# Chapter 8:Psychrometrics¶

## Ex8.1:PG-208¶

In [12]:
#initialization of variables
Ra=0.287 # specific gas constant for air
P=100.0 # pressure of room in kPa
V=150.0 # volume of room in m^3
T=25+273 # temperature of air in kelvin
phi=0.6 # relative humidity
Pg=3.29 # saturation vapour pressure in kPa at 25 *C from table C.1
Mv= 18 # molecular mass of water vapor
Ma=28.97 # molecular mass of air

Pv=Pg*phi # partial pressure of water vapour

Pa=P-Pv # partial pressure of air

w=0.622*(Pv/Pa) # humidity ratio in Kg of water/ Kg of dry air
Tdp=17.4 # dew point temperature from interpolation in table C.2 corresponding to partial pressure Pv=1.98 kPa

ma=Pa*V/(Ra*T) # mass of air
mv=w*ma # mass of water vapour in  kg

# now we find volume percentage
Nv=mv/Mv # moles of vapour
Na=ma/Ma # moles of air

Vw= Nv/(Na+Nv) # fraction of volume occupied by water vapour

print "The humidity ratio is",round(w,4),"kg water/ kg of dry air \n"
print "The dew point is",round(Tdp,1),"degree celsius \n "
print "The mass of water vapour in the air is",round(mv,3),"kg \n"
print "The volume percentage of the room that is water vapor is",round(Vw*100,2),"%"
# The answers are correct within given limits
# textbook while python is precise

The humidity ratio is 0.0125 kg water/ kg of dry air

The dew point is 17.4 degree celsius

The mass of water vapour in the air is 2.153 kg

The volume percentage of the room that is water vapor is 1.98 %


## Ex8.2:PG-209¶

In [18]:
#initialization of variables
Ra=0.287 # specific gas constant for air
P=100.0 # pressure of room in kPa
w1=0.0126 # old humidity ratio of example 8.1-
Pg=3.29 # saturation vapour pressure in kPa at 25 *C from table C.1
mv=2.17 # initial mass of water vapour in example 8.1
T=25+273 # temperature after reheat
V=150.0 # volume of room in m^3
Pv=1.228 # saturation vapour pressure in kPa @ 10 degree celsius from table C.1
Pa=P-Pv # partial pressure of air
w2=0.622*(Pv/Pa) # new humidity ratio in Kg of water/ Kg of dry air
deltaw=w1-w2 # difference in humidity ratio
ma=Pa*V/(Ra*T) # mass of air
deltamv=deltaw*ma # mass of water vapour condensed
X=deltamv*100/mv # percentage of water vapour condensed
print "The percentage that condenses is",round(X,2),"% \n"
# AFTER REHEATING
phi=1.608*w2*Pa/Pg
print "The relative humidity is",round(phi*100,3),"%"

# The answers are correct within given limits
# textbook while python is precise

The percentage that condenses is 38.85 %

The relative humidity is 37.332 %


## Ex8.3:PG-212¶

In [23]:
#initialization of variables
T1=40 # dry bulb temperature in degree celsius
T2=20 # wet bulb temperature in degree celsius
Cp=1.0 # specific heat
P=100 # pressure of air stream in kPa
pg1=7.383 #saturation pressure @ 40 degree celsius
hfg2=2454 # latent heat for 20 degree celsius
Pg2=2.338 # saturation pressure @ 20 degree celsius
w2=0.622*Pg2/(P-Pg2) # specific humidity for wet bulb condition
hg1=2574 # specific enthalpy of saturated vapour @ 40 degree celsius
hf2=83.9 #spedific enthalpy of saturated liquid @ 20 degree celsius
w1=((w2*hfg2)+Cp*(T2-T1))/(hg1-hf2)# specific humidity for 40 degree celsius
print "The humidity ratio is",round(w1,4),"kg water/ Kg dry air \n"
pv1=100*w1/(0.622+w1) # partial pressure of vapour
phi=pv1/pg1 # relative humidity
print "The relative humidity is",round(phi*100,1),"% \n"

hv=hg1 # temperature is at DBT=40 degree celsius
h=Cp*T1+w1*hv # specific enthalpy  of air
print "The specific enthalpy is",round(h,1),"kJ/kg dry air"

