# Ch-7 : Psychrometrics¶

## Ex - 7.1 : Pg - 290¶

In [1]:
#initialisation of variables
ps=0.024853 #at 21 degress
phi=0.34 #relative humidity
p=1.013 #pressure in bar
#CALCULATIONS
pv=ps*phi
w=0.622*(pv/(p-pv))
tdew=4.5 #at 0.00845 bar
#RESULTS
print 'specific humidity is %0.5f kg/kg of da'%(w)
print 'dew point temp is %0.2f degree C'%(tdew)

specific humidity is 0.00523 kg/kg of da
dew point temp is 4.50 degree C


## Ex - 7.2 : Pg - 291¶

In [2]:
#initialisation of variables
t1=26 #temp in degrees
t2=32 #temp in degrees
pvs=0.033597 #pressure in bar
ps=0.047534 #pressure in bar
p=1.013 #pressure in bar
a=6.6*10**-4
#CALCULATIONS
pv=pvs-(p*a*(t2-t1))
w=(0.622*pv)/(p-pv)
phi=pv/ps
#RESULTS
print 'specific humidity is %0.4f kg/kg of da'%(w)
print '\nrelative humidity is %0.3f '%(phi)
print 'dew point temp is 23.5 degree C' #from steam tables

specific humidity is 0.0187 kg/kg of da

relative humidity is 0.622
dew point temp is 23.5 degree C


## Ex - 7.3 : Pg - 291¶

In [3]:
#initialisation of variables
ps=0.042415 #under 30 degrees temp in bar
vg=32.929 #m*m*m/kg
phi=0.3 #relative humidity
p=1.01325 #bar
pv=0.012725**10**2 #pressure
rv=0.4615
t=313 #temp in k
pa=1.005*10**2
ra=0.287
#CALCULATIONS
pv=phi*ps
w1=0.622*(pv/(p-pv))
rhos=1/vg
rhov=phi*rhos
rho=pv/(rv*t)
pa=p-pv
rhoa=pa*100/(ra*t)
w2=rhov/rhoa
ds=phi*((p-ps)/(p-pv))
#RESULTS
print 'partial pressure of water vapour is %0.6f bar'%(pv)
print '\ndensity of dry air is %0.4f kg/m*m*m'%(rhoa)
print 'dew point temp is 10.5 degrees'
print '\nspecific humidity is %0.5f kg/kg of da'%(w2) #textbook answer slightly varies
print '\ndegree of saturation is %0.4f '%(ds)

partial pressure of water vapour is 0.012724 bar

density of dry air is 1.1138 kg/m*m*m
dew point temp is 10.5 degrees

specific humidity is 0.00818 kg/kg of da

degree of saturation is 0.2911


## Ex - 7.4 : Pg - 292¶

In [4]:
#initialisation of variables
ps=0.035636 #pressure in bar
pvw=0.018168 #pressure in bar
p=1.01325 #pressure in bar
a=6.6*10**-4
w=0.00667
td=27 #temparature in degrees
tw=16 #temparature in degrees
#CALCULATIONS
pv=pvw-(p*a*(td-tw))
w=0.622*(pv/(p-pv))
phi=pv/ps
h=(1.005*td+w*(2500+1.86*td))
#RESULTS
print 'humidity ratio is %0.5f kg/kg of da'%(w)
print '\nrelative humidity is %0.4f '%(phi)
print 'dew point temparature is 8 degrees'
print '\nenthalphy of moist air is %0.2f kg/kg of da'%(h)

humidity ratio is 0.00671 kg/kg of da

relative humidity is 0.3034
dew point temparature is 8 degrees

enthalphy of moist air is 44.24 kg/kg of da


## Ex - 7.5 : Pg - 293¶

In [5]:
#initialisation of variables
p=1.01325 #pressure in bar
pv=0.020 #pressure in bar at 21 degrees temp
ws=0.0154 #kg/kg of da
w=0.0123 #kg/kg of da
vs=0.86 #under 21 degrees temp m*m*m/kg
w1=0.0074
#CALCULATIONS
pa=p-pv
sr=w/ws
rho=1/vs
avc=0.0163-w1
#RESULTS
print 'partial pressure of vapour and dry air are %0.2f bar and %0.5f bar'%(pv,pa)
print 'dew point temp is 17.4 degrees'
print 'specific humidity is 0.0123 kg/kg of da'
print '\nsaturation ratio is %0.3f '%(sr)
print '\ndensity of misture is %0.3f kg/m*m*m'%(rho)
print '\namount of water vapour condensed is %0.4f kg/kg of da'%(avc)

