Chapter 5 : Humidification

Example 5.1

In [13]:
 

# Variable Declaration 
#dry bulb temperature=50  and  wet bulb temperature=35 
Tg=50.;                    #dry bulb temperature=50
To=0;                    #refrence temperature in degree celcius    
Mb=28.84;                #average molecular weight of air
Ma=18.;                   #average molecular weight of water

#part(i)
ybar=.0483                #0.003 kg of water vapour/kg of dry air
print "\n the humidity(from chart) is  \t\t:%f  percent"%ybar

#part(ii)
humper=35.;                #humidity percentage
print "\n the percentage humidity is(from chart) :%f  percent"%humper

# Calculation and Result
#part(iii)
pt=1.013*10**5;            #total pressure in pascal
molhum=0.0483;            #molal humidity =pa/(pt-pa)
pa=molhum*pt/(1+molhum);
#the vopour pressure of water(steam tables)at 50degree = .1234*10**5 N/m**2
relhum=(pa/(.1234*10**5))*100;      #percentage relative humidity =partial pressure/vapour pressure
print "\n the percentage relative humidity is \t percent:%f  "%relhum

#part(iv)
dewpoint=31.5;            #dew point temperature in degree celcius
print "\n the dew point temperature \t\t :%f degree celcius"%dewpoint

#part(v)
Ca=1.005;
Cb=1.884;
ybar=.03;                #saturation temperature inkg water vapour/kg dry air
Cs=Ca+Cb*ybar;            #humid heat  in kj/kg dry air degree celcius
print "\n we get humid heat as \t\t\t :%f kj/kg dry air degree celcius "%Cs

#part(vi)
d=2502;                    #latent heat in kj/kg
H=Cs*(Tg-0)+ybar*d;        #enthalpy for refrence temperature of 0 degree
print "\n we get H as \t\t\t\t :%f kj/kg"%H
Hsat=274.;                   #enthalpy of sturated air
Hdry=50.;                   #enthalpy of dry air in kj/kg
Hwet=Hdry+(Hsat-Hdry)*0.35;     #enthalpy of wet air in kj/kg
print "\n we get enthalpy of wet air  as \t:%f kj/kg"%Hwet

#part(vii)
VH=8315*((1./Mb)+(ybar/Ma))*((Tg+273.)/pt);        #humid volume in m**3mixture/kg of dry air
print "\n we get VH  as (a)\t\t\t :%f m**3/kg of dry air"%VH
spvol=1.055;              #specific volume of saturated air in m**3*kg
vdry=0.91;                #specific volume of dry air in m**3/kg
Vh=vdry+(spvol-vdry)*.35  #by interpolation we get Vh in m**3/kg of dry air 
print "\n by interpolation we get specific volume Vh as(b) :%f m**3/kg of dry air"%Vh

#end

 the humidity(from chart) is  		:0.048300  percent

 the percentage humidity is(from chart) :35.000000  percent

 the percentage relative humidity is 	 percent:37.822988  

 the dew point temperature 		 :31.500000 degree celcius

 we get humid heat as 			 :1.061520 kj/kg dry air degree celcius 

 we get H as 				 :128.136000 kj/kg

 we get enthalpy of wet air  as 	:128.400000 kj/kg

 we get VH  as (a)			 :0.963494 m**3/kg of dry air

 by interpolation we get specific volume Vh as(b) :0.960750 m**3/kg of dry air

Example 5.2

In [14]:
 
# Variable Declaration 

#dry bulb temperature=25  and  wet bulb temperature=22
Tg=25.;                    #dry bulb temperature=50
To=0;                    #refrence temperature in degree celcius    
Mb=28.84;                #average molecular weight of air
Ma=18.;                   #average molecular weight of water


# Calculation and Result
#part(i)
hum=.0145                #0.0145 kg of water/kg of dry air
print "\n the saturation humidity(from chart) is :%f  percent"%hum

#part(ii)
humper=57.;                #humidity percentage
print "\n the percentage humidity is  \t\t:%f  percent"%humper

