# Chapter 14 Psychrometrics¶

## Ex:14.1 Pg:659¶

In [1]:
#Initialization of variables
t1=80+460 #R
ps=0.5069 #psia
print "from steam tables,"
vs=633.1 #ft**3/lbm
phi=0.3
R=85.6
Ra=53.3
p=14.696
#calculations
tdew=46 #F
pw=phi*ps
rhos=1/vs
rhow=phi*rhos
rhow2= pw*144/(R*t1)
pa=p-pw
rhoa= pa*144/(Ra*t1)
w=rhow/rhoa
mu=phi*(p-ps)/(p-pw)
Ws=0.622*(ps/(p-ps))
mu2=w/Ws
#results
print "part a"
print "partial pressure of water = %.5f psia"%(pw)
print "\n dew temperature = %d F"%(tdew)
print "part b"
print "density of water = %.6f lbm/ft**3"%(rhow)
print "\n in case 2, density of water = %.6f lbm/ft**3"%(rhow2)
print "\n density of air = %.6f lbm/ft**3"%(rhoa)
print "part c"
print "specific humidity  = %.4f lbm steam/lbm air"%(w)
print "part d"
print "In method 1, Degree of saturation = %.3f"%(mu)
print "\n In method 2, Degree of saturation = %.3f"%(mu2)
from steam tables,
part a
partial pressure of water = 0.15207 psia

dew temperature = 46 F
part b
density of water = 0.000474 lbm/ft**3

in case 2, density of water = 0.000474 lbm/ft**3

density of air = 0.072765 lbm/ft**3
part c
specific humidity  = 0.0065 lbm steam/lbm air
part d
In method 1, Degree of saturation = 0.293

In method 2, Degree of saturation = 0.293

## Ex:14.2 Pg:659¶

In [2]:
#Initialization of variables
p=14.696 #psia
ps=0.0808  #psia
ps2=0.5069 #psia
phi2=0.5
phi=0.6
grain=7000
#calculations
pw=phi*ps
w1=0.622*pw/(p-pw)
pw2=phi2*ps2
w2=0.622*pw2/(p-pw2)
dw=w2-w1
dwg=dw*grain
#results
print "change in moisture content = %.6f lbm water/lbm dry air"%(dw)
print "\n in grains, change = %.2f grains water/lbm dry air"%(dwg)
print "The answers are a bit different due to rounding off error in textbook"
change in moisture content = 0.008857 lbm water/lbm dry air

in grains, change = 62.00 grains water/lbm dry air
The answers are a bit different due to rounding off error in textbook

## Ex:14.3 Pg:660¶

In [3]:
#Initialization of variables
t1=80 #F
t2=60 #F
p=14.696 #psia
ps=0.507 #psia
pss=0.256 #psia
cp=0.24
print "From steam tables,"
#calculations
ws=0.622*pss/(p-pss)
w=(cp*(t2-t1) + ws*1060)/(1060+ 0.45*(t1-t2))
pw=w*p/(0.622+w)
phi=pw/ps
td=46 #F
#results
print "\n humidity ratio = %.4f lbm/lbm dry air"%(w)
print "\n relative humidity  = %.1f percent"%(phi*100)
print "\n Dew point = %d F"%(td)
From steam tables,

humidity ratio = 0.0064 lbm/lbm dry air

relative humidity  = 29.7 percent

Dew point = 46 F

## Ex:14.4 Pg:661¶

In [4]:
#Initialization of variables
W=0.0065  #lbm/lbm of dry air
t=80 #F
td=60 #F
#calculations
H=0.24*t+W*(1060+0.45*t)
sig=H-W*(td-32)
Ws=0.0111
H2=0.24*td+Ws*(1060+0.45*td)
sig2=H2-Ws*(td-32)
#results
print "In case 1, enthalpy = %.2f Btu/lbm dry air"%(H)
print "\n In case 1, sigma function = %.2f Btu/lbm dry air"%(sig)
print "\n In case 2, enthalpy = %.2f Btu/lbm dry air"%(H2)
print "\n In case 2, sigma function = %.2f Btu/lbm dry air"%(sig2)
In case 1, enthalpy = 26.32 Btu/lbm dry air

In case 1, sigma function = 26.14 Btu/lbm dry air

In case 2, enthalpy = 26.47 Btu/lbm dry air

In case 2, sigma function = 26.15 Btu/lbm dry air

## Ex:14.5 Pg:662¶

In [5]:
#Initialization of variables
t1=30 #F
t2=60 #F
t3=80 #F
W1=0.00206
W2=0.01090
#calculations
cm1=0.24+0.45*W1
H1=cm1*t1+W1*1060
cm2=0.24+0.45*W2
H2=cm2*t3+W2*1060
hf=t2-32
dq=H2-H1-(W2-W1)*hf
#results
print "In case 1, Enthalpy = %.2f Btu/lbm dry air"%(H1)
print "\n In case 2, Enthalpy = %.2f Btu/lbm dry air"%(H2)
print "\n Heat added = %.2f Btu/lbm dry air"%(dq)
In case 1, Enthalpy = 9.41 Btu/lbm dry air

