#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
# The variation in answers is due to approximations made by
# textbook while python is precise
#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
# The variation in answers is due to approximations made by
# textbook while python is precise
#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"
# 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 "
# 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),"% "
# 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)
# 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 "
# 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"
# 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