#part(iii)
pt=1.;            #total pressure in atm
sathum=0.0255;            #molal humidity =pa/(pt-pa)
pa1=sathum*pt*(28.84/18)/(1+(sathum*(28.84/18)));
#the vopour pressure of water(steam tables)at 25 = .0393*10**5 N/m**2
pt=1;                     #total pressure in atm
molhum=0.0145;            #molal humidity =pa/(pt-pa)
pa2=molhum*pt*(28.84/18)/(1+(molhum*pt*(28.84/18)));
#the vopour pressure of water(steam tables)at 25 = .0393*10**5 N/m**2
relhum=(pa2/pa1)*100;      #percentage relative humidity =partial pressure/vapour pressure
print "\n the percentage relative humidity is \t :%f  "%relhum

#part(iv)
dewpoint=19.5;            #dew point temperature in degree celcius
print "\n the dew point temperature \t :%f degree celcius"%dewpoint

#part(v)
Ca=1005.;
Cb=1884.;
ybar=.0145;                # humidity inkg water /kg dry air
Cs=Ca+Cb*ybar;            #humid heat  in j/kg dry air degree celcius
d=2502300.;                    #latent heat in j/kg
H=Cs*(Tg-0)+ybar*d;        #enthalpy for refrence temperature of 0 degree
print "\n we get Humid heat H as \t :%f j/kg"%H
#the actual answer is 62091.3 bt in book it is given 65188.25(calculation mistake in book)
#end

 the saturation humidity(from chart) is :0.014500  percent

 the percentage humidity is  		:57.000000  percent

 the percentage relative humidity is 	 :57.842165  

 the dew point temperature 	 :19.500000 degree celcius

 we get Humid heat H as 	 :62091.300000 j/kg

Example 5.3

In [15]:
  
# Variable Declaration 

#part(i)
pt=800.;                    #total pressure in mmHg
pa=190.;                    #vapour pressure of acetone at 25 degree 
ys_bar=pa*(58./28)/(pt-pa)   #
#percentage saturation = y_bar/ys_bar *100
s=80;                      #percent saturation

# Calculation and Result
y_bar=ys_bar*s/100.;        #absolute humidity
print "\n the absolute humidity is \t :%f kg acetone/kmol N2 "%y_bar

#part(ii)
#y_bar=pa*(58/28)/(pt-pa) 
pa1=pt*y_bar*(28./58)/(1+(y_bar*(28./58)));
print "\n the partial pressure of acetone is:%f mmHg"%pa1

#part(iii)
y=pa1/(pt-pa1);              #absolute molal humidity
print "\n absolute molal humidity \t:%f kmol acetone/kmol N2"%y

#part(iv)
#volume of .249kmol acetone vapour at NTP =.249*22.14
#p1v1/T1 =p2v2/T2
p2=800.;                    #final pressure of acetone and nitrogen at 25 degree
p1=760.;                    #initial pressure of acetone and nitrogen at 25 degree
T2=298.;                    #final temperature of acetone and nitrogenat 25 degree
T1=273.;                   #initial temperature of acetone and nitrogen at 25 degree
vA1=5.581;                    #initial volume of acetone at 25 degree
vN1=22.414;                   #initial volume of nitrogen at 25 degree 
vA2=T2*vA1*p1/(T1*p2);        #final volume of acetone at 25 degree
vN2=T2*vN1*p1/(T1*p2);        #final volume of nitrogen at 25 degree
vtotal=vA2+vN2;            #total volume of the mixture
vper=vA2*100/vtotal;           #percentage volume of acetone
print "\n the percentage volume of acetone is :%f m**3"%vper
#end

 the absolute humidity is 	 :0.516159 kg acetone/kmol N2 

 the partial pressure of acetone is:159.580052 mmHg

 absolute molal humidity 	:0.249180 kmol acetone/kmol N2

 the percentage volume of acetone is :19.935703 m**3

Example 5.4

In [16]:
 

# Variable Declaration 

#part(i)
pa=13.3;                      #pressure in kpa
pa2=20.6;                    #vapour pressure at 60 degree
pt=106.6                     #total pressure in kpa
y=pa/(pt-pa);              #absolute molal humidity

# Calculation  and Result
y_bar=y*(18/28.84);          #relative humidity
print "\n absolute  humidity of mixture :%f kg water-vapour/kg dry air"%y_bar


#part(ii)
mf=pa/pt;                    #mole fraction
print "\n the mole fraction is :%f"%mf

#part(iii)
vf=mf;                            #volume fraction
print "\n the volume fraction is :%f"%vf

#part(iv)
Ma=18.;                      #molecular weight
Mb=28.84;                    #molecular weight
Tg=60.;                       #temperature of mixture
rh=(pa/pa2)*100.;            #relative humidity in pecentage        
print "\n we get relative humidity as  as :%f percent"%rh