The humidity ratio is 0.0066 kg water/ Kg dry air

The relative humidity is 14.3 %

The specific enthalpy is 57.1 kJ/kg dry air


## Ex8.5:PG-215¶

In [26]:
# initialization of variables

T1=40 # inlet temperature in degree celsius
T2=27 # outlet temperature in degree celsius
phi1= 10 # relative humidity at inlet
# as no heat transfer takes place thus isenthalpic process

#Thus following the enthalpy line at DBT=40 and Relative humidity=10

phi2=45 # by interpolation of constant enthalpy line
w1=0.0046# specific humidity @ T=40 and phi1=10
w2=0.010 # specific humidity at outlet
W=w2-w1 # amount of water added
Tmin=18.5 # minimum temperature at 100% relative humidity

print "The relative humidity is",round(phi2,1),"% \n "
print "The added water is",round(W,4),"kg water/kg dry air \n"
print "The lowest possible temperature is",round(Tmin,1),"*C "


The relative humidity is 45.0 %

The added water is 0.0054 kg water/kg dry air

The lowest possible temperature is 18.5 *C


## Ex8.6:PG-215¶

In [31]:
# initialization of variables
T1=5+273.0 # outside air temperature in kelvin
P=100.0 # pressure in kPa
Ra=0.287 # specific gas constant for air
phi=0.7 # relative humidity outside
Qf=50.0/60.0 # volume flow rate in m**3/sec
Pg1=0.872 # saturation pressure at 278 K
Pv1=phi*Pg1 # partial pressure of water vapour
Pa1=P-Pv1 # partial pressure of air

rhoa=Pa1/(Ra*T1) # density of dry air

mdota=Qf*rhoa # mass flow rate of dry air

# using psychrometric chart at T1=5*C and phi1=70%
h1=14 # inlet enthalpy in kJ/kg
h2=35 # enthalpy after heating in kJ/kg

Qdot=mdota*(h2-h1) # heat transfer rate
# from psychrometric chart for T=25 *C and 35 kJ/kg enthalpy
phi2=19 # realtive humidity

print "The heat transfer rate is",round(Qdot,1),"kJ/s \n"
print "The final relative humidity is",round(phi2,4),"% "

The heat transfer rate is 21.8 kJ/s

The final relative humidity is 19.0 %


## Ex8.7:PG-216¶

In [34]:
# initialization of variables
#DATA TAKEN FROM PSYCHROMETRIC CHART
T1=5+273.0 # outside temperature in kelvin
h1=10# enthalpy in kJ/kg @ T=5 *C and 40 % relative humidity
Pg1=0.872 # saturaion pressure in kPa for 5 degree celsius DBT
phi1=0.4
h2=33 # specific enthalpy at 25 *C and 40 % relatuve humidity
h3=45.0 # specific enthalpy at state 3
P=100.0 # atmospheric pressure in kPa
Ra=0.287 # specific gas constant for air
Qf=60.0/60.0 # volume flow rate in m**3/s
Pv1=phi1*Pg1 # partial presure of water vapour
Pa1=P-Pv1 # partial pressure of air
w2=0.0021 # specific humidity @ 40 % relative humidity and 25*C temperature
w3=0.008 # final specific humidity
rhoa1=Pa1/(Ra*T1) # air density
mdota=Qf*rhoa1 # mass flow rate of dry air

Qdot=mdota*(h2-h1) # heat transfer rate

# as the process is isothermal thus
mdots=mdota*(w3-w2)# mass flow rate of steam by conservation of mass
print "the rate of steam supplied is",round(mdots,4),"kg/s \n"
# also using energy balance
hs=(mdota*(h3-h2))/mdots # enthalpy of steam
hf=604.7 # enthalpy of saturated liquid @ 400 kPa
hg=2738.5 # enthalpy of saturated vapour @ 400 kPa
xs=(hs-hf)/(hg-hf)
print "The quality of steam is",round(xs,2)

the rate of steam supplied is 0.0074 kg/s

The quality of steam is 0.67


## Ex8.8:PG-217¶

In [38]:
# initialization of variables
# REFER TO FIG. 8.4
T1=30 # outside temperature in degree celsius
phi1=0.9 # outside relative humidity
T2=23 # room temperature in degree celsius
phi2=0.4 # relative humidity in room

