Chapter 14 - Psychrometrics¶

Example 1 - Pg 257¶

In [1]:
#calculate the partial pressures, dew temperature and density of air, water and specific humidity, Degree of saturation
#Initialization of variables
t1=80+460 #R
ps=0.5069 #psia
print '%s' %("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 '%s' %("part a")
print '%s %.5f %s' %("partial pressure of water =",pw,"psia")
print '%s %d %s' %("\n dew temperature =",tdew,"F")
print '%s' %("part b")
print '%s %.6f %s' %("density of water =",rhow,"lbm/ft^3")
print '%s %.6f %s' %("\n in case 2, density of water =",rhow2,"lbm/ft^3")
print '%s %.6f %s' %("\n density of air =",rhoa,"lbm/ft^3")
print '%s' %("part c")
print '%s %.4f %s' %("specific humidity  =",w,"lbm steam/lbm air")
print '%s' %("part d")
print '%s %.3f' %("In method 1, Degree of saturation = ",mu)
print '%s %.3f' %("\n In method 2, Degree of saturation = ",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

Example 2 - Pg 258¶

In [2]:
#calculate the change in moisture content and change in moisture content
#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 '%s %.6f %s' %("change in moisture content =",dw," lbm water/lbm dry air")
print '%s %.2f %s' %("\n in grains, change =",dwg,"  grains water/lbm dry air")
print '%s' %("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

Example 3 - Pg 264¶

In [3]:
#calculate the humidity ratio and relative humidity
#Initialization of variables
t1=80. #F
t2=60. #F
p=14.696 #psia
ps=0.507 #psia
pss=0.256 #psia
cp=0.24
#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*100
td=46. #F
#results
print '%s %.4f %s' %("\n humidity ratio =",w,"lbm/lbm dry air")
print '%s %.1f %s' %("\n relative humidity  =",phi," percent")
print '%s %d %s' %("\n Dew point =",td,"F")
humidity ratio = 0.0064 lbm/lbm dry air

relative humidity  = 29.7  percent

Dew point = 46 F

Example 4 - Pg 264¶

In [4]:
#calculate the enthalpy and sigma function
#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 '%s %.2f %s' %("In case 1, enthalpy =",H," Btu/lbm dry air")
print '%s %.2f %s' %("\n In case 1, sigma function =",sig," Btu/lbm dry air")
print '%s %.2f %s' %("\n In case 2, enthalpy =",H2," Btu/lbm dry air")
print '%s %.2f %s' %("\n In case 2, sigma function =",sig2," Btu/lbm dry air")
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

Example 5 - Pg 264¶

In [5]:
#calculate the Enthalpy and heat added
#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 '%s %.2f %s' %("In case 1, Enthalpy =",H1," Btu/lbm dry air")
print '%s %.2f %s' %("\n In case 2, Enthalpy =",H2," Btu/lbm dry air")
print '%s %.2f %s' %("\n Heat added =",dq," Btu/lbm dry air")
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

Example 6 - Pg 265¶

In [6]:
#calculate the partial pressure and dew temperature, density of air, water and specific humidity
#Initialization of variables
pw=0.15#psia
print '%s' %("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 '%s' %("part a")
print '%s %.2f %s' %("partial pressure of water =",pw," psia")
print '%s %d %s' %("\n dew temperature =",tdew,"F")
print '%s' %("part b")
print '%s %.6f %s' %("density of water =",rhow,"lbm/ft^3")
print '%s %.4f %s' %("\n density of air =",rhoa,"lbm/ft^3")
print '%s' %("part c")
print '%s %.5f %s' %("specific humidity  =",w,"lbm steam/lbm air")
using psychrometric charts,
part a
partial pressure of water = 0.15  psia

dew temperature = 46 F
part b
density of water = 0.000478 lbm/ft^3

density of air = 0.0728 lbm/ft^3
part c
specific humidity  = 0.00657 lbm steam/lbm air

Example 7 - Pg 266¶

In [7]:
#calculate the Enthalpy change
#Initialization of variables
W1=0.00206 #lbm/lbm dry air
W2=0.01090 #lbm/lbm dry air
t=60 #F
#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 '%s %.2f %s' %("Enthalpy change =",dq," Btu/lbm dry air")
Enthalpy change = 21.53  Btu/lbm dry air

Example 8 - Pg 267¶

In [8]:
#calculate the humidity and Temperature
#Initialization of variables
print '%s' %("From psychrometric charts,")
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
t=62.5# F
phi=0.83 #percent
#results
print '%s %.2f' %("humidity = ",phi)
print '%s %.1f %s' %("\n Temperature =",t," F")
From psychrometric charts,
humidity =  0.83

Temperature = 62.5  F

Example 9 - Pg 270¶

In [9]:
#calculate the dry bulb temperature and percent humidity
#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 '%s' %("from psychrometric charts,")
phi2=0.8
#results
print '%s %.2f %s' %("Dry bulb temperature =",tf," F")
print '%s %.2f' %("\n percent humidity  = ",phi2)
from psychrometric charts,
Dry bulb temperature = 71.76  F

percent humidity  =  0.80

Example 10 - Pg 271¶

In [10]:
#calculate the cooling range and Approach. Also calculate the amount of water cooled and percentage of water lost
#Initialization of variables
m=1. #lbm
t1=100. #F
t2=75. #F
db=65. #F
print '%s' %("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)*100
#results
print '%s %d %s' %("cooling range =",cr,"F")
print '%s %d %s' %("\n Approach =",appr,"F")
print '%s %.3f %s' %("\n amount of water cooled per pound of dry air =",mf4,"lbm dry air/lbm dry air")
print '%s %.2f %s' %("\n percentage of water lost by evaporation =",per,"percent")
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