#part(v)
VH=8315.*((1./Mb)+(y_bar/Ma))*((Tg+273)/pt)*10**-3;        #humid volume in m**3mixture/kg of dry air
x=y_bar/VH;                 #g water/m**3 mixture 
print "\n we get x i.e. gwater/m**3 mixture as :%f "%(x*1000)
#end

 absolute  humidity of mixture :0.088971 kg water-vapour/kg dry air

 the mole fraction is :0.124765

 the volume fraction is :0.124765

 we get relative humidity as  as :64.563107 percent

 we get x i.e. gwater/m**3 mixture as :86.460483

Example 5.5

In [17]:
 

# Variable Declaration 

#part(i)
y_bar=.0183;                #kg water vapour/kg dry air
print "\n we get humidity as(from chart)  :%f kg of water/kg dry air"%y_bar
print "\n we get saturation humidity as(from chart)  :%d percent"%67
Ma=18.;                      #molecular weight
Mb=28.84;                    #molecular weight
Tg=30.;   
pa = 13.3
pa2 = 20.6                    #temperature of mixture
rh=(pa/pa2)*100;            #relative humidity in pecentage        
pt=1.013*10**5;            #total pressure in pascal

# Calculation  and  Result
VH=8315*((1./Mb)+(y_bar/Ma))*((Tg+273)/pt);        #humid volume in m**3mixture/kg of dry air
print "\n we get humid volume as \t:%f  m**3/kg dry air"%VH

#part(ii)
Ca=1005.;
Cb=1884.;
Cs=Ca+Cb*y_bar;            #humid heat  in j/kg dry air degree celcius
print "\n we get humid heat as \t\t :%f j/kg dry air degree celcius "%Cs

#part(iii)
d=2502300.;                    #latent heat in j/kg
H=Cs*(Tg-0)+y_bar*d;        #enthalpy for refrence temperature of 0 degree
print "\n we get Enthalpy H as \t\t:%f j/kg dry air"%H

#part(iv)
dewpoint=23.5;            #dew point temperature in degree celcius
print "\n the dew point temperature \t :%f degree celcius"%dewpoint

#end

 we get humidity as(from chart)  :0.018300 kg of water/kg dry air

 we get saturation humidity as(from chart)  :67 percent

 we get humid volume as 	:0.887669  m**3/kg dry air

 we get humid heat as 		 :1039.477200 j/kg dry air degree celcius 

 we get Enthalpy H as 		:76976.406000 j/kg dry air

 the dew point temperature 	 :23.500000 degree celcius

Example 5.6

In [18]:
 
# Variable Declaration 

#part(i)
y=.048;                      #humidity kmol water vapour/kmol dry air
y_bar=y*(18/28.84);          #(from chart) absolute humidity
print "we get absolute humidity as  :%f kg of water/kg dry air"%(y_bar)
print "we get percentage humidity as(from chart)  :%f percent"%(25.5);

# Calculation and Result
y_bar=y*(18/28.84);          #relative humidity
Ma=18.;                      #molecular weight
Mb=28.84;                    #molecular weight
Tg=55.;                       #temperature of mixture
pt=1.013*10**5;            #total pressure in pascal
VH=8315*((1./Mb)+(y_bar/Ma))*((Tg+273)/pt);        #humid volume in m**3mixture/kg of dry air
print "\n we get VH  as \t :%f m**3/kg dry air"%VH

#part(ii)
Ca=1005.;
Cb=1884.;
Cs=Ca+Cb*y_bar;            #humid heat  in j/kg dry air degree celcius
print "\n we get humid heat as \t :%f j/kg dry air degree celcius "%Cs

#part(iii)
d=2502300.;                    #latent heat in j/kg
H=Cs*(Tg-0)+y_bar*d;        #enthalpy for refrence temperature of 0 degree
print "\n we get H as \t :%f j/kg dry air"%H

#end

we get absolute humidity as  :0.029958 kg of water/kg dry air
we get percentage humidity as(from chart)  :25.500000 percent

 we get VH  as 	 :0.978346 m**3/kg dry air

 we get humid heat as 	 :1061.441609 j/kg dry air degree celcius 

 we get H as 	 :133344.170596 j/kg dry air

Example 5.7

In [19]:
 

