi = (Pw1/Ps1) # Humidity ratio 

print "\n Example 15.2\n"
print "\n Humidity ratio of inlet mixture is ",W1 ," kg vap./kg dry air"
print "\n Relative humidity is ",fi*100 ," percent"
#The answers vary due to round off error
 Example 15.2


 Humidity ratio of inlet mixture is  0.0107221417941  kg vap./kg dry air

 Relative humidity is  39.9106245278  percent

Ex15.3:pg-633

In [3]:
import math
Psat = 2.339 # Saturation pressure in kPa
fi3 = 0.50 # Humidity ratio
P = 101.3 # Atmospheric pressure in kPa
cp = 1.005 # Constant pressure heat addition in kJ/kg
Pw3 = fi3*Psat # Vapor pressure
Pa3 = P-Pw3 # Air pressure
W3 = 0.622*(Pw3/Pa3) # Specific humidity
Psa1_1 = 0.7156 # Saturation pressure in kPa
Pw1 = 0.7156 # moister pressure in kPa 
Pa1 = P-Pw1 # Air pressure
W1 = 0.622*(Pw1/Pa1)  # Specific humidity
W2 = W1 # Constant humidity process
T3 = 293.0 # Temperature at state 3 in K
Ra = 0.287 # Gas constant
Pa3 = 100.13 # Air pressure at state 3
va3 = (Ra*T3)/Pa3 # volume of air at state 3
SW = (W3-W1)/va3 # spray water 
tsat = 9.65 # Saturation temperature in K
hg = 2518.0 # Enthalpy of gas in kJ/kg
h4 = 10.0 # Enthalpy at state 4 in kJ/kg
t3 = T3-273
t2 = ( W3*(hg+1.884*(t3-tsat))-W2*(hg-1.884*tsat) + cp*t3 - (W3-W2)*h4 )/ (cp+W2*1.884)
print "\n Example 15.3\n"
print "\n Mass of spray water required is ",SW ," kg moisture/m**3"
print "\n Temperature to which air must be heated is ",t2 ," degree celcius"
#The answers vary due to round off error
 Example 15.3


 Mass of spray water required is  0.00338125323083  kg moisture/m**3

 Temperature to which air must be heated is  27.0827212424  degree celcius

Ex15.4:pg-635

In [4]:
import math
h1 = 82.0 # Enthalpy at state 1 in kJ/kg
h2 = 52.0 # Enthalpy at state 2 in kJ/kg
h3 = 47.0 # Enthalpy at state 3 in kJ/kg
h4 = 40.0# Enthalpy at state 4 in kJ/kg
W1 = 0.020 # Specific humidity at state 1
W2 = 0.0115# Specific humidity at state 2 
W3 = W2 # Constant humidity process
v1 = 0.887 # Specific volume at state 1
v = 3.33 # amount of free sir circulated
G = v/v1 # air flow rate
CC = (G*(h1-h3)*3600)/14000 # Capacity of the heating Cooling coil
R = G*(W1-W3) # Rate of water vapor removal
HC = G*(h2-h3) #Capacity of the heating coil
print "\n Example 15.4\n"
print "\n Capacity of the cooling coil is ",CC ," tonnes"
print "\n Capacity of the heating coil is ",HC ," kW"
print "\n Rate of water vapor removal is ",R ," kg/s"
#The answers vary due to round off error
 Example 15.4


 Capacity of the cooling coil is  33.7880496054  tonnes

 Capacity of the heating coil is  18.7711386697  kW

 Rate of water vapor removal is  0.0319109357384  kg/s

Ex15.5:pg-636

In [5]:
import math
W1 = 0.0058 # Humidity ratio for first stream
W2 = 0.0187  # Humidity ratio for second stream
h1 = 35.0 # Enthalpy of first stream in kJ/kg
h2 = 90.0# Enthalpy of second stream in kJ/kg
G12 = 1.0/2.0 #ratio
W3 = (W2+G12*W1)/(1+G12) # Final humidity ratio of mixture
h3 = (2.0/3.0)*h2 + (1.0/3.0)*h1# Final enthalpy of mixture

print "\n Example 15.5 \n"
print "\n Final condition of air is given by"
print "\n W3 = ",W3 ," kg vap./kg dry air"
print "\n h3 =  ",h3 ," kJ/kg dry air"
#The answers vary due to round off error
 Example 15.5 