partial pressure of vapour and dry air are 0.02 bar and 0.99325 bar
dew point temp is 17.4 degrees
specific humidity is 0.0123 kg/kg of da

saturation ratio is 0.799

density of misture is 1.163 kg/m*m*m

amount of water vapour condensed is 0.0089 kg/kg of da


## Ex - 7.6 : Pg - 294¶

In [6]:
#initialisation of variables
p=1.01325 #pressure in bar
w1=0.01468
td=20 #temp in degrees
tw=40 #temp in degrees
#CALCULATIONS
ha=(1.005*td+w1*(2500+1.86*td))
w2=(ha-(1.005*tw))/(2500+1.86*tw)
#RESULTS
print 'humidity rate is %0.2f kg/kg of da'%(ha)
print '\nw2 is %0.5f kg/kg of da'%(w2)

humidity rate is 57.35 kg/kg of da

w2 is 0.00666 kg/kg of da


## Ex - 7.7 : Pg - 297¶

In [7]:
from __future__ import division
#initialisation of variables
ps1=0.006566 #bar pressure
phi1=0.6 #relative humidity
td2=21 #temp in degrees
td1=1 #temp in degrees
ps2=0.02486 #pressure in bar
td3=26 #temp in degrees
p=1.013 #pressure in bar
#CALCULATIONS
pv1=(phi1*ps1)
w=0.622*(pv1/(p-pv1))
q=(td2-td1)*(1.005+(1.86*w))
phi2=pv1/ps2
cbf=(td3-td2)/(td3-td1)
cf=1-cbf
#RESULTS
print 'heat supplied to air is %0.2f kg/kg of da'%(q)
print '\nfinal relative humidity is %0.4f kg/kg of da'%(phi2)
print '\ncoil bypass factor is %0.2f '%(cbf)
print '\ncontact factor is %0.2f '%(cf)

heat supplied to air is 20.19 kg/kg of da

final relative humidity is 0.1585 kg/kg of da

coil bypass factor is 0.20

contact factor is 0.80


## Ex - 7.8 : Pg - 298¶

In [8]:
#initialisation of variables
ps1=0.056216 #bar pressure
phi1=0.2 #relative humidity
td1=35 #temp in degrees
p=1.01325 #pressure in bar
td2=25 #temp in degrees
ps2=0.03166 #bar
#CALCULATIONS
pv1=phi1*ps1
w1=0.622*(pv1/(p-pv1))
ha=(1.005*td1+w1*(2500+1.86*td1))
w2=(ha-(1.005*td2))/(2500+1.86*td2)
pv2=(w2*p)/(w2+0.622)
phi2=pv2/ps2
#RESULTS
print 'relative humidity rate is %0.2f kg/kg of da'%(ha)
print '\nrelative humidity is %0.4f '%(phi2)
print '\namount of water to be added is %0.5f kg/kg of da'%(w2)

relative humidity rate is 53.08 kg/kg of da

relative humidity is 0.5550

amount of water to be added is 0.01098 kg/kg of da


## Ex - 7.9 : Pg - 300¶

In [9]:
#initialisation of variables
ps1=0.056216 #bar pressure
ps3=0.023366 #bar pressure
phi1=0.6 #relative humidity
td3=20 #temp in degress
td1=35 #temp in degrees
td2=12 #temp in degrees
r=0.287
p=1.01325 #pressure in bar
x1=90.12 #kj/kg
x2=34.08 #kj/kg
x3=42.25 #kj/kg
hf=0.4 #kj/kg
w1=0.02142
w2=0.00873
#CALCULATIONS
pv1=phi1*ps1
w1=0.622*(pv1/(p-pv1))
h1=(1.005*td1+w1*(2500+1.86*td1))
pv3=phi1*ps3
w3=0.622*(pv3/(p-pv3))
h3=(1.005*td3+w3*(2500+1.86*td3))
h2=(1.005*td2+0.0073*(2500+1.86*td2))
ma=((p-pv1)*100*2.5)/(r*(td1+273))
q1=ma*(x2-x1)+(w1-w2)*hf
q2=(ma*(x3-x2))
#RESULTS
print 'mass of dry air is %0.2f kg/s'%(ma)
print '\ncooler load on the dehumidyfier is %0.2f kw'%(q1) # ans wrong in the book
print '\nheating load of the heater is %0.2f kw'%(q2)