In case 2, Enthalpy = 31.15 Btu/lbm dry air

Heat added = 21.49 Btu/lbm dry air

## Ex:14.6 Pg:663¶

In [6]:
#Initialization of variables
pw=0.15#psia
print "using psychrometric charts,"
tdew=46 #F
#calculations
va=13.74 #ft**3/lbm dry air
rhoa=1/va
V=13.74
mw=46/7000
rhow=mw/V
w=0.00657
#results
print "part a"
print "partial pressure of water = %.2f psia"%(pw)
print "\n dew temperature = %d F"%(tdew)
print "part b"
print "density of water = %.6f lbm/ft**3"%(rhow)
print "\n density of air = %.4f lbm/ft**3"%(rhoa)
print "part c"
print "specific humidity  = %.5f lbm water/lbm air"%(w)
using psychrometric charts,
part a
partial pressure of water = 0.15 psia

dew temperature = 46 F
part b
density of water = 0.000000 lbm/ft**3

density of air = 0.0728 lbm/ft**3
part c
specific humidity  = 0.00657 lbm water/lbm air

## Ex:14.7 Pg:664¶

In [7]:
#Initialization of variables
W1=0.00206 #lbm/lbm dry air
W2=0.01090 #lbm/lbm dry air
t=60 #F
print "From humidity charts,"
#calculations
dw=W1-W2
hs=144.4
hs2=66.8-32
w1=14.4 #Btu/lbm
ws1=20 #Btu/lbm
w2=76.3 #Btu/lbm
ws2=98.5 #Btu/lbm
dwh1=-(w1-ws1)/7000 *hs
H1=9.3+dwh1
dwh2=(w2-ws2)/7000 *hs2
H2=31.3+dwh2
dwc=dw*(t-32)
dq=H2-H1+dwc
#results
print "Enthalpy change = %.2f Btu/lbm dry air"%(dq)
From humidity charts,
Enthalpy change = 21.53 Btu/lbm dry air

## Ex:14.8 Pg:665¶

In [8]:
#Initialization of variables
print "From psychrometric charts at 50 F and 80 F,"
va1=13 #ft**3/lbm dry air
va2=13.88 #ft**3/lbm dry air
flow=2000 #cfm
#calculations
ma1= flow/va1
ma2=flow/va2
print "The two initial states have been multiplied by 108/262 and distance 2-3 is located"
t=62.5# F
phi=0.83 #percent
#results
print "humidity = %.2f "%(phi)
print "\n Temperature = %.1f F"%(t)
From psychrometric charts at 50 F and 80 F,
The two initial states have been multiplied by 108/262 and distance 2-3 is located
humidity = 0.83

Temperature = 62.5 F

## Ex:14.9 Pg:666¶

In [9]:
#Initialization of variables
t=90 #F
ts=67.2 #F
phi=0.3
per=0.8
#calculations
dep=t-ts
dt=dep*per
tf=t-dt
print "from psychrometric charts,"
phi2=0.8
#results
print "Dry bulb temperature = %.2f F"%(tf)
print "\n percent humidity  = %.2f"%(phi2)
from psychrometric charts,
Dry bulb temperature = 71.76 F

percent humidity  = 0.80

## Ex:14.10 Pg:667¶

In [10]:
#Initialization of variables
m=1 #lbm
t1=100 #F
t2=75 #F
db=65 #F
print "From psychrometric charts,"
t11=82 #F
phi1=0.4
H1=30 #Btu/lbm dry air
w1=65 #grains/lbm dry air
w2=250 #grains/lbm dry air
#calculations
cr=t1-t2
appr=t2-db
dmf3=(w2-w1)*0.0001427
hf3=68
hf4=43
H2=62.2
H1=30
mf4= (H1-H2+ dmf3*hf3)/(hf4-hf3)
per=dmf3/(dmf3+mf4)
#results
print "cooling range = %d F"%(cr)
print "\n Approach = %d F"%(appr)
print "\n amount of water cooled per pound of dry air = %.3f lbm dry air/lbm dry air"%(mf4)
print "\n percentage of water lost by evaporation = %.2f percent"%(per*100)
From psychrometric charts,
cooling range = 25 F

Approach = 10 F

amount of water cooled per pound of dry air = 1.216 lbm dry air/lbm dry air

percentage of water lost by evaporation = 2.12 percent

## Ex:14.11 Pg:668¶

In [12]:
from math import log
#Initialization of variables
mfr=1
water=900 #gallons
t2=110 #F
t1=80 #F
cp1=1
#calculations
mfa=mfr*water*8.33*60
mfc=mfa/(60*0.075)
qa=mfa*(t2-t1)
dH=qa/(mfc*4.5)
dH2=mfr*cp1*(t2-t1)
H1=23.73
H2=5.08
f=3.309
lnmean=(H1-H2)/log(H1/H2)
dtt=(t2-t1)/lnmean
per=25
#results
print "flow rate of air = %d lbm/hr.It is equal to %d cfm"%(mfa,mfc)
print "\n Total heat transferred = %d Btu/hr"%(qa)
print "\n Enthalpy = %.1f Btu/lbm dry air"%(dH)
print "\n Using second method, Enthalpy = %.1f Btu/lbm"%(dH2)
print "\n Performance factor = %.3f "%(f)
print "\n logrithamic mean enthalpy difference = %.2f . Estimated low percentage = %d low"%(dtt,per)
print "The answers are a bit different due to rounding off error in textbook."
flow rate of air = 449820 lbm/hr.It is equal to 99960 cfm

Total heat transferred = 13494600 Btu/hr

Enthalpy = 30.0 Btu/lbm dry air

Using second method, Enthalpy = 30.0 Btu/lbm

Performance factor = 3.309

logrithamic mean enthalpy difference = 2.48 . Estimated low percentage = 25 low
The answers are a bit different due to rounding off error in textbook.