 Final condition of air is given by

 W3 =  0.0144  kg vap./kg dry air

 h3 =   71.6666666667  kJ/kg dry air

Ex15.6:pg-637

In [6]:
import math
# Given that
t = 21.0 # Temperature in degreee celsius
w = 20.0 # Relative humidity in percentage
t_ = 21.0 # Final temperature of air in degree celsius
print "\n Example 15.6 \n"
# From the psychrometric chart 
T2 = 38.5 # In degree celsius
h1_3 = 60.5-42 # In kJ/kg
fi3 = 53.0 # In percentage 
t4 = 11.2 # In degree celsius
W1_2 = 0.0153-0.0083 # In kg vap /kg dry air
print "\n The temperature of air at the end of the drying process is ",T2 ," degree celsius,\n Heat rejected during the cooling process is ",h1_3 ," kJ/kg,\n The relative humidity is ",fi3 ," percent,\n The dew point temperature at the end of drying process is ",t4 ," degree celsius,\n The moisture removed during the drying process is ",W1_2 ," kg vap/kg dry air"
 Example 15.6 


 The temperature of air at the end of the drying process is  38.5  degree celsius,
 Heat rejected during the cooling process is  18.5  kJ/kg,
 The relative humidity is  53.0  percent,
 The dew point temperature at the end of drying process is  11.2  degree celsius,
 The moisture removed during the drying process is  0.007  kg vap/kg dry air

Ex15.7:pg-638

In [7]:
import math
h1 = 57.0 # Enthalpy at state 1 in kJ/kg 
h2 = h1 # Isenthalpic process
h3 = 42.0 # Enthalpy at state 3 in kJ/kg
W1 = 0.0065 # Humidity ratio at sate 1
W2 = 0.0088 # Humidity ratio at sate 2
W3 = W2 # Constant humidity ratio process
t2 = 34.5 # Temperature at state 2
v1 = 0.896# Specific volume at state 1 in m**3/kg
n = 1500.0 # seating capacity of hall
a = 0.3 # amount of outdoor air supplied m**3 per person
G = (n*a)/0.896  # Amount of dry air supplied
CC = (G*(h2-h3)*60)/14000 # Cooling capacity 
R = G*(W2-W1)*60 # Capacity of humidifier

print "\n Example 15.7 \n"
print "\n Capacity of the cooling coil is ",CC ," tonnes"
print "\n Capacity of humidifier is ",R ," kg/h"
#The answers vary due to round off error
 Example 15.7 


 Capacity of the cooling coil is  32.2863520408  tonnes

 Capacity of humidifier is  69.3080357143  kg/h

Ex15.8:pg-639

In [8]:
import math
twb1 = 15.2# Wbt in degree Celsius 
twb2 = 26.7# Wbt in degree Celsius  
tw3 = 30  # Temperature at state 3 in degree Celsius 
h1 = 43 # Enthalpy at state 1 in kJ/kg
h2 = 83.5 # Enthalpy at state 2 in kJ/kg
hw = 84 # Enthalpy of water in kJ/kg
mw = 1.15 # mass flow rate of water in kg/s
W1 = 0.0088 # Humidity ratio of inlet stream 
W2 = 0.0213 # Humidity ratio of exit stream 
hw3 = 125.8 # Enthalpy of water entering tower in kJ/kg 
hm = 84 # Enthalpy of make up water in kJ/kg 
G = 1 # mass flow rate of dry air in kg/s
hw34 = (G/mw)*((h2-h1)-(W2-W1)*hw)  # Enthalpy change
tw4 = tw3-(hw34/4.19) # Temperature of water leaving the tower
A = tw4-twb1 #Approach of cooling water
R = tw3-tw4 #Range of cooling water
x = G*(W2-W1) #Fraction of water evaporated 

print "\n Example 15.8\n"
print "\n Temperature of water leaving the tower is ",tw4 ," degree celcius"
print "\n Range of cooling water is ",R ," degree Celsius"
print "\n Approach of cooling water is ",A ," degree celcius"
print "\n Fraction of water evaporated is ",x ," kg/kg dry air"
#The answers vary due to round off error
 Example 15.8