mass of dry air is 2.77 kg/s

cooler load on the dehumidyfier is -155.24 kw

heating load of the heater is 22.63 kw


## Ex - 7.10 : Pg - 302¶

In [10]:
#initialisation of variables
x1=90.12 #kj/kg
x3=42.25 #kj/kg
ps3=0.023366 #bar pressure
td3=35 #temp in degrees
phi1=0.6 #relative humidity
p=1.01325 #pressure in bar
#CALCULATIONS
pv3=phi1*ps3
w3=0.622*(pv3/(p-pv3))
h3=(1.005*td3+w3*(2500+1.86*td3))
qs=h3-x3
ql=x1-h3
shf=qs/(qs+ql)
#RESULTS
print 'sensible heat removed is %0.2f kj/kg of da'%(qs)
print '\nlatent heat removed is %0.2f kj/kg of da'%(ql)
print '\nsensible heat factor is %0.2f '%(shf)

sensible heat removed is 15.31 kj/kg of da

latent heat removed is 32.56 kj/kg of da

sensible heat factor is 0.32


## Ex - 7.11 : Pg - 302¶

In [11]:
#initialisation of variables
ps1=0.010720 #bar pressure
phi1=0.3 #relative humidity
td1=8 #temp in degrees
td2=32 #temp in degrees
td3=30 #temp in degrees
ps3=0.042415 #bar pressure
phi3=0.5 #relative humidity
hf=762.6 #kj/kg
hfg=2013.6 #kj/kg
p=1.01325 #pressure in bar
#CALCULATIONS
pv1=phi1*ps1
w1=0.622*(pv1/(p-pv1))
h1=(1.005*td1+w1*(2500+1.86*td1))
h2=(1.005*td2+w1*(2500+1.86*td2))
ha=h2-h1
pv3=phi3*ps3
w3=0.622*(pv3/(p-pv3))
h3=(1.005*td3+w3*(2500+1.86*td3))
wa=w3-w1
hw=(h3-h2)/(w3-w1)
x=(hw-hf)/hfg
#RESULTS
print 'heat added is %0.2f kJ/kg of da'%(ha)
print '\nwater added is %0.5f kg/kg of da'%(wa)
print 'temp os steam supplied is 179.88 degrees' #at 10 bar pressure
print '\nsteam required is %0.2f kj/kg of steam'%(hw)
print '\nquality of steam at 10 bar is %0.3f '%(x)

heat added is 24.21 kJ/kg of da

water added is 0.01132 kg/kg of da
temp os steam supplied is 179.88 degrees

steam required is 2377.53 kj/kg of steam

quality of steam at 10 bar is 0.802


## Ex - 7.12 : Pg - 305¶

In [12]:
#initialisation of variables
ps1=0.023366 #bar pressure
phi1=0.4#relative humidity
td1=20 #temp in degrees
m1=40 #kg/s
ps2=0.01227 #bar pressure
phi2=0.8#relative humidity
td2=10 #temp in degrees
m2=20 #kg/s
p=1.01325 #pressure in bar
#CALCULATIONS
pv1=phi1*ps1
w1=0.622*(pv1/(p-pv1))
h1=(1.005*td1+w1*(2500+1.86*td1))
ma1=m1/(1+w1)
pv2=phi2*ps2
w2=0.622*(pv2/(p-pv2))
h2=(1.005*td2+w2*(2500+1.86*td2))
ma2=m2/(1+w2)
w3=((ma1*w1)+(ma2*w2))/(ma1+ma2)
h3=((ma1*h1)+(ma2*h2))/(ma1+ma2)
td3=((ma1*td1)+(ma2*td2))/(ma1+ma2)
#RESULTS
print 'specific humidity is %0.5f kJ/kg of da'%(w3)
print '\ntemparature of air leaving chamber is %0.2f degrees'%(td3)