# Variable Declaration 

 
ft=46;                #final temperature in degree celcius
# Calculation and Result
print "\n final temperature is (from chart):%f degree celcius"%ft
y_bar=.0475;        # humidity of air
print "\n the humidity of air(from chart) :%f kg water vapour /kg dry air"%y_bar

#end

 final temperature is (from chart):46.000000 degree celcius

 the humidity of air(from chart) :0.047500 kg water vapour /kg dry air

Example 5.8

In [20]:
 
# Variable Declaration 

pa1=4.24            #data:vapour pressure of water at 30degree = 4.24 kpa
pa2=1.70            #vapour pressure of water at 30degree = 1.70 kpa

#part(i)
pt=100.;                         #total pressure

# Calculation  
ys_bar=pa1/(pt-pa1);              #kg water vapour/kg dry air
rh=.8;                          #relative humidity
pa3=rh*pa1;                       #partial pressure
y_bar=pa3*(18/28.84)/(pt-pa3);               #molal humidity
print "\n the molal humidity:%f kg/kg dry air"%y_bar

#part(ii)

pa=1.7;    
pt=200.;
ys=pa/(pt-pa);
ys_bar=ys*(18/28.84);

# Result
print "\n the molal humidity if pressure doubled and temp. is 15 :%f kg/kg dry air"%ys_bar

#part(iii) 
Ma=18.;                        #molecular weight
Mb=28.84;                    #molecular weight
Tg=30.;                       #temperature of mixture
rh=(pa/pa2)*100;            #relative humidity in pecentage        
pt=10**5;              #total pressure in pascal
VH=8315*((1./Mb)+(y_bar/Ma))*((Tg+273)/pt);        #humid volume in m**3mixture/kg of dry air
print "\n we get humid volume VH  as \t :%f m**3/kg of dry air"%VH
w=100/VH;                   #100 m**3 of original air 
wo= w*y_bar;                #water present in original air
wf= w*ys_bar;              #water present finally
wc=wo-wf;                   #water condensed from 100m**3 of original sample
print "\n the weight water condensed from 100m**3 of original sample:%f kg"%wc

#part(iv)
Tg=15.;                       #temperature of mixture       
pt=2*10**5;            #total pressure in pascal
VH=8315*((1./Mb)+(ys_bar/Ma))*((Tg+273)/pt);        #humid volume in m**3mixture/kg of dry air
vf=VH*110.6;                #final volume of mixture
print "\n we get VH final volume of mixture as \t :%f m**3"%vf

#end

 the molal humidity:0.021914 kg/kg dry air

 the molal humidity if pressure doubled and temp. is 15 :0.005351 kg/kg dry air

 we get humid volume VH  as 	 :0.904267 m**3/kg of dry air

 the weight water condensed from 100m**3 of original sample:1.831684 kg

 we get VH final volume of mixture as 	 :46.311825 m**3

Example 5.9

In [21]:
 
# Variable Declaration 

#part(i)
y_bar=.03;                   # humidity inkg water /kg dry air
pt=760.;                     #total pressure in pascal
pa2=118.;                    #final pressure

# Calculation  and  Result
y=y_bar/(18/28.84);        #humidity kmol water vapour/kmol dry air
pa=(y*pt)/(y+1);           #partial pressure
rh=pa/pa2;                  #relative humidity
sh=pa2/(pt-pa2);            #saturated humidity
ph=(y/sh)*100;                #percentage humidity
print "\n percentage humidity is :%f"%ph

#/part(ii)
Ma=18.;                      #molecular weight
Mb=28.84;                    #molecular weight
Tg=55.;                       #temperature of mixture
pt=1.013*10**5;            #total pressure in pascal
VH=8315*((1./Mb)+(y_bar/Ma))*((Tg+273)/pt);        #humid volume in m**3mixture/kg of dry air
print "\n we get VH humid volume as  :%f m**3/kg dry air"%VH


#part(iii)
Ca=1005.;
Cb=1884.;
Cs=Ca+Cb*y_bar;                #humid heat  in j/kg dry air degree celcius
print "\n we get humid heat as \t :%f j/kg dry air degree celcius "%Cs
d=2502300;                    #latent heat in j/kg
H=Cs*(Tg-0)+y_bar*d;         #enthalpy for refrence temperature of 0 degree
print "\n we get H enthalpy as \t :%f j/kg"%H