 Temperature of water leaving the tower is  21.8128048148  degree celcius

 Range of cooling water is  8.18719518522  degree Celsius

 Approach of cooling water is  6.61280481478  degree celcius

 Fraction of water evaporated is  0.0125  kg/kg dry air

Ex15.9:pg-639

In [9]:
import math
# Given that
DBT = 40.0 # Dry bulb temperature in degree celsius
DBT_ = 25.0 # Dry bulb temperature after cooling and dehumidification in degree celsius
RH = 70.0 # Relative humidity in percentage
f = 30.0 # Air flow rate in cmm
print "\n Example 15.9 \n"
# From the psychrometric chart 
v1 = 0.9125 # In m**3/kg
G = f/v1
h5 = 41.5 # In kJ/kg
W1 = 0.0182 # In kg vapor/kg dry air 
h1 = 86.0 # In kJ/kg d.a.
W2 = 0.0136 # In kg vapor/kg dry air 
h2 = 60.0 # In kJ/kg
L = G*(h1-h2)/3.5
Mo = G*(W1-W2)
x = (h2-h5)/(h1-h5)
print "\n Bypass factor of coolin coil is ",x
# Answer veries due to round off error
 Example 15.9 


 Bypass factor of coolin coil is  0.415730337079

Ex15.10:pg-641

In [10]:
import math
# Given that
c = 75.0 # Capacity of classroom in no of perasons
DBT1 = 10.0 # Outdoor Dry bulb temperature in degree celsius
WBT1 = 8.0 # Outdoor Wet bulb temperature in degree celsius
DBT2 = 20.0 # Indoor Dry bulb temperature in degree celsius
RH2 = 50.0 # Relative humidity in percentage
x =0.5 # Bypass factor
f = 0.3 # Air flow rate per person in cmm
print "\n Example 15.10 \n"
# From the psychrometric chart 
W1 = 0.0058 # In kg moisture/kg d.a.
h1 = 24.5 # In kJ/kg
h2 = 39.5 # In kJ/kg
h3 = h2
W3 = 0.0074 # In kg moisture/kg d.a.
t2 = 25.0 # In degree celsius
v1 = .81 # In m**3/kg d.a.
G = f*c/v1
C = G*(h2-h1)/60
t4 = (t2-x*DBT1)/(1-x)
ts = t4
C_H = G*(W3-W1)*60
print "\n Capacity of heating coil is ",C ," kW,\n Surface temperature of heating coil is ",ts ," degree celsius,\n Capacity of humidifier is ",C_H ," kg/h "
 Example 15.10 


 Capacity of heating coil is  6.94444444444  kW,
 Surface temperature of heating coil is  40.0  degree celsius,
 Capacity of humidifier is  2.66666666667  kg/h 

Ex15.11:pg-641

In [11]:
import math
# Given that
DBT = 31.0 # Dry bulb temperature in degree celsius
WBT = 18.5 # Wet bulb temperature in degree celsius
t = 4.4 # Effective surface temperature of coil in degree celsius
RE = 12.5 # Refrigeration effect by the coil in kW
f= 39.6 # Air flow rate in cmm
print "\n Example 15.11 \n"
# From the fig. given in the example
ws = 5.25 #In g/kg d.a.
hs = 17.7 #In kJ/kg d.a.
v1 = 0.872 # In m**3/kg d.a.
h1 = 52.5 # In kJ/kg d.a.
w1 = 8.2 # In g/kg d.a.
G = f/v1
h2 = h1-(RE*60)/G
w2 = w1-((h1-h2)/(h1-hs))*(w1-ws)
# From the psychrometric chart
t2 = 18.6 # In degree celsius
t_ = 12.5 # In degree celsius
x = (h2-hs)/(h1-hs)
print "\n DBT of air leaving the coil is ",t2 ," degree celsius,\n WBT of air leaving the coil is ",t_ ," degree celsius,\n Coil bypass factor is ",x  
# Answer veries due to round off error
 Example 15.11 