specific humidity is 0.00589 kJ/kg of da

temparature of air leaving chamber is 16.67 degrees


## Ex - 7.13 : Pg - 307¶

In [13]:
#initialisation of variables
ps1=0.062739 #bar pressure
phi1=0.9 #relative humidity
td1=37 #temp in degrees
td3=10.7 #dew point temparature
ps4=0.02366 #bar pressure
phi4=0.55 #relative humidity
td4=20 #temp in degrees
w12=1.5 #work input in kw
v4=50 #
t4=37+273 #temp  in k
Ra = 0.287
r= 1
w2= 1
w3= 1
hf3= 2
p=1.01325 #pressure in bar
#CALCULATIONS
pv1=phi1*ps1
w1=0.622*(pv1/(p-pv1))
h1=(1.005*td1+w1*(2500+1.86*td1))
pv4=phi4*ps4
w4=0.622*(pv4/(p-pv4))
h4=(1.005*td4+w4*(2500+1.86*td4))
h3=(1.005*td3+w4*(2500+1.86*td3))
pa4=p-pv4
ma=(pa4*v4*100)/(Ra*t4)
q12=(w12*60)/ma
h2=h1+q12
q23=((h3+(w2-w3)*hf3)-h2)
Q23=-1*q23*ma
q34=h4-h3
Q34=q34*ma
#RESULTS
#ans in the book are wrong.
print 'enthalpy rate 1 is %0.2f kJ/kg of da'%(h1)
print '\nenthalpy rate 4 is %0.2f kJ/kg of da'%(h4)
print '\nenthalpy rate 3 is %0.2f kJ/kg of da'%(h3)
print '\nmass of dry air is %0.2f kg/min'%(ma)
print '\nenthalpy rate 2 is %0.2f kJ/kg of da'%(h2)
print '\ncapacity od cooling coil q23 is %0.2f kJ/min'%(Q23)
print '\ncapacity od cooling coil q34 is %0.2f kJ/min'%(Q34)

enthalpy rate 1 is 131.48 kJ/kg of da

enthalpy rate 4 is 40.63 kJ/kg of da

enthalpy rate 3 is 31.14 kJ/kg of da

mass of dry air is 56.21 kg/min

enthalpy rate 2 is 133.08 kJ/kg of da

capacity od cooling coil q23 is 5730.05 kJ/min

capacity od cooling coil q34 is 533.25 kJ/min


## Ex - 7.14 : Pg - 310¶

In [14]:
#initialisation of variables
td3=15 #dew point temparature
ps3=0.017039 #bar pressure
phi3=0.55 #relative humidity
p=1 #bar pressure
ps4=0.029821 #bar pressure
phi4=1 #relative humidity
td4=24 #temp in degrees
mw1=1000 #kg/min
hf1=109 #kj/kg
hf2=50.4 #kj/kg
w4=0.01912
w3=0.00588
#CALCULATIONS
pv3=phi3*ps3
w1=0.622*(pv3/(p-pv3))
h3=(1.005*td3+w3*(2500+1.86*td3))
pv4=phi4*ps4
w4=0.622*(pv4/(p-pv4))
h4=(1.005*td4+w4*(2500+1.86*td4))
ma=mw1*(hf1-hf2)/(h4-h3-(w4-w3)*hf2)
x=ma*(w4-w3) #mw1-mw2
mf=ma+x
pl=(x/mw1)*100
#RESULTS
print 'mass of dry air is %0.1f kg/min'%(ma)
print '\nmass cooling water loss by evoporation is %0.2f kg/min '%(x)
print '\nmass flow of moist air is %0.2f kg/min '%(mf)
print '\npercentage loss by evoporation is %0.2f %%'%(pl)