#part(iv)
v=100.;                        #volume of air
mass=v/VH;                    #mass of dry air
Tg=110.;                       #temperature of mixture
d=2502300.;                    #latent heat in j
H_final=Cs*(Tg-0)+y_bar*d;    #enthalpy for refrence temperature of 0 degree
H_added=(H_final-H)*102.25;     #HEAT added  in kj     
print "\n we get heat added as \t :%f kj"%(H_added/1000)
#end

 percentage humidity is :26.151525

 we get VH humid volume as  :0.978409 m**3/kg dry air

 we get humid heat as 	 :1061.520000 j/kg dry air degree celcius 

 we get H enthalpy as 	 :133452.600000 j/kg

 we get heat added as 	 :5969.723100 kj

Example 5.10

In [1]:
 
# Variable Declaration 

%matplotlib inline
L=2000.;                    #flow rate of water to be cooled in kg/min
T1=50.;                    #temperature of inlet water
T2=30.;                    #temp. of outlet water
H1=.016;                  #humidity of incoming air
cp=4.18;                  #specific heat of water
cpair=1.005;              #specific heat capcity of air
cpwater=1.884;            #specific heat capcity of water
tg=20.;                    #temperature in degree
to=0;
ybar=0.016;               #saturated humidity at 20 degree
d=2502.;                   #latent heat
Ky_a=2500.;                #value of masstransfer coefficient in kg/hr*m**3*dybar

# Calculation  
E=cpair*(tg-to)+(cpwater*(tg-to)+d)*ybar;    #enthalpy
#similarly for other temperatures
T=[20,30,40,50,55]            #differnt temperature for different enthalpy calculation
i=0;
E= [0,0,0,0,0]
while(i<5):                   #looping for different enthalpy calculation of operating line
    E[i]=cpair*(T[i]-to)+(cpwater*(T[i]-to)+d)*ybar;
    print "\n the enhalpy at :%f is :%f"%(T[i],E[i]);
    i=i+1;

ES=[60.735,101.79,166.49,278.72,354.92]        #enthalpy of eqll condition
from matplotlib.pyplot import *

# Result
plot(T,E);
plot(T,ES);
title("Fig.5.10(b),Temperature-Enthalpy plot");
xlabel("X-- Temperature, degree celcius");
ylabel("Y-- Enthalpy ,kj/kg");
legend("operating line","Enthalpy at saturated cond")
show()

Hg1=71.09;                     #point on the oper. line(incoming air)
Hg2=253.;                       #point after drawing the tangent
slope=(Hg2-Hg1)/(T1-T2);       #we gt slope of the tangent
    #slope = (L*Cl/G)_min
Cl=4.18;
G_min=L*60*Cl/slope;                  #tangent gives minimum value of the gas flow rate
G_actual=G_min*1.3;                  #since actual flow rate is 1.3 times the minimum
slope2=L*Cl*60/G_actual;            #slope of operating line
Hg2_actual=slope2*(T1-T2)+Hg1;     #actual humidityat pt 2
Ggas=10000.;                       #minimum gas rate in kg/hr*m**2
Area1=G_actual/Ggas;             #maximum area of the tower(based on gas)
Gliq=12000.;                     #minimum liquid rate in kg/hr*m**2
Area2=60*L/Gliq;               #maximum area of the tower(based on liquid)
print "\n \n the maximum area of the tower(based on gas) is :%f m**2"%Area1
print "\n the maximum area of the tower(based on liquid) is :%f m**2"%Area2
dia=(Area1*4/3.14)**0.5;        #diameter of the tower in m


 
#table
T=[20,30,40,50,55]            #differnt temperature for different enthalpy calculation
#enthaly 
H_bar=[101.79,133.0,166.49,210.0,278.72]        #H_bar i.e. at equl.
Hg=[71.09,103.00,140.00,173.00,211.09]        #Hg i.e. of operating line
i=0;
y = [0,0,0,0,0]
while(i<5):          #looping for different enthalpy calculation of operating line
    y[i]=1./(H_bar[i]-Hg[i]);
    print "\n the enhalpy at :%f is :%f"%(T[i],y[i]);
    i=i+1;

plot(Hg,y,"o-");

#area under this curve gives Ntog =4.26
Ntog=4.26;                #no. of transfer unit
Gs=10000.;                  #gas flow rate
Htog=Gs/Ky_a;              # height of transfer unit
height=Ntog*Htog;          #height of the tower
print "\n \nthe tower height is :%f m"%height