 DBT of air leaving the coil is  18.6  degree celsius,
 WBT of air leaving the coil is  12.5  degree celsius,
 Coil bypass factor is  0.525426680599

Ex15.12:pg-641

In [12]:
import math
# Given that
c = 75.0 # Capacity of classroom in no of perasons
DBT1 = 35.0 # Outdoor Dry bulb temperature in degree celsius
RH1 = 70.0 # Outdoor relative humidity in percentage
DBT2 = 20.0 # Indoor Dry bulb temperature in degree celsius 
RH1 = 60.0 # Indoor relative humidity in percentage
DPT = 10.0 # Cooling coil dew point temperature in degree celsius
x =0.25 # Bypass factor
f = 300.0 # Air flow rate in cmm
print "\n Example 15.12 \n"
# From the psychrometric chart 
W1 = 0.0246 # In kg vap./kg d.a.
h1 = 98.0 # In kJ/kg
v1 = 0.907 # In m**3/kg d.a.
h3 = 42.0 # In kJ/kg
W3 = 0.0088 # In kg moisture/kg d.a.
h2 = 34.0 # In kJ/kg
hs = 30.0 # In kJ/kg
t2 = 12.0 # In degree celsius
G = f/v1
C = G*(h1-h2)/(60*3.5)
X = (h2-hs)/(h1-hs)
C_ = G*(h3-h2)/60
t4 = (DBT2-x*t2)/(1-x)
C_H = G*(W1-W3)
print "\n Capacity of cooling coil is ",C ," tonnes,\n Bypass factor of cooling coil is ",X ,",\n Capacity of heating coil is ",t4 ," kW,\n Surface temperature of heating coil is ",C_ ," degree celsius,\n Mass of water vapor removed is ",C_H ," kg/min "
#Answers veries due to round off error
 Example 15.12 


 Capacity of cooling coil is  100.803276106  tonnes,
 Bypass factor of cooling coil is  0.0588235294118 ,
 Capacity of heating coil is  22.6666666667  kW,
 Surface temperature of heating coil is  44.1014332966  degree celsius,
 Mass of water vapor removed is  5.22601984564  kg/min 

Ex15.13:pg-641

In [13]:
import math
# at 15 degree Celsius
Psat1 = 0.01705  # Saturation pressure in bar
hg1 = 2528.9 # Enthalpy in kJ/kg
# At 35 degree Celsius
Psat2 = 0.05628 # Saturation pressure in bar
hg2 = 2565.3 # Enthalpy in kJ/kg
fi1 = 0.55# Humidity ratio at state 1
Pw1 = fi1*Psat1 # water vapor pressure at state 1
fi2 = 1.0 # Humidity ratio at state 2
Pw2 = fi2*Psat2 # water vapor pressure at state 2 
P = 0.1 # Atmospheric pressure in MPa
W1 = (0.622*Pw1)/(P*10-Pw1)
W2 = (0.622*Pw2)/(P*10-Pw2)
MW = W2-W1 # unit mass flow rate of water
t2 = 35.0 # Air exit temperature in degree Celsius
t1 = 14.0 # make up water inlet temperature in degree Celsius 
m_dot = 2.78 # water flow rate in kg/s
cpa = 1.005 # Constant pressure heat capacity ratio in kJ/kg
h43 = 35*4.187 # Enthalpy change
h5 = 14*4.187 # Enthalpy at state 5in kJ/kg
m_dot_w = (-(W2-W1)*h5 - W1*hg1 + W2*hg2 + cpa*(t2-t1))/(h43) 
R = m_dot/m_dot_w 
MW = (W2-W1)*R #Make up water flow rate
RWA = R*(1+W1)
R = 0.287 # Gas constant 
V_dot = (RWA*R*(t1+273))/(P*1e03)  # Volume flow rate of air
print "\n Example 15.13\n"
print "\n Make up water flow rate is ",MW ," kg/s"
print "\n Volume flow rate of air is ",V_dot ," m**3/s"
#The answers vary due to round off error
 Example 15.13


 Make up water flow rate is  0.127715382722  kg/s

 Volume flow rate of air is  3.39095173631  m**3/s