mass of dry air is 1389.8 kg/min

mass cooling water loss by evoporation is 18.40 kg/min

mass flow of moist air is 1408.20 kg/min

percentage loss by evoporation is 1.84 %


## Ex - 7.15 : Pg - 311¶

In [15]:
#initialisation of variables
td3=17 #dew point temparature
ps3=0.019362 #bar pressure
phi3=0.6 #relative humidity
p=0.98 #bar pressure
t3=290 #temp in k
ps4=0.042415 #bar pressure
phi4=1 #relative humidity
td4=30 #temp in degrees
mw2=80
v=110 #volume
ma=127.98
w4=0.02814
w3=0.007464
r=0.287
hf1=209.3
#CALCULATIONS
pv3=phi3*ps3
w3=0.622*(pv3/(p-pv3))
h3=(1.005*td3+w3*(2500+1.86*td3))
pa3=p-pv3
m=(pa3*v*100)/(r*t3)
h2=h3+(240/ma)
pv4=phi4*ps4
w4=0.622*(pv4/(p-pv4))
h4=(1.005*td4+w4*(2500+1.86*td4))
mw1=mw2+ma*(w4-w3)
hf2=((mw1*hf1)+(ma*h2)-(ma*h4))/mw2
#RESULTS
print 'mass of dry air is %0.2f kg/min'%(m)
print '\nenthalpy rate 3 is %0.2f kJ/kg of da'%(h3)
print '\nenthalpy rate 2 is %0.2f kJ/kg of da'%(h2)
print '\nenthalpy rate 4 is %0.2f kJ/kg of da'%(h4)
print '\nenthalpy rate  is %0.1f kJ/kg of da'%(hf2)
print 'temparature of water leaving the tower is 27.1 degrees'

mass of dry air is 127.99 kg/min

enthalpy rate 3 is 35.98 kJ/kg of da

enthalpy rate 2 is 37.85 kJ/kg of da

enthalpy rate 4 is 102.07 kJ/kg of da

enthalpy rate  is 113.5 kJ/kg of da
temparature of water leaving the tower is 27.1 degrees


## Ex - 7.16 : Pg - 315¶

In [16]:
#initialisation of variables
uw=2.5
aw=127.82
to=34 #temp in degrees
tr=26 #temp in degrees
ur=1.5
ar=90
ag=8.68
clf1=100
pvwo=0.037782
p=1.013 #pressure in bar
a=6.66*10**-4
phi=0.5
#CALCULATIONS
shgw=uw*aw*(to-tr)
shgr=ur*ar*(to-tr)
sg=ag*clf1
pvo=pvwo-(p*a*(to-tr))
wo=0.622*(pvo/(p-pvo))
ho=(1.005*to+wo*(2500+1.86*to))
pvr=phi*pvo
wr1=0.622*(pvr/(p-pvr))
hr=(1.005*tr+wr1*(2500+1.86*tr))
#RESULTS
print 'recommended indoor conditions are 25.5-26.7 degrees and 50% rh and outdoor conditions are  26 degrees and 50%rh'
print 'area of the roof is 90 m*m'
print 'overall heat transfer coefficients are 2.5 w/m*m'
print '\nsensible heat gain through walls is %0.2f '%(shgw)
print '\nsensible heat gain through roofs is %0.2f '%(shgr)
print '\nsensible heat gain through windows is %0.2f '%(sg)
print 'sensible heat per adult male is 67.5 W and latent heat is 55.7 W'
print '\nenthalpy rate o is %0.2f '%(ho)
print '\nenthalpy rate r is %0.2f '%(hr)
print 'volume of air infiltered is 1.628 m*m*m/min'
print 'latent heat gain is 902.4 W'
print 'sensible heat gain is 257.2 W'
print 'room sensible heat factor is 0.803'

recommended indoor conditions are 25.5-26.7 degrees and 50% rh and outdoor conditions are  26 degrees and 50%rh
area of the roof is 90 m*m
overall heat transfer coefficients are 2.5 w/m*m

sensible heat gain through walls is 2556.40

sensible heat gain through roofs is 1080.00

sensible heat gain through windows is 868.00
sensible heat per adult male is 67.5 W and latent heat is 55.7 W

enthalpy rate o is 86.82

enthalpy rate r is 51.88
volume of air infiltered is 1.628 m*m*m/min
latent heat gain is 902.4 W
sensible heat gain is 257.2 W
room sensible heat factor is 0.803