# using psychrometric chart
w1=0.0245 # specific humidity @ 30 *C and relative humidity 0.9
h1=93 # specific enthalpy @ 30 *C and relative humidity 0.9
w2=w1 # during cooling humidity remains constant
w3=0.007 # specific humidity @ 23 *C and relative humidity 0.4
h4=41 # final specific enthalpy
h3=26 # specific enthalpy @ 23 *C and relative humidity 0.4
deltaw=w3-w2 # moisture removed
print " the amount of moisture removed is",round(deltaw,4),"kg \n"

qout=h3-h1 # heat removed F-G-H process

print " the heat removed is",round(qout,4),"kJ/kg \n "

qin=h4-h3 # heat added to bring to desired state

print " the heat added is",round(qin,4),"kJ/kg "

 the amount of moisture removed is -0.0175 kg

the heat removed is -67.0 kJ/kg

the heat added is 15.0 kJ/kg


## Ex8.9:PG-218¶

In [41]:
# initialization of variables
P=100 # atospheric pressure in kPa
R=0.287 # specific gas constant for air
T1=15+273 # outside temperature in kelvin
phi1=0.4# outside air relative humidity
Qf1=40 # outside air flow rate in m^3/min
T2=32+273 # inside temperature in kelvin
phi2=0.7 # inside air relative humidity
Qf2=20 # outside air flow rate in m^3/min
Ps1=1.7 # saturation pressure @ 15 degree celsius and 40% humidity
Ps2=4.9 # saturation pressure @ 32 degree celsius and 70% humidity

Pv1=Ps1*phi1 # partial pressure of water vapour outside

Pv2=Ps2*phi2 # partial pressure of water vapour inside

Pa1=P-Pv1 #partial pressure of dry air outside
Pa2=P-Pv2 #partial pressure of dry air inside

rhoa1=Pa1/(R*T1) # density of outside air
mdota1=Qf1*rhoa1 # mass flow rate of air outside

rhoa2=Pa2/(R*T2) # density of inside air
mdota2=Qf2*rhoa2 # mass flow rate of inside air
# using psychrometric chart locating state 1 and 2
h1=37 # specific enthalpy @ DBT 15*C and 40 % humidity
w1=0.0073 # specific humidity @ DBT 15*C and 40 % humidity
h2=110 # specific enthalpy @ DBT 32*C and 70 % humidity
w2=0.0302 # specific humidity @ DBT 32*C and 70 % humidity
ratio=mdota1/mdota2 # ratio of distance between states
#  using this ratio state 3 is located on psychrometric chart
T3=(mdota1*T1+mdota2*T2)/(mdota1+mdota2)-273 # final temparature in celsius

phi3=65# final relative humidity at T3 from psychrometric chart

print " The relative humidity is",round(phi3,4),"% \n"
print " The resultant temperature is",round(T3),"degree celsius"

 The relative humidity is 65.0 %

The resultant temperature is 20.0 degree celsius


## Ex7.10:PG-219¶

In [3]:
# initialization of variables
mdotw3=10000.0 # mass flow rate of water entering in cooling tower in kg/min
Tw1=40+273.0 # temperature of water entering cooling tower in kelvin
Ta1=20+273.0 # temperature of air entering cooling tower in kelvin
phi1=0.5# relative humidity of entering air
Tw2=25+273.0 # temperature of water leaving cooling tower in kelvin
Ta2=32+273 # temperature of air leaving cooling tower in kelvin
phi2=0.98 # relative humidity of leaving air
# from psychrometric chart
h1=37.0# specific enthalpy of air @ 20*C DBT and 50% humidity
w1=0.0073 # specific humidity of air @ 20*C DBT and 50% humidity
h2=110.0 # specific enthalpy of air @ 32*C DBT and 98% humidity
w2=0.030 # specific humidity of air @ 32*C DBT and 98% humidity

h3=167.5 # specific enthalpy of water from steam table at 40 degree celsius
h4=104.9 # specific enthalpy of water from steam table at 25 degree celsius

mdota=(mdotw3*(h4-h3))/(h1-h2+(w2-w1)*h4) # by energy balance

v1=0.84 # specific volume of air entering tower from psychrometric chart

Qf=mdota*v1 # volume flow rate in m^3/min
print "The volume flow rate of air into the cooling tower is",round(Qf)," m^3/min \n"

mdot4=mdotw3-(w2-w1)*mdota # by mass balance
print "The mass flow rate of water that leaves the cooling tower ",round(mdot4),"kg/min"
# The answers is slightly different in textbook due to approximations in calculations while in python solution is precise

The volume flow rate of air into the cooling tower is 7446.0  m^3/min

The mass flow rate of water that leaves the cooling tower  9799.0 kg/min