#M = E + B + W
W=.2/100 *L*60;                #windage loss(W)
B=0;                        #blow down loss neglected
E=G_actual*(.064-.016);     #assuming air leaves fully saturated
M = E + B + W;                #make up water is based onevaporation loss(E),blow down loss(B),windage loss(W)
print "\n make up water is based onevaporation loss(E),blow down loss(B),windage loss(W) is :%f kg /hr"%M
#end

Welcome to pylab, a matplotlib-based Python environment [backend: module://IPython.zmq.pylab.backend_inline].
For more information, type 'help(pylab)'.

 the enhalpy at :20.000000 is :60.734880

 the enhalpy at :30.000000 is :71.086320

 the enhalpy at :40.000000 is :81.437760

 the enhalpy at :50.000000 is :91.789200

 the enhalpy at :55.000000 is :96.964920

C:\Anaconda\lib\site-packages\matplotlib\legend.py:336: UserWarning: Unrecognized location "Enthalpy at saturated cond". Falling back on "best"; valid locations are
	right
	center left
	upper right
	lower right
	best
	center
	lower left
	center right
	upper left
	upper center
	lower center

  % (loc, '\n\t'.join(self.codes.iterkeys())))

 
 the maximum area of the tower(based on gas) is :7.169260 m**2

 the maximum area of the tower(based on liquid) is :10.000000 m**2

 the enhalpy at :20.000000 is :0.032573

 the enhalpy at :30.000000 is :0.033333

 the enhalpy at :40.000000 is :0.037750

 the enhalpy at :50.000000 is :0.027027

 the enhalpy at :55.000000 is :0.014786

 
the tower height is :17.040000 m

 make up water is based onevaporation loss(E),blow down loss(B),windage loss(W) is :3681.244571 kg /hr

Example 5.11

In [22]:
 
 
T1=30.;              #temperature at the inlet in degree celcius
T2=17.;              #temperature at the exit in degree celcius
f=100000.;           #flow rate of water in kg/hr
hi=.004;            #humidity of incoming air in kg/kg of dry air
hl=.015;            #humidity of leaving air in kg/kg of dry air
Hi=18.11;           #enthalpy of incoming air in kg/kg of dry air
Hl=57.16;            #enthalpy of leaving air in kg/kg of dry air
#w=mdry*(hl-hi) = mdry*0.011;                             -----equn 1st  
#mass of water evaporated

#making energy balance:  total heat in = total heat out
#heat in entering water + heat in entering air = heat in leaving water + heat in leaving air
#100000*1*(30-0) + mdry*Hi = (100000-w)*1*(17-0) + mdry*Hl  ----eqn 2nd

#substituting eqn 1st in 2nd we get;
a=14.4;                                   #cross sectional area of the tower in m**2

# Calculation  
mdry=(T1*f-T2*f)/(Hl-Hi-T2*.011);        #mass of dry air
velocity=mdry/a;                          #air velocity in kg/m**2* hr
x=mdry*.011;                             #make up water needed in kg/hr

# Result
print "\n the make up water needed is :%f kg /hr"%x
print "\n the velocity of air is as :%f kg/hr"%velocity
#end

 the make up water needed is :367.959241 kg /hr

 the velocity of air is as :2322.975009 kg/hr

Example 5.12

In [12]:
 
import math
 

T1=65;              #dry bulb temperature at the inlet in degree celcius
f=3.5;           #flow rate of air in m**3/s
hi=1.017;            #humidity of incoming air in kg/kg of dry air
hl=.03;            #humidity of leaving air in kg/kg of dry air
k=1.12;            #mass transfer coefficient in kg/m**3*s
y1=.017;           #molefraction at recieving end
y2=.03;            #molefraction at leaving end

#substituting eqn 1st in 2nd we get;
a=2;                                   #cross sectional area of the tower in m**2
d=1.113;                                  #density o fair in kg/m**3

# Calculation  
m=(f*d)                                   #mass flow rate  of  air
gs=m/hi;                                 #air velocity in kg/m**2* hr
ys_bar=.032;
#for recirculation humidifier
# Result
z=math.log((ys_bar-y1)/(ys_bar-y2))*gs/k;          #length of the chamber required
print "\n the length of the chamber required is :%f m"%z

#end

 the length of the chamber required is :6.890939 m